Based on my previous question here, lets us step back a little bit. The speed of light $c=1/\sqrt{\mu_0\epsilon_0}$ is assumed as a value that does not depend on the observer because it is just a product of two constants.

I am still wondering why Maxwell assumed that $\mu_0$ and $\epsilon_0$ are constant that do not depend on the frame of reference. In my understanding both these constants are obtained from experiments. Both these experimental constants are not like $\pi\approx3.14\ldots$ or $e\approx 2.71828\ldots$ which are constants obtained theoretically or geometrically.

So I think the derivation of Maxwell's electromagnetic wave equation should start by assuming that $\mu(x,y,z)$ and $\epsilon(x,y,z)$ first and then prove that both do not depend of any frame of references.


How to prove that both $\mu_0$ and $\epsilon_0$ do not depend of coordinate system of choices?

  • $\begingroup$ Shouldn't reference frame dependence depend on relative velocities rather than spatial coordinates? You can make materials such that $\mu$ and $\epsilon$ are in fact not constant over space, so $\mu(x,y,z)$ and $\epsilon(x,y,z)$ is completely valid. Can you please clarify what you mean? $\endgroup$ Jun 27, 2020 at 12:09
  • $\begingroup$ You can't "prove" anything when it come to physics. The speed of light is now a defined constant and $c ^2= (\mu_0 \epsilon_0)^{-1}$. The values of. $\mu_0$ and $\epsilon_0$ also used to be defined constants in the old SI system. They now can vary independently (though their product is fixed). Your question amounts to saying can we prove that Maxwell's equations work in all frames of reference. $\endgroup$
    – ProfRob
    Jun 27, 2020 at 12:34
  • 1
    $\begingroup$ These constants can be made to disappear just by choosing appropriate units (such as Gaussian units), so they have no physical significance. All they do is confuse students. $\endgroup$
    – G. Smith
    Jun 27, 2020 at 16:29
  • $\begingroup$ @G.Smith There are nothing in Gaussian units that corresponds to $\epsilon_0$ and $\mu_0$. But is it implicit and assumed in that fact that the relationship between $\bf D$ and $\bf E$ is frame independent? $\endgroup$
    – garyp
    Jun 28, 2020 at 18:54
  • $\begingroup$ @garyp The relationship is $\mathbf{D}=\mathbf{E}+4\pi\mathbf{P}$ in Gaussian units. I don’t understand your question about frame independence. $\endgroup$
    – G. Smith
    Jun 29, 2020 at 6:30

2 Answers 2


The way to this is to make the hypothesis at the outset that $\epsilon_0$ and $\mu_0$ are scalar invariant constants, and then check whether, under this hypothesis, the equations overall are Lorentz covariant. It turns out that they are. But it is easier to prove this by starting out with tensor notation which I am guessing you have not learned yet.

Now I will unpack the terminology used above.

  1. scalar = fully specified at each event by a single number

  2. invariant = the number you get is the same in all reference frames

  3. constant = the number you get is the same at all events in any given reference frame

So one is claiming quite a lot for these $\epsilon_0$ and $\mu_0$. It amounts to claiming that they are just numbers like 2 and $\pi$, except that they may have physical dimensions in the system of units being adopted. Having made the claim, the logic is, as I already said, that one now asks whether, if these quantities are indeed scalar invariant constants, then do the Maxwell equations survive unchanged from one frame to another? One can prove that they do by a rather lengthy calculation involving the transformation of force, or by a quicker calculation involving tensors.

(I expound this point fully and carefully in my own book on this subject; it is an undergraduate physics textbook.)


Maxwell DID NOT assume that $\mu_0$ and $\epsilon_0$ are constant. They were invented by an Italian engineer named Georgi much later. They have little to do with physics, so they won't change. $\mu_0/4\pi=10^{-7}$ is a conversion constant from physical units to SI. $10^{-5}$ comes from converting cgs to MKS. $10^{-2}$ comes from redefining the Ampere in the 1880's. $1/4\pi\epsilon_0$ is just $c^2$ in converted units.