Monday 21 August 2017

special relativity - Expression for the (relativistic) mass of the photon



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?



Answer



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 ($E=mc^2$)


$$mc^2=h\nu$$


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



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