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
/r−r
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 (BiotSavart 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 nonzero only on the curve.