# 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.

-
I just faced this situation with my Modern Physics class this week, and in simple truth fudged it. I used (1) and simply waved my hands and said "in the limit of low mass" once or twice. I too, would like a really clean motivation without requiring Hamiltonian mechanics (which first semester juniors have not usually seen). –  dmckee Sep 6 '13 at 20:28
Doesn't it follow that the particle's speed is c when its four-momentum is null (which is the case when $(mc^2)^2 = 0$)? –  Alfred Centauri Sep 6 '13 at 21:00
@dmckee: It seems perfectly reasonable to me to require that a particle with zero mass behave the same as the limiting behavior of a particle with finite mass. Otherwise we would have a magic method for detecting arbitrarily small masses such as $10^{-100000000000}$ kg. We don't actually know that the photon, for example, is massless. All we really have are upper limits on its mass: arxiv.org/abs/0809.1003 . Of course there are theoretical reasons to prefer its mass to be exactly zero, but it's not impossible to construct theories in which its mass is nonzero. –  Ben Crowell Sep 6 '13 at 21:42
@Ben I tried to ask your opinion in chat last week. Not getting it, I went ahead and gave those Modern physics students a simplified version of the argument from arxiv.org/abs/physics/0402024 for the relativistic mass and energy (because (1) I've always found the glancing asymmetric collision argument clunky and (2) I like to differ from the book more than follow it so the students get two different arguments for most things). Any thoughts? Do you have a favorite way (I'd already gotten the Lorentz transform and the velocity composition rule)? –  dmckee Sep 6 '13 at 22:05
@dmckee: Sorry, I didn't see the chat invitation. Very flattering to have my opinion solicited. Thanks for pointing out to me the Sonego and Pin paper. I like the flavor of it and the fact that it works in 1+1 dimensions. But they assume a relation that's equivalent to the work-kinetic energy, and although that's what Einstein did in the 1905 SR paper, I've never seen any coherent justification for why it's OK simply to assume its validity in SR. It also seems a bit cumbersome. The four-vector approach has the advantage that the techniques it introduces are useful in general. Another [...] –  Ben Crowell Sep 6 '13 at 23:28

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.

-
Clever. I don't think my students would like this any more that the limit bit, but it is very slick. –  dmckee Sep 6 '13 at 22:08
This is a cool argument, but doesn't it just replace the original question with a new question, namely "why does the energy-momentum-mass relation you write down hold for massless particles?" –  joshphysics Sep 6 '13 at 22:41
@joshphysics Now that one, I can handle. The relationship between energy, momentum and mass rises out of the transformation properties of the energy and momentum, and while we found those from the explicit forms for massive particles the transformation properties must hold for any other form as well. –  dmckee Sep 6 '13 at 22:58
@dmckee Hmm ok I want to believe this, but I don't entirely follow. I'll think about it more; thanks for the response. –  joshphysics Sep 6 '13 at 23:25
@joshphysics: Your point is valid, but this is a question about the properties of massless particles, and $m^2=E^2-p^2$ is usually taken to be the definition of mass. If we hadn't already established $m^2=E^2-p^2$ or didn't believe that it was applicable to massless particles, then we would need to do something else as a preliminary to define what we meant by "massless." –  Ben Crowell Sep 7 '13 at 0:00

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.

-
I don't think dimensional analysis suffices here. –  Ben Crowell Sep 7 '13 at 16:17