Deriving Finsler geodesic equations from the energy functional

Question

I'm struggling to derive the Finsler geodesic equations. The books I know either skip the computation or use the length functional directly. I want to use the energy. Let $(M,F)$ be a Finsler manifold and consider the energy functional $$E[\gamma] = \frac{1}{2}\int_I F^2_{\gamma(t)}(\dot{\gamma}(t))\,{\rm d}t\tag{1}$$evaluated along a (regular) curve $\gamma\colon I \to M$. We use tangent coordinates $(x^1,\ldots,x^n,v^1,\ldots, v^n)$ on $TM$ and write $g_{ij}(x,v)$ for the components of the fundamental tensor of $(M,F)$. We may take for granted (using Einstein's convention) that $$F^2_x(v) = g_{ij}(x,v)v^iv^j, \quad \frac{1}{2}\frac{\partial F^2}{\partial v^i}(x,v) = g_{ij}(x,v)v^j, \quad\frac{\partial g_{ij}}{\partial v^k}(x,v)v^k = 0.\tag{2} $$

Setting $L(x,v) = (1/2) F_x^2(v)$, and writing $(\gamma(t),\dot{\gamma}(t)) \sim (x(t),v(t))$, the Euler-Lagrange equations are $$0 = \frac{{\rm d}}{{\rm d}t}\left(\frac{\partial L}{\partial v^k}(x(t),v(t))\right) -\frac{\partial L}{\partial x^k}(x(t),v(t)),\quad k=1,\ldots, n=\dim(M).\tag{3}$$It's easy to see (omitting application points) that $$\frac{\partial L}{\partial x^k} = \frac{1}{2}\frac{\partial g_{ij}}{\partial x^k}\dot{x}^i\dot{x}^j\quad\mbox{and}\quad \frac{\partial L}{\partial v^k} = g_{ik}\dot{x}^i,\tag{4}$$so $$\frac{\rm d}{{\rm d}t}\left(\frac{\partial L}{\partial v^k}\right) = \frac{\partial g_{ik}}{\partial x^j}\dot{x}^j\dot{x}^i +{\color{red}{ \frac{\partial g_{ik}}{\partial v^j} \ddot{x}^j\dot{x}^i }}+ g_{ik}\ddot{x}^i\tag{5}$$ Problem: I cannot see for the life of me how to get rid of these $v^j$-derivatives indicated in red, even using the last relation in (2), as the indices simply don't match. I am surely missing something obvious. Once we know that this term does vanish, then (4) and (5) combine to give $$ g_{ik}\ddot{x}^i + \left(\frac{\partial g_{ik}}{\partial x^j} - \frac{1}{2}\frac{\partial g_{ij}}{\partial x^k}\right)\dot{x}^i\dot{x}^j =0\tag{6}$$as in the Wikipedia page.

Answer 1

Someone (not on this website) also pointed me to the Bao, Chern, Shen book, but namely, to Exercise 1.2.1 on page 11 and to relation (1.4.5) on page 23. Using the suggestive coordinate notation on the book, the exercise says that

(a) $y^iF_{y^i} = F$ (I already knew that)

(b) $y^iF_{y^iy^j} = 0$ (that one too)

(c) $y^iF_{y^iy^jy^k} = -F_{y^jy^k}$ (that was the missing piece for me! also a consequence of Euler's theorem)

while (1.4.5) says that $$y^i\frac{\partial g_{ij}}{\partial y^k} = y^j\frac{\partial g_{ij}}{\partial y^k} = y^k\frac{\partial g_{ij}}{\partial y^k} = 0.$$ In the notation of my original post, we just have to apply $v^i\partial_{v^j}$ to $g_{ik} = FF_{v^iv^k} + F_{v^i}F_{v^k}$ and use (c), and the conclusion follows easily.

Answer 2

OP's red term vanishes $$\frac{\partial g_{ik}}{\partial v^j} \ddot{x}^j\dot{x}^i ~\stackrel{\rm EOM}{\approx}~ v^i\frac{\partial g_{ik}}{\partial v^j} \dot{v}^j ~\stackrel{(C)}{=}~v^i\frac{\partial g_{jk}}{\partial v^i} \dot{v}^j ~\stackrel{(B)}{=}~0\tag{A}$$ because of the metric $g_{jk}$ has homogeneity weight 0: $$ v^i\frac{\partial g_{jk}}{\partial v^i} ~=~0 .\tag{B}$$ Eq. (B) is a consequence of the definition of the metric $$ g_{ij}~:=~\frac{1}{2}\frac{\partial (F^2)}{\partial v^i\partial v^j},\tag{C}$$ and that $F$ has homogeneity weight 1: $$v^i\frac{\partial F}{\partial v^i}~=~F ,\tag{D}$$ cf. the homogeneity property of the definition. [Eqs. (C) & (D) also imply the 1st equality in OP's eq. (2).]