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Suppose a black body as an enclosure of volume $V$ with a hole of section $A$. In the interior there is a photon gas, whose energy density $u$ is, at temperature $T$.

$$ u=cT^4$$

How can I show that the energy emitted per second is

$$E=\sigma A T^4$$

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2 Answers 2

up vote 1 down vote accepted

You just have to use the Stefan-Boltzmann Law. S-B Law is

$j=\sigma T^{4}$

the irradiance $j$ has dimensions of energy per area per time. So, to find the power (energy per time) you just have to multiply it with its surface area of emission, $A$.

$P=j A= A\sigma T^{4}$

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Hi, so It is possible to derive the Steffan-Boltzman law for a photon gas whose energy density is $u=cT^4$ ? because that answer you give to me doesn't use this fact –  Jorge Jun 5 '13 at 18:08
It is a calculus in a continuous version of grand-canonical systems, that is : $\bar E = 2\frac{V}{(2 \Pi)^3}\int \large \frac{|\vec k|~ d^3 \vec k}{e^{\beta |\vec k|} - 1}$ (in units $\hbar = c = 1$). So, you will see easily that $\bar E$ is proportionnal to $T^4$, $\frac{\bar E}{V} = \sigma T^4$ –  Trimok Jun 5 '13 at 18:56
Hi Trimok thanks. What you did is showing that $u \propto T^4 $ but that was given. My claim is that the result of Nijankowski V. seems to be independent of that fact. –  Jorge Jun 5 '13 at 21:46

Consider a volume of section $A$ and height $L=c_0\Delta t$ next to the hole. Here, $\Delta t$ is an arbitrarily small time.

In this volume $AL$, there is a photon energy $ALu\propto Au$

During $\Delta t$, all the photons in this volume which have a velocities aligned with the section normal will go through the hole. So their flux will also be $\propto Au$. You can then repeat this for all photons directions, meaning that there is always proportionality with $Au$. Hence your result. Now, if you want to find exactly the coefficient, then you have to carry out the integral over angles ...

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