# Total number of photons per unit volume in a box (extremely confusing)

This is a worked example from a text.

a) Find an expression for the number of photons per unit volume with energies between $E$ and $E+dE$ in a cavity at temperature $T$. $$n(E)dE = g(E)f(E)dE = \frac{8 \pi E^{2}dE}{(hc)^{3}(e^{(\frac{E}{K_{B}T})}-1)}$$

The second part confuses me very badly, to the point of it being disturbing.

b) Find an expression for the total number of photons per unit volume (all energies) $$\frac{N}{V} = \int_{\infty}^{0}n(E)dE = \frac{8 \pi ((K_{B}T))^{3}}{(hc)^{3}}\int_{0}^{\infty}\frac{(\frac{E}{K_{B}})^{2}(\frac{dE}{K_BT})} {e^{\frac{E}{K_{B}T}}-1}$$

Where did $$((K_{B}T))^{3}$$ come from and why is the numerator divided by $$K_{B}T$$ Obviously, the energy of a photon is $E=hf$ and not $$K_{B}T$$ and even if it was I can see that constants are moved outside of the integral operator.

Any help is going to be tremendously helpful!

• an Edit was made to a typing error – Physkid Jun 18 '15 at 9:37
• what is the text? Point b) looks like typo – aaaaaa Jun 18 '15 at 9:51
• Have you looked at Planck's law, eg on Wikipedia. That should help. – Dr Chuck Jun 18 '15 at 9:51
• Looks like a typo: there should be a factor $1/T$ in the $(E/k_B)^2$ bracket in the numerator of the integrand. The brackets will then contain $(E/k_BT)^2$, which coupled with the $1/(k_BT)$ underneath the $dE$ exactly cancels the factor of $(k_BT)^3$. – tok3rat0r Jun 18 '15 at 11:23
• (The point is, in some sense, to 'simplify' the integral so that it looks like the integral with respect to $x$ of $x^2/(e^x-1)$, where $x=E/(k_BT)$) – tok3rat0r Jun 18 '15 at 11:25

The $kT$ comes from the Bose-Einstein statistics. The photons are governed by this statistic, nothing suspicious is here. The power three appears when we go form the variable $E$ to the variable $E/kT$ in the integral.