## anonymous one year ago Why Photon has zero rest mass?

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1. anonymous

A photon does not exist at rest, so you can't use the usual definition of mass (the energy of a particle in its own proper reference frame). But the following relation holds (in natural units, so I'm dropping c's): E^2 = m^2 + p^2 For any particle in its own frame of reference, p=0, so E=m as advertised. For a photon, E=p as far as we can tell. Therefore, the mass is zero (or pretty close to it). The best evidence for the photon's lack of mass comes from the nature of electromagnetic interactions. If the photon has any mass whatsoever, the forces fall off faster than 1/r^2. We don't see this which allows us to place extremely small upper limits on the mass of the photon.

2. anonymous

Or simply - y Special Relativity: Nothing that has mass can travel at the speed of light. Photons travel at the speed of light. Therefore, the photon rest mass must be zero. Photons are pure energy. Think of a nuclear reaction, fission or fusion, some of the mass was converted to energy by E = Mc^2. That mass "disappeared," it cannot re-appear as mass. You can have one or the other but not both at the same time. It is a direct result of the Conservation of Matter/Energy. If it is energy it is NOT matter. If it is NOT matter, it has no mass.

3. anonymous

Medal if this helped!

4. anonymous

I am assuming you considered c = 1? You better mention that, cause usually you see equations like E = mc^2 + p^2 ... (just saying :P) Also its technically wrong to say the mass got converted to energy!

5. anonymous

$m=\frac{m_{o}}{\sqrt{1-\frac{v^2}{c^2}}}$ Where mo is rest mass of photon and m is relativistic mass of photon that varies with velocity but we know that a photon moves at the speed of light, so v=c $m=\frac{m_{o}}{\sqrt{1-\frac{c^2}{c^2}}}=\frac{m_{o}}{\sqrt{1-1}}=\frac{m_{o}}{0}$ $\implies m_{o}=0$ Although this is still full of mystery, divison by 0 is not well defined

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