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A Electromagnetic action in differential forms

  1. Oct 30, 2016 #1
    The electromagnetic action can be written in the language of differential forms as

    ##\displaystyle{S=-\frac{1}{4}\int F\wedge \star F.}##

    The electromagnetic action can also be written in the language of vector calculus as

    $$S = \int \frac{1}{2}(E^{2}+B^{2})$$

    How can you show the equivalence between the two formulations of the electromagnetic action?


    Here is my attempt:

    ##\displaystyle{S=-\frac{1}{4}\int F\wedge \star F}##

    ##\displaystyle{=-\frac{1}{4}\int \left(\sum_i E_i\,{\rm d}t\wedge{\rm d}x^i - \star\sum_i B_i\,{\rm d}t\wedge{\rm d}x^i\right)\wedge \star \left(\sum_j E_j\,{\rm d}t\wedge{\rm d}x^j - \star\sum_j B_j\,{\rm d}t\wedge{\rm d}x^j\right)}##

    ##\displaystyle{=-\frac{1}{4}\int \left(\sum_i E_i\,{\rm d}t\wedge{\rm d}x^i - \star\sum_i B_i\,{\rm d}t\wedge{\rm d}x^i\right)\wedge \left(\star \sum_j E_j\,{\rm d}t\wedge{\rm d}x^j - \sum_j B_j\,{\rm d}t\wedge{\rm d}x^j\right),}##

    since ##\displaystyle{**=(-1)^{p(n+p)+1}}## in Lorentzian space, where ##\star## is applied on a ##p##-form and ##n## is the number of spacetime dimensions, so that, in four dimensions for the ##2##-form ##\displaystyle{dt\wedge dx^{j}}##, ##\displaystyle{**=(-1)^{p(n+p)+1}=-1}##.

    What do you do next?
     
  2. jcsd
  3. Oct 31, 2016 #2

    vanhees71

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    First of all your ##S## in (1+3)-form is wrong. The integrand should be ##\propto (\vec{E}^2-\vec{B}^2)##. Then first write down more carefully ##F## and then ##*F## and then multiply out the forms.
     
  4. Oct 31, 2016 #3
    So, you mean that

    ##\displaystyle{F=\left(\sum_i E_i\,{\rm d}t\wedge{\rm d}x^i - \star\sum_i B_i\,{\rm d}t\wedge{\rm d}x^i\right)}##

    is wrong?
     
  5. Nov 2, 2016 #4

    vanhees71

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    It's much easier in components (as usual). The Lagrangian is
    $$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}.$$
    Now
    $$F_{0j}=E_j, \quad F_{jk}=-\epsilon_{ijk} B_i,$$
    and you can easily decompose the Lagrangian in temporal and spatial components to write it in terms of the ##(1+3)##-formalism. You must get something ##\propto (\vec{E}^2-\vec{B}^2)##. The other invariant of the Faraday tensor is ##{^\dagger}F^{\mu \nu} F_{\mu \nu} \propto \vec{E} \cdot \vec{B}##.
     
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