Subsections


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 $\skew3\vec{\cal E}$ (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

\begin{displaymath}
\varphi({\skew0\vec r}) = - \int_{{\skew0\vec r}_0}^{\skew...
...\underline{\skew0\vec r}}){\,\rm d}{\underline{\skew0\vec r}}
\end{displaymath} (D.47)

where ${\skew0\vec r}_0$ is some arbitrarily chosen reference point. You might think that the value of $\varphi({\skew0\vec r})$ would depend on what integration path you took from the reference point to ${\skew0\vec r}$, but the Stokes’ theorem of calculus says that the difference between integrals leading to the same path must be zero since $\nabla$ $\times$ $\skew3\vec{\cal E}$ is zero.

Now if you evaluate $\varphi$ at a neighboring point ${\skew0\vec r}+{\hat\imath}\partial{x}$ by a path first going to ${\skew0\vec r}$ and from there straight to ${\skew0\vec r}+{\hat\imath}\partial{x}$, the difference in integrals is just the integral over the final segment:

\begin{displaymath}
\varphi({\skew0\vec r}+{\hat\imath}\partial x) - \varphi({...
...\underline{\skew0\vec r}}){\,\rm d}{\underline{\skew0\vec r}}
\end{displaymath} (D.48)

Dividing by $\partial{x}$ and then taking the limit $\partial{x}\to0$ shows that minus the $x$-​derivative of $\varphi$ gives the $x$-​component of the electric field. The same of course for the other components, since the $x$-​direction is arbitrary.

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 $\varphi_m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$I\theta$$\raisebox{.5pt}{$/$}$$2\pi\epsilon_0c^2$. and as you know, an angle like $\theta$ is indeterminate by any integer multiple of $2\pi$.


D.73.2 The Laplace equation

The homogeneous Poisson equation,

\begin{displaymath}
\nabla^2 \varphi = 0 %
\end{displaymath} (D.49)

for some unknown function $\varphi$ is called the Laplace equation. It is very important in many areas of physics and engineering. This note derives some of its generic properties.

The so-called mean-value property says that the average of $\varphi$ over the surface of any sphere in which the Laplace equation holds is the value of $\varphi$ 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

\begin{eqnarray*}
0 & = & \int_{\rm sphere} \nabla^2 \varphi {\,\rm d}^3{\skew...
...-7pt\int}\nolimits \varphi\sin\theta {\,\rm d}\theta{\rm d}\phi
\end{eqnarray*}

the first equality since $\varphi$ 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 $\varphi$ is the same on all spherical surfaces centered around the origin. Since this includes as a limiting case the origin and the average of $\varphi$ over the single point at the origin is just $\varphi$ at the origin, the mean value property follows.

The so called maximum-minimum principle says that either $\varphi$ 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 $\varphi$ 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 $\varphi$ 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 $\varphi$ 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 $z$-​axis. Then the field lines in the $xz$-​plane satisfy

\begin{displaymath}
\frac{{\rm d}z}{{\rm d}x}= \frac{{\cal E}_z}{{\cal E}_x} = \frac{2z^2-x^2}{3zx}
\end{displaymath}

Change to a new variable $u$ by replacing $z$ by $xu$ to get:

\begin{displaymath}
x \frac{{\rm d}u}{{\rm d}x}= - \frac{1+u^2}{3u}
\quad\Lo...
...
\int \frac{3u{\,\rm d}u}{1+u^2} = - \int \frac{{\rm d}x}{x}
\end{displaymath}

Integrating and replacing $u$ again by $z$$\raisebox{.5pt}{$/$}$$x$ gives

\begin{displaymath}
(x^2+z^2)^{3/2} = C x^2
\end{displaymath}

where $C$ represents the integration constant from the integration. Near the origin, $x$ $\vphantom0\raisebox{1.5pt}{$\sim$}$ $z^{3/2}$$\raisebox{.5pt}{$/$}$$C$; 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 $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $C\sin^2\theta$ and $\phi$ constant, and that is also valid outside the $xz$-​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 ${\cal E}_i$ integrates to

\begin{displaymath}
\int_{\rm sphere} {\cal E}_i {\,\rm d}^3{\skew0\vec r}
=...
...ec r}
= - \int_{\rm sphere\ surface} \varphi n_i {\,\rm d}A
\end{displaymath}

where ${\hat\imath}_i$ is the unit vector in the $i$-​direction and the divergence theorem of calculus was used to convert the integral to an integral over the surface area $A$ of the sphere. Noting that the vector ${\vec n}$ normal to the surface of the sphere equals ${\skew0\vec r}$$\raisebox{.5pt}{$/$}$$r$, and that the potential is the ideal dipole one, you get

\begin{displaymath}
\int_{\rm sphere} {\cal E}_i {\,\rm d}^3{\skew0\vec r}= - ...
...c{\vec\wp\cdot{\skew0\vec r}}{r^3}
\frac{r_i}{r} {\,\rm d}A
\end{displaymath}

