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I've known about Noether's theorem for some time and reading some things about it recently I've realised I haven't completely understood it. In that case I've been trying to understand a more rigorous approach to it as presented in Spivak's Mechanics book.

There Spivak first derives a corollary from the derivation of the Euler-Lagrange's equation for the extremal of the functional

$$J(f) = \int_a^b F(f(t),f'(t),t)dt,$$

for $f : [a,b]\to \mathbb{R}^n$. This corollary is what he calls the boundary term corollary which states that if $f$ is a critical point of $J$ and if $\alpha : (-\epsilon,\epsilon)\times [a,b]\to \mathbb{R}^n$ is a variation of $f$ then if we set $\bar{\alpha}(u)=\alpha(u, \cdot)$ we have

$$\dfrac{dJ(\bar{\alpha}(u))}{du}\bigg|_{u=0}=\sum_{i=1}^n \dfrac{\partial \alpha^i}{\partial u}(0,t)\dfrac{\partial F}{\partial y^i}(f(t),f'(t),t)\bigg|_a^b.$$

Then, if $M$ is the configuration manifold of a system and if $\phi : (-\epsilon,\epsilon)\times M\to M$ is a one-parameter family of diffeomorphisms defining $\phi_s = \phi(s,\cdot)$ and $\Phi_s = (\phi_s)_{\ast}$ Spivak defines that $\phi_s$ preserves $L : TM\to \mathbb{R}$ if for all $v\in TM$ we have $L(\Phi_s (v)) = L(v)$.

He also defines $W = \partial \phi/\partial s$ and defines the quantity

$$\Phi_c(t) = \lim_{h\to 0}\dfrac{L(c'(t)+hW(c(t)))-L(c'(t))}{h}$$

After that Noether's theorem is stated and proven like this

If the $\phi_s$ preserve $L$, then $\Phi_c$ is constant along any solution $c$ of Lagrange's equations for $L$

PROOF: Since the $\phi_s$ preserve $L$, each of the curves

$$c_s(t)=\phi_s(c(t))=\phi(s,c(t)),$$

is also a solution of Lagrange's equations for $L$, and thus an extremal for $\int_a^bL(c(t),c'(t),t)dt$ for all $a,b$ in the interval under consideration.

The boundary term corollary then says that for all such $a$ and $b$ we have

$$0 = \sum_{i=1}^n \dfrac{\partial q^i}{\partial x}\dfrac{\partial L}{\partial \dot{q}^i}\bigg|_a^b,$$

so that

$$\sum_{i=1}^n \dfrac{\partial q^i}{\partial x}\dfrac{\partial L}{\partial \dot{q}^i}$$

is constant.

Now there are a couple of points. In summary I can't understand what is going on here. My doubts are

  1. That quantity $\Phi_c$ never entered the proof and I really don't see a connetion between the quantity proven constant and $\Phi_c$ at first.

  2. How the boundary term corollary applies here? Are we considering the variation $\alpha:(-\epsilon,\epsilon)\times [a,b]\to M$ given by $\alpha(u,t)=c_u(t)$? But this variation might not keep endpoints fixed. What is the meaning of this?

  3. Why we need each $c_s$ to be a solution of the equations? On the boundary term corollary we have that quantity is zero even if the variation is not yet a solution.

In summary what is the idea behind the proof of Noether's theorem as stated above? What is really going on?

Gold
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1 Answers1

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  1. $\Phi_c(t)$ is the directional derivative of $L(c(t),\dot{},t)$ along $W(c(t))$, so by the chain rule $$ \Phi_c(t) = \sum_i\frac{\partial L}{\partial \dot{q}^i}W^i = \sum_i\frac{\partial L}{\partial \dot{q}^i}\frac{\partial \phi^i}{\partial s}$$ There is then an inexplicable notational shift from $\frac{\partial\phi^i}{\partial s}$ to $\frac{\partial q^i}{\partial x}$, but it's meant to be the same, so the conserved quantity in the proof is indeed $\Phi_c$.

  2. Yes, we are considering the variation $\alpha(u,t) = c_u(t)$. Note that the boundary term corollary does not keep the endpoints fixed - for fixed endpoints the term on its r.h.s. is identically zero.

  3. Each of the paths needs to be a critical point of the action functional because the l.h.s. of the boundary term corollary needs to vanish for the proof to work.

As for "what's really going on": The quantity $\Phi_c(t)$ has a nice Hamiltonian interpretation: Since ${\partial L}/{\partial\dot{q}^i}$ is just the canonical momentum, $\Phi_c$ is the projection of the variation vector (the "direction" in which we vary the path) onto the momentum. Noether's theorem then states this projection is constant for variations which are symmetries.

ACuriousMind
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  • Really nice answer. Also, this may be useful: https://physics.stackexchange.com/a/754823/226902 – Quillo Mar 12 '23 at 15:43