Euler-Lagrange equations using tensor calculus

Question

Let $B_{\mu\nu}(\vec{x})$ be a symmetric and invertible tensor and let $\mathcal{L}$ be the following scalar function:

$$\mathcal{L}=-\left(\frac{dT}{d\lambda}\right)^{-1}B_{\mu\nu}(\vec{x})\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$$

Where the parameter $T$ is defined as:

$$T=\int dT,\quad dT^2=B_{\mu\nu}(\vec{x})dx^\mu dx^\nu$$

Note that Einstein's Summation Convention is used - both $\mu,\nu$ are summed from $0$ up to some natural number $n\in\mathbb{N}$. I am required to show that the following $n$ equations (Euler-Lagrange equations):

$$\frac{\partial\mathcal{L}}{\partial{x^\sigma}}=\frac{d}{d\lambda}\frac{\partial\mathcal{L}}{\partial\dot{x}^\sigma},\quad\quad \dot{x}^\sigma\triangleq \frac{dx^\sigma}{d\lambda},\sigma=0,1,2,...,n$$

Can be translated to the following $n$ equations:

$$\frac{d^2x^\alpha}{dT^2}=\Omega_{\ \ \mu\nu}^{\alpha}\frac{dx^\mu}{dT}\frac{dx^\nu}{dT}$$

Where $\Omega$ depends on the tensor $B$ but does not depend on $\lambda$.

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Fortunately, I am not frightened of the indices. However, I have no idea how to handle $dT/d\lambda$. For instance - does it depend on $x$? This is crucial in order to find the left hand side of the Euler-Lagrange equations. Additionally, even if I knew the answer to that question - how can I express that derivative? As much as it hurt, I tried to do the following:

$$\left(\frac{dT}{d\lambda}\right)^{-1}=\left[\left(\frac{dT}{d\lambda}\right)^2\right]^{-1/2}=\left[B_{\mu\nu}(\vec{x})\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}\right]^{-1/2}$$

But I have no idea if this is correct. The final answer should be:

$$\Omega_{\ \ \mu\nu}^{\alpha}=\frac 12(B^{-1})^{\alpha\sigma}\left(2\frac{\partial B_{\mu\nu}}{\partial x^\sigma}-\frac{\partial B_{\sigma\nu}}{\partial x^\mu}-\frac{\partial B_{\mu\sigma}}{\partial x^\nu}\right)$$

No matter what I tried (and I tried a lot of things) I couldn't get to that answer.

Thank you very much!

Answer 1

1. Define a Lagrangian $${\cal L}_0~:=~ B_{\mu\nu}(x)\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}~=:~\left(\frac{dT}{d\lambda}\right)^2,\tag{A}$$ where $B_{\mu\nu}$ are components of a pseudo-Riemannian metric tensor with Levi-Civita Christoffel symbols $-\Omega^{\lambda}_{\mu\nu}$, cf. above comment by user peek-a-boo. We will assume that the velocity vector of a virtual path has non-negative norm-square.

2. OP's Lagrangian becomes minus the square root: $${\cal L}~:=~-\left(\frac{dT}{d\lambda}\right)^{-1}{\cal L}_0~=~-\frac{dT}{d\lambda}~=~-\sqrt{{\cal L}_0}.\tag{B}$$ The corresponding functional $$T~=~-\int \! d\lambda~{\cal L}\tag{C}$$ is just the arc-length. Hence the stationary paths are all geodesics (in the sense that they satisfy the geodesic equation) and with arbitrary parametrization. See also this related Phys.SE post.