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This question already has an answer here:

I recently came across this theorem for the first time. As I understand it, what she showed was that conservation 'laws' are often simply an artifact of symmetry or invariance.

For example, the source I read said that conservation of momentum flows naturally if all the physics gives the same answer at all places - e.g., Newton's laws are the same here as they are there. It also said that conservation of energy derives from the laws of physics being invariant over time.

I sort of see the math in the proof, but don't quite get the intuition. Can anyone provide a brief explanation?

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marked as duplicate by Qmechanic classical-mechanics Jul 25 '18 at 3:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Possible duplicate of Can Noether's theorem be understood intuitively? $\endgroup$ – David Hammen Jul 24 '18 at 23:18
  • $\begingroup$ Just followed the link, and saw Lubos Motl's note at the end of his answer: "If your intuition doesn't find the comments intuitive enough, maybe you should train your intuition because your current intuition apparently misses the most important properties of time, space, angles, energy, momentum, and angular momentum. ;-)" LOL! $\endgroup$ – wcc Jul 24 '18 at 23:38
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    $\begingroup$ @IamAStudent Despite Lubos' unsympathetic phrasing there is a significant degree of truth in that. Some ideas are not obvious or intuitive unless you have internalized a significant body of theory. Nature does not care one way or the other if we are comfortable with the way things work and does not guarantee that things will agree with "common sense". $\endgroup$ – dmckee Jul 24 '18 at 23:52
  • $\begingroup$ Oddly I did not find the link above when I first searched, but it is the answer and the question I asked is, indeed, a duplicate. $\endgroup$ – eSurfsnake Jul 25 '18 at 18:52
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The essence of Noether's Theorem(s) is that if your system (with well defined generalized coordinates and energy) has a continuous symmetry then you are guaranteed to have a corresponding conserved quantity. By "continuous symmetry" I vaguely mean a real-valued, differentiable, linear transformation between the generalized coordinates.

Some examples:

  1. the energy functional of the system is conserved when time translation symmetry is respected.

  2. the linear momentum of the system is conserved when spatial translation symmetry is respected.

  3. the angular momentum of the system is conserved when angular translation symmetry is respected.

By "is respected" I mean that the dynamics of the system are unaffected by the respective transformation.

Edit: I did not understand the questioner was asking for intuition behind the $\bf{proof}$ of Noether's theorem. Of course, this depends on how the theorem is formally stated and on how one goes about proving that statement (there are numerous ways). But a simple proof is provided here beginning at the bottom of page 1. The intuition for this proof is similar to what is stated above: begin with the general statement of variation, $\delta L = 0$, and find a symmetry/transformation that depends on some parameter. Now apply the transformation to your variation of $L$, and try to $\it{find}$ a conserved quantity. The more abstract the proof becomes, the more abstract intuition is required, which I can provide a more detailed statement/proof of the theorem if the questioner desires.

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    $\begingroup$ This does not seem in any way to address the question, which asks for the intuition behind the proof. $\endgroup$ – WillO Jul 24 '18 at 22:48
  • $\begingroup$ @WillO Thank you, I was not sure if the questioner wanted intuition behind the principle or intuition behind the proof, but I edited my answer. $\endgroup$ – N. Steinle Jul 25 '18 at 5:11
  • $\begingroup$ Thanks for not taking (or at least not expressing) offense. It did (and still does) seem to me that it was clear what the OP was asking for, but I understand that a reasonable person might have read this differently. $\endgroup$ – WillO Jul 25 '18 at 6:13

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