# Integrating the deviation vector around in a loop?

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 $$s$$:

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

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

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 $$\delta \xi^\alpha$$.