a project exploring the Nahm equations, monopoles & more


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This page uses the Einstein summation convention: .

The matrices (Nahm or gauge field) on this page are hermitian.

The Nahm Equations: Monopoles

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Solutions of the Nahm equations are equivalent, via the Nahm transform to monopoles. On this page we'll give a short introduction to monopoles, and then show how the Nahm equations in the simplest, one-dimensional case, go together with the Nahm transform to produce an explicit expression for a monopole.


Dirac monopoles

Monopoles were first introduced by Paul Dirac (in 1931) as hypothetical particles that carry magnetic charge. His simple monopoles were given by a vector potential with singularities at the origin and along the positive -axis.

Dirac monopoles represent an extension of electromagnetism in which Maxwell's equations are completely symmetric between the electric and magnetic fields. As electromagnetism is invariant under gauge transformations (transformations of the electromagnetic potentials) from the gauge group , which is abelian, Dirac monopoles are sometimes called abelian monopoles.

Non-abelian monopoles

In the 1970s, 't Hooft and Polyakov showed that monopoles arose in non-abelian field theories. There are generalisations of electromagnetism in which the fields are no longer simple numbers, but matrices. This enlarges the internal structure and symmetries of the theory, and is what underpins all descriptions of elementary particles in the standard model, for instance.

We can describe this theory using Lagrangians very similar to those familiar from the Lagrangian formulation of electromagnetism. An example of such a Lagrangian is

The objects appearing here are as follows:

  • A gauge field , which is now a hermitian matrix.
  • The non-abelian generalisation of the electromagnetic field tensor
  • A field which is also a matrix (but is referred to as a scalar field because it has no vector indices). This field comes complete with a potential and is called the Higgs field (the same sort of field corresponding to the famed Higgs boson in fact).
  • The covariant derivative is

The gauge transformations - redefinitions of the potential leaving the action, and hence field equations, invariant - correspond to elements of some gauge group (in general a matrix group like , complex unitary matrices with unit determinant).

The Euler-Lagrange equations for this theory have solutions which, from a large distance, resemble Dirac monopoles. These monopoles though do have internal structure rather than being point-like, and their existence is actually predicted by the theory. These monopoles are termed non-abelian monopoles.

BPS Monopoles

In the Bogomolny-Prasad-Sommerfield limit, we let , i.e. we ignore the Higgs field's potential. We then search for monopoles in the form of static (time-independent) solutions to the field equations of this theory. Static solutions imply that we set the electric field equal to zero.

The crucial condition that gives us monopoles is that we specify a particular behaviour of the field at infinity; we assume that it does not vanish but rather satisfies constant (up to term of order ). This breaks the natural symmetry of the system, and if one also assumes that the total energy must be finite then one is led through a series of constraints on the fall-off of and that imply the existence of an observable magnetic field with a dependence - corresponding to a magnetic monopole.

A neat approach to studying this system is to impose what is known as a Bogomolny bound on the energy. The energy can be written as

We see that we have an equality in the above expression when

which is known as the Bogomolny equation. We can evaluate the integral on the right-hand side - essentially by using Gauss' theorem - to show

with equality when the Bogomolny equation is satisfied, where is the monopole charge.

A BPS monopole is then defined to be a pair of fields which have finite energy (i.e. appropriate fall-off at infinity) and which satisfy the Bogomolny equations. So one way of finding monopole solutions is to try to solve these equations. We will show below how one uses the Nahm transformation to do this.

The Bogomolny equations may also be expressed as the three-dimensional reduction of the self-duality equations.

Exercise. Beginning with the self-duality equations in :

identify and suppose independence to derive the Bogomolny equation (recall that ).

Nahm's approach

In this section we will work through the steps that take a solution of Nahm's equations to a solution of the Bogomolny equation - i.e. a magnetic monopole. We will do this for the simplest example, which is the BPS monopole solution for first found by Prasad and Sommerfield.

The hope of this exercise is to show you the power of the Nahm transform in action. It's quite a striking result when you come to the end of the calculation (which is surprisingly straightforward) and discover you've arrived at solutions to a difficult set of non-linear matrix partial differntial equations by a complete unusual method! If you want to learn more about why the Nahm transform works, there will be a page entitled Nahm transform for undergraduates soon.

