# 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?

• 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) May 28, 2016 at 18:29
• 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. May 29, 2016 at 8:35
• 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. May 29, 2016 at 9:49
• 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. May 29, 2016 at 10:29

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.

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 :)