Euler-Lagrange Equation for Purely Kinetic Lagrangian I was reading the set of notes for my course in Classical Dynamics and I came across the following statement that I have no idea how to show, any help is much appreciated.
Consider a particle described by a set of generalised coordinates $q^a$ where $a = 1,2,3,\cdots,n$. Let $L$ be a purely kinetic Lagrangian, i.e. $L$ has no potential energy term. Then the most general or of $L$ is $L = \frac{1}{2}g_{ab}(q_c)\dot{q}^a\dot{q}^b$ where $g_{ab}$ is an invertible symmetric matrix function of $q$ that depends on the generalised coordinates. E.g. in the Cartesian case we have $g_{ab}(q_c) = \delta_{ab}$. 
Then it is a simple exercise to see that the Euler-Lagrange Equation for $L$ is $\ddot{q}^a + \Gamma^a_{bc}\dot{q}^b\dot{q}^c = 0$ where $\Gamma^a_{bc} = \frac{1}{2}g^{ad}(\frac{\partial g_{bd}}{\partial q^c}+\frac{\partial g_{cd}}{\partial q^b}-\frac{\partial g_{bc}}{\partial q^d})$. 
Unfortunately I don't see how we get the Euler-Lagrange equation from that definition of the Lagrangian. Any hints or solutions much appreciated, especially as I don't think I have quite got the hang of tensors yet.
 A: Here's my solution:
$$
L=\dfrac{1}{2}g_{bc}\dot{q_b}\dot{q_c}
$$
So we calculate
\begin{alignat}{2}
&\bullet\dfrac{\partial L}{\partial q_a}&&=\dfrac{1}{2}\dfrac{\partial g_{bc}}{\partial q_a}\dot{q_b}\dot{q_c} 
\\ 
&\bullet\dfrac{\partial L}{\partial \dot{q_a}}&&=\dfrac{1}{2}g_{bc}\Big(\dot{q_b}\delta_{ac}+\dot{q_c}\delta_{ab}\Big)=\dfrac{1}{2}\big(g_{ba}\dot{q_b}+g_{ac}\dot{q_c}\big)
\\ 
&\bullet\dfrac{d}{dt}\Big(\dfrac{\partial L}{\partial \dot{q_a}}\Big)&&=\dfrac{1}{2}g_{ba}\ddot{q_b}+\dfrac{1}{2}g_{ac}\ddot{q_c}+\dfrac{1}{2}\Big(\dfrac{\partial g_{ba}}{\partial q_c}\dot{q_b}\dot{q_c}+\dfrac{\partial g_{ac}}{\partial q_b}\dot{q_b}\dot{q_c}\Big)
\\
&&&=g_{ab}\ddot{q_b}+\dfrac{1}{2}\Big(\dfrac{\partial g_{ba}}{\partial q_c}+\dfrac{\partial g_{ac}}{\partial q_b}\Big)\dot{q_b}\dot{q_c}.
\end{alignat}
So by the Euler Lagrange equations, we have 
\begin{align}
0&=\dfrac{d}{dt}\Big(\dfrac{\partial L}{\partial \dot{q_a}}\Big)-\dfrac{\partial L}{\partial q_a}
=g_{ab}\ddot{q_b}+\dfrac{1}{2}\Big(\dfrac{\partial g_{ba}}{\partial q_c}+\dfrac{\partial g_{ac}}{\partial q_b}\Big)\dot{q_b}\dot{q_c}-\dfrac{1}{2}\dfrac{\partial g_{bc}}{\partial q_a}\dot{q_b}\dot{q_c}
\\
&=g_{ab}\ddot{q_b}+\dfrac{1}{2}\Big(\dfrac{\partial g_{ba}}{\partial q_c}+\dfrac{\partial g_{ac}}{\partial q_b}-\dfrac{\partial g_{bc}}{\partial q_a}\Big)\dot{q_b}\dot{q_c}.
\end{align}
So we multiply through by $g^{-1}_{da}$ using the fact that $g^{-1}_{da}g_{ab}=\delta_{db}$ to find
\begin{align}
0&=g^{-1}_{da}g_{ab}\ddot{q_b}+g^{-1}_{da}\dfrac{1}{2}\Big(\dfrac{\partial g_{ba}}{\partial q_c}+\dfrac{\partial g_{ac}}{\partial q_b}-\dfrac{\partial g_{bc}}{\partial q_a}\Big)\dot{q_b}\dot{q_c}
\\&=\delta_{db}\ddot{q_b}+\Bigg[g^{-1}_{da}\dfrac{1}{2}\Big(\dfrac{\partial g_{ba}}{\partial q_c}+\dfrac{\partial g_{ac}}{\partial q_b}-\dfrac{\partial g_{bc}}{\partial q_a}\Big)\Bigg]\dot{q_b}\dot{q_c}
\end{align}
Therefore after relabeling indices (namely swapping $a$ and $d$), we have the answer
$$
0=\ddot{q_a}+\Gamma_{bc}^a\dot{q_b}\dot{q_c}
$$
where 
$$
\Gamma_{bc}^a=\dfrac{1}{2}g^{-1}_{ad}\Big(\dfrac{\partial g_{bd}}{\partial q_c}+\dfrac{\partial g_{dc}}{\partial q_b}-\dfrac{\partial g_{bc}}{\partial q_d}\Big).
$$
A: Euler-Lagrange equations are obtained as a result of extremizing a general Lagrangian $\mathcal{L}(\{q_{i}(t)\},\{\dot{q}_{i}(t),t\})$ and are given as :
$$\frac{d}{d t}\frac{\partial \mathcal{L}}{\partial \dot{q}_{i}}=\frac{\partial L}{\partial q_{i}}.$$
Applying this to the question in OP's Lagrangian $\mathcal{L}(\{q_{i}(t)\},\{\dot{q}_{i}(t),t\})=\frac{1}{2}\sum_{ij}^{} g_{ij}(\{q_{i}\})\dot{q}_{i} \dot{q}_{j}$ gives 
$\frac{d}{d t}\frac{\partial \mathcal{L}}{\partial \dot{q}_{i}}=0$ which is : 
$$\frac{d}{dt}[\frac{1}{2}\sum_{j}^{} [g_{ij}(\{q_{i}\})+g_{ji}(\{q_{i}\})]\dot{q}_{j}]=0$$
$$ \Rightarrow $$
$$ \sum_{j}^{} [g_{ij}(\{q_{i}\})+g_{ji}(\{q_{i}\})]\ddot{q}_{j}+\sum_{jk}^{}\frac{\partial [g_{ij}(\{q_{i}\})+g_{ji}(\{q_{i}\})]}{\partial q_{k}} \dot{q}_{j}\dot{q}_{k}=0$$
since metric tensor is a symmetric tensor :
$$ \sum_{j}^{} [g_{ij}(\{q_{i}\})]\ddot{q}_{j}+\sum_{jk}^{}\frac{\partial [g_{ij}(\{q_{i}\})]}{\partial q_{k}} \dot{q}_{j}\dot{q}_{k}=0$$
Now multiply both sides by elements $g(\{q_{i}\})_{}^{ki}$ of inverse of metric tensor  $g_{ij}(\{q_{i}\})$ , to get :
$$ \ddot{q}_{k}+\sum_{ijk}^{}g(\{q_{i}\})_{}^{ki}\frac{\partial [g_{ij}(\{q_{i}\})]}{\partial q_{k}} \dot{q}_{j}\dot{q}_{k}=0.$$
This is what i could reach so far (lacking my knowledge of tensor analysis), can some one simplify it.
