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Magnetic monopoles and Lorentz force law

  1. Jun 20, 2013 #1
    Hi buddies

    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations

    -

    3. 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
     
    Last edited: Jun 20, 2013
  2. jcsd
  3. Jun 20, 2013 #2

    TSny

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    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.]
     
    Last edited: Jun 20, 2013
  4. Jun 20, 2013 #3
    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
     
    Last edited: Jun 20, 2013
  5. Jun 20, 2013 #4

    TSny

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    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.
     
  6. Jun 20, 2013 #5
    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.
     
  7. Jun 20, 2013 #6

    TSny

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    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})##?
     
  8. Jun 20, 2013 #7
    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.
     
  9. Jun 20, 2013 #8

    TSny

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

    OK. Well, one thing you could do is take the Lorentz force for an electric charge ("electric monopole") qe 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 qe into a magnetic monopole qm? How do the fields transform for this value of θ? What does the Lorentz force equation look like after this transformation?
     
  10. Jun 20, 2013 #9
    Thanks for your patience:blushing:

    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.
     
  11. Jun 20, 2013 #10

    TSny

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    Yes

    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 μo, 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 qe and magnetic charge qm 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.
     
    Last edited: Jun 20, 2013
  12. Jun 21, 2013 #11
    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?
     
  13. Jun 21, 2013 #12

    TSny

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    I think it's just a matter of how you define the unit of magnetic charge qm. You can define it so that F = qmB or you could define it so that F = qmH = qmB/μo (for a magnetic charge at rest in a magnetic field). You'll get the same force either way.
     
  14. Jun 21, 2013 #13
    Thank you very much. I'll try to solve it finally tomorrow got to do some biochemistry first :eek:
     
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