# Tag Info

## Hot answers tagged magnetic-monopoles

26

Maxwell's equations do follow from the laws of electricity combined with the principles of special relativity. But this fact does not imply that the magnetic field at a given point is less real than the electric field. Quite on the contrary, relativity implies that these two fields have to be equally real. When the principles of special relativity are ...

20

Lubos Motl's answer is very good, but I think it's worth saying one or two additional things. You can regard magnetism as simply a byproduct of electricity, in the following sense: if you assume that Coulomb's Law is correct, and that special relativity is correct, and that charge is a Lorentz scalar (so that charge and current density form a 4-vector), ...

15

Well, in order for this splitting to be possible, the magnet would have to be made of two magnetic monopoles (like charged particles, but with "magnetic charge" instead of electric charge) bound together. No known magnet is actually constructed this way; all real magnets that have been studied are made of either little current loops, or particles that have a ...

14

This is basically how a magnet's atoms look like: so, when you split it into two, you do not change anything but the length of the magnet. As you can see the North poles(Black sides or "K" as it's in a different language) face north and south poles face south in each part of the magnet. IF there are such things as monopoles, then it is possible to define ...

12

Aqwis, it would help in the future if you mentioned something about your background because it helps to know what level to aim at in the answer. I'll assume you know E&M at an undergraduate level. If you don't then some of this explanation probably won't make much sense. Part one goes back to Dirac. In E&M we need to specify a vector potential ...

12

This is interesting. You would definitely have to 'nail down' the magnets to the sphere, because it will be an unstable configuration. Also in the real world, edge-effects will destroy any chance of perfect radial field lines, so let's assume we're in an ideal scenario. Outside the sphere, the magnetic field would be that of a source monopole placed at ...

10

The field would disappear completely. I think the simplest explanation is in terms of the surface currents that account for the field (assuming constant magnetization, which is reasonable for thin slices). For the initial torus magnet (your second image) the magnetic field is generated, in practice, by surface currents on the planar ends of the torus. One ...

10

Let me make quite clear that the recent experiment does NOT imply the detection of a true magnetic monopole. Somehow, in all the excitement, the word "synthetic" was dropped rather quickly from the phrase "synthetic magnetic field". A synthetic magnetic field is a physical quantity that obeys the same equations as a magnetic field, typically realized in ...

9

I think Emilio Pisanty's answer is good enough. But here is another longer, 'magnetic charge' approach. ( Let's specify the coordinates first (sorry I borrow your picture). It's obvious that the toroid is symmetrical under rotation along $\hat{\phi}$ direction. Thus we can't have magnetic field along $\hat{\phi}$. Which means it is sufficient for us to ...

9

Magnetic monopoles can be created according to numerous Grand Unified Theories (GUT). The idea is that at sufficiently high energies you can reach an energy range where three of the four fundamental forces (strong nuclear, weak nuclear, and electromagnetism) couple to one another and are the same force. Such a state existed in the universe a tiny fraction of ...

8

The article is pretty poorly written. As Siva said it doesn't even link to the original paper. So I just looked up the name mentioned in the article and found this which is probably what they're talking about (though this is just a guess). They measured the magnetic dipole moment of protons and antiprotons to ~4 parts per million (and verified the CPT ...

8

The symmetry between the exchange of the electric and magnetic fields would become more complete. Indeed, this is not a proof that the magnetic monopoles exist. Moreover, the symmetry is broken, for example because the mass and charges of the two types of charged particles substantially differ. But the actual state-of-the-art reasons why magnetic monopoles ...

8

Not a direct answer to your question but still a surprising derivation of Maxwells equations: Feynman's proof of the Maxwell equations (FJ Dyson - Phys. Rev. A, 1989) shows, that it is possible to derive Maxwells equations from Newtons second law of motion and the uncertainty principle.

8

There are no consequences concerning the quantization of the charge or the existence of real magnetic monopoles. The connection with the monopoles is only formal. What the experimentalists study is the (superfluid) velocity field $v$ and the corresponding vorticity $\Omega=\nabla \times v$ in the gas and the spin orientation of the atoms (the system is ...

8

For each $r>0$, the divergence of the magnetic field of the monopole is zero as you have already checked; \begin{align} \nabla\cdot\mathbf B(\mathbf x) = 0, \qquad \text{for all $\mathbf x\neq \mathbf 0$} \end{align} But what if we also want to find the divergence of this field at the origin? After all, that is where the point source presumably sits. ...

