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Lagrangian of electrodynamics

  1. Apr 16, 2008 #1

    haushofer

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    Hi, I have a computational question which concerns forms. I want to compute the variation of the electrodynamic Lagrangian, seen here as an n-form:

    [tex]L = -\frac{1}{2}F \wedge *F[/tex]

    with F=dA. I want to derive the Noether-current from this Lagrangian. The symmetrytransformation we are concerned with are coordinatetransformations induced by Lie-derivatives acting on A. A general variation of L can be composed as

    [tex]\delta L = E \cdot\delta A+ d\Theta[/tex]

    where [tex]\Theta[/tex] are the boundary terms and E are the equations of motion for the vector potential A. If we now have a vector field [tex]\xi[/tex] we can construct the Noether current

    [tex]
    \mathcal{J} \equiv \Theta -\xi\cdot L
    [/tex]

    (where the dot indicates contraction with the first index of L) such that

    [tex]d\mathcal{J} = - E\delta A[/tex]

    If the equations of motion hold, then there can be a Noether charge Q such that

    [tex] \mathcal{J} = dQ [/tex]

    I want to verify this for the electrodynamic Lagrangian given above, and I have the suspicion that for this particular Lagrangian we can't construct this Q ( so that the current [tex]\mathcal{J}[/tex] isn't exact, but it should be closed). But I'm a little stuck with the calculation. A variation of L gives me

    [tex] \delta L = -\frac{1}{2} (\delta F \wedge *F + F \wedge \delta *F)[/tex]

    which can be worked out, with F=dA, as

    [tex]
    \delta L = -\frac{1}{2}[d(\delta A \wedge *F) + \delta A \wedge d*F + F \wedge \delta *F ][/tex]

    I'm interested in the A-field. I thought that

    [tex]
    \delta * F = * \delta F + \frac{1}{2}(g^{\alpha\beta}\delta g_{\alpha\beta}) * F
    [/tex]

    and the metric-part is going to give me the energy-momentum tensor of the electromagnetic field, which we can disregard. I recognize in this variation

    [tex]
    \Theta = -\frac{1}{2}\delta A \wedge *F
    [/tex]

    So I would say that my Noether current is given by

    [tex]
    \mathcal{J} = -\frac{1}{2}\Bigr(\delta A - \xi\cdot F \Bigr)\wedge * F
    [/tex]

    but if I take the exterior derivative of this, it doesn't give me the form I want; It's not exact if the equations of motion for A hold.

    So my questions are :

    1)what is the corresponding Noether current for the electrodynamic Lagrangian associated with diffeomorphism-invariance of the action?

    2) Is this current exact?
     
  2. jcsd
  3. May 14, 2008 #2
    thanks.........................
     
  4. May 14, 2008 #3
    hi

    thanks..for all and i wait more...........
     
  5. May 15, 2008 #4
    Try consulting Eguchi,Gilkey and Hanson:"Gravitation,gauge theories and differential geometry",Physics Reports,Vol.66,6,pp.213-393,December 1980.This is a nice handbook-style article that you may already be familiar with.
     
  6. May 29, 2008 #5

    haushofer

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    Science Advisor

    Thanks for the advice ! A late reply, because I spend some time elsewhere :) I already solved this problem myself, but I will certainly take a glance at your article !
     
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