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Behavior of Magnetic Monopoles

Anonymous Fri 17 Jul 07:43:17 2020 No.11909467 View Reply Original Report

Quoted By: >>11909593 >>11911916

How exactly would magnetic monopoles interact with other matter? Let's stick with classical electrodynamics for the moment.
For starters, if I placed a magnetic monopole within the vicinity of a charged particle, what trajectories do the two particles take?

Anonymous

Anonymous Fri 17 Jul 2020 09:06:45 No.11909589 Report

Quoted By: >>11909598

You know they are impossible? Magnetic fields are all curl. And they are a result of a moving electric charge. Cant have monopoles.

Anonymous

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Anonymous Fri 17 Jul 2020 09:09:58 No.11909593 Report

Quoted By: >>11909638 >>11909717 >>11911215

>>11909467
the fully symmetric maxwell's equations are
$ \begin{cases} \nabla \times \vec{E} = -\vec{J}_{mf} - \frac{\partial \vec{B}}{\partial t} \\ \nabla \times \vec{H} = \vec{J}_{ef} + \frac{\partial \vec{D}}{\partial t} \\ \nabla \cdot \vec{D} = \rho_{ef} \\ \nabla \cdot \vec{B} = \rho_{mf} \\ \end{cases} $
where the e and m subscripts indicate electric and magnetic currents/charges, and the f subscript indicates free currents/charges (i.e. anything not due to polarization/magnetization).

for electric (point) charges the Lorentz force law is
$ \vec{F} = q_e (\vec{E} + \vec{v} \times \vec{B}) $
it's funny that i've never actually seen the theoretic force law for magnetic charges (i.e. magnetic monopoles), but by a principle called duality (where you swap electric and magnetic quantities in a prescribed way), the force law should be
$ \vec{F} = q_m (\vec{H} - \vec{v} \times \vec{D}) $
that should answer your question

why the hell would somebody introduce magnetic charges/currents if they don't physically exist? because they have a lot of theoretic value for advanced electromagnetic applications

t. EE PhD

Anonymous

Anonymous Fri 17 Jul 2020 09:15:17 No.11909598 Report

Quoted By: >>11909630

>>11909589
Magnetic monopoles are very much possible and in fact expected to exist. They are just thought to be extremely rare.

https://en.wikipedia.org/wiki/Magnetic_monopole#Grand_unified_theories_2

Anonymous

Anonymous Fri 17 Jul 2020 09:28:26 No.11909630 Report

Quoted By: >>11911215 >>11912003 >>11912504

>>11909598
the magnetic field can be interpreted as being the "relativistic part" of the electric field, so i wouldn't go as far as saying they are "expected to exist". it would be nice if they did for theoretical reasons, but you have to acknowledge to date there is no experimental evidence for their existence, and ultimately nature has the final word.

that being said, even if it's somehow shown they don't exist, there are still theoretically useful reasons why you'd want to introduce magnetic charge and currents into maxwell's equations. for instance, there are a whole class of techniques called equivalence principles used in antenna design. one particularly useful application is modeling an aperture using only magnetic currents

Anonymous

Anonymous Fri 17 Jul 2020 09:31:10 No.11909638 Report

Quoted By: >>11911215

>>11909593
I found it really neat how we replace electric fields in apertures with equivalent magnetic currents to solve far fields for antennas

Anonymous

Anonymous Fri 17 Jul 2020 10:29:41 No.11909717 Report

Quoted By:

>>11909593
nerd

Anonymous

Anonymous Fri 17 Jul 2020 21:44:37 No.11911215 Report

Quoted By:

>>11909593
>>11909630
>>11909638
OP here, didn't know they were useful in EE, always thought they super esoteric and only curiosities
thanks for the insight

Anonymous

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Anonymous Sat 18 Jul 2020 03:15:36 No.11911916 Report

Quoted By: >>11911967 >>11912403

>>11909467
\documentclass[12pt]{article}
\begin{document}
What I find interesting about the magnetic mono-pole is that it exposes an Achilles heel in Gausses law, that the sourciness of the electric field, the divergence in E is the total charge density per permittivity.

