# Why is it that in general geodesics are paths of stationary character

1. Mar 20, 2006

### Thrice

Well since the denizens of the relativity forum don't like me, I thought I might ask here see if I get better replies.

1) Why is it that in general geodesics are paths of "stationary character" rather than minimum?

2) http://img366.imageshack.us/img366/3280/math30016nx.jpg [Broken]

I can't quite follow equation 11.16. Specifically how they differentiate dxm/ds in the denominator.
Shouldn't be necessary, but for reference, the following page is http://img353.imageshack.us/img353/6488/math30024mk.jpg" [Broken].

Last edited by a moderator: May 2, 2017
2. Mar 20, 2006

### nrqed

They have been mean?

They impose that it's an extremum (functional derivative is zero) so it could be either a min or a max.

They do not differentiate dxm/ds..they differentiate with respect to dxm/ds. You must treat the *entire* combination dxm/ds as your variable and differentiate with respect to it (So, calling the variable x, L is essentially ${\sqrt{ g_{kn} x^k x^n}}$).

Pat

Last edited by a moderator: May 2, 2017
3. Mar 20, 2006

### Thrice

No they weren't mean. They just gave me the silent treatment.

Thanks. It's a lot clearer now.