Prove energy conservation using Noether's theorem I wonder how you prove that energy is conserved under a time translation using Noether's theorem. I've tried myself but without success. What I've come up with so far is that I start by inducing the following symmetry transformation
\begin{align}
\mathrm{h}_s:\ &q \rightarrow \mathrm{h}_s(q(t)) = q(t)\\
\hat{\mathrm{h}}_s:\ &\dot{q}(t) \rightarrow \hat{\mathrm{h}}_s(\dot{q}(t)) = \dot{q}(t)\\
&t \rightarrow t^\prime  = t+s\epsilon
\end{align}
$\mathrm{h}_s$ is a symmetry of the Lagrangian if:
$$
L(\mathrm{h}_s(q(t)),\hat{\mathrm{h}}_s(\dot{q}(t)),t^\prime) = L(x,\dot{x},t) + \frac{\textrm{d}}{\textrm{dt}}F_s
$$
Then I derivative with respect to $s$ and look for minimum.
$$
\frac{\partial}{\partial s}\Big(L(\mathrm{h}_s(q(t)),\hat{\mathrm{h}}_s(\dot{q}(t)),t^\prime) - \frac{\textrm{d}}{\textrm{dt}}F_s\Big)=0
$$
I find the derivative to be
$$
\frac{\partial L}{\partial \mathrm{h}_s(q(t))}\frac{\mathrm{h}_s(q(t))}{\partial s}+\frac{\partial L}{\partial \hat{\mathrm{h}}_s(\dot{q}(t))}\frac{\hat{\mathrm{h}}_s(\dot{q}(t))}{\partial s}+\frac{\partial L}{\partial t^\prime}\frac{\partial t^\prime}{\partial s}- \frac{\textrm{d}}{\textrm{dt}}\frac{\partial F_s}{\partial s}=0
$$
$$
\Rightarrow \frac{\partial L}{\partial t^\prime}\epsilon-\frac{\textrm{d}}{\textrm{dt}}\frac{\partial F_s}{\partial s} = 
\frac{\partial L}{\partial t}\frac{\mathrm{dt}}{\mathrm{dt^\prime}}\epsilon
-\frac{\textrm{d}}{\textrm{dt}}\frac{\partial F_s}{\partial s} 
= \frac{\partial L}{\partial t}\epsilon
-\frac{\textrm{d}}{\textrm{dt}}\frac{\partial F_s}{\partial s} = 0
$$
Here is the part where I get stuck. I don't know what to do next. I'm trying to find my Noether charge that corresponds to a time translation to be the Hamiltonian. Is there an easier or better way to do this? Please teach me, I'm dying to learn!
I found this book, Lanczos, The variational principles of mechanics, page 401, which explicit shows the energy conservation using Noether's theorem. Thou It seems that I can not follow  the step from equation 7 to 8. Can someone explain to me why the intregal looks the way it does? Have they taylor expanded the expression somehow?
 A: Comments to OP's post (v4):


*

*OP is trying to prove via Noether's theorem that no explicit time dependence of the Lagrangian leads to energy conservation.

*OP's transformation seems to be a pure horizontal infinitesimal time translation $$\tag{A} t^{\prime} - t ~=:~\delta t ~=~-\epsilon, \qquad \text{(horizontal variation)}$$
$$\tag{B} q^{\prime i}(t) - q^i(t)~=:~\delta_0 q^i ~=~0, \qquad \text{(no vertical variation)}$$
$$\tag{C} q^{\prime i}(t^{\prime}) - q^i(t)~=:~\delta q^i ~=~-\epsilon\dot{q}. \qquad \text{(full variation)}$$
It is explained in my Phys.SE answer here why this transformation (A)-(C) cannot be used to prove energy conservation.

*In eq. (1) on p. 401, the Ref. 1 is instead considering the following infinitesimal transformation
$$\tag{A'} t^{\prime} - t ~=:~\delta t ~=~-\epsilon, \qquad \text{(horizontal variation)}$$
$$\tag{B'} q^{\prime i}(t) - q^i(t)~=:~\delta_0 q^i ~=~\epsilon\dot{q}, \qquad \text{(vertical variation)}$$
$$\tag{C'} q^{\prime i}(t^{\prime}) - q^i(t)~=:~\delta q^i ~=~0. \qquad \text{(full variation)}$$
This is the same infinitesimal transformation as Section IV in my Phys.SE answer here, except for the fact that $\epsilon\equiv\alpha$ is allowed to be a function of time $t$. Therefore the variation of the action $S\equiv A$ is not necessarily zero, but of the form
$$ \tag{8} \delta S ~=~\int\! dt ~j \frac{d\epsilon}{dt}, $$
where the bare Noether current $j=h$ is the energy function, cf. eq. (8) on p. 402 in Ref. 1. The $t$-dependence in $\epsilon$ is tied to the Noether trick explained in this Phys.SE post. This in turn can be pieced together into a proof of the on-shell energy conservation
$$ \tag{9}\frac{dh}{dt}~\approx~0,$$
cf. eq. (9) on p. 402 in Ref. 1.
References:


*

*C. Lanczos, The variational principles of mechanics, 1970; Appendix II.

A: The easier way of doing this is to just consider a generic transformation, G, such that the canonical co-ordinates of the Hamiltonian are shifted as below:
$$ \delta p = \frac{\partial G}{\partial q} \delta \lambda$$ and $$ \delta q = - \frac{\partial G}{\partial p} \delta \lambda\,,$$
where $\lambda$ is the transformation parameter determining how much of the transformation you want to apply.
Now, consider a small change in the Hamiltonian, $H(p,q)$:
$$ \frac{\partial H}{\partial \lambda} = \frac{\partial{H}}{\partial q}\frac{\mathrm dq}{\mathrm d\lambda} + \frac{\partial{H}}{\partial p}\frac{\mathrm dp}{\mathrm d\lambda}$$
(^assume a time independent Hamiltonian).
Now using the transformation above , we see that:
$$\frac{\partial H}{\partial \lambda} = -\{H, G\} = -\frac{\mathrm dG}{\mathrm dt}$$
where the brackets used are Poisson brackets.
Thus, if the Hamiltonian is invariant under continuous transformation, then $G$ is a conserved charge.
If we let $G=H$, then it is easy to see that because $\{H,H\}=0$ then $\frac{\mathrm dH}{\mathrm dt}=0$.
Hope this helps :)
