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:

image

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 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.

The Failure of Equilibrium

image

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:

image

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.

image

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:

image

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:

image

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:

image

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

image

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

image

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

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

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!

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.

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

Alan Robertson

“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

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

“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

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

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

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

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

“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

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

Wayne2

@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

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

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

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

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

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

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

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

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?

Genghis

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

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

“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.

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

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

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 “

jhborn

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

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.

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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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.

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

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

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.

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)

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

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.

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

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

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.

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.

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

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.

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

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?

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

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

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

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.