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I started learning a bit ahead from an old physics book, and they were discussing the photoelectric effect and after that Planck's hypotheses and energy quantas.

The book said that the mass of a microscopic oscillator (what is that?) is not continuous, but discrete and the difference between states is an energy quanta:

$ \varepsilon = h\nu = E_k - E_i $

And since $ E = mc^2 $ then the (relativistic) mass of the photon is

$ m = \frac{h\nu}{c^2} $

How did they deduce that?

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closed as too localized by Qmechanic, Manishearth, Waffle's Crazy Peanut, Emilio Pisanty, Sklivvz Dec 26 '12 at 13:08

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It seems that in my book there has been a misprint, as the mass got rendered as $h\nu c^{-3}$ instead of $h\nu c^{-2}$. In that case I see how they deduced that. – andreas.vitikan Aug 13 '12 at 9:59
up vote 1 down vote accepted

This is probably related to the derivation of de-Broglie wavelength... Since photon has wave-particle duality,

We could equate Planck's quantum theory (wave nature) which gives the expression for energy of a wave of frequency $\nu$, ($E=h\nu$) with Einstein's mass-energy equivalence (particle nature) which gives relativistic energy for photon ($E=mc^2$)


The resultant mass gives the relativistic mass for a moving photon (since photon has zero rest mass)

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