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Em Variational Mathematical Prerequisites

1 · Prerequisites

2 · Summary

The variational core of the electromagnetic action layer, developed as pure mathematics on Euclidean space. The lemma The fundamental lemma of the calculus of variations for continuous coefficients states the fundamental lemma of the calculus of variations for continuous coefficients: under Countable Choice, an open set U⊆Rn and a continuous g:U→Rm; if ∫U⟨g,h⟩ dλ=0 for every compactly supported smooth test field h, then g=0 pointwise on U, with the scalar case the source of a contradiction argument that tests the nonnegative bump gχ and uses the contrapositive of Localization: vanishing integrals or fluxes force a continuous function to vanish. The lemma Compact variations of the quadratic field functional: first variation and the coefficient equation fixes the Minkowski index conventions on V=R4 and studies S[A]=∫U(−14κFμνFμν+JνAν) dλ for a bounded open U with closure inside an open neighbourhood N of the smooth fields: it proves S[A] finite, expands s↦S[A+sh] as the quadratic polynomial with first variation ∫U(−κ−1Fμν∂μhν+Jνhν)dλ, converts that first variation by the weak-derivative test identity into ∫U(κ−1∂μFμν+Jν)hν dλ, and shows that stationarity for the compactly supported class Cc∞(U;V) is equivalent to the coefficient equation ∂μFμν=−κJν on U. Existence of stationary fields and every physical identification of κ, A or S are explicitly not asserted.

All items are mathematics: they carry domain: mathematics, depend only on mathematical suppliers, and assert no physical conclusion. Countable Choice is declared and used exactly through the localization lemma, the classical-to-weak derivative comparison and the Borel measurability of continuous functions, as each item records. The companion examples page em-variational-mathematical-prerequisites-examples records the hypothesis and convention boundaries of these two lemmas.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

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The fundamental lemma of the calculus of variations for continuous coefficients

Statement

Assume the Axiom of Countable Choice (ACω). Let n,m≥1, let U⊆Rn be open, and let g:U→Rm be continuous. If ∫U⟨g,h⟩ dλ=0 for every h∈Cc∞(U;Rm), then g=0 on U. Here ⟨⋅,⋅⟩ is the Euclidean inner product and ∫U⟨g,h⟩ dλ is the finite sum ∑i<m∫Ugihi dλ of integrals of continuous compactly supported functions (Integral over a measurable subset). The scalar case is m=1: a continuous g with ∫Ugh dλ=0 for every h∈Cc∞(U;R) vanishes on U. The vector case is deduced from the scalar case componentwise, each component tested separately. The conclusion is pointwise on U, and the hypothesis is the full family of compactly supported smooth test functions. The Axiom of Countable Choice is declared because the localization lemma Localization: vanishing integrals or fluxes force a continuous function to vanish and the measurability fact Continuous functions on Euclidean spaces are Borel measurable used in the recorded proof assume it; no other choice principle enters.

Conventions. Cc∞(U;Rm) is the space of smooth Rm-valued functions with compact support in U, defined componentwise in the sense of Ck Euclidean maps and diffeomorphisms and Test function space d of an open set. Each product gihi is continuous on U and vanishes outside a compact subset of U, and ∫Ugihi dλ denotes the Lebesgue integral of its zero extension to Rn, where λ=λn is n-dimensional Lebesgue measure (Integrable real and complex functions, and their integrals); [F5] and [F6] below record that this is a finite real number and that for a nonnegative integrand it agrees with the nonnegative integral over the measurable set U of Integral over a measurable subset.

Facts & Assumptions

Given: The Axiom of Countable Choice ACω; naturals n,m≥1; an open set U⊆Rn; a continuous g:U→Rm with ∫U⟨g,h⟩ dλ=0 for every h∈Cc∞(U;Rm); and, inside the contradiction argument below, a point p∈U with g(p)≠0.

[F1]

Cutoff lemma, in ZF: for a compact set K⊆Ω with Ω⊆Rn open there is χ∈Cc∞(Ω) with 0≤χ≤1 and χ=1 on a neighborhood of K (Test function cutoffs and euclidean localization).

