Energy, momentum and angular momentum

Plain-language counterpart: Why energy and momentum are conserved.

Three cases, worked in full. Each is (N12) with a particular $\Delta x$ and $\Delta u$ substituted; nothing new is derived. The point of writing them out is that the machinery has to be seen to turn.

Throughout, one independent variable $x$, dependent variables $u_i(x)$, and $f$ containing first derivatives only, so that (N4) applies and

$A=-\sum \frac{\partial f}{\partial u{\text{'}}_i}\overline{\delta }{u}_i;B=A-f\cdot \Delta x\mathrm{(}\mathrm{1}\mathrm{)}$

and the Lagrange expressions are

${\psi }_i=\frac{\partial f}{\partial u_i}-\frac{d}{dx}\left(\frac{\partial f}{\partial u{\text{'}}_i}\right)\mathrm{(}\mathrm{2}\mathrm{)}$

Energy, from translation of the independent variable

Take $\Delta x=\epsilon$, $\Delta u_i=0$. By (N9),

$\overline{\delta }{u}_i=\Delta u_i-u{\text{'}}_i\Delta x=-u{\text{'}}_i\epsilon \mathrm{(}\mathrm{3}\mathrm{)}$

Substituting into (1),

$A=-\sum \frac{\partial f}{\partial u{\text{'}}_i}(-u{\text{'}}_i\epsilon )=\epsilon \sum \frac{\partial f}{\partial u{\text{'}}_i}u{\text{'}}_i\mathrm{(}\mathrm{4}\mathrm{)}$

$B=\epsilon (\sum \frac{\partial f}{\partial u{\text{'}}_i}u{\text{'}}_i-f)\mathrm{(}\mathrm{5}\mathrm{)}$

Now (N12), with the divergence in one variable a total differential:

$-\epsilon \sum {\psi }_{i}u{\text{'}}_i=\epsilon \frac{d}{dx}(\sum \frac{\partial f}{\partial u{\text{'}}_i}u{\text{'}}_i-f)\mathrm{(}\mathrm{6}\mathrm{)}$

Cancelling $\epsilon$ gives the identity

$-\sum {\psi }_{i}u{\text{'}}_i=\frac{d}{dx}(\sum \frac{\partial f}{\partial u{\text{'}}_i}u{\text{'}}_i-f)\mathrm{(}\mathrm{7}\mathrm{)}$

which holds whatever the $u_i$ are. On the variational problem, ${\psi }_i=0$, and

$\sum \frac{\partial f}{\partial u{\text{'}}_i}u{\text{'}}_i-f=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{(}\mathrm{8}\mathrm{)}$

This is the energy. Note what the invariance amounted to: by (N1), invariance under $\Delta x=\epsilon$ means $x$ does not appear explicitly in $f$. Noether states this in her sixth section, and calls the corresponding relations the energy relations and the $B^{(\lambda )}$ the energy components.

Check on the standard case. With $f=\frac{1}{2}u{\text{'}}^2-V(u)$: (2) gives $\psi =-V^{\prime }(u)-u^{″}$; the left of (7) is $(V^{\prime }(u)+u^{″})u^{\prime }$, which is $\frac{d}{dx}(V+\frac{1}{2}u{\text{'}}^2)$; the right of (7) is $\frac{d}{dx}(u{\text{'}}^2-\frac{1}{2}u{\text{'}}^2+V)$, the same. The constant in (8) is the total energy.

Momentum, from translation of a dependent variable

Take $\Delta x=0$, $\Delta u_k=\epsilon$ for one index $k$ and zero otherwise. By (N9), $\overline{\delta }{u}_k=\epsilon$. Then

$A=B=-\epsilon \frac{\partial f}{\partial u{\text{'}}_k}\mathrm{(}\mathrm{9}\mathrm{)}$

and (N12) reads

$\epsilon {\psi }_k=-\epsilon \frac{d}{dx}\left(\frac{\partial f}{\partial u{\text{'}}_k}\right)\mathrm{(}\mathrm{10}\mathrm{)}$

which by (2) is the statement that $\partial f/\partial u_k=0$ — that is, that $u_k$ is a cyclic coordinate, which is what the invariance said. On $\psi =0$,

$\frac{\partial f}{\partial u{\text{'}}_k}=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{(}\mathrm{11}\mathrm{)}$

the momentum conjugate to $u_k$.

Angular momentum, from a rotation of the dependent variables

Take $\Delta x=0$ and a rotation of two dependent variables into each other:

$\Delta u_1=-u_2\epsilon ;\Delta u_2=u_1\epsilon \mathrm{(}\mathrm{12}\mathrm{)}$

so that $\overline{\delta }{u}_i=\Delta u_i$. Then

$B=A=\epsilon (\frac{\partial f}{\partial u{\text{'}}_1}{u}_2-\frac{\partial f}{\partial u{\text{'}}_2}{u}_1)\mathrm{(}\mathrm{13}\mathrm{)}$

and (N12) gives the identity

$-{\psi }_1{u}_2+{\psi }_2{u}_1=\frac{d}{dx}(\frac{\partial f}{\partial u{\text{'}}_1}{u}_2-\frac{\partial f}{\partial u{\text{'}}_2}{u}_1)\mathrm{(}\mathrm{14}\mathrm{)}$

On $\psi =0$,

$\frac{\partial f}{\partial u{\text{'}}_1}{u}_2-\frac{\partial f}{\partial u{\text{'}}_2}{u}_1=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{(}\mathrm{15}\mathrm{)}$

which is the angular momentum in the $1$-$2$ plane, up to the overall sign fixed by the direction of rotation in (12).

What the three cases have in common

Energy came from moving the independent variable, so its current carried the $f\cdot \Delta x$ term. Momentum and angular momentum came from moving the dependent variables, so theirs did not. That is the whole structural difference between them, and it is visible in (5) against (9) and (13).

Provenance: the identification of translation invariance with the absence of explicit $x$ in $f$, and the naming of the energy relations and energy components, are Noether's, from her sixth section. The three derivations are this volume's own, carried out in her notation with her (N9), (N12) and (N4). No equation above is attributed to her except through those. See Sources.