91Ó°ÊÓ

Sometimes it is useful to know the free energy of a photon gas.

F= U-TS.

The Helmholtz free energy is:

$F=U-TS\left(1\right)$

The total energy is:

$U=\frac{8{\mathrm{Ï€}}^5(kT{)}^4}{15(hc{)}^3}V$

The entropy is:

$S=\frac{32{\mathrm{Ï€}}^5}{45}V{\left(\frac{kT}{hc}\right)}^{3}k$

Let,

$A=\frac{8{\mathrm{Ï€}}^2{k}^4}{15(hc{)}^3}$

The total energy and entropy is:

$U=AV{T}^4\text{and}S=\frac{4}{3}AV{T}^3$

Substitute into (1)

$$\begin{gathered} F=AV{T}^4-\frac{4}{3}AV{T}^4 \\ F=-\frac{1}{3}AV{T}^4 \\ F=-\frac{1}{3}U \end{gathered}$$

(b) At constant volume, the entropy equals the negative partial derivative of the Helmholtz energy with respect to temperature.

$S=-{\left(\frac{∂F}{∂T}\right)}_V$

Substitute F,

$$\begin{gathered} S=-\frac{∂}{∂T}\left(-\frac{1}{3}AV{T}^4\right) \\ S=\frac{4}{3}AV{T}^3 \end{gathered}$$

(c) At constant temperature, the pressure equals the negative partial derivative of the Helmholtz energy with respect to the volume, i.e.

$P=-{\left(\frac{∂F}{∂V}\right)}_T$

Substitute with F

$$\begin{gathered} P=-\frac{∂}{∂V}\left(-\frac{1}{3}AV{T}^4\right) \\ P=\frac{1}{3}A{T}^4 \\ P=\frac{1}{3}\frac{U}{V} \end{gathered}$$

(d)The Helmholtz free energy for one mode is given by:

$F=-kT\ln \left(Z\right)$

Where Z is partition function and is given as:

$Z=\frac{1}{1-e^{-\mathrm{Ï\mu }/kT}}$

Substitute in the above equation:

$$\begin{gathered} F=-kT\ln \left(\frac{1}{1-e^{-\mathrm{Ï\mu }/kT}}\right) \\ F=kT\ln \left(1-e^{-\mathrm{Ï\mu }/kT}\right) \end{gathered}$$

Because we have two polarisation modes, the total Helmholtz free energy equals the sum of F over all modes increased by a factor of two.

$F=2\underset{n_x}{∑}\underset{n_y}{∑}\underset{n_z}{∑}kT\ln \left(1-e^{-\mathrm{Ï\mu }/kT}\right)=2\underset{n_x,n_y,n_z}{∑}kT\ln \left(1-e^{-\mathrm{Ï\mu }/kT}\right)$

Consider we have a cubic box with volume of V and side width of L, the allowed energy for the photon is:

$\mathrm{Ï\mu }=\frac{hcn}{2L}$

Hence

$F=2\underset{n_x,n_y,n_z}{∑}kT\ln \left(1-e^{-hcn/2LkT}\right)$

Now we need to convert the total to an integral in spherical coordinates, which we can do by multiplying it by the spherical integration factor $n^2\sin \left(\mathrm{θ}\right)$, which gives us:

role="math" localid="1647765663139" $$\begin{gathered} F=2kT{∫}_0^{\mathrm{π}/2}d\mathrm{Φ}{∫}_0^{\mathrm{π}/2}\sin \left(\mathrm{θ}\right)d\mathrm{θ}{∫}_0^{∞}{n}^2\ln \left(1-e^{-hcn/2LkT}\right)dn \\ F=\mathrm{π}kT{∫}_0^{∞}{n}^2\ln \left(1-e^{-hcn/2LkT}\right)dn \end{gathered}$$

Let

$x=\frac{hcn}{2LkT}→dx=\frac{hc}{2LkT}dn$

Thus,

$F=\mathrm{π}kT{\left(\frac{2LkT}{hc}\right)}^3{∫}_0^{∞}{x}^2\ln \left(1-e^{-x}\right)dx$

Volume of the box is:

$F=8\mathrm{π}kTV{\left(\frac{kT}{hc}\right)}^3{∫}_0^{∞}{x}^2\ln \left(1-e^{-x}\right)dx$

Integrate by parts:

$F=8\mathrm{π}kTV{\left(\frac{kT}{hc}\right)}^3\left[{\left.\frac{x^3}{3}\ln \left(1-e^{-x}\right)\right|}_0^{∞}-{∫}_0^{∞}\frac{x^3}{3}\frac{e^{-x}}{1-e^{-x}}\right]$

Substituting values

$$\begin{gathered} F=-8\mathrm{π}kTV{\left(\frac{kT}{hc}\right)}^3{∫}_0^{∞}\frac{x^3}{3}\frac{e^{-x}}{1-e^{-x}} \\ F=-\frac{8\mathrm{π}}{3}V\frac{(kT{)}^4}{(hc{)}^3}{∫}_0^{∞}\frac{x^3{e}^{-x}}{1-e^{-x}} \\ F=-\frac{8\mathrm{π}}{3}V\frac{(kT{)}^4}{(hc{)}^3}{∫}_0^{∞}\frac{x^3}{e^x-1} \end{gathered}$$

Comparing with the equation,

$F=-\frac{1}{3}U$