3Magnetostatics

IB Electromagnetism

3.2 Vector potential
For general current distributions J, we also need to solve · B = 0.
Recall that for the electric case, the equation is
· E
=
ρ/ε
0
. For the
B
field, we have 0 on the right hand side instead of
ρ/ε
0
. This is telling us that
there are no magnetic monopoles, i.e. magnetic charges.
The general solution to this equation is B = × A for some A.
Definition (Vector potential). If B = × A, then A is the vector potential.
The other Maxwell equation then says
× B = −∇
2
A + ( · A) = µ
0
J. ()
This is rather difficult to solve, but it can be made easier by noting that
A
is
not unique. If A is a vector potential, then for any function χ(x),
A
0
= A + χ
is also a vector potential of B, since × (A + χ) = × A.
The transformation A 7→ A + χ is called a gauge transformation.
Definition
((Coulomb) gauge)
.
Each choice of
A
is called a gauge. An
A
such
that · A = 0 is called a Coulomb gauge.
Proposition. We can always pick χ such that · A
0
= 0.
Proof.
Suppose that
B
=
×A
with
·A
=
ψ
(
x
). Then for any
A
0
=
A
+
χ
,
we have
· A
0
= A +
2
χ = ψ +
2
χ.
So we need a
χ
such that
2
χ
=
ψ
. This is the Poisson equation which we
know that there is always a solution by, say, the Green’s function. Hence we can
find a χ that works.
If B = × A and · A = 0, then the Maxwell equation () becomes
2
A = µ
0
J.
Or, in Cartesian components,
2
A
i
= µ
0
J
i
.
This is 3 copies of the Poisson equation, which we know how to solve using
Green’s functions. The solution is
A
i
(r) =
µ
0
4π
Z
J
i
(r
0
)
|r r
0
|
dV
0
,
or
A(r) =
µ
0
4π
Z
J(r
0
)
|r r
0
|
dV
0
,
both integrating over r
0
.
We have randomly written down a solution of
2
A
=
µ
0
J
. However, this
is a solution to Maxwell’s equations only if it is a Coulomb gauge. Fortunately,
it is:
· A(r) =
µ
0
4π
Z
J(r
0
) ·
1
|r r
0
|
dV
0
=
µ
0
4π
Z
J(r
0
) ·
0
1
|r r
0
|
dV
0
Here we employed a clever trick — differentiating 1
/|rr
0
|
with respect to
r
is the
negative of differentiating it with respect to
r
0
. Now that we are differentiating
against r
0
, we can integrate by parts to obtain
=
µ
0
4π
Z
0
·
J(r
0
)
|r r
0
|
0
· J(r
0
)
|r r
0
|
dV
0
.
Here both terms vanish. We assume that the current is localized in some region
of space so that
J
=
0
on the boundary. Then the first term vanishes since it is
a total derivative. The second term vanishes since we assumed that the current
is steady ( · J = 0). Hence we have Coulomb gauge.
Law (Biot-Savart law). The magnetic field is
B(r) = × A =
µ
0
4π
Z
J(r
0
) ×
r r
0
|r r
0
|
3
dV
0
.
If the current is localized on a curve, this becomes
B =
µ
0
I
4π
I
C
dr
0
×
r r
0
|r r
0
|
3
,
since J(r
0
) is non-zero only on the curve.