In Stefan's law, the radiated energy of a black body with a temperature of $T$ is $q = \sigma T^4$ Where $\sigma$ is Stefan's constant.

However, if it is surrounded by a medium then the net radiated energy is $q = \sigma(T^4 -t^4)$ Where $t$ is the temperature of the surroundings.

So my question is, Why the energy received from the surrounding is $\sigma t^4$?

Even that air doesn't have a definite area, volume..etc. why do we assume it follows Stefan's law?


1 Answer 1


Typically for a situation like this we don't look at the air close by, but "the object at the other end of the air" (assuming that the emissivity of the air is not very large - that it is mostly transparent to EM radiation of interest).

But if you put any black body inside a container (which could be as large as the universe), there is radiation going out, and radiation coming back. The formula you give assumes that all of the surroundings are at the same temperature $t$; more realistically, you "see" a mixture of objects of different temperature, and receive "some" flux from each, proportional to the solid angle they subtend.

This is how Earth can have a temperature that is greater than the temperature of the universe, but less than the temperature of the Sun (which we can see, but it only fills part of our sky).

Does that help at all?

  • 1
    $\begingroup$ Also might be worth noting that you don't need air to transfer the radiant heat. That is why the sun can still transfer so much energy through the "vacuum" of space. $\endgroup$
    – JMac
    Oct 5, 2017 at 18:03
  • $\begingroup$ So the received heat is from other bodies surrounding the blackbody. But that doesn't make sense because the amount of heat radiated from these objects must depend on their emissitivity $\endgroup$ Oct 5, 2017 at 20:06
  • $\begingroup$ Yes it does; often this is omitted, but strictly speaking the expression should be something like $\sigma \int \epsilon T^4 d\Omega$ integrated over the full $4\pi$ solid angle, taking account of the relative size, temperature and emissivity of every point you "see" $\endgroup$
    – Floris
    Oct 5, 2017 at 20:22
  • $\begingroup$ Aha, Then what assumptions did they use to omit that? $\endgroup$ Oct 5, 2017 at 22:45

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