# Power for a blackbody radiation

1. Jun 24, 2014

### 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.

2. Jun 24, 2014

### UltrafastPED

Wouldn't it also depend upon the temperature?

3. Jun 24, 2014

### WannabeNewton

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

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.

4. Jun 25, 2014

### dEdt

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

5. Jun 25, 2014

### WannabeNewton

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

6. Jun 27, 2014

### mediray

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