What laws prevent massive particles to travel at the speed of light?

Question

I and a few of my friends have come across an interesting question.

Jackson talks about the case where photon has non-zero mass. By adding an extra term to the Lagrangian, he shows how Maxwell's equations and Coulomb's law will be changed if we assume photon to be massive. Relativity puts no constraint on the mass of the photon and non-zero mass for photon is consistent. So we are basically lead to this: If Coulomb's law is true, then we can say that mass of the photon is zero.

So, what we were wondering is that is there any other law if we assume to be right leads us to the fact that other massive particles must travel at speeds $<c$? I can think of a trivial law but I'm looking for something more satisfying.

I ask this question because I don't see any contradiction if a massive particle moves at $c$. Sure, if you want to accelerate it to $c$ from a lower velocity, the process requires infinite energy and momentum but we can easily circumvent this by assuming that they are created at that velocity.

So, basically the question is: why can't massive particles go at $c$? What law if assumed to be true constraints massive particle's speed to $<c$?

I hope nobody replies saying that it takes "infinite energy/momentum" if you put $c$ in denominator. That formula is valid only in the case $v<c$.

Answer 1

You seem to understand well the mathematical models that we use and describe these particles, and then we say wow, these models are perfectly describing reality, because they are all justified by our experiments, so massive particles can never travel at the speed of light in vacuum, that is what we see from the experiments, and the mathematical models show as you say a contradiction too when we talk about massive particles traveling at the speed of light in vacuum. But that is not what you are asking.

Then you see phrases where we say "you need infinite energy to speed up a massive particle to the speed of light in vacuum". It is true, and mathematically justified, but that is just not what you are asking for.

You are asking "what law prevents them from being created at c.", and what you are looking for, is called the Higgs mechanism.

The Higgs mechanism is a way of saying, that there is something, a field, that (just like other fields) permeates all of space, and interacts with certain particles. This mechanism (or a way of expressing another physical law that you are looking for), is what differentiates massive and massless particles, and interacts (couples to) with the former but not with the latter, creating a phenomenon that we see in our experiments as a law that says, massive particles cannot move at the speed of light in vacuum.

The Higgs field is another quantum field, and the Higgs field and electron field interact. That means you cannot just write an electron just as an excitation of the electron field, but instead it has to be written as an excitation of both the electron and Higgs fields together. Because the interaction is relatively small we can write the excitation as a slightly perturbed electron field excitation, that is we write it as an excitation of the electron field plus a bit of the Higgs field. If we now calculate how this excitation propagates we find it travels at less than the speed of light i.e. the excitation of the combined fields has a mass. The amount of mass is proportional to the strength of the interaction between the electron and Higgs fields.

Is this a good explanation of the Higgs mechanism?

https://en.wikipedia.org/wiki/Higgs_mechanism

Now we do not exactly know how, but this interaction with the Higgs field, this Higgs mechanism is somehow causing these particles (that we subsequently call massive) to travel always at speeds slower then the speed of light in vacuum.

Please note that:

1. for neutrinos, we still do not exactly know how (through what mechanism) they gain their rest masses

2. your question is (I assumed you are asking about vacuum) only true in vacuum. massive particles can and do travel faster then light in certain media

Answer 2

So, basically the question is: why can't massive particles go at c? What law if assumed to be true constraints massive particle's speed to <c?

I am not sure that this is a “law” in the sense you mean, it is actually just mathematics. In units where c=1 we have the following two equations that always hold for all particles (massive, massless, and even hypothetical tachyons): $$\vec v= \frac{\vec p}{E}$$ $$m^2=E^2-\vec p^2$$

If we set $v=1$ in the first equation then we get immediately $E=p$. Substituting that into the second gives $m=0$.

Although the infinite energy for accelerating it is the typical problem identified, it is not the only problem. A particle with $m>0$ and $v=1$ is mathematically inconsistent. But it sounds more exciting to talk about infinite energy than mathematical inconsistency. Hence the “coverage” goes to the more exciting reason. But again, that is not the only issue. The mathematical inconsistency is unavoidable, regardless of how it would arrive to that state.

