# Regarding the absorption property of a blackbody

Consider a blackbody of surface area $$S_b$$ and at temperature $$T_b$$. It is placed inside an evacuated chamber (to neglect all the effects of convection), with walls of chamber at temperature $$T_c$$ and surface area $$S_c$$.

Then, the energy emitted (radiated) per unit time by the blackbody is: ($$\sigma$$ being the Stefan-Boltzmann's constant) $$u_e = \sigma S_bT_b^4 \tag{i}$$

Now, when we come to the calculation of the energy absorbed per unit time by the blackbody in this case (or even when it is placed in a surrounding of temperature $$T_c$$), almost at all places I see it to be: $$u_a = \sigma S_bT_c^4 \tag{ii}$$

The problem is how is it able to absorb according to the same law by which it radiates?

Now, if we consider the chamber's inner wall to be blackbody and take its conductivity to $$0$$.The eq $$(ii)$$ starts making some sense:

As the walls would now radiate as a blackbody and since the blackbody kept in the chamber should absorb all the radiation, the energy absorbed should be: $$u_a' = \sigma S_cT_c^4 \tag{iii}$$

Another problem: It doesn't match with eq$$(ii)$$!!

How is the blackbody managing to absorb only a limited amount of radiation limited by its own area?(as indicated by the eq$$(ii)$$)

One reason I could think is that, maybe not all the radiations are falling on the blackbody. But, how is it possible that in each unit time only a specific amount of radiation is falling on it, given by eq$$(ii)$$.

One more question (a specific one), what would happen if the chamber is a hollow sphere and it's inner wall is a blackbody?

Would all the radiation emitted by the wall fall on the inside placed (at centre) blackbody. If yes, would all of them absorbed or the above law even applies there and more importantly, why?

Also , anyone please try to answer it , starting from basics and with enough use of proofs .

• I don't know what you mean by "How is the black body only absorbing a limited amount of radiation?", A black body absorbs all incoming radiation by definition. It will emit a limited amount of radiation that depends on its temperature. Where does the asymmetry come from? It comes from the thermal bath that is attached to the black body. All incoming radiation is absorbed by the thermal bath, but the thermal bath can only lose a limited amount of energy itself. Commented Jul 8 at 18:18
• @FlatterMann By the limited amount, I mean if you compare the eq$(iii)$ with eq$(ii)$ , you would find that the energy absorbed per unit time as suggested by eq$(ii)$ is less than that by eq$(iii)$ .And as mentioned all the places it is written that the eq$(ii)$ is correct. Commented Jul 8 at 18:35
• The radiation that is being absorbed by the black body does not have to have a black body spectrum itself, so there does not have to be an equilibrium in general. The statement that a black body can only absorb a limited amount of radiation is therefor a specialization to black body radiation. It does not hold in general. Commented Jul 8 at 18:52
• Could you add references from which various formulas come from? Otherwise, it is not clear whether the sources like "almost at all places I see it to be" are to be trusted. There is always a possibility of a typo in a book, so knowing the context in which formulas arise is crucial. Commented Jul 9 at 8:24
• Also, the conditions of the problem are not quite clear: the chamber and the body are not in equilibrium with each other. Calling them black bodies mean that they absorb all the radiation incident on them... but it doesn't mean that the radiation state is black body radiation, i.e., that it is radiation in thermodynamic equilibrium. See, e.g., here for some nuances. Commented Jul 9 at 8:28

how is it able to absorb according to the same law by which it radiates?

It must absorb according to the same law by which it radiates according to the second law of thermodynamics.

Suppose that a material emits at a lower rate than it absorbs under some specified equilibrium conditions. Then, if that material exchanges thermal radiation with a black body at the same temperature, net thermal energy will spontaneously transfer from the black body to the material. This will spontaneously decrease the entropy of the system.

How is the blackbody managing to absorb only a limited amount of radiation limited by its own area?

A black body, by definition, absorbs all thermal radiation that is incident on it. In many common cases the total amount of radiation incident on a surface depends on its area.

Would all the radiation emitted by the wall fall on the inside placed (at centre) blackbody.

No. If you trace rays you can easily see that some rays of light can leave the wall and reach another part of the wall without intersecting the object at all. The fraction of rays emitted by the wall that reach the object is called the view factor.

• I got it.Thanks! Commented Jul 10 at 12:35

Let us say that $$T_b < T_c$$. What will happen is that the central object will radiate less energy than it absorbs until it comes to a new equilibrium at some intermediate temperature. At the same time, the walls will radiate more than they absorb until they reach the same new equilibrium temperature.

If we assume the introduced blackbody is small then to first order we can say that initially it does not greatly change the radiation field it is sitting in. In which case, that would be an isotropic black body radiation field at $$T_c$$ and thus the flux per unit area in any direction is $$\sigma T_c^4$$ and the amount of radiation absorbed by the central introduced blackbody will be this multiplied by its surface area. The emission and absorption follow the same Stefan-Boltzmann law because both the emitted and absorbed radiation is that of a blackbody (at different temperatures).

Your second question is based on the mistaken premise that all radiation emitted by the walls are absorbed by the central blackbody. That isn't the case. Some fraction will be reabsorbed by other parts of the walls. The amount absorbed by the body is specifically given by equation (ii) because that is how much energy would be absorbed by any black object immersed in a radiation field with temperature $$T_c$$.

The last part of your question seems to want to distinguish between a sphere with an inner surface that is black and a cavity of arbitrary shape. There is no distinction. Both will equilibriate to have blackbody radiation inside their cavities and in both cases only some fraction of their emitted radiation will be absorbed by the central introduced black body object.

• Yeah , but would the absorption would happen according to Stephen -Boltzmann law , if the blackbody was placed in a large surrounding (temperature, $T_c$).By large , I mean that the temperature of the surrounding would not change significantly. Commented Jul 9 at 15:38
• @CPofPhysics yes, that is what I have assumed and the only situation in which your equation (ii) is accurate. If $T_c$ does not change "significantly" then the inserted body will eventiually reach an equilibrium at temperature $T_c$ - when it will emit and absorb energy at exactly the same rate. Commented Jul 9 at 16:37
• Thank you, I got it. Commented Jul 10 at 12:33