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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.)

Qmechanic
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Michael
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  • Possible duplicates: http://physics.stackexchange.com/q/91348/2451 and links therein. – Qmechanic Nov 29 '15 at 21:22
  • Can you add more details? You can't always choose a coordinate system where the geodesic equation is trivial everywhere. –  Nov 29 '15 at 21:30

1 Answers1

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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.

  • 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. – Valter Moretti Nov 30 '15 at 08:52
  • @ValterMoretti Are you referring to Fermi coordinates? – Ryan Unger Dec 01 '15 at 01:46
  • 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... – Valter Moretti Dec 01 '15 at 06:47