Question 2.164: By treating radiation in a cavity as a gas of photons whose ...
By treating radiation in a cavity as a gas of photons whose energy ε and momentum k are related by the expression ε = ck, where c is the velocity of light, show that the pressure p exerted on the walls of the cavity is one-third of the energy density.
With the above result prove that when radiation contained in a vessel with perfectly reflecting walls is compressed adiabatically it obeys the equation
P V^{\gamma}=\text { constant }.
\text { Determine the value of } \gamma .
Let n(\omega) d \omega denote the number of photons in the angular frequency interval \omega \sim \omega+d \omega . Consider the pressure exerted on the walls by such photons in the volume element dV at (r, \theta, \varphi) (Fig. 2.34). The probability that they collide with an area d A of the wall is d A \cdot \cos \theta / 4 \pi r^{2} , each collision contributing an impulse 2 k \cos \theta perpendicular to d A . Therefore, we have
d p_{\omega}=\frac{d f}{d A d t}.
=\frac{n d \omega \cdot d V \cdot \frac{d A \cdot \cos \theta}{4 \pi r^{2}} \cdot 2 k \cos \theta}{d A d t}.
=\frac{d \omega}{4 \pi d t} 2 n k \cos ^{2} \theta \sin \theta d r d \theta d \varphi.
p=\int_{r \leq c d t} d p_{\omega}=\int \frac{n}{3} k c d \omega=\int \frac{u(\omega)}{3} d \omega.
Integrating we get p=\frac{u}{3}=\frac{U}{3 V}, where u is the energy density and U is the total energy. From the thermodynamic equation
d U=T d S-p d V.
\text { and } p=U / 3 V \text {, we obtain } d U=3 p d V+3 V d p \text {. Hence } 4 p d V+3 V d p=T d S. For an adiabatic process d S=0,4 \frac{d V}{V}+3 \frac{d p}{p}=0 Integrating we have
p V^{4 / 3}=\text { const., } \quad \gamma=\frac{4}{3}.
