Special relativity and massless particles

Question

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.

Answer 1

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.

Answer 2

One way is to consider that Lorentz transformations apply more fundamentally to momentum/ energy, than to space/time. So, with a boost transformation along the $z$ axis, we will have :

$\begin {pmatrix} p'_z \\E' \end{pmatrix} = \gamma(v)\begin {pmatrix} 1 & -\frac{v}{c} \\-\frac{v}{c} &1\end{pmatrix}\begin {pmatrix} p_z \\E \end{pmatrix}$

It is not difficult to see that, if, $|\large \frac{p_zc}{E}|=1$, then $|\large \frac{p'_zc}{E'}|=1$

Now, by dimensional analysis, we have : $\frac{\vec Pc}{E} = \frac{\vec V}{c}$, where $\vec V$ has the dimension of a velocity. The most natural possibility is that $\vec V$ is the velocity of the particle (for instance, with $v=V_z$, you have $V'_z=0$). So, we see, that a particle with speed $|\vec V|=c$ in a Galilean frame, has also $|\vec V'|=c$ in another Galilean frame. We could also check that the quantity $E^2-\vec p^2c^2$ is conserved by Lorentz transformations, and call this quantity $m^2c^4$, where $m$ is the mass. So, particles who have $\left|\frac{\vec Pc}{E}\right| = \left|\frac{\vec V}{c}\right|=1$, are massless particles.