I am trying to understand the nature of Maxwell's equations a little more. Am I correct in stating that the constants $\epsilon_{0}$ and $\mu_{0}$ in Maxwell's equations must be positive? Is my reasoning below valid?

Consider Maxwell's equations: \begin{align*} \nabla\cdot\vec{E} &= \frac{\rho}{\epsilon_{0}}, \\ \nabla\cdot\vec{B} &= 0, \\ \nabla\times\vec{E} &= -\frac{\partial\vec{B}}{\partial t}, \\ \nabla\times\vec{B} &= \mu_{0}\vec{J} + \mu_{0}\epsilon_{0}\frac{\partial\vec{E}}{\partial t} \end{align*} along with the force equation $\vec{F} = q\vec{E} + q\vec{v}\times\vec{B}$.

If the signs of $\epsilon_{0}$ and $\mu_{0}$ differ, then the $\mu_{0}\epsilon_{0}$ factor is negative, and so the $1/c^{2}$ factor in the vacuum wave equations would be negative, leading to positive feedback loops where the E- and B-fields grow without bound, making the universe unstable.

Now suppose both $\epsilon_{0}$ and $\mu_{0}$ were negative. Then consider a loop of wire and you begin to drive the current in some direction more and more. Let's say the current is counterclockwise from the top view. Since $\mu_{0}$ is negative, the "reverse" of Ampere's law will create a magnetic field such that it goes into the loop from the top view. As current increases, so will $\vec{B}$, creating an electric field that points counterclockwise from the top view. By the force law written above, this will drive charges to go counterclockwise around the loop even more, increasing the current further. This again leads to a cascade effect.

Do both of these thought experiments demonstrate that Maxwell's theory is coherent only if $\epsilon_{0} > 0$ and $\mu_{0} > 0$?

This is really interesting, so it's really strange that I've never seen anyone talk about this. Is there any reference or theorem related to this result? Any material would be very appreciated.

  • $\begingroup$ $\mu_o$ and $\epsilon_o$ are both physical constants for a given material. They indicate how that material reacts to magnetic and electric fields respectively. If either (or both) were negative the electricity and magnetism would not exist. $\endgroup$
    – Michael
    Commented Dec 13, 2021 at 10:40
  • 1
    $\begingroup$ @Michael: I think the OP is talking about the vacuum values, not the values of permittivity and permeability in a linear medium. $\endgroup$ Commented Dec 13, 2021 at 16:05
  • 1
    $\begingroup$ Note that these constants are an artefact of the SI system of units - they do not appear in many other systems. $\endgroup$
    – Roger V.
    Commented Dec 13, 2021 at 16:13
  • 1
    $\begingroup$ Before everyone proves too conclusively that $\epsilon$ and $\mu$ have to be positive, remember that either, or both can be negative (at certain frequencies) in metamaterials en.wikipedia.org/wiki/Metamaterial $\endgroup$ Commented Dec 13, 2021 at 17:16
  • 1
    $\begingroup$ $\epsilon$ and $\mu$, as they depend on the polarisation of the medium, are frequency dependent. At some frequencies they can be negative. $\endgroup$ Commented Dec 13, 2021 at 18:11

2 Answers 2


Roughly speaking, yes; changing the signs of $\mu_0$ or $\epsilon_0$ (or both) can lead to "run-away" solutions.

The easiest way to see why changing the sign of $\mu_0$ and/or $\epsilon_0$ is problematic is to look at the potential energy stored in a set of electromagnetic fields: $$ U = \frac{1}{2} \iiint \left(\epsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) \, \mathrm{d} \tau $$ In the "real-world" case, both $\epsilon_0$ and $\mu_0$ are positive. This means that there is a bound on how big $|\vec{E}|$ and $|\vec{B}|$ can get (averaged over space.) There's only a certain amount of energy to go around in the system, and if I want to make a really really large magnetic field field, I am constrained by the fact that I only have a total amount of energy $U$ available to do it.

But if $\epsilon_0$ and $\mu_0$ are of different signs, then this constraint disappears. Let's suppose that $\epsilon_0 < 0$ and $\mu_0 > 0$. If that's the case, then I can make a very very strong magnetic field without violating conservation of energy, so long as I also make a large electric field! Since magnetic and electric fields would contribute to the integral with different signs, I could get arbitrarily large fields while still having the above integral being relatively small.

In fancy mathematical terms, we say that the above energy integral (with $\epsilon_0 \mu_0 < 0$) is "indefinite", since it can be either positive or negative. Models with indefinite energy often have this problem of run-away solutions. The "real-world" integral, with $\mu_0, \epsilon_0 > 0$, has a "positive definite" energy: the integral is always positive except in the case where both fields vanish, in which case the integral is zero. The positive-definiteness of the integral is important to placing bounds on the fields in the "real-world" case.

Finally, flipping both of $\epsilon_0$ and $\mu_0$ does not lead to runaway solutions in "vacuum" (i.e., due solely to the field energy.) In this case, the above integral is "negative definite" rather than "positive definite". But these fields can also (for example) do work on electric charges, giving them mechanical kinetic energy; and the kinetic energy of these charges is always positive definite. So in this case, there still exists the possibility of runaway solutions in which charged objects get very large kinetic energies (with a positive energy of great magnitude) and the fields become arbitrarily large (with a negative energy of great magnitude.)

  • $\begingroup$ I really like this comprehensive post. My only other question is, are there any papers, books, etc that discuss or mention this? This is interesting to me because this constrains the kinds of "field theories" that are "plausible." $\endgroup$ Commented Dec 13, 2021 at 17:54
  • $\begingroup$ I'm giving this a checkmark, because this is a good post, but I may change it if anyone else answers my question about references. I do admit this observation might be too simple to be explicitly discussed in literature. $\endgroup$ Commented Dec 13, 2021 at 18:28

As for as Maxwell's equations are concerned, changing the signs of both constants is equivalent to changing the signs of $\rho,\,\vec{J}$, i.e. charge conjugation. But, as @Javier & MaximalIdeal note, the Lorentz force law will reverse direction. This will lead positive and negative charges to separately clump, more like mass under gravity, albeit with clumps of opposite charges repelling. It certainly wouldn't give us the familiar electrons-around-nuclei result.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.