Categories: Electromagnetism, Optics, Physics.

Electromagnetic wave equation

Light, i.e. electromagnetic waves, consist of an electric field and a magnetic field, one inducing the other and vice versa. The existence and classical behavior of such waves can be derived using only Maxwell’s equations, as we will demonstrate here.

We start from Faraday’s law of induction, where we assume that the system consists of materials with well-known (linear) relative magnetic permeabilities μr(r)\mu_r(\vb{r}), such that B=μ0μrH\vb{B} = \mu_0 \mu_r \vb{H}:

∇×E=−∂B∂t=−μ0μr∂H∂t\begin{aligned} \nabla \cross \vb{E} = - \pdv{\vb{B}}{t} = - \mu_0 \mu_r \pdv{\vb{H}}{t} \end{aligned}

We move μr(r)\mu_r(\vb{r}) to the other side, take the curl, and insert Ampère’s circuital law:

∇×(1μr∇×E)=−μ0∂∂t(∇×H)=−μ0(∂Jfree∂t+∂2D∂t2)\begin{aligned} \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) &= - \mu_0 \pdv{}{t} \big( \nabla \cross \vb{H} \big) \\ &= - \mu_0 \bigg( \pdv{\vb{J}_\mathrm{free}}{t} + \pdvn{2}{\vb{D}}{t} \bigg) \end{aligned}

For simplicity, we only consider insulating materials, since light propagation in conductors is a complex beast. We thus assume that there are no free currents Jfree=0\vb{J}_\mathrm{free} = 0, leaving:

∇×(1μr∇×E)=−μ0∂2D∂t2\begin{aligned} \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) &= - \mu_0 \pdvn{2}{\vb{D}}{t} \end{aligned}

Having E\vb{E} and D\vb{D} in the same equation is not ideal, so we should make a choice: do we restrict ourselves to linear media (so D=ε0εrE\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}), or do we allow materials with more complicated responses (so D=ε0E+P\vb{D} = \varepsilon_0 \vb{E} + \vb{P}, with P\vb{P} unspecified)? The former is usually sufficient:

∇×(1μr∇×E)=−μ0ε0εr∂2E∂t2\begin{aligned} \boxed{ \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) = - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} } \end{aligned}

This is the general linear form of the electromagnetic wave equation, where μr\mu_r and εr\varepsilon_r both depend on r\vb{r} in order to describe the structure of the system. We can obtain a similar equation for H\vb{H}, by starting from Ampère’s law under the same assumptions:

∇×H=∂D∂t=ε0εr∂E∂t\begin{aligned} \nabla \cross \vb{H} = \pdv{\vb{D}}{t} = \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t} \end{aligned}

Taking the curl and substituting Faraday’s law on the right yields:

∇×(1εr∇×H)=ε0∂∂t(∇×E)=−ε0∂2B∂t2\begin{aligned} \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) &= \varepsilon_0 \pdv{}{t} \big( \nabla \cross \vb{E} \big) = - \varepsilon_0 \pdvn{2}{\vb{B}}{t} \end{aligned}

And then we insert B=μ0μrH\vb{B} = \mu_0 \mu_r \vb{H} to get the analogous electromagnetic wave equation for H\vb{H}:

∇×(1εr∇×H)=−μ0ε0μr∂2H∂t2\begin{aligned} \boxed{ \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) = - \mu_0 \varepsilon_0 \mu_r \pdvn{2}{\vb{H}}{t} } \end{aligned}

This is equivalent to the problem for E\vb{E}, since they are coupled by Maxwell’s equations. By solving either, subject to Gauss’s laws ∇⋅(εrE)=0\nabla \cdot (\varepsilon_r \vb{E}) = 0 and ∇⋅(μrH)=0\nabla \cdot (\mu_r \vb{H}) = 0, the behavior of light in a given system can be deduced. Note that Gauss’s laws enforce that the wave’s fields are transverse, i.e. they must be perpendicular to the propagation direction.

