# Photon Mass in E=MC^2

## Main Question or Discussion Point

A photon has “zero” rest mass, but it does have Energy,

But with zero Mass , E=MC^2 yields E=0

I am confused !

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ShayanJ
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PeroK
Homework Helper
Gold Member
A photon has “zero” rest mass, but it does have Energy,

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.

The energy of a massive particle is actually

$E = \gamma mc^2$

Whereas the energy of a massless photon is given by

$E= hf$

PeterDonis
Mentor
2019 Award
with zero Mass , E=MC^2 yields E=0
As the third article Shyan linked to says, $E = m c^2$ is not the correct general formula; the correct general formula is $E^2 = \left( m c^2 \right)^2 + \left( p c \right)^2$. $E = m c^2$ is then the obvious special case of this for which $p = 0$. But a photon can't have $p = 0$, because it can never be a rest; it always moves at the speed of light, so it always has momentum. In fact, as you can see from considering the general formula when $m = 0$, a photon's momentum is just its energy divided by $c$.

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).

PeroK
Homework Helper
Gold Member
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).
In the spirit of the OP's question:

If $p=\gamma mv$ then the momentum of a photon would likewise be 0 (or undefined).

ShayanJ
Gold Member
In the spirit of the OP's question:

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 $p=\frac E c=\frac{hf}{c}$.

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PeterDonis
Mentor
2019 Award
If $p=\gamma mv$ then the momentum of a photon would likewise be 0 (or undefined).
But $p = \gamma m v$ is not valid for a photon. Just to be clear: the fully general formula is $E^2 = \left( m c^2 \right)^2 + \left( p c \right)^2$. Any formula with $\gamma$ in it is not fully general; it only applies to objects with $v < c$, for which $\gamma$ is finite.

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}$. No, it isn't.$\infty \times 0$is not well-defined mathematically; there's no way to make it spit out a particular finite value, much less the correct finite value for$p$for a photon.$p = \gamma m v$doesn't apply to a photon, period. PeroK Science Advisor Homework Helper Gold Member But$p = \gamma m v$is not valid for a photon. Exactly. So, how are you defining the energy or momentum of a photon? You can't use$mc^2$or$mv$because they both have mass in them. Yes, you can relate the two in$E=pc$But that doesn't define either, except in terms of the other. PeterDonis Mentor 2019 Award you can relate the two in$E=pc$. But that doesn't define either, except in terms of the other. So what? The OP didn't ask how energy and momentum were defined for photons. He only asked how the photon's energy can be nonzero when it has zero invariant mass. That question has been answered. If you want to ask how energy and momentum are defined for photons, the answer is, the same way they are defined for anything else: energy is defined by the photon's ability to do work or transfer heat (for example, by being absorbed by an object and increasing its temperature), and momentum is defined by the photon's ability to exert force (for example, radiation pressure). The formula$E = p c$then simply relates the amount of heat or work a photon can produce, to the amount of force it can produce. The more general formula I wrote down does the same thing for any particle, whether it has zero or positive invariant mass. ChrisVer Gold Member But that doesn't define either, except in terms of the other Well the photon has definite momentum and energy. The problem is the definition of $\gamma$ when you go to null-worldlines. The problem there is that $\gamma$ is not definite. Thus the energy can't be written as $E= \gamma m$ and momentum $p_i= \gamma m u_i$ (or in other words define them with using the rest mass). The photon however can have energy (and has energy) and momentum (for the whole one can check the energy-stress tensor which can be obtained from Maxwell's theory itself). Now let's say it has energy $E$ and momentum $p$. It's 4 momentum is $P=(E, p)$ and the invariant object coming from this (as a null vector) is $P^2 = E^2 - p^2 =0 \Rightarrow E= pc$ (I inserted c to keep the SI units). If you apply the null-condition (or massless) for the four momentum, you will eventually reach the relation $p_0^2=\vec{p} \cdot \vec{p}$ no matter what you do. This is how a degree of freedom is lost/is not physical. 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.
Thanks, i think E=pc is more relevant answer to my question about the mass-less rest-less photon