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NT>NoetherVolumeBuild: Noether theorems volume, initial import (S-02)

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Noether theorems volume, initial import (S-02)

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{{DISPLAYTITLE:The second theorem stated and proved}}

''Plain-language counterpart: [[Noether's theorems/Explainers/The second theorem, and why it is not the first|The second theorem, and why it is not the first]].''

[[File:NT-F-13.png|thumb|right|360px|alt=Two groups of five bars representing field equations. On the left, without a local symmetry, all five are independent. On the right, one free function makes the fifth follow from the other four, shown by a curved arrow. A caption states the counting: one arbitrary function gives one identity and one equation that was never independent.|Identities among the field equations from a local symmetry]]

== The statement ==

'''Theorem II.''' If the integral <math>I</math> is invariant under a
<math>\mathfrak{G}_{\infty\varrho}</math>, in which the arbitrary functions occur
up to the <math>\sigma</math>th derivative, then there exist
<math>\varrho</math> identical relations between the Lagrange expressions and
their derivatives up to the <math>\sigma</math>th order; and here too the
converse holds.

Three words in that sentence do the work. '''Identical''' — the relations hold in
all arguments, not on solutions. '''Between the Lagrange expressions''' — not
between conserved quantities. '''Up to the <math>\sigma</math>th order''' — the
depth of the output is the depth of the group.

== The trunk shared with the first theorem ==

Everything up to and including (N12) is common to both proofs and is not repeated
here. It is on
[[Noether's theorems/First theorem/The first theorem stated and proved|the first
theorem's page]] and runs: the invariance condition (N1), the local form (N11)
obtained through the barred variation (N9), and then the identity (N3) applied to
<math>\bar{\delta}</math>, giving

<math>\sum\psi_{i}\bar{\delta}u_{i}=\mathrm{Div}\,B,\qquad\left(B=A-f\cdot\Delta x\right)\qquad\qquad\mathrm{(N12)}</math>

'''The two proofs diverge at exactly one point, and it is here.''' For a
<math>\mathfrak{G}_{\varrho}</math>, <math>\bar{\delta}u</math> is linear in
constant parameters, so (N12) splits into <math>\varrho</math> separate relations
by equating coefficients. For a <math>\mathfrak{G}_{\infty\varrho}</math>,
<math>\bar{\delta}u</math> is linear in arbitrary functions '''and their
derivatives''', which cannot be equated coefficient by coefficient because the
derivatives of a function are not independent of it. Removing them is the whole
work of this proof.

== The proof ==

'''Step 1. Write out the linear dependence on the arbitrary functions.''' With
<math>\mathfrak{G}</math> an infinite continuous group,
<math>\bar{\delta}u</math> and its derivatives, and hence <math>B</math>, are
linear in the <math>p(x)</math> and their derivatives. Substituting the values of
<math>\bar{\delta}u</math>, independently of (N12),

<math>\sum\psi_{i}\bar{\delta}u_{i}=\sum_{\lambda,i}\psi_{i}\left\{a_{i}^{(\lambda)}(x,\ u,\dots)p^{(\lambda)}(x)+b_{i}^{(\lambda)}(x,\ u,\dots)\frac{\partial p^{(\lambda)}}{\partial x}+\dots+c_{i}^{(\lambda)}(x,\ u,\dots)\frac{\partial^{\sigma}p^{(\lambda)}}{\partial x^{\sigma}}\right\}\qquad\qquad\mathrm{(1)}</math>

'''Step 2. Trade derivatives of <math>p</math> for derivatives of the
coefficients.''' Analogously to the formula for partial integration, the identity

<math>\varphi(x,u,\dots)\frac{\partial^{\tau}p(x)}{\partial x^{\tau}}=(-1)^{\tau}\cdot\frac{\partial^{\tau}\varphi}{\partial x^{\tau}}\cdot p(x)\qquad\qquad\mathrm{(2)}</math>

holds modulo divergences. Applying it term by term replaces the derivatives of
<math>p</math> by <math>p</math> itself, at the cost of divergences that are
themselves linear in <math>p</math> and its derivatives. Collecting those into a
single <math>\mathrm{Div}\,\Gamma</math>,

