Momentum of massless particles

Question

If $E=pc$ for massless particles, then it should be $p=0$ as $p=mv$ and $m=0$. Why do we use the equation $E=pc$ for massless particles?

Answer 1

The momentum of a massless particle is not $p=mv$. In special relativity, even for a massive particle, the 3-momentum is not given by mv, but rather

$$p^{a} = \frac{mv^{a}}{\sqrt{1 - \left(\frac{v}{c}\right)^{2}}}$$

The right hand side of this takes an indeterminate value (0/0 division) if $v$ goes to $c$ as $m$ goes to $0$.

All of this is a longwinded way of saying, "for a massless particle, the magnitude of the momentum is equal to the energy divided by $c$.

There are a large number of ways to work out why these formulae are the way that they are, but perhaps the most direct is to remember that we got Newtonian energy from the work-energy theorem, which said

$$\Delta KE = \int {\vec F}\cdot d{\vec s}$$

If we use this relationship, along with the appropriate generalization of Newton's second law:

$$F^{a} = \frac{dp^{a}}{dt}$$

you can work out these formulae.