# In a thermal equilibrium, why is the energy of individual photons proportional to the temperature?

In the book of The First Three Minutes, by Weinberg, he argues that

(p.65-66)when radiation and matter were in thermal equilibrium, the universe must have been filled with black-body radiation with a temperature equal to that of the material contents of the universe.



(p.68)The photon picture allows us easily to understand the chief qualitative properties of black-body radiation. First, the principles of statistical mechanics tell us that the typical photon energy is proportional to the temperature, while Einstein's rule tells us that any photon's wavelength is inversely proportional to the photon energy. Hence, putting these two rules together, the typical wavelength of photons in black-body radiation is inversely proportional to the temperature.

However, I do not understand how the conclusion is arrived at, that typical photons (i.e photons emitted with maximum intensity at given temperature) have an energy that is proportional to the temperature ? Is there any explanation that shows that $E \propto T$ for a photon in thermal equilibrium ?

Note that, I have seen this question, but that does not contain any answer to my question.

Please also do note that, the book is making arguments using the historical development of the subject, it is mostly based on the experimental facts, so please do not throw any quantum theoretical formulas at us.

The frequency at which the flux density (in frequency space) is maximised is given by Wien's law: $$f_{\rm max} = \frac{2.82}{h} k_{B}T$$
Thus a photon at the frequency at which the radiation spectrum peaks has an energy $h f_{\rm max} = 2.82 k_B T$.
You can do some similar analysis to come up with the average energy of a photon, and it comes to $2.70 k_B T$.