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Adam Rifai
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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.Adam Rifai said:A photon has “zero” rest mass, but it does have Energy,
But with zero Mass , E=MC^2 yields E=0
I am confused !
Adam Rifai said:with zero Mass , E=MC^2 yields E=0
PeroK said:the energy of a massless photon is given by
##E= hf##
PeterDonis said: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].PeroK said:In the spirit of the OP's question:
If ##p=\gamma mv## then the momentum of a photon would likewise be 0 (or undefined).
PeroK said:If ##p=\gamma mv## then the momentum of a photon would likewise be 0 (or undefined).
Shyan said: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}## .
PeterDonis said:But ##p = \gamma m v## is not valid for a photon.
PeroK said: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
PeterDonis said: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).
PeroK said: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.
E=MC^2 is an equation that relates energy (E) to mass (M) and the speed of light (C). It states that energy and mass are equivalent and can be converted into one another.
E=MC^2 is important because it revolutionized our understanding of the relationship between energy and mass. It also has practical applications, such as in nuclear energy and nuclear weapons.
A photon is a fundamental particle of light that carries energy and has no mass. It is the basic unit of all electromagnetic radiation, including visible light, radio waves, and X-rays.
The equation does not directly explain photon mass because, according to the equation, objects with mass cannot travel at the speed of light. However, the equation can be used to calculate the rest mass of a photon, which is zero, and the energy it carries.
Yes, E=MC^2 can be applied to all forms of energy because it is a universal equation that relates energy and mass. However, it is most commonly used in the context of nuclear reactions and the conversion of mass into energy.