Read SR1.
The formulas in SR1 are singular if \(v=c\), which is certainly the case for a photon, a quantum of light. We need to rewrite them by getting rid of the velocity. (Or take a careful limit. That's a nice alternative. But we won't do that).
In classical mechanics, this is already done when switching from the Lagrangian to the Hamiltonian. You want to write \(E(v)\) in function of \(p(v)\). So you invert \(p(v)\) as \(v(p)\) and then substitute \(E(v(p))\). Easier done than said:
\[ p_N = mv \Rightarrow\] \[ v = \frac{p_N}{m} \] \[ E_N(v) = \frac{1}{2} m v^2 \Rightarrow E_N(p) = \frac{p_N^2}{2m} \]
Which you'll recognize as a standard kinetic Hamiltonian if you're into Hamiltonian mechanics and you'll ignore this sentence if you don't.
We can do the same with the relativistic case. But first, a neat fact about \(\beta\) and \(\gamma\):
\[\gamma^2 - (\beta \gamma)^2 = 1\]
try it. It's very boring, but it's true. (It's cool because it implies that if \(\gamma = \cosh(\eta)\), then \(\sinh(\eta) = \beta \gamma\) and \(\tanh(\eta)=\beta\) and if you don't think hyperbolic functions are the shit then I don't know what to tell you).
So, we write
\[ \gamma = \sqrt{1+(\beta\gamma)^2} \] \[ E = \gamma m c^2 = \sqrt{1+(\beta\gamma)^2} m c^2 \] \[ = \sqrt{ (mc^2)^2 + (\beta\gamma m c^2)^2} \] \[ = \sqrt{ (mc^2)^2 + (pc)^2 } \]
So this is our \(E(p)\), the so-called "full expression" for the mechanical energy of a relativistic body.
(If you Taylor-expand \(E(p)\) around \(p=0\), you get \(E = mc^2 + \frac{p^2}{2m} + \ldots\;\). Go figure.)
Ok, so we've gotten rid of the gammas and betas. Just a last thing about them! The speed of an object is always recoverable from the energy and momentum:
\[\frac{p}{E} c^2 = \frac{\beta \gamma m c}{\gamma m c^2} c^2 = \beta c = v \]
And only now that we have built this architecture we plug in \(m=0\) to find about massless particles. We get
\[ E = pc \]
and
\[ v = c, \quad \gamma = \infty\]
So massless particles move always at the speed of light and have energy proportional to their momentum. In the limit where the momentum goes to zero, \(p\rightarrow 0\), the energy also goes to zero. Instead, for massive objects the energy tends to the rest energy \(mc^2\). Therefore it makes sense to extend the definition of the rest energy \(E_0 = mc^2\) to photons, with \(m=0\), even if the cannot ever be brought to rest.
The energy of a photon is entirely kinetical.
Note that the previous expressions \(E = \gamma(v) m c^2\) and \(p = \beta(v)\gamma(v) m c\) when \(m=0\) are both indeterminate forms (\(0\cdot\infty\)). This makes sense: we have many photons, all with \(v=c\), with different momentum and energy. We shouldn't be able therefore to determine the energy/momentum exactly just from the speed, so the math honourably breaks down.