Homogeneous linear media

In the special case where the medium is completely uniform, μr\mu_r and εr\varepsilon_r no longer depend on r\vb{r}, so they can be moved to the other side:

∇×(∇×E)=−μ0μrε0εr∂2E∂t2∇×(∇×H)=−μ0μrε0εr∂2H∂t2\begin{aligned} \nabla \cross \big( \nabla \cross \vb{E} \big) &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} \\ \nabla \cross \big( \nabla \cross \vb{H} \big) &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{H}}{t} \end{aligned}

This can be rewritten using the vector identity ∇×(∇×V)=∇(∇⋅V)−∇2V\nabla \cross (\nabla \cross \vb{V}) = \nabla (\nabla \cdot \vb{V}) - \nabla^2 \vb{V}:

∇(∇⋅E)−∇2E=−μ0μrε0εr∂2E∂t2∇(∇⋅H)−∇2H=−μ0μrε0εr∂2H∂t2\begin{aligned} \nabla (\nabla \cdot \vb{E}) - \nabla^2 \vb{E} &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} \\ \nabla (\nabla \cdot \vb{H}) - \nabla^2 \vb{H} &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{H}}{t} \end{aligned}

Which can be reduced using Gauss’s laws ∇⋅E=0\nabla \cdot \vb{E} = 0 and ∇⋅H=0\nabla \cdot \vb{H} = 0 thanks to the fact that εr\varepsilon_r and μr\mu_r are constants in this case. We therefore arrive at:

∇2E−n2c2∂2E∂t2=0\begin{aligned} \boxed{ \nabla^2 \vb{E} - \frac{n^2}{c^2} \pdvn{2}{\vb{E}}{t} = 0 } \end{aligned} ∇2H−n2c2∂2H∂t2=0\begin{aligned} \boxed{ \nabla^2 \vb{H} - \frac{n^2}{c^2} \pdvn{2}{\vb{H}}{t} = 0 } \end{aligned}

Where c=1/μ0ε0c = 1 / \sqrt{\mu_0 \varepsilon_0} is the speed of light in a vacuum, and n=μ0ε0n = \sqrt{\mu_0 \varepsilon_0} is the refractive index of the medium. Note that most authors write the magnetic equation with B\vb{B} instead of H\vb{H}; both are correct thanks to linearity.

In a vacuum, where n=1n = 1, these equations are sometimes written as □E=0\square \vb{E} = 0 and □H=0\square \vb{H} = 0, where □\square is the d’Alembert operator, defined as follows:

□≡∇2−1c2∂2∂t2\begin{aligned} \boxed{ \square \equiv \nabla^2 - \frac{1}{c^2} \pdvn{2}{}{t} } \end{aligned}

Note that some authors define it with the opposite sign. In any case, the d’Alembert operator is important for special relativity.

The solution to the homogeneous electromagnetic wave equation are traditionally said to be the so-called plane waves given by:

E(r,t)=E0eik⋅r−iωtB(r,t)=B0eik⋅r−iωt\begin{aligned} \vb{E}(\vb{r}, t) &= \vb{E}_0 e^{i \vb{k} \cdot \vb{r} - i \omega t} \\ \vb{B}(\vb{r}, t) &= \vb{B}_0 e^{i \vb{k} \cdot \vb{r} - i \omega t} \end{aligned}

Where the wavevector k\vb{k} is arbitrary, and the angular frequency ω=c∣k∣/n\omega = c |\vb{k}| / n. We also often talk about the wavelength, which is λ=2π/∣k∣\lambda = 2 \pi / |\vb{k}|. The appearance of k\vb{k} in the exponent tells us that these waves are propagating through space, as you would expect.

In fact, because the wave equations are linear, any superposition of plane waves, i.e. any function of the form f(k⋅r−ωt)f(\vb{k} \cdot \vb{r} - \omega t), is in fact a valid solution. Just remember that E\vb{E} and H\vb{H} are real-valued, so it may be necessary to take the real part at the end of a calculation.