For simplicity, take the $z$-​axis along the dipole moment; then $\vec\wp\cdot{\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\wp}z$. For the $x$-​component ${\cal E}_x$, $r_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x$ so that the integrand is proportional to $xz$, and that integrates to zero over the surface of the sphere because the negative $x$-​values cancel the positive ones at the same $z$. The same for the $y$-​component of the field, so only the $z$-​component, or more generally, the component in the same direction as $\vec\wp$, has a delta function. For ${\cal E}_z$, you are integrating $z^2$, and by symmetry that is the same as integrating $x^2$ or $y^2$, so it is the same as integrating $\frac13r^2$. Since the surface of the sphere equals $4{\pi}r^2$, 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 $\vec\jmath$ 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 $\int{r_i}^mr_{\overline{\imath}}^nj_i{\,\rm d}^3{\skew0\vec r}$ where $j_i$ is any component $j_1$, $j_2$, or $j_3$ of the current density, ${\overline{\imath}}$ is the index following $i$ in the sequence $\ldots123123\ldots$, $m$ and $n$ are nonnegative integers, and the integration is over all of space. By integration by parts in the $i$-​direction, and using the fact that the current densities vanish at infinity,

\begin{displaymath}
\int r_i^m r_{\overline{\imath}}^n j_i {\,\rm d}^3{\skew0\...
...\frac{\partial j_i}{\partial r_i}
{\,\rm d}^3{\skew0\vec r}
\end{displaymath}

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

\begin{displaymath}
\int r_i^m r_{\overline{\imath}}^n j_i {\,\rm d}^3{\skew0\...
...r_{\overline{\overline{\imath}}}}
{\,\rm d}^3{\skew0\vec r}
\end{displaymath}

where ${\overline{\overline{\imath}}}$ is the index preceding $i$ in the sequence $\ldots123123\ldots$. The final integral can be integrated in the ${\overline{\overline{\imath}}}$-​direction and is then seen to be zero because $\vec\jmath$ vanishes at infinity.

The first integral in the right hand side can be integrated by parts in the ${\overline{\imath}}$-​direction to give the final result:

\begin{displaymath}
\int r_i^m r_{\overline{\imath}}^n j_i {\,\rm d}^3{\skew0\...
...math}}^{n-1} j_{\overline{\imath}}{\,\rm d}^3{\skew0\vec r} %
\end{displaymath} (D.50)

It follows from this equation with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 that
\begin{displaymath}
\int r_i j_{\overline{\imath}}{\,\rm d}^3{\skew0\vec r}= -...
...nt {\skew0\vec r}\times\vec\jmath {\,\rm d}^3{\skew0\vec r} %
\end{displaymath} (D.51)

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

\begin{displaymath}
(i,{\overline{\imath}},{\overline{\overline{\imath}}}) \to...
...) \to ({\overline{\overline{\imath}}},i,{\overline{\imath}})
\end{displaymath}

because only the relative ordering of the indices in the sequence $\ldots123123\ldots$ 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 $m_{\rm {c}}$$\raisebox{.5pt}{$/$}$$q_{\rm {c}}$ of current carrier mass over charge. Then the angular momentum is

\begin{displaymath}
\vec L = \int {\skew0\vec r}\times \frac{m_{\rm c}}{q_{\rm...
...m d}^3{\skew0\vec r}
= \frac{2m_{\rm c}}{q_{\rm c}} \vec\mu
\end{displaymath}


D.73.6 Lorentz forces on a current distribution

Next is the derivation of the Lorentz forces on a given current distribution $\vec\jmath$ in a constant external magnetic field $\skew2\vec{\cal B}_{\rm {ext}}$. The Lorentz force law says that the force $\vec{F}$ on a charge $q$ moving with speed $\vec{v}$ equals

\begin{displaymath}
\vec F = q \vec v \times \skew2\vec{\cal B}_{\rm ext}
\end{displaymath}

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 ${\rm d}^3{\skew0\vec r}$ is:

\begin{displaymath}
{\rm d}\vec F = \vec\jmath \times \skew2\vec{\cal B}_{\rm ext} {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

The net force on the current distribution is therefore zero, because according to (D.50) with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, the integrals of the components of the current distribution are zero.

