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Geodesic Deviation

  1. Jan 24, 2004 #1
  2. jcsd
  3. Jan 24, 2004 #2
    i took a look at it, and did the calculation. i thought it was pretty straightforward. where did you get stuck? what extra terms do you have?

    remember that x is a geodesic. so there is a geodesic equation in x, and it therefore vanishes. and remember that χ is very small; drop any term with more than one χ in it.
     
  4. Jan 24, 2004 #3
    Re: Re: Geodesic Deviation

    I fingered it out :smile:

    One has to drop not only the term &chi*&chi but the term which is the product of &chi and a derivative of &chi. That was what I was missing.
     
  5. Jan 24, 2004 #4
    Re: Re: Re: Geodesic Deviation

    Thank you

    I believe that I've fingered it out :smile:

    One has to drop not only the term χ*χ but the term which is the product of χ and a derivative of χ. That was what I was missing.

    Again - thanks for the response

    Arcon
     
  6. Jan 25, 2004 #5
    Re: Re: Geodesic Deviation

    Seems that this small detail (drop term with χdχdT) has always tripped me up in that derivation. I guess I was just blind to it. But now that I know it then the derivation is simple.

    Just to make sure I understood the approximation can you check this for me?

    http://www.geocities.com/physics_world/gr/geodesic_deviation.htm

    I commented on the terms to drop right after Eq. (14) and right after Eq. (15)

    Thanks

    I don't know how I missed this before but the equation of geodesic deviation clearly shows that tidal forces are velocity dependant!

    Arcon
     
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