Inhomogeneous linear media

But suppose the medium is not uniform, i.e. it contains structures described by εr(r)\varepsilon_r(\vb{r}) and μr(r)\mu_r(\vb{r}). If the structures are much larger than the light’s wavelength, the homogeneous equation is still a very good approximation away from any material boundaries; anywhere else, however, they will break down. Recall the general equations from before we assumed homogeneity:

∇×(1μr∇×E)=−εrc2∂2E∂t2∇×(1εr∇×H)=−μrc2∂2H∂t2\begin{aligned} \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) &= - \frac{\varepsilon_r}{c^2} \pdvn{2}{\vb{E}}{t} \\ \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) &= - \frac{\mu_r}{c^2} \pdvn{2}{\vb{H}}{t} \end{aligned}

In theory, this is everything we need, but in most cases a better approach is possible: the trick is that we only rarely need to explicitly calculate the tt-dependence of E\vb{E} or H\vb{H}. Instead, we can first solve an easier time-independent version of this problem, and then approximate the dynamics with coupled mode theory later.

To eliminate tt, we make an ansatz for E\vb{E} and H\vb{H}, shown below. No generality is lost by doing this; this is effectively a kind of Fourier transform:

E(r,t)=E(r)e−iωtH(r,t)=H(r)e−iωt\begin{aligned} \vb{E}(\vb{r}, t) &= \vb{E}(\vb{r}) e^{- i \omega t} \\ \vb{H}(\vb{r}, t) &= \vb{H}(\vb{r}) e^{- i \omega t} \end{aligned}

Inserting this ansatz and dividing out e−iωte^{-i \omega t} yields the time-independent forms:

∇×(1μr∇×E)=(ωc)2εrE\begin{aligned} \boxed{ \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) = \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r \vb{E} } \end{aligned} ∇×(1εr∇×H)=(ωc)2μrH\begin{aligned} \boxed{ \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) = \Big( \frac{\omega}{c} \Big)^2 \mu_r \vb{H} } \end{aligned}

These are eigenvalue problems for ω2\omega^2, which can be solved subject to Gauss’s laws and suitable boundary conditions. The resulting allowed values of ω\omega may consist of continuous ranges and/or discrete resonances, analogous to scattering and bound quantum states, respectively. It can be shown that the operators on both sides of each equation are Hermitian, meaning these are well-behaved problems yielding real eigenvalues and orthogonal eigenfields.

Both equations are still equivalent: we only need to solve one. But which one? In practice, one is usually easier than the other, due to the common approximation that μr≈1\mu_r \approx 1 for many dielectric materials, in which case the equations reduce to:

∇×(∇×E)=(ωc)2εrE∇×(1εr∇×H)=(ωc)2H\begin{aligned} \nabla \cross \big( \nabla \cross \vb{E} \big) &= \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r \vb{E} \\ \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) &= \Big( \frac{\omega}{c} \Big)^2 \vb{H} \end{aligned}

Now the equation for H\vb{H} is starting to look simpler, because it only has an operator on one side. We could “fix” the equation for E\vb{E} by dividing it by εr\varepsilon_r, but the resulting operator would no longer be Hermitian, and hence not well-behaved. To get an idea of how to handle εr\varepsilon_r in the E\vb{E}-equation, notice its similarity to the weight function ww in Sturm-Liouville theory.

Gauss’s magnetic law ∇⋅H=0\nabla \cdot \vb{H} = 0 is also significantly easier for numerical calculations than its electric counterpart ∇⋅(εrE)=0\nabla \cdot (\varepsilon_r \vb{E}) = 0, so we usually prefer to solve the equation for H\vb{H}.

References

  1. J.D. Joannopoulos, S.G. Johnson, J.N. Winn, R.D. Meade, Photonic crystals: molding the flow of light, 2nd edition, Princeton.