# Refutation of Stable Thermal Equilibrium Lapse Rates

Guest post by Robert G. Brown

Duke University Physics Department

### The Problem

In 2003 a paper was published in Energy & Environment by Hans Jelbring that asserted that a gravitationally bound, adiabatically isolated shell of ideal gas would exhibit a thermodynamically stable adiabatic lapse rate. No plausible explanation was offered for this state being thermodynamically stable – indeed, the explanation involved a moving air parcel:

An adiabatically moving air parcel has no energy loss or gain to the surroundings. For example, when an air parcel ascends the temperature has to decrease because of internal energy exchange due to the work against the gravity field.

This argument was not unique to Jelbring (in spite of his assertion otherwise):

The theoretically deducible influence of gravity on GE has rarely been acknowledged by climate change scientists for unknown reasons.

The adiabatic lapse rate was and is a standard feature in nearly every textbook on physical climatology. It is equally well known there that it is a dynamical consequence of the atmosphere being an open system. Those same textbooks carefully demonstrate that there is no lapse rate in an ideal gas in a gravitational field in thermal equilibrium because, as is well known, thermal equilibrium is an isothermal state; nothing as simple as gravity can function like a “Maxwell’s Demon” to cause the spontaneous stable equilibrium separation of gas molecules into hotter and colder reservoirs.

Spontaneous separation of a reservoir of gas into stable sub-reservoirs at different temperatures violates the second law of thermodynamics. It is a direct, literal violation of the refrigerator statement of the second law of thermodynamics as it causes and maintains such a separation without the input of external work. As is usually the case, violation of the refrigeration statement allows heat engines to be constructed that do nothing but convert heat into work – violating the “no perfectly efficient heat engine” statement as well.

The proposed adiabatic thermal lapse rate in EEJ is:

where g is the gravitational acceleration (presumed approximately constant throughout the spherical shell) and cp  is the heat capacity per kilogram of the particular “ideal” gas at constant pressure. The details of the arguments for an adiabatic lapse rate in open systems is unimportant, nor does it matter what cp is as long as it is not zero or infinity.

What matters is that EEJ asserts that   in stable thermodynamic equilibrium.

The purpose of this short paper is to demonstrate that such a system is not, in fact, in thermal equilibrium and that the correct static equilibrium distribution of gas in the system is the usual isothermal distribution.

### The Failure of Equilibrium

In figure 1 above, an adiabatically isolated column of an ideal gas is illustrated. According to EEJ, this gas spontaneously equilibrates into a state where the temperature at the bottom of the column Tb is strictly greater than the temperature Tt at the top of the column. The magnitude of the difference, and the mechanism proposed for this separation are irrelevant, save to note that the internal conductivity of the ideal gas is completely neglected. It is assumed that the only mechanism for achieving equilibrium is physical (adiabatic) mixing of the air, mixing that in some fundamental sense does not allow for the fact that even an ideal gas conducts heat.

Note well the implication of stability. If additional heat is added to or removed from this container, it will always distribute itself in such a way as to maintain the lapse rate, which is a constant independent of absolute temperature. If the distribution of energy in the container is changed, then gravity will cause a flow of heat that will return the distribution of energy to one with Tb > Tt . For an ideal gas in an adiabatic container in a gravitational field, one will always observe the gas in this state once equilibrium is established, and while the time required to achieve equilibrium is not given in EEJ, it is presumably commensurate with convective mixing times of ordinary gases within the container and hence not terribly long.

Now imagine that the bottom of the container and top of the container are connected with a solid conductive material, e.g. a silver wire (adiabatically insulated except where it is in good thermal contact with the gas at the top and bottom of the container) of length  L . Such a wire admits the thermally driven conduction of heat according to Fourier’s Law:

where λ  is the thermal conductivity of silver, A is the cross-sectional area of the wire, and ΔT=TbTt . This is an empirical law, and in no way depends on whether or not the wire is oriented horizontally or vertically (although there is a small correction for the bends in the wire above if one actually solves the heat equation for the particular geometry – this correction is completely irrelevant to the argument, however).

As one can see in figure 2, there can be no question that heat will flow in this silver wire. Its two ends are maintained at different temperatures. It will therefore systematically transfer heat energy from the bottom of the air column to the top via thermal conduction through the silver as long as the temperature difference is maintained.

One now has a choice:

• If EEJ is correct, the heat added to the top will redistribute itself to maintain the adiabatic lapse rate. How rapidly it does so compared to the rate of heat flow through the silver is irrelevant. The inescapable point is that in order to do so, there has to be net heat transfer from the top of the gas column to the bottom whenever the temperature of the top and bottom deviate from the adiabatic lapse rate if it is indeed a thermal equilibrium state.
• Otherwise, heat will flow from the bottom to the top until they are at the same temperature. At this point the top and the bottom are indeed in thermal equilibrium.

It is hopefully clear that the first of these statements is impossible. Heat will flow in this system forever; it will never reach thermal equilibrium. Thermal equilibrium for the silver no longer means the same thing as thermal equilibrium for the gas – heat only fails to flow in the silver when it is isothermal, but heat only fails to flow in the gas when it exhibits an adiabatic lapse in temperature that leaves it explicitly not isothermal. The combined system can literally never reach thermal equilibrium.

Of course this is nonsense. Any such system would quickly reach thermal equilibrium – one where the top and bottom of the gas are at an equal temperature. Nor does one require a silver wire to accomplish this. The gas is perfectly capable of conducting heat from the bottom of the container to the top all by itself!

One is then left with an uncomfortable picture of the gas moving constantly – heat must be adiabatically convected downward to the bottom of the container in figure 1 in ongoing opposition to the upward directed flow of heat due to the fact that Fourier’s Law applies to the ideal gas in such a way that equilibrium is never reached!

Of course, this will not happen. The gas in the container will quickly reach equilibrium. What will that equilibrium look like? The answer is contained in almost any introductory physics textbook. Take an ideal gas in thermal equilibrium:

where N is the number of molecules in the volume V, k is Boltzmann’s constant, and T is the temperature in degrees Kelvin. n is the number of moles of gas in question and R is the ideal gas constant. If we assume a constant temperature in the adiabatically isolated container, one gets the following formula for the density of an ideal gas:

where M is the molar mass, the number of kilograms of the gas per mole.

The formula for that describes the static equilibrium of a fluid is unchanged by the compressibility (or lack thereof) of the fluid – for the fluid to be in force balance the variation of the pressure must be:

(so that the pressure decreases with height, assuming a non-negative density). If we multiply both sides by dz and integrate, now we get:

Exponentiating both sides of this expression, we get the usual exponential isothermal lapse in the pressure, and by extension the density:

where P0 is the pressure at z=0 (the bottom of the container).

This describes a gas that is manifestly:

1. In static force equilibrium. There is no bulk transport of the gas as buoyancy and gravity are in perfect balance throughout.
2. In thermal equilibrium. There is no thermal gradient in the gas to drive the conduction of heat.

If this system is perturbed away from equilibrium, it will quickly return to this combination of static and thermal equilibrium, as both are stable. Even in the case of a gas with an adiabatic lapse rate (e.g. the atmosphere) remarkably small deviations are observed from the predicted P(z) one gets treating the atmosphere as an ideal gas. An adiabatically isolated gas initially prepared in a state with an adiabatic lapse rate will thermally equilibrate due to the internal conduction of heat within the gas by all mechanisms and relax to precisely this state.

### Conclusion

As we can see, it is an introductory physics textbook exercise to demonstrate that an adiabatically isolated column of gas in a gravitational field cannot have a thermal gradient maintained by gravity. The same can readily be demonstrated by correctly using thermodynamics at a higher level or by using statistical mechanics, but it is not really necessary. The elementary argument already suffices to show violation of both the zeroth and second laws of thermodynamics by the assertion itself.

In nature, the dry adiabatic lapse rate of air in the atmosphere is maintained because the system is differentially heated from below causing parcels of air to constantly move up and down. Reverse that to a cooling, like those observed during the winter in the air above Antarctica, and the lapse rate readily inverts. Follow the air column up above the troposphere and the lapse rate fails to be observed in the stratosphere, precisely where vertical convection stops dominating heat transport. The EEJ assertion, that the dry adiabatic lapse rate alone explains the bulk of so-called “greenhouse warming” of the atmosphere as a stable feature of a bulk equilibrium gas, is incorrect.

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DanSanto
January 24, 2012 6:41 am

Do people really have to have stuff like this demonstrated? I saw, but didn’t read, the previous post on this topic. I thought it was a weird thing to have on a climate blog, though since it did deal with a gas and our climate is a gas, I didn’t think too much of it.
Now, I’ve gone back and read that previous post and glanced through the LONG bunch of comments, a disturbing number of which actually supported the impossible idea. This is basic physics, Second Law of Thermodynamics sort of stuff – you can’t get perpetual-anything.
I thought everyone realized you can’t magically use gravity or magnets to generate perpetual energy machines. It blows my mind that there are actually people who think a thermally graded column would result. It’s nothing but a variation on a perpetual energy machine.
Kudos to WUWT for spending some time debunking this sort of nonsense. It’s sad that there are apparently so many people who swallow this sort of nonsense in the first place.

steveta_uk
January 24, 2012 6:42 am

There seem to be a lot of people who think that a lapse rate is a proprty of the matter itself, rather than a convenient description of the behaviour of the gas.
I’m not sure that Dr Brown will convince them otherwise.
But maybe Hans Jelbring can convince us all that the heat in figure 2 cannot rise up the silver bar because graivity is holding it down!

January 24, 2012 6:43 am

Sometimes something like this comes along that refutes a common idea so well and simply that it just makes a guy go ‘holy crap!’. Thanks for this excellent post as for me it was a revelation.

January 24, 2012 6:48 am

‘what goes up, has to come down,spinning wheel has to go round’ blood sweat and tears. it’s a bit warm !

Alan Robertson
January 24, 2012 6:58 am

“If this system is perturbed away from equilibrium, it will quickly return to this combination of static and thermal equilibrium, as both are stable.
_______________________
This is also perfectly expressed by Le Chatelier’s Principle.

Bryan
January 24, 2012 6:59 am

What would complement this theoretical explanation is if an experiment backed it up.
So far as I know no experiment has ever been carried out.
All suggested proposals seem to run into problems when real components and physically accurate numbers are used.

Joules Verne
January 24, 2012 7:04 am

“Spontaneous separation of a reservoir of gas into stable sub-reservoirs at different temperatures violates the second law of thermodynamics. It is a direct, literal violation of the refrigerator statement of the second law of thermodynamics as it causes and maintains such a separation without the input of external work.”
No, Robert. The second law requires that no energy gradient can be maintained without input of work. It requires the reservoir of gas to be isoenergetic not isothermal. A horizontal layer may be a different temperature than another if the cooler layer has its lesser kinetic energy balanced by greater gravitational energy. This is obviously the case since it goes without a shadow of a doubt that a molecule of air in a higher layer has more gravitational energy than a molecule in a lower layer. Thus the second law actually demands a temperature difference of equal and opposite polarity to compensate for the difference in gravitational energy. An isothermal atmosphere in a gravity field is the one that violates the second law.
My name is Joules because I know how to find and count them no matter how they try to hide.
Thanks for playing.

Richard
January 24, 2012 7:18 am

Hmmm. Temperature is the integral of the number and energy of particles seen at the measuring surface. Less particles equal lower temperature given equal velocity profiles of the particles. Less particles also equals lower pressure. Temperature and pressure are thus directly linked. Hence the dry lapse rate. Gravity and pressure are also directly linked. Hence the dry lapse rate.

Alan Millar
January 24, 2012 7:22 am

Well thanks for taking the time to post Doc.
I gave up trying to convince Tallbloke in the other thread as he is clearly in ‘La La La’ mode.
His reply as to what was going to stop heat flow in a system with a temperature gradient was that at the interface between hotter and colder the actual temperature was the same and therefore no heat would flow!
Lets see, take a rod with a temperature gradient and cut it into the thinnest possible slices allowed in physics and apparently adjoining slices are the same temperature!
However, that would mean the rod was actually the same temperature all the way through as not only are those two touching slices (call them a & b) the same temperature, then obviously, the two slices c & d touching them are also at the same temperature and the two slices e & f touching c & d are the same as c & d and a & b also. Well you get the picture.
Anyway I am sure he will be along soon to show that gravity has some magical quality that allows work to be done indefinitly in a closed system.
Alan

kdk33
January 24, 2012 7:22 am

The dry adiabatic lapse rate is derived several ways, each of which assumes an isentropic atmosphere.
Example 1. A packet of dry air rises. Let’s first assert that it is adiabatic (hopefully not controversial), as it rises it expands and does work on it’s environemnt. But that work is reversible. Thus the process is adiabatic and reversible, hence isentropic. By definition.
Example 2. We know from calculus that dx/dy(w)*dy/dw(x)*dw/dx(y) = -1. Let us apply this identity to dT/dp(s), the temperature pressure relationship at constant entropy. We will find that dT/dp=-dS/dP * dT/ds. the second RHS term is Cp/T by the definitoin of entropy. The first RHS term is, by a Maxwell relationship, -dV/dT, which for an ideal gas is R/P (you can google Maxwell relationship). Making the substitutions: dT/dp = RT/PCp = V/Cp for an ideal gas. Lastly we recognize that the gravity imposed pressure gradient is dP/dz = rho*g, but rho is 1/V, so when we multiply: dT/dp * dp/dz = V/Cp * g/V = g/Cp.
Example 3. The state of an ideal gas is fully specified when two variables are specified. We shall choose T and p. The total entropy differential is: dS = dS/dT * dT + dS/dp * dp. at constant entropy, dS=0, then dT/dp = dS/dp * dT/dS and we proceed as above.
So, it seems quite clear to me that the near surface atmosphere is isentropic.
The moist adiabatic lapse rate MALR differs from the DALR because it must account for the latent heat release as water condenses. This extra heat slows the rate of cooling of a rising air packet, so the MALR is less than the DALR – but it is still isentropic as it satisfies all the condition of my example 1, but the math is kinda nasty.
The adiabatic lapse rate need not be an equilibrium critereia, but instead a steady state condition if heat transfer in the near surface atmosphere is dominated by convection.
Let’s set moisture aside for the moment and set up a simple dry ideal gas atmosphere. Incoming radiation heats the planet surface. Air at the surface is heated and begins to rise, thus convection begins. Convection is attempting to return the atmospheric temperature gradient to zero, but as it rises it experiences an isentropic expansion, which causes it to cool (the reverse also occurs). If convection is the dominate mode of heat transfer (conduction negligible), it cannot drive the temperature gradient to zero, but only to the DALR.
Thus, I think the DALR is a steady state condition that arises because, in the near surface atmosphere, convection is the dominate heat transfer mode and it is a compressible fluid with a gravity imposed pressure gradient, hence convection is constrained by the DALR.
Lastly, if I assume the entire atmosphere is isentropic – all the way to the tippy top – then T1/T2 = P1/P2^0.4. If I solve for T2 letting T1 and P1 be the conditions at the tippy top of the atmosphere, I calculate an enormous value for T2, the surface temperature. Clearly the atmosphere cannot be isentropic all the way up. At some point it becomes non-isentropic.
I think the isentropic condition breaks down when convection ceases to be the dominant mode of heat transfer. As you rise, the atmosphere becomes less dense, convection less effective, until eventually heat transfer is dominated by radiative heat transfer. Radiative heat transfer is not constrained by the DALR and can drive the temperature gradient to zero. Hence the planet surface temperature is not enormous.
So, the non-radiative atmospheric thermal effect becomes an exercise in identifying at what point in the atmosphere does convection cease to dominate, which is also the point where the isentropic assumption breaks down. If that point is known (yes I know it won’t be a sharpt break between convection and radiation, but let’s keep it simple), then the equation above can be used to find the increase in surface temperature resulting from an isentropic near surface atmosphere.
Now, I could very well be wrong, or have made mistakes along the way (I do that sometimes – maybe even more than sometimes). But please don’t wave your hands at me and tell me to “check the meaning of entropy” – I find that frustrating. I think I’ve shown sufficient work…
Show me my mistake. Anybody. I won’t be offended.
T = temperature
p = pressure
V = olume
R = ideal gas constant
Cp = ideal gas heat capacity = 5/2R
g = gravitational constant
rho = density = 1/V
z = vertical spatical coordinate.

John Marshall
January 24, 2012 7:26 am

Very interesting but really irrelevant given that our atmosphere is not an ideal gas in a cylinder. Gas, ideal or otherwise, is a very poor heat conductor compared to Silver so perhaps your example above is not very good and enrtopy will still increase. It can be demonstrated by observation that convecting gas does rise, due to the density difference between the rising gas and that of the surroundings. If we consider real air it can be saturated with water vapour and rise to form clouds. The rising air will cool adiabatically at the SALR (4c/1000m rise) but will descend, as it must as dried air having the water vapour removed in the cloud formation, and warm at the DALR (9.8C/1000m) and end up with more heat than it started with. Rather like a Foehne Wind in a vertical loop. Katabatic winds are warmer at the bottom of their descent than the cold mountain start. The atmosphere is never in equilibrium because the planet rotates, there is a non-uniform surface and moving clouds which alter the solar energy falling on the surface. It is the lack of atmospheric equilibrium that gives us weather.
I am not being obtuse here but there is so much wrong with the GHG theory, cooling planet and rising GHG’s, lack of the predicted tropospheric heat, and the violation of 2nd law with this heat transfer from a cool troposphere to the warmer surface. If you could do that you really would have a PMM. So an alternative mechanism must be found to explain the BB heat anomaly, assuming that this is correct.

Ed Caryl
January 24, 2012 7:36 am

“In nature, the dry adiabatic lapse rate of air in the atmosphere is maintained because the system is differentially heated from below causing parcels of air to constantly move up and down.”
I live half-way up a mountain at 6100 feet. The valley below is at 4500 feet. The temperature difference is nearly always the dry lapse rate 8 degrees F, whether it is calm or windy. Only if it is raining or snowing will it be different. It then goes to the moist lapse rate. Most of the time it is sunny, heating the ground equally, both in my back yard and in the valley below. What maintains the lapse rate temperature difference?

Bryan
January 24, 2012 7:38 am

In the real atmosphere the Earth surface is heated and radiation cools at top of atmosphere.
Convection is the major method of heat transfer.
Is convection always present?
The answer according to textbooks is no.
We can have the interesting situation where there is little or no convection, still air in other words.
This condition is called the Neutral Atmosphere.
This atmospheric condition is known as the neutral atmosphere and can be stable particularly at night.
See the near Neutral RESIDUAL LAYER page 31
What happens then?
Robert Brown says
“What maintains the adiabatic lapse rate is convection”
Nick Stokes and Joel Shore would agree.
So if convection is absent presumably the lapse rate disappears!
Well not in the real world.
If the air is dry, the lapse rate is at its maximum of g/Cp = – 9.8K/km
Climate Science define convection as an UNSTABLE vigorous vertical exchange of air. .
See bottom of page 13.
The stable condition (hydrostatic approximation) is used to derive the DALR. See page 12
This condition holds for still air and air parcels moving up and down at constant speed (no unbalanced force) will track the DALR.
These air parcels are assumed not to exchange heat with their surroundings.
On going up expansion work PdV is stored by the surroundings(temperature dropping by 9.8K/km)
At TOA there will be a loss of heat by radiation to space causing the down phase
On going down the surroundings do work on the parcel (PdV) (temperature increasing by 9.8K/km)
Stationary parcels will not change temperature.
These two idealised adiabatic processes (like the adiabatic stages in the Carnot Cycle) will result in the parcel returning to Earth with nearly the same temperature as leaving (the slight drop being accounted for by radiation at TOA).
http://www-as.harvard.edu/education/brasseur_jacob/ch2_brasseurjacob_Jan11.pdf

January 24, 2012 7:39 am

@Joules Verne: “This is obviously the case since it goes without a shadow of a doubt that a molecule of air in a higher layer has more gravitational energy than a molecule in a lower layer.” Does the fact that the air in a higher layer also has fewer molecules matter? That is, are we talking individual molecules or rather volumes of molecules here?

Douglas Hoyt
January 24, 2012 7:50 am

Two points:
1) If the atmosphere were isothermal, then a unit mass of atmosphere would have greater total energy (thermal + potential) the higher it is located and the further it is away from the surface heating source. That is not a stable situation.
2) The silver wire will transport heat from warmer region to the cooler region, but in so doing it short circuits the transport of heat by convection. So with the wire present, convection will be less, but the net transport of heat will remain the same.

Trick
January 24, 2012 7:52 am

This head post now makes me want to more fully question all the textbooks to which I’ve ever been exposed.
Consider this simplest and no simpler demonstration:
Robert Brown’s wire is U shaped. This sudden U-turn enables the wire to enter a 2nd thermal energy reservoir (another control volume that happens to be a gray colored one). The wire, in Robert’s example in his words, is: “adiabatically insulated except where it is in good thermal contact with the gas at the top and bottom of the container”.
Thus the wire is adiabatically insulated from the temperature field of the gas in the white colored area. Heat will indeed flow until the gray reservoir is in thermal equilibrium with the white reservoir. This just shows why there are no perfect insulators – Perpetuum Mobiles could be constructed in gas in a gravity field. This IS textbook stuff.
Why did Robert Brown have to go to the trouble of constructing a second gray reservoir with the U-turn? Robert Brown needed a second thermal body.
Trick’s view is Robert Brown should run this analysis again with the wire not leaving thermal contact w/white colored control volume gas and report back with only one thermal body or one energy reservoir or one thermodynamic system. Call it what you will.
Meaning Robert Brown is allowed only one heat reservoir to demonstrate his proposed isothermal gas column where the wire stays in thermal contact with the white colored gas everywhere – no U-turns as here to a 2nd thermal reservoir. Trick’s view is Robert will be unable to do so – the gas column will not be isothermal – there will be a temperature lapse rate.
Trick’s view remains that Robert Brown’s proper application 0th, 1st,2nd Thermo Laws will enable Robert Brown to eventually see the one thermodynamic system GHG-free gas column w/gravity is not isothermal in theory since Robert Brown is smart and the thermo grand masters are right.
NB: I am posting here b/c I have had a miserable head cold last few days and was looking for a way to pass the time. It has been interesting & fun to re-learn about thermo. I have to thank Robert Brown (and Willis) for a more enjoyable few days than I would have had otherwise .

Joules Verne
January 24, 2012 7:52 am

steveta_uk says:
January 24, 2012 at 6:42 am
“I’m not sure that Dr Brown will convince them otherwise.”
Brown won’t convince him this way.
“But maybe Hans Jelbring can convince us all that the heat in figure 2 cannot rise up the silver bar because graivity is holding it down!”
The device in figure 2 doesn’t work because it’s a closed system and the work extracted will reduce the total energy of the column until eventually there’s no more energy to extract at which point the gas reaches a temperature of absolute zero and has presumably vanished from this universe being totally converted to kinetic energy in the extracted useful work. In the real world the gas will collapse to the surface as a liquid before it gets to absolute zero and this will shut off further extraction of energy because the cold side of the thermocouple no longer has any cold gas to cool it.

Eilert
January 24, 2012 7:54 am

When doing thought experiments it is wise to think about what might have been assumed.
I see two assumptions above:
1. It does not matter what the density of the gas is. It will equally conduct heat at the bottom into silver wire, as the wire will be able to conduct its heat into the gas at the top, even though the density at the bottom and top is different, due to the gravitational effect on the gas.
2. The cross-section of the wire will stay the same, which means the ability of the wire to conduct the heat, which depends on its cross-section, is the same at the bottom and top.
The gravitational field will actually pull down a considerable part of the mass to the bottom, making it far wider at the bottom then the top (depending on the length of the wire and its tensile strength), deforming it more into a tear drop shape.
With your setup you may be able to change the lapse rate, but I doubt that you achieve an isothermal state in this way.

The iceman cometh
January 24, 2012 7:58 am

I find the analysis quite reasonable – but it is so idealized as to be useless. A more interesting thought experiment has a spherical planet heated by a remote star, rotating on an axis roughly normal to the line to the star, with an atmosphere of non-greenhouse gases. The equator would be warmer than the poles, so there would be Hadley-type circulation that would cool the equator and warm the poles. Would there then be a vertical thermal gradient? I think there would be, but I’m sure someone would like to argue to the contrary.

Archonix
January 24, 2012 8:00 am

I see a lot of people talking about heat and temperature as if they’re the same thing here, then basing their arguments on that false premise.

pat
January 24, 2012 8:03 am

Isn’t that, for the most part. why we have variable wind? And the reason there is not a pocket of ‘missing’ heat 800 meters below the oceans surface, as is commonly conjectured by Hansen, et al.

Marc77
January 24, 2012 8:04 am

It clearly shows that without an already present differential of temperatures, gravity cannot create one. But the atmosphere is a different problem. It is heated at the bottom and it loses its heat in altitude. So the question is, what can impede the flow of heat from the warmer ground to the cooler “layer of emissions”. It seems to me that both greenhouse gases and gravity would enhance the lapse rate or impede the flow of heat between those 2 layers.
I often hear people say that nights would not be as warm if IRs were not coming from the atmosphere, but what about gas particles falling and hitting them all of the time?

January 24, 2012 8:06 am

This is funny : ) Almost everyone is right.
Take a single gas molecule and put it at the top of the tube. It has zero kinetic energy and zero temperature. Let it fall and just before it hits the bottom it will have a lot of kinetic energy and heat. That is the lapse rate and it isn’t in equilibrium by definition.
Now place the atom at the bottom of the tube, it is now in isothermic equilibrium, its kinetic energy and temperature is zero.
If we let the atom bounce up and down in the tube, and don’t allow any energy to be extracted, it will stay in perpetual motion (and gravity will be continuously accelerating it) and will have a lapse rate (as long as the lapse rate isn’t measured).

