Damien Horton

2022-07-20

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:
$\epsilon =h\nu ={E}_{k}-{E}_{i}$
And since $m=\frac{h\nu }{{c}^{2}}$ then the (relativistic) mass of the photon is
$m=\frac{h\nu }{{c}^{2}}$
How did they deduce that?

Jorge Franklin

Expert

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 ($E=m{c}^{2}$)
$m{c}^{2}=h\nu$
The resultant mass gives the relativistic mass for a moving photon (since photon has zero rest mass)

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