How can we conclude from Maxwell's wave equation that the speed of light is the same regardless of the state of motion of the observers?

Question

I am reading a book titled "Relativity Demystified --- A self-teaching guide by David McMahon".

He explains the derivation of electromagnetic wave equation. $$ \nabla^2 \, \begin{cases}\vec{E}\\\vec{B}\end{cases} =\mu_0\epsilon_0\,\frac{\partial^2}{\partial t^2}\,\begin{cases}\vec{E}\\\vec{B}\end{cases} $$

He then compares it with

$$ \nabla^2 \, f =\frac{1}{v^2}\,\frac{\partial^2 f}{\partial t^2} $$

and finally find

$$ v=\frac{1}{\sqrt{\mu_0\epsilon_0}}=c $$

where $c$ is nothing more than the speed of light.

The key insight to gain from this derivation is that electromagnetic waves (light) always travel at one and the same speed in vacuum. It does not matter who you are or what your state of motion is, this is the speed you are going to find.

Now it is my confusion. The nabla operator $\nabla$ is defined with respect to a certain coordinate system, for example, $(x,y,z)$. So the result $v=c$ must be the speed with respect to $(x,y,z)$ coordinate system. If another observer attached to $(x',y',z')$ moving uniformly with respect to $(x,y,z)$ then there must be a transformation that relates both coordinate systems. As a result, they must observe different speed of light.

Questions

Let's put aside the null result of Michelson and Morley experiments because they came several decades after Maxwell discovered his electromagnetic wave derivation.

I don't know the history of whether Maxwell also concluded that the speed of light is invariant under inertial frame of reference. If yes, then which part of his derivation was used to base this conclusion?

Answer 1

Your question is an excellent one and you are right about the $\nabla$ operator. And you are also right about the insufficiency of the argument you report in the book you are reading.

To make the argument more carefully, there are two options. The first would be to work out how the Maxwell equations themselves change as you go to another inertial frame. That would take a lot of calculating if you start from first principles. (And by the way, they don't change---you get back the same equations but now in terms of ${\bf E}', {\bf B}', \rho', {\bf j}', {\bf \nabla}', \partial/\partial t'$).

A second option, mathematically easier but still requiring some work if you are not familiar with it, is to show that the $\nabla$ operator and the $\partial/\partial t$ operator have a special property: when you combine them in the combination $$ \nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} $$ then their effect is the same as $$ \nabla'^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t'^2} $$ All the changes when moving from unprimed to primed coordinates cancel out. If you are familiar with partial differentiation then you could try checking this. When you learn the subject more fully, it becomes an example that can be handled more easily using the language of 4-vectors.

I think that McMahon might possibly have not thought carefully enough about what he was deriving and what he was assuming in his argument. He might for example have been taking it for granted that the Maxwell equations themselves take the same form in all inertial frames. But if he did not first prove that in his book then he ought not to claim that the derivation of waves of given speed from them proves that the wave speed will be independent of the motion of the source.

Answer 2

If Maxwell's equations have the same form in all frames of reference, then the wave speed is defined by the product of two physical constants, irrespective of coordinate system. i.e. Your book just implicitly assumes that, but of course it requires experimental testing - i.e. Michelson-Morley etc.

Answer 3

Your observation is correct, Maxwell's equations alone do not imply an invariant speed of light. One can make a Galilean transformation and get an observer-dependent speed of light as shown in the answer to this question. However, the derivation of Maxwell's equations makes no assumption of a privileged reference frame: $\varepsilon_0$ and $\mu_0$ are assumed to be properties of the vacuum. Yes, a coordinate system must be chosen, but from the point of view of derivation of the equations this is totally arbitrary. In order to keep a non-constant speed of light one would have to retroactively make the assumption after the fact that the coordinates chosen happened to be stationary coordinates with respect to the aether.

Answer 4

Without experimental evidence the constancy of the speed of light cannot be concluded. If space contained a medium, the aether, for electromagnetic waves one would expect the velocity of light in wrt to the aether to be constant. The aether theory was disproved by the experiment of Michelson and Morley. That left special relativity as an alternative.

Answer 5

Maxwell originally assumed that the speed of light would vary depending on the frame of reference. This would imply that Maxwell's equations only hold with respect to some kind of universal coordinate system. When experiments (like Michelson-Morley) indicated that the speed of light did not vary between inertial reference frames, physicists like Hendrik Lorentz figured out how to transform Maxwell's equations in a way that would keep the speed of light constant when moving from one reference frame to another. This required all kinds of weird concepts like length contraction and time dilatation. In 1905, Einstein demonstrated that these weird ideas could be derived in a very natural way by throwing away old ideas about space and time being absolute, and starting with the assumption that the laws of physics (including Maxwell's equations) are equally valid in all inertial reference frames. Your book apparently just adopts this view from the start. There is certainly an aesthetic argument to be made for adopting this point of view, but obviously, any scientific idea needs to be backed up by experimental evidence. Therefore, any book that tries to "derive" scientific laws without reference to experiment is really just feeding you misleading arguments like this one.

Answer 6

A related point, which seems very little known, is that Maxwell's electromagnetic theory does not imply the speed of light is $c$ in all directions! It is only because we implicitly input that assumption (of isotropic $c$) when formulating the equations, that it pops out at the end. Anderson, Vetharanium, & Stedman (1998) $\S2.3.3$ formulate "Electromagnetism in a more general synchronization" (that is, a different simultaneity convention). Another paper which does this is Rizzi, Ruggiero, & Serafini (2004) $\S A2$.

Having said this, it still seems the most natural choice that $c$ is the same in all directions, for all observers. It's just that Maxwell doesn't prove this, nor does any other theory or experiment prove the one-way speed of light.