# Why doesn't magnetic monopole found in spin ice modify the Maxwell's Equations?

The magnetic monopole predicted by Dirac nearly a century ago(1) was found in spin ice as a quasi-particle(2). My question is: why doesn't the magnetic monopole found in spin ice modify Maxwell's Equations? (I know they are not elementary particles but quasi-particles.)

(1) Dirac, P. A. M. Quantised singularities in the electromagnetic field. Proc. R. Soc. A 133, 60–72 (1931). DOI: 10.1098/rspa.1931.0130.

(2) Castelnovo, C., Moessner, R. & Sondhi, S. Magnetic monopoles in spin ice. Nature 451, 42–45 (2008). arXiv: 0710.5515 [cond-mat.str-el].

## 3 Answers

These are "fake monopoles", in the sense that the north and south poles are not actually separated. They are the ends of thin tubes which behave like Dirac strings - like long thin twisted magnets. The tubes are formed due to geometrical frustration, which forces the magnetic field to be orientated either toward the outside or toward the inside of the tetrahedral cells. The tetrahedral cells end up by having two faces with "spin in" orientation and two faces with "spin out" orientation. Two tetrahedral cells couple one another like tiny magnets, so that tiny magnetic tubes form. The ends seem to be separated north and south magnetic poles, while in fact they are connected through the tube.

Had they been genuine monopoles, they would have indeed modified Gauss's law for magnetism, and Faraday's law of induction, by adding terms corresponding to magnetic charge density and current.

There are two questions.. 1. What is the concept of magnetic monopole here? 2. Why Maxwell's eqns remain unmodified?

1. They are not elementary particles as anticipated by P.M.Dirac. But the concept comes from the non-zero divergence of magnetization field. In electrostatics when you have $\nabla\cdot \mathbf{P}$ (polarization vector) not equal to zero, you have a concept of bound electric charge since $\rho = -\nabla\cdot\mathbf P$. Similarly when you have non-zero divergence of Magnetization vector , you have concept of magnetic monopole like $\rho=-\nabla\cdot \mathbf M$. In spin ice w.r.t. center of each tetrahedra when you have 3in1out or 3out1in spin configuration, net divergence of magnetization is not zero locally, and you have a monopole or an antimonopole respectively.

2. Even you have a monopole you don't have to modify Maxwell's eqn because here magnetic intensity and magnetization serve as source and sink of magnetic line of force. Thats why even $\nabla \cdot \mathbf B = 0$ you have $\nabla\cdot \mathbf H=-\nabla \cdot\mathbf M$ (because $\mathbf B = \mu_0\cdot(\mathbf H+\mathbf M)$). Furthermore, you don't have a concept of isolated monopole or antimonopole there, with a flip of single spin always generate a monopole-antimonopole there which interacts through long range dipolar interaction or Coloumb's Law.

According to this review paper, in the ground state, the 2in2out spin arrows are considered "pseudo magnetic fields" because of the zero divergences everywhere. As one spin arrow (or dumbbell) is flipped, that pseudo magnetic field has nonzero divergences at two sites. It's not a real magnetic field, however, the magnetization of one rare-earth atom is proportional to such a pseudo magnetic field.

(Below is my understanding. Please correct me if I am wrong!) Notice it's the magnetization of one atom, not the whole crystal, that is proportional to the pseudo magnetic field. That means the magnetic field induced by other atoms is considered as "H", while only the field induced by the local atom is considered as "M". And when you move to the next atom, again M is the field induced by that next atom, not the previous one. Such M has a nonzero divergence. And no matter at which site you consider M and H, you find the total B = M + H still has a zero divergence.