So this has been bothering me for quite a time:

$$\frac{dq}{dt}=e \sigma A(T_1^4-T_0^4)$$

Net radiation is $emission-radiation$. According to that if you observe:

Consider a situation where a body is at temperature $T_1$ and the surrounding is at temperature $T_0$. The rate of emission is

$$\frac{dq}{dt} =e \sigma A T_1^4.$$

So my question is how is the absorption of radiation equal to

$$\frac{dq}{dt} =e \sigma A T_0^4?$$

Is there a proof for this?

I don’t understand how the absorption of radiation is dependent on the surrounding temperature.

Please suggest edits.


1 Answer 1


I feel the opposite that absorption is fitted to the black body radiation law, but emission is doubltful.

The black body radiation spectrum was derived for a closed cavity, where the emission and absorption reached detailed balance (absorption = emission). Then you integrate the whole spectrum to obtain the Stefan-Boltzmann law $e\sigma A T^4$.

The atmophere is considered as a black-body cavity, and the object is among the boundary of the cavity, therefore a thermal equilibrium absorption rate, absorb radiation from the "cavity" under thermal equlibrium: $$ P_{abs} = e \sigma A T_o^4 $$

On the other hand, the object is emiting raditions to the atomosphere. But the emission is running into the vast open space, it is not under thermal equilibrium with temperature $T_1$, therefore the black-body radition power can only be adopted as an approximation.


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