Magnetic monopoles and Lorentz force law

[Lindsayyyy]

Hi buddies

Homework Statement

Show that the Lorentz Force Law generalizes to

[tex] F=q_e (\vec E+ \vec v \times \vec B)+ \frac {q_m}{\mu _0} (\vec B - \frac{\vec v}{c^2} \times \vec E)[/tex][/tex]

when there are magnetic monopoles

Homework Equations

-

The Attempt at a Solution

Well, not much to say here. I don't know where to start. I thought I could approach everything with [tex] \vec F=q \vec E[/tex] wheres [tex] \vec E=-grad \Phi -\frac {\partial \vec A}{\partial t}[/tex]
and then show that for qe and qm and use the superposition principle

But that idea didn't go as planned.

Can anyone help me out?

Thank you very much in advance

 

[TSny]

I'm not sure what you're suppose to start with here. Can you use ideas of relativity? In particular, can you use the transformation equations for force and the E and B fields between two inertial frames?

[Typo: [itex] \vec F=-q \vec E[/itex] should not have a negative sign.]

 

[Lindsayyyy]

Thanks for the quick reply. What do mean with transformation? Do you mean the dual transformation? That's the only thing he mentioned in our lecture in that context.

I just have no idea how I am supposed to show this, when we assume there are magnetic charges.

edit: yes thanks, I edited my first post

 

[TSny]

I was thinking of starting in a reference frame in which the magnetic monopole is instantaneously at rest. Then, by definition it seems to me, the force on the monopole would be ##\vec{F} = \frac{q_m}{\mu_o}\vec{B}## where the ##μ_o## is an artifact of how you define the units of the magnetic "charge" in the SI system.

To see what the force on the monopole would be in a different frame of reference where the monopole is in motion (say along the x-axis), you can just use the well-known transformation equations for the components of ##\vec{F}## and ##\vec{B}## to rewrite ##\vec{F} = \frac{q_m}{\mu_o}\vec{B}## in the frame in which the monopole is moving.

 

[Lindsayyyy]

Thanks again.
Do you mean the Lorentz transformation? I'm not familiar with transformations because we didn't have them discussed yet during the lecture.

 

[TSny]

Thanks again.
Do you mean the Lorentz transformation? I'm not familiar with transformations because we didn't have them discussed yet during the lecture.

Well, the Lorentz transformation is for transforming 4-vectors. ##\vec{F}##, ##\vec{E}##, and ##\vec{B}## are not 4-vectors, but there are transformation equations for these quantities that are derived in the study of relativity. Since you haven't covered these yet in your course, I would suspect that this is not the way you are meant to approach the problem.

You mentioned using a duality transformation. So, I guess you have covered how the fields and charges transform under these types of transformations. Have you tried applying these transformations to the Lorentz force for an electric charge ##\vec{F} = q_e(\vec{E} + \vec{v} \times \vec{B})##?

 

[Lindsayyyy]

I know that Lorentz transformation will be a topic soon. I know that our lecturer gives us some tasks which doesn't have to do anything with the lecture. I just read about the transformations you mentioned ( https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity ) if that's it what you mean.

back to the dual transform:

[tex] \begin{pmatrix}
f_e \\
f_m\\

\end{pmatrix}
=
\begin{pmatrix}
cos \theta & -sin \theta \\
sin \theta & cos \theta\\

\end{pmatrix}

\begin{pmatrix}
{f_e}' \\
{f_m}'\\

\end{pmatrix}
[/tex]

That's the only thing he gave us. I know that f stands for B E or qe/qm but I think it's in cgs units and I have trouble converting it into SI units. Furthermore I read in David Grifftiths book and the dual transform there looks different to the one our lecturer mentioned.

 

[TSny]

I know that Lorentz transformation will be a topic soon. I know that our lecturer gives us some tasks which doesn't have to do anything with the lecture. I just read about the transformations you mentioned ( https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity ) if that's it what you mean.

Yes. You would also need to know how the components of ##\vec{F}## change when switching frames.

back to the dual transform:

[tex] \begin{pmatrix}
f_e \\
f_m\\

\end{pmatrix}
=
\begin{pmatrix}
cos \theta & -sin \theta \\
sin \theta & cos \theta\\

\end{pmatrix}

\begin{pmatrix}
{f_e}' \\
{f_m}'\\

\end{pmatrix}
[/tex]

That's the only thing he gave us. I know that f stands for B E or qe/qm but I think it's in cgs units and I have trouble converting it into SI units. Furthermore I read in David Grifftiths book and the dual transform there looks different to the one our lecturer mentioned.

OK. Well, one thing you could do is take the Lorentz force for an electric charge ("electric monopole") q_e and apply the duality transformation to convert it to the force on a magnetic monople.

What should θ be to completely change an electric monopole q_e into a magnetic monopole q_m? How do the fields transform for this value of θ? What does the Lorentz force equation look like after this transformation?

 

[Lindsayyyy]

Thanks for your patience

theta must be pi/2 I guess?

so qe=-qm and E=-B

But I'm still uncertain about my units. I'm not comfortable at all when it comes so cgs units.

edit:

[tex] \vec F = q_m (\vec B -\frac {1} {c^2} \vec v \times \vec E)[/tex]

That relates to the right side, but I'm missing the left term.

 

[TSny]

theta must be pi/2 I guess?

Yes

so qe=-qm and E=-B

But I'm still uncertain about my units. I'm not comfortable at all when it comes so cgs units.

edit:

[tex] \vec F = q_m (\vec B -\frac {1} {c^2} \vec v \times \vec E)[/tex]

That relates to the right side, but I'm missing the left term.

You now have the expression for the force on a magnetic monopole (a particle with magnetic charge but no electric charge). To get this result, you must have used Griffiths' transformation, where the matrix

[itex] \begin{pmatrix}
f_e \\
f_m\\

\end{pmatrix}[/itex] can represent [itex] \begin{pmatrix}
cq_e \\
q_m\\

\end{pmatrix}[/itex] or [itex] \begin{pmatrix}
\textbf{E} \\
c\textbf{B}\\

\end{pmatrix}[/itex] with some factors of c. You still don't get the denominator of μ_o, but that must mean that Griffiths is using a different choice of unit for the magnetic monopole charge than your lecturer.

If a particle were to carry both electric charge q_e and magnetic charge q_m then the force would be the sum of the force on the electric charge and the force on the magnetic charge. This would give the expression that you want to show.

 

[Lindsayyyy]

Thanks for the help, concerning the mu0 I don't understand exactly why it is there. I took a look at
http://en.wikipedia.org/wiki/Magnetic_monopole#In_SI_units there are two entries: weber convention and ampere convention. They are both the same with the exception that one got the mu0 in it. But that would mean that the force gets another value. How does that even work?

 

[TSny]

I think it's just a matter of how you define the unit of magnetic charge q_m. You can define it so that F = q_mB or you could define it so that F = q_mH = q_mB/μ_o (for a magnetic charge at rest in a magnetic field). You'll get the same force either way.

 

[Lindsayyyy]

Thank you very much. I'll try to solve it finally tomorrow got to do some biochemistry first