Joules Verne
January 24, 2012 8:07 am

Wayne2 says:
January 24, 2012 at 7:39 am
“@Joules Verne: “This is obviously the case since it goes without a shadow of a doubt that a molecule of air in a higher layer has more gravitational energy than a molecule in a lower layer.” Does the fact that the air in a higher layer also has fewer molecules matter? That is, are we talking individual molecules or rather volumes of molecules here?”
No. The fewer molecules must have sufficient gravitational energy that, if it were converted to kinetic energy, would be able to raise the temperature of a larger number of molecules in the lower layer the temperature in that lower layer. It MUST be isoenergetic to satisfy the second law. It need not be isothermal. In politics they say to follow the money to arrive at the truth. In physics you want to follow the jewels joules.

markus
January 24, 2012 8:10 am

“Follow the air column up above the troposphere and the lapse rate fails to be observed in the stratosphere, precisely where vertical convection stops dominating heat transport. The EEJ assertion, that the dry adiabatic lapse rate alone explains the bulk of so-called “greenhouse warming” of the atmosphere as a stable feature of a bulk equilibrium gas, is incorrect”.
So your wonderful assertion, is that the radiative forcing of Co2, occur after its entry into the thermostats of the tropopause, and that extra radiative forcing, causes that missing hot spot, increasing the temperature back through the stratosphere and down again through the thermostat of the tropopause.
Been there, done that.

January 24, 2012 8:14 am

My name is Joules because I know how to find and count them no matter how they try to hide.
Well then, by all means go patent your perpetual motion machine of the second kind or explain heat flow in the second diagram, Joules.
And I’m ever so sorry, but in an ideal gas the temperature is not determined by the total energy. That’s an absurd idea, given that one can perform a gauge transformation — change the zero of the total energy — without changing any of the physics. What matters is the distribution of energy in degrees of freedom. The number of degrees of freedom in an ideal gas does not depend on whether or not the air is in a gravitational field. Take a sealed jar full of air at temperature T and gently carry it upstairs, and it is still at T.
But all of this is too difficult for you, so stick with explaining why figure 2 — based on absolutely trivial physical principles would not occur as described, given a thermal lapse rate in the gas. Is there something miraculously interesting in the thermal contact between silver and air that keeps heat from being conducted from hot to cold — in just this one special circumstance? I’m all ears.
rgb

Marc77
January 24, 2012 8:15 am

Also, a constant temperature with altitude means that particles at the top of the atmosphere have more momentum than particles at the bottom. Can you show that this sorting will happen at the molecular level? Or can this sorting happen only by convection of masses of air?

Steeptown
January 24, 2012 8:18 am

Robert:
I would be interested in your take on the paper at http://arxiv.org/PS_cache/arxiv/pdf/0812/0812.4990v3.pdf where it is conjectured that “….If one could concede that the true equilibrium state may be isentropic instead of isothermal “

Joe Born
January 24, 2012 8:21 am

That Dr. Brown has it wrong is readily demonstrated by a thought experiment nearly any layman can perform.
If an ideal monatomic gas subjected to gravity in a thermally isolated container consists of only a single molecule, its kinetic energy K–and thus the mean translational kinetic energy–at any altitude z is given by K = mg(z_max -z), where m is molecular mass, g is the acceleration of gravity, and mgz_max is the total (kinetic + potential) energy of the gas. This is true no matter how long you’ve allowed the gas to “equilibrate.” In other words, temperature depends on altitude at equilibrium: there’s a non-zero temperature lapse rate.
Extending this result to any number N of moleculles yields
K = 3 mg(5N-2)(1-z/z_max),
an equation that I’ve adapted from Equation 8 of the Velasco et al. paper, to which I was introduced here: http://tallbloke.wordpress.com/2012/01/04/the-loschmidt-gravito-thermal-effect-old-controversy-new-relevance/. Note that temperature still depends on altitude.
As a practical matter, this result differs only negligibly from the isothermality for which Dr. Brown argues if the number of molecules is large. As Dr. Brown states, though, “[t]he magnitude of the [temperature difference, and the mechanism proposed for this separation are irrelevant,” to his attempted refutation. So the fact that there is any non-zero lapse rate at all at equilibrium establishes that Dr. Brown’s attempted refutation is invalid.
I hasten to add that the lapse rate that does prevail at equilibrium is much smaller than that for which Jelbring contends, so Jelbring is still wrong. .
I should also state that I was not able to follow each and every step of Velasco et al. and the Román et al. paper on which it relies. But its result is consistent with the thought experiment above, whereas Dr. Brown’s isothermality theory is not. Moreover, the Román et al. paper starts from a statistical-mechanics basis,, i.e., from first principles, rather than being based based on blindly accepting equations as received truth without double-checking their ranges of applicability.
I would welcome the assistance of any true physicists out there in examining those papers’ equations further.

kdk33
January 24, 2012 8:21 am

says: These two idealised adiabatic processes (like the adiabatic stages in the Carnot Cycle) will result in the parcel returning to Earth with nearly the same temperature as leaving (the slight drop being accounted for by radiation at TOA).
Carnot cycles are ISENTROPIC. By definition. Isentropic means adiabatic AND reversible.

January 24, 2012 8:23 am

Hmmm. Temperature is the integral of the number and energy of particles seen at the measuring surface. Less particles equal lower temperature given equal velocity profiles of the particles. Less particles also equals lower pressure. Temperature and pressure are thus directly linked. Hence the dry lapse rate. Gravity and pressure are also directly linked. Hence the dry lapse rate.
Sure, sure, sure. But no. I provide the explicit algebra that shows that an isothermal gas is perfectly happy supporting itself. If you want to discuss the temperature of the gas, learn what microscopic temperature is, because it does not depend on the number of particles. 10^18 particles of gas in a container can have any temperature you like. So can 10^23. If those two containers have the same temperature, they have the same average kinetic energy per particle (for a monatomic gas). This doesn’t even depend on the mass of the particles.
However, the reason I drew the pictures is so you could all stop pretending that you can do stat mech computations in your head without even knowing what molecular temperature actually is, and concentrate on easier stuff, like heat flow. If the stable thermal equilibrium of the gas in figure 2 has a lapse rate, heat has to be resorted by gravity from the top to the bottom order to maintain the lapse rate as heat flows in the silver! Heat will definitely flow in the silver, right? It’s just a chunk of metal that’s an excellent conductor of heat. Put it in good thermal contact in between gases at two different temperatures, heat will flow because there isn’t any bullshit about gravity that you can invoke without understanding it. It’s just like Newton’s Balls — whack it on one end and the whack is transmitted, more or less undiminished, to the other end until thermal equilibrium is reached.
Only it is never reached, is it? As fast as you warm the top, gravity has to move the heat to the bottom to restore the lapse rate, which means that it keeps flowing through the silver to the top, where it flows back to the bottom, where it flows to the top — perpetual motion — of naked heat, absolutely predicted by high school physics.
You want to assert otherwise, you tell me what the equilibrium state is of figure 2.
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January 24, 2012 8:25 am

Joules Verne – I think you are getting at the crux of the matter here.
These “idealized” descriptions are REALLY DEADLY in this debate.
WHY? What would be the “equilibrium explanation of my coming into LAX 10 years ago, from Hawaii, and watching the temperature INCREASE until 6,000 Feet ASL, where it was 80 F. Then from that point on, until we hit the ground there was an INVERSE lapse rate, going to 65 F on the ground.
Certainly this is a demonstration that “equilibrium thermodynamics” is NOT a proper way to approach any correct modeling of the atmosphere.
It become a difficult, and intractable problem which does not yield to simple differential equations applied to idealized columns of gasses in “ideal” states.

January 24, 2012 8:28 am

Show me my mistake. Anybody. I won’t be offended.
No, I think you are generally quite right, and this agrees rather well with Caballero’s argument. Isentropic because it is dominated by convection, not conduction, in an open system heated at the bottom. Isolate the system, or heat it at the top and explain to me how the bottom will end up warmer than the top.
Yeah, right. Just like the oceans. I wonder why the argument fails for the oceans? They seem to come into thermal equilibrium at, well, thermal equilibrium (constant temperature, independent of pressure, density, “gravity” etc), below the convection-dominated thermocline.
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commieBob
January 24, 2012 8:29 am

Ignoring conduction and radiation for the time being and considering only convection:

A column of gas which is sufficiently tall that there is a pressure difference between top and bottom will also have a temperature difference between top and bottom.

For convection to occur, there must be a difference in density. If there is no density gradient, there will be no convection.
The ideal gas law is:

PV = NkT

where:

P is the absolute pressure of the gas measured in atmospheres; V is the volume (in this equation the volume is expressed in liters); N is the number of particles in the gas; k is Boltzmann’s constant relating temperature and energy; and T is the absolute temperature.

Density in molecules per unit volume would be:

density = N/V

Rearranging, we get:

N/V = P/(kT)

So, there will be an evenly varying gas column with no convection because:

Ntop/Vtop = Ptop/(kTtop)
=
Nbottom/Vbottom = Pbottom/(kTbottom)

and

Ptop/(kTtop) = Pbottom/(kTbottom)

So, we would expect the atmosphere at the surface of a planet to be warmer that it is at the top of the atmosphere. Of course, we can’t totally ignore conduction and radiation but, compared with convection, they are second order effects.

Greg Elliott
January 24, 2012 8:33 am

Clearly there are two schools of thought. One school believes that the temperature will be lower at the top due to kinetic energy being changed to gravitational potential energy. The other school believes this will not happen. The GHG controversy rests largely on this point.
Where are the learned papers where someone has actually conducted experiments to test this? Measured the temperatures in a gravitationally bound column to determine if we have built a 150 years of science on a faulty assumption.
I am troubled on one point. The argument that a continuous flow in a cycle is not equilibrium and thus is some sort of proof favoring one school over the other. Surely dynamic systems can be in “equilibrium” in that there is no net flow into or out of the system, but still allow a cyclical flow within the system.
The only definitive test is observation. Otherwise we could turn science over to theorists with computers. We cannot “test” this question with calculations that are in any way based on the same underlying assumption. They will by necessity confirm the school of thought they are based on.
Human beings have an infinite capacity to rationalize. History shows that a single faulty assumption does not in the least prevent us from building a huge body of self-confirming science in support of the assumption. In the end however, nature has an infinite capacity to surprise.
As our technology improves we gain the ability to replace assumption with observation and uncover in which direction the truth lies. Thus the development of Relativity to explain small observed variations in the orbit of Mercury as compared to the predictions of Newton.
On average temperatures are warmer at sea level than at mountain tops. We could run a silver wire between the two and heat would run through it indefinitely. This in itself does not appear to favor one school of thought over the other. I’d like to see the observation evidence.

January 24, 2012 8:36 am

I agree that a column of Ideal Gas in the Dr. Brown’s column above will be isothermal at equilibrium.
Now, let’s fill the column with a GHG, say CO2.
Case C0: Leave it in the dark. It has a pressure gradient, more CO2 at the bottom. If the Bottom of the column is at temperature T1b, what will be the top temp T1h? Since the bottom is at T1b, the column is receiving and emitting energy according to SB theory. If necessary, put the column in a room where all the walls are held at T1b.
Case C1: Leave it in the room, but turn on the lights. Bathe it in 240 w/m^2 from all directions. Will it be isothermal, or establish a gradient. I think it will stay isothermal, Despite more CO2 at the bottom of the column. Will the result depend upon T1b?
Case C2: Leave the light on, but replace the CO2 with O2 of the same mass (higher pressure). What will change?
Case C3: Same as Case C2 but use the same molarity (same number of gas molecules) as C1.
T1b is remains the temp of the base of the column, the floor, walls and ceiling of the room in all cases.
Case C1D: Return to case C1. Now turn on the lights for 1000 sec, Turn the lights off for 1000 sec. Repeat. This is an optical pumping, but T1b says the same.
Case Pxx: Same as above except now we leave the room at T1b, but pump the base of the column with T1b(t) = T1b(0) + 10 * ( int (time_sec)/1000 mod 2)
(i.e. every 1000 sec, raise or remove 10 deg K in a square wave)

January 24, 2012 8:39 am

I am not being obtuse here but there is so much wrong with the GHG theory, cooling planet and rising GHG’s, lack of the predicted tropospheric heat, and the violation of 2nd law with this heat transfer from a cool troposphere to the warmer surface. If you could do that you really would have a PMM. So an alternative mechanism must be found to explain the BB heat anomaly, assuming that this is correct.
Please, if you heat your house when it is cold outside, and then add insulation (defined to be anything that slows the transfer of heat), is the second law of thermodynamics violated if the house gets warmer? I don’t think so.
I’ll just repeat what I’ve posted on many other threads. The Greenhouse Effect itself is positively confirmed by the actual measurements of the IR spectra from above the atmosphere. Asserting that it doesn’t exist is just plain stupid when you can measure the actual radiation being given off by the CO_2 and the surface.
If you want to complain about the “upwelling” and “downwelling” radiation arguments, well, I find them unconvincing as well, but that has nothing to do with whether or not the GHE exists! The primary place the atmosphere cools is up at the top of the troposphere — via radiation from a single optical path length thickness of the optically thick (in a selected band) CO_2.
Why is it that you want to fight over physics that you can actually see with IR eyes? Save your energy for useful things, like arguing about the magnitude of the GHE, the sensitivity of it to changes in CO_2 concentration, the sign and nature of climate feedback or albedo modulation or the complex effects of atmospheric convection on local heating or cooling rates, or the ocean’s effect. The IR spectra render arguing about GH warming per se moot.
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Joules Verne
January 24, 2012 8:41 am

P = T*V helps to understand what’s going on. One must constantly keep in mind that in the gravitationally bound column of gas pressure is constant while temperature and volume are the variables. As its temperature goes up and down its volume goes up and down. Surface pressure is determined by gravitational constant and mass of the gas which do not vary. Temperature is not coupled to pressure therefore pressure is not coupled to temperature. So raising the surface pressure will not cause a rise in equilibrium temperature. It will cause a rise in volume and the gas law wil be satisfied by the change in volume.

January 24, 2012 8:44 am

Robert Brown says
“What maintains the adiabatic lapse rate is convection”

What maintains the adiabatic lapse rate is energy transfer between reservoirs at different temperatures. Read your own description. Turn off the radiative cooling of top of atmosphere, and the radiative heating at the bottom, and you get — not overnight, but you get — inversion.
You might try looking at the “DALR” over antarctica around July. Oooo, top of atmosphere hotter than the bottom! How did that happen?
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Alan Robertson
January 24, 2012 8:48 am

Joules Verne says:
January 24, 2012 at 7:04 am
“My name is Joules because I know how to find and count them no matter how they try to hide.
Thanks for playing.

______________________
How.s the family, Joules?

A physicist
January 24, 2012 8:48 am

It seems to me Robert Brown’s analysis implicitly makes the following claim: if all the greenhouse gases (mainly H20 and C02) were cleansed tonight from earth’s atmosphere, then the atmosphere would evolve toward a more nearly isothermal equilibrium
But how would that work, exactly? Because on the first day following the cleansing, the sun would still warm the earth during the mornings, thermal currents would still rise during the afternoons, and these afternoon parcels of air would still adiabatically cool as they rose …
So upon this cleansed-of-GHG planet Earth, how exactly would the atmosphere’s temperature profile evolve toward a more nearly isothermal profile, in which rising thermal updrafts were weaker than in the present atmosphere?
If Robert Brown answered this question clearly (and it is a subtle question IMHO), then it seems to me that his theoretical ideas would prevail.

January 24, 2012 8:49 am

I thought the compressed gas at the bottom in relation to the less compressed gas at the top simply contained more heat energy/volume even though all molecules in the column would have the same level of excitation.

January 24, 2012 8:55 am

The person who observes that temperatures, in general, decrease with height in the troposphere, and who notes that pressure does too, might well be inclined to link one with the other in a causal way not least because of course you can raise the temperature of a gas by compressing it. It is tempting. But a source of great complexity in our spinning, turbulent, inhomogeneous air is the fact that it is largely heated from below, differentially by latitude, by surface properties, by cloud cover, and by time of day and time of year. This simple exposition cuts through all that befuddling complexity to highlight the role of gravity in thermal isolation, i.e. with no heat transfer at the base, or anywhere else on the bounds of the containing surface. The thought-experiment with the silver wire applies the coup-de-grace. I also liked the notion that gravity is a poor candidate to be Maxwell’s demon! I conclude that a thermally isolated container of gas in zero gravity and at uniform temperature, will not see any temperature change if some other demon could switch a gravitational field on at will. Merely the creation or a density gradient. Is that right?

Schrodinger's Cat
January 24, 2012 8:57 am

The air at the bottom has a higher temperature as a consequence of being compressed by the mass of air above being acted upon by gravity. Over time, I would expect the system to reach thermal equilibrium. I am assuming that there is no introduction or loss of energy from the system.
In the real world (our atmosphere) the ground heats the air above continuousy during daylight hours and there are many other processes such as convection, radiation and evaporation that lead to energy leaving the system, so the process is never at rest and the temperature gradiant is maintained.

January 24, 2012 9:01 am

Take a single gas molecule and put it at the top of the tube. It has zero kinetic energy and zero temperature. Let it fall and just before it hits the bottom it will have a lot of kinetic energy and heat. That is the lapse rate and it isn’t in equilibrium by definition.
No, it won’t have any “heat”. You are conflating work, organized kinetic energy, and heat. Drop a jar of air. Are you asserting that a thermometer placed inside will go up as it falls? Of course not. It is when it inelastically collides at the bottom, and the organized kinetic energy (which is quite capable of doing reversible work still) becomes disorganized, moving into the far more probable state with the same total energy but with the particles of gas moving every which way, that we might talk about “heat”, but even that is really a false idea.
Temperature of a monatomic ideal gas is one thing, and one thing only. It is a direct measure of its internal, disorganized (equilibrated) average kinetic energy. Not its kinetic energy in a moving frame, not its potential energy in a moving frame or otherwise. Only its plain old kinetic energy. To be very specific, the gas will be in equilibrium at a given temperature T when the distribution of the molecular kinetic energies (or by transformation speeds) is given by the Maxwell-Boltzmann distribution.
The problem is that if you drop the molecules as you describe in a real tube, they won’t just bounce up and down. They’ll bounce sideways and quickly “thermalize” as they collide, eventually sharing the kinetic energy around so that the probable transfer of energy in every collision is the same in both (all) directions. That’s the rub. An ideal gas collides instantly — a hard sphere approximation, like perfectly elastic pool balls. Gravity has no time to act during the collision. The solution for the pressure at a constant temperature above indicates that a gas is perfectly happy to support its own weight and density/pressure profile at a constant temperature, and at a constant temperature all collisions have equal probabilities of heat transfer in all collisions in all directions. That simply isn’t the case if the MB distributions (and hence temperatures) vary with height.
But the basic point of my paper is that Jelbring is wrong not because of any possible microscopic description of a lapse rate. A lapse rate itself is wrong in thermal equilibrium, because figure 2 is very, very easy to understand. There is no question that the silver will conduct heat between reservoirs at different heights exactly the same way it does any other time. If you doubt me, put a pan on the stove, put your fingers on the pan, turn on the heat below, or hold onto a piece of solder while you heat one end of it. Or read wikipedia articles on the heat equation or Fourier’s Law. Or take a course in stat mech.
If any lapse rate is stable, the system violates the second law, as heat will flow through the silver for any difference in temperature until there is no difference in temperature, and therefore any steady state that still has a lapse rate must transport heat down the gas column on the left from colder to hotter (violating the second law all by itself, but it is so difficult for people to understand this, alas). And we’re done. No, heat will not flow forever in any physical system.
Done.

Alan Robertson
January 24, 2012 9:02 am

Joules Verne says:
January 24, 2012 at 8:07 am
…” The fewer molecules must have sufficient gravitational energy that, if it were converted to kinetic energy, would be able to raise the temperature of a larger number of molecules in the lower layer the temperature in that lower layer. It MUST be isoenergetic to satisfy the second law. It need not be isothermal. In politics they say to follow the money to arrive at the truth. In physics you want to follow the jewels joules.”
_______________________________
How can the conversion be isoenergetic while remaining isothermal? Is it adiabatic with an increase in pressure?

January 24, 2012 9:07 am

No. The fewer molecules must have sufficient gravitational energy that, if it were converted to kinetic energy, would be able to raise the temperature of a larger number of molecules in the lower layer the temperature in that lower layer. It MUST be isoenergetic to satisfy the second law. It need not be isothermal. In politics they say to follow the money to arrive at the truth. In physics you want to follow the jewels joules.
You leave me — almost — speechless.
I can only reiterate — you tell me what the heat flow will be in figure 2 above. Which is violated — the heat equation in silver or your absurd assertion that gravity can stably sort out a gas into a hotter temperature and a colder one?
One or the other. Only one makes you look silly, though.
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MDR
January 24, 2012 9:32 am

Two distinct scenarios are being discussed here in a somewhat confusing fashion. One scenario involves an adiabatically stratified system and the other involves an isothermal system. It is important to realize that thermodynamic equilibrium does not necessarily mean the gas is isothermal. It is possible for a system to be in thermodynamic equilibrium and *not* be isothermal, as in the case discussed here where gravity stratifies the gas with height.
Here is a more detailed explanation. In the system described above [without the wire], there are only two sources of energy: gravitational potential energy and internal energy. Because matter located at lower z [height] will have a lower amount of gravitational potential energy, it then follows that matter located at lower z also has a greater amount of internal energy. As a result, the total energy [that is, the sum of gravitational potential energy and internal energy] with height is a constant, and the system can be said to be in thermodynamic equilibrium. If this weren’t the case, then energy transfer would occur, in the direction so as to equilibrate the total energy.
Now, we just showed that internal energy decreases with height, as explained above. Since internal energy [of an ideal gas] is directly proportional to temperature, this must mean that temperature also decreases with height. The gradient of temperature with height is of course the lapse rate. This is an example of a system that is both in thermodynamic equilibrium and possesses a gradient in temperature.

Jeremy
January 24, 2012 9:42 am

Thank you Dr Brown.
As many have commented here, this blog has become a haven for rather foolish comments.
It proves that skeptics are just as gullible to poor logic and bad science as the CAGW.
When something fits with your world view there is a tendency to embrace it.
However, Science and Physics is NOT about anyone’s world view – as James Brown would say “It is what it is”.
I don’t buy the scaremongering CAGW nonsense about man-made CO2 because the Science and Physics clearly do NOT support such claims. However, for the very same reason, I cannot support some of the wildly inaccurate nonsense science being discussed here lately in these forums.

Greg Elliott
January 24, 2012 9:49 am

Consider sunlight reaching the earth’s surface. This heats the surface and energy is radiated back to space. The incoming and outgoing energy must balance.
Add GHG to the atmosphere and some of the outgoing radiation will be intercepted and prevented from reaching space. Thus the surface temperature of the earth must increase to increase the out-flowing radiation and restore the balance.
Now, consider what happens if at the moment a molecule of the earth’s surface is about to emit a photon to space, instead a molecule of N2 comes into contact with the surface and the energy from the surface is instead conducted into the molecule of N2. This flow of energy through conduction will reduce the surface temperature and prevent the photon from being radiated to space.
This will have the effect of decreasing the out-flowing radiation from the surface, in a manner that is for all intents and purposes indistinguishable from the effects of GHG. Thus, the temperature of the surface must rise to restore the out-flowing radiation that is being lost to conduction. The greater the atmospheric pressure, the more N2 molecules, the greater the likely-hood that conduction will take place limiting radiation from the surface to space, the more surface temperatures must rise.
But what about the energy absorbed by the N2 some might ask. Indeed and what about the energy intercepted by the CO2? Both must either heat the atmosphere or be returned to the surface and thus are indistinguishable in their effects.
However, the GHG theory tells us that only radiative transfer is responsible for warming the surface, that conduction cannot have this effect. Yet it is clear to see that convection limits radiation to space in a manner that is for all intents and purposes equivalent to GHG. The greater the pressure, the more likely this becomes, the more surface temperatures must rise.

January 24, 2012 9:51 am

Meaning Robert Brown is allowed only one heat reservoir to demonstrate his proposed isothermal gas column where the wire stays in thermal contact with the white colored gas everywhere – no U-turns as here to a 2nd thermal reservoir. Trick’s view is Robert will be unable to do so – the gas column will not be isothermal – there will be a temperature lapse rate.
Excuse me? I have no idea what you could possibly be talking about. Look, grab a copper wire by one end. Hold the other end in the flame of your stove. I don’t care what shape it has, you will burn the hell out of your fingers (and keep burning them until your fingers are at the same temperature as the flame).
My picture shows a wire insulated on the sides so that any heat that goes into the wire can’t come out anywhere but the end. It just makes the wire a one dimensional conductor of heat. Put the damn wire (insulated on the sides) right into the container, perfectly straight if you like. As long as the bottom UNinsulated end is in contact with the gas at the bottom at T_b, and the top UNinsulated end is in contact with the gas at the top at T_t, and T_b > T_t, heat will flow in the wire from the bottom to the top.
The point is that heat will flow in this system forever if you postulate that gravity will maintain a lapse between the bottom and the top stably. That means that any small packet of heat that is moved around in the gas has to eventually settle back down into equilibrium, and you are asserting that equilibrium has a lapse. So when the wire carries heat from the bottom to the top — which it will — gravity has to sort it back down to the bottom, because you assert that a lapse is the stable equilibrium.
Only it won’t. If it did, the second law wouldn’t be satisfied, heat would flow forever.
The real point is that you don’t need the silver wire to make this argument. The gas itself conducts heat from the bottom to the top as long as the temperatures are different. It’s what systems do. Conduct heat from hotter places to colder places, unless you do work to prevent it. Gravity does no work in this problem, not in steady state. So what makes the heat go round and round?
It doesn’t.
Of course.
It evolves to the isothermal state where no heat flows.
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Nick Shaw
January 24, 2012 9:52 am

I am not a scientist and never claimed to be so, could someone explain why the gas, or atmosphere in this case, should be colder on top than on the bottom assuming convection works in all cases (cold air falls while hot air rises) Yes, I can figure, as air gets closer to outer space (in really simple terms) it would get mighty cold but, cold air is more dense and as such it should fall more rapidly. Exactly where does gravity enter the picture? It is exerted equally on all temperature states of air, right?
Or should I up my meds? 😉

January 24, 2012 9:54 am

RGB
thank you for this elegant demonstration that a gas in a gravitational field is isothermal when in equilibrium. I lost count of how many times I pointed this out on Willis Eschenbach’s original thread (the one that caused all the controversy). Even Willis didn’t get it at the time. I hope he does so now.
[REPLY: Indeed you did, Paul, and you were right and I was wrong. Thanks for your contribution in fighting my ignorance. I mean this quite seriously. That’s how I learn. –w.]

