Finding Christoffel Symbols using via variational method.

I'm trying to find the Christoffel Symbols for the Lorentz metric $${\rm d}s^2 = \cos(2\pi x)({\rm d}x^2-{\rm d}y^2) - 2\sin(2\pi x)\,{\rm d}x\,{\rm d}y$$by looking at the Euler-Lagrange equations for $$L(x,\dot{x},y,\dot{y}) = \cos(2\pi x)(\dot{x}^2-\dot{y}^2) - 2\sin(2\pi x)\,\dot{x}\,\dot{y}.$$I have already done my fair share of computations like this, but I must be making some algebraic mistake that I cannot find for the life of me. If we write \begin{align}\frac{\partial L}{\partial x} - \frac{{\rm d}}{{\rm d}t}&\left(\frac{\partial L}{\partial \dot{x}}\right) = -2\pi\sin(2\pi x)(\dot{x}^2-\dot{y}^2)-4\pi\cos(2\pi x)\dot{x}\dot{y} \\ &\qquad - \frac{{\rm d}}{{\rm d}t}\left(2\dot{x}\cos(2\pi x) - 2\dot{y}\sin(2\pi x)\right),\end{align}and we will have a term with $$\ddot{y}$$. This is a problem, since as far as I understand the geodesic equation corresponding to the $$x$$-coordinate should have the form $$\ddot{x} + \Gamma(\dot{x},\dot{y})=0,$$maybe after dividing by something. What is going on?

• I see it unavoidlable due to the existence of the cross term. The Euler-Lagrange equations aren't the geodesic ones. You need to rearrange the terms of the former to get the latter.. A practical example – Rafa Budría Nov 29 '18 at 18:57
• I understand and I don't, at the same time. I understand that the coefficient of $\ddot{x}$ need not be $1$, which requires rearranging. And while I agree that it seems unavoidable because of the $\dot{x}\dot{y}$ will always generate a $\ddot{y}$, I think that it should not appear there and I still don't see what is wrong. Or are you saying that that in $$g_{\ell k}\ddot{x}^k + g_{\ell k}\Gamma_{ij}^k \dot{x}^i\dot{x}^j = 0$$ we could still have $\ddot{x}_m$ terms for $m \neq \ell$ since the metric coefficient matrix is not diagonal? – Ivo Terek Nov 29 '18 at 19:23
• I don't see any reason E-L should decouple for you. Maybe you should "complete the square" and change coordinates in the metric first. It looks like it won't be pretty — but, having done it, it's not that bad, assuming you're working where $\cos(2\pi x)>0$. – Ted Shifrin Nov 29 '18 at 19:55

1. The infinitesimal variation of the Lagrangian $$L(x,\dot{x})~=~ g_{ij}(x)~ \dot{x}^i\dot{x}^j \tag{1}$$ is $$\frac{1}{2}\delta L~=~ -\left\{ \color{blue}{ g_{k\ell}\ddot{x}^{\ell}} +\color{red}{\Gamma_{k,ij} \dot{x}^i\dot{x}^j}\right\}\delta x^k +\frac{\mathrm{d}}{\mathrm{d}t}\left\{ \color{green}{ g_{k\ell}\dot{x}^{\ell} \delta x^k}\right\},\tag{2}$$ where we have introduced the lowered Levi-Civita Christoffel symbols $$\Gamma_{k,ij}~:=~g_{k\ell}\Gamma^{\ell}_{ij}. \tag{3}$$ Note that eq. (2) contains three different types of terms (displayed in different colors), which are uniquely characterized by how the $$t$$-derivatives are distributed.
2. Example. OP's Lagrangian reads $$L~=~c(\dot{x}^2-\dot{y}^2) - 2s\dot{x}\dot{y}, \qquad c~:=~\cos(2\pi x), \qquad s~:=~\sin(2\pi x) ,\tag{4}$$ corresponding to the metric $$\begin{pmatrix} g_{xx} & g_{xy} \cr g_{yx} & g_{yy} \end{pmatrix} ~=~\begin{pmatrix} c & -s \cr -s & -c \end{pmatrix}. \tag{5}$$ We calculate the infinitesimal variation: \begin{align}\frac{1}{2} \delta L ~=~&\left\{ \color{blue}{s\ddot{y}-c\ddot{x}} + \color{red}{\pi s (\dot{x}^2+\dot{y}^2)} \right\} \delta x + \left\{ \color{blue}{s\ddot{x}+c\ddot{y}} + \color{red}{2\pi (c\dot{x}^2-s\dot{x}\dot{y})} \right\} \delta y \cr & + \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \color{green}{(c\dot{x}-s\dot{y})\delta x -(s\dot{x}+c\dot{y})\delta y}\right\}, \end{align}\tag{6} which should be compared with the general formula (2).
From the red terms in eq. (6) we can read off the non-zero lowered Christoffel symbols $$\Gamma_{x,xx}~=~-\pi s ~=~\Gamma_{x,yy}~=~-\Gamma_{y,xy}, \qquad \Gamma_{y,xx}~=~-2\pi c,\tag{7}$$ cf. OP's title.