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I have come across two statements of Noether's theorem which are given below:

Statement 1

For every continuous symmetry of the Lagrangian there is a conserved quantity.

Statement 2

If the Lagrangian is invariant under the transformation $\vec q\rightarrow \vec q+\epsilon \vec K$ then $\frac{\partial L}{\partial \vec{\dot{q}}}\cdot \vec K$ is a conserved quantity.

Now it is clear that statement 1 implies statement 2 (just without saying what the conserved quantity is) but is it true that statement 2 implies statement 1. If not which is the "more correct" way to state Noethers theorem?

Note: I am using a symmetry of the Lagrangian to mean the same as invariant under the transformation. My main concern is whether statement 2 allows for e.g. transformations of time leaving the energy invariant etc.

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  • $\begingroup$ Whether statement 2 implies statement 1 depends on your definitions of "continuous symmetry of the Lagrangian" and the Lagrangian being "invariant under the transformation". If a "symmetry" is precisely a transformation of that form that leaves the Lagrangian invariant, then they're the same, otherwise not. But how are we to tell you which of the possible and subtly different notions of "symmetry" for a Lagrangian (or action functional) you're using here? $\endgroup$
    – ACuriousMind ♦
    May 22, 2017 at 14:28
  • $\begingroup$ @ACuriousMind I have edited the question :). $\endgroup$
    – Quantum spaghettification
    May 22, 2017 at 14:33

2 Answers 2

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Comments to the OP's post (v2):

  • It seems that OP is trying to say that the symmetry transformation in statement 2 is a special case of the symmetry transformation in statement 1. This is true.

  • On one hand, since statement 2 concerns a special case (which excludes e.g. time transformations), it does not imply the general case of statement 1.

  • On the other hand, since no formula is given in statement 1, it does not imply the formula in statement 2.

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You've got the logic backwards; statement 2 implies statement 1. I think you're confused by the difference between statements being true, and statements implying each other.

That is, assuming that "continuous symmetry of the Lagrangian" is defined as "a vector $\vec K$ such that the substitution $\vec q \mapsto \vec q + \epsilon~\vec K$ for arbitrarily small $\epsilon$ leaves $L(q,\dot q, t)$ invariant," the second statement is an explicit construction of an object which is asserted to exist in the first statement.

Explicit constructions are more specific than general statements of existence. The first statement does not specify how the construction comes about, therefore if we were doing something that was not quite Lagrangian mechanics but had significant overlap, you might still have a Lagrangian in your system and might even be able to prove that its symmetries are one-to-one with conserved quantities: but possibly, that construction would not generate them.

This circumstance of explicit constructions being more specific than general statements of existence reaches a very fine point in some of the discussions around the axiom of choice in pure mathematics. For example the axiom of choice implies that there exists a well-order of the real numbers, in other words, a way to sort the real numbers so that every subset contains a least element. (The obvious ordering is not a well ordering because of intervals like $(0, 1)$ which do not contain their endpoints and intervals like $(-\infty, 0)$ which go off to negative infinity.) Nobody knows what such a well-ordering would look like; there is no plausible construction of the well-order, and the best we can do is to prove that the existence of the well-order is not necessarily unconstructable... it gets to be a bit of a mess. In addition to the mathematicians who don't like the axiom of choice because of this, there are some even-more-reactionary mathematicians who in addition don't like the idea of proof by contradiction for similar reasons; to their mind you should not be able to reason from "not-(not P)" to "P". This is not as bad as it sounds because there are some places where things like logic look very familiar, for example in type theory, where there is no easy equivalent to the "not-" predicate and those that do exist probably cannot admit a function of type forall x. Negate (Negate x) -> x.

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  • $\begingroup$ Thanks, for your answer. Playing devil's advocate here, is $t\mapsto t+\epsilon t_0$ not generally considered a continuous symmetry which would not seem to fit in with the definition $\vec q \mapsto \vec q + \epsilon~\vec K$? $\endgroup$
    – Quantum spaghettification
    May 22, 2017 at 14:59
  • $\begingroup$ It actually can fit in with that definition; what you need is to smear out the Lagrangian over points in space, so that it becomes a Lagrangian density and your "action integral" is not just an integral with respect to time but becomes a full 4D volumetric integral: then it makes sense to use a 4-dimensional vector $\vec q$ that incorporates a time component and your 4-momentum contains a time-component of energy, so the time-translation corresponds to conservation of energy. $\endgroup$
    – CR Drost
    May 22, 2017 at 15:29

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