Magnetic monopole

dRic2

Gold Member
Problem Statement
Assuming that "Coulomb's law" for magnetic charges ($q_m$) reads
$$\mathbf F = \frac { \mu_0 } {4 \pi } \frac {q_{m1} q_{m2}} {r^2} \mathbf{ \hat r}$$
work out the force law for a monopole $q_m$ moving with velocity $v$ through electric and magnetic fields.
Relevant Equations
.
For the magnetic fields it is obvious that $F = q_m B$, but I don't get why the final result is
$$\mathbf F = q_m(\mathbf B -\frac 1 {c^2} \mathbf v \times \mathbf E)$$
The second part is like a "counterpart" of Faraday's Law, but I do not understand why it should be there... For what reason? How do I know that? Isn't Faraday's Law an "empirical law"? Why should I expect a "counterpart" of Faraday's Law for magnetic monopoles?

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TSny

Homework Helper
Gold Member
Are you allowed to use the transformation laws for E and B between two inertial reference frames?

dRic2

Gold Member
I'm not following any E&M course, so I'm allowed to do whatever I want  But I must say I do not have a very strong background in special relativity... I found this exercise in Griffiths' book. In the solutions he (or whoever wrote them) doesn't talk about transformations: he just says that one might "expect" this relation to hold (in view of some kind of physical intuition). I did some digging in the web and I found that indeed we can derive this expression using duality transformation (for example), but I though there might be a simpler explanation. If there isn't I should come back to this when I'll be more comfortable with special relativity.

sysprog

DRic2 said:
Sorry, I don't see the connection.
Maybe I misunderstood what you were asking.
Isn't Faraday's Law an "empirical law"?
Yes, it is. Faraday's first law of electrical induction says that when a (dipole) magnet moves (its field lines are caused to change their positions over time) relative to a conducting loop (closed circuit), electrical current is induced in the loop. It's abundantly verified empirically.
Why should I expect a "counterpart" of Faraday's Law for magnetic monopoles?
A magnetic monopole is to a magnetic dipole as an open electrical circuit is to a closed electrical circuit, but what we call a magnetic monopole has never been observed, whereas what we call an open electrical circuit is commonplace, but it's not observed to be really a circuit if it's open.

dRic2

Gold Member
Ok, but why do you expect the force acting on a magnetic monopole to be like $F = -q_m \frac 1 {c^2} \left( v \times E \right)$ ?

robphy

Homework Helper
Gold Member
The electromagnetic force of an electric charge
is the electrostatic force when Faraday's Law has a zero-curl Electric Field.
When the curl is nonzero, we get the full Faraday's Law and the full Lorentz force on an electric charge.

By analogy,
consider the zero-electric-current Ampere-Maxwell Law to form the analogue of the Lorentz force for a magnetic charge. Note the sign in the Faraday Law vs. the zero-electric-current Ampere-Maxwell Law.

• sysprog

dRic2

Gold Member
I'm sorry, I'm an idiot. When I wrote "Faraday's Law" I meant "Lorentz Force". I'm no good with names and I always confuse them (I don't know why I usually call it like "force equation" or something similiar...). I'm very sorry if I wasted your time.

My questione Is: why is the force exerted on a magnetic charge moving through an electric field $F= \frac{q_m} {c^2}(v \times E)$? It looks like the Lorentz Force for an iphotetical magnetic charge.

I'm sorry again for my mistake. I just realized it.

sysprog

dRic2 said:
... When I wrote "Faraday's Law" I meant "Lorentz Force". I'm no good with names and I always confuse them (I don't know why I usually call it like "force equation" or something similiar...)...
Well, if we're being fussy about about whose law is whose, we should probably notice too that Lorentz was not the only person who had something to do with the dimensional analysis that brought in the necessity of the $1/c^2$ term: from a W.D. Bauer paper entitled, "The Maxwell equations including magnetic monopoles"(Bauer apparently disregards in this paper how we spell 'years' in Englisch):
The Maxwell equations are about 150 Jahre old. They are the mathematical compilation of the experiments and considerations based on the original work of Cavendish, Coulomb, Poisson, Ampère, Faraday and others . Mathematically they are partial differential equations. Different notations exist for them: most popular is the vector notation (O. Heaviside), which replaced the original notations in quaternions (J.C. Maxwell). More modern is the tensor notation (H. Minkowski, A. Einstein), which is able to describe situations which are discussed in the theory of relativity . All notations are equivalent in the nonrelativistic limit.
Its easy when swimming in these waters to forget the names of the vanguards when we just want them to help us to get to another safety zone. This paper by F. Moulin might help to make that metaphor more palpable: Magnetic monopoles and Lorentz force

Did you perhaps disregard what @robphy said about the difference in sign (+ or -) between the formulations of Faraday and Ampere-Maxwell?

In the equations you've presented, you've used both (+ by default in the latter):
$F = -q_m \frac 1 {c^2} \left( v \times E \right)$
$F= \frac{q_m} {c^2}(v \times E)$
I trust that you recognize that in both equations, $q_m$ is the magnetic monopole force, regardless of sign (we have (+ or -) $e_m$ for electrical ions, but we don't have a correspondingly specific conventional sign (e.g. N or S) $q_m$ for magnetic monopoles, as it would (presumably) spoil the arithmetic if we did).
It looks like the Lorentz Force for an iphotetical magnetic charge.
Assuming I'm right about what you mean by "iphotetical" (that's not a commonly used word in English unless it's merely an alternate spelling of 'hypothetical'), yes, it is precisely like that.

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• dRic2

dRic2

Gold Member
I trust that you recognize that in both equations, qmqmq_m is the magnetic monopole force, regardless of sign (we have (+ or -) ememe_m for electrical ions, but we don't have a correspondingly specific conventional sign (e.g. N or S) qmqmq_m for magnetic monopoles, as it would (presumably) spoil the arithmetic if we did).
Yes, I'm sorry. I was in hurry when writing the post.

Assuming I'm right about what you mean by "iphotetical"

it is precisely like that.
Thanks, for the article although I'm not familiar with the derivation of Maxwell equations using the principle of least action. If anyone can give a more "physical explanation" for the general form of the Lorentz Force I'd be very happy, otherwise I'll be back when I can understand the article (give me a couple of years though ).

sysprog

dRic2 said:
I was in hurry when writing the post.
I intended that remark only to support my suggestion that fully considering what @robphy had written might help you to better understand.
Thanks, for the article although I'm not familiar with the derivation of Maxwell equations using the principle of least action. If anyone can give a more "physical explanation" for the general form of the Lorentz Force I'd be very happy, otherwise I'll be back when I can understand the article (give me a couple of years though ).
If you do a search on Lorentz here, even with no other keywords, you'll find a wealth of good material.

Note: When you want to quote only part of a post, if you use the quote or reply button, and then excise the part you don't need, instead of just using copy and paste, you won't mangle the $\TeX$ content. Alternatively, you can use: [highlight] > [right click] > Show Math As > TeX Commands, then copy and paste the $\TeX$ expression separately from the rest of the text. Last edited:
• dRic2

"Magnetic monopole"

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