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I am studying a bit of general relativity and differential geometry in Schutz's book "A first course in general relativity" and I don't understand an equation regarding the "geodesic deviation". At page 162, he writes Eq. (6.84)

$ \nabla_V \nabla \xi^{\alpha} = \nabla_V ( \nabla_V \xi^{\alpha}) = \frac{d}{d \lambda} (\nabla_V \xi^{\alpha} ) = \Gamma_{\beta 0}^{\alpha} (\nabla_V \xi^{\beta})$

where $\nabla$ denotes the covariant derivative and $\xi$ is the connecting vector between the two geodesics. What I fail to understand is the last equality. Schutz writes "we can use Eq. (6.48)", which is the definition of parallel transport of a vector $V$ along a curve $U$:

$ U^{\beta} V^{\alpha}_{; \beta} = 0 \iff \frac{d}{d \lambda} V = \nabla_{U} V =0$.

If I were to use that equation, I would simply write

$ \nabla_V \nabla \xi^{\alpha} = \nabla_V ( \nabla_V \xi^{\alpha}) = \frac{d}{d \lambda} (\nabla_V \xi^{\alpha} )$.

How can we retrieve a Christoffel symbol from the derivative?

On the other hand, I was also thinking about using the definition of covariate derivative, which reads as:

$ (\nabla V)_{\beta}^{\alpha} = (\nabla_{\beta} V)^{\alpha} = V_{; \beta}^{\alpha} = V_{,\beta}^{\alpha} + V^{\mu} \Gamma_{\mu \beta}^{\alpha}$.

Then, I would guess

$ \nabla_V \nabla \xi^{\alpha} = \nabla_V ( \nabla_V \xi^{\alpha}) = \frac{d}{d \lambda} (\nabla_V \xi^{\alpha} ) + \Gamma_{\beta 0}^{\alpha} (\nabla_V \xi^{\beta})$.

Either way, I am quite confused. Could you please help me?

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I am 99% sure this is an erratum in Schutz, and that the correct equation is your second version above: $$ \begin{align} \nabla_V \nabla_V \xi^\alpha &= \nabla_V (\nabla_V \xi^\alpha) \\ &= \frac{d}{d\lambda}(\nabla_V \xi^\alpha) + \Gamma^{\alpha} {}_{\beta 0} (\nabla_V \xi^\alpha). \end{align} $$ Note that in the unnumbered equation following Eq. (6.84), Schutz substitutes in $\Gamma^{\alpha} {}_{\beta 0} = 0$ in the second term of this expression, and substitutes in $\nabla_V \xi^\alpha = \frac{d}{d\lambda} \xi^\alpha + \Gamma^{\alpha} {}_{\beta 0} \xi^\beta$ in the first term. The equation also appears in this form in the first edition of the text; the error appears to have been introduced in the second edition.

That said, this does not appear in the list of known errata that were corrected in the second printing in 2011. You might want to contact Prof. Schutz and ask him whether this is an erratum; textbook authors don't always reply to such e-mails, but it can't hurt.

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  • $\begingroup$ Thanks a lot for your reply! At this point, though, how would you interpret the expression "we can use Eq. (6.48)"? Where is the need to use the definition of parallel transport? $\endgroup$ – Fred G. Mar 9 '17 at 22:05
  • $\begingroup$ @FredG.: Honestly, I'm not sure. The whole point of the geodesic deviation equation is that the separation vector $\xi^\alpha$ between nearby geodesics does not satisfy the parallel-transport equation. $\endgroup$ – Michael Seifert Mar 9 '17 at 22:25
  • $\begingroup$ I've had a reply from Prof. Schutz on a similar matter. In my experience, authors really appreciate such contact. $\endgroup$ – WetSavannaAnimal Mar 9 '17 at 22:26

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