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This page http://mit.edu/16.unified/www/FALL/thermodynamics/notes/node133.html explains how reflected radiation is broken into two components …
(1) The specular (mirror-like) reflection where angle of incidence = angle of reflection.
(2) The “diffuse” reflection which is deflected at any random angle..
Here the second component is what I prefer to call deflected (or scattered) radiation, as the process is very different from specular reflection. But whatever you call it, it obviously does exist.
This is the process I have been talking about all along. As far as energy is concerned, as I have always said, it is the same as reflection and thus has no effect whatsoever on the temperature of the target, in this case the surface.
This diffuse reflection is what happens when the target is warmer than the source. If such radiation were absorbed and converted to thermal energy there would be a violation of the Second Law of Thermodynamics. That is why you need to know the temperatures of both source and target before you can know the overall absorptivity and emissivity. These factors will be affected by this diffuse reflection, which cuts in when the target starts to get warmer than the source.
Thus, all radiation from a cooler atmosphere undergoes diffuse reflection when it strikes a warmer surface. This is why an atmospheric radiative greenhouse effect is a physical impossibility.
(continued from my last post)
I am of course aware that “diffuse” reflection is usually applied to reflection of light which appears to be diffused because of a rough surface. On a microscopic scale, the rough surface may well comprise many small smooth surfaces which simply produce specular reflection at various angles.
So, strictly speaking, the “rejection” of low frequency radiation (from a cooler source) which meets a warmer surface and then resonates and is scattered is yet another different process. But it helps to think of it as diffuse reflection because the end result is the same. My main point is that it can have no effect on the surface temperature because it is as good as reflected.
When IR from the atmosphere strikes a rough Earth surface, you would not be able to distinguish between this scattered radiation and diffuse reflection. In general you will measure inflated values of emissivity as a result. I would suggest that true absorptivity should be expressed as a function of both source and target temperatures. Then its measure should reflect the proportion of the radiation from a source at that temperature which is actually absorbed and converted to thermal energy. In the case of a target temperature greater than a source temperature the absorptivity would thus always be zero.
So this is where models go wrong because they use mean absorptivity measurements which disregard the temperatures of source and target and probably include a lot of scattered radiation anyway. Thus the models end up assuming thermal energy is transferred to the warmer atmosphere simply because they do not rate absorptivity for the relevant temperatures as being zero. Thus they assume violation of the Second Law of Thermodynamics and are of course wrong as a result. There can be no radiative Greenhouse effect.
(continued)
In calm conditions on a clear day we have the Sun (and only the Sun) warming the surface. Its high frequency (high energy) radiation right across its spectrum can be converted to thermal energy (if not already reflected) and this energy flows by conduction deeper into the land surfaces. The radiation itself already penetrates a fair way into the oceans before it is converted to thermal energy in the depths. Meanwhile thermal energy is also escaping the surface more or less as fast as conduction in the land and convection in the oceans will allow it to get back to the surface.
In the morning the rate of warming exceeds the rate of cooling, and vice versa in the evening. In summer the longer hours of daylight may trap some energy that cannot escape before the next morning. This trapped energy may build up as the middle of summer approaches, but escape by the next winter as daylight hours reduce and there is more time for cooling at night.
Now, looking at the cooling process, at least half (maybe 70%) of the thermal energy escapes to the atmosphere by evaporation, chemical processes and diffusion, which involves molecular collisions between the surface and the adjoining air, as in conduction in solids. The remaining energy will be radiated.
However, experiments in such conditions show that the lower atmosphere is always cooler than the surface, and cools faster than the surface at night. Radiation can never transfer heat from a cooler source to a warmer target and neither can diffusion. So these processes can never make the atmosphere warmer than the surface. The Sun also usually warms the surface faster than the lower atmosphere, so it is only very unusual weather conditions which might leave the surface cooler than the adjoining air. The air which is warmed in the morning will rise by convection.
The radiation from the surface may escape to space, but most will be absorbed by some molecules in the atmosphere. These molecules are likely to be warmed and may share some of the thermal energy with other molecules, or simply radiate it in all directions in small bursts.
The radiation which is emitted by the cooler atmosphere will have frequencies which are generally lower than the original radiation from the surface. If some of this radiation gets to the warmer surface it cannot be converted to thermal energy. Instead it is simply scattered by the surface without leaving any energy behind. Its energy cannot be converted back to thermal energy until it collides with something cooler than the original layer of the atmosphere from which it was emitted. Such cooler air will usually be higher up. It may even escape to space and only warm some cooler object in space maybe years later.
So each time any radiation goes back to the surface it will have absolutely no effect on the surface temperature, but will instead make it further towards space on the next trip up, if indeed it doesn’t escape altogether. Clearly there can be no Greenhouse effect.