Consider modeling the magnetic field B bold as the curl of the magnetic vector potential A bold
$\textbf{B} = \nabla \times \textbf{A} \\ $

With this in mind let us sub in this new definition of the magnetic field into the Electric field equation.
Noting the induction relationship in the electric field equation, let us reexpress the electric field with the A vector, the magnetic potential.
$ \textbf{E} = \frac{1}{\frac{1}{\sqrt{\rho \epsilon}}} \frac{\partial \textbf{A}}{\partial t } - \nabla \phi \\ $
Now let us calculate the divergence of the electric field, will it form gausses law?, namely will \\
$ \frac{\rho_2}{\epsilon} = \nabla \cdot \textbf{E} = \frac{1}{c} \frac{\partial \nabla \cdot \textbf{A}}{\partial t } - \nabla \cdot \nabla \phi \\ no divergence in scalars \nabla \cdot \textbf{E} = \frac{1}{c} \frac{\partial \nabla \cdot \textbf{A}}{\partial t } \\ $
This part used to be super easy, Pardon me, but the literature has only gone so far as to say, A can be anything, as long as it derives the appropriate B, which by the Way has zero divergence, pretty easy to Find an A than, where its time derivative provides the appropriate units to balance 1/c (1/1/sqrt('ability*'ttivity')) to satisfy the experimental observations of Gauss, but now its tricker...

(1/2)

good overview https://arxiv.org/ftp/physics/papers/0608/0608051.pdf

Anonymous

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Anonymous Sat 18 Jul 2020 03:45:24 No.11911967 Report

Quoted By: >>11912403

>>11911916
(2/2)

Now $\textbf{assuming}$ that this monopole bears the same physics as the electric monopole, save only in orthogonal rotation to it. \\
$\textbf{B} = q_m \frac{\textbf{r}}{r^3} \\$

We can see that the array of viable A fields has become alot stricter.

$\nabla \cdot \textbf{B} = \nabla \cdot (\nabla \times \textbf{A}) = 0 = 4 \pi \rho_m$

by definition of vector calculus the divergence of a curl must be zero, otherwise you didn't abstract the orthogonal pieces of your field correctly, you got x components in your y array. so what is too be done? It appears many have gotten close, dirac himself posited the dirac string, but in the end their choice of A is not defined on some bound, leaving the theory unable to account for divergence where we 'aught' to observe magnetic monopoles (the same places as E)

anyways, my way at arriving at that sources statement that 'In another word the Maxwell's equations do not have the duality symmetry with respect to the ordinary potentials. ' with myself reminding the readers, in a classic mag-monopole model.

Perhaps one can see what I see, that magnetic monopoles, demand extended electrodynamics

Anonymous

Anonymous Sat 18 Jul 2020 04:12:07 No.11912003 Report

Quoted By: >>11912403

>>11909630
I've been reading lately around Bohm-Aharonov and this idea of Primacy of the Potentials. Its has a comedic element in fact. Our science is deeply rooted in materialism and action at un-distance. But in your electrodynamics we're content to treat the fields, the effects generated at a distance, as primal. Maybe the magnetic current really is physically there, but declines to be viewed optically

Anonymous

Anonymous Sat 18 Jul 2020 08:19:51 No.11912403 Report

Quoted By: >>11913022

>>11911916
>>11911967
anon, first, completely relearn how to write and use latex, because you write like a mong. second, your potential formulation is only half complete, and it's the half that assumes no magnetic sources. derive the symmetric electric potentials when you assume no electric charge, and you'll see that Gauss' law holds when you add everything together.

>>11912003
potentials seem to be the more fundamental concept, but until you break into quantum, the field formulation is completely equivalent. indeed, the Aharonov-Bohm experiment is some wacky shit

Anonymous

Anonymous Sat 18 Jul 2020 09:23:52 No.11912504 Report

Quoted By:

>>11909630
>the magnetic field can be interpreted as being the "relativistic part" of the electric field, so i wouldn't go as far as saying they are "expected to exist". it would be nice if they did for theoretical reasons, but you have to acknowledge to date there is no experimental evidence for their existence, and ultimately nature has the final word.
you fail to understand that scientists are not empricists but rationalists

Anonymous

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Anonymous Sat 18 Jul 2020 14:57:44 No.11913022 Report

Quoted By: >>11914913

>>11912403
putting ad homs like this doesn't do the thread much good.

