4Electrodynamics
IB Electromagnetism
4.5 Electromagnetic waves
We now look for solutions to Maxwell’s equation in the case where
ρ
= 0 and
J = 0, i.e. in nothing/vacuum.
Differentiating the fourth equation with respect to time,
µ
0
ε
0
∂
2
E
∂t
2
=
∂
∂t
(∇ × B)
= ∇ ×
∂B
∂t
= ∇( ∇ · E
|
{z}
=ρ/ε
0
=0
) + ∇
2
E by vector identities
= ∇
2
E.
So each component of E obeys the wave equation
1
c
2
∂
2
E
∂t
2
− ∇
2
E = 0.
We can do exactly the same thing to show that B obeys the same equation:
1
c
2
∂
2
B
∂t
2
− ∇
2
B = 0,
where the speed of the wave is
c =
1
√
µ
0
ε
0
Recall from the first lecture that
– ε
0
= 8.85 × 10
−12
m
−3
kg
−1
s
2
C
2
– µ
0
= 4π ×10
−6
m kg C
−2
So
c = 3 × 10
8
m s
−1
,
which is the speed of light!
We now look for plane wave solutions which propagate in the
x
direction,
and are independent of y and z. So we can write our electric field as
E(x) = (E
x
(x, t), E
y
(x, t), E
z
(x, t)).
Hence any derivatives wrt
y
and
z
are zero. Since we know that
∇ · E
= 0,
E
x
must be constant. We take
E
x
= 0. wlog, assume
E
z
= 0, ie the wave propagate
in the
x
direction and oscillates in the
y
direction. Then we look for solutions of
the form
E = (0, E(x, t), 0),
with
1
c
2
∂
2
E
∂t
2
−
∂
2
E
∂x
2
= 0.
The general solution is
E(x, t) = f (x − ct) + g(x + ct).
The most important solutions are the monochromatic waves
E = E
0
sin(kx − ωt).
Definition (Amplitude, wave number and frequency).
(i) E
0
is the amplitude
(ii) k is the wave number.
(iii) ω is the (angular) frequency.
The wave number is related to the wavelength by
λ =
2π
k
.
Since the wave has to travel at speed c, we must have
ω
2
= c
2
k
2
So the value of k determines the value of ω, vice versa.
To solve for B,we use
∇ × E = −
∂B
∂t
.
So B = (0, 0, B) for some B. Hence the equation gives.
∂B
∂t
= −
∂E
∂x
.
So
B =
E
0
c
sin(kx − ωt).
Note that this is uniquely determined by
E
, and we do not get to choose our
favorite amplitude, frequency etc for the magnetic component.
We see that
E
and
B
oscillate in phase, orthogonal to each other, and
orthogonal to the direction of travel. These waves are what we usually consider
to be “light”.
Also note that Maxwell’s equations are linear, so we can add up two solutions
to get a new one. This is particularly important, since it allows light waves to
pass through each other without interfering.
It is useful to use complex notation. The most general monochromatic takes
the form
E = E
0
exp(i(k · x − ωt)),
and
B = B
0
exp(i(k · x − ωt)),
with ω = c
2
|k|
2
.
Definition (Wave vector). k is the wave vector, which is real.
The “actual” solutions are just the real part of these expressions.
There are some restrictions to the values of
E
0
etc due to the Maxwell’s
equations:
∇ · E = 0 ⇒ k · E
0
= 0
∇ · B = 0 ⇒ k · B
0
= 0
∇ × E = −
∂B
∂t
⇒ k × E
0
= ωB
0
If
E
0
and
B
0
are real, then
k, E
0
/c
and
B
0
form a right-handed orthogonal triad
of vectors.
Definition
(Linearly polarized wave)
.
A solution with real
E
0
, B
0
, k
is said to
be linearly polarized.
This says that the waves oscillate up and down in a fixed plane.
If
E
0
and
B
0
are complex, then the polarization is not in a fixed direction.
If we write
E
0
= α + iβ
for α, β ∈ R
3
, then the “real solution” is
Re(E) = α cos(k · x − ωt) − β sin(k ·x −ωt).
Note that
∇·E
= 0 requires that
k ·α
=
k ·β
. It is not difficult to see that this
traces out an ellipse.
Definition
(Elliptically polarized wave)
.
If
E
0
and
B
0
are complex, then it is
said to be elliptically polarized. In the special case where
|α|
=
|β|
and
α ·β
= 0,
this is circular polarization.
We can have a simple application: why metals are shiny.
A metal is a conductor. Suppose the region
x >
0 is filled with a conductor.
E
inc
A light wave is incident on the conductor, ie
E
inc
= E
0
ˆ
y exp(i(kx + ωt)),
with ω = ck.
We know that inside a conductor,
E
= 0, and at the surface,
E
k
= 0. So
E
0
·
ˆ
y|
x=0
= 0.
Then clearly our solution above does not satisfy the boundary conditions!
To achieve the boundary conditions, we add a reflected wave
E
ref
= −E
0
ˆ
y exp(i(−kx − ωt)).
Then our total electric field is
E = E
inc
+ E
ref
.
Then this is a solution to Maxwell’s equations since it is a sum of two solutions,
and satisfies E ·
ˆ
y|
x=0
= 0 as required.
Maxwell’s equations says ∇ × E = −
∂B
∂t
. So
B
inc
=
E
0
c
ˆ
z exp(i(kx − ωt))
B
ref
=
E
0
c
ˆ
z exp(i(−kx − ωt))
This obeys B ·
ˆ
n = 0, where
ˆ
n is the normal to the surface. But we also have
B ·
ˆ
z|
x=0
−
=
2E
0
c
e
−iωt
,
So there is a magnetic field at the surface. However, we know that inside the
conductor, we have
B
= 0. This means that there is a discontinuity across the
surface! We know that discontinuity happens when there is a surface current.
Using the formula we’ve previously obtained, we know that the surface current
is given by
K = ±
2E
0
µ
0
c
ˆ
ye
−iωt
.
So shining a light onto a metal will cause an oscillating current. We can imagine
the process as the incident light hits the conductor, causes an oscillating current,
which generates a reflected wave (since accelerating charges generate light — cf.
IID Electrodynamics)
We can do the same for light incident at an angle, and prove that the incident
angle is equal to the reflected angle.