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In a photon gas, we know that pressure, $P$, and energy density, $u$, are related by: $$P=\frac{u}{3}$$ We also know from relativity that the momentum of a photon is $$p=\frac{E}{c}$$

Finally, the pressure can usually be thought of as the flux of momentum through a surface, but that would imply that the flux of energy through that surface is: $$E_{e}=Pc=\frac{u}{3} c$$

However, the standard formula, from kinetic theory, is:

$$E_{e}=\frac{u}{4} c$$

I guess the assumption which is most probably wrong is identifying this sort of thermodynamic pressure with momentum flux, but I'm not sure how this comes about.

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Actually, you are on the right track. If you take into account that momentum is not a scalar, so you consider the magnitude of the component orthogonal to the surface area (and also make sure the photons are reflected instead of moving straight through to yield a pressure), then you will get the correct formula for the pressure using the same method that you use to derive the formula for the energy flux:

$\frac{1}{4} = \frac{1}{4\pi}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\sin(\theta)\cos(\theta)d\theta d\phi$

$\frac{1}{3}=\frac{2}{4\pi}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\sin(\theta)\cos^2(\theta)d\theta d\phi$

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  • $\begingroup$ Thanks! That was helpful. Of course, the momentum transfer from particles bouncing of a wall isn't the same as just flowing through a hole, as you say! $\endgroup$ – guillefix May 30 '14 at 18:02

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