EM Differential Form version of Lagragian
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 LaTeX Code: \\int 1/2dALaTeX Code: \\wedge *dA+ALaTeX Code: \\wedge 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?
|All times are GMT -5. The time now is 03:42 AM.|
Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums