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I Schutz: question regarding geodesic deviation

  1. Mar 21, 2017 #1
    I have some problem understanding the section on "Geodesic deviation" in schutz, more specifically I'm confused by eq. 6.84:

    Eq 6.84 reads (ξ is the 'connecting vector' from one geodesic to Another, V is the tangent vector):

    We can use (6.48) to obtain
    VVξα = ∇V(∇Vξα) = (d/dλ)(∇Vξα) = Γαβ0(∇Vξα)

    (Eq 6.48 gives the second equality, but I fail to see why the last equality is true)​

    Eq 6.48 says the following:
    UβVα = 0 ⇔ (d/dλ)V = ∇UV = 0
    (U is tangent to the curve, λ is the parameter along it)

    Can someone please help me and explain what's going on?
  2. jcsd
  3. Mar 21, 2017 #2


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    In the book it is

    ##\nabla_V(\nabla_V\xi^\alpha)=\frac{d}{d\lambda}(\nabla_V\xi^\alpha)+\Gamma^\alpha_{\beta 0}(\nabla_V\xi^\beta)##
  4. Mar 21, 2017 #3
    Nope. At least not in my book. (Photo attached)
    So, this is a misprint then?

    Attached Files:

  5. Mar 21, 2017 #4
    No, the relaevant equation, equivalent to the one martinbn wrote, is the next one:6.85
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