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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?

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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.

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  • $\begingroup$ -@ACuriousMind. How do you make the link between $\Phi_{c}(t)$ and the Hamiltonian ? because Hamiltonian is equal to : $$H=\sum_{i} \dot{q_{i}}p_{i}-L$$ and above is written : $$\Phi_{c}(t)=\sum_{i}p_{i}W^{i}$$. As you can see, Lagrangian has disappeared in this last expression. $\endgroup$ – youpilat13 Feb 2 '18 at 1:44

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