To introduce magnetic monopoles in Maxwell equations, Dirac uses special strings, that are singularities in space, allowing potentials to be gauge potentials. A consequence of this is the quantization of charge.

Okay, it looks great. But is this the only way to introduce magnetic monopoles?


4 Answers 4


If you have a source of radial magnetic field $B\sim Q_M/r^2$, then one may prove that the vector potential $\vec A$ can't be single-valued. It's because $\vec B={\rm curl}\vec A$ for a well-defined $\vec A$ automatically satisfies ${\rm div}~\vec B=0$. However, $Q_M/r^2$ has a curl proportional to the delta-function at the origin.

Still, this delta-function vanishes everywhere except for a Dirac string (and the space minus the Dirac semiinfinite string is simply connected), so with the Dirac string, $\vec A$ may be defined everywhere. $\vec A$ still changes under the loop around the Dirac string. In this way, the magnetic monopole is replaced by a very long magnetic dipole. Two poles are connected with a thin solenoid and one of the monopoles is sent to infinity and becomes irrelevant. The Dirac string, i.e. a very thin solenoid, becomes unobservable as well, even for interference experiments (as long as the confined magnetic flux is properly quantized).

The arguments above are waterproof and you can't circumvent them. So if you're asking whether there is a way to introduce a magnetic monopole so that the vector potential would be single-valued, the answer is a resounding No, much like if you ask if it is possible to introduce the number 4 so that it isn't equal to 2+2.

However, one may try to search for solutions to similar, less singular problems. In theories with Higgses, one may "dilute" the delta-function a little bit and find non-singular solutions of Yang-Mills theories with Higgs fields, the so-called


't Hooft-Polyakov monopole that is non-singular but indistinguishable from the Dirac monopole when you're very far from the center of the solution, relatively to its characteristic length scale. This solution has various generalizations, too.


Introducing magnetic charge into Maxwell equations is not a problem at all, and it does not require any strings etc. Moreover, it makes Maxwell equations symmetric w.r.t. magnetic and electric fields/charges. The equations are as follows:

\begin{split} \mathop{\mathrm{curl}} E + \frac{\partial H}{\partial t} &= -J_m \\ \mathop{\mathrm{curl}} H - \frac{\partial E}{\partial t} &= J_e \\ \mathop{\mathrm{div}} E &= J_e^0 \\ \mathop{\mathrm{div}} H &= J_m^0 \end{split}

However, introduction of magnetic charge leads to non-zero divergence of magnetic fields, hence making it impossible to represent the magnetic field as a curl of vector potential. As you stated in your question, Dirac introduced a singularity to preserve gauge potentials description of the phenomenon.

Use of gauge potentials is justified by an excellent fit of QED predictions and experimental data in all aspects that are not related to identification of particle's mass and charge values from the theory. The straightforward calculation leads to infinities, and re-normalization procedure does not help with identification of mass and charge values.

Elementary particles that were observed experimentally are known to have zero magnetic charge but non-zero magnetic moment. Hence there might be a non-zero distribution of magnetic charge density "inside" the particles, if you admit that particles are not point-like. This approach can be developed using Gordon decomposition of the vector current constructed from Dirac spinors. See for instance here.

  • 1
    $\begingroup$ And we expect the magnetic field to be the curl of a vector potential (even in presence of magnetic monopoles which no one has ever seen): because they are fun or for a fundamental reason? $\endgroup$
    – Isaac
    Commented Apr 1, 2012 at 7:53
  • $\begingroup$ According to Helmholtz's theorem, any 3D vector field (such as magnetic field $B$ as well as electric field $E$) can be represented as a sum of gradient of the scalar field and curl of a vector field. Hence if $div B$ is non-zero, there is a need to introduce additional scalar field to be added to the curl of (real-valued) vector potential: $B = curl A + grad \Phi$, where $\Phi$ is not the same as scalar potential $\phi$ used to define electric field (in a static case $E=-grad \phi$). $\endgroup$ Commented Apr 1, 2012 at 13:40
  • 1
    $\begingroup$ It makes the theory a bit more complicated. You need to either use complex-valued 4-potential (see e.g. Mignani and Recami, Il Nuovo Cimento, vol. 30 (1975), p. 533), or a pair of real-valued 4-potentials (see arxiv.org/pdf/math-ph/0203043). $\endgroup$ Commented Apr 1, 2012 at 13:49
  • $\begingroup$ I don't think that there is a fundamental reason to avoid introducing magnetic charges etc. It is only because QED works very well for a specific class of physical problems (when particles considered point-like, and due to zero (total) magnetic charge of known particles). I'm sure that non-zero magnetic charge density need to be introduced in the theory that will describe intrinsic structure of particles and explain the values of their masses/charges. $\endgroup$ Commented Apr 1, 2012 at 13:56
  • $\begingroup$ Tell me if I am wrong: electric monopoles and magnetic monopoles are not really symmetric as we would have liked, the former needing only a point as a singularity, the latter needing a Dirac string; so we cannot permute E and B by any linear combination of themselves because of the Dirac string, can we? $\endgroup$
    – Isaac
    Commented Apr 4, 2012 at 8:52

There is also the geometric approach to a Dirac-type magnetic monopoles with a singular core, pioneered by Wu and Yang, which avoids the Dirac string by introducing non-trivial topology, fiber bundles, and locally defined gauge potentials. It leads to the same physical predictions, e.g. charge quantization, as Dirac's method.

(The above should not be conflated with a 't Hooft-Polyakov-type magnetic monopoles with a regular core, cf. Lubos Motl's answer and this Phys.SE post.)


  1. T.T. Wu and C.N. Yang, Concept of non-integrable phase factors and global formulation of gauge fields, Phys. Rev. D 12 (1975) 3845.

  2. T.T. Wu and C.N. Yang, Dirac Monopole without Strings: Classical Lagrangian theory, Phys. Rev. D 14 (1976) 437.

  3. M. Nakahara, Geometry, Topology and Physics, 1990.


Below it is explained how a string free monopole theory can be constructed.

Monopoles have not been found in experiments. Therefore the main issue to be defined are the principles used as cornerstones for the required monopole theory.

The first step is to formulate a theoretical definition of monopoles. This is done by the well known duality transformation. Applying this transformation to the Maxwellian system of electric charges and electromagnetic fields, one obtains a Maxellian-like system of monopoles and electromagnetic fields (without charges). The next step is to construct a unified charge-monopole theory.

Let us examine two postulates that pertain to the required charge-monopole theory:

(A) For chargeless systems the unified theory must take a form which is completely dual to the theory of charges and fields and for systems without monopoles it must take the form of Maxwellian electrodynamics.

(B) Electromagnetic fields of a system of monopoles and those of a system of charges have identical dynamical properties.

One may be tempted to use both postulates (A) and (B) as fundamental elements of the theory. However, it turns out that this course is unattainable because different sets of equations of motion are obtained from postulate (A) (without (B)) and from postulate (B) (without(A)).

It can be proved that postulate (A) yields a string-free charge-monopole theory which is consistent with the variational principle and postulate (B) yields Dirac's strings.

Information on scientific articles that discuss these issues can be found in the page pointed out by the following link.


E. Comay


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.