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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?

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    $\begingroup$ Your transformation is the wrong one for "time translation". Qmechanic explains here why and gives the correct derivation. (The other answers are also worth reading) $\endgroup$ – ACuriousMind May 28 '16 at 18:29
  • $\begingroup$ Thanks for the help, but there's one part in the derivation that I don't understand -> "The (bare) Noether current (multiplied with ϵ) becomes...". I can't seem to find where on wiki the bare Noether current is stated. $\endgroup$ – Turbotanten May 29 '16 at 8:35
  • $\begingroup$ It's not stated on Wiki as "bare" (because Wiki doesn't consider quasi-symmetries, i.e. those that only leave the Lagrangian invariant up to total derivative). The "bare" Noether current is the current if the transformation is a symmetry of the Lagrangian, while the "full" Noether current is the bare current + contributions from the boundary terms from the total derivative. $\endgroup$ – ACuriousMind May 29 '16 at 9:49
  • $\begingroup$ Okey, so in my case the $F_s$ term correspond to the contributions from the boundary terms and the rest is what you would call "bare" Noether current? I'm just very confused at the moment. $\endgroup$ – Turbotanten May 29 '16 at 10:29
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Comments to OP's post (v4):

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

  2. 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.

  3. 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:

  1. C. Lanczos, The variational principles of mechanics, 1970; Appendix II.
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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 :)

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