From Euler-Lagrange equation to non-affine geodesic equation

Question

1. I have some problems to demonstrate the non-affine geodesic equation from Euler-Lagrange's equations.

2. I start defining the square root Lagrangian $$L=\sqrt{ g_{ij}(x) \dot{x}^i \dot{x}^j},$$ but then I'm not able to find the Christoffel symbol's expression. Can anybody help me?

Answer 1

Let $L = \sqrt {g_{\mu\nu}(x) \dot x^\mu \dot x^\nu}$, where $\dot x^\mu = \frac {d}{ds} (x^\mu)$

Not that : $\frac {d}{ds} = (\frac {d}{ds} (x^\beta)) \partial_\beta = \dot x^\beta \partial_\beta$

Euler-Lagrange equations give :

$$\frac {1}{2L}(g_{\mu\nu,\alpha}\dot x^\mu \dot x^\nu) = \frac {1}{2L}2 \frac {d}{ds}(g_{\mu\alpha} \dot x^\mu)$$

(Here, the notation $g_{\mu\nu,\alpha}$ means $\frac {\partial g_{\mu\nu}}{\partial x^\alpha}$)

Assuming here that $L \neq0$ (we consider here massive particles):

$$g_{\mu\nu,\alpha}\dot x^\mu \dot x^\nu = 2[(\frac {d}{ds}g_{\mu\alpha})\dot x^\mu + g_{\mu\alpha} \frac{d}{ds}\dot x^\mu]$$

$$g_{\mu\nu,\alpha}\dot x^\mu \dot x^\nu = 2[ g_{\mu\alpha,\beta}\dot x^\beta \dot x^\mu + g_{\mu\alpha} \frac{d}{ds}\dot x^\mu]$$

Renaming the indices in the left side, we get : $$g_{\beta\mu,\alpha}\dot x^\beta \dot x^\mu = 2[ g_{\mu\alpha,\beta}\dot x^\beta \dot x^\mu + g_{\mu\alpha} \frac{d}{ds}\dot x^\mu]$$

Symmetrizing in the right side goes to :

$$g_{\beta\mu,\alpha}\dot x^\beta \dot x^\mu = (g_{\mu\alpha,\beta} + g_{\beta\alpha,\mu}) \dot x^\beta \dot x^\mu + 2 g_{\mu\alpha} \frac{d}{ds}\dot x^\mu$$

That is :

$$ g_{\mu\alpha} \frac{d}{ds}\dot x^\mu + \frac{1}{2} ( g_{\mu\alpha,\beta} + g_{\beta\alpha,\mu} - g_{\beta\mu,\alpha})\dot x^\beta \dot x^\mu = 0$$

Now, multiplying by $g^{\gamma \alpha}$ the two sides, we get:

$$\frac{d}{ds}\dot x^\gamma + \frac{1}{2} g^{\gamma \alpha}( g_{\mu\alpha,\beta} + g_{\beta\alpha,\mu} - g_{\beta\mu,\alpha})\dot x^\beta \dot x^\mu = 0$$

Which is nothing but :

$$\frac{d}{ds}\dot x^\gamma + \Gamma^{\gamma}_{\beta \mu} \dot x^\beta \dot x^\mu = 0$$

Answer 2

Concerning OP's 1st subquestion:

1. On one hand, the EL equations $$ \frac{\partial L_0}{\partial x^i}-\frac{d}{d\lambda}\frac{\partial L_0}{\partial \dot{x}^i}~\approx~0 \tag{1}$$ for a non-square root Lagrangian $$ L_0(x,\dot{x})~:=~ g_{ij}(x) \dot{x}^i \dot{x}^j~\geq~0\tag{2} $$ leads to the affine geodesic equation $$ \nabla_{\dot{\gamma}} \dot{\gamma}~=~0 , \tag{3}$$ whose solutions are affinely parametrized geodesic, i.e. the arc length $s=a\lambda +b$ is an affine function of the parameter $\lambda$.

2. On the other hand, the EL equations $$\begin{align} \frac{\partial \sqrt{L_0}}{\partial x^i}-\frac{d}{d\lambda}\frac{\partial \sqrt{L_0}}{\partial \dot{x}^i}~\approx~&0 \cr\cr \Updownarrow~& \cr\cr \frac{\partial L_0}{\partial x^i}-\frac{d}{d\lambda}\frac{\partial L_0}{\partial \dot{x}^i} ~\approx~&-g_{ij}(x) \dot{x}^j\frac{d \ln L_0}{d\lambda}\end{align} \tag{4}$$ for a square root Lagrangian $\sqrt{L_0}$ leads to the (non-affine) geodesic equation, whose solutions are arbitrarily parametrized geodesics.

See also this related Phys.SE post.