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

\begin{displaymath}
\vec M = \int {\skew0\vec r}\times
\left(\vec\jmath\times \skew2\vec{\cal B}_{\rm ext}\right) {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

According to the vectorial triple product rule, that is

\begin{displaymath}
\vec M =
\int \left({\skew0\vec r}\cdot \skew2\vec{\cal ...
...right) \skew2\vec{\cal B}_{\rm ext} {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

The second integral is zero because of (D.50) with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. What is left is can be written in index notation as

\begin{displaymath}
M_i =
\int r_i {\cal B}_{\rm ext,i} j_i {\,\rm d}^3{\ske...
...{\overline{\overline{\imath}}}} j_i {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

The first of the three integrals is zero because of (D.50) with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. The other two can be rewritten using (D.51):

\begin{displaymath}
M_i =
- \mu_{\overline{\overline{\imath}}}{\cal B}_{\rm ...
...ine{\imath}}{\cal B}_{\rm ext,{\overline{\overline{\imath}}}}
\end{displaymath}

and in vector notation that reads

\begin{displaymath}
\vec M = \vec \mu \times \skew2\vec{\cal B}_{\rm ext}
\end{displaymath}

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

\begin{displaymath}
- M {\,\rm d}\alpha = - \mu {\cal B}_{\rm ext} \sin\alpha {\,\rm d}\alpha
= {\,\rm d}(\mu {\cal B}_{\rm ext} \cos\alpha)
\end{displaymath}

where $\alpha$ is the angle between $\vec\mu$ and $\skew2\vec{\cal B}_{\rm {ext}}$. By integration, it follows that the work done corresponds to a change in energy for an energy given by

\begin{displaymath}
E_{\rm ext} = - \vec\mu \cdot \skew2\vec{\cal B}_{\rm ext}
\end{displaymath}


D.73.7 Field of a current dipole

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

\begin{displaymath}
\nabla \cdot \skew2\vec{\cal B}= 0 \qquad
\nabla \times \skew2\vec{\cal B}= \frac{1}{\epsilon_0c^2} \vec\jmath
\end{displaymath}

where $\skew2\vec{\cal B}$ vanishes at infinity assuming there is no additional ambient magnetic field.

A magnetic vector potential $\skew3\vec A$ will now be defined as the solution of the Poisson equation

\begin{displaymath}
\nabla^2 \skew3\vec A= - \frac{1}{\epsilon_0c^2} \vec\jmath
\end{displaymath}

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

\begin{displaymath}
\skew2\vec{\cal B}= \nabla \times \skew3\vec A
\end{displaymath}

because it satisfies the equations for $\skew2\vec{\cal B}$: 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 $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, the asymptotic expansion (13.31) of the Green’s function integral shows that at large distances, the components of $\skew3\vec A$ behave as a dipole potential. Specifically,

\begin{displaymath}
A_i \sim \frac{1}{4\pi\epsilon_0c^2r^3}
\sum_{{\underlin...
...ne r}_{\underline i}j_i {\,\rm d}^3{\underline{\skew0\vec r}}
\end{displaymath}

Now the term ${\underline i}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $i$ in the sum does not give a contribution, because of (D.50) with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. The other two terms are

\begin{displaymath}
A_i \sim \frac{1}{4\pi\epsilon_0c^2r^3}
\left[
r_{\ove...
...{\imath}}}j_i {\,\rm d}^3{\underline{\skew0\vec r}}
\right]
\end{displaymath}

with ${\overline{\imath}}$ following $i$ in the sequence $\ldots123123\ldots$ and ${\overline{\overline{\imath}}}$ preceding it. These two integrals can be rewritten using (D.51) to give

\begin{displaymath}
A_i \sim - \frac{1}{4\pi\epsilon_0c^2r^3}
\left[
r_{\o...
...overline{\overline{\imath}}}\mu_{\overline{\imath}}
\right]
\end{displaymath}

Note that the expression between brackets is just the $i$-​th component of ${\skew0\vec r}$ $\times$ $\vec\mu$.