8

The joshpysics's answer is good. I'm only want to tell about some details. Let's have field $$\mathbf A = g\frac{\mathbf r}{|\mathbf r|^{3}}. \qquad (.1)$$ It has singularity at zero. We can eliminate it by modification $(.1)$ by $$\mathbf A = g\frac{\mathbf r}{|\mathbf r|^{3}} \to g\frac{\mathbf r}{(r^{2} + a^{2})^{\frac{3}{2}}}.$$ Then we can take ...

7

The distribution approach nicely described by joshphysics and PhysiXxx wholly answer your question and show why your proof doesn't work, but there another way to reason with the correct part of your proof. It is, of course, ultimately mathematically equivalent. Simply work out the flux through a spherical shell $\mathcal{S}$ centred on the origin; from the ...

7

The answer to your question, is yes, it has indeed been considered. The bound state has even been given a name "monopolium". Here is a paper discussing prospects for detection and production. I should add the caveat that they're not strictly, in your words "confined together like quarks". You could separate them if you input enough energy, unlike the ...

7

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 ...

7

Monopoles are still created in inflationary models. They're just created before (or during) inflation, so that the rapid expansion thereafter dilutes their density to unobservably low levels. At the time when the monopoles are created, they're created at a density of order 1 per Hubble volume -- that is, there's one in each "observable Universe" at that ...

7

As Mark M says in his answer,you cannot have a monopole magnet. You can simulate one. After all when you are at the north pole of earth, to all intents and purposes that is a monopole for magnets in the area. By spreading the magentic lines of one of the poles on a large area and concentrating the other to a very small one. Look at the images here. If you ...

6

No, it isn't. While it is possible to create geometrically asymmetric magnetic fields, the net magnetic flux through any bounded surface must be zero. While you could, for instance, construct a magnetic where the N pole had much lower magnetic field intensity than the S pole, the N pole face would have to be much larger than the S face, and the total ...

6

Gravitational monopoles are forbidden by the positive mass theorem--- any configuration of GR has positive mass, and therefore is an ordinary "pole" in the analogy with electromagnetism. The analogy is not very good, because the energy is always positive, unlike charge. The reason magnetic monopoles make sense in EM is because of electric-magnetic duality, ...

6

i) First of all, the Dirac quantization rule $$\tag{1} \frac{qg}{2\pi\hbar} ~\in~ \mathbb{Z}$$ for magnetic monopoles can be generalized to the Dirac-Zwanziger-Schwinger quantization condition $$\tag{2} \frac{q_1g_2-q_2g_1}{2\pi\hbar} ~\in~ \mathbb{Z}$$ for dyons. (In a slight misuse of terminology, we shall in the following also include purely ...

6

Another option, besides modifying the potential $A_\mu = (A_i, \phi)$ in some way, is to introduce another 4-potential $C_\mu = (C_i, \psi)$. Then the electric and magnetic field are given by $$E = - \nabla \times C - \frac{\partial A}{\partial t} - \nabla \phi$$ $$B = \nabla \times A - \frac{\partial C}{\partial t} - \nabla \psi$$ More on this 2-potential ...

6

I think this kind of set up is similar to Halbach Cylinders and Halbach Spheres.

6

By analogy (between $\mathbf{E}$ and $\mathbf{B}$ as they are pretty much equivalent) then the divergeance of $\mathbf{B}$ field wouldn't be 0 anymore, instead: $$\nabla \cdot \mathbf{B}= \frac{\rho_{\rm magnetic}}{\mu_0}$$ With $\rho_{\rm magnetic}$ the magnetic charge density, and $\mu_0$ the permeability in vacuum, to interpret it, the divergence of the ...

5

No, I believe the Standard Model does not predict monopoles as a result of symmetry breaking. This is because the symmetry breaking $\mathrm{SU(2)} \times \mathrm{U(1)} \rightarrow \mathrm{U(1)}$ does not allow for topological solitons to exist. Edit: $\pi_2(\mathrm{SU(2)} \times \mathrm{U(1)}/\mathrm{U(1))}=\pi_2(S^3)=0$ Source: To be or not to be? ...

5

The news release from NASA does mention this. Apparently in this transitional phase the field is not well represented as a dipolar field. Right now it may be a Quadra polar or higher (even numbered) field. Think of the total field being the superposition of multiple non-axial non-centered dipoles. The field is weakening in intensity too and may become weaker ...

5

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 ...

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