Today in Physics 218: the Maxwell equations
‰ Beyond magnetoquasistatics
‰ Displacement current,
and Maxwell’s repair
of Ampère’s Law
‰ The Maxwell
equations
‰ Symmetry of the
equations: magnetic
monopoles?
14 January 2004
Rainbow over the Potala Palace,
Lhasa, Tibet, by Galen Rowell.
Physics 218, Spring 2004
1
Beyond magnetoquasistatics
In PHY 217, we came up with the basic equations for
electrodynamics, namely Gauss’s law, the “no magnetic
monopoles” law, Faraday’s law and Ampère’s law:
ρ
—⋅Ε =
ε0
—⋅B = 0
∂B
—× E = −
∂t
— × B = µ0 J
— ⋅ Ε = 4πρ
1 ∂B
—× E = −
c ∂t
14 January 2004
—⋅B = 0
4π
—×B =
J
c
in MKS units, or
in our preferred cgs units.
Physics 218, Spring 2004
2
Beyond magnetoquasistatics (continued)
As has no doubt been mentioned to you, these equations are,
strictly speaking, false. Why? Because we know that the
divergence of a curl has to be zero, yet from these equations,
1 ∂
— ⋅ (— × E) = −
—⋅B = 0
( — ⋅ B always = 0 ) , but
c ∂t
4π
4π ∂ρ
∂ρ
⎛
⎞
0
continuity:
+
⋅
=
— ⋅ (— × B) =
—⋅J = −
—
J
⎜
⎟
c
c ∂t
t
∂
⎝
⎠
∂ρ
only if
=0
= 0.
∂t
Thus these equations are only an approximation, good only
when the rate of change of charge density is small enough.
We call this approximation magnetoquasistatics.
14 January 2004
Physics 218, Spring 2004
3
Beyond magnetoquasistatics (continued)
The problem, of course, is Ampère’s law. We derived this law
from the Biot-Savart law, using along the way the
magnetostatic condition — ⋅ J = 0.
Here, I’ll remind you how it went; please consult your notes
from PHY 217, or view
http://www.pas.rochester.edu/~dmw/phy217/Lectures/Lect_27b.pdf
for the context of the derivation.
14 January 2004
Physics 218, Spring 2004
4
Flashback: Derivation of Ampère’s Law
Any vector field is uniquely specified by its divergence and
curl. What are the divergence and curl of B?
Consider a volume V to contain current I, current density J ( r ′ ) :
1 J ( r ′ ) × rˆ
′
τ
B (r ) =
d
c
r2
∫
Denote gradient with respect to
the components of r and r’ by
— and —′. Now note that
⎛1⎞
⎛1⎞
— ⎜ ⎟ = −—′ ⎜ ⎟ (because r = r − r ′),
⎝r⎠
⎝r⎠
rˆ
⎛1⎞
and — ⎜ ⎟ = − 2 .
⎝r⎠
r
14 January 2004
Physics 218, Spring 2004
S
V
P
J ( r ′)
r
r
r’
dτ ′
5
Flashback (continued)
With these,
1
B (r ) = −
c
∫
V
rˆ
1
1⎞
⎛
× J ( r ′ ) dτ ′ =
— ⎜ ⎟ × J ( r ′ ) dτ ′
2
c
⎝r⎠
r
V
∫
J ( r ′)
1
dτ ′
= —×
c
r
∫
( remember, J ≠ f ( r ) ) .
V
This is a useful form for B, which we will use a lot next
lecture too (the integral turns out to be the magnetic vector
potential, A). Take its divergence:
⎞
J ( r ′)
1 ⎛⎜
dτ ′ ⎟ = 0 .
— ⋅ B (r ) = — ⋅ — ×
⎟
c ⎜
r
V
⎝
⎠
∫
14 January 2004
Physics 218, Spring 2004
The divergence
of any curl is
zero, remember.
6
Flashback (continued)
Integrate this last expression over any volume:
∫ — ⋅ B ( r ) dτ = v∫ B ⋅ da = 0
.
Compare these to the expressions for E in electrostatics, and
we see that magnetostatics involves no counterpart of charge:
there’s no “magnetic charge.”
