A guest post by Leonard Weinstein. The original document is here and has more legible equations.
Leonard Weinstein
July 18, 2012
The argument is frequently made that back radiation from optically absorbing gases heats a surface more than it would be heated without back radiation, and this is the basis of the socalled Greenhouse Effect on Earth. The first thing that has to be made clear is that a suitably radiation absorbing and radiating atmosphere does radiate energy out based on its temperature, and some of this radiation does go downward, where it is absorbed by the surface (i.e., there is back radiation, and it does transfer energy to the surface). However, heat (which is the net transfer of energy, not the individual transfers) is only transferred down if the ground is cooler than the atmosphere, and this applies to all forms of heat transfer. While it is true that the atmosphere containing suitably optically absorbing gases is warmer than the local surface in some special cases, on average the surface is warmer than the integrated atmosphere effect contributing to back radiation, and so average heat transfer is from the surface up. The misunderstanding of the distinction between energy
transfer, and heat transfer (net energy transfer) seems to be the cause of much of the confusion about back radiation effects.
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Before going on with the back radiation argument, first examine a few ideal heat transfer examples, which emphasize what is trying to be shown. These include an internally uniformly heated ball with either a thermally insulated surface or a radiationshielded surface. The ball is placed in space, with distant temperatures near absolute zero, and zero gravity. Assume all emissivity and absorption coefficients for the following examples are 1 for simplicity. The bare ball surface temperature at equilibrium is found from the balance of input energy into the ball and radiated energy to the external wall:
To=(P/σ)^0.25 (1)
Where To (oK) is absolute temperature, P (Wm2) is input power per area of the ball, and
σ=5.67Ε−8 (Wm2 T4) is the StefanBoltzmann constant.
Now consider the same case with a relatively thin layer (compared to the size of the ball) of thermally insulating material coated directly onto the surface of the ball. Assume the insulator material is opaque to radiation, so that the only heat transfer is by conduction. The energy generated by input power heats the surface of the ball, and this energy is conducted to the external surface of the insulator, where the energy is radiated away from the surface. The assumption of a thin insulation layer implies the total surface area is about the same as the initial ball area. The temperature of the external surface then has to be the
same (=To) as the bare ball was, to balance power in and radiated energy out. However, in order to transmit the energy from the surface of the ball to the external surface of the insulator there had to be a temperature gradient through the insulation layer based on the conductivity of the insulator and thickness of the insulation layer. For the simplified case described, Fourier’s conduction law gives:
qx=k(dT/dx) (2)
where qx (Wm2) is the local heat transfer, k (Wm1 T1) is the conductivity, and x is distance outward of the insulator from the surface of the ball. The equilibrium case is a linear temperature variation, so we can substitute DT/h for dT/dx, where h is the insulator thickness, and DT is the temperature difference between outer surface of insulator and surface of ball (temperature decreasing outward). Now qx has to be the same as P, so from (2):
DT=(ToT’)=Ph/k (3)
Where T’ is the ball surface temperature under the insulation, and thus we get:
T’=(Ph/k)+To (4)
The new ball surface temperature is now found by combining (1) + (4):
T’=(Ph/k)+(P/σ)0.25 (5)
The point to all of the above is that the surface of the ball was made hotter for the same input energy to the ball by adding the insulation layer. The increased temperature did not come from the insulation heating the surface, it came from the reduced rate of surface energy removal at the initial temperature (thermal resistance), and thus the internal surface temperature had to increase to transmit the required power. There was no added heat and no back heat transfer!
An alternate version of the insulated surface can be found by adding a thin conducting enclosing shell spaced a small distance from the wall of the ball, and filling the gap with a highly optically absorbing dense gas. Assume the gas is completely opaque to the thermal wavelengths at very short distances, so that he heat transfer would be totally dominated by diffusion (no convection, since zero gravity). The result would be exactly the same as the solid insulation case with the correct thermal conductivity, k, used (derived from the diffusion equations). It should be noted that the gas molecules have a range of speeds, even at a specific temperature (Maxwell distribution). The heat is transferred only by molecular collisions with the wall for this case. Now the variation in speed of the molecules, even at a single temperature, assures that some of the molecules hitting the ball wall will have higher energy going in that leaving the wall. Likewise, some of the molecules hitting the outer shell will have lower speeds than when they leave inward. That is, some energy is transmitted from the colder outer wall to the gas, and some energy is transmitted from the gas to the hotter ball wall. However, when all collisions are included, the net effect
is that the ball transfers heat (=P) to the outer shell, which then radiates P to space. Again, the gas layer did not result in the ball surface heating any more than for the solid insulation case. It resulted in heating due to the resistance to heat transfer at the lower temperature, and thus resulted in the temperature of the ball increasing. The fact that energy transferred both ways is not a cause of the heating.
