This section gives various derivations for the electromagnetostatic solutions of chapter 13.3.
This subsection shows that if the curl of the electric field (or of any other vector field, like the magnetic one or a force field), is zero, it is minus the gradient of some potential.
That potential can be defined to be
Now if you evaluate at a neighboring point
by a path first going to and from there
straight to , the difference in integrals
is just the integral over the final segment:
Note that if regions are multiply connected, the potential may not be quite unique. The most important example of that is the magnetic potential of an infinite straight electric wire. Since the curl of the magnetic field is nonzero inside the wire, the path of integration must stay clear of the wire. It then turns out that the value of the potential depends on how many times the chosen integration path wraps around the wire. Indeed, the magnetic potential is . and as you know, an angle like is indeterminate by any integer multiple of .
The homogeneous Poisson equation,
The so-called mean-value property says that the average of over the surface of any sphere in which the Laplace equation holds is the value of at the center of the sphere. To see why, for convenience take the center of the sphere as the origin of a spherical coordinate system. Now
The so called maximum-minimum principle says that either is
constant everywhere or its maximum and minimum are on a boundary or at
infinity. The reason is the mean-value property above. Suppose there
is an absolute maximum in the interior of the region in which the
Laplace equation applies. Enclose the maximum by a small sphere.
Since the values of would be less than the maximum on the
surface of the sphere, the average value on the surface must be less
than the maximum too. But the mean value theorem says it must be the
same. The only way around that is if is completely constant
in the sphere, but then the
maximum is not a true
maximum. And then you can start
show that is constant everywhere. Minima go the same way.
The only solution of the Laplace equation in all of space that is zero at infinity is zero everywhere. In more general regions, as long as the solution is zero on all boundaries, including infinity where relevant, then the solution is zero everywhere. The reason is the maximum-minimum principle: if there was a point where the solution was positive/negative, then there would have to be an interior maximum/minimum somewhere.
The solution of the Laplace equation for given boundary values is unique. The reason is that the difference between any two solutions must satisfy the Laplace equation with zero boundary values, hence must be zero.
The egg shape of the ideal dipole field lines can be found by assuming
that the dipole is directed along the -axis. Then the field
lines in the -plane satisfy
Next is the delta function in the electric field generated by a charge
distribution that is contracted to an ideal dipole. To find the
precise delta function, the electric field can be integrated over a
small sphere, but still large enough that on its surface the ideal
dipole potential is valid. The integral will give the strength of the
delta function. Since the electric field is minus the gradient of the
potential, an arbitrary component integrates to
In subsequent derivations, various integrals of the current density are needed. In all cases it is assumed that the current density vanishes strongly outside some region. Of course, normally an electric motor or electromagnet has electrical leads going towards and away from of it; it is assumed that these are stranded so tightly together that their net effect can be ignored.
Consider an integral like where
is any component , , or of the current
density, is the index following in the sequence
, and are nonnegative integers, and
the integration is over all of space. By integration by parts in the
-direction, and using the fact that the current densities
vanish at infinity,
The first integral in the right hand side can be integrated by parts
in the -direction to give the final result:
In quantum applications, it is often necessary to relate the dipole
moment to the angular momentum of the current carriers. Since the
current density is the charge per unit volume times its velocity, you
get the linear momentum per unit volume by multiplying by the ratio
of current carrier mass over charge. Then the
angular momentum is
Next is the derivation of the Lorentz forces on a given current
distribution in a constant external magnetic field
. The Lorentz force law says that the force
on a charge moving with speed equals
In terms of a current distribution, the moving charge per unit volume
times its velocity is the current density, so the force on a volume
The moment is not zero, however. It is given by
When the (frozen) current distribution is slowly rotated around the
axis aligned with the moment vector, the work done is
A current density creates a magnetic field because of
Maxwell’s second and fourth equations for the divergence and curl
of the magnetic field:
A magnetic vector potential will now be defined as the
solution of the Poisson equation
You might of course wonder whether there might not be more than one magnetic field that has the given divergence and curl and is zero at infinity. The answer is no. The difference between any two such fields must have zero divergence and curl. Therefore the curl of the curl of the difference is zero too, and the vectorial triple product shows that equal to minus the Laplacian of the difference. If the Laplacian of the difference is zero, then the difference is zero, since the difference is zero at infinity (subsection 2). So the solutions must be the same.
Since the integrals of the current density are zero, (D.50)
with 0, the asymptotic expansion (13.31) of the
Green’s function integral shows that at large distances, the
components of behave as a dipole potential. Specifically,
The magnetic field is the curl of , so
However, for an ideal current dipole, the delta function at the origin
will be different than that derived for a charge dipole in the first
subsection. Integrate the magnetic field over a sphere large enough
that on its surface, the asymptotic field is accurate:
In the previous section, it was noted that the magnetic field of a
current distribution is the curl of a vector potential .
This vector potential satisfies the Poisson equation
Now assume that the current distribution is limited to one or more
thin wires, as it usually is. In that case, a volume element of
nonzero current distribution can be written as