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According to Noether's theorem, systems that are not time-symmetric have $\frac{\mathrm{d}E}{\mathrm{d}t}\neq0$. I have essentially two questions, then:

  1. Are there any real systems (discovered or thought to perhaps exist) such that $\frac{\mathrm{d}E}{\mathrm{d}t}>0$?
  1. Regardless of whether they exist or not, what kind of behavior would time have in said systems?
agaminon
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1 Answers1

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Noether's theorem implies that if $L$ does not depend on time, then $H$ (Hamiltonian) is conserved. But $H$ is not always energy in the physics sense. See

When is the Hamiltonian of a system not equal to its total energy?

Any system can be described by Lagrangian/Hamiltonian that is time-dependent: the Hamiltonian $H = H_0 +kt$ has the same equations of motion as the Hamiltonian $H_0$.

So whether $dH/dt$ is positive or zero can sometimes be controlled by adding a suitable term in the Lagrangian, without changing the behavior of the system in any way. So

  1. $dH/dt>0$ can be in any system, even system that conserves energy, by choosing suitable Hamiltonian;
  2. time is not influenced by position or time in classical mechanics, time is always an independent parameter.
hft
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Ján Lalinský
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  • Oh, I didn't remember that the Noether current for time symmetry was not just energy but the Hamiltonian. In that case, is there any condition/theorem that determines when total energy (let's just limit ourselves to point particles and say T+V=E) is conserved, and not some other quantity that might or might not be equal to E? – agaminon Apr 27 '22 at 18:52
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    Yes. If all forces are conservative (gradient of the same time-independent function of coordinates), then total energy $T+V$ is conserved. – Ján Lalinský Apr 27 '22 at 19:26