Different definitions of acceleration : Geodesic eq. and from intrinsic derivative?

Question

I was reading the deduction of the geodesic eq. and at the end the expression for acceleration was expressed by \begin{equation} \ddot{x}^{m}=-\Gamma_{ij}^{m}\dot{x}^{j}\dot{x}^{i} \end{equation} In another chapter that explains the equations of motion for a particle it reads \begin{equation} a^i=\frac{\delta}{\delta t}v^i =\frac{d}{dt}v^i+\Gamma_{jk}^i v^jv^k =\frac{d^2}{dt^2}x^i+\Gamma_{jk}^i\frac{d}{dt}x^j\frac{d}{dt}x^k \end{equation} Shouldn't both agree? I can't conciliate both as true becase one has the same term but as negative. Does the first one only works for geodesic paths? What considerations am I missing?

I need to resolve this conflict because later one I want to have the expression for the Jolt/jerk which reads \begin{equation} \dddot{q}\,^k=-\left(\frac{\partial}{\partial q^\gamma}\Gamma_{\alpha\beta}^k-2\Gamma_{m\beta}^k\Gamma_{\alpha\gamma}^m\right)\dot{q}^\alpha\dot{q}^\beta\dot{q}^\gamma \end{equation}

which am also having trouble to derive. I could only develop until the following:
Since \begin{equation} a^i=\frac{\delta}{\delta t}v^i =\frac{d}{dt}v^i+\Gamma_{jk}^i v^jv^k =\frac{d^2}{dt^2}x^i+\Gamma_{jk}^i\frac{d}{dt}x^j\frac{d}{dt}x^k \end{equation} I differentiate it to have \begin{equation} \begin{split} J^i&=\frac{d}{dt}a^i+v^l\Gamma_{lm}^i a^m \\ &=\frac{d}{dt}\left[\frac{d}{dt}v^i+\Gamma_{jk}^i v^jv^k \right]+v^l\Gamma_{lm}^i \left[\frac{d}{dt}v^m+\Gamma_{jk}^m v^jv^k \right] \\ &=\frac{d^2}{dt^2}v^i+\frac{d}{dt}\left[\Gamma_{jk}^i v^jv^k \right]+v^l\Gamma_{lm}^i \frac{d}{dt}v^m + v^l\Gamma_{lm}^iv^k\Gamma_{jk}^mv^j \end{split} \end{equation} which is not completely the same equation but I fell it's getting there but I believe both of my questions are intimately related.

Thanks in advance!

Answer 1

The equation $\ddot{x}^{m}=-\Gamma_{ij}^{m}\dot{x}^{j}\dot{x}^{i}$ can be written $0 = \ddot{x}^{m} + \Gamma_{ij}^{m} \dot{x}^{i} \dot{x}^{j}.$

Likewise $a^i=\frac{d^2}{dt^2}x^i+\Gamma_{jk}^i\frac{d}{dt}x^j\frac{d}{dt}x^k$ can be written $a^m = \ddot x^m + \Gamma_{ij}^m \dot x^i \dot x^j.$

Thus, for a geodesic we have $a^m = 0.$

Answer 2

The acceleration is indeed given by$$a=a^m e_m,\quad a^m=\ddot{x}^m+\Gamma_{ij}^m\dot{x}^i\dot{x}^j,$$as claimed in your second equality. So, the curve given by $x^1,\ldots,x^n$, is a geodesic if and only if we have$$\ddot{x}^m+\Gamma_{ij}^m\dot{x}^i\dot{x}^j=0,$$that is, if and only if your first equality holds. In other words, your first equality is the defining condition for geodesics, and all the facts mentioned in your post are coherent.