Friday, 22 April 2016

general relativity - 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?




No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...