# 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.

• Step one would be to apply the Euler-Lagrange equations to this lagrangian. – Triatticus Oct 20 '17 at 19:01
• Right, So how do I do that? Which index am I taking the partial derivatives w.r.t $q$ and $\dot{q}$ with? – Abdul Hadi Khan Oct 20 '17 at 19:04
• I see what you mean now, is the set of Euler lagrange equations defined in wherever you're using it? – Triatticus Oct 20 '17 at 19:13
• I am not sure, I think it is the usual definition as it was proved that the EL equations take the same form in any coordinate system. You can see the notes at page 27 of damtp.cam.ac.uk/user/tong/dynamics/two.pdf – Abdul Hadi Khan Oct 20 '17 at 19:23

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

$$\Gamma_{bc}^a=\dfrac{1}{2}g^{-1}_{ad}\Big(\dfrac{\partial g_{bd}}{\partial q_c}+\dfrac{\partial g_{dc}}{\partial q_b}-\dfrac{\partial g_{bc}}{\partial q_d}\Big).$$

• Seems to be correct, although giving the indices of the tensors as superscripts and subscripts may have been easier to follow. – Abdul Hadi Khan Oct 22 '17 at 16:42

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.