# Tag Info

## Hot answers tagged magnetic-monopoles

45

There is no theoretical reason why magnetic monopoles cannot exist and indeed there are good reasons for supposing that they should exist. It's just that we have never observed one. In the past there have been various experiments to detect magnetic monopoles, though I think everyone has given up on the idea by now. If you're asking why we can't get ...

22

No, a magnetic monopole a la the Dirac string does not "violate" gauge symmetry. Rather, the statement "we have a magnetic monopole" means only that we are forced to consider the gauge theory not on the whole spacetime, but on the spacetime with the location of the magnetic monopole removed. Why? Because, at the location of the magnetic monopole, the ...

16

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 ...

14

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 ...

14

In the absence of magnetic monopoles, Maxwell's equations are \begin{align} \text d F &= 0 ,\\ \text d{\star F} &= J_e , \end{align} where $J$ is the 4-current 3-form due to electric charges (assuming a metric with signature $(-,+,+,+)$). For cohomological reasons, from the first equation one can asserts that there exists a 1-...

12

Describing a magnetic monopole is a problem in Classical Electromagnetism: essentially, if you have a magnetic monopole, then $\nabla \cdot \mathbf{B} \neq 0$, meaning that we can no longer define a vector potential $\mathbf{A}$, since -- if you remember -- it was precisely the fact that the divergence of $\mathbf{B}$ was always zero in Maxwell's Equations ...

11

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 find ...

11

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 sits. We might ...

11

The mathematical model for classical electromagnetism just doesn't forbid magnetic monopoles by construction. Consider an arbitrary vector field $X$ in 3d. Such a vector field is totally characterized by its divergence and curl. Suppose the following is true: $$\nabla \cdot X = \sigma, \quad \nabla \times X = Y$$ Then knowing $\sigma$ and $Y$ everywhere,...

11

The difference between the two arises because Maxwell's equations, while looking perfectly "equal", actually are not all of the same nature when we phrase electromagnetism in terms of a potential. If you think of $F$ as the dynamical variable, then $$\mathrm{d}F = 0 \quad \mathrm{d}{\star}F = 0$$ in vacuum look perfectly symmetric, and you might imagine ...

10

Dirac's discovery of the quantization of the magnetic charge is distinct from the Aharonov-Bohm effect. These effects depend on different topological properties of the manifold on which a charged particle moves. The Aharonov-Bohm effect appears on manifolds with a nonvanishing first cohomology group $H^1(M)$, while the Dirac quantization condition takes ...

10

You can find a solution on most of space, but not all of space. To see why, imagine taking the flux integral of $\vec{B} = \hat{r}/r^2$ over the surface of a sphere $S$ of radius $r$. Doing this integral is straightforward, and yields $$\iint_S \vec{B} \cdot d\vec{a} = 4\pi.$$ Now let's assume that there exists a vector field $\vec{A}$ such that $\vec{\... 9 The article is pretty poorly written. As Siva said it doesn't even link to the original paper. So I just looked up the name mentioned in the article and found this which is probably what they're talking about (though this is just a guess). They measured the magnetic dipole moment of protons and antiprotons to ~4 parts per million (and verified the CPT ... 9 There are no consequences concerning the quantization of the charge or the existence of real magnetic monopoles. The connection with the monopoles is only formal. What the experimentalists study is the (superfluid) velocity field$v$and the corresponding vorticity$\Omega=\nabla \times v$in the gas and the spin orientation of the atoms (the system is ... 9 You cannot just add a term to the Lagrangian to give the usual electromagnetic gauge theory magnetic charge. The reason is rather simple: The equation of motion for a magnetic four-current$j_m$is$\mathrm{d}F = j_m$. But$\mathrm{d} F = \mathrm{d}\mathrm{d} A = 0$independently of the equations of motion. So simply adding a term doesn't work. The first ... 8 It should perhaps be stressed that the magnetic monopoles that many GUTs predict are generalized 't Hooft Polyakov monopoles (as opposed to e.g. the Dirac/Wu-Yang monopoles, which are singular in a point/exclude a point). Once a GUT action$S[A,\phi,\psi]$is adapted, then the 't Hooft Polyakov monopoles do in principle not constitute a new independent ... 8 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 ... 8 The English Wikipedia article on magnetic monopoles has the following equation for the 'extended' Lorentz-Force of a magnetic field on a electrically and magnetically charged particle: $$\vec{F}=q_{\mathrm e}\left(\vec{E}+\vec{v}\times\vec{B}\right) + q_{\mathrm m}\left(\vec{B}-\vec{v}\times\frac{\vec{E}}{c^2}\right)$$ Under time reversal ($t$is ... 8 What do we mean with magnetic monopole and dipole? I can not find a way to relate magnetic monopoles and dipoles with electric ones. I do not understand their outcomes. Luckily, there exists a truly amazing one-to-one correspondence between magnetism and electricity. Monopole in magnetism is analogous to charge in electrostatics/electricity. Just like ... 8 No free magnetic monopoles have ever been confirmed to exist. There have been experimental results that are consistent with the existence of monopoles, most famously on Valentine's Day, February 14 in 1982, but that observation has never been repeated and is now dismissed as an artifact of some kind. No experiment has confirmed the existence of magnetic ... 7 You don't hear about gravitomagnetic monopoles because they, unlike magnetic monopoles, cannot be well defined within the context of GR. Gravitomagnetism is only a weak field approximation of GR which isn't Lorentz-invariant, so it is not to be expected that everything from electromagnetism has an analogy in GR. In electromagnetism, there are two ways one ... 7 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? ... 7 I am not sure if I know the correct answer (as I am a student my self), but I will try (and if I am wrong, someone please correct me). The first thing that took me some time to figure out is what they mean by adjoint representation. In Georgi's book he defines the adjoint representation of a generator as: [T_i]_{jk} \equiv -if_{ijk} \end{... 7 A good starting point to see that gluing small magnets to a sphere doesn't work is seeing what the field of just two magnets opposed gives you. A single dipole magnet has a field that falls off as$\frac{1}{r^3}$, while the two opposed has a field that dies faster, falling off as$\frac{1}{r^4}\$. As you add more and more magnets (pointing in different ...

7

Within the framework of standard model (SM) magnetic monopoles are non-existent. It is quite subtle as to why this is not the case. To begin, look at the Dirac's famous charge quantization condition. It was first pointed out by Dirac that on the quantum level the existence of the monopole will lead to the qg = 2\pi n \qquad\qquad \text{...

7

Both of these follow from desirable properties of this hypothetical magnetic charge, namely: Magnetic charge is conserved. Magnetic field lines radiate outwards from positive magnetic charges. The net force between two magnetic charges moving at constant speed along parallel tracks is less than that between two stationary charges. All three of these ...

7

The answer depends on the assumptions. When you ask how to derive something you must ask yourself from what you want to derive it. For example: if you experimentally find out that there are no magnetic monopoles, since you simply don't observe them, and you state this as a law of Physics, then Gauss's law for magnetism is the mathematical way to express ...

7

First, have a look at the papers https://arxiv.org/abs/1602.04251 and https://arxiv.org/abs/1605.02391 where the authors (Seiberg and Witten) have a more careful analysis of the properties of the monopole operators, which is extremely useful in the paper you have cited. Now go back to your question. 1&3. Short answer, yes. But let me clarify a little ...

6

The answer is that you can build such a sphere. Assuming the sphere is built perfectly, then the magnetic field will cancel everywhere.

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