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A photon has “zero” rest mass, but it does have Energy,
But with zero Mass , E=MC^2 yields E=0
I am confused !
But with zero Mass , E=MC^2 yields E=0
I am confused !
Your confusion is based on the false assumption that there is only one form of energy.A photon has “zero” rest mass, but it does have Energy,
But with zero Mass , E=MC^2 yields E=0
I am confused !
with zero Mass , E=MC^2 yields E=0
the energy of a massless photon is given by
##E= hf##
This is correct, but not relevant to the OP's question. The relevant formula for the OP's question, for a photon, is ##E = p c## (see above).
We have [itex] \lim_{v\rightarrow c} \gamma=\infty [/itex] and so for a photon, we'll have [itex] p=\infty \times 0 [/itex]. So this formula is compatible with [itex] p=\frac E c=\frac{hf}{c} [/itex].In the spirit of the OP's question:
If ##p=\gamma mv## then the momentum of a photon would likewise be 0 (or undefined).
If ##p=\gamma mv## then the momentum of a photon would likewise be 0 (or undefined).
We have ##\lim_{v\rightarrow c} \gamma=\infty## and so for a photon, we'll have ##p=\infty \times 0## . So this formula is compatible with ##p=\frac{E}{c} = ##\frac{hf}{c}## .
But ##p = \gamma m v## is not valid for a photon.
you can relate the two in ##E=pc## . But that doesn't define either, except in terms of the other.
But that doesn't define either, except in terms of the other
As the third article Shyan linked to says, ##E = m c^2## is not the correct general formula;
This is correct, but not relevant to the OP's question. The relevant formula for the OP's question, for a photon, is ##E = p c## (see above).
Exactly. So, how are you defining the energy or momentum of a photon? You can't use mc2mc^2 or mvmv because they both have mass in them. Yes, you can relate the two in E=pcE=pc But that doesn't define either, except in terms of the other.