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E&M if there were monopoles (Lagrangian?)

  1. Feb 26, 2010 #1
    I've seen in textbooks before, there if there were monopoles, Maxwell's equations would look more "symmetric" between the E and B fields. Such that (in cgs units):

    [tex]\begin{align*}
    \nabla \cdot \mathbf{E} &= 4 \pi \rho_e \\
    \nabla \cdot \mathbf{B} &= 4 \pi \rho_m \\
    -\nabla \times \mathbf{E} &= \frac{1}{c}\frac{\partial \mathbf{B}} {\partial t} + \frac{4 \pi}{c}\mathbf{j}_m \\
    \nabla \times \mathbf{B} &= \frac{1}{c}\frac{\partial \mathbf{E}} {\partial t} + \frac{4 \pi}{c} \mathbf{j}_e
    \end{align*}
    [/tex]

    And the Lorentz force law becomes:
    [tex]\mathbf{F}=q_e\left(\mathbf{E}+\frac{\mathbf{v}}{c}\times\mathbf{B}\right) + q_m\left(\mathbf{B}-\frac{\mathbf{v}}{c}\times\mathbf{E}\right)[/tex]



    My question is, How can we write electrodynamics in a Lagrangian now?

    Using non-relativistic momentum, and without magnetic monopoles it is:
    define the electric and magnetic fields in terms of a scalar and vector potential as such
    [tex] \mathbf{E} = -\nabla \phi - \frac{\partial}{\partial t} \mathbf{A}[/tex]
    [tex] \mathbf{B} = \nabla \times \mathbf{A}[/tex]
    Using four vector notation to collect and define some things
    [tex] A^\mu = (\phi,A_x,A_y,A_z)[/tex] (the potential four-vector)
    [tex] j^\mu = (\rho c,j_x,j_y,j_z)[/tex] (the current four-vector)
    [tex] F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu[/tex] (the electromagnetic tensor)

    The lagrangian is then:
    [tex] \mathcal{L} = \frac{1}{2} m \dot{\mathbf{x}}^2 + j^\mu A_\mu + \frac{1}{4} \int \mathrm{d}^3r \ F^{\mu\nu}F_{\mu\nu} [/tex]
    Where the degrees of freedom to vary are [itex]x[/itex] (the particle position), and the potential components [itex]A^\mu[/itex].
    (I may have messed up signs and constants in this one.)


    How in the world would I write this for the case with magnetic monopoles?
    Since I should recover this when there aren't monopoles, I'm guessing I just need to add stuff?
    I can easily write the case where there aren't electric charges, and only magnetic charges by analogy (creating another set of four-vectors). But then I need to "couple" these four-vectors and the charges to both of them, in order for the combination to work. I can't figure out how to do it right. In particular, I need to get a force q_m . B How the heck do you write than in terms of the usual vector potential?


    Does anyone know how to do this?
     
  2. jcsd
  3. Feb 26, 2010 #2

    Born2bwire

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    Yeah, I think I've asked this question and got nowhere. There is a small article in some Italian physics journal that I found that derived the Lagrangian. I really do not know the reputation of the journal though. I can't find a soft copy on my computer but my recollection was that you simply added the terms that represented the dual of the appropriate electric sources.
     
  4. Feb 27, 2010 #3

    clem

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    You have to introduce another 4-vector potential and a new EB field tensor to accommodate the new rho_m and j_m.
     
  5. Feb 27, 2010 #4
    I have played around with this as well, but never fully resolved it. I think the answer is in adding a second four potential as you mentioned, and of course the question is how. The way I proposed is to define it in terms of a multi-vector potential, and the only way I really know how to describe that is with geometric algebra, which does not seem to be very common, so I will give a brief rundown.

