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Bending of light

  1. Apr 10, 2010 #1
    Can the bending of light because of gravity be derived from the Maxwell equations written in curved space time, i.e.,

    [tex]\frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}F^{\mu\nu}})=0 [/tex]

    In all the examples the bending of light is treated as a massless particle travelling on a light-like geodesic (if I understand the examples correctly) with no reference to the electromagnetic field at all.
  2. jcsd
  3. Apr 10, 2010 #2


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    Yes, it is possible to show that if light is assumed to be plane waves, then paths that are normal to the planes are null geodesics.

    The plane wave ansatz is implemented by setting
    [tex]A_\mu=\hat{A}_\mu e^{i\omega S}[/tex]

    The rest of the derivation is rather long. See 'geometrical optics'.
  4. Apr 10, 2010 #3


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    I don't know if this has been done directly.

    I think usually one shows there is a ray approximation to a solution of the Einstein-Maxwell equations, and these are null geodesics, and then lives with that.

    http://relativity.livingreviews.org/Articles/lrr-2004-9/ [Broken]
    Last edited by a moderator: May 4, 2017
  5. Apr 10, 2010 #4
    So if you have a massless particle, say a graviton, which is a 2nd-ranked tensor, or a neutrino which is a spinor, then would the bending be exactly the same as for light which is a vector? I would add scalar particles but I'm unaware of any that are massless.

    So somehow the mass=0 free solutions of the Dirac, the Maxwell, and spin-2 (is there a name for the spin 2 equation?) equations all have the same solutions?

    Also, a quick question on gravitons. It seems to me that the graviton is a very special spin-2 particle. All spin 1/2 particles obey the same equation, the Dirac equation. All spin 0 particles obey the Klein-Gordan equation. But the action for the graviton [tex]g_{\mu\nu} [/tex] is:

    [tex]S=\int d^4x\sqrt{-g}R [/tex]

    where [tex]R=R(g_{\mu \nu}) [/tex] is the scalar curvature which is a function of the graviton field.

    For a generic spin 2 field [tex]z_{\mu\nu} [/tex] that is not the graviton, the action is:

    [tex]S=\int d^4x \sqrt{-g}R [/tex]

    where R has the same form as the scalar curvature, but the [tex]g_{\mu\nu} [/tex]'s are replaced with [tex]z_{\mu\nu} [/tex]'s.

    So it seems the EOM for the graviton and other spin 2 particles are different. Is this right?
    Last edited: Apr 10, 2010
  6. Apr 10, 2010 #5


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    Well, the classical graviton is a gravitational wave which should travel on null geodesics of the background spacetime.

    For other spin 2 particles, I might try searching for bimetric theories http://relativity.livingreviews.org/Articles/lrr-2005-5/ [Broken]. Lubos Motl had interesting comments on whether it was possible to have another massless spin 2 particle http://motls.blogspot.com/2008/07/bimetric-pseudoscience.html (as you probably know, ignore the rhetoric about "pseudoscience", he usually has good physics comments).
    Last edited by a moderator: May 4, 2017
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