Proper time invariance and conformal symmetry group

Question

Special relativity is based on the fact that the proper time is always the same in any inertial frame:

$$ds^2=(cdt)^2-dx^2=ds'^2=(cdt')^2 -dx'^2 $$

If I understand it correctly this is based on the outcome of the Michelson-Morley experiment. The Michelson-Morley experiment showed that light has the same speed in any inertial frame. And therefore we have $ds^2=0=ds'^2$.

However if we do not have light but a particle which does not propagate with the speed of light we have $ds\neq 0$ and in this case it is not clear to me why we should have $ds^2=ds'^2$.

Weinberg comments about this in his book ... page 27 and 28:

Incidentally, if we had only assumed that the transformations $x\to x'$ leave $ds^2$ invariant when $ds^2=0$, that is, for a particle moving at the speed of light, then we would have found that these transformations are in general non-linear, and form a 15-parameter group, the conformal group, which contains the Lorentz transformations as a subgroup. But the statement that a free particle moves at constant velocity would not be an invariant statement unless the velocity were that of light, and since there are massive particles in the world, we must reject the conformal group as a possible invariance.

I don't understand this... Weinberg is saying in the last sentence that $ds^2$ is not invariant a particle moves not at the speed of light. This is clear but what tells us that $ds^2$ has to be invariant in the case when $ds\neq 0$? What experiment shows this? Weinbergs first sentences sound like that when we use the conformal group then $ds^2=ds'^2$ if $ds^2=0$ but when $ds^2\neq 0$ the we do not have necessarily that $ds^2=ds'^2$...

Answer 1

I will try to rephrase a bit how I understand what Weinberg is saying. A priori we only know what for two different inertial reference frames we have $ds^2 = ds'^2$ when $ds^2 = 0$. We then try to find the most general coordinate transformations of spacetime which possess this property. This way we discover conformal transformations. However, when we try to look what happens under a general conformal transformation to a worldline of a massive particle we find an unpleasant "surprise": it might be that in another "conformal frame" (obtained from the initial frame via a conformal transformation) a free massive particle does not move with constant velocity. But we would like to have the usual property of inertial frames that free massive particles move with constant velocity. The whole conformal group does not satisfy this, so we restrict ourselves to its subgroup which has the desired property, namely Lorentz group which turns out to have $ds'^2 = ds^2$ in any case.

$\textbf{Edit}$: Let us show explicitly that conformal transformations could make constant velocity not constant. For simplicity we will work in 2d (one time and one space dimension) but we will consider conformal transformation which is fully generalizable to higher dimensions (2d is kind of special when it comes to conformal transformations, but we won't focus on these specialities). As an example of such a conformal transformation we will take inversion:

$$x'^\mu = \frac{x^\mu}{x^2}.$$

Assume that in the $x^\mu$ coordinates our particle has the following worldline, parametrized by some $\lambda$:

$$X^\mu = (\lambda, x_0 + \beta \lambda)$$

where $\beta$ is speed of the particle in units of $c$ and $x_0$ is its' position at the moment $t=0$. The above worldline simply means $x(t) = x_0 + \beta t$, so the particle is moving at constant velocity. After a conformal transformation the worldline becomes:

$$X'^\mu = \frac{X^\mu}{X^2} = \frac{X^\mu}{\lambda^2 (1-\beta^2)} = \frac{1}{\lambda^2 (1-\beta^2)} (\lambda, x_0 + \beta \lambda) = (\frac{1}{(1-\beta^2)\lambda}, \beta \frac{1}{(1-\beta^2)\lambda} + \frac{x_0}{(1-\beta^2)\lambda^2}) = (\lambda', \beta \lambda' + x_0 (1-\beta^2)\lambda'^2)$$

where we introduced a new notation: $\lambda' = \dfrac{1}{(1-\beta^2)\lambda}$ which simply has the interpretation of time in the transformed coordinates. So after a conformal transformation our particle moves with acceleration:

$$x'(t') = \beta t' + (1-\beta^2)x_0 t'^2.$$

Answer 2

Let me say some words about how we determine symmetries in physics. If we were to imagine closing our eyes and forgetting we ever saw the world, we would have no idea what symmetry groups might be found when we open our eyes. The flip side of this is that, before we open our eyes, any symmetry group is possible. As soon as we open our eyes and look at things, we can start ruling possibilities out.

So, if we accept that experiment has informed us that $ds^2=0$ is preserved by our symmetry group (but, importantly, the experiment does not tell us anything about whether $ds^2$ is invariant if it isn't zero), then we can ask "what is the largest group of symmetries which preserve $ds^2=0$." The answer, as Weinberg notes, is the conformal group. However, since we don't really know about whether $ds^2$ should be invariant when it's non-zero, all we can say is that our symmetry group should be the conformal group or some subgroup thereof.

There are only so many (continuous) subgroups of the conformal group. Really, the only one is the Poincare group (and subgroups thereof, like Lorentz, rotation, etc). Since there aren't that many options, we can just start working out the consequences of each possibility. Once you do that, you quickly find that conformal is too restrictive and, for example, would not allow QED for instance. So we conclude that we don't have conformal symmetry but rather have it's subgroup, Poincare symmetry.

We can do the same thing for the discrete symmetries and ask whether they are in the symmetry group or not, which is why a lot of particle physics around the 60's and 70's was focused on parity, time reversal, and charge conjugation.