8
$\begingroup$

As I was reading the book The Trouble With Physics, I encountered a small paragraph which seemed bit confusing. The paragraph goes as follows:

Picture field lines, like the lines of magnetic field running from the north to south pole of a magnet. The field lines can never end, unless they end on the pole of a magnet, this is one of the Maxwell's laws. But they can make closed circles, and those circles can tie themselves in knots. So perhaps atoms are knots in magnetic field lines.

My questions are:

  1. What exactly is the knot being described here?

  2. And how are atoms related to such knots in magnetic field lines?

$\endgroup$
6
$\begingroup$

A magnetic field configuration corresponds to a knot when for two magnetic field lines given by the parametric curves: $\mathbf{x}_1(s)$ and $\mathbf{x}_2(s)$, the Gauss linking number

$$L\{x_1, x_2\} = \int ds_1 ds_2 \frac{d \mathbf{x_1}(s_1)}{ds_1} .\frac{\mathbf{x_2}(s_1) - \mathbf{x_2}(s_2)}{|\mathbf{x_1}(s_1) - \mathbf{x_2}(s_2|^3}\times \frac{d\mathbf{x_2}(s_2)}{ds_2}$$

is nonvanishing. This integral is an knot invariant and does not depend on a smooth deformation of the magnetic field lines (i.e., without cutting and reconnection of the field lines). Knotted configuration solutions of Maxwell equations are described by: Irvine and Bouwmeester in the following review article, based on a previous work by: Ra$\tilde{\rm{n}}$ada.

In order to answer your second question, let me describe a brief history of the application of knots to physics:

The possible connection of knots to elementary particles was originally suggested by Lord Kelvin in 1867 who speculated that atoms might be knotted vortex tubes in ether. This suggestion was the motivation of the mathematical theory of knot theory treating the analysis and classification of knots. Physicists returned to consider knots and knot invariants in the 1980s. Let me mention the two seminal works by Polyakov and Witten. Both works treat the relation of knots to the Chern-Simons theory. This subject has applications in string theory and condensed matter physics but not directly in particle physics.

However, the situation significantly changed due to the discovery of knotted stable and finite energy solutions in many models of classical field theories used in particle physics. This direction was initiated by Faddeev and Niemi, where they describe a knotted solution in the $O(N)$ sigma model in $3+1$ dimensions. Please, see the following review by Faddeev. Later, they argued that such solutions might also play an important role in low energy QCD. There are many works following Faddeev and Niemi's pioneering work.

Now, as very well known, stable, finite energy solutions of nonlinear classical field theories are called solitons. The most famous types of solitons in gauge field theories are monopoles and instantons.

The soliton solutions are not unique, for example a translation of a soliton in a translation invariant theory is also a soliton (remains a solution of the field equation). Also, there exist rotationally invariant solutions which when rotated in space or around certain directions, and there are also internal degrees of freedom (which correspond for example to isospin). The collection of these degrees of freedom is called the moduli space of the soliton. Thus the soliton can move and rotate and change its internal state, this is why it corresponds to a particle.

These degrees of freedom (moduli) can be quantized and solitons after quantization, can describe elementary and more generally non-elementary particles. A wide class of solitons are associated with topological invariants (topological quantum numbers) which are responsible for its stability. One of the sucessful soliton models is known as the Skyrme model and its solitons approximate the proton and the neutron and also heavier nuclei.

Thus, in summary, these knotted solutions correspond to particles because they are solitons.

$\endgroup$
  • $\begingroup$ either $\to$ ether? $\endgroup$ – Emilio Pisanty Oct 22 '13 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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