[Physics] Noether Theorem and Energy conservation in classical mechanics

Question

I have a problem deriving the conservation of energy from time translation invariance. The invariance of the Lagrangian under infinitesimal time displacements $t \rightarrow t' = t + \epsilon$ can be written as
\begin{equation} \delta L = L\left( q(t),\frac{dq(t)}{dt},t\right) – L\left( q(t+ \epsilon),\frac{dq(t+ \epsilon)}{dt},t+\epsilon \right) = 0.
\end{equation}
Using Taylor series, keeping only first order terms this gives
\begin{equation}\rightarrow \delta L =- \frac{\partial L }{\partial q} \frac{\partial q}{\partial t} \epsilon- \frac{\partial L }{\partial \dot{q}} \frac{\partial \dot{q}}{\partial t} \epsilon – \frac{\partial L }{\partial t} \epsilon = 0.
\end{equation}
Using the Euler-Lagrange equation and assuming that the Lagrangian does not depend explicitly on time we get
\begin{equation}\rightarrow \delta L =- \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \right) \frac{\partial q}{\partial t} \epsilon- \frac{\partial L }{\partial \dot{q}} \frac{\partial \dot{q}}{\partial t} \epsilon =0.
\end{equation}
Which we can write as
\begin{equation}\rightarrow \delta L = – \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \frac{\partial q}{\partial t} \right) \epsilon = – \frac{d}{d t} \left(p \frac{\partial q}{\partial t} \right) \epsilon = 0. \end{equation}
But unfortunatly this is not the Hamiltonian. This computation should yield
\begin{equation} \rightarrow \frac{d}{dt} \left( p \dot{q} – L \right) = 0. \end{equation}
But I can't find no reason why and how the the extra $-L$ should emerge. I can see that this term can be written at the place where it is written because we have $\delta L = – \frac{d L}{dt } \epsilon$ and therefore
\begin{equation} \rightarrow \delta L = – \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \frac{\partial q}{\partial t} \right) \epsilon = – \frac{d L}{dt } \epsilon.
\end{equation}
And then the desired equation would only say $0-0=0$. Any idea where i did a mistake would be much appreciated.

Answer 1

Reiterating pppqqq's answer, your error is right at the beginning where you set $\delta L = 0$. The Lagrangian is not a constant of motion, so this equation is fallacious.

Instead, you want

$\frac{dL}{dt} = \frac{\partial L}{\partial q} \dot{q} + \frac{\partial L}{\partial \dot{q}}\ddot{q}$

which assumes $\frac{\partial L}{\partial t} = 0$.

When you apply the Euler-Lagrange equation, you get

$\frac{dL}{dt} = \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}} \dot{q})$

which is just a short algebra step from showing that the Hamiltonian is conserved.

Your original derivation simply shows that if the Lagrangian is time-independent and if also it is a constant of motion, then $p \dot{q}$ is also a constant of motion.