BFP Rauch: 1 revision imported

2026-09-10T13:29:55Z

1 revision imported

← Older revision		Revision as of 13:29, 10 September 2026
(No difference)

NT>NoetherVolumeBuild: Noether theorems volume, initial import (F-11)

2026-09-10T00:00:00Z

Noether theorems volume, initial import (F-11)

New page

{{DISPLAYTITLE:Energy, momentum and angular momentum}}

''Plain-language counterpart: [[Noether's theorems/Explainers/Why energy and momentum are conserved|Why energy and momentum are conserved]].''

Three cases, worked in full. Each is (N12) with a particular
<math>\Delta x</math> and <math>\Delta u</math> 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 <math>x</math>, dependent variables
<math>u_{i}(x)</math>, and <math>f</math> containing first derivatives only, so
that (N4) applies and

<math>A=-\sum\frac{\partial f}{\partial u'_{i}}\bar{\delta}u_{i};\qquad B=A-f\cdot\Delta x\qquad\qquad\mathrm{(1)}</math>

and the Lagrange expressions are

<math>\psi_{i}=\frac{\partial f}{\partial u_{i}}-\frac{d}{dx}\left(\frac{\partial f}{\partial u'_{i}}\right)\qquad\qquad\mathrm{(2)}</math>

== Energy, from translation of the independent variable ==

Take <math>\Delta x=\varepsilon</math>, <math>\Delta u_{i}=0</math>. By (N9),

<math>\bar{\delta}u_{i}=\Delta u_{i}-u'_{i}\Delta x=-u'_{i}\varepsilon\qquad\qquad\mathrm{(3)}</math>

Substituting into (1),

<math>A=-\sum\frac{\partial f}{\partial u'_{i}}\left(-u'_{i}\varepsilon\right)=\varepsilon\sum\frac{\partial f}{\partial u'_{i}}u'_{i}\qquad\qquad\mathrm{(4)}</math>

<math>B=\varepsilon\left(\sum\frac{\partial f}{\partial u'_{i}}u'_{i}-f\right)\qquad\qquad\mathrm{(5)}</math>

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

<math>-\varepsilon\sum\psi_{i}u'_{i}=\varepsilon\frac{d}{dx}\left(\sum\frac{\partial f}{\partial u'_{i}}u'_{i}-f\right)\qquad\qquad\mathrm{(6)}</math>

Cancelling <math>\varepsilon</math> gives the identity

<math>-\sum\psi_{i}u'_{i}=\frac{d}{dx}\left(\sum\frac{\partial f}{\partial u'_{i}}u'_{i}-f\right)\qquad\qquad\mathrm{(7)}</math>

which holds whatever the <math>u_{i}</math> are. On the variational problem,
<math>\psi_{i}=0</math>, and

<math>\sum\frac{\partial f}{\partial u'_{i}}u'_{i}-f=\mathrm{const}\qquad\qquad\mathrm{(8)}</math>

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

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

== Momentum, from translation of a dependent variable ==

Take <math>\Delta x=0</math>, <math>\Delta u_{k}=\varepsilon</math> for one
index <math>k</math> and zero otherwise. By (N9),
<math>\bar{\delta}u_{k}=\varepsilon</math>. Then

<math>A=B=-\varepsilon\frac{\partial f}{\partial u'_{k}}\qquad\qquad\mathrm{(9)}</math>

and (N12) reads

<math>\varepsilon\psi_{k}=-\varepsilon\frac{d}{dx}\left(\frac{\partial f}{\partial u'_{k}}\right)\qquad\qquad\mathrm{(10)}</math>

which by (2) is the statement that <math>\partial f/\partial u_{k}=0</math> —
that is, that <math>u_{k}</math> is a cyclic coordinate, which is what the
invariance said. On <math>\psi=0</math>,

<math>\frac{\partial f}{\partial u'_{k}}=\mathrm{const}\qquad\qquad\mathrm{(11)}</math>

the momentum conjugate to <math>u_{k}</math>.

== Angular momentum, from a rotation of the dependent variables ==

Take <math>\Delta x=0</math> and a rotation of two dependent variables into each
other:

<math>\Delta u_{1}=-u_{2}\varepsilon;\qquad\Delta u_{2}=u_{1}\varepsilon\qquad\qquad\mathrm{(12)}</math>

so that <math>\bar{\delta}u_{i}=\Delta u_{i}</math>. Then

<math>B=A=\varepsilon\left(\frac{\partial f}{\partial u'_{1}}u_{2}-\frac{\partial f}{\partial u'_{2}}u_{1}\right)\qquad\qquad\mathrm{(13)}</math>

and (N12) gives the identity

<math>-\psi_{1}u_{2}+\psi_{2}u_{1}=\frac{d}{dx}\left(\frac{\partial f}{\partial u'_{1}}u_{2}-\frac{\partial f}{\partial u'_{2}}u_{1}\right)\qquad\qquad\mathrm{(14)}</math>

On <math>\psi=0</math>,

<math>\frac{\partial f}{\partial u'_{1}}u_{2}-\frac{\partial f}{\partial u'_{2}}u_{1}=\mathrm{const}\qquad\qquad\mathrm{(15)}</math>

which is the angular momentum in the <math>1</math>-<math>2</math> 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
<math>f\cdot\Delta x</math> 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 <math>x</math> in <math>f</math>, 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
[[Noether's theorems/Sources|Sources]].''

Noether's theorems/First theorem/Energy, momentum and angular momentum - Revision history

BFP Rauch: 1 revision imported

NT>NoetherVolumeBuild: Noether theorems volume, initial import (F-11)