Now, I understand how there are no magnetic monopoles, and hence no analogue of magnetic charge in terms of a scalar field, but is there a similar concept that instead uses a vector or spinor field to represent infinitesimal orientations of magnetic dipoles (as an approximation of course, much as general charge distributions are used to approximate actual, discrete collections of charged particles), which can be used to calculate an induced magnetic field?

This seems to me like it would be a natural concept, given that diagrams like this are often used when discussing magnetism, but I haven't been able to find a way to formalize this intuition.

  • $\begingroup$ You mean "spin" and "magnetic moment"? We don't know if there are magnetic monopoles, or not. We certainly haven't given up looking for them. $\endgroup$
    – CuriousOne
    Jan 24, 2016 at 4:16

1 Answer 1


Yes there is.

To me, the best way to understand this is through the idea of multipole expansion. You can even do this for an electrostatic potential --- and then, via the symmetry apparent with Maxwell equations, figure out a similar expansion for a magnetic (vector) potential.

To see the derivation of a multipole expansion, you can consult Griffith's Electrodynamics. This pdf seems to offer a similar, if not identical, derivation.

In short, the multipole expansion for an electrostatic potential, achieved with little more than a Taylor series expansion, reads, in SI:

$$V(r) = \frac{1}{4 \pi \varepsilon_0}\frac{Q}{r}\sum_{i=0}^{\infty}\left(\frac{r'}{r}\right)^n P_n(\cos \theta)$$


  • $Q$ is the total charge of the system,
  • $r$ is a distance from a point of interest to the origin (you choose the origin arbitrarily; generally, your expansion depends on the choice of origin),
  • $r'$ is a distance from any given charge to the origin, and
  • $P_n(\cos \theta)$ are Legendre polynomials.

Note that if the total charge of the system is zero, our monopole (zeroth) moment term, $\frac{Q}{r}$, is also zero. The term that follows the zeroth term is a familiar dipole term.

By symmetry, you can easily verify that the corresponding multipole expansion for a magnetic (vector) potential is as follows:

$$\vec{A}(r) = \frac{\mu_0}{4 \pi} \sum_{i=0}^{\infty} \left( \frac{1}{r} \right)^{n+1} \oint (r')^n P_n (\cos \theta) \vec{I}(r') dl' $$

Mathematically speaking, I just transcribed $\frac{1}{\varepsilon_{0}} \Rightarrow \mu_{0}$, and replaced a small piece of charge $dq$ with $I(r')dl'$. Note that here the monopole term is always zero. The rest gives an expansion of a magnetic (vector) potential. You will have a dipole term, a quadrupole term, an octopole term, etc.

For a detailed derivation, you can, again, read Griffiths or consult this source.

Now that you have a magnetic potential, you can, of course, calculate the magnetic field. It is usually cumbersome now that we don't have a scalar quantity, but is seldom useful in certain applications.


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