What exactly would it take to show (hypothetically) that the speed of light (in vacuum) is not constant?

Question

Consider Minkowski space, which is determined by the spacetime line element $$ ds^{2} = -c^{2}dt^{2} + dx^{2} + dy^{2} + dz^{2}. $$ Now suppose we modify this so that the $c$ constant is not a constant, but a function of time with respect to a fixed reference frame, so instead of the above we have $$ ds^{2} = -c(t)^{2}dt^{2} + dx^{2} + dy^{2} + dz^{2}. $$ Now an interesting observation here is that we can rescale the time coordinate so that the prefactor for the time element can be made constant again. In other words, we can define another notion of time in which the speed of light is constant.

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In more detail, let $c_{0}$ be the usual speed of light constant and let $c(t)$ be the speed of light function from above. Then let $f$ be any solution of the first-order ODE $$ f'(t)t + f(t) = \frac{c(t)}{c_{0}}. $$ Then take the change of coordinates by $$ T = f(t)t, \qquad X = x, \qquad Y = y, \qquad Z = z. $$ Then $$ \frac{dT}{dt} = f'(t)t + f(t) \implies dT = (f'(t)t + f(t))dt \implies c_{0}dT = c(t) dt. $$ In the new coordinates, we find $$ ds^{2} = -c_{0}^{2} dT^{2} + dX^{2} + dY^{2} + dZ^{2} $$ and in these new coordinates it seems as though the speed of light is constant.

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This brings me to the question, what exactly would it even mean for the $c$ constant to vary? It seems like even if we allow $c$ to vary, there is still a coordinate system where $c$ is constant. I find this to be very confusing, and I am wondering if anyone can help clear this confusion.

Answer 1

You are correct and in good modern treatments, people are careful to say that it's only meaningful to say that dimensionless quantities change with time. The "grown up version" of looking at variations of the speed of light is to look at variations in the fine structure constant $\alpha$, which (in SI units) is given by \begin{equation} \alpha = \frac{1}{4\pi \epsilon_0} \frac{e^2}{\hbar c} \end{equation} According to wikipedia, there is a bound on the time variation of $\alpha$ to be less than one part in $10^{17}$ per year based on precision measurements of optical clocks. I do not claim this is the best available bound at the time I am writing this answer, however; just one that I could find with a quick google search. To answer your question more directly, of course those experiments placing an upper bound of the time variation of $\alpha$ could also (theoretically) detect time variation of $\alpha$, which would imply that at least one of $e$, $\hbar$, $\epsilon_0$, or $c$ must change with time (we can hopefully agree that $1$, $4$, and $\pi$ are safe :))

Answer 2

In the Minkowski metric the $dt$ is the time I measure on my clock and the $dx$, $dy$ and $dz$ are the distances I measure with my ruler. That's how those coordinates are defined.

Suppose we choose our axes so the light is travelling along the $x$ axis then I can measure the distance $dx$ that the light travels in a time $dt$ then use the fact light travels on a null geodesic so:

$$ 0 = c^2 dt^2 - dx^2 $$

and find:

$$ \frac{dx}{dt} = c $$

And since $c$ is a constant the speed I measure is constant. We can of course do the same with your new metric and we'd get:

$$ \frac{dX}{dT} = c_0 $$

But $dT$ is not the time I measure on my clock so $c_0$ is not the speed I measure. Although you've managed to find coordinates where $dX/dT$ is a constant that doesn't mean the speed observers will measure for a light beam is constant. To make the measured speed a constant $c_0$ I would need to be continually changing my definition of the second, which is the point made by Andrew.

Answer 3

The answer was already said by others, but I will rephrase it a little differently:

Spacetime metric is not the end. You also need other physics than just structure of spacetime, like, for example, equations of motion.

Yes, you can define time by keeping speed of light constant (assuming the speed of light depends only on time), but what would this do to the rest of the physics? Check for example this answer.