Deriving the speed of the propagation of a change in the Electromagnetic Field from Maxwell's Equations

Question

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?

Answer 1

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

Answer 2

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.

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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.

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

Answer 3

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