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The action integral for the EM field in a dielectric medium

  1. Jan 8, 2007 #1
    Show that the action integral for the electromagnetic field in a matter with dielectric constant [tex]\epsilon[/tex] is given by [tex]S_m=\frac{1}{8 \pi} \int dt d^3 \bf x \left[ \epsilon \bf E^2 - \bf B^2 \right][/tex]

    where the standard expressions for the E field and B field [tex]E=-\frac{1}{c}\partial_t \bf A - \nabla \phi[/tex] and [tex]\bf \nabla \times \bf A[/tex] are used.

    I'm familiar with the Lagrangian principle L = T - V, where T and V are kinetic and potential energy respectively, and I'm familar with the concept that minimising the action integral of a system leads us to finding the equations of motion for that system. In this case [tex]d^3 \bf x \left[ \epsilon \bf E^2 - \bf B^2 \right][/tex] is the Lagrangian density of the system. I can see that the expression for Lagrangian density has units of energy per unit volume as one would expect.. But why does the Lagrangian density take that particular form?

    Thank you for taking the time to read this; any help would be much appreciated.
    Last edited: Jan 8, 2007
  2. jcsd
  3. Jan 10, 2007 #2
    For a system of particles, finding the Lagrangian is quite easy, since then you know that [tex] L = T - V [/tex] as you wrote, and the kinetic and potential energy are often known.

    Now, however, you are working with a Lagrangian for the electromagnetic field. If I have understood things correctly, there is no general method of finding a Lagrangian.

    To solve your problem, I would have started from the action and shown that the Euler-Lagrange equations then reproduces the Maxwell equations for fields in matter. This will then prove that the action [tex] S_m [/tex] reproduces the correct equations of motion and hence that it is the correct one.
  4. Jan 10, 2007 #3
    Thanks Jezuz. I think I'm happy with the problem now :smile:

    Now for a related problem. I have

    [tex]S_m=\frac{1}{8 \pi} \int dt d^3 \bf{x} \left[ \frac{\epsilon}{c^2} (\partial_t \bf{A})^2 - (\nabla \times \bf{A})^2 \right][/tex]

    and apparently this can somehow be written as

    [tex]S_m=\frac{1}{8 \pi} \int \frac{dw}{2 \pi} \int d^3r A_i(-w,r)[\epsilon(w,r) (w/c)^2 \delta_{ij} + \partial_k \partial_k \delta_{ij} - \partial_i \partial_j] A_j (w,r)[/tex]

    where w is angular frequency I think.

    to me this suggests using fourier transformation and the expression for curl but damned if I know how.
  5. Jan 11, 2007 #4
    OK so I was correct in thinking I need to do a Fourier transform and expand the curl, in particular using the Levi-Civita tensor for the curl and Parseval's theorem for the transform.

    First I should concentrate on the curl and use the following expression

    [tex](\nabla \times A)_i = \varepsilon_{ijk} \partial_j A_k[/tex]


    Anyway I still have lots of avenues to explore, just thought I'd update the thread a bit especially since it helps in terms of organising.

    Please feel free to jump in at any time and make this a bit less painful and slow :P

    Last edited: Jan 11, 2007
  6. Jan 18, 2007 #5
    Yepp, a Fourier transform should be used. I don't know if there is any differences between the variables [tex] x [/tex] and [tex] r [/tex], perhaps not.

    There is a way to rewrite the levi-civita symbol in terms of delta's, but I can't remember it. This might simplify your calculations somewhat. I will post the expression if I find it.
  7. Jan 19, 2007 #6
    ah no worries jezuz got there in the end :) (should've said so earlier sorry)
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