Now for the curl:
J ( r ′)
1
— × B (r ) = — × — ×
dτ ′ .
r
c
V
Use Product Rule #10:
∫
— × — × A = — ( — ⋅ A) − ∇2 A :
14 January 2004
Physics 218, Spring 2004
7
Flashback (continued)
⎞ 1
J ( r ′)
J ( r ′)
1 ⎛⎜
2
dτ ′ ⎟ − ∇
dτ ′
— × B (r ) = — — ⋅
⎟ c
r
r
c ⎜
V
⎝ V
⎠
⎞ 1
J ( r ′)
1 ⎜⎛
⎛1⎞
= — —⋅
dτ ′ ⎟ −
J ( r ′ ) ∇ 2 ⎜ ⎟ dτ ′ .
⎟ c
r
r⎠
c ⎜
⎝
V
⎝V
⎠
Now use your old friend Product Rule #5,
∫
∫
∫
— ⋅ ( fA ) = f ( — ⋅ A ) + A ⋅ ( — f ) ,
to write
—⋅
J ( r ′)
14 January 2004
∫
r
=0 (J independent of r)
1⎞
1⎞
1⎞
⎛
⎛
⎛
dτ ′ = ⎜ ⎟ — ⋅ J ( r ′ ) + J ( r ′ ) ⋅ — ⎜ ⎟ = J ( r ′ ) ⋅ — ⎜ ⎟
⎝r⎠
⎝r⎠
⎝r⎠
Physics 218, Spring 2004
8
Flashback (continued)
⎛ rˆ
⎛1⎞
∇ ⎜ ⎟ = — ⋅—⎜ ⎟ = — ⋅⎜ 2
⎝r⎠
⎝r⎠
⎝r
2 ⎛1⎞
Also,
⎞
3
=
πδ
4
(r) ,
⎟
⎠
so — × B ( r ) = 1 — J ( r ′ ) ⋅ — ⎛ 1 ⎞ dτ ′ + 4π
⎜ ⎟
c
c
⎝r⎠
∫
V
3
′
J
r
δ
(
)
( r − r ′ ) dτ ′
∫
V
1
4π
⎛1⎞
= − — ∫ J ( r ′ ) ⋅ —′ ⎜ ⎟ dτ ′ +
J (r )
c
c
⎝r⎠
V
Here’s where
we assumed
.
statics:
Use Product Rule #5 again, on the first term:
⎛ J ( r ′) ⎞ 1
⎛ J ( r ′) ⎞
1⎞
⎛
J ( r ′ ) ⋅ —′ ⎜ ⎟ = —′ ⋅ ⎜
⎟ − —′ ⋅ J ( r ′ ) = —′ ⋅ ⎜
⎟
r
r
r
r
⎝ ⎠
⎝
⎠
⎝
⎠
=0 in magnetostatics
14 January 2004
Physics 218, Spring 2004
9
Flashback (continued)
⎛ J ( r ′ ) ⎞ ⎞ 4π
1 ⎛⎜
— × B ( r ) = − — ∫ —′ ⋅ ⎜
J (r )
⎟ dτ ′ ⎟ +
c ⎜
r ⎠ ⎟ c
⎝
V
⎝
⎠
⎞ 4π
1 ⎛ J ( r ′)
= − —⎜ v
⋅ da ′ ⎟ +
J (r ) .
∫
⎟ c
c ⎜
r
⎝S
⎠
But by definition J = 0 on the surface, so the integral vanishes:
So,
4π
— × B (r ) =
J (r ) .
c
14 January 2004
Ampère’s Law
Physics 218, Spring 2004
10
Beyond magnetoquasistatics (continued)
We could go back and fix this:
‰ substitute − ∂ρ ∂t for ∇ ′ ⋅ J ( r ′ ) ,
‰ do another integration by parts,
‰ arguing that two more surface integrals vanish, and
‰ substitute ∇ ⋅ E 4π for ρ ,
and we’d naturally get a more general form of Ampère’s law
that is valid for any time variation in the charge density.
I encourage you to do this, by way of review; here, let’s just
take a shortcut to the answer, and demonstrate that it works:
4π
1 ∂E
—×B =
J+
c
c ∂t
14 January 2004
Physics 218, Spring 2004
11
Beyond magnetoquasistatics (continued)
Does this work? Yes, because
4π
— ⋅ (— × B) =
c
4π
=
c
=0
so
—×B =
1 ∂
—⋅ J +
—⋅E
c ∂t
∂ρ ⎞
⎛
⎜—⋅ J +
⎟
∂
t
⎝
⎠
,
4π
1 ∂E
J+
c
c ∂t
Use Gauss’s law…
Use continutity…
∂E
⎡
⎤
—
B
J
×
=
+
µ
µ
ε
in
MKS
.