Answer 3

Einstein's law $E = m_0 c^2/\sqrt{1 - \frac{v^2}{c^2}}$ only tells what is the energy of a massive particle at a particular velocity $v$. where $m_0$ is the rest mass and $c$ is the speed of light in vacuum. Surely this does not say that a massive particle cannot move with velocity $c$. It simply tells that the energy of such a massive particle is infinite. Thus I can agree with you that special relativity does not prevent a massive particle to have velocity $c$. However the law that does prevent this is the good old energy conservation law. Consider an inelastic collision process, $a + b \rightarrow m + n$. Let us assume particle $a,b,n$ are of finite mass and are moving with velocity $v<c$. But the particle $m$ has finite mass and moves with velocity $c$. So the energy conservation law says,

$$M_a c^2 + M_b c^2 = M_m c^2 + M_n c^2.$$

This implies,

$$M_m c^2 = M_a c^2 + M_b c^2 - M_n c^2.$$

But by Einstein's mass energy equivalence law, we have the energy of the particle $m$ given by, $E = M_mc^2 = \infty$. Thus the L.H.S. of last equation is infinite whereas, the R.H.S. is bound to be finite. This certainly cannot be and thus violates the mass-energy conservation law. Therefore unless you start with a particle of infinite energy, you cannot balance the equation and hence cannot create any particles with your desired specification subsequently.

Not sure if this answer meets your criterion. But this is the best I could do.

Answer 4

If particle has mass and still moves at c, the four momentum vector is timelike and hence there exists a frame where it's momentum is zero. This means when that particle is created from some reaction, there exists a frame where it's momentum is zero but travels with speed c. So then direction of it's motion can be anything. This seems really weird (though i can't see any inconsistency).

This seems like a dead end. So maybe the following can be thought of a law: Particles should have fixed direction of motion when created (when all products have a fixed arbitrary momentum). This is an experimentally verifiable law. This is probably just a restatement of the question but this seems more satisfying. This is what I think without the knowledge of qft.

Answer 5

We can see that massive particles should have $v<c$ from two perspectives.

1. Let us define the velocity as the infinitesimal change in position in infinitesimal time: $dx=vdt$. Then the Lorentz invariant $dx_\mu dx^\mu=-c^2dt^2+dx^2=-(c^2-v^2)dx^2$ is zero or not depending on whether $v=c$ or $v\ne c$.

Let us now consider 4-momenta $p^\mu$. On physical grounds, we expect momenta to be parallel to the infinitesimal change in position hence $p^\mu\sim dx^\mu$ (or rather $p^\mu d\tau\sim dx^\mu$ for proper time $\tau$), hence we get $$p_\mu p^\mu\sim dx_\mu dx^\mu=-(c^2-v^2)dx^2$$

If we define mass $m$ by the equation $p_\mu p^\mu=-c^2m^2$, then requiring $m>0$ imposes that $v<c$.

2. Let us define velocity as $$v_i=\frac{\partial E}{\partial p^i}$$ from $p_\mu p^\mu=-c^{-2}E^2+p^2=-c^2m^2$, we get $E=\sqrt{c^4m^2+c^2p^2}$ (ignore negative energy solution for our purposes), hence $$v=\frac{\partial E}{\partial p}=\frac{p}{\sqrt{c^2m^2+p^2}}c$$ We see that $m>0$ forces $v<c$ with the limit $m\rightarrow 0$ yielding $v=c$.

For small momenta, we can Taylor-expand the expression above to find $$v=\frac{p}{m}\left(1-\frac{p^2}{2c^2m^2}+\mathcal{O}\left(\frac{p}{mc}\right)^4\right)$$ which matches the standard relation between velocity and momentum, i.e. $v=\frac{p}{m}$, in low energy limit.