    In geometric algebra, you define the field strength tensor as a bivector, like so:

    [tex]F = E + IB[/tex]

    Where E and B are themselves bivectors. If you are not familiar with this form, the basis vectors are the space-time basis:

    [tex]\gamma_0, \gamma_1, \gamma_2, \gamma_3[/tex]

    The "pseudo scalar" I is defined as

    [tex]I = \gamma_0\gamma_1\gamma_2\gamma_3[/tex]

    the E bivector is:

    [tex]E_1\gamma_1\gamma_0 + E_2\gamma_2\gamma_0 + E_3\gamma_3\gamma_0[/tex]

    Where IB becomes

    [tex]-B_1\gamma_2\gamma_3 + B_2\gamma_1\gamma_3 -B_3\gamma_1\gamma_2[/tex]

    So, you can see this is an equivalent expression for F when you notice the components of the bivectors are simply the indices of the F matrix.

    I will try to skip to the punch line, but basically, you can express maxwells equations as:

    [tex]\nabla F = J[/tex]

    and the field strength in terms of the four-potential as

    [tex] F = \nabla A[/tex]

    And so the wave-equations as

    [tex]\nabla^2 A = J[/tex]

    So, now, when you add magnetic sources, maxwells equations become:

    [tex]\nabla F = J_e + J_mI[/tex]

    where we now have two four currents (one electric and one magnetic)

    Now, the field strength, I believe, will become

    [tex] F = \nabla (A_e + A_mI)[/tex]

    If you were to write this out in classic form, using only the (3)vector and scalar potential form, it would be:

    [tex] E = -\nabla \Phi_e + \partial_t A_e - \nabla \times A_m[/tex]

    [tex] B = -\nabla \Phi_m + \partial_t A_m + \nabla \times A_e[/tex]

    Ok, so now what about the Lagrangian? It is typically written with the interaction term:

    [tex] J \cdot A[/tex]

    which I will write here, specifically using the electric current and potential, as:

    [tex] <J_eA_e>_0[/tex]

    which will pick out the scalar part of the geometric product. If we replace with only magnetic sources the interaction term is of course:

    [tex] <J_mA_m>_0[/tex]

    The question is now what about the cross interactions, which will NOT give the correct if we simply put something like:

    [tex] <J_mA_e>_0[/tex]

    Because that would just be traditional electric force, but on a magnetic particle, which is incorrect. Also, it that for would not obey parity and time reversal, which is bad for our Lagrangian.

    I have not really been able to show this to my satisfaction, yet, but I think it is really something like this:

    [tex] <J_mA_e>_0I = <J_mA_eI>_4[/tex]

    This combination now obeys parity and time reversal, which is good, because the pseudo-scalar does not. The funny business is now the Lagrangian would not be a pure scalar! However, the interaction term would simply be:

    [tex] <(J_e + J_mI)(A_e + A_mI)>_{0,4} = <JA>_{0,4}[/tex]

    which is fairly elegant, where the current and potential are now multi vectors. If a theorist could take this further I would love to see it.
     
    Last edited: Feb 27, 2010
  6. Feb 28, 2010 #5
    Your suggestions look great, but since I don't fully understand the notation, I'm only getting vague hints of what actually is going on. Would you mind answering some mundane math questions to help me understand your comments a bit better?

    I don't understand what that is. If a vector uses one basis for each component, like
    [tex]\vec{r} = r_0 \hat{e}_0 + r_1 \hat{e}_1 + r_2 \hat{e}_2 + r_3 \hat{e}_3[/tex]
    And a rank two tensor uses two bases
    [tex]T = T_{00} \hat{e}_0 \hat{e}_0 + T_{10} \hat{e}_1\hat{e}_0 + ... [/tex]
    Then what exactly is multiplying all four together, and how is it like a scalar?

    Can you show how I would actually work out IB to get that last part?
    If I understood that notation, then I can play with the rest of it myself and see if your coupling ideas work out. They look promising.

    Thanks.

    EDIT:
    Actually, one last thing
    How does that notation work out to mean the same as
    [tex]F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu[/tex]
    ??
     
  7. Feb 28, 2010 #6
    Yes. I know this form is not used often, but I think it more elegant than tensor notation once you get used to it.

