# What is the net radiative balance between two blackbodies with different temperature?

Context

In trying to understand a crop productivity model, I want to figure out how to derive the equation for the net exchange of longwave radiation between two adjacent blackbodies (soil and atmosphere) over some discrete time step.

In order to calculate the net energy balance, longwave radiation ($L$) is calculated first (and can easily be calculated from first principles), so we can go from:

$$S_\downarrow + L_\downarrow = L_\uparrow+H+\lambda E+G$$

To here:

$$S_\downarrow + (L_\downarrow - L_\uparrow) = H+\lambda E+G$$

where $S$ is solar flux, $L$ is longwave radiation, $\lambda E$ is latent heat and $G$ is conduction.

Problem

This is the equation I am trying to derive:

$$L_{\text{soil}\rightarrow\text{air}}=4\sigma T^3\Delta T$$

Where

• $\sigma$ is the Stefan-Boltzman constant
• $L_{\text{soil}\rightarrow\text{air}}=L_\uparrow - L_\downarrow$ is the net heat flux over time step $\Delta t$
• $\Delta T=T_\text{air}-T_\text{soil}$

I am starting with the Stefan-Boltzman law:

$$L_\downarrow=4\sigma T^4$$

And I recognize the following:

• I am trying to calculate $L_\downarrow-L_\uparrow$ over some $\Delta t$ and
• $dL/dt = 4\sigma T^3$

However, I can't get back to the SB law. I have gotten as far as

$$L_\uparrow-L_\downarrow =\sigma T_\text{soil}^4-\sigma T_\text{air}^4$$

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I think it's very unlikely that heat transfer between the ground and atmosphere is dominated by radiative transfer. It seems far more like to be dominated by convection/conduction, in which case Newton's law would be a good approximation. –  John Rennie Sep 11 '13 at 14:14
Even if the transfer was dominated by radiation (and I agree with John on that), treating the atmosphere as a blackbody is at beast a rough approximation and is probably just a plain ol' bad idea. –  dmckee Sep 11 '13 at 14:53
@JohnRennie I should have been more clear that this is only one term; I will update my answer. –  Abe Sep 11 '13 at 15:07
@dmckee I have updated my answer to provide context - at one point I had used the term "heat" where I meant "longwave radiation" –  Abe Sep 11 '13 at 15:16
Ah. Not sure that my comment has any validity in that band. –  dmckee Sep 11 '13 at 16:09

Firstly, I should point out that this is a mistake: $$dL/dt = 4\sigma T^3.$$ $L$ is the flux - it isn't changing in time. Or rather, it does (it's different during the day than during the night, for instance), but in doing this sort of calculation we normally assume it's constant, because the time scale over which it changes is long compared to the flux itself. If the flux was changing by such a large amount, it would mean that the ground was losing heat at a rapidly increasing rate, which would be odd.
All you need to do is note that the ground loses heat to the air at a rate $\sigma T_\text{soil}^4$ (note there is no factor of 4 here) whereas the air loses heat to the ground at a rate $\sigma T_\text{air}^4$. Thus this difference is $$L_\downarrow = \sigma (T_\text{air}^4 - T_\text{soil}^4).$$
You could stop there, but it's inconvenient to work with those powers of 4, and it's not really necessary because $T_\text{air}$ and $T_\text{soil}$ are quite close in value (they're both around $300\:\mathrm{K}$). So let's let $T_\text{soil} = T$ and $T_\text{air} = T+\Delta T$. Then we have $$L_\downarrow = \sigma (T+\Delta T)^4 - \sigma T^4 \\ = \sigma (T^4 + 4T^3\Delta T + 6T^2(\Delta T)^2 + 4T(\Delta T)^3 + (\Delta T)^4 \quad) -\sigma T^4.$$ The $T^4$ terms cancel out, and if we assume $\Delta T$ is small then we can ignore the $(\Delta T)^2$ term and all the other higher powers of $\Delta T$. This leaves us with $$L_\downarrow = 4\sigma T^3\Delta T,$$ which is the expression you were trying to derive.
We could have let $T_\text{air} = T$ and $T_\text{soil} = T-\Delta T$ instead and it would have made no difference. The point is that we can do this trick whenever the temperatures are close enough that they can both be considered as small deviations from a single 'ambient' value.
thanks for taking the time to work through this. Assuming $4\sigma T^3$ was a derivative sent me down the wrong track. –  Abe Sep 11 '13 at 19:22
Well, it kind of is a derivative, just with respect to temperature rather than time. Another way to derive it is to plot $\sigma T^4$ against $T$ and approximate it with a straight line using the derivative - but the method I showed amounts to the same thing, and I find it clearer. –  Nathaniel Sep 12 '13 at 5:36