0
0 0
⎢⎣
⎥
∂t
⎦
must therefore be the correct generalization of Ampère’s law
for time-variable charge and current densities.
14 January 2004
Physics 218, Spring 2004
12
Displacement current
Maxwell was, of course, the first to get this result. He didn’t
do it this way, though; he put in the extra term because it was
the only way to get a wave equation by combining the four
differential equations of electrodynamics, that resembled the
equations for elastic waves in matter. He noted afterward
that it fixed the div-curl-B problem. Maxwell thought of this
extra term as related to a source he called the displacement
current density.
The role of this term as a current density is made clearer in
integral form, and applied to the simple example of a
parallel-plate capacitor charging up:
14 January 2004
Physics 218, Spring 2004
13
Displacement current (continued)
S2 ( enclosing capacitor plate )
C
E
I
S1
A
w
+
Consider the capacitor plates to be closely spaced, even
though they’re not drawn that way, and consider two
surfaces S1 and S2, both bounded by circle C, with S2
ballooning to enclose the nearer plate.
14 January 2004
Physics 218, Spring 2004
14
Displacement current (continued)
Integrate the new “corrected” form of Ampère’s law over
either of these surfaces:
4π
1 ∂E
⋅ da ,or, by Stokes’s
J ⋅ da +
( — × B ) ⋅ da =
theorem,
c
c ∂t
∫
∫
S
v∫
∫
S
B ⋅ dA =
C
S
4π
1 ∂E
⋅ da .
I encl +
c
c ∂t
∫
S
Considering the circle C to be an Ampèrean loop, we could use
this to calculate B. Most would use S = S1 for the area integral,
and note that the enclosed current is just the current I in the
wire. But the enclosed current for S = S2 is zero, so that term
must vanish. All S2 intercepts is electric field, E = V/w:
14 January 2004
Physics 218, Spring 2004
15
Displacement current (continued)
∂E 1 ∂V
1 ∂q
1
4π
=
=
=
I=
I
wC ∂t w A
A
∂t w ∂t
4π w
( !)
Since the electric field is constant between the plates and very
small outside them,
1 ∂E
1 4π
4π
⋅ da =
IA =
I ,
c ∂t
c A
c
∫
S2
4π
just like the other surface; so B ⋅ dA =
I no matter which
c
surface is used.
C
v∫
14 January 2004
Physics 218, Spring 2004
16
Displacement current (continued)
If we therefore define the displacement current density as
1 ∂E
Jdisp =
,
4π ∂t
then there is a “displacement current” between the capacitor
plates that is exactly equal to I, and there is a more general
“current” that is continuous throughout the circuit.
14 January 2004
Physics 218, Spring 2004
17
The Maxwell equations
So here are the Maxwell equations, in vacuum, in final form:
— ⋅ Ε = 4πρ
1 ∂B
—× E = −
c ∂t
ρ
—⋅Ε =
ε0
—× E = −
14 January 2004
—⋅B = 0
4π
1 ∂E
J+
—×B =
c
c ∂t
—⋅B = 0
∂B
∂t
— × B = µ0 J + µ0 ε 0
∂E
∂t
Physics 218, Spring 2004
in cgs units, or
in MKS units.
18
Magnetic monopoles
The only remaining sense in which these equations may still
be approximate is if magnetic charges (monopoles) exist. We
will see a powerful argument for searching for magnetic
monopoles in the first homework set (Griffiths problem 8.12);
they would also symmetrize the Maxwell equations. Note
that if there are no electric charges or currents, the Maxwell
equations are symmetrical:
—⋅Ε = 0
1 ∂B
—× E = −
c ∂t
14 January 2004
—⋅B = 0
1 ∂E
—×B =
c ∂t
Physics 218, Spring 2004
19
Magnetic monopoles (continued)
If, on the other hand, there were magnetic as well as electric
monopoles, with magnetic charge density η and magnetic
current density K, then we’d have
— ⋅ Ε = 4πρ
4π
1 ∂B
K−
—× E = −
c
c ∂t
— ⋅ B = 4πη
4π
1 ∂E
J+
—×B =
c
c ∂t
where, if both electric and magnetic charge were conserved,
∂ρ
+—⋅J = 0
∂t
∂η
+—⋅K = 0
∂t
14 January 2004
Physics 218, Spring 2004
20