Ed Fix
January 24, 2012 9:55 am

Robert, it seems that you have completely missed the fact that gravity causes a pressure and density gradient in your air column.
Your equilibrium air column is NOT isothermic, as you assert–that could only happen in the absence of gravity. Because the density and pressure decrease with altitude, the temperature at the top is much lower than at the bottom. The bulk of the mass and heat energy of the air column is at the bottom.
Remember what heat energy is: it’s defined by the kinetic energy of the individual air molecules. Temperature is defined by both that molecular kinetic energy and by the density of the atmosphere. The pressure gradient leads to a sorting; the more energetic molecules tend to be at the top of the air column–more space to allow a longer mean free path above than below. However, because the density and pressure decreases with altitude faster than the thermal energy of individual molecules increases, total temperature decreases with altitude.
Your thesis may indeed be correct, but you can’t prove it by considering only temperature and convection without considering density and pressure and conduction as well. It’s the pressure and density gradient that is alleged to cause gravitational heating of the lower atmosphere.
This is a much more complex problem than a quick, partial recitation of a freshman physics text can handle.

January 24, 2012 9:58 am

The device in figure 2 doesn’t work because it’s a closed system and the work extracted will reduce the total energy of the column until eventually there’s no more energy to extract at which point the gas reaches a temperature of absolute zero and has presumably vanished from this universe being totally converted to kinetic energy in the extracted useful work. In the real world the gas will collapse to the surface as a liquid before it gets to absolute zero and this will shut off further extraction of energy because the cold side of the thermocouple no longer has any cold gas to cool it.
Work? What work? Are you crazy? Collapse to a liquid?
Let’s try again. I-s-o-l-a-t-e-d S-y-s-t-e-m means that no energy enters or leaves. No mass transport means no work is being done In the real world, the system will evolve to an isothermal state precisely as I described it because it is in equilibrium. In any imaginary world where gravity acts on “heat” or does “work” on a gas that is in static force equilibrium and not moving, you can make it come out any way that you like, but please understand that it is nothing but a fantasy on your part.
The point is that heat will not cycle indefinitely — you can see that that makes no sense. No work is done. No energy enters or leaves — where would it go? How would it get there? That’s what the adiabatic walls around the gas and wire prevent. If gravity maintains a constant lapse rate in steady state, the second law and common sense are massively violated by the enternal heat flow. All other solutions mean that equilibrium is isothermal.
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son of mulder
January 24, 2012 9:59 am

Heat is Energy is mass by M=E/c^2 so said Einstein.
So what force causes Mass to rise up the silver conductor against gravity ie work has to be done?
The silver conductor is little different from the gas in a column in this respect. The top will be colder than the bottom and heat will not flow up the silver conductor unless a heat source (work) is supplied from the bottom..

kdk33
January 24, 2012 10:02 am

If the atmosphere was heated from the top there would be no convection, hence no lapse rate.
The lapse rate doesn’t apply to the ocean because water is incompressible. Hot water doesn’t expand as it rises, hence does not do work on the surroundings, hence does not change temperaure, hence no lapse rate.

January 24, 2012 10:03 am

I conclude that a thermally isolated container of gas in zero gravity and at uniform temperature, will not see any temperature change if some other demon could switch a gravitational field on at will. Merely the creation or a density gradient. Is that right?
Not quite. “Turning on the field” is like “a collision” and the gas will rearrange, releasing a bit of gravitational potential energy as heat. But then it will “thermalize” to an isothermal temperature, one a tiny (and I do mean tiny, generally speaking) higher than before.
This is not unlike the mechanism that heats protostars as the gas they are made up of falls inward and stops (on average) converting their infalling KE into heat. Or what happens when a big asteroid hits the earth and stops
The key elements are movement and inelastic stopping.
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JKrob
January 24, 2012 10:04 am

