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Consider a $1$-parameter family of timelike geodesics $x_s(\tau)$, where $s$ labels each geodesic in the family whilst $\tau$ is an affine parameter along each $x(\tau)$. Then the vector field $\xi≡\partial/\partial s$ is tangent to curves of constant $\tau$ and is interpreted as a deviation vector between neighboring geodesics. In 't Hooft's GR notes (page $23$) he introduces the the deviation $ \delta \xi^\alpha$:

$$ \delta \xi^\alpha = \oint d \tau \frac{d \xi^\alpha(x(\tau))}{ d \tau} = \frac{1}{2}(\oint x^\beta \frac{d x^\lambda}{d \tau} d \tau) R^\alpha_{\beta \kappa \lambda} \xi_\kappa $$

where $R^\alpha_{\beta \kappa \lambda}$ is the Riemann tensor. Now, I ask the same question but going around a loop of $s$:

$$ \tilde \delta \xi^\alpha = \oint d s \frac{d \xi^\alpha(x(\tau))}{ d s} = ?$$

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    $\begingroup$ Intuitively I think the answer is $0$ $\endgroup$ Nov 1, 2022 at 15:07

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Actually, you can write $x_s(\tau)$ as $x(\tau,s)$, and you will find that $s$ and $\tau$ is symmetric. So it is just $$\xi^\alpha :=\partial/\partial_\tau$$ and calculate the predefined $\delta \xi^\alpha$.

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