The electromagnetic wave equation describes
the propagation of light through various media.
Since an electromagnetic (light) wave consists of
an electric field
and a magnetic field,
we need Maxwell’s equations
in order to derive the wave equation.
Uniform medium
We will use all of Maxwell’s equations,
but we start with Ampère’s circuital law for the “free” fields H and D,
in the absence of a free current Jfree=0:
∇×H=∂t∂D
We assume that the medium is isotropic, linear,
and uniform in all of space, such that:
D=ε0εrEH=μ0μr1B
Which, upon insertion into Ampère’s law,
yields an equation relating B and E.
This may seem to contradict Ampère’s “total” law,
but keep in mind that Jbound=0 here:
∇×B=μ0μrε0εr∂t∂E
Now we take the curl, rearrange,
and substitute ∇×E according to Faraday’s law:
Using a vector identity, we rewrite the leftmost expression,
which can then be reduced thanks to Gauss’ law for magnetism ∇⋅B=0:
−μ0μrε0εr∂t2∂2B=∇(∇⋅B)−∇2B=−∇2B
This describes B.
Next, we repeat the process for E:
taking the curl of Faraday’s law yields:
∇×(∇×E)=−∂t∂(∇×B)=−μ0μrε0εr∂t2∂2E
Which can be rewritten using same vector identity as before,
and then reduced by assuming that there is no net charge density ρ=0
in Gauss’ law, such that ∇⋅E=0:
−μ0μrε0εr∂t2∂2E=∇(∇⋅E)−∇2E=−∇2E
We thus arrive at the following two (implicitly coupled)
wave equations for E and B,
where we have defined the phase velocity v≡1/μ0μrε0εr:
∂t2∂2E−v21∇2E=0∂t2∂2B−v21∇2B=0
Traditionally, it is said that the solutions are as follows,
where the wavenumber ∣k∣=ω/v:
E(r,t)B(r,t)=E0exp(ik⋅r−iωt)=B0exp(ik⋅r−iωt)
In fact, thanks to linearity, these plane waves can be treated as
terms in a Fourier series, meaning that virtually
any function f(k⋅r−ωt) is a valid solution.
Keep in mind that in reality E and B are real,
so although it is mathematically convenient to use plane waves,
in the end you will need to take the real part.
Non-uniform medium
A useful generalization is to allow spatial change
in the relative permittivity εr(r)
and the relative permeability μr(r).
We still assume that the medium is linear and isotropic, so:
D=ε0εr(r)EB=μ0μr(r)H
Inserting these expressions into Faraday’s and Ampère’s laws
respectively yields:
∇×E=−μ0μr(r)∂t∂H∇×H=ε0εr(r)∂t∂E
We then divide Ampère’s law by εr(r),
take the curl, and substitute Faraday’s law, giving:
∇×(εr1∇×H)=ε0∂t∂(∇×E)=−μ0μrε0∂t2∂2H
Next, we exploit linearity by decomposing H and E
into Fourier series, with terms given by:
H(r,t)=H(r)exp(−iωt)E(r,t)=E(r)exp(−iωt)
By inserting this ansatz into the equation,
we can remove the explicit time dependence:
∇×(εr1∇×H)exp(−iωt)=μ0ε0ω2μrHexp(−iωt)
Dividing out exp(−iωt),
we arrive at an eigenvalue problem for ω2,
with c=1/μ0ε0:
∇×(εr(r)1∇×H(r))=(cω)2μr(r)H(r)
Compared to a uniform medium, ω is often not arbitrary here:
there are discrete eigenvalues ω,
corresponding to discrete modesH(r).
Next, we go through the same process to find an equation for E.
Starting from Faraday’s law, we divide by μr(r),
take the curl, and insert Ampère’s law:
∇×(μr1∇×E)=−μ0∂t∂(∇×H)=−μ0ε0εr∂t2∂2E
Then, by replacing E(r,t) with our plane-wave ansatz,
we remove the time dependence:
∇×(μr1∇×E)exp(−iωt)=−μ0ε0ω2εrEexp(−iωt)
Which, after dividing out exp(−iωt),
yields an analogous eigenvalue problem with E(r):
∇×(μr(r)1∇×E(r))=(cω)2εr(r)E(r)
Usually, it is a reasonable approximation
to say μr(r)=1,
in which case the equation for H(r)
becomes a Hermitian eigenvalue problem,
and is thus easier to solve than for E(r).
Keep in mind, however, that in any case,
the solutions H(r) and/or E(r)
must satisfy the two Maxwell’s equations that were not explicitly used:
∇⋅(εrE)=0∇⋅(μrH)=0
This is equivalent to demanding that the resulting waves are transverse,
or in other words,
the wavevector k must be perpendicular to
the amplitudes H0 and E0.
References
J.D. Joannopoulos, S.G. Johnson, J.N. Winn, R.D. Meade,
Photonic crystals: molding the flow of light,
2nd edition, Princeton.