Step 1: Solving Nahm's equations

We will use a form of the Nahm equations in which the matrices are hermitian and the equations are

The are defined on some interval or intervals parametrised by . There is a gauge transformation acting on these matrices by

Exercise: How can we use this gauge freedom to get rid of ?

In practice one often considers the Nahm equations just as

Now, the size of the Nahm matrices determines the charge of the monopole you get out of the Nahm transform. For a charge one monopole then the Nahm matrices are just numbers, and the equations are trivial:

So a solution to the Nahm equations is just a triplet of real numbers.

Exercise: The Nahm transformation is going to produce a monopole solution. What interpretation do you expect we will find for ? What about in the case when each is a matrix rather than a scalar?

Step 2: Forming the Weyl operator

The next step is to use our solution of the Nahm equations to write down a linear differential operator which is

where the are a representation of the quaternionic algebra; that is, they are a set of matrices satisfying

Exercise: Show that taking to be the two-by-two identity matrix and where are the Pauli matrices:

satisfy these relationships; you may want to use (or show) that .

The symbol "" appearing means the tensor product; we don't really need to worry about this in our example because we are dealing with size one Nahm matrices - the tensor product of a scalar with a matrix is just the same as multiplication by the scalar. So we have

This acts on two-component objects which we denote by . We define an inner product

This allows us to define the adjoint operator

Exercise: Show that

(hint: the sigma matrices are hermitian i.e. self-adjoint, integrate by parts).

Exercise: In general one has

Show that requiring be real (i.e. proportional to ) is equivalent to imposing the Nahm equations on the . This seeming coincidence is in fact one of the most crucial parts of the construction.

Step 3: Twisting the operator

From now on we will work only with our charge one example. This means we have

and we also take where is some positive real number.

We now introduce coordinates on , and twist the operator with respect to them:

We can write and let the relative coordinate be , so


Exercise: We will need to use some neat properties of the sigma matrices soon. Show that

(where we understand there to be an unwritten two-by-two identity matrix on the right-hand-side, and .

Exercise: Using the result of the previous exercise show that

(You might do this by considering the power series expansions of these functions.)

Step 4: Solving the Weyl equation

We proceed by solving the equation

Solutions are given by

for a two-component object independent of , chosen to give us a basis of solutions. The easiest thing to do is take and . Nahm's construction requires us to form the normalisable solutions into a matrix which we call . So in our case we just have

Exercise: Is this normalisable? Compute

You should find that . It may be helpful to remember that cosh is even and sinh is odd.

The final part of Nahm's transformation then uses the normalised solution

to determine and .

Step 5: Computing

The Nahm transformation result for is

Exercise: Compute this! You should find

Step 6: Computing

The Nahm transformation result for is

where the differential is .

Exercise: Show that this definition means is hermitian.

Exercise: Compute using . If you pay attention you should only have to integrate one term, for which the identity might be useful. The result should be

Exercise: Use properties of the sigma matrices to show this means

These monopole potentials were first found (using "shimmying") by Prasad and Sommerfield.

Exercise: If you don't believe the Nahm transformation really works, either show explicitly that and found above satisfy the Bogomolny equation, or look up the paper by Prasad and Sommerfield.

Exercise: Investigate the behaviour of this monopole solution at the monopole position (), and at infinity, ().

Step 7: Learn more

A more technical description of the Nahm transform will be found here. Further reading might include learning about the ADHM construction of instantons on , which inspired the Nahm transform, or learning how the Nahm transform itself can be generalised beyond monopoles, to objects such as a calorons or instantons on curved spaces.

Exercise: Carry out the Nahm transformation on the interval . Note that in this case you might not have as many normalisable solutions of as before. What monopole do you get from this construction?

Exercise (for the incredibly enthusiastic): Consider the charge two monopole in which case the Nahm matrices are two-by-two. Make the ansatz (no sum on ) which implies plus permutations of . On this page we see that the solutions are elliptic functions. Construct and explicitly.

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This page was last modified on 7 November 2011, at 16:33. - This page has been accessed 19,098 times. - About Islands