I have been following these threads, lurking from the sideline, for a while. Well, the time has come to add my $.02 worth. I have been studying meteorology for 40 some-odd years now & I shake my head in seeing some of the most common properties of the atmosphere being missed in these threads as it applies to these ‘thought experiments’. 1) if the several km-long tube is horizontal & the perfectly dry air is at a constant temperature throughout & is moved to the vertical, the dry adiabatic gradient will be produced (warm at the bottom, cool at the top w/ approx 8C/1000m gradient in between) due to the ‘work’ of gravity creating a pressure gradient to the compressible gas. Notice, no gradient will be produced if water is used instead of gas because water is non-compressible so no work will be done. If no heat is added or removed to the gas, the column will be in a neutral buoyant state (and will stay that way!!) – if a parcel of air is moved vertically by an outside force, it’s temperature will change to reflect the change in pressure but will still be the same temperature as it’s surroundings. 2) as to the experiment with the thermal conductive wire at the base & top of the tube, the author here is incorrect. If the wire moves heat from the bottom of the tube (the base cools) to the top of the tube ( the top heats), presuming, as the author says, “…save to note that the internal conductivity of the ideal gas is completely neglected.”, the heat from the *local* area of the wire is all that will be moved from the bottom to the top ***and nothing else*** . Why, you ask?? In moving the heat from the bottom of the tube to the top is causing the lapse rate to become **more stable** – cool at the bottom with warm air above is an inversion which inhibits vertical mixing!! THAT is why the engine will not work as it is set up. Just a few thoughts… Jeff January 24, 2012 10:15 am Brown: To do LaTeX in WordPress, do$\latex n^2$(except leave out the backslash in front of “latex”. It’s just like going into math mode in LaTeX, except that you add the word “latex ” after the opening dollar sign. Greg Elliott January 24, 2012 10:15 am nothing as simple as gravity can function like a “Maxwell’s Demon” There is nothing simple about gravity. it is the least understood force in the universe with many unresolved questions. Frank January 24, 2012 10:15 am Professor Brown: To some extent, two of your arguments start by assuming what you want to prove. You and the introductory textbooks start by assuming a isothermal column of gas. The situation is far more complicated if you consider a column with a temperature gradient. For a thin layer of gas in a non-isothermal cylinder, the pressure difference across that layer is produced (at a molecular level) by differences in the vertical impulse provided by the gas molecules at the top and bottom of the layer. The density of molecules and pressure in an isothermal column both change following the same exponential, -mgh/kT. This means that the difference in impulse between the top and bottom of a layer is due only to the difference in density and the average speed of the molecules moving up and down must be the same. This is consistent with the original postulate that the column is isothermal. In a non-isothermal column, however, the density of molecules and pressure don’t change in parallel. In this case, the speed of the molecules at the top and bottom of a thin layer will not be the same and energy will flow up or down. Which way is the flux? If the flux reduces the temperature gradient (and I presume that it will), does the flux persist until isothermal or are other stationary states possible? To prove that heat flow in a cylinder of gas is unaffected by a gravitational field, you assume that heat flow in the silver conductor is also unaffected by the same gravitational field. IF exchanging kinetic for potential energy were important to energy flux in a gas, it would probably also be important in a solid. In a sense, you are assuming what you want to prove. You can use the 2LoT to eliminate the possibility that a lapse rate of g/Cp develops spontaneous. If you have two equally tall columns in thermal contact with the ground which are filled with gases with different Cp’s, you could use the temperature difference which would develop spontaneous at a given height in the gravitational field to produce perpetual motion. However, this argument doesn’t work when the lapse rate that hypothetically forms spontaneously in a gravitational field is independent of composition and only depends on height. Tim Folkerts January 24, 2012 10:18 am I feel for you Robert. It seems the more challenging and subtle the physics, the more experts there are. And the more sure they are that they are right. It is a lot like playing whack-a-mole — every time you think you have explained something so well that it couldn’t be clearer, someone will find a new objection (or more likely, recycle an old objection). For example, the “it loses KE on the way up so the temperature must go down” is convincing unless you have a subtle understanding of thermodynamics. And we have seen it rear its head a dozen times in the last few days in these threads. I hate to admit that I, like Willis, even fell for this argument for a brief time until the obvious flaws were pointed out. But we both quickly reformed. I wish you patience and persistence in your efforts to bring correct science to WUWT. You will need it! George Turner January 24, 2012 10:20 am As an aside, it’s trivial to design a machine that seems to violate the Second Law and pumps heat from cold to hot with no input of work. On a sheet of paper draw two horizontal lines, one at the top and one at the bottom, that represent radiating surfaces at temperatures Ttop and Tbottom. Then draw a cute little flight of stairs ascending from left to right, with a smooth bottom and the usual treads on top. Cover the stairs with mylar and let them sweep from right to left at nearly the speed of light. (You can also make the stairs steeper and sweep them slower). There are a very large number of flights of stairs, looking in 3-dimensions like a venician blind with treads on one side. Assume the top and bottom emitting surfaces are very distant and finite, so photons are traveling between them in a roughly vertical direction. Photons from the bottom don’t hit the stairs because the stairs are moving out of the way as fast as the photons are traveling upward. (The stairs dodge upward moving photons). Photons moving downward cannot find a clear path between flights of stairs and always slam into a tread, getting reflected back towards the top. So photons emitted from the top surface return to the top surface, and photons emitted from the bottom surface travel freely to the top surface, regardless of temperature. The bottom surface cools and the top surface warms, even if the bottom is already cooler than the top, and the stairs aren’t doing anything but freely moving along between. At most, the stairs might extract a little work from the momentum of the photons being reflected from the top, but this would happen regardless of the difference between the top and bottom temperatures. Ged January 24, 2012 10:21 am The problem here is that heat is not temperature. Heat is energy, temperature is average kinetic motion, which is one -type- of energy. A glass of water and a bathtub full of water can have the same temperature, but the bathtub has more heat than the glass of water, as more heat is required to bring the tub up to the same temperature as the glass. Moreover, heat can be added to a system without temperature changing. When ice melts, continuous heat is required to continue the melting, yet the temperature will remain the same until a significant amount of the ice has melted. Still, the system has far more heat content now as a liquid, even at the same temperature, than it did as a solid. For instance: “This heat in turn may lift mountains, via plate tectonics and orogenesis. This slow lifting of terrain thus represents a kind of gravitational potential energy storage of the heat energy. The stored potential energy may be released to active kinetic energy in landslides, after a triggering event. Earthquakes also release stored elastic potential energy in rocks, a kind of mechanical potential energy which has been produced ultimately from the same radioactive heat sources. Thus, according to present understanding, familiar events such as landslides and earthquakes release energy which has been stored as potential energy in the Earth’s gravitational field, or elastic strain (mechanical potential energy) in rocks.” So, this action of heat pushing against a gravity well and from kinetic to potential energy is part of what drives plate tectonics itself. Here we see heat being transformed from kinetic (TEMPERATURE) energy to potential energy by moving against a gravity field. The resulting raised land mass has lower temperature as it as lower kinetic energy, but it contains similar amounts of heat. This effect of changing kinetic energy to potential energy (E = mgh) won’t be seen measurably in a gas over a small distance, but over the miles of the atmosphere? This is why it’s hard for me to understand your explanation, which doesn’t mean you aren’t correct, Dr. Brown. But it seems like your equations are all missing the big picture: potential energy. When h (height) is very big, even though gasses have low m (mass), you’re going to have a transfer of energy into potential energy that is considerable. The only source of energy that can go into the potential energy (since conservation of energy demands transformation not creation) is the kinetic energy of the molecules. Just like bouncing balls. And if kinetic energy drops, -so too does apparent temperature-. Again, -temperature is not energy-. Temperature is not heat. Temperature is just our observation of heat in the form of -kinetic energy-. And kinetic energy can be changed to potential energy while keeping the total energy of the system -the same-. Like a bouncing ball, or a stone rolling up hill a ways. Or plate tectonics! You’re looking at far too small a case. An ideal, not realistic, case. Or maybe i just don’t get it. Joe Born January 24, 2012 10:23 am Robert Brown: “There is no question that the silver will conduct heat between reservoirs at different heights exactly the same way it does any other time.” In my post above, where I disputed Dr. Brown’s isothermality conclusion, I was silent about his proof; I concentrated on mine. For the sake of completeness, though, I’ll mention that, if the Velasco et al. paper that I’ve been evangelizing is correct, the silver actually will not conduct heat once the temperature difference that Velasco et al. specify is imposed across it. This is a result of the fact that, according to Velasco et al., entropy is maximized, not by an isothermal configuration, but by a configuration whose (again, quite small) temperature lapse rate is the one that Velasco et al.’s paper prescribes. If you think of heat transfer in the silver as a diffusion phenomenon and recognize that concentration gradients prevail all the time when maintained by forces from, e.g., electric fields, this is not as hard a proposition to swallow as it may at first sound. Tim Folkerts January 24, 2012 10:23 am Greg Elliott says: “But what about the energy absorbed by the N2 some might ask. Indeed and what about the energy intercepted by the CO2? Both must either heat the atmosphere or be returned to the surface and thus are indistinguishable in their effects.” But this is the flaw in your thinking. N2 can only transfer energy to the surface or the atmosphere, as you say. But CO2 can ALSO transfer energy to space via IR photons. That is ultimately the cause of the greenhouse effect. Ged January 24, 2012 10:24 am Oh, I should point out you cannot -generate heat- with gravity, but over a very large distance, gravity should maintain a -temperature- gradient if the heat is in equilibrium by necessity due to the change of kinetic to potential energy as one moves far enough up a gravity well. Notice that the energy of the gas column will be in equilibrium, but temperature is only the measure of kineitc not potential energy. That fact is why I can’t wrap my head around such a small case explanation as yours, which is correct on the small scale, when applied to the entire planet. Again, maybe I’m completely wrong. But I feel we’re missing out on one entire half of the equation. Gasses are still subject to potential energy as far as I know! DavidB January 24, 2012 10:25 am I think one reason that some people are taken in by Jelbring’s theory is that they are vaguely aware that if a gas is compressed, e.g. the air in a tire being compressed by a pump, its temperature goes up, so in their minds they form the vague association ‘higher pressure = higher temperature’. But the increase in temperature is a temporary effect, due to the transfer of kinetic energy from the piston of the pump to the air molecules. If you stop pumping, the tire will cool down to the ambient temperature, as it loses heat by conduction and radiation. As the air in the tire cools, it will also reduce in pressure, but not to the orginal level (otherwise there would be no point in pumping up a tire!) There is no necessary connection between high pressure and high temperature; a hot gas can have low pressure and a cold gas can have high pressure (unless it is so cold as to liquify). Incidentally, I wonder if someone could explain the opening quote from Jelbring: “An adiabatically moving air parcel has no energy loss or gain to the surroundings. For example, when an air parcel ascends the temperature has to decrease because of internal energy exchange due to the work against the gravity field”. I don’t understand the second sentence of this. If an air parcel ascends, it is surely because it has first expanded due to heating (usually from the sun). As it expands, it does work against the surrounding or covering air, compressing or displacing it, and loses some kinetic energy (heat) in the process, but it does not immediately cool to the ambient temperature. (If it did, it would not ascend at all, contrary to experience.) The expanded parcel of air is less dense, and therefore less heavy, than the surrounding air, and the entire column of air above the parcel (including the parcel itself) is lighter than the surrounding columns. At this point the parcel begins to rise as the heavier surrounding air forces it up. During the ascent the air parcel itself is not ‘doing work against the gravity field’, it is having work done on it by the surrounding air, which is ‘repaying’ the work done on it during the initial phase of heating and expansion. Does Jelbring suppose that a hot-air balloon would rise in a vacuum? I really can’t make sense of his second sentence at all. D. J. Hawkins January 24, 2012 10:31 am Joules Verne says: January 24, 2012 at 7:04 am “Spontaneous separation of a reservoir of gas into stable sub-reservoirs at different temperatures violates the second law of thermodynamics. It is a direct, literal violation of the refrigerator statement of the second law of thermodynamics as it causes and maintains such a separation without the input of external work.” No, Robert. The second law requires that no energy gradient can be maintained without input of work. It requires the reservoir of gas to be isoenergetic not isothermal. A horizontal layer may be a different temperature than another if the cooler layer has its lesser kinetic energy balanced by greater gravitational energy. This is obviously the case since it goes without a shadow of a doubt that a molecule of air in a higher layer has more gravitational energy than a molecule in a lower layer. Thus the second law actually demands a temperature difference of equal and opposite polarity to compensate for the difference in gravitational energy. An isothermal atmosphere in a gravity field is the one that violates the second law. My name is Joules because I know how to find and count them no matter how they try to hide. Thanks for playing. You’ve very coyly avoided answering the objections raised by Dr Brown’s Gedankenexperiment. So according to you, we could take a perfectly insulated container several miles high in a gravitational field under a hard vacuum, fill it with gas and that gas will self-organize so that it’s warmer at the bottom and colder at the top. Now I can construct a heat engine which extracts useful work based on the temperature gradient and gravity will continue to organize the air column forever and my heat engine will never run out of “fuel”? Really?? January 24, 2012 10:32 am To Joules, According to Newton, gravitational force is expressed as F=m1*m2/R^2 where m1 and m2 are the mass of two attracting bodies and R is the distance between the centers of mass. The molecules at TOA have less gravitational energy than molecules at sea level but not by much because the change in R is relatively small. So the pressure at sea level is essentially the mass of the earth times the sum of the mass of all those molecules in the atmosphere divided by the radius of the earth. In a steady state condition, the pressure at sea level will be constant and the number of molecules causing that pressure will be constant. Pressure decreases with altitude because the number of molecules per volume decreases. Now apply this knowledge to the perfect gas law expressed as pv/nt= constant by integrating from TOA to the surface and see what happens to temperature as a function of altitude. tallbloke January 24, 2012 10:38 am Joules Verne says: January 24, 2012 at 7:04 am “Spontaneous separation of a reservoir of gas into stable sub-reservoirs at different temperatures violates the second law of thermodynamics. It is a direct, literal violation of the refrigerator statement of the second law of thermodynamics as it causes and maintains such a separation without the input of external work.” No, Robert. The second law requires that no energy gradient can be maintained without input of work. It requires the reservoir of gas to be isoenergetic not isothermal. A horizontal layer may be a different temperature than another if the cooler layer has its lesser kinetic energy balanced by greater gravitational energy. This is obviously the case since it goes without a shadow of a doubt that a molecule of air in a higher layer has more gravitational energy than a molecule in a lower layer. Thus the second law actually demands a temperature difference of equal and opposite polarity to compensate for the difference in gravitational energy. An isothermal atmosphere in a gravity field is the one that violates the second law. Joules is correct, and to push the point home a bit further I will be posting a new paper by Hans Jelbring on my website later this evening which demonstrates the dynamic situation. This will complement and supplement the earlier 2003 paper setting out the static situation which Robert refers to in this article. Hans has been working steadily on the new paper over the last few weeks and now seems to be the apposite time to publish it. Q. Daniels January 24, 2012 10:43 am Basically, the Second Law is used to prove the second law isn’t violated. That’s circular logic. It’s like using the Carnot Cycle to prove the Second Law. As I said at Tallbloke’s, I think the silver wire is a terrible design. You’d be lucky to extract kbT from it. IIRC, Feynmann showed it was possible to extract nearly that much from a Brownian Ratchet, without anyone getting excited. A better design would be to use a single column containing two gases, one of which is heavy with a low boiling point, and the other is light with a high boiling point. A reservoir of the light gas in condensed form at the bottom of the column also serves as a thermal reservoir. At the bottom, the light gas will be in equilibrium with its condensate. Fill the column with a significant majority of the heavier gas, such that its gravitational lapse rate dominates. The temperature/pressure profile of the lighter gas (being a lifting gas in context) will be saturated at all elevations above the reservoir. A series of basins and catchments can be arranged to collect the condensing lighter gas. By George (Westinghouse), it might work! tallbloke January 24, 2012 11:02 am People being invited to consider an argument should have the opposing argument easily available to them. This enables people to form their own judgment as to whether the opposing argument has been correctly and fairly represented. Especially considering the paper in question was written by a Phd Meteorologist. Since Robert Brown is setting a refutation of Hans Jelbring’s 2003 paper, it would be a common courtesy to provide a link to that paper in the headline post. It is available with a new preface written at the time of publication on my website here: http://tallbloke.wordpress.com/2012/01/01/hans-jelbring-the-greenhouse-effect-as-a-function-of-atmospheric-mass/ If WUWT has some problem with providing a link to my site, the paper without the 2012 preface is available here: ruby.fgcu.edu/courses/twimberley/EnviroPhilo/FunctionOfMass.pdf [Thanks, tallbloke. I’ve added the link up in the head post. -w.] David L. Hagen January 24, 2012 11:07 am Alan Robertson January 24, 2012 11:08 am Roger, Your site is linked on the right column of this page, as always. David L. Hagen January 24, 2012 11:12 am For context, Jelbring (2003) is cited by: Gravitation and Gas Laws: An Alternative Approach to Climatology W Brune – Energy & Environment, 2009 – Multi-Science and The thermodynamic relationship between surface temperature and water vapour concentration in the troposphere WC Gilbert – Energy & Environment, 2010 – Multi-Science Kasuha January 24, 2012 11:17 am Dear Mr. Brown, I’m sorry but your conclusion about figure 2 is completely wrong as you expect the air reservoir to conduct heat in gravitational field and the silver rod to conduct heat in zero gravity. I’m sorry to say that this assumption is incorrect and the example you are building your conclusions on is wrong. Even atoms of silver have the property of gaining potential energy and losing kinetic energy (therefore losing temperature) when traveling “up” in a gravitational field while transferring their temperature in that direction. The result is, the rod would demonstrate exactly the same temperature gradient the air container does. Your thermal engine would not work not because there is no temperature gradient but because it would not transfer any heat. Bryan January 24, 2012 11:23 am kdk33 says “if the atmosphere was heated from the top there would be no convection, hence no lapse rate. The lapse rate doesn’t apply to the ocean because water is incompressible. Hot water doesn’t expand as it rises, hence does not do work on the surroundings, hence does not change temperaure, hence no lapse rate.” Do you not remember the school experiment with a beaker of water and a single copper sulphate crystal showing fairly rapid liquid convection? How do you think an electric kettle heats water? Editor January 24, 2012 11:29 am Bryan says: January 24, 2012 at 6:59 am What would complement this theoretical explanation is if an experiment backed it up. So far as I know no experiment has ever been carried out. All suggested proposals seem to run into problems when real components and physically accurate numbers are used. For some of us, a thoughtful examination of the flame suffices. We don’t have to actually do the experiment and burn our fingers to know that it is hot. If you wish to do the experiment, however, be my guest. I’ll just have a quick beer while I wait for your experimental refutation of a couple centuries of scientific thought … w. tallbloke January 24, 2012 11:31 am Joules Verne says: January 23, 2012 at 4:25 pm Gravity maintains TWO energy gradients. One kinetic and one potential. The kinetic gradient decreases with altitude and the potential gradient increases with altitude. The two opposing gradients cancel out and the column is isogenergetic. This is how you can have a perpetual temperature gradient yet not be able to extract any work from it for a perpetual motion machine – a temperature gradient can be nullified by an equal but opposite gradient of energy in a different form. You can’t connect the cold and hot sides of the atmosphere without climbing up in a gravity well and the useful energy represented by the change in temperature is exactly used up by the energy required to climb uphill against gravity. The books thus balance and conservation of energy is once again safe from the abuses of junk science. Neat comment Joules. I can’t wait for the mad inventors to put their money where their mouth is and build one of the erroneously designed machines they propose. Problem is, when it fails to work they’ll come to the equally erroneous conclusion that it failed because the atmosphere is isothermal. Hope no-one catches a nasty chill while bolting thermopiles together at the top of their 10km high rig. While they’re up there, they might notice how much thinner the air is too. That might give a bit of pause for thought about density and its effect on the ability of air packets at different altitudes to retain the heat of the Sun. Paul Potter January 24, 2012 11:33 am Look the adiabatic lapse rate exists this is basic physics equal partition of energy. it is evident in the atmosphere but is upset by clouds at low level and UV absorption at higher altitudes. If we cannot get even the basics correct we might as well give up. For those who say there is no back radiation why is the sky temperature at night not 4K For those who do not believe an warmer but still cold atmospheric layer cannot cause the surface to warm clearly do not understand the basic physics of radiative heat transfer. Why are layers of mylar used to thermally insulate space craft? Unless we get the basic physics correct then the skeptical community will be undermined. Alan Millar January 24, 2012 11:35 am A gas giant planet like Saturn radiates more than twice the amount of radiation than it recieves from the Sun. How do these planets ever die if gravity is constantly maintaining a heat gradient in the atmosphere? The Sun will die and become a cold white dwarf. What is going to kill these gas giants? If gravity is constantly maintaining hotter gases at the bottom then convection will move gases around and therefore we seem to have an everlasting living planet. What kills it and when, if Jelbring is correct? Alan hotrod (larry L) January 24, 2012 11:37 am The last few days have certainly been interesting regarding the debate of lapse rate in a gravity field. I think the point that is getting missed, but I see as intuitively obvious, is that in a real planet with a gas atmosphere, it is not and never can be in a state of thermal equilibrium. The atmospheric column is always in direct contact with a massive heat sink(source) in the form of the planetary body, and the top of the column (at its effective radiating altitude) is a nearly ?? infinite heat sink at the temperature of interstellar space. Until or unless the planetary body is at the same temperature as deep space there will always be energy input at the bottom of the atmospheric column (and a temperature gradient) and there will always be heat loss by radiation (or some other means like boiling off of the atmosphere) at the top of the column. In that sense, all these discussions regarding physically impossible constraints on the model are merely debating how many angels can stand on the head of a pin. Lets look at real planets which are always warmer than interstellar space, and real atmospheres where at the base (troposphere) heat transport by convection dominates all other mechanisms by a huge margin. Under those conditions, there will always be a lapse rate in the atmosphere (in the troposphere) as long as it can find some means to lose energy to interstellar space by radiation or loss of mass. The only discussion, is are there any realistic conditions where an atmospheric shell could not radiate heat to interstellar space? If and only if that condition can be achieved do the discussions of a system in equilibrium have any meaning. In all other cases (with or without radiant heating by a near by star) the planet will always be losing heat to space, and will always have a troposphere layer (if it has an atmosphere of non-condensing gases) where heat transport is dominated by convection. In all cases the presence of green house gases can only improve heat loss to interstellar space at the effective radiating height over the theoretical condition of no ability to radiate energy in the electromagnetic spectrum (note I did not limit energy loss to IR band width). It is my view that the atmosphere will always find a way to lose energy to deep space by some means at or near its top, and will thus always have a troposphere lapse rate which implies that the planetary body will always be warmer with an atmosphere (regardless of gas mix) than it would be without one. One other note that no one has mentioned. As I understand it, the average effective radiating altitude here on earth is around 4.5 km (14763 ft), that means that here where I live in Colorado at the summit of Pikes peak or Mt. Elbert I could just about throw a rock to the mean altitude where heat is lost to space by radiation. In short high altitude areas like the Himalayas, Andes, and the Rockies and a few volcanic peaks are “short circuits” in the troposphere, where energy absorbed by the ground from solar heating can be almost immediately radiated back to deep space directly from the ground. It would be interesting to see high resolution IR measurements of the heat loss from these areas directly to outer space. Larry MaxL January 24, 2012 11:38 am I have a question along this topic which will likely show my ignorance. If gravity alone cannot induce a thermal gradient in a gas, how then are stars formed from gases where there is only gravity as an external force? Ged January 24, 2012 11:42 am Let’s do some equation work. Let’s say we take one mole of N2. That’s 0.028 kg. Now we take that mole at 15 C and move it up from the sea surface to the average top of the troposphere which is 17,000 meters high. Gravity constant will be 9.8 m/s^2. The total joules needed to do this is around 4660 J (E = mgh). Now the heat capacity of N2 is 29.124 J·mol^−1·K^−1, so if we transformed 4660 J of heat kinetic energy, that would reduce the temp by -160 degrees, or so, or to -145 C. The real temp at the top of the troposhere is about -55 C, so we’re off by almost a factor of 3. Obviously, the atmosphere is also taking in energy from the sun (radiatively, and solar wind), there’s more than just N2, and the conversion of kinetic temperature energy to potential is probably more complex, to make up for this difference. If there was no energy from the sun, the atmosphere would shrink as energy was lost to space, and along with it potential energy, dropping molecules down till they settled on the surface of the planet and solidified (e.g. Pluto). This is what I see. And again, I may be completely off base. But conservation of energy tells me we must be changing forms as we move up the atmosphere. So for the temperature to stay steady (which it doesn’t in real life), we’ve have to have even more energy at the higher altitudes in the air column than is at the surface. And would not this energy fall back down to the surface to equilibrate, driving the surface temperature even higher? The way I see it, there will always be a -temperature- difference at such large gravity distances if energy is in equilibrium. Again, I could have done this all wrong. Editor January 24, 2012 11:47 am John Marshall says: January 24, 2012 at 7:26 am Very interesting but really irrelevant given that our atmosphere is not an ideal gas in a cylinder. It’s the disproof of Jelbring’s theory, which lately has attracted many supporters. As such it is an interesting and very relevant exercise. w. alex January 24, 2012 11:47 am School physics. The atmosphere stability condition is ds/dz>0, where s is the entropy density and z is the hight. Atmosphere with ds/dz 0 are stable (including the isothermal atmosphere!) What is all the ridiculous discussion about??? Graeme W January 24, 2012 11:50 am Dr. Brown, I’ve been reading this thread and the earlier thread with a great deal of interest and I understand the point you’ve raised. However, I note that it relies on the Ideal Gas Law being applicable. My question is: Is the Ideal Gas Law applicable? I’ve just done some quick research and it seems that the derivation of the Ideal Gas Law assumes that any gravitational field in question will have a negligible impact on the behaviour of the gas. That’s true in almost all reasonable cases, but it’s not true in the example we’re discussing. I would therefore ask for your comments on the underlying assumptions in the derivation of the Ideal Gas Law in this respect and if they are still true. In particular, looking at the Empirical Derivation from the Wikipedia article: http://en.wikipedia.org/wiki/Ideal_gas_law#Empirical I noticed that there is a constant C that is directly proportional to the amount of gas. But in a gravitationally stratified column, the amount of gas is not constant throughout the column, and hence the “constant” is not really constant. Similarly the derivation from statistical mechanics in that article assumes that there is no gravitational force involved. For small columns, that’s a reasonable assumption,Is it a reasonable assumption for large columns? I don’t understand the theoretical derivation well enough, but again it seems to assume that there is no conversion of kinetic energy into gravitational potential energy – something that is core to this issue. Again, thank you for your posts. I have enjoyed reading and learning from them. Trick January 24, 2012 11:50 am Robert Brown says at 1/24 8:14am: “…in an ideal gas the temperature is not determined by the total energy. That’s an absurd idea, given that one can perform a gauge transformation — change the zero of the total energy — without changing any of the physics.” Ahhh..this is the big deal. Consider this is equivalent given Caballero’s ref. in the Perpetuum Mobile thread w/gravity where in 2.1 find “Temperature is just another name for the mean kinetic energy density of molecular motion.” to Trick saying “in an ideal gas the mean kinetic energy density is not determined by the total energy”. If Trick said this, Trick would be going against 1st Law & Trick has not written any such thing. So in Trick’s view, what Robert Brown says at 8:14am is inconsistent with 1st Law. I am pretty sure but not perfectly sure that Trick’s view is consistent with all the thermo grand master Laws. Robert Brown’s 8:14am violation of the 1st Law above, Robert Brown doesn’t grok yet. He will eventually! (Trick predicts the life of the universe is longer) b/c Robert’s smart and the thermo grand master’s are right. This is all like the slow progress of thermal conduction thru a near perfect insulator. I admit this PE setting with h is actually a difficult preconceived notion to get over. It took me many years of practice to really come to grips with it. Notice that every time you see total energy equation written in thermo theory – it looks something like this: total energy = TE = PE + KE = constant. That’s all the 1st Law really can do, it doesn’t tell us what that constant equals – it can’t and be general theory. 