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I don’t understand Noether’s theorem… there is nothing to prove?

If I understand Noether’s theorem correctly it says: if there is coordinate where the Lagrangian is invariant, then the conjugate momentum is conserved. However, this follows almost immediately from the Euler–Lagrange equation:

$$\frac{d}{dt}\frac{\partial L}{\partial\dot{q}}=\frac{\partial L}{\partial q}$$

If the Lagrangian is does not change in the direction of the coordinate $q$, we can describe this as:

$$\frac{\partial L}{\partial q}=0$$

According to the Euler–Lagrange equation, we then have;

$$\frac{d}{dt}\frac{\partial L}{\partial\dot{q}}=\frac{\partial L}{\partial q}=0$$

$$\frac{d}{dt}\frac{\partial L}{\partial\dot{q}}=0$$ $$\frac{d}{dt}p_q=0$$ $$p_q=\mathrm{constant}$$

In other words we have already shown that if the Lagrangian does not change in the coordinate $q$, the related momentum to $q$ which is $p_q$ is conserved. But isn't this then already Noether's theorem?

A problem here is if you choose a coordinate system that does not contain a coordinate basis vector that points in the direction where the Lagrangian is invariant. We could have chosen a wrong basis. However, you can just change the basis and the proof is similar.

Now imagine we have a coordinate system ${q_1,q_2,\ \ldots,\ q_i}$ where for every direction, the Lagrangian isn’t invariant. Then we have

$$\frac{\partial L}{\partial q_1}\neq0$$ $$\frac{\partial L}{\partial q_2}\neq0$$ $$\ldots$$

So there isn't any conserved conjugate momentum, or any conserved Noether’s charge. Now we are told that there is a certain direction ${\hat{q}}^\ast\ $ that the Lagrangian is invariant. However, our coordinate system ${q_1,q_2,\ \ldots,\ q_i}$ does not have a unit vector that points in that exact direction.

Then you can just perform a change of basis in such a way that we have ${q^\ast,q_2^\prime,q_3^\prime,..}$ and then perform the following again.

Then how to find the form of the Noether’s charge you can just project the total momentum vector in the first q system onto the direction of $\hat{q}$: $$Q=\vec{p}\cdot{\hat{q}}^\ast$$

$$Q=\begin{bmatrix} p_x \,\\ p_y \\ \ldots \end{bmatrix}^T \cdot {\hat{q}}^\ast $$

So my problem is basically that the conclusion of Noether's theorem really trivially falls out of setting the Euler-Lagrange equations equal to $0$. In the worst case you would have to do a coordinate transformation before you can do that. This suggests that I am missing something important, because this seems too easy for a proof.

I am probably misunderstanding something fundamental of the theorem. Where does my reasoning fail?

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    $\begingroup$ If there was nothing to prove, why so long a text full of implications? Also, Noether's theorem allows for Noether symmetries involving combination of time $t$ and coordinates $q_k$. arxiv.org/abs/1812.03682 $\endgroup$ Commented Aug 14, 2023 at 10:15
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    $\begingroup$ Noether's theorem allows for broader class of symmetries than just $\partial L/\partial q_k = 0$ (Qmechanic's 2nd point). $\endgroup$ Commented Aug 14, 2023 at 12:54
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    $\begingroup$ So if I understand correctly I am just looking at a special simple case and the importance of Noether's theorem lies in the fact that it also is true for broader symmetries that cannot be expressed in cyclic coordinates? $\endgroup$ Commented Aug 14, 2023 at 13:25
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    $\begingroup$ Yes yes yes yes. $\endgroup$ Commented Aug 14, 2023 at 13:35
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    $\begingroup$ One key property of Noether's theorem that makes it so useful is that it is constructive. If you know the the symmetry transformation (which can be complicated), then Noether's theorem tells you exactly what the associated conserved charge is -- not just that it exists. The general procedure that goes from an arbitrary symmetry transformation to the corresponding charge is meat of the proof. $\endgroup$
    – Andrew
    Commented Aug 16, 2023 at 0:46

2 Answers 2

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Comments to the post (v3):

  1. Noether's theorem is just one method to determine conservation laws. If you have another, that's totally fine.

  2. Not all quasi-symmetry transformations (which in principle can depend functionally on both generalized positions $q^j$, generalized velocities $\dot{q}^k$ and time $t$) can be reduced to the case of a cyclic/ignorable coordinate via a pertinent coordinate transformation, cf. OP's trick.

    Counterexamples:

    • Technically speaking, time $t$ is never a cyclic/ignorable coordinate, so OP's trick does not apply to the case of energy conservation. (Although it is a consequence of the Beltrami identity.)

