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?


2 Answers 2


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.

  • $\begingroup$ 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, 2017 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. $\endgroup$ Mar 9, 2017 at 22:25
  • $\begingroup$ I've had a reply from Prof. Schutz on a similar matter. In my experience, authors really appreciate such contact. $\endgroup$ Mar 9, 2017 at 22:26

The symbol $\nabla_V$ is not the covariant derivative. The symbol for that would be $\nabla_\alpha$. In Eq. (6.1) on page 143, Schutz defines the symbol $\nabla_V$ (where V is the tangent vector to the curve $\phi$) acting on $\phi$ as

\begin{equation} \nabla_V\phi = V \cdot \tilde{d}\phi = \frac{dx^{\alpha}}{d\lambda} \frac{\partial\phi}{\partial x^{\alpha}} = \frac{d\phi}{d\lambda} \end{equation}

Here $\phi$ is a scalar, but in Eq. (6.84) on page 162, $\xi$ is a vector. Below I will derive the correct equation for $\nabla_V (W)$ for a vector $W = W^{\alpha}\overrightarrow{e}_{\alpha}$. I have looked for this equation in Schutz, but did not find it. My derivation is similar to his derivation of Eq. (5.50) on page 128. I start by defining $\tilde{d}W$

\begin{equation} \begin{split} (\tilde{d}W)_{\mu} & = \left(\frac{\partial W^{\alpha}}{\partial x^{\mu}}\right)\overrightarrow{e}_{\alpha} + W^{\alpha}\frac{\partial}{\partial x^{\mu}}\overrightarrow{e}_{\alpha} \\ & = \left(\frac{\partial W^{\alpha}}{\partial x^{\mu}}\right)\overrightarrow{e}_{\alpha} + W^{\alpha}\Gamma^{\nu}_{\alpha \mu}\overrightarrow{e}_{\nu} \\ & = \left(\frac{\partial W^{\alpha}}{\partial x^{\mu}} + W^{\beta}\Gamma^{\alpha}_{\beta \mu}\right)\overrightarrow{e}_{\alpha} \end{split} \end{equation}

where, on the 2nd line I have used Eq. (5.44) on page 127, and on the 3rd line I have relabeled some dummies. Finally, we have

\begin{equation} \begin{split} (\nabla_V(W))^{\alpha} & = (V \cdot \tilde{d}W)^{\alpha} = V^{\mu} \left(\frac{\partial W^{\alpha}}{\partial x^{\mu}} + W^{\beta}\Gamma^{\alpha}_{\beta \mu}\right) \\ & = \frac{dx^{\mu}}{d\lambda}\left(\frac{\partial W^{\alpha}}{\partial x^{\mu}} + W^{\beta}\Gamma^{\alpha}_{\beta \mu}\right) \\ & = \frac{dW^{\alpha}}{d{\lambda}} + V^{\mu} \Gamma^{\alpha}_{\beta \mu}W^{\beta} \end{split} \end{equation}

In our case, $V = (1, 0, 0, 0)$ and W = $\nabla_V\xi$, so we have

\begin{equation} (\nabla_V(\nabla_V\xi))^{\alpha} = \frac{d}{d{\lambda}}(\nabla_V\xi^{\alpha}) + \Gamma^{\alpha}_{\beta 0}(\nabla_V\xi^{\beta}) \end{equation}

I prefer my placement of $\alpha$ on the l.h.s. to his because I think of $\xi^{\alpha}$ as a scalar for a particular value of $\alpha$, but our meanings are the same.

On the 1st line of Eq. (6.85) on page 163, we see "+ 0" which is the 2nd term set to zero since $\Gamma$ is zero. We also see my equation for $\nabla_V(W)$ used on $\xi$ itself. Here the derivative of $\Gamma$ is not zero.


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