Scott Hughes
28 April 2005
Massachusetts Institute of Technology
Department of Physics
8.022 Spring 2005
Lecture 20:
Wave equation & electromagnetic radiation
20.1
Revisiting Maxwell’s equations
In our last lecture, we finally ended up with Maxwell’s equations, the four equations which
encapsulate everything we know about electricity and magnetism. These equations are:
Gauss’s law:
~
∇ ·
~
E
= 4
πρ
Magnetic law:
~
∇ ·
~
B
= 0
Faraday’s law:
~
∇ ×
~
E
=

1
c
∂
~
B
∂t
Generalized Ampere’s law:
~
∇ ×
~
B
=
4
π
c
~
J
+
1
c
∂
~
E
∂t
.
In this lecture, we will focus on the source free versions of these equations: we set
ρ
= 0 and
~
J
= 0. We then have
~
∇ ·
~
E
=
0
~
∇ ·
~
B
=
0
~
∇ ×
~
E
=

1
c
∂
~
B
∂t
~
∇ ×
~
B
=
1
c
∂
~
E
∂t
.
The source free Maxwell’s equations show us that
~
E
and
~
B
are
coupled
: variations in
~
E
act
as a source for
~
B
, which in turn acts as a source for
~
E
, which in turn acts as a source for
~
B
,
which ... The goal of this lecture is to fully understand this coupled behavior.
To do so, we will find it easiest to first
uncouple
these equations. We do this by taking
the curl of each equation. Let’s begin by looking at
~
∇ ×
‡
~
∇ ×
~
E
·
=
~
∇ ×

1
c
∂
~
B
∂t
.
The curl of the lefthand side of this equation is
~
∇ ×
~
∇ ×
~
E
=
~
∇
‡
~
∇ ·
E
·
 ∇
2
~
E
=
∇
2
~
E .
(We used this curl identity back in Lecture 13; it is simple to prove, albeit not exactly
something you’d want to do at a party.) The simplification follows because we have restricted
ourselves to the source free equations — we have
~
∇ ·
~
E
= 0. Now, look at the curl of the
righthand side:
~
∇ ×

1
c
∂
~
B
∂t
=

1
c
∂
∂t
‡
~
∇ ×
~
B
·
=

1
c
2
∂
2
~
E
∂t
2
.
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Putting the left and right sides together, we end up with
∂
2
~
E
∂t
2

c
2
∇
2
~
E
= 0
.
Repeating this procedure for the other equation, we end up with something that is essentially
identical, but for the magnetic field:
∂
2
~
B
∂t
2

c
2
∇
2
~
B
= 0
.
As we shall now discuss, these equations are particularly special and important.
20.2
The wave equation
Let us focus on the equation for
~
E
; everything we do will obviously pertain to the
~
B
equation
as well. Furthermore, we will simplify things initially by imagining that
~
E
only depends on
x
and
t
. The equation we derived for
~
E
then reduces to
∂
2
~
E
∂t
2

c
2
∂
2
~
E
∂x
2
= 0
.
20.2.1
General considerations
At this point, it is worth taking a brief detour to talk about equations of this form more
generally. This equation for the electric field is a special case of
∂
2
f
∂t
2

v
2
∂
2
f
∂x
2
= 0
.
This equation is satisfied by
ANY
function whatsoever
1
provided that the argument of the
function is written in the following special way:
f
=
f
(
x
±
vt
)
.
This is easy to prove. Let
u
=
x
±
vt
, so that
f
=
f
(
u
). Then, using the chain rule,
∂f
∂x
=
∂u
∂x
∂f
∂u
=
∂f
∂u
;
it follows quite obviously that
∂
2
f
∂x
2
=
∂
2
f
∂u
2
.
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 Spring '10
 ScottHughes
 Light, wave equation, Electromagnetic wave equation

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