> and it's the half that assumes no magnetic sources

really highlights you failed to comprehend my text, which is all about introducing the magnetic charge, I am assuming magnetic sources, as per OP.

>derive the symmetric electric potentials when you assume no electric charge
Why would i assume away the electric charge?, Q = kq/r^2 is like the first assumption in electrostatics?

As long as EM admits the induction term in the electric field, and magnetic fields have sources, the divergence of the electric field won't simply be charge density per unit permittitvity, but that + whatever a nearby magnetic charge is inducing upon it.

>but until you break into quantum

I'm not convinced. http://physics.princeton.edu/~mcdonald/examples/lewin.pdf provides a good example. It admits the the physicality of A is a stronger argument in QM it exists in CM.

"Vector Potential, electromagnetic inductions and 'physical meaning' https://arxiv.org/pdf/1005.2350.pdf

A note which wrestles with above source, provides first link.
http://physics.princeton.edu/~mcdonald/examples/blondel.pdf
Classically there is power flow when regions of the field where $\textbf{E} = \textbf{B} = \nabla \times \textbf{A} = 0$ Combine that with

Anonymous

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Anonymous Sun 19 Jul 2020 05:12:50 No.11914913 Report

Quoted By: >>11915953 >>11916000

>>11913022
when you derive the potential formulation, you use superposition to break the problem up into two parts; part 1 is to derive the potentials only due to electric sources, part 2 is to derive the potentials only due to magnetic sources. you can add the results together to describe the full fields. denote electric/magnetic source dependent quantities with e/m. Assume linear materials so the constitutive relations $\vec{B}=\mu\vec{H}$ and $\vec{D}=\epsilon\vec{E}$

In this way you define and derive the equations you know for electric sources
$\vec{B}_e=\mu\vec{H}_e=\nabla\times\vec{A}_e$
and
$\vec{E}_e=-\nabla\phi_e-\frac{\partial\vec{A}_e}{\partial t}$
in this case $\nabla\cdot\vec{B}_e=0$ holds, which is good because that's what was assumed.

for magnetic sources (assuming no electric sources) you define
$\vec{D}_m=\epsilon\vec{E}_m=-\nabla\times\vec{A}_m$
and derive
$\vec{H}_e=\mu^{-1}\vec{B}_m=-\nabla\phi_m-\frac{\partial\vec{A}_m}{\partial t}$
take the divergence of the last equation
$\nabla\cdot\vec{B}_m=-\nabla\cdot(\mu\nabla\phi_m)-\nabla\cdot\left(\mu\frac{\partial\vec{A}_m}{\partial t}\right)$
none of these terms have to be zero, so checking your assumptions
$\nabla\cdot\vec{B} = \rho_{mf}$
you find no contradiction.

I'll point out that gauss's law doesn't apply to $\vec{D}$ or $\vec{B}$ because they are defined in the equations [\math]\vec{D}\equiv\epsilon_0\vec{E} + \vec{P} $and [\math]\vec{H}\equiv\frac{1}{\mu_0}\vec{B} - \vec{M}<script type="math/tex">, which don't have zero curl because the curl of the polarization/magnetization densities don't have to be 0. of course, when there is no matter (or in a homogenous isotropic linear material), that's a different story. no contradiction. in the equations.$

Anonymous

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Anonymous Sun 19 Jul 2020 16:09:41 No.11915953 Report

Quoted By:

>>11914913
Ah okay, I see what your doing now, this is like comparative statics, considering the electric component than the magnetic component.

But seriously though, please view https://arxiv.org/ftp/physics/papers/0608/0608051.pdf page 2 of this paper between equations 6 and 8, the statement where it says "Then the Gauss' law for the electric charges would no longer hold because ??B=??(?×A)=0=4??m"

what would you say to this author?

Anonymous

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Anonymous Sun 19 Jul 2020 16:37:49 No.11916000 Report

Quoted By:

>>11914913
Your approach reminds me of this paper as well https://vixra.org/pdf/1007.0054v2.pdf they go full decomposition

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