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Question about EM in GR

  1. Mar 21, 2013 #1

    quasar987

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    So apparently (wiki), the manisfestly covariant form of Maxwell's equations is dF=0 and d*F=µJ for F the Faraday 2-form and J the current 3-form. My question will probably seem silly to you but I am simply wondering how does this field affect the motion of a particle of charge q in GR? Is it just [itex]m_0\nabla_{\dot{\gamma}}\dot{\gamma}=f[/itex] for some 4-force vector field f? If so, what is the 4-force f corresponding to a Faraday 2-form F? Thanks again! :)
     
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  3. Mar 21, 2013 #2

    WannabeNewton

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    ##u^a\triangledown_au^b = \frac{q}{m}F^{b}{}{}_cu^c##.
     
  4. Mar 22, 2013 #3

    tom.stoer

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  5. Mar 22, 2013 #4

    quasar987

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    Ok so now I think I understand better what Einstein wanted to do in his later years. He simply wanted to get rid of F=ma altogether! He had done it for gravity and now he wanted to incorporate EM into the geometric structure as well.
     
  6. Mar 22, 2013 #5

    WannabeNewton

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    quasar you might also be interested in this: http://en.wikipedia.org/wiki/Gravitoelectromagnetism

    IMO it is one of the coolest things about GR and there is such a beautiful parallelism between EM and gravity in this context.
     
  7. Mar 23, 2013 #6

    quasar987

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    That is very cool WBN. Thanks for that link. :)
     
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