Scott Hughes
27 April 2004
Massachusetts Institute of Technology
Department of Physics
8.022 Spring 2004
Lecture 19:
Displacement current. Maxwell’s equations.
19.1
Inconsistent equations
Over the course of this semester, we have derived 4 relationships between the electric and
magnetic fields on the one hand, and charge and current density on the other. They are:
Gauss’s law:
~
∇ ·
~
E
= 4
πρ
Magnetic law:
~
∇ ·
~
B
= 0
Faraday’s law:
~
∇ ×
~
E
=

1
c
∂
~
B
∂t
Ampere’s law:
~
∇ ×
~
B
=
4
π
~
J
c
As written, these equations are slightly inconsistent.
We can see this inconsistency very
easily — we just take the divergence of both sides of Ampere’s law. Look at the left hand
side first:
Left hand side:
~
∇ ·
‡
~
∇ ×
~
B
·
= 0
.
This follows from the rule that the divergence of the curl is always zero (as you proved on
pset 3). Now look at the right hand side:
Right hand side:
~
∇ ·
4
π
~
J
c
=

4
π
c
∂ρ
∂t
.
Here, I’ve used the continuity equation,
~
∇ ·
~
J
=

∂ρ/∂t
.
Ampere’s law is inconsistent with the continuity equation except when
∂ρ/∂t
= 0
!!!
This
is actually a fairly common circumstance in many applications, so it’s not too surprising
that we can go pretty far with this “incomplete” version of Ampere’s law. But, as a matter
of principle — and, as we shall soon see, of practice as well — it’s just not right. We need
to fix it somehow.
19.2
Fixing the inconsistency
We can fix up this annoying little inconsistency by inspired guesswork. When we take the
divergence of Ampere’s left hand side, we get zero — no uncertainty about this whatsoever,
it’s just ZERO. But, we should be able to add a function to the right hand side such that
the divergence of the right side is forced to be zero as well.
Let’s suppose that our “generalized Ampere’s law” takes the form
~
∇ ×
~
B
=
4
π
~
J
c
+
~
F
.
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Our goal now is to figure out what
~
F
must be. Taking the divergence of both sides, we find
0
=
4
π
~
∇ ·
~
J
c
+
~
∇ ·
~
F
→
~
∇ ·
‡
c
~
F
·
=
4
π
∂ρ
∂t
.
Our mystery function has the property that when we take its divergence (and multiply by
c
), we get the rate of change of charge density.
This looks a lot like Gauss’s law! If we take the time derivative of Gauss’s law, we have
∂
∂t
~
∇ ·
~
E
=
4
π
∂ρ
∂t
→
4
π
∂ρ
∂t
=
~
∇ ·
∂
~
E
∂t
On the last line, I’ve use the fact that it is OK to exchange the order of partial derivatives:
∂
∂t
~
∇
=
ˆ
x
∂
∂t
∂
∂x
+ ˆ
y
∂
∂t
∂
∂y
+ ˆ
z
∂
∂t
∂
∂z
=
ˆ
x
∂
∂x
∂
∂t
+ ˆ
y
∂
∂y
∂
∂t
+ ˆ
z
∂
∂z
∂
∂t
=
~
∇
∂
∂t
.
(This actually works in any coordinate system; Cartesian coordinates are good enough to
demonstrate the point.)
Substituting in for 4
π ∂ρ/∂t
, we have
~
∇ ·
‡
c
~
F
·
=
~
∇ ·
∂
~
E
∂t
which tells us
~
F
=
1
c
∂
~
E
∂t
.
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 Spring '08
 Hughes
 Current, Mass, Magnetic Field, Ampere, displacement current density

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