I find it hard to understand that time-translation invariance necessarily implies conservation of energy. As I understand it, Noether's theorem says that there is an energy conservation because the laws of nature do not depend on time.

The following statement is logically true: "Because there is an energy conservation and the laws of nature do not change over time, it doesn't matter which point in time we choose, the energy content of the universe is always the same".

The statement: "Because the laws of nature do not change over time, it doesn't matter which point in time we choose, the energy content of the universe is always the same" is not necessarily true in my view.

From a philosophical perspective I can imagine a possible world where (a) time exists and (b) the laws of nature do not change over time, but there is no conservation of energy. Imagine a world where a black hole is constantly creating energy or a fundamental force that is not conservative. I do not see that energy conservation is a priori given, something that Noether's theorem implies.

Is it that Noether's theorem in relation to energy conservation is circular reasoning in some way? The theorem is based on the assumption that the laws of nature are energy conserving - when these laws don't change over time - there is an universal energy conservation. But hypothetically speaking - if the laws of nature were not energy conserving and they wouldn't change over time - there would be no universal energy conservation.

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    $\begingroup$ A world in which energy is not conserved would not have time translation invariant laws. Noether's theorem is as simple as that. (Note: How do you define energy if not as "that which is conserved by time translation?) $\endgroup$
    – ACuriousMind
    Jan 3, 2015 at 20:11
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    $\begingroup$ It's not clear what you're asking. Are you asking how conservation of energy follows from time translation invariance (NB time translation invariance not time invariance)? If so I don't know of an intuitive explanation. You need to go through the maths. $\endgroup$ Jan 3, 2015 at 20:17
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    $\begingroup$ Just to let you know that the Universe has a whole is not time translational invariant, and its energy is not at all conserved. In fact, its expansion is accelerating which means that extra energy is being put in. All these arguments are valid locally $\endgroup$
    – SuperCiocia
    Jan 3, 2015 at 20:41
  • $\begingroup$ The Universe is an open system, we don't know its borders. All the information that we get from the far galaxies tells us what happened to them billions of years ago. So, what we know of the laws of the universe? We know a couple of laws in our part of the universe. As a small example: it is not known whether the light velocity was always he same as in our time. $\endgroup$
    – Sofia
    Jan 3, 2015 at 20:48
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    $\begingroup$ @Hypnosifl: That's the inverse Lagrangian problem, and the answer is: No, not all laws (in the form of differential equations) must arise through Lagrangian mechanics. Noether's theorem needs the assumption that at least one of Lagrangian and Hamiltonian exists, I think. Yet, if you are going to assume that the fundamental laws of nature are not Hamiltonian/Lagrangian, you're very far off the mainstream physics path. $\endgroup$
    – ACuriousMind
    Jan 3, 2015 at 22:37

1 Answer 1


The derivation of conservation of energy from Noether's theorem is somewhat pointless. The problem is that in Lagrangian mechanics (in which Noether's theorem applies), the definition of the Lagrangian involves assuming the existence potential energy. This means, that by definition, we assume there are no dissipative forces nor there are "energy-gaining" loops.

As a counter-example, consider the following hypothetical universe, in which $\vec F = \vec C \times \vec x$, where $\vec C = \textrm{const}\;.$ Of course, a particle could gain some kinetic energy by just doing some loops with respect to origin of coordinates (if we would further restrict it's motion to a circle, the particle would come to the very same point after some time, just with a larger kinetic energy). However note that potential energy can not be defined, thus Lagrangian can't be too.

However, it seems very likely, that the fundamental law's of physics can be defined in terms of Lagrangian.


Actually, non-conservative forces could be simulated by varying the potential, however that would obviously conflict with the assumption that the law's of nature do not vary in time (because then the Lagrangian function does vary).

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    $\begingroup$ The answer here says that it's possible to have a non-conservative Lagrangian dealing with a combination of forces, some of which are conservative and so are associated with a potential function V, but others of which are not. Not sure what Noether's theorem would say about such a case, but there can be combinations of conservative and non-conservative forces that conserve energy, like the electric and magnetic forces in classical electromagnetism. $\endgroup$
    – Hypnosifl
    Jan 3, 2015 at 22:12
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    $\begingroup$ @Hypnosifl: Yes, but Noether's theorem assumes $L = T - U$, not some modified Lagrangian as in the answer in the link: $L = T - U + F$. $\endgroup$
    – kristjan
    Jan 3, 2015 at 22:18
  • $\begingroup$ @kristjan: "This means, that by definition, we assume there are no dissipative forces nor there are "energy-gaining" loops." This is the same conclusion that I made. Noether's theorem is based on the assumption that one of the qualities of the laws of nature is that they are energy-conserving. When these laws - which by defintion are energy conserving - do not change over time, then there is indeed an universal energy conservation. But you can't say that for all possible worlds in my view. $\endgroup$
    – Chris
    Jan 4, 2015 at 10:52
  • $\begingroup$ @Chris: Yes, indeed, as I said in my answer, it is just likely that everything can be formulated in terms of Lagrangians. However, of course, this is not proven (and quite probably can never be proven). I have to say that maybe I didn't understand your question totally (thus my answer might not answer directly your question), though I gave my best. $\endgroup$
    – kristjan
    Jan 4, 2015 at 16:44
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    $\begingroup$ @kristjian : Why do you say that Noether theorem assumes L = T - U ? I don't see where this assumption is made $\endgroup$
    – agemO
    Jan 11, 2015 at 12:39

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