Power for a blackbody radiation

[dEdt]

If a blackbody is in equilibrium with the surrounding electromagnetic field, the power emitted by the surface of the blackbody will be related to the energy density of the electromagnetic field by $P=\frac{cu}{4}$. Try as I might, I haven't found a good derivation for this equation (the Hyperphysics one has several problems). I thought that it shouldn't be too hard to derive, but playing with Poynting's theorem led me nowhere, so now I humbly turn to physicsforums for help.

 

[UltrafastPED]

Wouldn't it also depend upon the temperature?

 

[WannabeNewton]

Try as I might, I haven't found a good derivation for this equation (the Hyperphysics one has several problems).

What's wrong with the standard derivation in texts? See e.g. pp. 385-388 of Reif.

Wouldn't it also depend upon the temperature?

The usual temperature dependence of the blackbody power spectrum is contained in the expression for the energy density of the photon gas in the cavity.

 

[dEdt]

What's wrong with the standard derivation in texts? See e.g. pp. 385-388 of Reif.

I just checked Reif's book and found his derivation to be very clear. Thanks.

 

[WannabeNewton]

I just checked Reif's book and found his derivation to be very clear. Thanks.

No problem! It's a brilliant gem of a book.

 

[mediray]

When the temperature of a blackbody radiator increases, the overall radiated energy increases