[Physics] Special relativity and massless particles


I encountered an assertion that a massless particle moves with fundamental speed c, and this is the consequence of special relativity. Some authors (such as L. Okun) like to prove this assertion with the following reasoning:

Let's have $$ \mathbf p = m\gamma \mathbf v ,\quad E = mc^{2}\gamma \quad \Rightarrow \quad \mathbf p = \frac{E}{c^{2}}\mathbf v \qquad (.1) $$ and
$$ E^{2} = p^{2}c^{2} + m^{2}c^{4}. \qquad (.2) $$ For the massless
case $(.2)$ gives $p = \frac{E}{c}$. By using $(.1)$ one can see that
$|\mathbf v | = c$.

But to me, this is non-physical reasoning. Relation $(.1)$ is derived from the expressions of impulse and energy for a massive particle, so its scope is limited to massive cases.

We can show that a massless particle moves with the speed of light by introducing the Hamiltonian formalism: for a free particle

H = E = \sqrt{p^{2}c^{2} + m^{2}c^{4}},
for a massless particle
H = pc,
and by using Hamilton's equation, it's easy to show that
\dot {|r|} = \frac{\partial H}{\partial p} = c.
But if I don't want to introduce the Hamiltonian formalism, what can I do to prove an assertion about the speed of a massless particle? Maybe the expression $\mathbf p = \frac{E}{c^{2}}\mathbf v$ can be derived without using the expressions for the massive case? But I can't imagine how to do it by using only SRT.

Best Answer

For the reasons given in the comment above, I think the argument from the $m\rightarrow 0$ limit is valid. But if one doesn't like that, then here is an alternative. Suppose that a massless particle had $v<c$ in the frame of some observer A. Then some other observer B could be at rest relative to the particle. In that observer's frame of reference, the particle's three-momentum $\mathbf{p}$ is zero by symmetry, since there is no preferred direction for it to point. Then $E^2=p^2+m^2$ is zero as well, so the particle's entire energy-momentum four-vector is zero. But a four-vector that vanishes in one frame also vanishes in every other frame. That means we're talking about a particle that can't undergo scattering, emission, or absorption, and is therefore undetectable by any experiment.

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