Footnore: In situations when the relative humidity is high, the moist adiabatic lapse rate is lower than the dry one, so such humidity (as well as clouds) can slow the rate of cooling of the atmosphere, but this can never lead to any thermal energy going back into the surface, so the rate of cooling of the surface need not be slower. In a sense, thermal energy is falling over a smaller temperature step, but it still falls over at the same rate. The air we stand in may well feel warmer partly because there is less evaporation off our skin. In any event, these are just weather conditions which average out and do not relate to or affect climate.
From Doug Cotton on February 20, 2012 at 10:10 pm:
I’m thinking I live in the real world, your experiment is outdoors and there’s something called wind which can move the warm air sideways and under your cover.
Putting the wind issue aside, I can see an additional flaw. The outgoing longwave infrared originates from the ground, as it gives off its retained thermal energy. You said of those windshield sunscreens: “These have low emissivity and will generally disperse most upward radiation from the surface because of their rough surface.” But the dispersal pattern will be downward. Near the center of the cover, where your sand-filled Thermos resides, most to nearly all of the longwave will be returned to the same general area.
So you are attempting to show that back radiation cannot have a warming effect, actually a slowing of the net cooling rate, by blocking the returning longwave, with a setup that is preventing much of the longwave from escaping in the first place thus slowing down the net cooling rate. Thus your cover is producing about the same effect as the back radiation should have, and you are not noticing any great temperature difference between each Thermos. Which would be expected, and is not disproving the “warming” effect of back radiation at all.
You were talking about returning a portion of the emitted thermal radiation to the source by using a mirror. I was talking about returning the emitted thermal radiation to the source by using a full and a partial thermally reflective enclosure. If you wish to bring convection into this, the half of a box would allow it, and I would still expect the radiator to reach maximum temperature sooner than without it.
And why the specification of a mirror? Common mirrors have the reflective coating applied to the back of the glass, which would overwhelmingly be common window glass (float glass). Thus the spectral qualities of the glass are the first thing to consider. Window glass is opaque to longwave infrared with a transmittance of only 0.02 (ref) for 4mm (5/32″) thickness, and a thermal emissivity given as 0.95 (ref) with thickness not specified.
So with your little radiator and mirror experiment, the glass of the mirror will be absorbing virtually all the longwave and emitting it in a diffuse pattern. Thus your proposed experiment would not work as advertised and cannot be used to disprove what you have stated it would disprove. The correct experiment would use a true thermally reflective material with a low thermal emissivity such as aluminum foil which is specified as 0.03 to 0.05.
What would be the point? You’re not discussing, virtually every reply sums up to “You’re ignorant of the physics, the Second Law forbids it.”
And a total of four reply posts later, you still have not deigned to respond to my vacuum chamber experiment.
Regarding the backyard experiment, I measured the air temperature and it was always cooler than the surface, and cooling faster. So it could not transfer thermal energy by diffusion to the warmer sand. If it were transferring any by radiation, so too would similar air above the unshielded sand. But the unshielded sand still had far more radiation from the rest of the atmosphere within view.
The shield was significantly convex from below which would tend to reflect more of the upward raiation away from the samd, though I accept that some would have been reflected onto the sand. However, considering the area of the local ground under the shield compared with the whole area of the atmosphere above, I suggest the radiative flux from that small part of the ground would be negliglible in comparison.
I accept your point about the glass in the mirror. So go and try it with perhaps a shiny sheet of aluminium. It still won’t have any effect on the rate at which the radiator warms.
Regarding your vacuum chamber experiment, yes it does sound right to me. You can have any amount of radiation reflecting around containing any amount of radiative flux without that energy being converted to thermal energy. Any vacuum flask with hot coffee and internal reflective walls proves my point because the coffee does not get any hotter.
Now, maybe you should consider my two points:
(1) Take a parabolic car headlamp and remove the globe and front glass. Suspend a very small piece of metal at the focal point where the globe would have been. Cut a flat circular piece of the same metal material to fit where the front glass was. Ensure everything is at the same temperature (say 25 deg.C) in an air conditioned room. Then fit the circular piece of metal so it covers the hole where the glass was. The radiation from that sheet should be focused mostly towards the small piece of metal at the focal point. Clearly there is a greater flux of radiation from the larger piece of metal than from the small piece, so there is a net radiative flux towards the small piece. Will it get warmer? I say no because that would violate the Second Law of Thermodynamics. So what happens to the radiation which hits it? I say it must be scattered and not converted to thermal energy. Now, cool the circular plate by setting the air conditioner to, say, 22 deg.C. Heat will transfer from the small warmer piece to the larger metal one, but net radiation will still be from the much larger piece towards the smaller one. So net radiative flow does not necessarily have the same direction as heat flow.
