91影视

Number of photons in a photon gas.

Number of photons in equilibrium in a box of volume V at temperature T is

$N=8蟺V{\left(\frac{kT}{hc}\right)}^3{鈭�}_0^{鈭�}\frac{x^2}{e^x-1}dx$

Because we have two polarisation modes, the number of photons equals the total of Planck distributions over all modes ${\mathit{n}}_x,{\mathit{n}}_y\text{and}{\mathit{n}}_z$, multiplied by a factor of two, that is:

$N=2\underset{n_x}{鈭�}\underset{n_y}{鈭�}\underset{n_z}{鈭�}{\stackrel{炉}{n}}_{\mathrm{Pl}}\left(系\right)=\underset{n_x,n_y,n_z}{鈭�}{\stackrel{炉}{n}}_{\mathrm{Pl}}\left(系\right)$

But, ${\stackrel{炉}{n}}_{\mathrm{Pl}}\left(系\right)=\frac{1}{e^{系/kT}-1}$

Consider a cubic box with a volume of V and a side width of L; the photon's permitted energy is:

$系=\frac{hcn}{2L}$

Substitute this the above equation:

$N=2\underset{n_x,n_y,n_z}{鈭�}\frac{1}{e^{hcn/2LkT}-1}$

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 $n^2\sin \left(胃\right)$which gives us:

$N=2{鈭�}_0^{蟺/2}d桅{鈭�}_0^{蟺/2}\sin \left(胃\right)d胃{鈭�}_0^{鈭�}\frac{n^2}{e^{hcn/2LkT}-1}dn$

So,

$N=蟺{鈭�}_0^{鈭�}\frac{n^2}{e^{hcn/2LkT}-1}dn$

Now, let

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

Thus,

$N=蟺{\left(\frac{2LkT}{hc}\right)}^3{鈭�}_0^{鈭�}\frac{x^2}{e^x-1}dx$

The volume of the box is $V=L^3$

$N=8蟺V{\left(\frac{kT}{hc}\right)}^3{鈭�}_0^{鈭�}\frac{x^2}{e^x-1}dx$

Evaluate the integral:

${鈭�}_0^{鈭�}\frac{x^2}{e^x-1}dx$

We get

$x=2.4041$

The number of photons is:

$N=8蟺(2.404)V{\left(\frac{kT}{hc}\right)}^3\left(1\right)$

The entropy is:

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

Hence, entropy per photon is:

$$\begin{gathered} \frac{S}{N}=\frac{\frac{32{蟺}^5}{45}V{\left(\frac{kT}{hc}\right)}^{3}k}{8蟺(2.404)V{\left(\frac{kT}{hc}\right)}^3}=3.60k \\ \frac{S}{N}=3.60k \end{gathered}$$

(c) To find the number of photons per cubic meter at T= 300 K, substitute the values

$$\begin{gathered} {\left.\frac{N}{V}\right|}_{300}=8蟺(2.404){\left(\frac{\left(1.38脳{10}^{-23}\mathrm{J}/\mathrm{K}\right)(300\mathrm{K})}{\left(6.626脳{10}^{-34}\mathrm{J}路\mathrm{s}\right)\left(3.0脳{10}^8\mathrm{m}/\mathrm{s}\right)}\right)}^3 \\ {\left.\frac{N}{V}\right|}_{300}=5.458脳{10}^{14}{\mathrm{m}}^{-3} \\ {\left.\frac{N}{V}\right|}_{1500}=8蟺(2.404){\left(\frac{\left(1.38脳{10}^{-23}\mathrm{J}/\mathrm{K}\right)(300\mathrm{K})}{\left(6.626脳{10}^{-34}\mathrm{J}路\mathrm{s}\right)\left(3.0脳{10}^8\mathrm{m}/\mathrm{s}\right)}\right)}^3 \\ {\left.\frac{N}{V}\right|}_{1500}=6.82脳{10}^{16}{\mathrm{m}}^{-3} \\ {\left.\frac{N}{V}\right|}_{2.73}=8蟺(2.404){\left(\frac{\left(1.38脳{10}^{-23}\mathrm{J}/\mathrm{K}\right)(300\mathrm{K})}{\left(6.626脳{10}^{-34}\mathrm{J}路\mathrm{s}\right)\left(3.0脳{10}^8\mathrm{m}/\mathrm{s}\right)}\right)}^3 \\ {\left.\frac{N}{V}\right|}_{2.73}=4.11脳{10}^8{\mathrm{m}}^{-3} \end{gathered}$$