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My question is about the field theoretic version of Noether's theorem. I am deeply troubled by one of the hypotheses of the theorem.

As it is the standard textbook for Lagrange mechanics, I'll follow Goldstein's account (starting p. 588 in the second edition of "Classical Mechanics").

I have no problem with condition 1 since I work in Minkowski space. I am completely okay with condition 2, which amounts to asking that the equations of motion be the same for two observers who use different systems of coordinates to describe the same spacetime and different functions to describe the same fields.

However, I can't make any sense of condition 3. I don't see what its physical meaning can be. I haven't seen it explained convincingly anywhere, and can't seem to figure it out for myself.

For those who don't have any access to Goldstein's book but feel they might be able to help, condition 3 is the requirement that the action integrals be equal for the two aforementioned observers.

I hope someone has some fantastic insight on this! :-)

Vinsanity
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  • Well probabily is something among the line that, condition 3 ensures that the action is invariant under Lorentz transformations. – iiqof Dec 18 '13 at 16:05
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    My copy of the 3rd edition says below Condition 3 that [this condition] will be called the condition of $scale$-$invariance$. Does your copy not have this sentence? – Kyle Kanos Dec 18 '13 at 16:11
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    @KyleKanos Well... No. The name is instructive, but it does not really tell me why the equality of actions is imposed. Maybe it tells smarter people, but at this point it's just a name to me. – Vinsanity Dec 18 '13 at 16:21
  • Well I'm somewhat familiar with scale invariance in statistical field theory. I don't see how it solves the problem, though. It seems to me scale invariance is a symmetry of certain physical theories just like rotational or translational invariance. If a theory is scale invariant then one can apply Noether's theorem. I'm still highly unclear, however, on why it should be featured in the proof of the theorem. – Vinsanity Dec 18 '13 at 18:03
  • I seem not to be the only one wondering about this, as can be seen here:

    http://www.physicsforums.com/showthread.php?t=193613

    I would hardly say the matter is settled, though.

     – Vinsanity Dec 18 '13 at 18:06

1 Answers1

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The condition 3 is the main assumption that goes into Noether's (first) theorem. It states that the action functional is invariant under a (global, continuous, off-shell) symmetry transformation (of the fields and spacetime).

Referring to condition 3 as scale invariance, as Goldstein does, is non-standard terminology, and probably a bit confusing, since the pertinent symmetry transformation does not have to be a dilation.

Qmechanic
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  • Thanks for the answer. I think I've understood what the assumption is. I don't get why it is made. – Vinsanity Dec 19 '13 at 13:40
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    @Vinsanity : Without a symmetry assumption, there is no reason to conclude a conservation law. In plain English: There ain't no such thing as a free lunch. – Qmechanic Dec 19 '13 at 13:48
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    :-) I totally agree with your statement. This assumption is needed to prove the theorem. But, as much as I understand why we ask that equations of motion have the same form, I don't see what requiring the conservation of the action truly means. This defines a symmetry transformation, okay.

    Maybe that's just it, and that's what everything points to. Maybe there's no better answer than "This is necessary to have a conservation law".

     – Vinsanity Dec 19 '13 at 16:10
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    Perhaps your real question is What is the physical meaning of the action functional? – Qmechanic Dec 19 '13 at 16:38
  • It's a pretty close question to mine indeed. – Vinsanity Dec 20 '13 at 09:30
  • Noether's theorem shows that there is a conserved quantity for every physical symmetry. The action being invariant under a transformation is the definition of a symmetry transformation. You could just as easily state the theorem as "for every transformation preserving the action, there is a conserved quantity". Is that useful? – Jess Riedel Jan 30 '14 at 18:52
  • @JessRiedel yeah, but the question as I understand it is, why is it the action that's preserved by symmetry transformations in this formalism, as opposed to, say, the energy, or just the equations if motion. After all, it seems easy to imagine that the equations of motion could stay the same while the action changes. I think it's a good question! – N. Virgo Mar 19 '14 at 12:27