18
$\begingroup$

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.


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.


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.

$\endgroup$
6
  • 1
    $\begingroup$ Your expression for dT does not look right. T is a product of two functions of time. $\endgroup$
    – nasu
    Commented Jun 9, 2022 at 10:17
  • $\begingroup$ @nasu Thank you. Please tell me whether my new edited post has any mistakes. $\endgroup$ Commented Jun 9, 2022 at 19:02
  • $\begingroup$ It seems like … intuitively it is a more complicated situation then minikowski space, where typical constant light speed mjnikowski space, is now a section of the new space. $\endgroup$ Commented Jun 10, 2022 at 11:07
  • 1
    $\begingroup$ @Andrew Morton Yes, the speed of light is varied in materials, but I was particularly referring to the speed of light in vacuum. My fault for not making this explicit. As far as we understand, the speed of light in vacuum does not vary. $\endgroup$ Commented Jun 10, 2022 at 23:29
  • 1
    $\begingroup$ @MaximalIdeal I suspect that this kind of thing is somewhere that being explicit is necessary, even if it may be tedious to declare that you mean $c$ from $\mu _0$ and $\epsilon _0$, so as to avoid hassle from people like me who pop up into the aether sometimes. $\endgroup$ Commented Jun 10, 2022 at 23:41

3 Answers 3

24
$\begingroup$

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 :))

$\endgroup$
10
  • 3
    $\begingroup$ @MaximalIdeal There's a nice quote from John Barrow here: en.wikipedia.org/wiki/… $\endgroup$
    – Andrew
    Commented Jun 9, 2022 at 4:42
  • 3
    $\begingroup$ @MaximalIdeal We have some good posts on that topic here, eg physics.stackexchange.com/q/78684/123208 $\endgroup$
    – PM 2Ring
    Commented Jun 9, 2022 at 5:47
  • 1
    $\begingroup$ Someone told they weren't sure if constancy of $\alpha$ was enough to keep physics the same, because there could be other "coupling constants" that might change. $\endgroup$ Commented Jun 9, 2022 at 19:26
  • 3
    $\begingroup$ @MaximalIdeal It's true that $c$ can appear in multiple dimensionless quantities that can vary independently. For example, besides the fine structure constant, there is the ratio of the speed of gravitational waves and the speed of light (which is known to be very close to $1$ because of the multimessenger detection of GW170817). However, it just doesn't mean anything to ask if the speed of light itself changes, because you can always choose units where $c=1$. $\endgroup$
    – Andrew
    Commented Jun 9, 2022 at 19:34
  • 3
    $\begingroup$ @Seb The choice of units does not impact the numerical value of $\alpha$ (nor its physical dimension). But in other unit systems (notably the CGS / Gaussian system of units for electromagnetism, which does not use $\epsilon_0$) the expression for $\alpha$ is different. $\endgroup$ Commented Jun 10, 2022 at 11:15
12
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ I see. This post makes a lot of sense. Maybe I should add that based off of Emilio Pisanty's post here I surmise that we have a choice in terms of whether to say that the speed of light varies or whether some atomic process is making clocks and/or measuring rods vary. Based off of the current BIPM SI unit standards, people would say that our measuring rods would be at fault rather than our clocks or the speed of light. This has been confusing to me, but now I think I understand things much better. $\endgroup$ Commented Jun 9, 2022 at 19:16
  • $\begingroup$ It should also be noted that if such a scenario did happen, we would probably change our SI unit standards anyways. $\endgroup$ Commented Jun 9, 2022 at 19:18
  • $\begingroup$ I would say you don't necessarily need to continuously redefine the second; the issue is more whether you choose to define the second in a way that assumes the speed of light is constant or if you define it without referencing the speed of light directly. The fact that whether the speed of light is constant or not depends on what convention you use to define the second is why I would argue that it's physically meaningless to talk about changes in the speed of light. $\endgroup$
    – Andrew
    Commented Jun 9, 2022 at 19:44
  • $\begingroup$ But isn't this very problematic because the length of your ruler and the period of your atomic clock themselves depend on interactions which are governed by the fundamental constants? $\endgroup$
    – Ryder Rude
    Commented Nov 27, 2022 at 7:20
4
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ I see. This rephrasing is actually is very helpful. Also I really like the linked post. $\endgroup$ Commented Jun 9, 2022 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.