1st Law tells us (go look!) energy can be neither created or destroyed, but can be changed from one form to another, we have to do the deducing what that means. The constant specific value is not set by nature; it is not a natural constant. TE is set in the specific experiment. Whatever the total energy in the white space top post is, it has to be constant to obey the 1stnd Law. In Robert Brown’s top post we actually can just (out of thin air so to speak!) announce that our h=0 to be at the bottom of the white cylinder of gas. Here we know at h=0 that TE=PE +KE. Always. Cite the 1st Law! At the bottom of the cylinder we just announce PE is 0 there. This is the tough nut to crack to mix my metaphors. PE is a potential energy – as such nature and the thermo grand masters allow this construct. So we can move on. At Trick’s announced h=0: TE of the molecule(s) will be constant = TE = PE + KE = mgh + 1/2mv^2 = m*g*0 + 1/2mv(0)^2. Voila we have figured the TE constant. In white area TE = constant = m*g*0 + 1/2mv(0)^2 = 0 + 1/2mv(0)^2 = 1/2mv(0)^2 where v is the velocity and m the mass of the molecule(s) at h=0. The TE constant is ½*m(0)*v(0)^2. KE varies with PE(h) in a gravity field… (can you say non-isothermal?) (Say it again: Deduce from 1st Law: molecule(s) KE or temperature must vary in a gravity field with PE, for total energy to be constant, OMG!) But we are ok, what do you know, we have a formula consistent with 1st Law that found the total energy constant of the white area*. The constant = 1/2mv(0)^2 comes from finding the v of the molecule at h=0. For that, all we need do is find the KE or temperature (= kinetic energy density) at h=0 and we will know the temperature anywhere else in the one reservoir or one thermodynamic ideal system of the white area (i.e. at any h) from bottom to top thru TE = constant = ½*m(0)* v(0) = (mgh + 1/2mv^2). A thermometer placed at h=0 will work just great for this experiment. Note 0 location IS arbitrary, I can pick ANY h and put my calibrated thermometer there and find m(h) and v(h) and get the same TE constant, it is just easier to think thru at h=0. Robert Brown needs to not read FAST but s-l-o-w-l-y to grok this; to really come to terms with it, it will take some practice. The thermo grand masters did a fine job, Robert Brown can do one too. Another big deal: Robert Brown cannot use, in Trick’s view, 1/24 8:14 statement to argue against Hans Jelbring’s 2003 paper. There may be other arguments but this Robert Brown one doesn’t work with the 1st Law. Robert Brown continues: “Is there something miraculously interesting in the thermal contact between silver and air that keeps heat from being conducted from hot to cold — in just this one special circumstance? I’m all ears.” Nope there isn’t any Trick violation of 0thLaw equilibrium or 2nd Law entropy being allowed ideally constant, either. Robert Brown’s top post just needs to get rid of 1) the second thermodynamic system of the top post – that 2nd system is not in Willis’ Perpetuum Mobile original premise and 2) the perfect ideal insulator for compliance to all three Laws. Then Robert can do the easy math & will grok the big deal. Eventually Trick & Robert Brown will be in (Tallbloke’s) violent agreement. Trick’s head cold is receding, I won’t be hanging here forever. Robert Brown should try to make this happen in a few days. If not, I predict violent equilibrium will eventually happen, that’s the 0th Law, LOL. *barring typo’s and consistent with mass conservation, the molecule(s) mass does not change thus can assume E to mc^2 in total energy is justifiably being ignored here by all posters on the original Willis’ premise GHG-free air column. Ged January 24, 2012 11:54 am “Now I can construct a heat engine which extracts useful work based on the temperature gradient and gravity will continue to organize the air column forever and my heat engine will never run out of “fuel”? Really??” Nope! Impossible. As you took energy out, the molecules would not be able to move up as high in the gravity well (the pressure of the gas would drop), as there wouldn’t be enough energy to go into potential energy, and eventually the gas would condense and settle at the bottom of the container. It would cease being gas. In short, the temperature at the surface level would steadily decrease as you took out energy, as potential energy and kinetic energy are -interchangable-, but temperature is -only- kinetic energy. And heat can -only- be transferred by kinetic collisions (if we ignore radiation). So your engine would take out the kinetic portion, leaving less to be turned into potential, making the air column decrease in height until what I said above happened. Same as what happens on Pluto. Editor January 24, 2012 11:54 am Wayne2 says: January 24, 2012 at 7:39 am @Joules Verne: “This is obviously the case since it goes without a shadow of a doubt that a molecule of air in a higher layer has more gravitational energy than a molecule in a lower layer.” Does the fact that the air in a higher layer also has fewer molecules matter? That is, are we talking individual molecules or rather volumes of molecules here? Excellent insight, Wayne. That is exactly what happens. In an isothermal column of air, individual molecules at high altitude have more energy because of gravity. But for exactly that same reason, there are fewer molecules at high altitude. As a result, and as we would expect, in the isothermal condition the energy is spread out evenly through space (equal energy per volume) rather than equal energy per molecule as Hans Jelbring and Mr. Verne assert. w. Bryan January 24, 2012 11:56 am Willis Eschenbach says “If you wish to do the experiment, however, be my guest. I’ll just have a quick beer while I wait for your experimental refutation of a couple centuries of scientific thought ” Its always better to do the experiment! If all the qualified scientists here were honest they would admit that this is an example of an experiment where prejudgement was entirely wrong. http://en.wikipedia.org/wiki/Erasto_Mpemba hotrod (larry L) January 24, 2012 11:57 am Refutation of Stable Thermal Equilibrium Lapse Rates Posted on January 24, 2012 by Anthony Watts Guest post by Robert G. Brown Duke University Physics Department What matters is that EEJ asserts that image in stable thermodynamic equilibrium. The purpose of this short paper is to demonstrate that such a system is not, in fact, in thermal equilibrium and that the correct static equilibrium distribution of gas in the system is the usual isothermal distribution. I think it is also important to note that Jelbring does not assert thermal equilibrium, they assert “energetic equilibrium” considering both heat energy and gravitational potential energy. THE “GREENHOUSE EFFECT” AS A FUNCTION OF ATMOSPHERIC MASS Hans Jelbring 2003 “The energy content in the model atmosphere is fixed and constant since no energy can enter or leave the closed space. Nature will redistribute the contained atmospheric energy (using both convective and radiative processes) until each molecule, in an average sense, will have the same total energy. In this situation the atmosphere has reached energetic equilibrium. The crucial question is what temperature difference (GE) will exist between A and S?” Larry Editor January 24, 2012 11:57 am The iceman cometh says: January 24, 2012 at 7:58 am I find the analysis quite reasonable – but it is so idealized as to be useless. It is the formal disproof of Jelbring’s theory. It is idealized by its very nature. w. eyesonu January 24, 2012 12:02 pm Thank you Dr. Brown. Editor January 24, 2012 12:09 pm A physicist says: January 24, 2012 at 8:48 am It seems to me Robert Brown’s analysis implicitly makes the following claim: if all the greenhouse gases (mainly H20 and C02) were cleansed tonight from earth’s atmosphere, then the atmosphere would evolve toward a more nearly isothermal equilibrium But how would that work, exactly? Although that may be true, I don’t see why, I don’t see Robert Brown making the claim, and I If Robert Brown answered this question clearly (and it is a subtle question IMHO), then it seems to me that his theoretical ideas would prevail. Wait, wait, you claim to be a physicist, answer the question. Does heat flow forever in the silver wire or not? His ability to answer your random question about possible implications is immaterial to whether his proof is correct. Does heat flow forever or not? w. Mike January 24, 2012 12:12 pm Bryan/Willis: I could be missing the whole point here but isn’t the Graeff paper (ref. http://tallbloke.files.wordpress.com/2012/01/graeff1.pdf) an example experiment? h/t Lucy Skywalker Ed Fix January 24, 2012 12:12 pm Robert Brown says: January 24, 2012 at 9:01 am Take a single gas molecule and put it at the top of the tube. It has zero kinetic energy and zero temperature. Let it fall and just before it hits the bottom it will have a lot of kinetic energy and heat. That is the lapse rate and it isn’t in equilibrium by definition. No, it won’t have any “heat”. You are conflating work, organized kinetic energy, and heat. Robert, you’re making an inappropriate distinction here. When that hypothetical molecule hits another, and they both bounce off in random directions, how is that gravitationally induced motion distinguishable from thermal motion? Answer: they’re the same thing. Gravitationally induced motion within a gas is heat energy. Drop a jar of air. Are you asserting… Avoiding the question by changing the premise. However, that jar WILL disturb air on the way down, and add a little thermal energy to the column. Septic Matthew January 24, 2012 12:12 pm Robert G. Brown, thank you again for a good presentation and determinedly addressing the criticisms. January 24, 2012 12:20 pm Temperature depends on the altitude, because more dense atmosphere means more molecules hit the thermometer’s mercury. To assign this to gravity is maybe not correct, but truth is the gravity causes denser, and thus warmer air. When talking about Moon SURFACE temperature, what is the daily temperature there in 2m altitude? Alan Millar January 24, 2012 12:22 pm “MaxL says: January 24, 2012 at 11:38 am I have a question along this topic which will likely show my ignorance. If gravity alone cannot induce a thermal gradient in a gas, how then are stars formed from gases where there is only gravity as an external force?” Gravity can indeed do that. However, that is perfectly allowable under the second Law of Thermodynamics if entropy is increased in another system. Like a fridge I can decrease its entropy but only by increasing the entropy of the room it sits in with the waste heat. Overall the entropy of the combined system will have increased and the second Law is preserved. Stars are part of the universe and you can be certain that overall entropy is increasing with the flow of time even if parts of the system are seeing entropy decreasing. The arguement here is Jelbring is trying to say he can prevent entropy increasing in a CLOSED system whilst work is being performed. There is a problem with Gedanken experiments, they can allow you to construct something that seems viable and yet breaches agreed physical laws. I could invent a closed system that was initially composed of a diffuse cloud of particles in thermodynamic equilibrium. Now that system would be near maximum entropy. However the addition of gravity starts to cause the particles to compress and voila I now have a system like the solar system or a galaxy or the universe even and I have reduced the entropy of the system without the addition of energy or increasing the entropy of another system. Perhaps someone could say that this proves gravity can reverse the flow of entropy in a closed system and the Thermodynamic laws need revising. However, in the real universe we inhabit we cannot create such a system in such an inital state. Perhaps a supreme, all powerful being could but I am not holding my breath! The existence of such a being would invalidate all known physical laws in any event. So we have to be careful with gedanken experiments. I tend to the view that if such a system is proposed, that is in breach of thermodynamic laws, that it is either in error or could not ever be created in the universe we inhabit. Alan Editor January 24, 2012 12:23 pm Folks, a lot of you here don’t seem to get it. The beauty of Robert’s proof is that there is only one question in it—does heat flow forever in the silver wire or not? So all of your claims of deep insights into where the joules are, and all of you talking about some mechanism or other that you are absolutely sure will make the air temperature at the top and bottom different, that has NOTHING TO DO WITH THE PROOF. The proof is about the outcome, what the mechanism is that you say results in that outcome doesn’t matter. IF you are correct and any of your hotly defended mechanisms work, gravity will make the air at the bottom of the tall cylinder warmer than the air at the top. (Your claim, not mine, just following it to see where it leads.) IF there is a temperature difference in the air top to bottom, heat will flow in the silver wire. Gravity can’t stop that. IF heat flows in the silver wire, it will move heat from the bottom to the top, and thus cool the bottom air and warm the top air. Duh. IF you are correct and any of your hotly defended mechanisms work, gravity will once again make the air at the bottom warmer than the air at the top, and the cycle will continue forever. So forget about your mechanisms, forget about the joules, forget about the lapse rates and how they are maintained, and just ANSWER THE FREAKIN’ QUESTION: Will heat flow in the silver wire forever? Me, I say no, and I say Roberts thought experiment elegantly proves that the answer is no. w. Alan Millar January 24, 2012 12:26 pm Should have said ‘decreasing’ in my third para of course. Doh!! Now if one of these new hotrod physicists can explain how and when Saturn dies in the Jelbring universe. Alan markus January 24, 2012 12:29 pm They just don’t get it. Or, as Bono would say, “stuck in a moment we can’t get out off.” To the tune of whatever ditty you like. —————————————————————————————– I’m a little radiation, radiation, radiation, I’m a little radiation, all day long Down through the mesopause , mesopause , mesopause, down through the , menopause, all day long. Down through tropopause, tropopause, tropopause, down through,tropopause, all day long. Now I’m a little kinetic, kinetic, kinetic I’m a little kinetic, all day long Up through the pressure, pressure, pressure, up through the pressure, all day long. Back to a little radiation, radiation, radiation, I’m a little radiation, all day long. —————————————————————————————————————————- Consider how a refrigerator works – 2 thermostats going down, and a heat pump going up. What happens to pressured gas through a condenser and then a separation device? Co2 forcing, what dribble. Pick a system – greenhouse or refrigerator. hotrod (larry L) January 24, 2012 12:53 pm just ANSWER THE FREAKIN’ QUESTION: Will heat flow in the silver wire forever? Yes but to no avail, as the heat flow would be exactly canceled by adiabatic heating of the gas at the bottom of the tube. The model above postulates only a temperature gradient from top to bottom in the tube but leaves out the pressure gradient developed in a sufficiently long vertical tube in a gravity field. In the case of a sufficiently long tube, where both gradients exist, the silver wire would try to transport heat from the warmer bottom of the tube to the cooler top of the tube as it must due to thermodynamic laws. The gas at the top of the tube would be warmed (thus increasing its pressure slightly (ideal gas laws temperature change constant volume tube) and like a piston this pressure increase would propagate down the tube at the local speed of sound in the gas causing adiabatic heating of the gas in each subsequent layer until it reached the bottom of the tube, instantly replacing the heat lost to the silver wire. Net effect – no heat loss from the bottom of the tube, and no net heat gain to the top of the tube as the two actions will exactly cancel each other out. The gas in the tube would never reach thermal equilibrium but would be in equilibrium energetically (PE+KE) Larry Q. Daniels January 24, 2012 12:55 pm Willis wrote: Will heat flow in the silver wire forever? If you extract energy from the system, it will shut down as the entire system cools. Energy is conserved. If you extract energy, then it has to come from somewhere, and that somewhere is the thermal energy of the system. If you do not extract energy, then yes, it will. It’s a terrible design for such. See my comment above. You seem to have rejected Graeff’s work as being insufficient proof. What would you consider a sufficient proof? Editor January 24, 2012 1:02 pm Alan Millar says: January 24, 2012 at 12:26 pmShould have said ‘decreasing’ in my third para of course. Doh!! Now if one of these new hotrod physicists can explain how and when Saturn dies in the Jelbring universe. Alan [I think I fixed it, there were two, better check and see if I got it right. -w.] Graeme W January 24, 2012 1:02 pm Willis Eschenbach says: January 24, 2012 at 12:23 pm Folks, a lot of you here don’t seem to get it. The beauty of Robert’s proof is that there is only one question in it—does heat flow forever in the silver wire or not?, Well… I don’t know the answer, but I do need to point out that a closed system with a changing level of kinetic energy in different parts doesn’t violate conservation of energy. Consider on object in an elliptical orbit around a gravitational point source. Its kinetic energy changes as it orbits, but the system is in ‘equilibrium’ with no external input of energy. Editor January 24, 2012 1:04 pm Q. Daniels says: January 24, 2012 at 12:55 pm Willis wrote: Will heat flow in the silver wire forever? … If you do not extract energy, then yes, it will. It’s a terrible design for such. See my comment above. I hate it when my perpetual motion machines are poorly designed … w. Q. Daniels January 24, 2012 1:13 pm Willis wrote: I hate it when my perpetual motion machines are poorly designed … A curious thing. I’ve noticed that engineering tends to be particularly bad when people want it to fail. MDR January 24, 2012 1:14 pm @Willis Heat is indeed conducted upward in the column forever. Why? There is a temperature gradient, and as long as there is a temperature gradient, conduction will occur. It doesn’t matter that the conduction is via the wire, or within the gas itself. It sounds like you are expecting the temperature profile would eventually smooth out [i.e., become isothermal] over time, and indeed if there we no gravity this is exactly what would happen – the gas would become isothermal, isobaric, and have a constant density. But in the presence of gravity, you have to take into account the potential energy imparted to the gas as a function of height. Gravity has dome more work on the gas at the bottom of the column than at the top. By virtue of being at the bottom, some of that gas’ gravitational potential energy has been spent [that is, gravity has done work on that gas parcel]. As a result, with no other outlet, this work energy has been converted into thermal energy. This is what the First Law of Thermodynamics is saying. Thus, the temperature of the gas at the bottom is higher than the gas at the top in the presence of gravity, and indeed this is a stable arrangement in thermodynamic equilibrium. Joe Born January 24, 2012 1:28 pm Willis Eschenbach: “So forget about your mechanisms, forget about the joules, forget about the lapse rates and how they are maintained, and just ANSWER THE FREAKIN’ QUESTION: Will heat flow in the silver wire forever?” As I said explicitly and a couple have implied (by noting that Dr. Brown has begged the question), heat will not even begin to flow, despite the temperature gradient, if that temperature gradient is the one that Velasco et al. specify, at least if Velasco et al. are correct. As I mentioned above, this is not hard to understand if you look at heat transport as a diffusion phenomenon, i.e., as flow in accordance with the laws of probability from a region characterized by a higher concentration (of fast molecules or fast electrons) to one with a lower concentration. Superimposed upon that diffusion flow is a contrary drift flow from gravity that cancels it out and thereby maintains a gradient. I’m told that an analogous effect occurs when a semiconductor diode is fabricated. At the instant two differently doped semiconductor materials join to make a diode, there exists across the resultant junction a hole gradient in response to which a cross-junction hole current begins to flow that tends to eliminate the gradient. But that current stops flowing before the gradient disappears. The reason is that the charge thereby transported sets up an electric field that opposes the cross-junction hole current. So the electric field maintains a gradient that the laws of probability (diffusion) would otherwise eliminate. kdk33 January 24, 2012 1:29 pm Brian, What is the lapse rate in the ocean? markus January 24, 2012 1:35 pm Am I one of the boys yet Willis? Regards, Markus. Jordan January 24, 2012 1:39 pm Gulp! I have my doubts about the above analysis. IMO, the tidy demonstration of the exponential pressure profile does not advance either position. It is merely astatesment of the profile to expect at isothermal conditions (i.e. “if we assume constant temperature”). The crux of the issue is whether the thought experiment justifies the invitation to accept the assumption. The silver may conduct energy, but this doesn’t lead to the conclusion that we have devised PM as there is no energy leaving the system (and no case is made to say that it could be removed indefinitely). The container has a mass of molecules jostling around forever (so long as we don’t take their energy away). The silver looks like an extension to the container – a somewhat circuitous route in which to carry out their mutual exchanges. If (for now) all moleculecular collisions were to transact a fixed quantum of energy, the flow through the silver (‘Q’) would be limited by the frequency of exchanges at the lower pressure end. Any additional collisions at the higher pressure end of the silver would have no potential to increase ‘Q’. It is then unclear whether the silver makes any difference to an assumed isothermal end state. If we remove the fixed quantum constraint, things may improve for ‘Q’, as the lower frequency transactions at the low pressure end may then be more energetic. But all this seems to be saying is that the silver may be a better conductor than the gas (a preferred route for mutual jostling). This alone doesn’t support the leap to an isothermal end state any more than leaving the gas to its own devices. Finally, there is the significant point that the lower pressure molecules have the greatest total energy in the isothermal state. It is this question that makes me reluctant to go with the assumption. markus January 24, 2012 1:40 pm Sorry about the rapid fire. Mr Willis Eschenbach, Sir, you have done us a great service, you have helped us to reason. Thank you very much. Joe Born January 24, 2012 1:41 pm I misspoke slightly when I said that heat will not even begin to flow in the wire. Whether it does or not initially depends on the temperature distribution that prevails in the wire before it is connected across the gas column. And there may be an initial transfer of energy between the air and the wire. But heat flow will stop before the temperature gradient disappears. Editor January 24, 2012 1:42 pm When I illustrated my post called “Perpetuum Mobile“, I chose a photo of a Civil War era perpetual motion machine, because that’s what they are in my mind—a relic of a time long ago when people hadn’t grasped that such a machine is an impossibility. However, I’m starting to see that perpetual motion still maintains its historical death grip on the scientific illiterati. And I can kinda see why, everyone wants something for nothing. And truly, people, I hate to bust your bubble and maybe I can’t do so in any case, but the heat can’t flow in the silver wire forever. That would be perpetual motion, and the laws of thermodynamics don’t allow that. Look, I know that we all break various laws all the time, someone once estimated that Americans break one to three laws every day. But the laws of thermodynamics aren’t like that. They are not just good ideas, or regulations put in place to protect us from each other or from ourselves. As far as anyone has ever been able to determine (and lots have tried) those laws simply can’t be broken. That’s why they are called the Laws of Thermodynamics, and not the Good Ideas of Thermodynamics. Those laws say we can’t have a perpetual motion machine driven by gravity. So if you want to continue to believe that heat will flow through the silver wire forever and ever without end, and that the mystery power that will make it do that is gravity, or lapse rates, or unicorns, or density-driven molecular interactions, or “gravito-thermal forces” or anything else, be my guest. As I pointed out with my Civil War machine, that mistake has a long and storied history, you’re not the first to believe in energetic fairies and Maxwell’s demons. Just don’t expect your belief, that gravity can do continuous unending work forever and ever amen, to be widely shared in the scientific community … w. Alan Millar January 24, 2012 1:47 pm “MDR says: January 24, 2012 at 1:14 pm Thus, the temperature of the gas at the bottom is higher than the gas at the top in the presence of gravity, and indeed this is a stable arrangement in thermodynamic equilibrium.” So MDR A gas giant planet, like Saturn, radiates more than twice the amount of radiation than it receives from the Sun. Presumably this little gravity induced energy engine is at work here according to you. How do these planets ever die, if gravity is constantly maintaining a heat gradient in the atmosphere? The Sun will die and become a cold white dwarf with no solar wind to blow any atmosphere away. What is going to kill these gas giants? If gravity is constantly maintaining hotter gases at the bottom then convection will move gases around and therefore we seem to have an everlasting living planet. What kills it and when, if Jelbring is correct? Alan Q. Daniels January 24, 2012 1:49 pm Willis wrote: Just don’t expect your belief, that gravity can do continuous unending work forever and ever amen, to be widely shared in the scientific community … Such a demonstration would be worth at least a handshake from Carl Gustaf. MDR January 24, 2012 1:49 pm @Willis Implicit in this discussion is that gravity is an unvarying external force being applied to the column of gas. That is what provides the [apparently infinite] source of energy. The difference between this scenario and a perpetual motion machine is that, for a perpetual motion machine to work, it cannot rely on infinite external sources of energy. Given that one is assuming here that gravity exists as an external agent, and is capable of doing work on the gas, I stand my my comments above. MDR January 24, 2012 2:07 pm Millar Sorry, I can’t directly answer your question. But it’s probably worth considering that Saturn is probably not in thermodynamic equilibrium [it has seasons, weather, storms, etc.]. It may in fact still be settling, that is, not enough time has passed since Saturn’s formation to reach an equilibrium state where the heavier matter is underneath the lighter matter. If settling is occurring, gravity would still be doing work on the gas, converting gravitational potential energy into other forms of energy, and some of this energy may radiate into space making it appear that Saturn has an internal heat source. glen martin January 24, 2012 2:07 pm “Willis Eschenbach says: January 24, 2012 at 12:23 pm Folks, a lot of you here don’t seem to get it. The beauty of Robert’s proof is that there is only one question in it—does heat flow forever in the silver wire or not? . . . IF there is a temperature difference in the air top to bottom, heat will flow in the silver wire. Gravity can’t stop that.” Actually it can and does, heat in the wire is being transmitted via the interaction of moving particles, gravity will cause the particles to slow slightly as its height increases thus slightly less energy is will be transferred to the atom above a particular atom than was received from the atom below it. This results in a gravitationally induced thermal gradient in the wire. Ged January 24, 2012 2:10 pm @Willis, Gravity doesn’t make the air warmer at the surface, it makes the air -colder- above the surface. A silver wire has a high thermal conductivity, so it’s easy for heat to flow through its length. But what would happen if you stretched that silver wire over a mile? 12 miles? Would the heat flux be the same over its length? No. You would get microdomains, fluctuations where some areas get randomly distributed with more heat than others, and those domains will flow around. You can see this easily with objects that have very, very low thermal conductivity. You have microlattice vibrations. But also realize gas is -not a solid lattice-. I don’t understand, Willis. Before hand you were so concerned with the conservation of energy. Now are you claiming you can take a molecule with a certain kinetic energy (which is what temperature is a measurement of), and raise it 17 kilometers above the Earth (increasing its potential energy by 4664 Joules if we’re talking about a mole of N2) without inputting more energy, and have it maintain that same kinetic energy, that same temperature? Answer that question, Willis. Peter Spear January 24, 2012 2:14 pm A physicist says: January 24, 2012 at 8:48 am “It seems to me Robert Brown’s analysis implicitly makes the following claim: if all the greenhouse gases (mainly H20 and C02) were cleansed tonight from earth’s atmosphere, then the atmosphere would evolve toward a more nearly isothermal equilibrium But how would that work, exactly?” The answer is easy. By conduction. Convection will shut down quite quickly as the heat transported upwards by convection reduces the environmental lapse rate. Once the lapse rate is below the moist adiabatic lapse rate even cloud formation cannot drive convection anymore. Oh, except you removed all the H2O. Therefore once the lapse rate is below the dry adiabatic lapse rate all convection will stop. Without GHG to cool the upper atmosphere there is no way to regain a lapse rate suitable to drive convection. All vertical motion will cease. Conduction will become the dominant mode of heat transport. That too will stop when the atmosphere becomes isothermal. January 24, 2012 2:15 pm Joe Born says: January 24, 2012 at 8:21 am “…If an ideal monatomic gas subjected to gravity in a thermally isolated container consists of only a single molecule, its kinetic energy K–and thus the mean translational kinetic energy–at any altitude z is given by K = mg(z_max -z), where m is molecular mass, g is the acceleration of gravity, and mgz_max is the total (kinetic + potential) energy of the gas. This is true no matter how long you’ve allowed the gas to “equilibrate.” In other words, temperature depends on altitude at equilibrium: there’s a non-zero temperature lapse rate.” As I pointed out in the other thread, this is not a thermal system at all – or even a gas! – unless the atom is allowed to thermalise with the walls of the container. In which case the temperature (as measured by its average kinetic energy as it passes through a given level) is independent of height. I have now read the Velasco et al article, and it agrees with what I said: in either the microcanonic (totally isolated) ensemble (with a reasonable number of particles in the gas) or the canonic ensemble (in thermal equilibrium with the surface or walls, irrespective of the number of particles), the gas is isothermal. You are trying to make a haystack out of the negligible point that, for a tiny number of isolated particles, the statistics aren’t precisely the same as for the usual smooth distribution – they’re “lumpy”. However, even in this extreme case, the temperature at equilibrium will still be the same throughout the entire height, in the crucial sense that no net work could be extracted from the gas by connecting different levels, by any means whatsoever. The “lapse rate” is still zero. Note by the way that any thermometer capable of measuring the the temperature of two levels of any such system alternately is itself either extracting work from any measured temperature difference, or has to have work done on it to obviate this happening. Blue Sky January 24, 2012 2:18 pm All these tree people making a joke of climate science. You have Mann with his cherry picked tree proxies and Ned Nikolov, Ph.D. Ph.D. in forestry making a joke of Thermodynamics. Any way the 600mb temp is going cold right now. http://discover.itsc.uah.edu/amsutemps/execute.csh?amsutemps Graeme W January 24, 2012 2:20 pm Ged says: January 24, 2012 at 2:10 pm I don’t understand, Willis. Before hand you were so concerned with the conservation of energy. Now are you claiming you can take a molecule with a certain kinetic energy (which is what temperature is a measurement of), and raise it 17 kilometers above the Earth (increasing its potential energy by 4664 Joules if we’re talking about a mole of N2) without inputting more energy, and have it maintain that same kinetic energy, that same temperature? Answer that question, Willis. I’ll take a stab to see if I can answer it correctly. The short answer is it will not maintain the same kinetic energy, the same temperature. This has never been postulated. What’s stated is that eventually the average kinetic energy (the temperature) of all molecules will be the same throughout the gas. That’s not necessarily the same kinetic energy as any individual molecule will have at any given time – it’s the average that’s the temperature. Individual molecules can be hotter or cooler than this average. Jordan January 24, 2012 2:22 pm “Will heat flow in the silver wire forever?” Will the molecules continue to jostle forever? If we never take any energy away from the system, what is the difference between jostling in the gas and flow through the wire? The questions that remain are: Is isothermal state the lowest energy state for a compressible gas in a gravity field? And on what measure is there a stable (mininum energy) outcome if we are asked to accept that molecules at higher altitude have more total energy than molecules at low altitude? Just asking. Tim Folkerts January 24, 2012 2:25 pm Ged says: I don’t understand, Willis. Before hand you were so concerned with the conservation of energy. Now are you claiming you can take a molecule with a certain kinetic energy (which is what temperature is a measurement of), and raise it 17 kilometers above the Earth (increasing its potential energy by 4664 Joules if we’re talking about a mole of N2) without inputting more energy, and have it maintain that same kinetic energy, that same temperature? This is a very subtle point, but one worth understanding. “Temperature” is a measure of the average thermal energy at some location. As implausible as it might seem at first, the average kinetic energy of the molecules that make it 17 km will be the same as the average KE of the molecules at the bottom. You see, only a very few molecules will make it 17 km high. The rest are pulled back by gravity before they get that high, so they never get counted. The self-selected molecules that DO make it 17 km high are the ones that had LOTS of KE to start with. Sure they lose a bunch on the way up, but that ends up leaving this subset 17 km high with the same average KE as the ENTIRE set had at ground level. (The same is true at any other level up thru the atmosphere). This has been discusses MANY times in related threads recently. kuhnkat January 24, 2012 2:32 pm Willis Eschenbach wrote: Just don’t expect your belief, that gravity can do continuous unending work forever and ever amen, to be widely shared in the scientific community … So Willis, when will we start flying off the planet??? When will the pressure on my feet from standing in one place stop?? When will the oceans boil from lack of pressure? Oh yeah, gravity is apparently an unending source of energy that counteracts centrifugal force. If not unending, we haven’t yet measured its reduction. kuhnkat January 24, 2012 2:33 pm Alan Millar, “The Sun will die and become a cold white dwarf with no solar wind to blow any atmosphere away.” This is an assumption without much empiraical proof. Willy January 24, 2012 2:34 pm MDR says: January 24, 2012 at 9:32 am “…Now, we just showed that internal energy decreases with height, as explained above. Since internal energy [of an ideal gas] is directly proportional to temperature, this must mean that temperature also decreases with height…” The correct statement would be ” that ALL ELSE BEING EQUAL temperature also decreases with height”. But the internal energy of an ideal gas is also directly proportional to density, which decreases with height. So all else is not equal. The argument fails. In other words the decrease of internal energy with height manifests itself as decreasing density, not as decreasing temperature. (In case someone does not see why internal energy is proportional to density it is because if stuff has energy, and you have more of the same kind of energetic stuff, you must have more energy.) January 24, 2012 2:36 pm loads of politics going on here. Sceptical AGW blogs still trying to prove that planets are at temperatures dictated by their atmospheric gas composition rather than their distance from the sun. The Greenhouse effect still rules concensus thinking be you an aye or a nay for AGW. quondam January 24, 2012 2:37 pm In 1964, Manabe and Strickler published “Thermal Equilibrium of the Atmosphere with a Convective Adjustment”: 2b. Thermal Equilibrium with convective adjustment. The procedure of convective adjustment is to adjust the lapse rate to the critical lapse rate whenever the critical lapse rate is exceeded in the course of the numerical integration of the initial value problem. The observed lapse rate of temperature is approximately 6.5 deg km-1. The explanation for this fact is rather complicated. It is essentially the result of a balance between (a) the stabilizing effect of upward heat transport in moist and dry convection on both small and large scales and (b), the destabilizing effect of radiative transfer. Instead of exploring the problem of the tropospheric lapse rate in detail, we here accept this as an observed fact and regard it as a critical lapse rate for convection. In a nutshell, this adjustment eliminates the possibility that a greater potential gradient might compensate for increased transport resistance by GHGs. Instead, only flux changes are allowed which can be nullified by positive feedback. Et voilà, CAGW! Ged January 24, 2012 2:37 pm @Tim Folkerts, Maybe that is the case. Maybe you are totally right. But I still find that hard to believe form observations. As you move up in altitude, the temperature, the measurable, observable temperature drops steadily. In fact, form my calculation I was less than 3x off just looking at KE turned into PE from the actual temperature at that height. So the problem is, the average thermal energy you are talking about includes -all- altitudes, and this is a value that cannot change unless absolute energy is given or take from the system. But it does not -follow- the temperature change that occurs as any molecules moves from low to high against gravity. And that’s what I’m talking about. That’s what this whole discussion is actually about. Yes, the molecules that make it 17 km up had more KE when they started than those that don’t (well, to a degree, as it’s a random walk). But once they MAKE IT 17 km up, they’ve lost a great deal of kinetic energy, temperature, into potential energy, versus what they had at sea level, as energy most be conserved. Correct? In that way, gravity does NOT DO WORK, but gravity TRANSFORMS ENERGY from one type (kinetic) to another (potential). This must drop the temperature of the molecules in question, since temperature is only a measure of the KINETIC ENERGY side of the TOTAL ENERGY equation. So, if air rises, it must lose temperature as a consequence of moving against gravity. Average kinetic energy of the entire system means little, as that will not change unless energy is removed from the system or inputted to the system. Average kinetic energy of the cohort of air that moves from low to high, now that does mean something, and that is going to be changed into potential energy. So then, gravity seems to be able to maintain a temperature gradient -in this way-, as it seems to me, and maybe I’m wrong and your way of looking at it is right. But, what predictions would we make for the real world from what I’m seeing? For starters, we’d expect that if the average energy of the entire system increased, so there was more kinetic energy at the surface, that the atmosphere would “puff up” from the Earth and it’s edge would get higher due to more KE being available for transforming into PE. And this is -exactly what we see- in the real world. And the converse is true. Which is what we see with Martian poles in the winter, or Pluto’s entire atmosphere whenever it gets far enough away from the Sun. January 24, 2012 2:38 pm “Why is it that you want to fight over physics that you can actually see with IR eyes? Save your energy for useful things, like arguing about the magnitude of the GHE, the sensitivity of it to changes in CO_2 concentration, the sign and nature of climate feedback or albedo modulation or the complex effects of atmospheric convection on local heating or cooling rates, or the ocean’s effect. The IR spectra render arguing about GH warming per se moot.” Thank you Robert. I sometimes wonder why skeptics waste their time an energy fighting against working science. The real question is sensitivity. Think of all the energy and time devoted here on WUWT to clearly false theories. Imagine if that effort were put to better purposes. Like the surface stations project expanded on a global basis. sad to see so much human energy, curiousity and intelligence wasted on crap like this and N&Z George Turner January 24, 2012 2:40 pm Actually Willis, heat flowing through the silver wire forever doesn’t mean it’s impossible, as heat always flows forever in any system above absolute zero. Take any object and an arbitrary plane that defines it. The two parts will never be in exact thermal equilibrium because atomic collisions are discrete, so half the time one side is hotter and half the time it is colder. Thus heat flows back and forth across the boundary – forever. That doesn’t mean the existance of an object above absolute zero is impossible. Or, take the case of the Maxwell Demon I constructed in my first comment in the thread. It’s made of nothing but pine and mylar but should maintain an imbalance of temperatures between two objects forever, because photon transmission and reflection can be made to be asymmetrical or unidirectional by having mirrors moving to create a path that is only valid in one direction. The concept is like the way volleyball players set the ball to each other. Given the timing of their motions, a volleyball can’t travel the reverse path and find the right hands in the right places at the right times. You can do the same with light if your mirrors are moving very fast. So if you can construct a system that never reaches the isothermal state, and attach another path for heat flow, the heat will travel from hot to cold forever. This changes the stable temperature difference between the two objects, turning the Demon into a less efficient Demon. You can’t extract anything from the permanent heat flow without having the energy dissipate. As for thoughts on the problems of heat flow through a long silver wire, the equations we use have never needed to include any gravity component because it’s unimportant to most purposes. It’s only fairly recently that physicists got irked enough to tweak heat flow equations so the heat couldn’t travel at infinite velocity, exceeding the speed of light. Equations for electricity, from Ohms law to Maxwell’s equations, likewise completely neglect gravity, implying that I can connect a copper wire to a black hole with a potential of -24 VDC and have electrons pour right up out of the gravity well. Or I could show that an atmosphere of electrons and protons (hydrogen) wouldn’t have more pressure at the bottom than the top because Ohm’s law or some other handy formula says the electrons would be evenly distributed, as gravity was never factored in to the completely accepted electrical formulas we use every day. son of mulder January 24, 2012 2:41 pm I shall repeat in different words the argument i provided at son of mulder says: January 24, 2012 at 9:59 am which no one challenged (or read). If I’m wrong please challenge. Heat cannot flow up the silver conductor unless more heat is pushed in at the base. Heat is energy and hence by Einstein mass. If you consider all the heat in the silver conductor as a distribution of mass then that mass has a center of gravity in the gravity field. If heat were to flow from the base to the top (without input of new heat at the base) then there would be reducing heat (mass) a the bottom and increasing heat (mass) at the top. Overall the centre of gravity of that heat (mass) would rise. ie work would have to be done against gravity. Put another way a force against gravity would have to be applied. If no new energy is being pumped into the base then the law of conservation of energy means that heat cannot be conducted up in a gravity field and so the top will be colder than the base and remain so ie no force can be applied. The same logic can be applied to the atmosphere. You just have to ensure no new heat is applied and no heat is removed. If no new heat is applied and energy is allowed to be radiated away then eventually all temeprature would be equalised only at absolute