Before we discuss massless particles in particular, let’s first discuss general statements within the context of special relativity that are true for all particles regardless of their mass.
A particle is moving with some velocity vector given by \vec{v} = c \vec{\beta} where c is the speed of light in vacuum and \beta measures the particle’s speed relative to the speed of light in vacuum. A stationary particle has \beta = 0 whereas a particle moving at the speed of light has \beta = 1.
The gamma factor is given by \gamma = 1/\sqrt{1 - \beta^2}. A stationary particle has \gamma = 1 and a particle moving at the speed of light has \gamma \rightarrow \infty.
The 4-momentum of a particle is given by p^\mu = (E/c,\vec{p}) where E is the energy of the particle and \vec{p} is the regular 3-momentum vector of the particle. The invariant “length” squared of the particle’s 4-momentum is given by p^\mu p_\mu = (E/c)^2 - \vec{p}\cdot \vec{p} = (m c)^2 where m is the rest mass of the particle. The energy of a particle is the quadrature sum of its rest mass and its momentum taking in account the appropriate factors of c needed to make the units work out. The kinetic energy of a particle T is energy of the particle E minus the energy corresponding to its rest mass: T = E - mc^2. Up to this point, all of the above equations are true for all particles regardless of their rest mass m.
Now let’s focus on the definitions for the energy and momentum of a particle in terms of their dependence on the particle’s speed. The energy and momentum of a massive particle (i.e. nonzero rest mass m > 0) can be written as E = \gamma m c^2 and \vec{p} = \gamma m c \vec{\beta}. What happens to these equations when the particle is massless (i.e. the rest mass is zero m=0). If we plug in m=0 into these two equations, then they seem to imply that the energy and momentum for a massless particle are both zero! This should feel wrong to you and, if you feel this way, then you are correct! From quantum mechanics, we know that the photon (a massless particle) has an energy related to its wavelength \lambda given by E = hc/\lambda. How do we reconcile these two equations for the energy?
To catch a glimpse that we need to be careful, we’ll first note that a photon is a massless particle which travels a the speed of light. This implies that \beta = 1 and \gamma \rightarrow \infty for a photon. Both the energy and momentum equations involve the product \gamma m and herein lies the problem. For a photon, this product \gamma m equals infinity times zero which is formally mathematically undefined. In a physics context, where we are free to take off our math nerd hat, we argue that the product \gamma m is just a nonzero constant \gamma m \rightarrow (\infty) (0) ``='' \mathrm{constant} \ge 0 . But what is the value of this constant? According to our energy equation, E = \gamma m c^2. Let us now isolate the \gamma m product and we find that it is equal to \gamma m = E/c^2. We will use this equation to define the desired constant \lim_{m \rightarrow 0} \lim_{\gamma \rightarrow \infty} \gamma m = \mathrm{constant} \equiv E/c^2. If we plug this into our energy equation, we find E = \gamma m c^2 = (E/c^2)c^2 = E which really feels like a circular waste of time. Don’t worry about it - let’s just charge ahead and plug this into the momentum equation: \vec{p} = \gamma m c \vec{\beta} = (E/c^2) c \vec{\beta} = (E/c) \vec{\beta}. Note that photons travel at the speed of light so \beta = 1 and consequently their momentum is given by \vec{p} = (E/c) \vec{\beta} = (E/c) \beta \hat{\beta} = (E/c) \hat{\beta}. The \hat{\beta} is just a unit vector that points in the direction of the photon’s motion.
Now what if our massless particle was not a photon? It turns out that, in the context of special relativity, a massless particle by definition travels at the speed of light in all reference frames. Think about this last statement which is actually quite profound! A photon is just a special case of a massless particle.
In summary, the following are always true for all particles regardless of their rest mass m:
p^\mu = (E/c,\vec{p})
p^\mu p_\mu = (E/c)^2 - \vec{p}\cdot \vec{p} = (m c)^2
\vec{\beta} = c \vec{p}/E
T = E - mc^2
For massive particles only (i.e. rest mass is nonzero m>0):
0 \le \beta < 1
1 \le \gamma < \infty
\vec{p} = \gamma m c \vec{\beta}
E = \gamma m c^2
T = E - mc^2
For massless particles only (i.e. rest mass is zero m=0):
\beta = 1
\gamma \rightarrow \infty
\vec{p} = (E/c) \hat{\beta}
E = pc
T = E