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EM Differential Form version of Lagragian

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haloempty
#1
Feb27-11, 11:48 PM
P: 2
Given that we have the one form A we can get the F=dA, and I see how dF=0 and d*F=J, give maxwell's equations. But if we write the Action as [tex]\int[/tex]1/2dA[tex]\wedge[/tex]*dA+A[tex]\wedge[/tex]J how do we go about doing a variational procedure to get d*dA=J? I tried taking d(Lagrangian)=0 but that seemed to be off by a factor of 2. I guess I'm not sure how you vary an action when your lagrangian is composed of forms. Any ideas?
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