zero. January 24, 2012 2:42 pm Willis, Seems like folks here want to argue that the laws of thermo are not settled science. same with the laws of radiative transfer kuhnkat January 24, 2012 2:45 pm Let me toss my in my idea to show what is wrong with this thought experiment. Take the earth system and remove gravity. What happens to the GHG’s, the water, loose objects, the adiabatic lapse rate… Take gravity away from the thought experiment and what happens. Nothing except a very slight redistribution of mass as the pressure equalizes through the column? pidge January 24, 2012 2:53 pm For a volume of gas to have a thermal gradient requires a heat source and heat sink at each end of the gradient, so that heat is transported along the gradient from source to sink. Remove the heat source and the heat sinks from the gas, and the thermal gradient will disappear. In the case of Earth’s atmosphere, the heat source is the heating of the surface of the earth by the sun, and the heat sink is radiative to Deep Space. So it is not appropriate to dismiss Robert’s because it’s not the atmosphere – Robert is just showing the mechanism in the paper being criticised is not correct. A physicist January 24, 2012 2:57 pm A physicist says: “It seems to me Robert Brown’s analysis implicitly makes the following claim: if all the greenhouse gases (mainly H20 and C02) were cleansed tonight from earth’s atmosphere, then the atmosphere would evolve toward a more nearly isothermal equilibrium. But how would that work, exactly?” Peter Spear says: The answer is easy. By conduction. Convection will shut down quite quickly as the heat transported upwards by convection reduces the environmental lapse rate. Once the lapse rate is below the moist adiabatic lapse rate even cloud formation cannot drive convection anymore. Oh, except you removed all the H2O. Therefore once the lapse rate is below the dry adiabatic lapse rate all convection will stop. Without GHG to cool the upper atmosphere there is no way to regain a lapse rate suitable to drive convection. All vertical motion will cease. Conduction will become the dominant mode of heat transport. That too will stop when the atmosphere becomes isothermal. Peter Spear, (IMHO) your scenario is correct. The resulting no-GHG Earth would have nearly isothermal, hence stratified atmosphere, for the physical reason that every thermal would carry heat into the atmosphere that could never be radiated away, making it ever-harder for thermals to rise on subsequent days. The no-GHG weather would be freezing no-wind nights followed by still cold days. As (relatively) warmer tropical air slowly circulated (colder poles), first Earth’s polar oceans would freeze, then the mid-latitudes, then even the equatorial oceans. Some folks we’ve seen this scenario even here on earth, way back when the sun was cooler and GHG’s were scarcer, a world of low GHG’s and frozen seas … This hypothesis is called Snowball Earth. Elevator Summary: GHG’s prevent Snowball Earth. Tim Folkerts January 24, 2012 2:58 pm kuhnkat says: “Oh yeah, gravity is apparently an unending source of energy that counteracts centrifugal force.” Repeat after me: Force is not energy; energy is not force. Gravity can and does provide a continuing force. That does not mean that it is providing any continuing energy. Gravity only does work when there is a net movement inward. Since the atmosphere is not continually falling, gravity is not doing work. (NOTE, on the gas planets, there is no solid surface to stop contraction, so there the planets are indeed contracting and generating continued thermal energy of the sort many people seem to think exists on earth. This is also how protostars warm as they collapse inward.) Camburn January 24, 2012 3:01 pm steven mosher says: January 24, 2012 at 2:38 pm 100% correct Sir. Sensativity is the question and there have been no answers with acceptable confidence from a model yet. The present parameters do not hindcast with enough accuracy to forcast with any confidence. Physiscs, as we presently know it, is physics. Unless a new law of thermodynamics is found, Dr. Robert Brown is 100% correct. January 24, 2012 3:04 pm Nick Shaw says: January 24, 2012 at 9:52 am as air gets closer to outer space (in really simple terms) it would get mighty cold Space has no temperature. To have a temperature there must be kinetic energy which means that there must be mass. Things in space get cold because they radiate heat away. Look at photos (or drawings) of the space station – look for the radiators. These are positioned so that the Sun does not shine on them. In the case of an atmosphere without IR radiators, it will have no way to cool itself. Chas January 24, 2012 3:06 pm Do I have this correct : the temperature is dependent on the kinetic energy of individual molecules(?) When I throw a single ball upwards it’s velocity decreases with height. Do gas particle’s velocities not decrease as they go higher and higher,? Can someone explain why, if their velocities do not decrease, there fewer of them at height. Sorry if I am being dim! LongCat January 24, 2012 3:18 pm While I agree with the underlying point, I’m not sure why the wire would necessarily violate the laws of thermodynamics if it continuously transferred heat. Under normal circumstances, it would radiate some of this energy away and otherwise be an imperfect conductor. If, however, we’re assuming a closed system with a perfect conductor surrounded by a perfect insulator, why would any energy be lost? To put another way, assume I have a wheel with a frictionless axle at rest in a vacuum. If I spin it, it will spin endlessly. The conclusion that it will have perpetual motion doesn’t violate thermodynamics; the assumption that there is no friction does. Likewise, the wire would not violate any physical laws by endlessly transferring heat; those laws were broken by the assumption of a closed system with a perfect conductor/insulator. I know I’m disputing people far above my pay-grade, so I’m assuming that I’m wrong in this. I’m just curious as to why. hotrod (larry L) January 24, 2012 3:22 pm What the recent theories regarding gravity induced temperature gradient are really saying is that the ideal gas law as commonly stated is incomplete. A factor is left out because in most terrestrial situations it is irrelevantly small. They are implying the ideal gas law should be stated as: PV = NkT +(delta PEg) where PEg = the change in gravitational potential energy. If you radically increase the gravitational potential energy of a mass of gas, you have changed the total energy in the system unless you give up an equivalent amount of energy in the form of temperature. Larry Stephen Wilde January 24, 2012 3:26 pm “The real question is sensitivity.” Yes indeed it is but the lapse rate issue goes to the heart of it. If the lapse rate and thus the surface temperature is set by pressure and solar input alone then the effect of GHGs is zero. GHGs do introduce more radiative and conductive energy around themselves by virtue of their thermal characteristics so, if there is more energy in the air but the surface temperature fails to change then something else has to give. I propose a miniscule change in the surface air pressure distribution instead. That would simply change the rate of energy flow through the system from surface to space and redistribute the energy at the surface as necessary for no global change in surface temperature at all. Tim Folkerts January 24, 2012 3:29 pm “Heat is energy and hence by Einstein mass.” I hate to even bring this up, but I am rather sure that gravitational red-shifting will indeed have a theoretical affect on things (but I don’t think in the way imagined by the earlier poster). A photon will be red-shifted when it rises from the surface. It would be a straightforward task to estimate the change in wavelength and hence the change in “temperature” for thermal IR photons arriving high above the earth from the surface. However, typically such effects are only noticeable very close to very massive objects. I am sure that the “relativistic lapse rate” would be microkelvins at most, and hence not important in this discussion. But, hey, if anyone wants to calculate the actual “relativistic lapse rate” and its effects — go for it. As a warm-up, I would suggest calculating the time correction for GPS satellites to make sure you know enough to get started. Joules Verne January 24, 2012 3:30 pm Ged says: January 24, 2012 at 2:10 pm @Willis, Gravity doesn’t make the air warmer at the surface, it makes the air -colder- above the surface. ____________________________________________________ Bingo! Joules Verne January 24, 2012 3:39 pm Ged says: January 24, 2012 at 2:10 pm “Gravity doesn’t make the air warmer at the surface, it makes the air -colder- above the surface.” Yes. And if you remove the source of heat the column will cool and as it cools it shrinks and as it shrinks the molecules fall toward the surface and as they fall they gain back the kinetic energy they lost in making the ascent. If the temperature of the column drops enough the gas turns into an incompressible liquid or solid, completely collapses to the surface, and the gravity induced gradients are history. MDR January 24, 2012 3:42 pm @Willy Your statement that “the internal energy of an ideal gas is also directly proportional to density” is incorrect. Internal energy, a term with a very specific meaning in thermodynamics, is proportional only to temperature for an ideal gas. For starters, see http://en.wikipedia.org/wiki/Ideal_gas , and in particular the section entitled “Classical thermodynamic ideal gas” where it states that the internal energy $U$ for an ideal gas is $U = c_V n R T$ where $c_v$ is the specific heat at constant volume, $n$ is the number of moles of the gas, $R$ is the gas constant, and $T$ is the temperature. None of these quantities depend on volume or density. dlb January 24, 2012 3:42 pm Willis at 11.54am wrote: Excellent insight, Wayne. That is exactly what happens. In an isothermal column of air, individual molecules at high altitude have more energy because of gravity. But for exactly that same reason, there are fewer molecules at high altitude. As a result, and as we would expect, in the isothermal condition the energy is spread out evenly through space (equal energy per volume) rather than equal energy per molecule as Hans Jelbring and Mr. Verne assert. Although I agree with Dr Brown, I disagree with what Willis has said here. Consider a cubic metre of soil and a cubic metre of air above it, although both are at the same temperature, they certainly have different amounts of enegy due to differing densities. January 24, 2012 3:45 pm equilibrate the total energy This is a major misunderstanding. Thermal equilibrium does not equate the total energy. Read the equipartition theorem. Open a standard introductory physics textbook. Learn what temperature is. Then return. Besides — and I’m going to make this a standard answer for all off topic replies. The presentation above challenges you to do just one thing. Tell me whether or not the system in figure 2 permits energy to flow in a circle forever. If you answer “no, of course not” you are quite right, and you have conceded that thermal equilibrium is isothermal, because the silver wire is just a proxy for thermal conductivity in the air itself that makes it clear why not (since detailed balance computations get confused when you add an utterly irrelevant process that you dream up involving gravity to them and then try to do them in your head without the faintest idea of how statistical mechanics actually works). If you answer yes, I’ve got this bridge in Brooklyn you might want to look at, right after I convince you to invest in the machine we can build that will turn heat into energy at the rate of 100%, because I could stick a heat engine into the thermal pathway of the silver conduction and the resulting system would convert 100% of any energy added to the fixed-lapse air on the left into work. The choice is yours, of course. rgb January 24, 2012 3:45 pm My head hurts. Is someone keeping score? If so, who is wining, the For team or the Against team? Confused minds need to know. MDR January 24, 2012 3:48 pm @Willy Note that I am *not* arguing that there *isn’t* a variation of density [or pressure] with height. It’s simply that once one knows the internal energy of a gas, one also knows its temperature, irrespective of either the density or pressure. Of course, the profiles of both density and pressure must still satisfy the ideal gas law, and as you intuitively expect both decrease with height [as does the temperature]. January 24, 2012 3:49 pm 2) The silver wire will transport heat from warmer region to the cooler region, but in so doing it short circuits the transport of heat by convection. So with the wire present, convection will be less, but the net transport of heat will remain the same. It won’t short circuit convection — in stable equilibrium there is no convection. Convection itself is a kind of heat engine driven by temperature differences that transports heat (on average) from a hot reservoir to a cold one. In stable equilibrium nothing moves, because there is always dissipation associated with movement that will slow it down, right down to the extreme quantum regime. I assume you aren’t talking about superfluid circulation and calling it “convection”. There is no input energy, also, so there cannot be net transport of heat. That’s the bit about “violating the second law of thermodynamics” in spades. A system with interminable flow of heat in a circle is a textbook violation of the second law, and it permits one to design any number of perpetual motion machines of the second kind. rgb wayne January 24, 2012 3:52 pm I thought metals had free electrons and electrons were matter and gravity accelerates all matter and metals conduct heat so well due to the free electrons and …. and how again is this proposed perpetual machine supposed to refute a lapse rate? Seems lifting the electrons against gravity in the metal bar from the warm to the cool would cancel if the gradient became -0.0098C/m. The only real thermal motion it seems would be with an isothermal air column, but that would just run, backwards to the arrows on the diagram, until all relevant potential energy was minimized. I think that point would be called the DALR. http://en.wikipedia.org/wiki/Energy_minimization We sure could use a proper experiment. I have already designed one that should work fine if anyone should ever want to consider doing one. Basically Graeff’s only designed correct this time with many sensors, multiple layers of insulation, large, and with air. Why he chose a liquid I’ll never understand. “In the modern world the stupid are cocksure while the intelligent are full of doubt.” ~ Bertrand Russell EthicallyCivil January 24, 2012 3:56 pm Clearly a non-isothermal dry adiabatic column isn’t a minimum energy state (as energy can be extracted, imagining an efficient thermal conduction system between top and bottom layers. Convection requires that some parcel of gas is already at an elevated energy state (or it wouldn’t be rising *due* to convection), and thus convection doesn’t violate conservation or 2nd law. The adiabatic column is (ignoring radiation) the minimum energy state with differentially heated (cool top, hot bottom) boundary conditions. I don’t have a clear picture how a heat engine connecting the top and bottom would look in T-S space, or what the minimum energy solution would look like were one to take an established dry adiabatic column, insulate top and bottom and run a heat engine within the column. It seems like it would have to go to an isothermal state as minimum energy A physicist January 24, 2012 3:58 pm A physicist says: It seems to me Robert Brown’s analysis implicitly makes the following claim: if all the greenhouse gases (mainly H20 and C02) were cleansed tonight from earth’s atmosphere, then the atmosphere would evolve toward a more nearly isothermal equilibrium But how would that work, exactly? Willis Eschenbach says: Wait, wait, you claim to be a physicist, answer the question. Does heat flow forever in the silver wire or not? Willis, the short answer is “Yes”, the medium answer is “IMHO Robert Brown basically has got it right”, and the long answer is “The logical next step is discuss whether day-versus-night temperature swings alone — in the absence of GHG effects — suffice to mix the Earth’s atmosphere.” Because that claim —foreseeably (and strictly IMHO) — is going to emerge as the primary fallback position of GHE skeptics. January 24, 2012 4:04 pm I see two assumptions above: 1. It does not matter what the density of the gas is. It will equally conduct heat at the bottom into silver wire, as the wire will be able to conduct its heat into the gas at the top, even though the density at the bottom and top is different, due to the gravitational effect on the gas. 2. The cross-section of the wire will stay the same, which means the ability of the wire to conduct the heat, which depends on its cross-section, is the same at the bottom and top. The gravitational field will actually pull down a considerable part of the mass to the bottom, making it far wider at the bottom then the top (depending on the length of the wire and its tensile strength), deforming it more into a tear drop shape. With your setup you may be able to change the lapse rate, but I doubt that you achieve an isothermal state in this way. It does not matter what the density of the gas is, or how good the contact of the silver with the gas is (as long as there is thermal contact, or how thick the wire is. The point is that if any heat enters at the top ever and the system spontaneously restores the lapse rate (which is constant, recall, for a container of fixed size independent of gas density so we can make the gas nice and thick with great thermal contact with the silver) then you’ve violated the second law, because any thermal pathway between the bottom and the top will deliver heat from the bottom to the top. All I’ve done with the wire is show you a pathway that is clearly completely independent of the supposed lapse rate in the gas, one that will conduct heat in any direction without prejudice, so that you can see why a lapse rate is impossible. If heat to the top goes back to the bottom because of “gravity”, and there is a pathway to the top that must conduct heat in the direction of a thermal gradient (the wire) you’re done. The system will circulate heat indefinitely. But no system can circulate heat indefinitely, it’s absurd. So the message is, stop trying to do statistical mechanics in your head without understanding it. Stick to thermodynamics. It is simpler, safer, and you can’t make mistakes with it as long as you remember TANSTAAFL. There ain’t no such thing as a free lunch. A nonzero thermal lapse is a Maxwell’s Demon working the free lunch counter for the suckers. Here’s the deal, really. Every book I’ve ever read on statistical mechanics or thermodynamics is wrong — and you can safely assume I’ve read a few, since I did numerical simulations of both static and dynamic critical phenomena that actually were published in places like Physical Review, with referees and everything — or figure 2 above makes it clear that there is no possibility that figure 1 is correct. That is, assuming that you can’t believe the actual words of the second law that tell you that you can’t take a system and create a permanent thermal gradient in it without doing work to maintain it, because heat will flow from the cold side to the hot side to neutralize temperature differences otherwise. That’s why you have to pay to refrigerate or air condition your house. It is why you can’t build a perfectly efficient heat engine. Tanstaafl, man. rgb January 24, 2012 4:10 pm I find the analysis quite reasonable – but it is so idealized as to be useless. A more interesting thought experiment has a spherical planet heated by a remote star, rotating on an axis roughly normal to the line to the star, with an atmosphere of non-greenhouse gases. The equator would be warmer than the poles, so there would be Hadley-type circulation that would cool the equator and warm the poles. Would there then be a vertical thermal gradient? I think there would be, but I’m sure someone would like to argue to the contrary. Not at all. It is merely specific. I am specifically proving that EEJ, a specific paper written by Jelbring and published in a journal (God help the referees, absent that day on vacation or something), violates the zeroth but especially the second law of thermodynamics when it asserts that there will be a thermal lapse rate in an adiabatically isolated column of ideal gas in thermal equilibrium in a gravitational field. No, there won’t. Is it stupid to have to prove this? Sure, given that nearly any introductory physics textbook — I’m not talking about thermodynamics text, just things like Tipler and Mosca, or Halliday, Resnick and Walker — teach enough thermodynamics for one to be able to see that the spontaneous appearance of a stable thermal gradient in any system is impossible, because it is a direct violation of the second law, and indirectly the first, which more or less says that equilibrium is isothermal (in order to permit the definition of thermometry in the first place). If thermometers “work” to measure temperature, equilibrium is isothermal. Period. rgb January 24, 2012 4:25 pm So your wonderful assertion, is that the radiative forcing of Co2, occur after its entry into the thermostats of the tropopause, and that extra radiative forcing, causes that missing hot spot, increasing the temperature back through the stratosphere and down again through the thermostat of the tropopause. Been there, done that. No, my “wonderful assertion” is that EEJ, Jelbring’s paper, is obviously incorrect because a stable, isolated atmosphere cannot support a thermal gradient. There is no such thing, in other words, as “gravitational heating” for a system in static equilibrium, nor is there any such thing as “gravitational heating” for a system in dynamic equilibrium. There is such a thing as gravitational heating in a collapsing system, which is what raises the temperature of protostars to the ignition point and provides the heat outflow from brown dwarfs. As the greenhouse effect is concerned, look at the IR spectrum from over the top of the atmosphere. I don’t care how you think heat gets to the top of the troposphere; the point is that one chunk of the outgoing spectrum observed from satellites comes from a gas that radiates in the CO_2 band that happens to be at top of troposphere temperatures. The net radiation in the water window therefore has to be higher (than it would otherwise be) in order to keep the Earth in detailed balance (on average). End of story. I don’t give a rat’s ass where the extra radiation comes from, or how it gets there, it is there. You can see it. It is emitted/transmitted at the blackbody temperature of the ground give or take a bit. Here, let’s use our fingers and toes. Total outflow in all frequencies has to be the same. Outflow in one band of frequencies is smaller because it comes from colder molecules. In order to keep total outflow the same, the energy radiated in the other frequencies has to: a) Go up. b) Go down c) Remain the same. That’s it. Come up with any mechanism you like for heat absorption and transportation, they’re all the same to me. Just don’t forget the incontrovertible experimental IR spectroscopy data and the finger and toe arithmetic involved. rgb ikh January 24, 2012 4:26 pm Lets try a slightly different explanation of this closed static system. Of course, a closed static thermodynamic system is, per see a thought experiment that can not exist in reality. We have a quantity of gas in a sealed cylinder and the cylinder is a closed system thermodynamically. That means energy can not enter or leave the cylinder. The cylinder is *not* in a gravitational field. Therefore, the mean pressure of the gas is the same through out the tube and the temperature of the gas is isothermal. This is thermodynamic equilibrium and entropy can not change. Now, we switch on our gravitational field. The gas settles into a pressure gradient with higher pressure at the bottom and lower pressure at the top of the cylinder. Because of the gas laws, the gas at the bottom heats up and the gas at the top is cooler. We will also assume that the gas does not absorb or emit at the frequencies associated with these temps ( i.e. we can ignore radiative transfer). So convection and conduction set in and start to mix the gas. The total energy in the gas is still constant. As the gas mixes energy is transferred from the warmer gas to the cooler gas and the temperature differential reduces. convection slows. This continues until T.E ( Thermal Equilibrium ) is reached. At the new T.E. there is no more convection, the gas is isothermal and the pressure gradient remains. The gravitational field is no longer performing work on the gas so it can not add energy to the system no matter what potential it exerts on any particular molecule. Just imagine if this were not the case and convection carried on at T.E. We could have a wind turbine driven by the convection current and connected to outside the system. We would now be removing a continuous source of energy from at system at T.E. Good luck with your patent applications ;-). /ikh January 24, 2012 4:36 pm Also, a constant temperature with altitude means that particles at the top of the atmosphere have more momentum than particles at the bottom. It means nothing of the kind. Momentum (magnitude) is p = mv. The distribution of v at the top and the bottom is identical — the Maxwell-Boltzmann distribution. Oh, do you mean more total momentum (in any given general direction) at the bottom than at the top because there are more particles? Sure — that’s why the pressure at the bottom is greater than the pressure at the top in a compressible fluid (where there are more molecules at the bottom than at the top). See “Kinetic Theory of Gases” at your friendly wikipedia outlet: http://en.wikipedia.org/wiki/Kinetic_theory rgb Alan Millar January 24, 2012 4:46 pm So again you post modern hotrod physicists please answer the question. This is the real world it IS going to happen. The stupid people here, who believe in the Laws of Thermodynamiics, know what will happen in the universe ruled by the Laws. But do the clever people, who know that the Laws are wrong, know how it happens in the Jelbring universe. Now don’t be shy, it doesn’t really matter if you show that you are a complete idiot more than once, does it? A gas giant planet, like Saturn, radiates more than twice the amount of radiation than it receives from the Sun. Presumably this little gravity induced energy engine is at work here according to you. How do these planets ever die, if gravity is constantly maintaining a heat gradient in the atmosphere? The Sun will die and become a cold white dwarf with no solar wind to blow any atmosphere away. What is going to kill these gas giants? If gravity is constantly maintaining hotter gases at the bottom then convection will move gases around and therefore we seem to have an everlasting living planet. What kills it and when, if Jelbring is correct? Trick January 24, 2012 4:47 pm Folks – Fellow interested posters, I am going to attempt to have a conversation with many of /y’all & all at once, skim for your handle if interested in engaging. All in time sort. Joules Verne says at 1/24 7:52am: “The device in figure 2 doesn’t work…the gas will collapse to the surface as a liquid before it gets to absolute zero and this will shut off further extraction of energy because the cold side of the thermocouple no longer has any cold gas to cool it.” Good one. Joules groks this stuff. Now by Robert Brown definition, the gas is isothermal, meaning the gas can’t cool – the Perpetuum Mobile w/isothermal gas & perfect insulator is born in Fig. 2. A neat design for Willis’ dream machine. Patent pending. Eilert sats at 1/24 7:54am: “…I doubt that you achieve an isothermal state in this way.” This actually adds to the discussion in a good way. A good thought experiment is to turn up the gravity field and see what would happen. The silver rod could react in the manner you describe. Interesting. Genghis says at 8:06am: “Take a single gas molecule and put it at the top of the tube. It has zero kinetic energy and zero temperature. Let it fall and just before it hits the bottom it will have a lot of kinetic energy and heat.” Very good one. At the top, the molecule also has lotsa’ PE. At the bottom, it also has 0 PE. No isothermal molecule down the tube, it is isoenergetic (I learned a new term) – where there is no wire! Robert Brown says at 8:14am: “Well then, by all means go patent your perpetual motion machine of the second kind or explain heat flow in the second diagram, Joules.” Robert – YOUR machine beat Joules. That is your machine design, it is still running. It will run tomorrow. It will run forever w/isothermal gas & you just have developed the perfect insulator patent. Go for it. The Perpetuum Mobile design, fig. 2 in the top post is worth A LOT!! And I’m ever so sorry, but in an ideal gas the temperature IS determined by the total energy. I cite the 1st Law consistent with oth law in one reservoir and the 2nd law constant entropy. Robert continues: “That’s an absurd idea, given that one can perform a gauge transformation — change the zero of the total energy — without changing any of the physics.” The datum concept enables many good & proper science in many texts. Insert the thermometer at any datum to compute the invariant total energy value, say TE = C, & thereafter allow temperature and PE to vary with each other from that datum thermometer reading and you will be ok. Just place the thermometer at any different h. The temperature or mean kinetic energy of the molecules it will measure will be different: KE(h) = C – mgh. Robert Brown says at 8:23am: “Heat will definitely flow in the silver, right? It’s just a chunk of metal that’s an excellent conductor of heat.” Right. This is why Robert Brown had to invent the perfect insulator design in fig. 2. How’s the patent pending process coming along Robert? Robert Brown continues: “You want to assert otherwise, you tell me what the equilibrium state is of figure 2.” I have tried elsewhere. In fig. 2, let’s say white body starts with Twhite and gray body Tgray>Twhite. Cite 0th law that says heat will flow to a paler shade of gray, both bodies at single reservoir Tavg with silver tube! Or with moving jars, dipping bird, etc. At least while Robert’s perfect insulator patent is pending. Robert Brown says at 8:28: “No, I think you are generally quite right, and this agrees rather well with Caballero’s argument.” Remember the Caballero text in the ref. given in Perpetuum Mobile thread makes an argument temperature is non-isothermal in a gravitation field section 2.3. Caballero does write: 2.1 No gravity: “Note that pressure is due to only to the local properties of the gas and not to anything going on far away.” 2.3 w/gravity “Mean velocities (insert his defn. Temperature for mean velocities here) will be greater near the bottom of the box than near, the top: in other words, pressure decreases with height. We will now work out an equation giving the precise rate of decrease.” Note Caballero defined “mean velocities” as temperature – I am just inserting for clarity, no meaning change. Thus temperature will be “greater near the bottom of the box”. This is not consistent with Robert Brown’s as yet unproven pronouncement that it is isothermal. Other than the fig. 2 w/perfect insulator inconsistent with 0th law driving an inconsistency with 1st Law. Schodinger’s Cat says at 8:57am: “Over time, I would expect the system to reach thermal equilibrium.” Welcome back from the quantum world I see you survived after all. Yes, by gosh, the system in fig. 2 with real non-perfect insulator will reach thermal equilibrium over time by 0th law. The gas temperature will be non-isothermal. Robert Brown says at 9:01am: “No, it won’t have any “heat”” Right, but Genghis molecule will have the increased kinetic energy and per Caballero’s 2.3 ref. increased temperature. Rap Genghis on the knuckles for saying “heat”. Robert Brown continues: “But the basic point of my paper is that Jelbring is wrong not because of any possible microscopic description of a lapse rate. A lapse rate itself is wrong in thermal equilibrium, because figure 2 is very, very easy to understand.” Robert Brown fig. 2 is wrong to have a perfect insulator which cannot exist by 0th Law. Jelbring cannot be proven wrong by fig. 2 which is not physical. At least until Robert Brown’s patent for the perfect insulator is granted. Robert Brown says at 9:07am: “…in figure 2 above. Which is violated — the heat equation in silver or your absurd assertion that gravity can stably sort out a gas into a hotter temperature and a colder one? One or the other.” The heat equation in silver is not violated in fig. 2. MDR says at 9:32am: “This is an example of a system that is both in thermodynamic equilibrium and possesses a gradient in temperature.” Good. This is consistent with 0th, 1st and 2nd thermo laws. Fig 2. is not & runs forever. Once fig. 2 has a non-perfect insulator admitted per 0th law, it will eventually be in equilibrium, non-isothermal gas, and not run forever. Rober Brown says at 9:51am: “Excuse me? I have no idea what you (Trick) could possibly be talking about.” Believe me I grok this. You are smart though and the thermo master’s were right in the 0th, 1st, and 2nd law development. They have stood the test of time right up to fig. 2. Robert Brown continues: “heat will flow in the wire from the bottom to the top.” Yes. Cite 0th Law. “The point is that heat will flow in this system forever…” Yes, fig. 2 is a Perpetuum Mobile. To win the patent for it though Robert Brown is going to first have to win the patent for the perfect insulator. If Robert cannot patent a perfect insulator, cite the 0th and 1st law tell us the system will stop flowing heat when it reaches thermal equilibrium and with 1st law non-isothermal gas (see Caballero 2.3 ref. telling us gravity results in non-isothermal gas column) thru the non-perfect conductor, this is less than forever but can be quite long because there are some quite close to perfect insulators. “The real point is that you don’t need the silver wire to make this argument…..Gravity does no work in this problem, not in steady state.” I whole heartedly agree. Wire is sort of useful to understand the concepts of the thermo laws. The ideal gas molecules in fig 2 lose just as much energy going up against gravity as energy they gain coming back down. Fig. 2 modified with real non-perfect rl insulator is ideal isentropic reversible process with non-isothermal gas (Per Caballero 2.3 – check it out). “So what makes the heat go round and round?” The perfect insulator. The perfect insulator is non 0th thermo law compliant. Your patent still pending? “Of course. It doesn’t.” Heat does stop flowing round and round with a patentable real non-perfect insulator. The gas will be non-isothermal per Caballero ref. 2.3. Robert Brown says at 9:58am: “Are you crazy?” No Joules Verne is eminently sane, Robert Brown misses the other poster (Joules Verne) said the work extracted was not from the system but from the ideal gas column with the perfect insulator (patent pending). Robert continues: “In the real world, the system will evolve to an isothermal state precisely as I described it because it is in equilibrium.” Not isothermal unless you think Caballero is crazy too (my view Caballero is not crazy), since his reference 2.3, disagrees with you. Caballero shows ideal gas column is non-isothermal in the presence of gravity. Caballero also shows ideal gas column is isothermal w/o gravity. JKrob says at 10:24: ?In moving the heat from the bottom of the tube to the top is causing the lapse rate to become **more stable** – cool at the bottom with warm air above is an inversion which inhibits vertical mixing!! THAT is why the engine will not work as it is set up.” A good .02 added viewpoint. Thanks for de-lurk. D.J. Hawkins: “Now I can construct a heat engine which extracts useful work based on the temperature gradient and gravity will continue to organize the air column forever and my heat engine will never run out of “fuel”? Really??” Robert Brown will demand a royalty for using his patent pending perfect insulator. Other than that cost, yeah fig. 2 works for free. Graeme W says at 11:50am: “I don’t understand the theoretical derivation well enough…” In the WUWT Perpetuum Mobile thread at the top post, there is a Rodrigo Caballero link cited by Robert Brown to the theory of ideal gas without and with gravity field. In 2.3 gravity is added to a non-isothermal gas column and this condition is theoretically justified – it is pretty simple derivation, I recommend it. Fig 2 perfect insulators needed only for a Perpetuum Mobile. Robert Brown for some reason does not grok Caballero 2.3 even though he cites it as good ref. Willis Eschenbach says at 11:54am: “In an isothermal column of air, individual molecules at high altitude have more energy because of gravity.” Bzzzt! Isothermal column of air is for no gravity case ref. Chapter 2.1 Caballero link in your Perpetuum Mobile post. Bzzzzt! “…because of gravity”. Ok, with gravity Rodrigo Caballero (bless him) tells us in 2.3 same link the air column becomes non-isothermal. Bzzzzt! Fail 1st law: individual molecules must have constant energy. Note the period. Willis “more energy” fails. More energy is not constant energy. Willis tries to knock down Jelbring paper with these fails. Not possible to knock down Jelbring with these 3 arguments. Willis Eschenbach says at 11:57am: “It is the formal disproof of Jelbring’s theory. It is idealized by its very nature.” By Willis use of “it”, my view Willis means top post where Willis summarizes his understanding of “it” at 11:54am. Willis at 11:54am fails any disproof on all 3 points of his points, see right above. Gosh, I hope Robert Brown groks Caballero & 1st law energy conservation quickly. Willis Eschenbach at 12:09pm (? to another poster): “Does heat flow forever or not?” Heat flows forever in fig. 2. The reason is a grand master thermal law is broken, the easiest one, the zeroth law. There can be no perfect insulator. Body A placed in contact with body B will always eventually equilibrate temperature. This is so easy. Geez. If Robert Brown can patent a perfect insulator, I will change my view. Patent granted: Perpetuum Mobiles will be possible. Until the patent is granted, heat will NOT flow forever in fig. 2 b/c there is no perfect insulator. Willis should grok this on his own (read Caballero slowly). Robert Brown is going to take longer I think, to reach a state of equilibrium grokness. LOL, I feel better my head cold is receding faster. Willis Eschenbach at 12:09pm: “Folks, a lot of you here don’t seem to get it.” No kidding. Including Willis viz: Will heat flow in the silver wire forever? “Me, I say no, and I say Roberts thought experiment elegantly proves that the answer is no.” For fig. 2 silver wire heat will flow forever like Robert Brown says it will in top post, his words:´ One is then left with an uncomfortable picture of the gas moving constantly.. At least until poster Joules Verne situation is reached and the gas cools & liquefies under a vacuum. That’s good thinking but beyond the intention of top post. My view is the insulator in non-perfect to be 0th law compliant and in that cas heat will not flow forever and gas per Caballero 2.3 will be non-isothermal. Willis Eschenbach at 1:04pm: “I hate it when my perpetual motion machines are poorly designed …” That fig. 2 is a WELL designed Perpetuum Mobile w/isothermal air & perfect insulator. Just gotta’ pay royalties for those: (patents pending). – – – – – – – Out of breath and w/blisters on my fingers I stopped at 1:04pm. I will happily rengage if more interested parties appear thereafter. Typo’s are possible in this post. Probable really. Walter January 24, 2012 4:51 pm Robert Brown and Willis: You will never convince some people. These same people are quite sure that the reason we don’t have machines capable of giving free energy forever is due to “poor engineering”, or its a conspiracy by Big Oil or Big Gubmint to fleece us all. If only the shackles of these evil doers were thrown off we could live a life of luxury forever, with energy too cheap to meter, made in our own basements be a nice perpetual motion machine – yours for$499 from your local hardware store. But of course, us incompetent engineers and physicists wouldn’t want that… line us up against wall now because clearly thats what we deserve.
/sarc

