Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 35A / iii

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- Solution iii

Solution

The first law of thermodynamics is

d E = T d S - P d V,

so, regarding

S

as a function of

(T, V)

,

(\frac{\partial S}{\partial T})_{V} = \frac{1}{T} (\frac{\partial E}{\partial T})_{V}, (\frac{\partial S}{\partial V})_{T} = \frac{1}{T} [(\frac{\partial E}{\partial V})_{T} + P] .

Equality of mixed partial derivatives gives the thermodynamic derivation of blackbody radiation pressure

(\frac{\partial E}{\partial V})_{T} = T (\frac{\partial P}{\partial T})_{V} - P .

a = π^{2} / (15 ℏ^{3} c^{3})

, we have

E = aV T^{4}

, and hence

T \frac{d P}{d T} - P = a T^{4} .

Equivalently,

d (P / T) / d T = a T^{2}

, so

P = \frac{a}{3} T^{4} + CT .

The condition

P = o (T)

T \to 0

C = 0

. Therefore the radiation pressure is

P = \frac{π ^{2} T ^{4}}{45 ℏ ^{3} c ^{3}} = \frac{E}{3 V} .

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