= Radiation pressure of an isotropic photon gas
{title2=$p_{\rm rad}=u/3$}
For radiation in <vacuum> with <isotropy>, <energy> is distributed equally per <solid angle>: write $u_\Omega=u/(4\pi)$ for <energy density> per unit <solid angle>. Let $\theta$ be the <angle> between a ray's path and the wall normal. Its normal <velocity> $c\cos\theta$ determines the arrival rate, while normal <momentum> per <energy> is $\cos\theta/c$. Their product is $\cos^2\theta$. For a perfectly reflecting cavity wall,
$$
p_{\rm rad}=2\int_{\rm incoming\ hemisphere}u_\Omega\cos^2\theta\,d\Omega
=\frac{2u}{4\pi}\int_0^{2\pi}d\varphi\int_0^{\pi/2}\cos^2\theta\sin\theta\,d\theta
=\frac u3.
$$
The leading two is the normal <momentum> reversal; the incoming <hemisphere> contains half of the field's directional <energy density>. This is also the normal component of the bulk <pressure tensor>, $P_{nn}=\int_{4\pi}u_\Omega\cos^2\theta\,d\Omega=u/3$. <Isotropy> makes all three diagonal components equal, while their sum is $u$ because the squared direction components sum to one. The factor one third therefore depends on three-dimensional angular geometry, not on the shape of the <blackbody radiation> spectrum.
\Image[/radiation-pressure-isotropic-angular-average.png]
{title=<Isotropic> radiation: angular <energy> weights and normal <momentum flux>}
{description=Original angular model of a reflecting cavity wall. The left panel shows representative incoming paths, not a literal two-dimensional <photon gas>. The right panel uses the correct three-dimensional <solid angle> weight, normalized on an incoming <hemisphere>: $\sin\theta$. The normal <pressure> contribution adds $\cos^2\theta$, giving total weight $\int_0^{\pi/2}\sin\theta\cos^2\theta\,d\theta=1/3$.}
A perfectly absorbing wall exposed to this incoming <hemisphere> receives only $u/6$ from absorption. If it is in <thermal equilibrium> with the <isotropic> radiation, its outgoing emission supplies another $u/6$ of recoil, recovering $u/3$. A cold absorber before equilibrium is a different boundary condition. Similarly, the one-way incident <energy flux> is $cu/4$, whereas the net <energy flux> across a plane in the undisturbed <isotropic> field is zero. Dividing $cu/4$ by $c$ misses the extra normal-<momentum> projection and does not give the cavity <pressure>.
For a stipulated isotropic radiation <energy density> $u=6\,\mu\mathrm{J\,m^{-3}}$, the bulk <pressure> is $2\,\mu\mathrm{Pa}$ regardless of its spectral shape. In <thermal equilibrium>, <blackbody radiation> supplies $u=a_{\rm rad}T^4$ with <radiation constant> $a_{\rm rad}=4\sigma/c$, so $p_{\rm rad}=a_{\rm rad}T^4/3$; <Stefan–Boltzmann law> defines $\sigma$. The equilibrium <photon gas> derives this same <equation of state> from its <free energy>.<Statistical Physics by David Tong/3 Quantum Gases>[§3.2, pressure immediately after the photon free-energy formula: $p=E/(3V)$]{ref}
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