Mike McMillan
January 24, 2012 4:51 pm

Great theory. Baloney, but very entertaining.
The reason the silver thingy won’t generate perpetual motion is that the exposed ends will assume whatever the air temp is at that altitude. A temperature difference of 1 degree will not move any heat in a silver rod 100 meters long.

dscott
January 24, 2012 4:51 pm

adiabatically isolated column of gas in a gravitational field cannot have a thermal gradient maintained by gravity.
And yet we see this stratification in the atmospheres of Venus, Earth and Jupiter. In the real world, it’s hotter at the ground because the air is heated from the ground up and rising air columns draw this warmer air up… you know, the thermopiles birds ride up instead of having to beat their wings furiously. Imagine that, it’s colder on top of a mountain than at it’s base. It’s called the triumph of empirical science over clever explanations that claim otherwise.
The details of the arguments for an adiabatic lapse rate in open systems is unimportant, nor does it matter what cp is as long as it is not zero or infinity.
What matters is that EEJ asserts that in stable thermodynamic equilibrium.

Which does not occur in the real world. That’s why we call it weather. Sorry, no sale. This is just as unacceptable as the nonsense about CO2, a trace gas having more effect than water vapor on the planet’s temperature.

January 24, 2012 4:58 pm

I hasten to add that the lapse rate that does prevail at equilibrium is much smaller than that for which Jelbring contends, so Jelbring is still wrong.
If you want to bet against the laws of thermodynamics, I wouldn’t advise it.
I do not care about what generates the lapse rate. If the lapse rate is stable, so that heat delivered to the top redistributes to maintain a constant equilibrium temperature lapse between the top and the bottom — the sole case examined in the article above — then it violates the second law of thermodynamics.
Let me put it bluntly. If somebody presents a statistical mechanical computation that suggests that the second law is violated, I would knee jerk assume that the authors had made a terrible mistake unless and until proven otherwise, especially if I “could not understand” everything that they did.
Even then I would be doubtful.. To be honest, I would be doubtful if I did the work myself. I think that the paper you link has the right idea, and you will note that on other threads I propose precisely the same experiment. Show me, in other words. I’m a theorist, but I’m no fool. Experiments trump theory every time, but the Earth’s atmosphere does not exhibit a DALR:
* Uniformly. See “troposphere”, “stratosphere”, “thermosphere” etc. Why exactly doesn’t the DALR extend to the top of the exosphere again?
* Ubiquitously. See “thermal inversion”, or for extreme cases “thermal profile of the atmosphere above Antarctica in July” (there’s an IR spectrum in Caballero with the data you need). Hmmm, thermal inversion, with the upper troposphere hotter than the ground? Could that be, maybe, because the ground is cooler than the air instead of warmer? Regardless, it is nearly impossible to explain if there was an intrinsic DALR that didn’t depend on differential heating.
Nor is the atmosphere ever even crudely static. Air is always moving up, down, sideways. Even when it is “windless” down near the ground, there are damn few places on Earth where the air isn’t moving up and down and sideways from a kilometer on up, on any given day. So it isn’t really all that surprising that the air (a decent insulator and hence quasi-adiabatic a parcel at a time) has a DALR when it is differentially heated and moving all of the time, keeping it crudely “well-mixed”.
Thermal static equilibrium is something else. No mixing. Conduction matters — it can be slower, but there is plenty of time to reach equilibrium. That’s why the thermodynamic argument is so powerful — it is very difficult to explain how heat delivered to the top of any truly equilibrated air column would spontaneously redistribute to maintain a vertical lapse rate and not enable a heat loop and/or any number of PMM2Ks.
But show me the “high precision” experimental result, done with a dewar in a centrifuge filled with maybe Xenon gas at a G value such that there is sufficient pressure at the top of the vessel to justify the thermodynamic assumptions, with recording high-precision, carefully calibrated thermometers.
Just bear in mind that if it works, and there is a lapse rate, the second law itself is done. I’ll stick a thermocouple between the top and the bottom of said gas, and it will sit in there spinning forever.
rgb

Neil
January 24, 2012 5:01 pm

Dr Brown is not correct with his explanation of Fig2
In a cylinder with gas at the usual DALR, all that his conducting wire will achieve is a infintesimally thin layer or hotter gas at the top plus an infintesimally thin layer of cooler gas at the bottom.
Both of these would reverse the lapse rate ( inversion) and thus no further heat can be exchanged without adding work to the cylinder (His statement that the system would reorganise itslf into an adiabatic column is wrong)
This situation would also apply to an attempt to use a themocouple to derive any work from the system. (Or any other heat engine)
It is thus clear that the adiabatic column cant be used as a perp -motion device so the proof that it breaks thermo laws has gone away.. The reverse holds true for an isothermal column where the bottom must be cooler than a surrounding DALR atmosphere with the opposite at the top– In this case heat can be removed at the top and added at the bottom. This would comprise a real-perp motion machine as the column would then reoganise itself into the isothermal state
Clearly it is the isothemal condition that breakes the 2nd Law
PS Much has been said about the need in the real atmosphere to maintain a DALR by pumping from below with energy from the Sun. In the real world the column is subject to all sorts of energy lossed and gains. The list is very large as Anthony has enumerated is is recent post– The water cycle is one of the most important
Most contributors seem to have ignored the fact that in the DALR cylinder all the heat losses and gains have been elininated and thus it takes an infintesimally small amount of energy to lift a parcel of air fron the bottom to the top hence an infintesimally small amount of energy will maintain the DALR.
We just need to overcome conductive and radiative transfers which have already been assumed to be zero

January 24, 2012 5:02 pm

Oops, I mixed replies to the previous two comments. The paper was a separate suggestion. The answers are the same, however. Experiments talk, bullshit walks. And please, address the actual content of my argument above instead of invoking obscure stat mech or nonstandard axioms. Explain how a nonzero lapse rate does not enable second law violating, perpetual heat flow, if you would, right up to the point where real thermal equilibrium is achieved. Also remember, gravity is doing no net work while all of this is going on. If it were, things would actually be worse — you’d start violating the first law of thermodynamics as well.
rgb

DeWitt Payne
January 24, 2012 5:04 pm

Joe Born,
Please post a calculation of the Velasco lapse rate for a gas column with the surface at STP (101325Pa and 273.15K) and g = 9.81 m/s^2. The number density/cubic meter is ~6E23 molecules/mole/0.0224m^3/mole = 2.7E25.
One minor point. Since statistical mechanics is all about probabilities, it is possible to violate the Second Law. It’s just extremely unlikely. All the molecules of air in a room could migrate into one half of the room but the time for the probability of this event reaching 50% is really, really long. It’s many orders of magnitude longer than the age of the universe, depending on the size of the room and the air pressure in the room. I suspect Velasco’s calculation is something like that. For one molecule there’s a 50% chance it will be in one half of the room at any given time.

Q. Daniels
January 24, 2012 5:08 pm

Robert Brown,
I’ve read a number of thermodynamics texts myself.
Not once did I see a proof of the Second Law that did not rely upon circular logic. I’ve seen plenty of empirical proof, that people have been unable to violate it, but no direct proof.
Relying upon the Carnot Cycle is pretty clearly circular logic. It’s more difficult to show, but assuming that the MB distribution remains uniform under gravity may also be circular.
If you have a proof of the Second Law that is not based on circular logic, I’d be happy to read it.
Failing that, all we have is that there are no publicly recognized or understood violations of the Second Law. It is empirical, and nothing more.
Poor engineering is not proof of impossibility.

gbaikie
January 24, 2012 5:12 pm

” “In nature, the dry adiabatic lapse rate of air in the atmosphere is maintained because the system is differentially heated from below causing parcels of air to constantly move up and down.”
I live half-way up a mountain at 6100 feet. The valley below is at 4500 feet. The temperature difference is nearly always the dry lapse rate 8 degrees F, whether it is calm or windy. Only if it is raining or snowing will it be different. It then goes to the moist lapse rate. Most of the time it is sunny, heating the ground equally, both in my back yard and in the valley below. What maintains the lapse rate temperature difference?”
Mostly gravity.
If you lived planet with 1/2 the gravity, the the temperature difference would about 1/2.
You could say you would be warmer. You could also say lower elevation would be cooler there would less difference temperature.
Less difference in temperature is undeniable.
With a silver rod, one could transfer heat from lower elevation to your elevation. Or one could use other material which cheaper and conducts heat better than air. But silver is one of best conductor of heat and air is one the poorest conductors of heat.
One can generate energy from differences in heat. Venus isn’t particularly good place to generate energy- despite being very hot.
Venus doesn’t even “have a lot of energy”- in terms of total joules of heat in it atmosphere, assuming one find a lot something which is colder one couldn’t create an extraordinary amount of energy.
Over a period of time, say 24 hours, earth absorbs more of the sun’s energy than Venus does- despite being closer to sun and getting around 2700 watts per square meter compared to Earth’s 1300 watts per square meter.
This can quickly seen: Venus is hot, it’s temperature near surface does not vary over time, and since Venus is hot, it can not be made hotter by the sun. The earth more distance and receiving less sunlight, is much cooler. And because it is cooler it can heated by the sun and there is daily variation in temperature. And on planetary scale, earth has large variation in equator and polar temperature. So earth is churning heat engine, whereas Venus is more stagnant.
So in terms of power, in terms of horsepower, earth is a more powerful engine.
It seems to me that in subject of climatology, there is measurement of the engine temperature, and there should more attention to how much power the engine produces- and what’s earth miles per gallon.

January 24, 2012 5:21 pm

Robert G. Brown says

In nature, the dry adiabatic lapse rate of air in the atmosphere is maintained because the system is differentially heated from below causing parcels of air to constantly move up and down.

The dry adiabatic lapse rate determines how high thermals will go – usually, only a few kilometers. The actual lapse rate is normally significantly different from the DALR.
As this article argues, the lapse rate without IR emitters would be zero. It is the greenhouse gases that move the actual lapse rate (ELR) from zero to -6.5 K/km. The DALR is -9.8 K/km. To claim that anything “maintains” the DALR simply means that you have not looked at the data.
Related to the DALR is the concept of “potential temperature” – the temperature that a parcel of atmosphere would have if it moved adiabatically from some altitude to sea level. For instance, if the atmosphere at 10 km was -55C, its potential temperature is

10km * 9.8 C/km – 55C = 43C (109 F)

The other meaning of this is that for air to convect from sea level to an altitude of 10 km, it must be at least 43C. A more typical value in the sub-tropics would be -70C at 17km

17km * 9.8 C/km – 70C = 96.6C (206 F)

Since the surface temperature is typically a bit cooler than this, the bulk of the atmosphere does not follow the DALR. (Near the surface, the DALR is seen almost every day. But only near the surface.)
I suggest that everyone look at some real data.

Alan Millar
January 24, 2012 5:22 pm

“kuhnkat says:
January 24, 2012 at 2:33 pm
Alan Millar,
“The Sun will die and become a cold white dwarf with no solar wind to blow any atmosphere away.”
This is an assumption without much empiraical proof”
I hope your Mum isn’t reading that because even she would be embarrassed!!
Is this the level people will sink to just because they would like to have a new theory of Thermodynamics which will disprove the theory of CAGW?
Get a grip folks. These sort of statements are going to hold this well respected site up to ridicule.
Stick to facts and real physics and CAGW will be shown to be nonsense in any event.
It will take a bit longer but I will live to see it and I ain’t young! Have patience, trust in Gaia!
Alan

January 24, 2012 5:25 pm

@ Dr. Brown –
I’m curious, what if the gravity isn’t constant, but is fluctuating?

robr
January 24, 2012 5:28 pm

“Willis Eschenbach says:
Robert Brown says:
I have been seriously trying to follow this conversation. I have my Thermo text, my Heat Transfer text, my CRC handbook, the internet, and my calculator. For a dumb ass like me (tau beta phi), I wonder if one of you would answer give me your opinion to the following questions (2):
Does the atmospheric pressure, no GH gases, play a role in the dry adiabatic lapse rate?
Does the atmospheric pressure, no GH gases, effect the overall equilibrium temperature of the near surface of a planet?
Thank You,
Robert S Rider

robr
January 24, 2012 5:30 pm

I’m so dumb I can’t even spell Pi.

January 24, 2012 5:34 pm

says: These two idealised adiabatic processes (like the adiabatic stages in the Carnot Cycle) will result in the parcel returning to Earth with nearly the same temperature as leaving (the slight drop being accounted for by radiation at TOA).
Carnot cycles are ISENTROPIC. By definition. Isentropic means adiabatic AND reversible.

Hmm, maybe we read different textbooks. Any cyclic heat engine — being cyclic — returns to its original state at the end of a cycle. All cyclic heat engines — being heat engines — increase the entropy of the Universe while operating. In particular the Carnot cycle absorbs heat \Delta Q at temperature T_H for a change of entropy of the hot-side reservoir of -\Delta Q/T_H, and then rejects it into the cold-side reservoir for a change of entropy of the cold reservoir of +\Delta Q/T_C.
The total entropy change — per cycle — is thus:
\Delta S = \Delta Q (1/T_C – 1/T_H) > 0
To correct you — and please, bear in mind that I’ve taught this for 30 years — Carnot cycles are NOT isoentropic. Isoentropic DOES mean adiabatic (\Delta Q = 0) and reversible, but Carnot cycles contain two isothermal expansion/compressions (also reversible). The isothermal parts are not isoentropic, as explicitly shown above.
Now, what drives the circulation cycle you describe? How about absorbing heat at constant temperature from the hot ground (reservoir) and delivering it to the upper troposphere where the heat is lost to radiation at much colder temperatures. Hmm, sounds like something that increases the entropy of the Universe to me. The air that is rising and the air that is falling are not at the same temperature. They may both be isoentropic processes, but they don’t occur at the same place and the same time, and it is the thermal variations of density with temperature that ultimately provides the lift (or lack thereof) that drives the cycle, positive or negative net buoyancy.
With all that said (filling in details), I think we agree. So what is the point?
rgb

Joe Born
January 24, 2012 5:52 pm

Paul Birch:
“I have now read the Velasco et al article, and it agrees with what I said: in either the microcanonic (totally isolated) ensemble (with a reasonable number of particles in the gas) or the canonic ensemble (in thermal equilibrium with the surface or walls, irrespective of the number of particles), the gas is isothermal.”
Of course, what we’re talking about is the microcanonic ensemble, to which Equations 5-8 apply. If you read Velasco et al.’s Equation 8 for mean single-molecule kinetic energy K as a function of altitude z, you’ll see that the expression for K is the product of a constant and (1-mgz/E), where m is molecular mass, g is the acceleration of gravity, and E is total system energy. To me that looks as though K decreases with altitude z: the temperature decreases with altitude. Is there some different way you interpret that factor? As I read it, it says there will be a lapse rate that’s small for large numbers of molecules but stll finite and non-zero so long as the number of molecules is not infinite.
Presumably, you are basing your interpretation of Levasco et al. on its penultimate paragraph, in which they made an execrable attempt to state verbally what the equations express mathematically. Unfortunately, that passage is so abysmally opaque that any exegesis thereof matching the mathematical result is doomed to appear hopelessly strained. So I will forgo the attempt. The real question is, Does Equation 8 define an altitude-dependent temperature or not? If so, there’s a non-zero lapse rate at equilibrium.
If you can’t reach the answer by considering Equation 8 itself, consider the lead-up to it, where Velasco et al point out that the state density as a function of both velocity and altitude (Equation 5) is not the product of state density as a function of altitude alone (Equation 6) and state density of a function of velocity alone (Equation 7)–as it would be if temperature were independent of altitude. They also observed that the density distribution as a function of velocity is not, as one would expect of an isothermal configuration, the Maxwell-Boltzmann distribution.
That was the wind-up, and the pitch was Equation 8, which says that, indeed, the temperature is not isothermal.
Do you interpret those equations differently?

A physicist
January 24, 2012 6:25 pm

Mike McMillan says: Great theory. Baloney, but very entertaining.
The reason the silver thingy won’t generate perpetual motion is that the exposed ends will assume whatever the air temp is at that altitude. A temperature difference of 1 degree will not move any heat in a silver rod 100 meters long.

Mike, it was to address this specific objection that the Beach-House Block Story was conceived … this story shows that the length of the column is irrelevant.

Q. Daniels says: Robert Brown, I’ve read a number of thermodynamics texts myself. Not once did I see a proof of the Second Law that did not rely upon circular logic. I’ve seen plenty of empirical proof, that people have been unable to violate it, but no direct proof.

In particular, that page’s geometric theorem “the symplectic form ω is preserved by Hamiltonian flow” is equivalent to the dynamical principle “the state-space volume of a dynamical ensemble never decreases”, which in turn is equivalent to the thermodynamic principle “the entropy of a dynamical system never decreases”.
Whether or not this geometric reasoning conveys belief … at least it is not circular reasoning.

Billy
January 24, 2012 6:31 pm

What a fun thread! I should be working but I’m reading it instead. (That is okay, it is 8:22 PM and I’m self-employed, so it’s not like the boss doesn’t know what I am doing.)
My thanks to Dr. Brown and Willis for their patience. I would have given up long ago. I liked Dr. Brown’s simple example.
My next comment is going to offend many people so, if you are sensitive, quit reading. The discussion in this thread reminded me of many of the discussions on RealClimate. Only the role of the climate scientists was taken by those who opposed Dr. Brown’s explanation—they knew what the truth was and they could figure out some theory to disprove it.
This thread is probably a healthy process—although somewhat painful to watch. But, it does show that one need to think hard about these issues.
I feel that this discussion is a reasonable model of how science often advances. Someone throws out a good idea. Many dump on it with facile but incorrect criticisms. A few offer support. Finally, after a long time and much confusion there is reasonable agreement that the idea is right (or wrong, depending). We now believe that bacteria cause many ulcers but that N-rays don’t exist. But, when the two theories were offered, the bacteria/ulcer theory was dumped on but the N-ray theory was not.
My guess it that a few who participate in this thread will actually learn something, which is probably more than you can say for most students in freshman physics in college.
Billy

Trick
January 24, 2012 6:32 pm

Robert Brown says at 3:45pm:
“Tell me whether or not the system in figure 2 permits energy to flow in a circle forever.”
Robert Brown says in a top post verbatim quote, search on the text for context of his answer:
Yes. “Heat will flow in this system forever; it will never reach thermal equilibrium.”
Robert Brown at 3:45pm:
“If you answer “no, of course not” you are quite right.”
Now Robert Brown is going so fast he is not right with himself. S-l-o-w down again Robert, when you do, you are quite good. Maybe you really are struggling & working to line up with the past thermo masters to grok this stuff better like Joules Verne and Tallbloke handles.
Robert continues to struggle forward to equilibrium grokness by reading up on the subject:
“Thermal equilibrium does not equate the total energy. Read the equipartition theorem. Open a standard introductory physics textbook. Learn what temperature is. Then return.”
I have returned. Here is a quote from my standard introductory physics text book: “Equipartition gives the total average kinetic and potential energies for a system at a given temperature.”
Robert – You are advancing in your studies! This is good. You must now grok potential energy better. It is equipartitioned with kinetic energy at a given temperature. I have learned (long ago) that temperature is mean kinetic energy (reading your Caballero ref. was a cool refresher…).
So Robert is right here thermal equilibrium does not equate to the total energy. Thermal energy is equipartitioned with potential energy for the total energy. Good going Robert, I see advancement in this 3:45 post toward the thermo master’s laws.
Robert gave a little back with the two different answers to the same question but still see some progress.
Robert will be way better off moving to grokness equilibrium understanding the non-isothermal gas column upon reading Caballero section 2.3. Then return.

gbaikie
January 24, 2012 6:40 pm

Venus atmosphere is ~4.8 x 10^20 kg
Average temperature: 737 K (464 C)
http://nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html
Specific heat of CO2: [kJ/kgK]
600 K 1.075
650 K 1.102
700 K 1.126
750 K 1.148
800 K 1.168
http://www.engineeringtoolbox.com/carbon-dioxide-d_974.html
To lower one degree of K requires 1.148 times 4.8 x 10^20 kg
Earth ocean:
1.4 x 10^21 kg
278 K [5 C] 4.204 (kJ/kgK)
To lower or increase one degree of K requires 4.204 times 1.4 x 10^21 kg
To freeze or melt into ice requires 334 kJ/kg
To vaporize water 2,270 kJ/kg
http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html
So how many joules does it take heat a Venus from say 20 K to it’s present
temperature?
The chart starts at:
175 K 0.709
So roughly .7 times 4.8 x 10^20 kg times [175minus 20K] 155 equals KJ
And from 175 to 375 .8 times 4.8 x 10^20 kg times 200 equals KJ
375 K to 600 K: 1.0 times 4.8 x 10^20 kg times 225 equals KJ
600 to 737 K: 1.1 times 4.8 x 10^20 kg times 137 equals KJ And sums:
5.2 x 10^22
7.68 x 10^22
10.8 x 10^22
7.23 x 10^22
Which is 30.9 x 10^22 kJ
With earth melt 1.4 x 10^21 kg ice is 334 kJ/kg times 1.4 x 10^21 which is
46.7 x x 10^22 KJ
So to heat ice from 273 K to 274 K require more joules than the atmosphere of Venus
requires to heat from 20 K to 737 K.
To warm the ocean from 276 [3 C] to 13 C
Requires 10 times 4.2 times 1.4 x 10^21 which is 5.88 x 10^22
In terms of ten millions of year, the earth oceans have had 10 C increase or difference
in ocean temperature. I don’t think many consider that Venus has had such swing
in temperature: 7.23 x 10^22 is amount joules required to heat from 600 to 737 K.
So about 100 K [or 100 C] change in Venus temperature.
In terms of ice age to interglacial periods there is about 10 C difference in average global
temperature [not ocean change in temperature- but atmospheric]. So in terms couple tens of thousands, earth temperature changes, still roughly close to comparable temperature of somewhere around 100 C change in Venus.
So it requires real absorption of energy [joules] to change “average temperature”, locally daily one can easily have more than 10 C change in temperature. Such changes reflect dynamic, and powerful heat engine.

January 24, 2012 6:41 pm

So, we would expect the atmosphere at the surface of a planet to be warmer that it is at the top of the atmosphere. Of course, we can’t totally ignore conduction and radiation but, compared with convection, they are second order effects.
We might expect this if we didn’t understand convection. You’re getting things backwards, and amazingly, even though I derive the density of an ideal gas in thermal equilibrium above, you don’t even bother to read and understand it.
At a uniform temperature there is an exponential pressure and temperature gradient. If you warm the system at the bottom, the density of the fluid there decreases and there is a net buoyant force that lifts it up. Convection is cone-head complicated — I mean seriously complicated — in the general case. The equations that describe it are the Navier-Stokes equations, which are nonlinear partial differential equations so complex that mathematicians haven’t even been able to prove that a solution exists in the general case, let alone solve it. The idea is simple enough, and in simple geometries with e.g. uniform thermal gradients across the fluid one can predict some of the structure observed, e.g. convective rolls. But your base assumptions are completely false, and you have cause and effect completely reversed.
Undriven convection does not lead to warmer air on the bottom. Not ever. Heat a gas or liquid uniformly on top and you will have no vertical convection, because the density profile is stable. The stable undriven density profile is precisely what I derive, and is isothermal, with no convection. Only if you differentially heat a system on the bottom or on the sides (which is still part way to the bottom) do you get convection, and that only because the bottom is already warmer because of something else. Convection, in fact, cools the bottom!
A secondary comment is that radiation and convection make (from what I understand) roughly equal contributions to surface cooling. Either one can be dominant under different circumstances. During the day, convective cooling can be important — convection driven breezes can pick up a fair bit of heat from the surface. On a clear dry night, however, the ground temperature quickly inverts (becomes cooler than the bulk of the air immediately overhead) and radiation becomes far more important than convection.
The desert heats up to 45C or so during the day, but can still cool to 0C overnight. Not (primarily) from convection, not at all. By radiation.
Way back in boy scouts, winter camping in upstate New York, I learned about radiation and temperature. Cloudy nights are far warmer than clear nights. Clouds reflect a lot of heat back down and slow the cooling of the ground considerably. Water is a powerful contributor to the greenhouse effect. I also learned about a space blanket. A teensy thin layer of aluminized plastic, and yet it keeps you amazingly warm, because it reflects back your body’s radiant heat. You can feel it instantly if you put your hand into it. Your body loses roughly half of its heat from conduction/convection, and half from radiation. Air is a lousy conductor of heat, and often there is little or no wind. Radiation can easily be dominant, not just “second order”. I have no idea which one is dominant overall as far as surface cooling is concerned, but neither of them is negligible.
rgb

Myrrh
January 24, 2012 6:42 pm

Robert Brown says:
January 24, 2012 at 8:28 am
Show me my mistake. Anybody. I won’t be offended.
[kdk33 says:
January 24, 2012 at 7:22 am ]
No, I think you are generally quite right, and this agrees rather well with Caballero’s argument. Isentropic because it is dominated by convection, not conduction, in an open system heated at the bottom. Isolate the system, or heat it at the top and explain to me how the bottom will end up warmer than the top.
Yeah, right. Just like the oceans. I wonder why the argument fails for the oceans? They seem to come into thermal equilibrium at, well, thermal equilibrium (constant temperature, independent of pressure, density, “gravity” etc), below the convection-dominated thermocline.
========================
The thermocline exists because of these, heated surface waters warmer and therefore less dense will sit on top of colder, denser water.

