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This is similar to this question: Is the converse of Noether's first theorem true: Every conservation law has a symmetry?. However, the answer given there is very technical and general. I am only interested in the specific case of energy conservation (mostly because dark energy seems to break energy conservation / time invariance).

Qmechanic
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gardenhead
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    How are you defining "energy" other than as "the charge of time translation"? If we define energy as the charge of time translation, then it is a tautology that its conservation implies time invariance. Also, "dark energy" suggests you are thinking about cosmology, which as a general relativistic theory is far from ordinary applications of Noether's theorem in classical Lagrangian mechanics. – ACuriousMind Jun 05 '20 at 15:04
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    @ACuriousMind I was defining energy in as the sum of the kinetic and potential energies of the system. To be more precise, the system consists of $N$ particles each with kinetic energy $\frac12 m_i v_i^2$ and pairwise potential energies $V(q_i, q_j), i \neq j$. Is this an unsatisfactory definition of energy? – gardenhead Jun 06 '20 at 04:48
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    Well, that definition doesn't work for field theories (EM fields can store energy, too!) and certainly excludes the dark energy term, so I'm not sure how it is relevant to your question – ACuriousMind Jun 06 '20 at 09:00
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    Assuming you wan t to know that is the converse true, Goldstein in his book classical mechanics say that the converse is not true, he doesn't explicitly proves his statement but he gives a counter example: korteweg de vries equations – Soham Patil Oct 14 '21 at 04:25

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You may possibly want this explained with regards to Noether's theorem or something similar, but the answer is yes if we're dealing with quantum mechanics. In quantum mechanics the time-translation operator is given as

\begin{equation} \hat{T}(t) = \exp{(-i\hat{H}t/\hbar)} \end{equation}

such that $T(t_{0})|\psi(t)\rangle = |\psi(t + t_{0})\rangle$.

In the Heisenberg picture we know

\begin{equation} \frac{\text{d}\hat{A}(t)}{\text{d}t} = \frac{i}{\hbar}[\hat{H}, \hat{A}] \end{equation}

Energy conservation imples $\frac{\text{d}\hat{H}}{\text{d}t} = 0$ which is easily shown by the above equation:

\begin{equation} \frac{\text{d}\hat{H}}{\text{d}t} = [\hat{H}, \hat{H}] = 0 \end{equation}

We can rephrase this in terms of the time-translation operator as

\begin{equation} \begin{split} [e^{i\hat{H}t/\hbar}, \hat{H}] &= \\ [\hat{T}(t), \hat{H}] &= 0 \end{split} \end{equation}

We can start with this final equation and work through the above derivation backwards to show that time-translational symmetry does indeed imply energy conservation.

Niall
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It is actually the other way around. Time translation symmetry refers to Energy conservation. We define the Hamiltonian as $$\normalsize {H} = \Large{\Sigma_i}\normalsize{p_i\overset{.}{q_i} - L} $$ This says that the Hamiltonian in other words the energy is conserved when the Lagrangian has no explicit time dependance. i.e. $$\frac{dH}{dt}=\frac{\partial L}{\partial t}$$

This means as long as the laws of motion are time translation invariant, the energy of the system in consideration is conserved.

The converse is true as well. As you can see the equations say that

$$\frac{dH}{dt}=\frac{\partial L}{\partial t}$$

Which means that if energy is conserved it means that the Lagrangian has no explicit time dependance. Now even though it might seem in certain systems that the energy is not conserved, we must remember that the system is not necessarily isolated, so when we see that the energy is not conserved it just means that the flow of energy is from the surroundings

SK Dash
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    I think the OP is aware of Noether's theorem and is asking if the converse is also true (for energy/time specifically). – Rococo Jun 05 '20 at 17:08