I've been told that, from Maxwell's equations, one can find that the propagation of change in the Electromagnetic Field travels at a speed $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$ (the values of which can be empirically found, and, when plugged into the expression, yield the empirically found speed of light)

I'm really not sure how I would go about finding $v = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$ simply from Maxwell's equations in the following form, in SI units --

$$\nabla \cdot \mathbf{E} = \frac {\rho} {\epsilon_0}$$

$$\nabla \cdot \mathbf{B} = 0$$

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$$

$$\nabla \times \mathbf{B} = \mu_0 \left( \mathbf{J} + \epsilon_0 \frac{\partial \mathbf{E}} {\partial t} \right)$$

Is what I believe true? (that the speed of propagation is derivable from Maxwell's Equations)

If not, what else is needed?

If so, can you provide a clear and cogent derivation?

  • $\begingroup$ It's a good question. One thing I'd recommend is to learn CGS units. The speed of light is made explicit and there, and the equations look a little more symmetric. Engineers and some physicists are more familiar with the SI units you've posted, so it's worthwhile to learn both. $\endgroup$ Commented Dec 3, 2010 at 7:18
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    $\begingroup$ If you want to learn more, a good book to read is David J. Griffiths' Introduction to Electrodynamics, chapter 9. $\endgroup$
    – ptomato
    Commented Dec 3, 2010 at 9:00
  • $\begingroup$ @Mark I think that the CGS units of the equations make them a bit..."too" explicit? In that the answer probably wouldn't be as interesting as if they were in SI units. In any case, from what I was taught, my understanding is that CGS units suppose that $|\mathbf{E}| = c |\mathbf{B}|$, which is a part of what we are trying to prove anyway. Is this correct? $\endgroup$
    – Justin L.
    Commented Dec 6, 2010 at 9:04
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    $\begingroup$ I don't think we're trying to prove $|E| = c|B|$. If we take that as an assumption in CGS units, then $c$ is just a constant with no particular meaning to begin with. We then need to show that electromagnetic signals propagate at $c$. The answer is just as interesting, since we would have to go through the same derivation. The only difference is that instead of $v = 1/\sqrt{\mu_0\epsilon_0}$ we'd have $v = c$ at the end. $\endgroup$ Commented Dec 6, 2010 at 9:11

3 Answers 3


Although this is a standard derivation, you frequently don't see it in introductory electromagnetism courses, maybe because those courses shy away from the heavy use of vector calculus. Here's the usual approach. We'll find a wave equation from Maxwell's equations.

Start with

$\nabla \times \vec{E} = -\frac{\partial\vec{B}}{\partial t}$.

Take a partial derivative of both sides with respect to time. The curl operator has no partial with respect to time, so this becomes

$\nabla \times \frac{\partial\vec{E}}{\partial t} = -\frac{\partial^2\vec{B}}{\partial t^2}$.

There's another of Maxwell's equations that tells us about $\partial\vec{E}/\partial t$.

$\nabla \times \vec{B} = \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}$

Solve this for $\partial\vec{E}/\partial t$ and plug into the previous expression to get

$\nabla \times \frac{(\nabla \times \vec{B})}{\mu_0\epsilon_0} = -\frac{\partial^2 \vec{B}}{\partial t^2}$

the curl of curl identity lets us rewrite this as

$\frac{1}{\mu_0 \epsilon_0}\left(\nabla(\nabla \cdot \vec{B}) - \nabla^2\vec{B}\right) = -\frac{\partial^2 \vec{B}}{\partial t^2}$

But the divergence of the magnetic field is zero, so kill that term, and rearrange to

$\frac{-1}{\mu_0 \epsilon_0}\nabla^2\vec{B} + \frac{\partial^2 \vec{B}}{\partial t^2} = 0$

This is the wave equation we're seeking. One solution is

$\vec{B} = B_0 e^{i (\vec{x}\cdot\vec{k} - \omega t) }$.

This represents a plane wave traveling in the direction of the vector $\vec{k}$ with frequency $\omega$ and phase velocity $v = \omega/|\vec{k}|$. In order to be a solution, this equation needs to have

$\frac{\omega^2}{k^2} = \frac{1}{\mu_0\epsilon_0}$.

Or, setting $v = 1/\sqrt{\mu_0\epsilon_0}$

$\frac{\omega}{k} = v$

This is called the dispersion relation. The speed that electromagnetic signals travel is given by the group velocity

$\frac{d\omega}{d k} = v$

So electromagnetic signals in a vacuum travel at speed $c = 1/\sqrt{\mu_0\epsilon_0}$.

Edit You can follow the same steps to derive the wave equation for $\vec{E}$, but you will have to assume you're in free space, i.e. $\rho = 0$.

Edit The curl of the curl identity was wrong, there's a negative number in there

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    $\begingroup$ Out of curiosity, how do you go from "one solution is..." to generalize to all solutions to the wave equation? $\endgroup$
    – Justin L.
    Commented Dec 3, 2010 at 10:06
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    $\begingroup$ Basically, the form of the equation. As Marek puts it in his answer, it's the d'Alembert wave operator. It's a thoroughly studied operator and the spectrum is well-known. In other words, the behaviour of the solutions is known to be waves moving at velocity the square root of the coefficient in front of the Laplacian (if the coefficient in front of the double time derivative is 1). $\endgroup$ Commented Dec 3, 2010 at 10:43
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    $\begingroup$ @Justin L.: by weighted-integrating over all $\vec k$'s that fulfil the dispersion relation $\endgroup$ Commented Dec 3, 2010 at 11:18
  • $\begingroup$ @Justin: another (more mathematical way) to say what Raskolnikov and Tobias have already said is that plane waves form a basis for the space of solutions to the wave equation. $\endgroup$ Commented Dec 3, 2010 at 20:38
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    $\begingroup$ You only derived half the story... $\endgroup$
    – Sklivvz
    Commented Dec 3, 2010 at 21:31

Mark's answer is correct but it is way too long and hides the punchline. So let me show a shorter derivation using more advanced math. Not too advanced though, just tensor formalism in Minkowski space-time for Special Relativity and differential forms. You will need all of this sooner of later, so it should be useful to learn (at least a little) about it already. This answer would be just a few lines if you already knew the formalism but it will be a little longer because I'll try to teach you also about the formalism.