Next we look at the bare ball, but with an enclosure of a very small thickness conductor placed a small distance above the entire surface of the ball (so the surface area of the enclosure is still essentially the same as for the bare ball), but with a high vacuum between the surface of the ball and the enclosed layer. Now only radiation heat transfer can occur in the system. The ball is heated with the same power as before, and radiates, but the enclosure layer absorbs all of the emitted radiation from the ball. The absorbed energy heats the enclosure wall up until it radiated outward the full input power P. The final temperature of the enclosure wall now is To, the same as the value in equation (1). However, it is also radiating inward at the same power P. Since the only energy absorbed by the enclosure is that radiated by the ball, the ball has to radiate 2 P to get the net transmitted power out to equal P. Since the only input power is P, the other P was absorbed energy from the enclosure. Does this mean the enclosure is heating the ball with back radiation? NO. Heat transfer is NET energy transfer, and the ball is radiating 2 P, but absorbing P, so is radiating a NET radiation heat transfer of P. This type of effect is shown in radiation equations by:
Pnet=σ (T^4hotT^4 cold) (6)
That is, the net radiation heat transfer is determined by both the emitting and absorbing surfaces. There is radiation energy both ways, but the radiation heat transfer is one way. This is not heating by back radiation, but is commonly also considered a radiation resistance effect. There is initially a decrease in net radiation heat transfer forcing the temperature to adjust to a new level for a given power transfer level. This is directly analogous to the thermal insulation effect on the ball, where radiation is not even a factor between the ball and insulator, or the opaque gas in the enclosed layer, where there is
no radiation transfer, but some energy is transmitted both ways, and net energy (heat transfer) is only outward. The hotter surface of the ball is due to a resistance to direct radiation to space in all of these cases.
If a large number of concentric radiation enclosures were used (still assuming the total exit area is close to the same for simplicity), the ball temperature would get even hotter. In fact, each layer inward would have to radiate a net P outward to transfer the power from the ball to the external final radiator. For N layers, this means that the ball surface would have to radiate:
P’=(N+1)Po (7)
Now from (1), this means the relative ball surface temperature would increase by:
T’/To=(N+1)^0.25 (8)
Some example are shown to give an idea how the number of layers changes relative absolute temperature:
N 
T’/To 


1 
1.19 
10 
1.82 
100 
3.16 
Change in N clearly has a large effect, but the relationship is a semilog like effect.
Planetary atmospheres are much more complex than either a simple conduction insulating layer or radiation insulation layer or multiple layers. This is due to the presence of several mechanisms to transport energy that was absorbed from the Sun, either at the surface or directly in the atmosphere, up through the atmosphere, and also due to the effect called the lapse rate. The lapse rate results from the convective mixing of the atmosphere combined with the adiabatic cooling due to expansion at decreasing pressure with increasing altitude. The lapse rate depends on the specific heat of the atmospheric gases,
gravity, and by any latent heat release, and may be affected by local temperature variations due to radiation from the surface directly to space. The simple theoretical value of that variation in a dry adiabatic atmosphere is about 9.8 C per km altitude on Earth. The effect of water evaporation and partial condensation at altitude, drops the size of this average to about 6.5 C per km, which is the called the environmental lapse rate.
The absorbed solar energy is carried up in the atmosphere by a combination of evapotransporation followed by condensation, thermal convection and radiation (including direct radiation to space, and absorbed and emitted atmospheric radiation). Eventually the conducted, convected, and radiated energy reaches high enough in the atmosphere where it radiates directly to space. This does require absorbing and radiating gases and/or clouds.
The sum of all the energy radiated to space from the different altitudes has to equal the absorbed solar energy for the equilibrium case. The key point is that the outgoing radiation average location is raised significantly above the surface. A single average altitude for outgoing radiation generally is used to replace the outgoing radiation altitude range. The temperature of the atmosphere at this average altitude then is calculated by matching the outgoing radiation to the absorbed solar radiation. The environmental lapse rate, combined with the temperature at the average altitude required to balance incoming and outgoing energy, allows the surface temperature to be then calculated. The equation for the effect is:
T’=To GH (9)
Where To is the average surface temperature for the nonabsorbing atmospheric gases case, with all radiation to space directly from the surface, G is the lapse rate (negative as shown), and H is the effective average altitude of outgoing radiation to space. The combined methods that transport energy up so that it radiated to space, are variations of energy transport resistance compared to direct radiation from the surface. In the end, the only factors that raise ground temperature to be higher than the case with no greenhouse gas is the increase in average altitude of outgoing radiation and the lapse rate. That is all
there is to the socalled greenhouse effect. If the lapse rate or albedo is changed by addition of specific gases, this is a separate effect, and is not included here.
The case of Venus is a clear example of this effect. The average altitude where radiation to space occurs is about 50 km. The average lapse rate on Venus is about 9 C per km. The surface temperature increase over the case with the same albedo and absorbed insolation but no absorbing or cloud blocking gases, would be about 450 C, so the lapse rate fully explains the increase in temperature. It is not directly due to the pressure or density alone of the atmosphere, but the resulting increase in altitude of outgoing radiation to space. Changing CO2 concentration (or other absorbing gases) might change the outgoing
altitude, but that altitude change would be the only cause of a change in surface temperature, with the lapse rate times the new altitude as the increase in temperature over the case with no absorbing gases. One point to note is that the net energy transfer (from combined radiation and other transport means) from the surface or from a location in the atmosphere where solar energy was absorbed is always exactly the same whatever the local temperature. For example, the hot surface of Venus radiated up (a very short distance) over 16 kWm2. However, the total energy transfer up is just the order of absorbed solar energy, or about 17 Wm2, and some of the energy carried up is by conduction and convection. Thus the net radiation heat transfer is to be almost exactly the same as radiation up. The back radiation is not heating the surface; the thermal heat transfer resistance from all causes results in the excess heating.
In the end, it does not matter what the cause of resistance to heat transfer is. Th total energy balance and thermal heat transfer resistance defines the process. For planets with enough atmosphere, the lapse rate defines the lower atmosphere temperature gradient, and if the lapse rate is not changed, the distance the location of outgoing radiation is moved up by addition of absorbing gases determines the increase in temperature effect. It should be clear the back radiation did not do the heating.