Question 4.8.1: A black body is an object at equilibrium with the radiation ...

A black body is an object at equilibrium with the radiation it emits. This radiation is characterised by the fact that the internal energy density depends only on the temperature at thermal equilibrium. The internal energy of this radiation is given by,

U(S,V)=\frac{3}{4} \biggl(\frac{3c}{16\sigma } \biggr)^{1/3} S^{4/3}V^{-1/3}.

1. Determine the free energy F (T, V) of this radiation.
2. Show that the internal energy U(S, V) of the radiation can be obtained by performing an inverse Legendre transformation on the free energy F (T, V).
3. Find the expressions p (T, V) and p (S, V) of the radiation pressure.

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1. The black body temperature (2.9) is defined as, i.e.

T(S,V,N_1,……N_r)≡\frac{\partial U(S,V,N_1,…,N_r)}{\partial S}.

T(S,V)= \frac{\partial U(S,V)}{\partial S}\biggl(\frac{3c}{16\sigma } \biggr)^{1/3} S^{1/3}V^{-1/3}.

When inverting this relation, we obtain the radiation entropy S (T, V) as a function of
the temperature T and of the volume V, i.e.
S (T, V) =\biggl(\frac{16\sigma}{3c } \biggr) T^{3}V.

When substituting this result into the expression for the internal energy of the radiation U(S, V), we find,

U=\frac{4\sigma}{c}T^4V.

The free energy F (T, V) is obtained by Legendre transformation (4.22) of the internal energy U(S, V) with respect to the entropy S. Using the two previous equations this transformation is written explicitly as, i.e.

F = U − TS

F(T,V)=U-TS=-\frac{4\sigma}{3c}T^4V.

2. For the black body radiation, the entropy defined by (4.26) is found to be,

S(T,V,\left\{N_A\right\} )=-\frac{\partial F(T,V,\left\{N_A\right\})}{\partial T}.

S(T,V)=-\frac{\partial F(T,V)}{\partial T}=\frac{16\sigma}{3c}T^3V.

When inverting this relation, we obtain the radiation temperature T (S, V) as a function of the entropy S and of the volume V, i.e.

T(S,V)= \biggl(\frac{3c}{16\sigma } \biggr)^{1/3} S^{1/3}V^{-1/3}.

When substituting for T in the radiation free energy F (T, V), we find,

F=- \frac{1}{4}\biggl(\frac{3c}{16\sigma } \biggr)^{1/3} S^{4/3}V^{-1/3}.

The internal energy U(S, V) is obtained by Legendre transformation (4.22) of the free energy F (T, V) with respect to the temperature T. Using the two previous equations, this transformation is written explicitly as,

F = U − TS

U(S,V)=F+ST= \frac{3}{4}\biggl(\frac{3c}{16\sigma } \biggr)^{1/3} S^{4/3}V^{-1/3}.

3. According to definition (2.10), the black body radiation pressure p (S, V) is expressed in terms of S and V as,

p(S,V,N_1,……N_r)≡-\frac{\partial U(S,V,N_1,…,N_r)}{\partial V}.

p(S,V)=–\frac{\partial U(S,V)}{\partial V}= \frac{1}{4}\biggl(\frac{3c}{16\sigma } \biggr)^{1/3} S^{4/3}V^{-4/3}.

According to definition (4.27), the black body radiation pressure p (T, V) is expressed in terms of T and V as,

p(T,V,\left\{N_A\right\} )=-\frac{\partial F(T,V,\left\{N_A\right\})}{\partial V}.

p(T,V)=-\frac{\partial F(T,V)}{\partial V}=\frac{4\sigma}{3c }T^4.

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