Why is the emissivity same for both absorption and emission?

Question

While learning about ways in which heat can be transferred I came to the section on Radiation. I noticed one thing peculiar which wasn't that obvious.

A body at a temperature $T_b$ emits thermal energy via electromagnetic radiation. The rate of heat transfer is given by the following formula:

$$H_{\text {rad}} = \varepsilon_{rad} \sigma A T_b^4$$

A body also absorbs radiation from surrounding which is at a temperature $T_{s}$. The rate of absorption of heat is given by:

$$H_{\text {abs}} = \varepsilon_{abs} \sigma A T_s^4$$

And for the fact $\varepsilon_{\text {abs}} = \varepsilon_{\text {rad}} = \varepsilon$.

-Why is this so? I mean why is it necessary for these two to be one and the same?

One reason that I can think of is that if that wasn't the case then it's would violate the Zeroth law of Thermodynamics. But

Since $\varepsilon_{\text {abs}} $ and $\varepsilon_{\text {rad}} $ are properties of materials. Therefore I think that there should be an explanation if we look at it from this angle.

• So what it is?

Answer 1

Assume a body in equilibrium. You shine light on it, let it heat up, and at some point it will reach a constant temperature. At that point input and output have to be the exact same. This can be argued at every temperature. So whenever the first equation is valid, the second must be too.

Answer 2

I think that I got the answer, but I would like to know of what others think of this. So here it is.

As soon as I saw the comment by Jon Custer, it became crystal clear to me why these two quantities might be one and the same. As he points out in his comment :

Time reversal symmetry is sufficient.

So since the process of emission would look the same like absorption if we reverse the time. Therefore $\varepsilon_{\text {abs}} = \varepsilon_{\text {rad}}$.