(2) Carbon dioxide absorbs IR radiation coming from the Sun and there will thus be some backradiation to space. As about half the Sun’s radiation is in the IR spectrum, this causes a quite significant cooling effect.
PS I will re-do my backyard experiment with a sheet of glass with the windscreen shields on top of it just to help overcome objections such as yours. The glass should absorb upward radiation from the ground. I will slope the glass just a little to allow escape of air by convection. When are you going to try it yourself? Seeing is believing.
Guys, you can stop worrying about endless warming going up forever with carbon dioxide levels. It can’t happen.
The whole Earth system (including atmosphere) has to emit very close to the flux it receives from the Sun. So there will be some temperature – let’s say 255K – which is a mean and is somewhere up in the atmosphere.
The natural adiabatic lapse rate is determined, not by carbon dioxide, but by the acceleration due to gravity, the mass of the atmosphere and, to some extent, relative humidity which mostly averages out. The lapse rate sets the gradient of the atmospheric temperature plot which has to swivel about the 255K mean, so the surface end is warmer and the TOA colder. The drop in temperatures between the surface and the tropopause has been very close to constant in all the years of records shown on the NASA site since the end of 2002.
My point is that the mean surface temperature is dictated by these two values – the 255K (or whatever the exact figure is) and the lapse rate. Carbon dioxide cannot affect either, so neither are under mankind’s control.
Both the surface and the atmosphere will simply shed energy faster if they get a little warmer, thus tending back to the mean.
Don’t try to tell me there is a long term TOA net radiative flux difference. The net radiative flux varies between about 99.5% and 100.5% of incoming radiation. This is just random noise or short-term cycles. Longer natural cycles may have to do with variations in the effective power from solar radiation (affecting that 255K figure) and maybe the thermal energy generated under the surface. Small variations in the latter over many thousands of revolutions of the Earth could have a cumulative effect. The very fact that the terrestrial heat flow is low means that the massive quantity of thermal energy from the surface down to the core stays fairly much the same and brings about a stabilising effect as I have explained on the ‘Explanation’ page of my website. . This also is an additional comfort, so relax!
I just wrote this reply to a comment on another website, and I feel it may help people here …he wrote:
The obvious effect of this is that the radiative cooling of the hotter body will be slowed down by the presence of the cooler body, because some of the energy is being returned. This is completely different than there being a NET transfer of heat from the cooler body to the hotter body, which I agree is not possible.
No it’s not completely different. While the energy is still in the radiation it is not equivalent to thermal energy because it has not yet been converted. It cannot affect the temperature of the target unless and until it is converted to thermal energy. Only thermal energy can be added to other thermal energy with a resulting temperature change. You can only slow a rate of cooling by adding thermal energy. Hence the original thermal energy from the cooler body would have to end up being thermal energy in the warmer one before having any effect on temperature. Hence the 2nd Law would be violated.
My funnel experiment focuses more radiation from a large object onto a small one at the same temperature and made of the same metal material. What happens? Think about it. The only way the 2nd Law can apply is if you always disregard the radiation from cold to hot and only consider the radiation from hot to cold.
Any radiation from cold to hot merely resonates – as it can because the hotter body can always itself radiate at all the frequencies in the cooler body emission. It is as good as if diffuse reflection had happened – no energy is left behind and there is no effect on the temperature of the hotter body.
In case I am still not convincing someone, the AGW models assume backradiation works 24 hours a day – right? So they assume it also “works” when the temperature of the surface is getting hotter on a clear sunny morning and net radiation is into the surface. The assume that, not only does it slow the rate of cooling in the evening, but it still does something in the morning, namely increases the rate of warming.
Now if that is not adding thermal energy from a cooler atmosphere to a warmer surface then I’m a monkey’s uncle. So it is violating the Second Law. It is also doing so if it slows the rate of cooling as I explained above.
Is anyone aware of published experiments of absorptivity which use sources of spontaneous emission which are themselves around 10 deg.C to -60 deg.C, like typical atmospheric temperatures?
This* indicates that absorptivity is actually determined not by any warming effect, but by measuring reflected radiation in the visible spectrum.
I am saying that absorptivity cuts out (ie goes to zero) when the source becomes cooler than the target, which of course is not the case by a long shot when making these measurements using much higher frequency radiation than that contained in all spontaneous radiation from the atmosphere.
Hence, for proponents of the radiative greenhouse conjecture to use such measures of absorptivity (close to unity) and thus assume the surface absorbs “backradiation” is a complete abuse of physics.
* http://naca.central.cranfield.ac.uk/reports/arc/cp/0601.pdf.