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Quick one-line working on Geodesic Deviation Equation

  1. Feb 28, 2015 #1
    Taken from my lecturer's notes on GR:

    GR9.png

    I'm trying to understand what goes on from 2nd to 3rd line:

    [tex]N^\beta \nabla_\beta (T^\mu \nabla_\mu T^\alpha) - N^\beta \nabla_\beta T^\mu \nabla_\mu T^\alpha = -T^\beta \nabla_\beta N^\mu \nabla_\mu T^\alpha [/tex]

    Using commutator relation ## T^v \nabla_v N^\alpha = N^v\nabla_v T^\alpha ## we swap the 'N' for the 'T' in the last term of the second line:

    [tex] N^\beta \nabla_\beta (T^\mu \nabla_\mu T^\alpha) - T^\beta \nabla_\beta N^\mu \nabla_\mu T^\alpha = -T^\beta \nabla_\beta N^\mu \nabla_\mu T^\alpha [/tex]

    [tex] N^\beta \nabla_\beta \left(\frac{D T^\alpha}{D\lambda}\right) - T^\beta \nabla_\beta N^\mu \nabla_\mu T^\alpha = -T^\beta \nabla_\beta N^\mu \nabla_\mu T^\alpha [/tex]


    This means that ## \nabla_\beta \left(\frac{D T^\alpha}{D\lambda}\right) = 0 ##. Why is this so?
     
  2. jcsd
  3. Feb 28, 2015 #2

    ChrisVer

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    May I ask what is T and N?
    Nevertheless, it looks like the thing you have in the parenthesis [itex](T^a \nabla_a T^b)[/itex] should be the geodesic equation?
     
  4. Feb 28, 2015 #3

    George Jones

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    Use the product rule on the second term of the second line.
     
  5. Feb 28, 2015 #4
    That would just take me back to the first line.
     
  6. Feb 28, 2015 #5
    ##T## and ##N## are tangent vectors and normal vectors correspondingly, with ##T## describing movement along a geodesic and ##N## describing movement from one geodesic to another.

    I thought the geodesic equation is ##\frac{d^2x^\beta}{d\tau^2} + \Gamma^\beta _{\mu v} \frac{d x^\mu}{d\tau} \frac{d x^v}{d\tau} ##?

    I reduced the term in the brackets to ## \nabla_\beta \left(\frac{D T^\alpha}{D\lambda}\right) ##. I'm trying to understand why it is zero.
     
  7. Feb 28, 2015 #6

    ChrisVer

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    The geodesic equation by writting [itex]U^a = \frac{dx^a}{d \tau}[/itex] (the velocity vector/ is it tangent to the trajectory??) can be written in the form : [itex]U^a D_a U^b=0[/itex] (that's why that expression even though I didn't know what it stands for reminded me of the geodesics equation)
     
  8. Feb 28, 2015 #7
    For a curve on space-time ##x^\alpha = x^\alpha (\lambda)##, we define the absolute derivative of a vector ##V^\mu## along that path to be expression ## \frac{D V^\mu}{D\lambda} = T^\alpha \nabla_\alpha V^\mu ##. For ##V^\mu = T^\mu##, we get ## \frac{D T^\mu}{D\lambda} = T^\alpha \nabla_\alpha T^\mu ##

    If the vector ##T^\mu## is parallely transported along the path, ##\frac{D T^\mu}{D\lambda} =0 ##. This is obviously true for the tangent vector ##T^\alpha## ,so ##\frac{D T^\alpha}{D\lambda} =0 ##

    Using ## T^\alpha = \frac{dx^\alpha}{d\lambda} ## and expanding, it simply gives the geodesic equation:

    [tex] \frac{d^2 x^\alpha}{d \lambda^2} + \Gamma^{\alpha}_{\mu v} \frac{dx^\mu}{d\lambda} \frac{dx^v}{d\lambda} = 0 [/tex]
     
    Last edited: Feb 28, 2015
  9. Feb 28, 2015 #8

    George Jones

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    Oops, I misread your first post.
     
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