[Physics] the distance between photons

electromagnetic-radiationphotonsthermal-radiationthermodynamicswavelength

In the book of The First Three Minutes by Weinberg, at page 69, he states that

We also saw that the decrease in the energy density of black-body
radiation at long wavelengths is due to the diffi- culty of putting
radiation in any volume whose dimensions are smaller than a
wavelength. In fact, the average distance between photons in
black-body radiation is roughly equal to the typical photon wavelength

However, how can we talk about "the distance between photons" ? What is the author referring to ? and what is the physical explanation of his claim ?

Best Answer

The average distance between photons can be estimated in the following conceptual way.

First, calculate the energy density of the blackbody radiation field.

Then divide by the average energy of a photon (which is $2.7 k_BT$ - refer to my answer to In a thermal equilibrium, why do individual photons energy is proportional to the temperature?).

The result is the number density (per unit volume) of photons. The inverse of this gives the volume occupied by one photon and so the cube root of this gives an estimate for the average separation between photons.

More formally:

The energy density of a blackbody radiation field is $$u_{\nu} = \frac{4\pi}{c} B_{\nu}\ d\nu = \frac{8\pi h \nu^{3}}{c^3} \frac{d\nu}{\exp(h\nu/k_BT) - 1}$$ The number density of photons is therefore $$ n = \int \frac{u_{\nu}}{h\nu}\ d\nu = \frac{8\pi}{c^3} \int \frac{\nu^2\ d\nu}{\exp(h\nu/k_BT) - 1} $$ This integral needs the help of a set of integral tables or Wolfram integrator to give $$ n = 8\pi \left(\frac{k_BT}{hc}\right)^3 \times 2.404 $$

But from Wien's law we know that a "typical" photon wavelength (well, the wavelength of a photon at the peak of the Planck distribution) has $$\lambda_{\rm Wien} = \frac{2.9\times 10^{-3}}{T}$$

Putting these together we can express number density in terms of the Wien wavelength, rather than temperature, as $$ n = 8\pi (\frac{0.20}{\lambda_{\rm Wien}})^3 \times 2.404$$
Taking the inverse cube root we finally arrive at a typical photon separation of $$ l_{\rm photon} \simeq 1.27 \lambda_{\rm Wien}$$