    Each rank represents a directed area/volume. 1=length,2=area, 3=volume, 4=hypervolume. For example, in 3D cartesian e_1e_2 represent a unit area whos normal is along e_3. It is like a cross product, but it is rank two, not rank one. Also, you can see there are 3 basis for rank 2 vectors (ie, all rank 2 vectors are linear combinations of e_1e_2, e_2e_3, e_3e_1). For the 3D cartesian case, e_1e_2e_3 is called the pseudoscalar because all rank three vectors are a linear combination of only this single element, just like there is only one unit scalar element (1). The use of I is indicative of the similarity to the imaginary unit, but that is another story.

    In 4D, the number of basis are 1 of rank 0 (1), 4 of rank 1, 6 of rank 2, 4 of rank 3, and 1 of rank 4 (I).

    [tex]
    B = B_1\gamma_1\gamma_0 + B_2\gamma_2\gamma_0 + B_3\gamma_3\gamma_0
    [/tex]

    Just looking at the first component of IB

    [tex]
    IB_1\gamma_1\gamma_0 = B_1\gamma_0\gamma_1\gamma_2\gamma_3\gamma_1\gamma_0 = B_1\gamma_1\gamma_2\gamma_3\gamma_1 = -B_1\gamma_2\gamma_3
    [/tex]

    this uses the identities of

    [tex]
    \gamma_\mu\gamma_\nu = -\gamma_\nu\gamma_\mu
    [/tex]

    And, the (1,-1,-1,-1) spacetime metric of

    [tex]
    \gamma_i\gamma_i = -1, (i = 1,2,3)
    [/tex]

    [tex]
    \gamma_0\gamma_0 = 1
    [/tex]

    Yeah, this is the tricky one and you do have to be careful with indices here, which I have not been so far.

    [tex]
    \nabla = \gamma^\mu \partial_\mu = \gamma_\mu \partial^\mu (\mu= 0,1,2,3)
    [/tex]
    where you sum the indices.

    and

    [tex]
    A = \gamma_\mu A^\mu
    [/tex]

    When you you perform the geometric product, you get a rank 0 term, and a rank 2 term. The rank 0 term is:

    [tex] \partial_\mu A^\mu[/tex]

    Which if you notice is the divergence of A, and should be zero.

    the rank two term is, once you collect like terms, (notice the change of index from lower to upper on the derivative)

    [tex] (\partial^\mu A^\nu - \partial^\nu A^\mu)\gamma_\mu\gamma_\nu = F[/tex]

    edit: P.S.

    It occurs to me that if we do not restrict the rank of the Lagrangian, why not throw in the other Lorentz invariant term of FF.

    [tex] <FF>_{0,4} = <(E+IB)(E+IB)>_{0,4} = E^2 - B^2 + 2 E\cdot B I[/tex]

    What the dynamical consequence of that would be idk.
     
    Last edited: Feb 28, 2010
  8. Feb 28, 2010 #7

    samalkhaiat

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    Have you looked at Dirac's original work? For a proper understanding, I suggest the following papers:

    1) Dirac, Phys.Rev. 74(1948),817. or
    1') Dirac, Proc.Roy.Soc.A,133(1931),60.
    2) Schwinger, Phys.Rev.144(1966),1087.
    3) Zwanziger, Phys.Rev.176(1968),1489.
    4) Zwanziger, Phys.Rev.D3(1971),880.
    5) Zwanziger, Phys.Rev.D6(1972),458.

    Dirac formulation of monopole theory was a nonlocal one and it contained non-physical dynamical variable(to describe his string). In 1966 J. Schwinger[2] constructed a quantum field theory of magnetic and electric charge which avoid the singular string of Dirac, it is based on Hamiltonian density expressed nonlocally in terms of field variables. Relativistic invariance of his theory was shown to be a consequence of the charge quantization condition. Zwanziger[3-5], brought the theory of monopole close to the standard form of Lagrangian field theory. His local Lagrangian depends on a pair of 4-potentials (in contrast with Dirac's single-potential formulation) and it also depends on a fixed 4-vector ( this expresses the fact that Bianchi identity is lost in the generalized Maxwell equations).Because of this fixed 4-vector, manifest isotropy of spacetime is lost (price one has to pay for avoiding the string).However Isotropy is regained only for quantized values of coupling constants.


    regards

    sam
     
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