How to find a coordinate system whose geodesic equation does not have the "Christoffel symbol" term? (i.e. free particle - generalized Newton's second law.)
-
$\begingroup$ Possible duplicates: physics.stackexchange.com/q/91348/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Nov 29, 2015 at 21:22
-
$\begingroup$ Can you add more details? You can't always choose a coordinate system where the geodesic equation is trivial everywhere. $\endgroup$– user12029Commented Nov 29, 2015 at 21:30
1 Answer
Suppose you're in a coordinate system where the Christoffels don't vanish at some point.
To choose a coordinate system where the Christoffel symbols vanish at a given point $p$, you must apply a Christoffel symbol change of variables:
$$0={\bar\Gamma}^k{}_{ij} = \frac{\partial x^p}{\partial y^i}\, \frac{\partial x^q}{\partial y^j}\, \Gamma^r{}_{pq}\, \frac{\partial y^k}{\partial x^r} + \frac{\partial y^k}{\partial x^m}\, \frac{\partial^2 x^m}{\partial y^i \partial y^j}$$
For simplicity, maybe $\frac{\partial x^a}{\partial y^b}=\delta^a_b$ (evaluated at point $p$ and point p only, so this says nothing about the second derivatives), in which case the equation becomes:
$$0= \Gamma^k_{ij} + \frac{\partial^2 x^k}{\partial y^i \partial y^j}$$
if $x^k=y^k+C^k_{i j} y^i y^j$ and $p$ is the origin, this tells you immediately that if you choose $C^k_{ij}=-\Gamma^k_{ij}$ then you're in a frame where all of the Christoffels vanish.
-
$\begingroup$ It is possible to improve this result, obtaining $\Gamma_{ij}^k=0$ all along a timelike geodesic in a suitable coordinate system defined in a open tube centered on the geodesic. This is a better mathematical description of the equivalence principle than the coordinate system in the answer (mentioned in Landau-Lifsits' textbook). This is because here the "gravitational field" is (approximately) cancelled in a free falling coordinate system (the one constructed in the mentioned tube around the timelike geodesic) and not only around a fixed event. $\endgroup$ Commented Nov 30, 2015 at 8:52
-
$\begingroup$ @ValterMoretti Are you referring to Fermi coordinates? $\endgroup$ Commented Dec 1, 2015 at 1:46
-
$\begingroup$ I do not think so, I as far as I remember Fermi coordinates also include the notion of non-rotating frame, here unnecessary. I simply referred to so-called normal (or Gaussian) coordinates adapted to a submanifold, the graph of the geodesic in this case. The construction is elementary, it can be found, e.g. in O'Neill's textbook... $\endgroup$ Commented Dec 1, 2015 at 6:47