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Dirac Monopoles

Consider a electric charge $e$ at the origin and an monopolar charge $g$ at an arbitrary point on the $z$ axis. From the generalized form of MEs, we expect the electric field to be given by the well-known:

\begin{displaymath}
\mbox{\boldmath$E$}= \frac{e \hat{\mbox{\boldmath$r$}}}{4\pi\epsilon_0 r^2}
\end{displaymath} (8.125)

at an arbitrary point in space. Similarly, we expect the magnetic field of the monopolar charge $g$ to be:
\begin{displaymath}
\mbox{\boldmath$B$}= \frac{g \hat{\mbox{\boldmath$r$}}'}{4\pi\mu_0 r'^2}
\end{displaymath} (8.126)

where $\mbox{\boldmath$r$}= \mbox{\boldmath$z$}+ \mbox{\boldmath$r$}'$.

The momentum density of this pair of fields is given as noted above by:

\begin{displaymath}
\mbox{\boldmath$g$}= \frac{1}{c^2}(\mbox{\boldmath$E$}\times \mbox{\boldmath$H$})
\end{displaymath} (8.127)

and if one draws pictures and uses one's right hand to determine directions, it is clear that the field momentum is directed around the $e-g$ axis in the right handed sense. In fact the momentum follows circular tracks around this axis in such a way that the field has a non-zero static angular momentum.

The system obviously has zero total momentum from symmetry. This means one can use any origin to compute the angular momentum. To do so, we compute the angular momentum density as:

\begin{displaymath}
\frac{1}{c^2}\mbox{\boldmath$r$}\times \left(\mbox{\boldmath$E$}\times \mbox{\boldmath$H$}\right)
\end{displaymath} (8.128)

and integrate it:
$\displaystyle \mbox{\boldmath$L$}_{\rm field}$ $\textstyle =$ $\displaystyle \frac{1}{c^2} \int \mbox{\boldmath$r$}\times (\mbox{\boldmath$E$}\times \mbox{\boldmath$H$}) dV$  
  $\textstyle =$ $\displaystyle \frac{\mu_0 e}{4 \pi} \int \frac{1}{r} \hat{\mbox{\boldmath$n$}}
\times (\hat{\mbox{\boldmath$n$}}\times \mbox{\boldmath$H$}) dV$  
  $\textstyle =$ $\displaystyle -\frac{\mu_0 e}{4 \pi} \int \frac{1}{r} \left\{ \mbox{\boldmath$H...
...x{\boldmath$n$}}(\hat{\mbox{\boldmath$n$}}\cdot\mbox{\boldmath$H$}) \right\} dV$ (8.129)

over all space. Using the vector identity:
\begin{displaymath}
(\mbox{\boldmath$a$}\cdot\mbox{\boldmath$\nabla$})\hat{\mbox...
...th$n$}}\cdot\mbox{\boldmath$a$})\frac{\partial f}{\partial r}
\end{displaymath} (8.130)

this can be transformed into:
\begin{displaymath}
\mbox{\boldmath$L$}_{\rm field} = -\frac{e}{4 \pi} \int (\mb...
...B$}\cdot \mbox{\boldmath$\nabla$})\hat{\mbox{\boldmath$n$}}dV
\end{displaymath} (8.131)

Integrating by parts:

\begin{displaymath}
\mbox{\boldmath$L$}_{\rm field} = \frac{e}{4 \pi} \int (\mbo...
...th$n$}}(\mbox{\boldmath$B$}\cdot\hat{\mbox{\boldmath$n$}}')dA
\end{displaymath} (8.132)

The surface term vanishes from symmetry because $\hat{\mbox{\boldmath$n$}}$ is radially away from the origin and averages to zero on a large sphere. $\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$B$}=
g\delta(\mbox{\boldmath$r$}- \mbox{\boldmath$z$})$ Thus we finally obtain:
\begin{displaymath}
\mbox{\boldmath$L$}_{\rm field} = \frac{eg}{4 \pi}\hat{\mbox{\boldmath$z$}}
\end{displaymath} (8.133)

There are a variety of arguments that one can invent that leads to an important conclusion. The arguments differ in details and in small ways quantitatively, and some are more elegant than this one. But this one is adequate to make the point. If we require that this field angular momentum be quantized in units of $\hbar$:

\begin{displaymath}
\frac{eg}{4 \pi}\hat{\mbox{\boldmath$z$}}= m_z \hbar
\end{displaymath} (8.134)

we can conclude that the product of $eg$ must be quantized. This is an important conclusion! It is one of the few approaches in physics that can give us insight as to why charge is quantized.

This conclusion was originally arrived at by (who else?) Dirac. However, Dirac's argument was more subtle. He created a monopole as a defect by constructing a vector potential that led to a monopolar field everywhere in space but which was singular on a single line. The model for this vector potential was that of an infinitely long solenoid stretching in from infinity along the $-z$ axis. This solenoid was in fact a string - this was in a sense the first quantum string theory.

The differential vector potential of a differential magnetic dipole $d\mbox{\boldmath$m$}= gd\mbox{\boldmath$\ell$}$ is:

\begin{displaymath}
d\mbox{\boldmath$A$}(\mbox{\boldmath$x$}) = -\frac{\mu_0 }{...
...}{\vert\mbox{\boldmath$x$}-
\mbox{\boldmath$x$}'\vert}\right)
\end{displaymath} (8.135)

so
\begin{displaymath}
\mbox{\boldmath$A$}(\mbox{\boldmath$x$}) = -\frac{\mu_0 g}{4...
...}{\vert\mbox{\boldmath$x$}-
\mbox{\boldmath$x$}'\vert}\right)
\end{displaymath} (8.136)

This can actually be evaluated in coordinates for specific lines $L$, e.g. a line from $-\infty$ to the origin along the $-z$ axis (to put a ``monopole'') at the origin. If one takes the curl of this vector potential one does indeed get a field of:
\begin{displaymath}
\mbox{\boldmath$B$}= \frac{\mu_0 }{4\pi} \frac{\hat{\mbox{\boldmath$r$}}}{r^2}
\end{displaymath} (8.137)

everywhere but on the line $L$, where the field is singular. If we subtract away this singular (but highly confined - the field is ``inside'' the solenoid where it carries flux in from $-\infty$) we are left with the true field of a monopole everywhere but on this line.

Dirac insisted that an electron near this monopole would have to not ``see'' the singular string, which imposed a condition on its wavefunction. This condition (which leads to the same general conclusion as the much simpler argument given above) is beyond the scope of this course, but it is an interesting one and is much closer to the real arguments used by field theorists wishing to accomplish the same thing with a gauge transoformation and I encourage you to read it in e.g. Jackson or elsewhere.


next up previous contents
Next: Plane Waves Up: Magnetic Monopoles Previous: Magnetic Monopoles   Contents
Robert G. Brown 2014-08-19