Wiki on Thermocline: ” The warm layer is called the epilimnion and the cold layer is called the hypolimnion. Because the warm water is exposed to the sun during the day, a stable system exists, and very little mixing of warm water and cold water occurs, particularly in calm weather.
One result of this stability is that as the summer wears on, there is less and less oxygen below the thermocline, as the water below the thermocline never circulates to the surface, and organisms in the water deplete the available oxygen. As winter approaches, the temperature of the surface water will drop as nighttime cooling dominates heat transfer. A point is reached where the density of the cooling surface water becomes greater than the density of the deep water, and overturning begins as the dense surface water moves down under the influence of gravity. This process is aided by wind or any other process (currents for example) that agitates the water. This effect also occurs in Arctic and Antarctic waters, bringing water to the surface which, although low in oxygen, is higher in nutrients than the original surface water. This enriching of surface nutrients may produce blooms of phytoplankton, making these areas productive.
As the temperature continues to drop, the water on the surface may get cold enough to freeze and the lake/ocean begins to ice over. A new thermocline develops where the densest water (4 °C) sinks to the bottom, and the less dense water (water that is approaching the freezing point) rises to the top. Once this new stratification establishes itself, it lasts until the water warms enough for the ‘spring turnover,’ which occurs after the ice melts and the surface water temperature rises to 4 °C. During this transition, a thermal bar may develop.”

Because going through 4°C to lower temperatures water gets dense enough to sink and doesn’t freeze, freezing water floats, so not independent of density and pressure and as these are due to gravity, not independent of that either. Salt water has lower temperature before freezing, -1.9°C.
Some extra bits.
In this description of how oil is formed in the oceans, increased pressure means increased temperature:
“Other sediments continued to be deposited and further buried the oganic-rich sediment layer to depths of thousands of feet, compressing the layers into a rock that would become the source for oil. Over the years, as the depth of the burial increased, pressure increased, along with the temperature.”
“Water pressure at the deepest point in the ocean is more than 8 tons per square inch, the equivalent of one person trying to hold 50 jumbo jets.”
“Atlantic sea water is heavier than Pacific sea water due to its higher salt content.”
“The Antarctic ice sheet that forms and melts over the ocean each year is nearly twice the size of the United States”
“90% of all volcanic activity on Earth occurs in the ocean. The largest known concentration of active volcanoes (approximately 1,133) on the sea floor is located in the South Pacific”
“Under the enormous pressures of the deep ocean, sea water can reach very high temperatures without boiling. A water temperature of 400 degrees C has been measured at one hydrothermal vent.”
“The top ten feet of the ocean hold as much heat as our entire atmosphere”
The above from: http://www.savethesea.org/STS%20ocean_facts.htm
“90% of the total volume of ocean is found below the thermocline in the deep ocean. The deep ocean is not well mixed. The deep ocean is made up of horizontal layers of equal density.”
http://www.windows2universe.org/earth/Water/temp.html
And bearing in mind heat rises.., http://www.esr.org/outreach/glossary/insulation.html
“Ice is a great insulator. A lot of what causes climate and weather involves the exchange of heat and fresh water between the ocean and atmosphere. If the ice cover is high, very little heat escapes from the warm ocean to the cold polar atmosphere in winter. But the heat loss through open water is so high, maybe 10-100 times more than through ice, that even a small fraction of open water has a big effect on area-averaged heat loss. Typical heat loss values are ~10 W/m2 through thick sea ice, and ~1000 W/m2 in winter through open water (depending on wind speed, air temperature, etc.)”

Q. Daniels
January 24, 2012 6:43 pm

Robert Brown wrote:
I do not care about what generates the lapse rate. If the lapse rate is stable, so that heat delivered to the top redistributes to maintain a constant equilibrium temperature lapse between the top and the bottom — the sole case examined in the article above — then it violates the second law of thermodynamics.
Let me put it bluntly. If somebody presents a statistical mechanical computation that suggests that the second law is violated, I would knee jerk assume that the authors had made a terrible mistake unless and until proven otherwise, especially if I “could not understand” everything that they did.
Even then I would be doubtful.. To be honest, I would be doubtful if I did the work myself. I think that the paper you link has the right idea, and you will note that on other threads I propose precisely the same experiment. Show me, in other words. I’m a theorist, but I’m no fool. Experiments trump theory every time, …

I agree with this in detail, and find it a completely reasonable attitude. I will note that some people have difficulty understanding particular pieces of math, independent of their complexity. The Lorenz Transform is one such.
Graeff’s work is probably not sufficient proof for you. He believes he has successfully measured said temperature differential. http://www.firstgravitymachine.com
I also don’t think that a PMM2 machine violates TANSTAAFL. For one thing, the energy must come from somewhere, even if it’s just being exhausted as waste. For another, such a device would require an insight and a good deal of skill to build and use, even if it was as simple as rain. Doing takes effort, even if you’re just recycling waste.

glen martin
January 24, 2012 6:50 pm

The essence of this Robert Brown’s ‘refutation’ is that if the wire and the gas do not have the same thermal gradient i.e. if the wire is isothermal and the gas is not then, there is a violation of the laws of thermodynamics.
The problem with the refutation is that the wire is not isothermal for essentially the same reason as the gas is not: the atoms at the top of the wire will move slower than those at the bottom due to gravity and will therefore be at a lower temperature.
The atoms of the wire will lose velocity as they rise in the gravitational field just as those in the gas, thus there is less energy available transferred in interactions this will produce an gradient in the kinetic energy of atoms that make up the wire resulting in a temerature gradient due to gravity. That the distance covered between interactions is much smaller in the solid than it would be in the gas and that there are other interactions in a solid does not change this fact.

Editor
January 24, 2012 6:50 pm

glen martin says:
January 24, 2012 at 2:07 pm

“Willis Eschenbach says:
January 24, 2012 at 12:23 pm

Folks, a lot of you here don’t seem to get it. The beauty of Robert’s proof is that there is only one question in it—does heat flow forever in the silver wire or not?

IF there is a temperature difference in the air top to bottom, heat will flow in the silver wire. Gravity can’t stop that.”

Actually it can and does, heat in the wire is being transmitted via the interaction of moving particles, gravity will cause the particles to slow slightly as its height increases thus slightly less energy is will be transferred to the atom above a particular atom than was received from the atom below it. This results in a gravitationally induced thermal gradient in the wire.

Phew … gravity slowing electrons in a wire … thought I’d heard everything.
In any case, even if your interesting theory about gravity slowing heat transfer in wires were correct, it doesn’t matter. That just slows down the transmission of heat, it doesn’t stop it. So it doesn’t mater for the disproof.
w.

MDR
January 24, 2012 6:52 pm

Brown
OK, after pondering some more, I have identified a flaw in my reasoning. I think you and Willis and the others are correct – the column of gas does indeed eventually relax to an isothermal state if the column is in fact thermally isolated. It is still stratified, of course, but it would end up isothermal. Apologies for confusing everyone.
Here is where I was going wrong. I mentioned that gas near the bottom of the column has a smaller amount of potential energy than gas the top. While this is undoubtedly true, that potential energy only comes into play if the gas is being mixed or is otherwise dynamic, that is, if such energy is being released as a result of changing the height of some of the gas parcels. But of course the equilibrium state does not have any such mass motions, and so no work is being done on the fluid. As a result, the internal energy of the gas is the same everywhere, and thus the gas has the same temperature everywhere.
Of course, for a column of gas subject to heat exchanges at the top and at the bottom [such as a column of gas in a planetary atmosphere] but is otherwise thermally isolated, there will be a temperature gradient established in accordance with those boundary conditions. But that scenario is apparently off-topic in this thread.

Trick
January 24, 2012 6:55 pm

Robert Brown says at 4:10pm:
“I am specifically proving that EEJ, a specific paper written by Jelbring and published in a journal (God help the referees, absent that day on vacation or something), violates the zeroth but especially the second law of thermodynamics when it asserts that there will be a thermal lapse rate in an adiabatically isolated column of ideal gas in thermal equilibrium in a gravitational field.”
Robert Brown needs to read Caballero sec. 2.3 that proves the zeroth is not violated & the 2nd is not violated for “adiabatically isolated column of ideal gas in thermal equilibrium in a gravitational field” which Caballero proves is non-isothermal & there will be a thermal lapse rate.
I imagine Caballero is just like every other thermo text book (but I have not read them all like Robert) & the thermo grand masters assert. Thus Jelbring EEJ is not in violation & not refuted if Jelbring asserts same as Caballero and Caballero is right.

Khwarizmi
January 24, 2012 6:55 pm

Q. Daniels says:
Willis wrote: Will heat flow in the silver wire forever?
If you extract energy from the system, it will shut down as the entire system cools. Energy is conserved. If you extract energy, then it has to come from somewhere, and that somewhere is the thermal energy of the system.
If you do not extract energy, then yes, it will.
=======
LongCat says:
While I agree with the underlying point, I’m not sure why the wire would necessarily violate the laws of thermodynamics if it continuously transferred heat. Under normal circumstances, it would radiate some of this energy away and otherwise be an imperfect conductor. If, however, we’re assuming a closed system with a perfect conductor surrounded by a perfect insulator, why would any energy be lost?
To put another way, assume I have a wheel with a frictionless axle at rest in a vacuum. If I spin it, it will spin endlessly. The conclusion that it will have perpetual motion doesn’t violate thermodynamics; the assumption that there is no friction does. Likewise, the wire would not violate any physical laws by endlessly transferring heat; those laws were broken by the assumption of a closed system with a perfect conductor/insulator.
I know I’m disputing people far above my pay-grade, so I’m assuming that I’m wrong in this. I’m just curious as to why.
==============
Trick
Heat flows forever in fig. 2. The reason is a grand master thermal law is broken, the easiest one, the zeroth law. There can be no perfect insulator.
==============
Correct.
All real heat engines increase entropy. Engines do work. Figure 2 doesn’t.

A physicist
January 24, 2012 7:03 pm

Robert Brown says: But show me the “high precision” experimental result, done with a dewar in a centrifuge filled with maybe Xenon gas at a G value such that there is sufficient pressure at the top of the vessel to justify the thermodynamic assumptions, with recording high-precision, carefully calibrated thermometers.

Robert, thank you for making this excellent point.
Precisely the situation you outline is present in all of the world’s high-speed centrifuges containing uranium hexafluoride (UF6) for isotope separation.
If an adiabatic lapse were present in the centrifuges, between the high-pressure rim and the low-pressure central axis, then solid UF6 would condense at the cold central axis … which needless to say, is not observed.
Elevator Summary: Gas centrifuges prove that gravito-thermal theory is wrong.

Editor
January 24, 2012 7:03 pm

kuhnkat says:
January 24, 2012 at 2:32 pm

Willis Eschenbach wrote:

Just don’t expect your belief, that gravity can do continuous unending work forever and ever amen, to be widely shared in the scientific community …

So Willis, when will we start flying off the planet??? When will the pressure on my feet from standing in one place stop?? When will the oceans boil from lack of pressure? Oh yeah, gravity is apparently an unending source of energy that counteracts centrifugal force. If not unending, we haven’t yet measured its reduction.

kuhnkat, you are conflating a constant force with unending work. Gravity is just there all the time, a force pulling in one direction, keeping you on the planet. When you move upwards against gravity, it takes energy to do that. When you move with it you get the energy back.
But when you come back to where you started, it’s a zero sum game (less with friction). No matter how many times you go up and down the hill, you don’t gain any energy at all, despite the constant presence of gravity. Which is another way of saying that in all of those trips up and down the hill, the lazy bum gravity hasn’t done a net lick of work. It did do work, but what gave with one hand, it took with the other by requiring the exact same amount of work in the other direction.
It’s like running a waterwheel by continuously filling the headrace up with buckets of water from below the wheel. Sure, you can get work out of the wheel … but that’s work that you are putting in by continuously lifting the water, not work that’s coming from gravity. Stop lifting the water and see what your friend gravity does for you … nothing.
Hope that helps.
w.

Tim Folkerts
January 24, 2012 7:04 pm

Wayne says: “Seems lifting the electrons against gravity in the metal bar from the warm to the cool would cancel if the gradient became -0.0098C/m. ”
Very insightful.
IF the gradient for air and electrons and all other materials were the same 0.0098C/m, THEN maintaining a lapse rate would not lead to a perpetual motion machine.
However, since the lapse rate is given by C_p / g, and since different gases have different values for C_p, then different gases will have different lapse rates. So the premise is indeed wrong, and a stable lapse rate in gases would indeed violate the laws of thermodynamics.

January 24, 2012 7:04 pm

Clearly there are two schools of thought. One school believes that the temperature will be lower at the top due to kinetic energy being changed to gravitational potential energy. The other school believes this will not happen. The GHG controversy rests largely on this point.
No, it doesn’t. There isn’t any controversy. Gravity is incapable of providing net heat to the Earth’s energy budget, and the GHE only deals with the rate at which the Earth loses heat in that budget. Nobody argues that there is a lapse rate in the actual atmosphere. I have just definitively proven above that it is not a feature of static equilibrium, it is a dynamic phenomena caused by differential and irregular time dependent heating and cooling, where the bulk of the heating is at the surface, but where heat loss occurs to some extent very high up in the atmosphere as well.
This has nothing to do with “schools of thought”.
I am troubled on one point. The argument that a continuous flow in a cycle is not equilibrium and thus is some sort of proof favoring one school over the other. Surely dynamic systems can be in “equilibrium” in that there is no net flow into or out of the system, but still allow a cyclical flow within the system.
Surely they cannot, not as long as they are thermally connected at the microscopic scale to allow internal energy transfer within the system. This is precisely the point. Heat never “flows in cycles” in a system in thermal equilibrium. It flows from hot to cold. It never ever spontaneously flows from cold to hot as an steady state thermal process unless one does work on the system. This is what the second law of thermodynamics is all about. The only systems that can “move”, transporting energy around in cycles are ones without mechanisms for energy sharing or dissipation, like planets going around the sun in the limit that you ignore tidal heating and gravity waves and light pressure. In enough time, even those weak effects move energy around and damps periodic motions.
In the case of the gas in the figures above, one could take the silver wire, replace the middle of it with a thermoelectric junction, use the electricity to drive a fan or light a bulb. This, too, would violate the second law — the heat content of the container would systematically lower (as some of the heat in the cycle was converted to work). The net effect would be that all of the energy lost from the container would be converted to work — its temperature would drop as the work appeared in the outside world, quite independent of the temperature out there. This, too, is a direct, textbook case of the violation of the second law of thermodynamics, both the refrigerator statement and the heat engine statement.
At least you are troubled by the right things — you should be troubled by this because your mere common sense tells you that heat, which is basically random motion, cannot remain organized enough to flow around in a circle without something reorganizing it. The microscopic form of the second law says that basically, systems evolve in time from less probable states to more probable states. Take a jar full of identical marbles, some blue and some red, organized with the blue on top and red on the bottom and shake it. You can shake a long, long time before you can expect to see all of the blue on the bottom and the red on top — there are a near-infinity of ways for the marbles to be mixed; only one for them to be cleanly separated red on blue.
That’s the sad thing about this — people don’t understand how much of a law the second law is. Your odds of winning 150 million dollars in the lottery and having the IRS forget to charge you taxes on it due to a clerical error are a gazillion times better than the odds of shaking that jar and getting even a very modest number of mixed-color marbles sorted out by random chance, and that is precisely the reason that heat flows from hot to cold and not the other way around. There are way more states where the energy (per degree of freedom in the system) is approximately equally shared than there are states where it is split up.
Otherwise, your observations about ability to rationalize and so on are well made, but — my advice is don’t bet against the second law. You’ll just lose.
rgb

Bill Illis
January 24, 2012 7:07 pm

For 13.7 billion years, …
… Gravity fields have been heating up matter.
That is long enough to be called perpetual.
Just look out at your night sky and see the proof of that. Or maybe even the day-time. The big bright white thing in the sky started shining because of gravity fields heating up matter.
We can extract energy from that we figure out it works exactly. Just like the other ten sources of energy that we didn’t have clue about in times past.
Anyone disputing Jelbrings hypothesis needs to prove that gravity does not provide a minimum heat/energy level in matter when that matter is being held back from falling further through the gravity field by the electro-magnetic and strong forces of the atoms in the rocks at the surface of the planet. The matter is still being pulled through the gravity field, it is just being stopped by the other forces in the atoms of the rocks/liquid.
Has anyone proved that? How could you? We don’t even know how the force of gravity extends its pull. Maybe Higgs bosons increasingly accumulate/stick to matter as it moves into a gravity field and they are attracted to each other. Maybe the Higgs provides heat energy to the matter. If you don’t how it works, you cannot say it has no impact.

Editor
January 24, 2012 7:07 pm

George Turner says:
January 24, 2012 at 2:40 pm

Actually Willis, heat flowing through the silver wire forever doesn’t mean it’s impossible, as heat always flows forever in any system above absolute zero. Take any object and an arbitrary plane that defines it. The two parts will never be in exact thermal equilibrium because atomic collisions are discrete, so half the time one side is hotter and half the time it is colder. Thus heat flows back and forth across the boundary – forever. That doesn’t mean the existance of an object above absolute zero is impossible.

George, if you do not know from the context that we are talking about net heat flow through the wire, and not freakin’ brownian movement of electrons, you are not paying enough attention.
w.

January 24, 2012 7:10 pm

If the column of air is not isothermal, that is an emergent phenomenon in the presence of several things:
– A planet
– An active star
– An atmosphere
By this excessive logic, a solar panel can’t possibly be a perpetual motion machine either, so there is no such thing as renewable energy. Especially after the sun fizzles out…

Trick
January 24, 2012 7:11 pm

Robert Brown says at 4:36pm:
“The distribution of v at the top (of the atmosphere) and the bottom is identical — the Maxwell-Boltzmann distribution.”
Maxwell-Boltzmann is not applicable in a gravity or electrostatic field. M-B quite clearly limit their derivation of particle velocity to special case of particles with no external forces like no gravity, no electrostatic etc. M-B applies to our famous gas column when it actually is isothermal – in the no gravity case. Caballero 2.2 shows that is the fact and in my view Caballero is correct.

Joe Born
January 24, 2012 7:20 pm

Dewitt Payne: “Please post a calculation of the Velasco lapse rate for a gas column with the surface at STP (101325Pa and 273.15K) and g = 9.81 m/s^2. The number density/cubic meter is ~6E23 molecules/mole/0.0224m^3/mole = 2.7E25.”
I’m a little pressed for time this evening, so I’ll just give you what I have handy, without using your particular values or double-checking. Note that this assumes a monatomic gas. I’ll revise it tomorrow for diatomic (after I read up on rotational degrees of freedom; I’m a layman). As you can see, the lapse rate would be hard to measure experimentally.
f = 3 ; # degrees of freedom
E = 2.2e9; # guesstimate of total energy in a meter-square gas column arbitrarily high
VRW_LapseRate = function(f, E){
k = 1.38e-23; # Boltzmann’s constant
N_0 = 6.023e23; # Avogadro’s number
w_m = 29; # “molecular weight”
m = w_m / N_0 /1000; # molecular mass
g = 9.8; # acceleration of gravity
P_0 = 1.01e5; # atmospheric pressure at sea level
M = P_0 / g; # atmospheric mass per unit earth-surface area
N_m = 1000 * M / w_m; # moles of atmosphere per unit earth-surface area
N = N_m * N_0; # number of molecules per unit earth-surface area
– (2/3) * f * E / (f * N + 2 * N – 2) * m * g / E / k;
}
VRW_LapseRate = function(f, E)
[1] -6.389722e-32
The last line of the function is the result of differentiating Velasco et al’s Equation 8 and converting from kinetic energy to temperature.
As I said, I haven’t double-checked, so there’s likely an error here. You may want to check it yourself.

Editor
January 24, 2012 7:20 pm

dlb says:
January 24, 2012 at 3:42 pm

Willis at 11.54am wrote:

Excellent insight, Wayne. That is exactly what happens. In an isothermal column of air, individual molecules at high altitude have more energy because of gravity. But for exactly that same reason, there are fewer molecules at high altitude. As a result, and as we would expect, in the isothermal condition the energy is spread out evenly through space (equal energy per volume) rather than equal energy per molecule as Hans Jelbring and Mr. Verne assert.

Although I agree with Dr Brown, I disagree with what Willis has said here. Consider a cubic metre of soil and a cubic metre of air above it, although both are at the same temperature, they certainly have different amounts of enegy due to differing densities.

Um … er … well … I must confess, I’m picking my jaw up off the floor.
Here’s the first thing. Nobody is talking about a cubic metre of soil here but you. Why?
Because it has nothing to do with the energy distribution of an ideal gas in a cylinder connected with a piece of silver wire.
I can only shake my head in amazement, dlb. Perhaps if you read the head post again?
w.

Editor
January 24, 2012 7:24 pm

A physicist says:
January 24, 2012 at 3:58 pm

Willis Eschenbach says:

Wait, wait, you claim to be a physicist, answer the question. Does heat flow forever in the silver wire or not?

Willis, the short answer is “Yes”

Dang. Well, can’t say I’m really surprised. Turn in your PhD at the door, or demonstrate it will flow forever and ever amen, and win the Nobel Prize.
w.

January 24, 2012 7:29 pm

P = T*V helps to understand what’s going on. One must constantly keep in mind that in the gravitationally bound column of gas pressure is constant while temperature and volume are the variables. As its temperature goes up and down its volume goes up and down. Surface pressure is determined by gravitational constant and mass of the gas which do not vary. Temperature is not coupled to pressure therefore pressure is not coupled to temperature. So raising the surface pressure will not cause a rise in equilibrium temperature. It will cause a rise in volume and the gas law wil be satisfied by the change in volume.
OK, Joules, you’re scaring me. P is absolutely, categorically never equal to V*T. Have you ever heard of “units”? You might look them up some time. Nor is pressure proportional to T*V in an ideal gas (or any gas I can think of). PV = NkT, so P = NkT/V. Nor is pressure constant in a gravitationally bound column of ideal gas. Don’t be absurd — pressure is never constant in any vertical fluid column in a gravitational field. The equation for static force equilibrium is:
$dP/dz = - \rho g$
Density is (in any situation where this might apply) a strictly positive number, so the pressure must vary with height. I actually derive its isothermal variation with height at the top of this thread. Surface pressures on the Earth vary all of the time, by a few percent. Pressure variations help cause “weather” — that’s why God invented “barometers”, because falling air pressure often warns of a storm, while high pressure usually indicates a fair, sunny day.
Temperature isn’t necessarily “coupled to pressure” — one can certainly have different contains of fluids at any pressure and any temperature — but that doesn’t mean that in various thermodynamic systems:
$\partial P/\partial T = 0$
as a general rule. If you think that, take a soda bottle, screw the lid on good and tight, and put it into some boiling water. Hell, I have a problem just taking those large plastic refillable water bottles back to the store — if you put them into a car with the lids tight you’ll get there with them (often irreversibly) blown up like a balloon. Works the other way too — screw the lid down when it is hot and come back to them partially collapsed.
As for “raising the surface pressure” causing changes in temperature — well, if the surface in question is the cylinder of a gas piston that is being compressed, I beg to differ, especially if it is done rapidly.
Here’s a cute toy that I like to teach my students about:
http://www.practicalsurvivor.com/firepiston
http://www.phy.duke.edu/~rgb/Class/review_53/review_53/node64.html
Lost the figure that went with the problem, sorry, but the pictures in the first one should give you the idea. The point is that whether or not increasing pressure changes the temperature depends on where and how you do it. It certainly can raise the temperature — it depends on the path followed on the P-V curve. Only if you follow an isothermal path does it not change the temperature, but isothermal paths are just one of a myriad of possible paths between a myriad of possible pairs of state points.
I’m also waiting to hear you acknowledge that a static lapse rate in an ideal gas at thermal equilibrium violates the second law of thermodynamics as per the example given above. Since you actually tried to make fun of the textbook physics I’m presenting, it might be appropriate as it sounds like you might be coming to grips with the truth of it.
rgb

Bill Illis
January 24, 2012 7:38 pm

The wire suffers from the same loss in gravitational potential energy as it goes higher in the column.

Joseph Murphy
January 24, 2012 7:42 pm

Dear Lord this thread is entertaining. I moved some joules reading through it. Thank you to everyone.

January 24, 2012 7:44 pm

Q. Daniels said @ January 24, 2012 at 5:08 pm

Robert Brown,
I’ve read a number of thermodynamics texts myself.
Not once did I see a proof of the Second Law that did not rely upon circular logic. I’ve seen plenty of empirical proof, that people have been unable to violate it, but no direct proof.
Relying upon the Carnot Cycle is pretty clearly circular logic. It’s more difficult to show, but assuming that the MB distribution remains uniform under gravity may also be circular.
If you have a proof of the Second Law that is not based on circular logic, I’d be happy to read it.
Failing that, all we have is that there are no publicly recognized or understood violations of the Second Law. It is empirical, and nothing more.
Poor engineering is not proof of impossibility.

I have commenced an extremely empirical experiment what is designed as a empirical disproof of the Second Law of Thermodynamics. This very morning, I mixed 400 ml sterile H2O with 400 ml of the best Italian balsamic vinegar in a bottle and sealed it. Having sealed it with much twisting of the Stelvin seal, I shook it vigorously. Seven times seven times did I shake the Stelvin sealed bottle. When the balsamic vinegar spontaneously separates out from the H2O, verily I shall know that I have violated (desecrated even) the Second Law of Thermodynamics and fully expect to be deported to the US of A for having engaged in a successful conspiracy theory contrary to US law.