Radiation pressure. One may think of radiation as a gas of photons and apply many of the results from kinetic theory and thermodynamics to the radiation gas. (a) Prove that the pressure exerted by an isotropic radiation field of energy density u on a perfectly reflecting wall is p = u/3.

Question

Radiation pressure.

One may think of radiation as a gas of photons and apply many of the results from kinetic theory and thermodynamics to the radiation gas.

(a) Prove that the pressure exerted by an isotropic radiation field of energy density u on a perfectly reflecting wall is p = u/3.

(b) Blackbody radiation is radiation contained in, and in equilibrium with, a cavity whose walls are at a fixed temperature T. Use thermodynamic arguments to show that the energy density of blackbody radiation depends only on T and is independent of the size of the cavity and the material making up the walls.

(c) From (a) and (b) one concludes that for blackbody radiation the pressure depends only on the temperature, p = p(T), and the internal energy U is given by U = 3p(T)V where V is the volume of the cavity.
Using these two facts about the gas, derive the functional form of p(T), up to an unspecified multiplicative constant, from purely thermodynamic reasoning.

Accepted Answer

(a) Consider an area element dS of the perfectly reflecting wall and the photons impinging on dS from the solid angle [latex]d \Omega=\sin heta d heta d \varphi[/latex]. The change of momentum per unit time in the direction perpendicular to dS is u. [latex]\sin heta d heta d \varphi \cdot d S \cos heta \cdot 2 \cos heta / 4 \pi[/latex]. Hence the pressure on the wall is

[latex]p=\left(\frac{u}{2 \pi} ight) \int_{0}^{\pi / 2} d heta \int_{0}^{2 \pi} d \varphi \cos ^{2} heta \sin heta=\frac{u}{3}[/latex].

(b) Consider the cavity as consisting of two arbitrary halves separated by a wall. The volumes and the materials making up the sub-cavities are different but the walls are at the same temperature T. Then in thermal equilibrium, the radiations in the sub-cavities have temperature T but different energy densities if these depend also on factor other than temperature. If a small hole is opened between the sub-cavities, there will be a net flow of radiation from the sub-cavity of higher u because of the pressure difference. A heat engine can then absorb this flow of heat radiation and produce mechanical work. This contradicts the second law of thermodynamics if no other external effect is involved. Hence the energy density of black body radiation depends only on temperature.

(c) Since the free energy F is an extensive quantity and

[latex]\left(\frac{\partial F}{\partial V} ight)_{T}=-p=-\frac{1}{3} u(T)[/latex].

we have

[latex]F=-\frac{1}{3} u(T) V[/latex].

From thermodynamics we also have F = U - TS, where U = uV is the internal energy, S is the entropy, and

[latex]S=-\left(\frac{\partial F}{\partial T} ight)_{V}[/latex].

[latex]=\frac{1}{3} \frac{d u(T)}{d T} V[/latex].

Hence

[latex]\frac{d u}{u}=4 \frac{d T}{T}[/latex].

[latex] ext { giving } u=a T^{4}, p=\frac{1}{3} a T^{4} ext {, where } a ext { is a constant. }[/latex]

Question 2.165: Radiation pressure. One may think of radiation as a gas of p...

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