- D.73.1 Existence of a potential
- D.73.2 The Laplace equation
- D.73.3 Egg-shaped dipole field lines
- D.73.4 Ideal charge dipole delta function
- D.73.5 Integrals of the current density
- D.73.6 Lorentz forces on a current distribution
- D.73.7 Field of a current dipole
- D.73.8 Biot-Savart law

D.73 Various electrostatic derivations.

This section gives various derivations for the electromagnetostatic solutions of chapter 13.3.

D.73.1 Existence of a potential

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

(D.47) |

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:

(D.48) |

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 .

D.73.2 The Laplace equation

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 first equality since satisfies the Laplace equation, the second because of the divergence theorem, the third because the integral is zero, so a constant factor does not make a difference, and the fourth by changing the order of integration and differentiation. It follows that the average of is the same on all spherical surfaces centered around the origin. Since this includes as a limiting case the origin and the average of over the single point at the origin is just at the origin, the mean value property follows.

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 sphere-hopping

to
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.

D.73.3 Egg-shaped dipole field lines

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

Change to a new variable by replacing by to get:

Integrating and replacing again by gives

where represents the integration constant from the integration. Near the origin, ; therefore the field line has infinite curvature at the origin, explaining the pronounced egg shape. Rewritten in spherical coordinates, the field lines are given by and constant, and that is also valid outside the -plane.

D.73.4 Ideal charge dipole delta function

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

where is the unit vector in the -direction and the divergence theorem of calculus was used to convert the integral to an integral over the surface area of the sphere. Noting that the vector normal to the surface of the sphere equals , and that the potential is the ideal dipole one, you get

For simplicity, take the -axis along the dipole moment; then . For the -component , so that the integrand is proportional to , and that integrates to zero over the surface of the sphere because the negative -values cancel the positive ones at the same . The same for the -component of the field, so only the -component, or more generally, the component in the same direction as , has a delta function. For , you are integrating , and by symmetry that is the same as integrating or , so it is the same as integrating . Since the surface of the sphere equals , the delta function included in the expression for the field of a dipole as listed in table 13.2 is obtained.

D.73.5 Integrals of the current density

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,

Now use the fact that the divergence of the current density is zero since the charge density is constant for electromagnetostatic solutions:

where is the index preceding in the sequence . The final integral can be integrated in the -direction and is then seen to be zero because vanishes at infinity.

The first integral in the right hand side can be integrated by parts
in the -direction to give the final result:

with the current distribution’s dipole moment. In these expressions, you can swap indices as

because only the relative ordering of the indices in the sequence is relevant.

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

D.73.6 Lorentz forces on a current distribution

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
element is:

The net force on the current distribution is therefore zero, because according to (D.50) with 0, the integrals of the components of the current distribution are zero.

The moment is not zero, however. It is given by

According to the vectorial triple product rule, that is

The second integral is zero because of (D.50) with 1, 0. What is left is can be written in index notation as

The first of the three integrals is zero because of (D.50) with 1, 0. The other two can be rewritten using (D.51):

and in vector notation that reads

When the (frozen) current distribution is slowly rotated around the
axis aligned with the moment vector, the work done is

where is the angle between and . By integration, it follows that the work done corresponds to a change in energy for an energy given by

D.73.7 Field of a current dipole

A current density creates a magnetic field because of
Maxwell’s second and fourth equations for the divergence and curl
of the magnetic field:

where vanishes at infinity assuming there is no additional ambient magnetic field.

A magnetic vector potential will now be defined as the
solution of the Poisson equation

that vanishes at infinity. Taking the divergence of this equation shows that the divergence of the vector potential satisfies a homogeneous Poisson equation, because the divergence of the current density is zero, with zero boundary conditions at infinity. Therefore the divergence of the vector potential is zero. It then follows that

because it satisfies the equations for : the divergence of any curl is zero, and the curl of the curl of the vector potential is according to the vectorial triple product its Laplacian, hence the correct curl of the magnetic field.

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,

Now the term in the sum does not give a contribution, because of (D.50) with 1, 0. The other two terms are

with following in the sequence and preceding it. These two integrals can be rewritten using (D.51) to give

Note that the expression between brackets is just the -th component of .

The magnetic field is the curl of , so

and substituting in for the vector potential from above, differentiating, and cleaning up produces

This is the same asymptotic field as a charge dipole with strength would have.

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:

Using the divergence theorem, the right hand side becomes an integral over the surface of the sphere:

Substituting in the asymptotic expression for above,

The integrals of and are zero, for one because the integrand is odd in . The integrals of and are each one third of the integral of because of symmetry. So, noting that the surface area of the spherical surface is ,

That gives the strength of the delta function for an ideal current dipole.

D.73.8 Biot-Savart law

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

The solution for the vector potential can be written explicitly in terms of the current density using the Green’s function integral (13.29):

The magnetic field is the curl of ,

or substituting in and differentiating under the integral

In vector notation that gives the Biot-Savart law

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

where in the right hand side describes the position of the centerline of the wire and is the current through the wire. More specifically, is the integral of the current density over the cross section of the wire. The Biot-Savart law becomes

where the integration is over all infinitesimal segments of the wires.