From Euler-Lagrange equation to non-affine geodesic equation 
*

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

*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?
 A: 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 :
$$\frac{d}{ds}\dot x^\gamma + \Gamma^{\gamma}_{\beta \mu} \dot x^\beta \dot x^\mu = 0$$
A: Concerning OP's 1st subquestion:

*

*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 $\lambda$.


*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.
