- #1

Pencilvester

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- TL;DR Summary
- I’m having trouble with a statement in Poisson’s “A Relativist’s Toolkit” in regards to the Lie derivative along a geodesic congruence.

Hello PF, here’s the setup: we have a geodesic congruence (not necessarily hypersurface orthogonal), and two sets of coordinates. One set, ##x^\alpha##, is just any arbitrary set of coordinates. The other set, ##(\tau,y^a)##, is defined such that ##\tau## labels each hypersurface (and presumably also represents the proper time of a specific geodesic, ##\gamma##, to which all the hypersurfaces are orthogonal) and ##y^a## is assigned to label the geodesics themselves, with the basis vectors ##\partial_{y^a}## lying tangent to the hypersurfaces. Poisson then defines $$u^\alpha = \left({\frac{\partial x^\alpha}{\partial \tau}}\right)_{y^a}$$and$$e^\alpha_a = \left({\frac{\partial x^\alpha}{\partial y^a}}\right)_{\tau}$$and then says that ##\mathcal{L}_u e^\alpha_a =0##.

This is where I

This is where I

*was*having trouble, but I think I just realized what I was missing as I was typing, and I just want to make sure. I was thrown by the fact that he equates the basis vector ##\partial_\tau## with a vector ##\mathbf u##—a letter that is typically reserved for 4-velocity. But in general, ##\partial_\tau## at a point P will not equal the 4-velocity of the geodesic passing through P, yes? Does anyone know why he chose the letter u?