    • The conservation of the Laplace-Runge-Lenz vector in the Lagrangian (as opposed to the Hamiltonian) formulation of the Kepler problem. In this case Noether's theorem uses a velocity-dependent quasi-symmetry transformation.

  3. Yes, Noether's theorem requires a proof, see e.g. the original 1918 paper.

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    $\begingroup$ Thank you for your comment. Where is my reasoning false? Could you help me understand this? I hope this is possible in undergraduate level physics $\endgroup$ Commented Aug 14, 2023 at 12:37
  • $\begingroup$ There exist more general quasi-symmetry transformations. $\endgroup$
    – Qmechanic
    Commented Aug 14, 2023 at 12:58
  • $\begingroup$ Thanks for your reaction. I see that I did not incorporate that. But even leaving those quasi-symmetry transformations out and just evaluating the regular symmetries that can be made cyclic with a proper coordinate transformation... the proof seems so trivial and short. It's just equationg EL equation to $0$. I seem to be missing something even in this case. $\endgroup$ Commented Aug 14, 2023 at 13:03
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    $\begingroup$ @bananenheld One effect that also could play a role here is that often when an important result is discovered, later formalisms and teaching are restructured around that result, making it a central piece. So in an intellectual tradition that chose Noether's theorem as a key result, we wouldn't be surprised if Noether's theorem looks quite obvious in retrospect, even if it was hard to see for people educated in the tradition of early 1900s. (I know too little about the relevant history to claim this actually is the case, just noting that it could be based on analogies from elsewhere). $\endgroup$ Commented Aug 14, 2023 at 17:53
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    $\begingroup$ historical aside: the Noether theorems are often presented in simplified form, for teaching purposes. Unfortunately numerous authors published "generalisations" of these simplifications, without reading Noether's original papers. It wasn't until the 1970s that actual generalisations were published (Kosmann-Schwarzbach 2011). Hence the Noether theorems are deeper than is often assumed $\endgroup$ Commented Aug 17, 2023 at 1:52
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Invariant Lagrangian and Conservation Laws:

You're correct that if the Lagrangian is invariant under a transformation, it leads to a conserved quantity according to Noether's theorem. However, the Lagrangian's invariance doesn't directly imply conservation of the conjugate momentum (often referred to as generalized momentum). Instead, it leads to a conserved current, which, when integrated over space or time, gives rise to a conserved charge. The conserved charge is related to the conserved quantity you're looking for.

Change of Coordinates:

Your point about changing coordinates is somewhat correct, but there's a subtlety. Noether's theorem deals with symmetries in the action (the integral of the Lagrangian over time). If you change coordinates, it might alter the form of the Lagrangian, but if the action remains invariant under the coordinate transformation, Noether's theorem can still be applied. Your idea of changing coordinates to align with the invariant direction is useful, but remember that the transformed Lagrangian should still exhibit the same invariance.

Conservation Laws in Absence of Invariance:

Your assertion that if the Lagrangian is not invariant in any direction, then there are no conserved quantities, might not always hold. It's true that Noether's theorem is directly applicable when there's an explicit symmetry in the Lagrangian, but there are situations where conservation laws can arise even without an explicit symmetry, through other mathematical considerations.

Projection onto Invariant Direction:

While projecting the total momentum vector onto the invariant direction is a reasonable approach, remember that Noether's theorem is more general and doesn't always involve just momentum conservation. For example, in a gauge theory, the conserved current can be more complex than simple momentum, involving other variables as well.

Physical Significance:

Finally, Noether's theorem not only provides a mathematical framework for deriving conservation laws from symmetries but also carries important physical implications. It links fundamental symmetries (like translations, rotations, gauge symmetries, etc.) to the preservation of physical quantities, contributing to our understanding of the underlying principles governing physical systems.

By these points you can understand it.

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  • $\begingroup$ Thank you very much for your clear explanation. So if I understand correctly that if the Lagrangian isn't invariant, but the action IS invariant under that transformation, there is still a conserved noether current? So I often see that the invariance of the lagrangian leads to conserved noether charge, but that is not strictly the correct condition of the theorem? $\endgroup$ Commented Aug 16, 2023 at 5:40
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    $\begingroup$ Is this answer written by AI? The way it is written and the way it explains things reminds me of AI. $\endgroup$
    – Noone
    Commented Aug 19, 2023 at 13:37
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    $\begingroup$ None, I doesn't written by aI $\endgroup$ Commented Aug 19, 2023 at 13:41

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