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Recently I came across a video, which states that it is impossible to experimentally measure the one-way speed of light. It was said that any attempts at directly measuring the one-way speed were actually measuring the two-way speed of light.

But I believe that the one way speed of light, being a scalar (independent of reference frame) arises from Electrodynamics. Classical Electrodynamics states that the speed of light must be $$c=\frac{1}{\sqrt{\mu_o\epsilon_o}}.$$ And hence if the one-way speed of light has to be a scalar, it means that both $\mu_o$ and $\epsilon_o$ have to be scalars.

My question is that if there is experimental proof of $\mu_o$ and $\epsilon_o$ being scalars, and if there are such experiments, can they be considered as proof for the one-way speed of light to be a scalar.

Qmechanic
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SK Dash
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2 Answers2

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But I believe that the one way speed of light, being a scalar(independent of reference frame) arises from Electrodynamics. Classical Electrodynamics states that the speed of light must be

You are saying that it is a scalar and independent of frames, but this thing cannot be said by using only electrodynamics. You have to supplement Maxwell's equations with some other kinematical laws to speak about this. For example $$\frac{E^2-(pc)^2}{c^4}=m^2 $$ is a scalar according to special relativity but it is not a scalar according to Newton's laws (supplemented with Galilean relativity). Using Newton's laws(supplemented with Galilean relativity) we expect that the 1-way and 2-way speed's both should be same. But Newton's laws are inconsistent with Maxwell's equations. To make it consistent we need to use special relativity. But due to the way we define synchronisation in special relativity we cannot find the 1-way speed of light.

My question is that if there is experimental proof of $μ_o$ and $ϵ_o$ being scalars, and if there are such experiments, can they be considered as proof for the one-way speed of light to be a scalar.

Even if we know that Maxwell's equations are 100% correct we cannot expect that 1-way speed of light is equal to 2-way speed of light.

Edit: If you are thinking that since electromagnetic wave equations are of the form $$\ddot{\textbf{E}}=c^2\nabla^2 {\textbf{E}}$$ $$\ddot{\textbf{B}}=c^2\nabla^2 {\textbf{B}}$$ and these 2 can be obtained from Maxwell's equations in vacuum for $c=\frac{1}{\sqrt{\mu_0\epsilon_0}}$ then you are already assuming that the wave is such that the 1-way speed and 2-way speed are both same. Since the wave equations are linear we can add several solution of it and find a solution moving at different speed. For example if we add two wave equations which are moving in opposite directions, we can get a solution which is a standing wave, that doesn't move at the speed $c$. So even though we got these wave equations directly form Maxwell's equations we can't say that it admits only solutions which are moving such that the 1-way speed of light is $c$. Of course all such solutions satisfy the above wave equations. But they are not the only solutions. And it is perfectly possible that the solutions which corresponds to the physical electromagnetic waves don't have 1-way speed as $c$ but satisfy the above wave equations.

K. Sreeman Reddy
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The highlighted answer says that solutions to the E&M wave equations exist that don't go at $c$. Technically, the wave equation permits solutions that don't have a 1-way speed $c$, but these solutions must time evolve in some other way to compensate. Put mathematically, if we posit a solution to the wave equation $$f_{xx} - \frac{1}{c^2}f_{tt}$$ of the form $f(x-vt)$, we find $$f'' - \frac{v^2}{c^2}f'' = 0$$ I.e. $v=\pm c$. If you want a standing wave, or a wave moving at velocity $v \neq c$, you need a solution of the following form: $$f(x,t) = g(x\pm v)h(t)$$ Then the wave equation becomes $(c^2-v^2)f''g + (c^2-1)fg'' - 2(c^2-v)f'g' = 0$, so that gives you a new set of constraints to follow.

If regular waves of the form $f(x\pm vt)$ traveled at different speeds in different directions, you'd get some strange stuff. For example, if the speed of light in one direction were $c_1$ and in the other were $c_2$, the (1D) wave equation for $\mathbf{E}$ would become $$(\partial_t - c_1\partial_x)\cdot (\partial_t - c_2 \partial_x)\mathbf{E}=0$$ The same thing holds for $\mathbf{B}$. If you work back from this to modify Maxwell's equations, this suggests a bunch of whacky stuff (as far as I can tell), including stationary electric fields generating stationary magnetic fields (or perhaps the reverse, I don't think you can fully determine this from the wave equations alone).

TLDR; as far as I can tell, Maxwell's equations fully require that the one way speed of light is the same in both directions in any reference frame. This (plus experimental issues related to the Aether) is why 19th physicists came up with what would become special relativity in the first place. I hope someone comes along to read this and correct me if I've said anything wrong.

Nathan A
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