Massless particles

From the expression for the total energy

$$E = \frac{m_0c^2}{\sqrt{1-v^2/c^2}}$$ (53)

and the momentum

$$p = \frac{m_0v}{\sqrt{1-v^2/c^2}}$$ (54)

of a relativistic particle, it is clear that

$$E = p = o \qquad \mbox{if} \qquad m_0 = 0 \qquad \mbox{and} \qquad c\ne 0$$ (55)

except if the particle travels with the speed of light ($v=c$), as then

$$E = \frac{0}{0} \qquad \mbox{and} \qquad p = \frac{0}{0}$$ (56)

which are indeterminate (can have any value). Thus, the only particles that can have zero rest mass are those which travel at the speed of light. In addition we can rewrite the last two equations as the single equation

$$E^2 - p^2c^2 = m_0^2c^4$$ (57)

Exercise 1.14
Show this.

Thus

$$\mbox{\bf Massive particles} \qquad E^2 = m_0^2c^4 + p^2c^2$$ (58)

and

$$\mbox{\bf Massless particles} \qquad E = pc$$ (59)

The photon ( $E_{\gamma} = hf$) is a particle with zero rest mass but which nonetheless carries a momentum $E_{\gamma} = pc$.

Exercise 1.15
Show the consistency of this with de Broglie's relation and ruminate on your introduction to quantum mechanics via the wave particle-duality in your 1st year of physics.