The magnetic field is the curl of $\skew3\vec A$, so

\begin{displaymath}
{\cal B}_i = \frac{\partial A_{\overline{\overline{\imath}...
...overline{\imath}}}{\partial r_{\overline{\overline{\imath}}}}
\end{displaymath}

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

\begin{displaymath}
{\cal B}_i = \frac{3(\vec\mu\cdot{\skew0\vec r}){\skew0\vec r}- \vec\mu r^2}{4\pi\epsilon_0c^2r^5}
\end{displaymath}

This is the same asymptotic field as a charge dipole with strength $\vec\mu$ 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:

\begin{displaymath}
\int {\cal B}_i {\,\rm d}^3{\skew0\vec r}=
\int \frac{\p...
...l r_{\overline{\overline{\imath}}}} {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

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

\begin{displaymath}
\int {\cal B}_i {\,\rm d}^3{\skew0\vec r}=
\int A_{\over...
...\imath}}\frac{r_{\overline{\overline{\imath}}}}{r} {\,\rm d}A
\end{displaymath}

Substituting in the asymptotic expression for $A_i$ above,

\begin{displaymath}
\int {\cal B}_i {\,\rm d}^3{\skew0\vec r}= - \frac{1}{4\pi...
...math}}}) r_{\overline{\overline{\imath}}}{\,\rm d}A
\right]
\end{displaymath}

The integrals of $r_{i}r_{{\overline{\imath}}}$ and $r_{i}r_{{\overline{\overline{\imath}}}}$ are zero, for one because the integrand is odd in $r_i$. The integrals of $r_{{\overline{\imath}}}r_{\overline{\imath}}$ and $r_{{\overline{\overline{\imath}}}}r_{{\overline{\overline{\imath}}}}$ are each one third of the integral of $r^2$ because of symmetry. So, noting that the surface area $A$ of the spherical surface is $4{\pi}r^2$,

\begin{displaymath}
\int {\cal B}_i {\,\rm d}^3{\skew0\vec r}= \frac{2}{3\epsilon_0c^2} \mu_i
\end{displaymath}

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 $\skew3\vec A$. This vector potential satisfies the Poisson equation

\begin{displaymath}
\nabla^2 \skew3\vec A= - \frac{1}{\epsilon_0c^2} \vec\jmath
\end{displaymath}

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

\begin{displaymath}
A_i = \frac{1}{4\pi\epsilon_0c^2}
\int \frac{1}{\vert{\s...
...erline{\skew0\vec r}}) {\,\rm d}^3 {\underline{\skew0\vec r}}
\end{displaymath}

The magnetic field is the curl of $\skew3\vec A$,

\begin{displaymath}
{\cal B}_i = \frac{\partial A_{\overline{\overline{\imath}...
...overline{\imath}}}{\partial r_{\overline{\overline{\imath}}}}
\end{displaymath}

or substituting in and differentiating under the integral

\begin{displaymath}
{\cal B}_i = - \frac{1}{4\pi\epsilon_0c^2}
\int \frac{r_...
...erline{\skew0\vec r}}) {\,\rm d}^3 {\underline{\skew0\vec r}}
\end{displaymath}

In vector notation that gives the Biot-Savart law

\begin{displaymath}
\skew2\vec{\cal B}= - \frac{1}{4\pi\epsilon_0c^2}
\int \...
...t^3} \times \vec\jmath {\,\rm d}^3 {\underline{\skew0\vec r}}
\end{displaymath}

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

\begin{displaymath}
\vec\jmath {\,\rm d}^3 {\underline{\skew0\vec r}}= I {\,\rm d}{\underline{\skew0\vec r}}
\end{displaymath}

where in the right hand side ${\underline{\skew0\vec r}}$ describes the position of the centerline of the wire and $I$ is the current through the wire. More specifically, $I$ is the integral of the current density over the cross section of the wire. The Biot-Savart law becomes

\begin{displaymath}
\skew2\vec{\cal B}= - \frac{1}{4\pi\epsilon_0c^2}
\int \...
...underline{\skew0\vec r}}) {\,\rm d}{\underline{\skew0\vec r}}
\end{displaymath}

where the integration is over all infinitesimal segments ${\,\rm d}{\underline{\skew0\vec r}}$ of the wires.