Photon Mass in E=MC^2

The point is that formula is applicable to rest energy. Photons have zero rest energy but its better to say their rest energy is not defined because there is no frame in which they're at rest.
See threads below:
https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/
https://www.physicsforums.com/threads/what-is-relativistic-mass-and-why-is-it-not-used-much.783220/ [Broken]
https://www.physicsforums.com/threads/what-is-the-massenergy-equivalence.763067/

 

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##

 

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 [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].

 

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.

 

Well the photon has definite momentum and energy. The problem is the definition of [itex]\gamma[/itex] when you go to null-worldlines. The problem there is that [itex]\gamma[/itex] is not definite. Thus the energy can't be written as [itex]E= \gamma m [/itex] and momentum [itex]p_i= \gamma m u_i[/itex] (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 [itex]E[/itex] and momentum [itex]p[/itex]. It's 4 momentum is [itex]P=(E, p)[/itex] and the invariant object coming from this (as a null vector) is [itex]P^2 = E^2 - p^2 =0 \Rightarrow E= pc[/itex] (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 [itex]p_0^2=\vec{p} \cdot \vec{p}[/itex] no matter what you do. This is how a degree of freedom is lost/is not physical.