- #1

AriAstronomer

- 48

- 1

This can be proved with the geodesic equation:

[tex] \ddot{x^a} + \Gamma^a_{bc} \dot{x^b} \dot{x^c}=0 [/tex]

Locally we can find a coordinate system such that [tex] \Gamma =0 [/tex], and thus:

[tex] \ddot{x^a} =0 [/tex]

So along a geodesic at some point P, acceleration = 0 and there is constant velocity.

But geodesics are the paths that freely falling objects will take, and freely falling objects accelerate according to Newton's law F=ma. But from the geodesic equation we get that a=0 and thus F=0 ... what gives?

Also, a geodesic always a local path (extending off a point), right?? I.e. a freely falling rock dropped off a cliff will traverse many different geodesics on its way to Earth vs. a single geodesic on its way to Earth?

Thanks,

Ari