# Geodesic Deviation between Test Particles from Gravitational Wave

I'm having trouble understanding how Carroll (Spacetime and Geometry, p.296) explains the effect of a passing gravitational wave on test particles.

If we have two geodesics with tangents $\vec{U}$, $\vec{U'}$ that begin parallel and near each other, and $\vec{S}$ is a vector connecting one geodesic to another at equal affine parameter values, then the equation of geodesic deviation is: \begin{align*} \frac{D^2}{d\tau^2}S^\mu = R_{\ \ \nu\rho\sigma}^\mu U^\nu U^\rho S^\sigma. \tag{7.103} \end{align*} We work in the weak-field limit and the transverse-traceless gauge. If we assume our particles on the geodesics are moving slowly, then $$\vec{U} \approx (1,0,0,0),\tag{7.104}$$ so: \begin{align*} \frac{D^2}{d\tau^2}S^\mu = R_{\ \ 00\sigma}^\mu S^\sigma.\tag{*} \end{align*} Now the bit I don't understand is how Carroll is able to turn the double covariant derivative on the left into a simple double-derivative with respect to $t$: \begin{align*} \frac{\partial^2}{\partial t^2}S^\mu = R_{\ \ 00\sigma}^\mu S^\sigma.\tag{**} \end{align*} Carroll's reasoning is that "for our slowly moving particles we have $\tau = x^0 = t$ to lowest order", but I don't know what he means. I just don't understand why the Christoffel symbols vanish in the covariant derivatives. I have read several books about this. Some say the Christoffel symbols vanish because we work in a local inertial frame. But then why doesn't the Riemann tensor on the RHS also vanish?

• Did you try computing it explicitly and use $S^{\mu}=U^{\prime \mu}-U^{\mu}$.
– user7757
Commented Jun 8, 2014 at 17:26
• I would say, that the $\frac{D^2}{d \tau^2}=\frac{\partial d^2}{dt^2}$ and $\tau=t$ is the $\Gamma \Gamma$ term is neglected, while the Riemann curvature tensor in still non zero under this approximation.
– user7757
Commented Jun 8, 2014 at 18:23

As a sanity check, note first of all, that the two total $D$-derivatives on the lhs. of eq. (7.103) in Ref. 1 is what is causing the curvature on the rhs. in the first place, see e.g. Ref. 1 p. 146.

Carroll is on the rhs. eq. (7.107) changing notation for the two total $D$-derivatives to two $\partial$-derivatives, but they are still total derivatives.

In fluid-dynamical language, one may say that Carroll is going from an Eulerian to a Lagrangian picture.

He is considering linearized gravity, so the Riemann curvature tensor is proportional to $\epsilon$, and we can (to the order that we are calculating, namely to first order in $\epsilon$) interpret $S^{\sigma}$ on the rhs. as following the flow.

References:

1. Sean Carroll, Spacetime and Geometry: An Introduction to General Relativity, 2003; Chapter 7.

2. Sean Carroll, Lecture Notes on General Relativity, Chapter 6. The pdf file is available here.

3. R. Wald, GR, 1984; p. 81, eq. (4.4.40).