Undergrad Geodesics with arbitrary parametrization

[wrobel]

Let $x=(x^1,\ldots,x^m)$ be local coordinates in a manifold $M$; and let $\{\Gamma^i_{jk}(x)\}$ be a connection. Assume that we have a curve $x=x(t),\quad \dot x\ne 0$. Is this curve geodesic or not?
My guess is that the answer is "yes" iff for all $k,n$ the function $x(t)$ satisfies the following system
$$\dot x^k(\ddot x^n+\Gamma^n_{rj}\dot x^r\dot x^j)=\dot x^n(\ddot x^k+\Gamma^k_{rj}\dot x^r\dot x^j).$$
In my taste this system looks strange. Or not?

 

[martinbn]

Let $x=(x^1,\ldots,x^m)$ be local coordinates in a manifold $M$; and let $\{\Gamma^i_{jk}(x)\}$ be a connection. Assume that we have a curve $x=x(t),\quad \dot x\ne 0$. Is this curve geodesic or not?
My guess is that the answer is "yes" iff for all $k,n$ the function $x(t)$ satisfies the following system
$$\dot x^k(\ddot x^n+\Gamma^n_{rj}\dot x^r\dot x^j)=\dot x^n(\ddot x^k+\Gamma^k_{rj}\dot x^r\dot x^j).$$
In my taste this system looks strange. Or not?

Well, when it comes to your taste only you can say if it is strange or not. It is not strange of you notice the following: a curve is geodesic if the acceleration is proportional to the velocity i.e. $a^i=\lambda v^i$. Equivalently (eliminating lambda) you have $v^ja^i=v^ia^j$, which is your "strange" formula.

 

[wrobel]

Ok, thanks. I employed the following definition of geodesic: the geodesic is a curve that can be parametrized such that $x=x(s),$
$$x^i_{ss}+\Gamma^i_{kj} x^k_s x^j_s=0.$$ Here $\Gamma$ is not obliged to be generated by a Riemann metric.
I suspect that your remark about proportionality of the acceleration and the velocity is not an independent fact but follows from this definition by the same argument as I used. Anyway that is good that the formula from #1 does not contradict to your intuition. Thus it must be correct.

 

[martinbn]

Yes, they are equivalent definitions. If you have the proportionality, you can reparametrize to get that the acceleration is zero.