- #1

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I have done some reading but cannot find an answer. From this reading I have got the following:

We have the following standard equation:

E

^{2}= p

^{2}c

^{2}+ m

_{0}

^{2}c

^{4},

where p is the momentum and m

_{0}is the rest mass of the object.

Now for a photon travelling in a vacuum at the speed of light, its energy is hf, where h is Planck's constant and f is the frequency. Also we have that the momentum of a photon in a medium is

p = (hf) / (cn),

where n is the refractive index, given by

n = c / v.

Therefore, by looking at the energy of the photon in a vacuum and in a medium we would get the following:

h

^{2}f

^{2}= h

^{2}f

^{2}c

^{2}/ (c

^{2}n

^{2}) + m

_{0}

^{2}c

^{4}

h

^{2}f

^{2}= h

^{2}f

^{2}v

^{2}/ c

^{2}+ m

_{0}

^{2}c

^{4}

Rearranging for m

_{0}, gives

m

_{0}= (hf / c

^{2}) * sqrt(1 – v

^{2}/c

^{2}).

Hence this implies that when a photon is travelling at the speed of light it would have zero mass. However if it was not travelling at the speed of light, then it would have mass, with its maximum occuring when v=0, i.e. just before the photon is destroyed.

So what are you thoughts regarding this?

P.S. hopefully I have posted this in the correct section.