<math>\sum\psi_{i}\bar{\delta}u_{i}=\sum_{\lambda}\left\{\left(a_{i}^{(\lambda)}\psi_{i}\right)-\frac{\partial}{\partial x}\left(b_{i}^{(\lambda)}\psi_{i}\right)+\dots+(-1)^{\sigma}\frac{\partial^{\sigma}}{\partial x^{\sigma}}\left(c_{i}^{(\lambda)}\psi_{i}\right)\right\}p^{(\lambda)}+\mathrm{Div}\,\Gamma\qquad\qquad\mathrm{(N14)}</math>

'''Step 3. Combine with the trunk.''' Setting (N14) against (N12),

<math>\sum\left\{\left(a_{i}^{(\lambda)}\psi_{i}\right)-\frac{\partial}{\partial x}\left(b_{i}^{(\lambda)}\psi_{i}\right)+\dots+(-1)^{\sigma}\frac{\partial^{\sigma}}{\partial x^{\sigma}}\left(c_{i}^{(\lambda)}\psi_{i}\right)\right\}p^{(\lambda)}=\mathrm{Div}\,(B-\Gamma)\qquad\qquad\mathrm{(N15)}</math>

'''Step 4. Kill the right-hand side by choosing the arbitrary functions.''' Form
the <math>n</math>-fold integral of (N15) over any region, and choose the
<math>p(x)</math> so that they vanish on the boundary together with all
derivatives occurring in <math>B-\Gamma</math>. The integral of a divergence
reduces to a boundary integral, so the right-hand side integrates to zero.
Therefore the integral of the left-hand side vanishes, for arbitrary
<math>p(x)</math> subject only to that boundary condition. By the standard
argument, the integrand vanishes for every <math>p(x)</math>, and the
<math>\varrho</math> relations follow:

<math>\sum\left\{\left(a_{i}^{(\lambda)}\psi_{i}\right)-\frac{\partial}{\partial x}\left(b_{i}^{(\lambda)}\psi_{i}\right)+\dots+(-1)^{\sigma}\frac{\partial^{\sigma}}{\partial x^{\sigma}}\left(c_{i}^{(\lambda)}\psi_{i}\right)\right\}=0\quad(\lambda=1,2\dots\varrho)\qquad\qquad\mathrm{(N16)}</math>

These are the sought dependencies between the Lagrange expressions and their
derivatives. Their linear independence follows as in the finite case, since the
converse leads back to (N12).

'''Step 5. A corollary that comes free.''' From (N15) and (N16) together,

<math>\mathrm{Div}\,(B-\Gamma)=0\qquad\qquad\mathrm{(3)}</math>

so a function system that does not vanish identically has an identically
vanishing divergence. This is the non-uniqueness noted on
[[Noether's theorems/First theorem/The Noether current and the Noether charge|the
current page]], arriving here as an output rather than a caveat.

== What Step 4 required, and what it forbids ==

Step 4 is where the arbitrary functions earn their name. The choice of
<math>p(x)</math> vanishing on the boundary but arbitrary inside is available
only because they are functions. It is unavailable for a parameter, and that is
why no amount of increasing the parameter count of a
<math>\mathfrak{G}_{\varrho}</math> produces this theorem.

'''Nothing in this proof used the field equations.''' (N16) is an identity, and
it stays one. Setting <math>\psi_{i}=0</math> does not convert it into a
conservation law; it converts it into a statement about the field equations
themselves, which is
[[Noether's theorems/Second theorem/Fewer independent field equations than fields|a
different sentence with a different meaning]].

'''Audit.''' The proof used: (N12) from the first theorem's page; (N3) from the
Lagrange-expressions page, through (N12); the definition of
<math>\mathfrak{G}_{\infty\varrho}</math> and the linearity in
<math>p</math> from
[[Noether's theorems/Second theorem/Groups depending on arbitrary functions|the
previous page]]; and nothing else.

''Provenance: the theorem statement is a paraphrase of Noether's Satz II. (N14),
(N15) and (N16) are her equations (14), (15) and (16), and displays (1), (2), (3)
are hers, set as running text in the original. Reproduced in her notation from
the public-domain German original. '''Two deviations from the electronic
text.''' Its (14) carries the unbarred variation where the surrounding argument
and its own derivation require the barred one, and its sign factor in (14) is
punctuated with the exponent inside the parenthesis, where the same factor in
(16) is punctuated outside. Both are written here as the argument requires;
whether the slips are the 1918 printing's or the transcription's needs the page
images. See [[Noether's theorems/Sources|Sources]].''

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