You probably know that Lorentz transformations mix $\mathbf E$ and $\mathbf B$. So they are not really independent and it turns out that they are just a part of rank 2 antisymmetric 4-dimensional tensor (this really means $4 \times 4$ antisymmetric matrix) $\mathbf F$. Now, it should be at least dimensionally clear that such a matrix has 6 independent components which precisely coincides with 3+3 degrees of freedom of $\mathbf E$ and $\mathbf B$.

You should probably also know that both $\mathbf E$ and $\mathbf B$ can be expressed in terms of potentials. In our formalism it translates into ${\mathbf F} = {\rm d} {\mathbf A}$ where ${\rm d}$ is the exterior derivative $\mathbf A$ is the four-potential that combines scalar $\phi$ and three-vector $\mathbf A$ potentials you should already know and love.

Now it turns out that Maxwell equations in vacuum have really simple form in this formalism $${\rm d}{\mathbf F} = 0$$ $${\rm \delta}{\mathbf F} = 0$$ with $\delta$ being the codifferential which is dual to $\rm d$. The first equation actually tells us that four-potential exists (because ${\rm d}^2 = 0)$ and the second one is the actual evolutionary equation that would contain four-current $\mathbf j$ if we weren't in vacuum. Now whenever we have a solution to these equations, they will also solve $$\square {\mathbf F} = ({\rm d\delta + \delta d}) {\mathbf F} = 0$$ But this $\square$ is precisely d'Alembert wave operator and so indeed $\mathbf F$ propagates at the speed of light.

Reference: Wikipedia article on the covariant or four-vector formalism

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    $\begingroup$ This notation is beautiful (even though I haven't understood a thing :) I would like to learn this. Where do I begin? $\endgroup$ Commented Dec 3, 2010 at 16:58
  • $\begingroup$ @Charmer: Well, I am not really sure what are the good books (especially introductory ones); I learned both physics and math for this at my standard university courses. But later I happened to come across one book that covers some of this (and lots of other great stuff from theoretical physics): Fecko. $\endgroup$
    – Marek
    Commented Dec 3, 2010 at 17:07
  • $\begingroup$ @Charmer: but I guess you could ask that question of yours Where do I begin? on this site and I am sure you'll get lots of good answers and references ;-) $\endgroup$
    – Marek
    Commented Dec 3, 2010 at 17:08
  • $\begingroup$ @Marek Thanks! But I don't exactly know what to ask for here :-) So, would you ask it for me? $\endgroup$ Commented Dec 3, 2010 at 21:06
  • $\begingroup$ @Marek If we don't have books then we can list topics one should study and order in which those should be studied. Something like syllabus of standard university courses. What do you say? $\endgroup$ Commented Dec 3, 2010 at 21:10

Start by taking the curl of Maxwell's third equation (for vacuum), and substituting $\vec{B}=\mu_0\vec{H}$ one can obtain,

$$ \nabla^2.\vec{E} = \mu_0 \epsilon_0 \frac{\partial^2}{\partial t^2} \vec{E} $$

Similarly, by taking curl of Maxwell's fourth equation, substituting

$$ \nabla\times\vec{E} = -\frac{\partial}{\partial t}\mu_0\vec{H} $$

one can obtain

$$ \nabla^2.\vec{H} = \mu_0\epsilon_0 \frac{\partial^2}{\partial t^2} \vec{H} $$

The solution of the two equations are of form

$$ E = E_o e^{\iota(\omega t + \beta z)}\ \ \ \ ;\ \ \ \ H = H_oe^{\iota(\omega t+\beta z)} $$

Taking double time derivative of these results yield

$$ \frac{\partial}{\partial t} = \iota\omega\ \ \ \ ;\ \ \ \ \frac{\partial^2}{\partial t^2}=-\omega^2 $$

If we insert these results in our $\nabla^2$ equations, we'll get the Helmholtz Equation for $\vec{E}$ and $\vec{H}$ as

$$ (\nabla^2 + \omega^2 \mu_0 \epsilon_0)\vec{E} = 0 = (\nabla^2 + \omega^2 \mu_0 \epsilon_0)\vec{H} $$

Here the expression $\omega^2 \mu_0 \epsilon_0 = \beta^2$ which is the wave number. On solving this expression we can get the above mentioned equation.

$$ \frac{\omega}{\beta} = \frac{1}{\sqrt{\mu_0 \epsilon_0}} $$

Also, $\omega = 2\pi f$ and $\beta = 2\pi/\lambda$ which brings us to the desired equation,

$$ c = \frac{1}{\sqrt{\mu_o \epsilon_0}} $$

Since, $mu_0 = 4\pi\times10^{-7}$ H/m and $\epsilon_0 \approx 8.85\times10^{-12}$ F/m this equation yields $ c = 299 792 458$ m/s


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