# Geodesic deviation in Schutz's book: a typo?

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

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

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.
• @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. – Michael Seifert Mar 9 '17 at 22:25