[F2]

Localization, under ACω: if G:U→Rm is continuous on an open set U and ∫VG dλn=0 for every ball V compactly contained in U, then G=0 on U (Localization: vanishing integrals or fluxes force a continuous function to vanish, where ∫VG is the componentwise integral over the measurable set V). Below we use only its contrapositive form: if G(q)≠0 for some q∈U, then some ball V compactly contained in U has ∫VG dλn≠0.

[F3]

Order and scalar rules for the nonnegative integral (Monotonicity and nonnegative homogeneity of the nonnegative integral): for nonnegative measurable f≤h one has ∫f dλ≤∫h dλ, and for c>0 one has ∫cf dλ=c∫f dλ, while the zero function has integral 0; the nonnegative integral over a measurable set of Integral over a measurable subset is an instance of this integral.

[F4]

Continuity of pairings: if f,w:U→Rm are continuous then x↦⟨f(x),w(x)⟩ is continuous (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions, clause 3); for m=1 this is the ordinary product of two continuous real functions under The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn. In particular the components gi of g are continuous, and for ψ∈Cc∞(U;R) the function ψei lies in Cc∞(U;Rm) and has ⟨g,ψei⟩=giψ.

[F5]

Signed integral (Integrable real and complex functions, and their integrals): for measurable real F with at most one of ∫F+ dλ, ∫F− dλ equal to +∞, one sets ∫F dλ:=∫F+ dλ−∫F− dλ; F is integrable when both parts are finite. If F≥0 then F−=0 and the signed integral of F equals its nonnegative integral, which over a measurable set E is the integral of FχE (Integral over a measurable subset); and ∫(−F) dλ=−∫F dλ is definitional from (−F)+=F− and (−F)−=F+.

[F6]

Compactly supported continuous functions have finite integrals: let F:Rn→R be continuous with compact support K. If K=∅ then F=0 and ∫F dλ=0; otherwise K is a nonempty compact metric space, so M:=max⁡K∣F∣ is finite (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value), K is bounded (A compact subset of a metric space is closed and bounded) hence contained in a closed box Q with λ(Q)<+∞ (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included), F is Borel measurable (Continuous functions on Euclidean spaces are Borel measurable, under ACω), and ∣F∣≤MχQ, so ∫∣F∣ dλ≤∫MχQ dλ=Mλ(Q)<+∞ by Monotonicity and nonnegative homogeneity of the nonnegative integral, The nonnegative integral agrees with the simple integral on simple functions and The integral of a nonnegative simple function; thus F is integrable by [F5]. In particular each integrand gihi of the hypothesis has a continuous compactly supported zero extension by [F4], and each pairing in the hypothesis is a finite real number.

[F7]

Continuity in balls (Continuity of a map between metric spaces, at a point and globally, in the ε-δ form, Open ball, closed ball and sphere in a metric space): u is continuous at p when for every real ε>0 there is δ>0 with u[B(p,δ)]⊆B(u(p),ε); and B(p,δ)={x:d(p,x)<δ} for δ>0.

Proof

technique · contradiction; assume a point where $g\ne0$, reduce to a scalar continuous function with a vanishing test pairing, and construct a nonnegative compactly supported product of that coefficient with an admissible smooth test function, whose integral is both $0$ and positive
1.1givenF4F5assume-contra

Suppose the conclusion fails: g(p)≠0 for some p∈U. If m=1, put u:=g; then u is continuous and the hypothesis reads ∫Uuψ dλ=0 for every ψ∈Cc∞(U;R). If m≥2, choose i<m with gi(p)≠0 and put u:=gi; for every ψ∈Cc∞(U;R) the function ψei lies in Cc∞(U;Rm) with ⟨g,ψei⟩=giψ by [F4], so the hypothesis gives ∫Uuψ dλ=0. In both cases u:U→R is continuous with u(p)≠0, and replacing u by −u preserves the vanishing pairing because ∫U(−u)ψ dλ=−∫Uuψ dλ by [F5]; hence we may assume u(p)>0.

2.1step 1.1F7algebra

Continuity of u at p with u(p)>0 yields δ>0 with B(p,δ)⊆U and u(x)≥u(p)/2 for every x∈B(p,δ): taking ε:=u(p)/2>0 in [F7] gives δ0>0 with ∣u(x)−u(p)∣<u(p)/2 for d(p,x)<δ0, and since U is open one may shrink δ0 to a δ>0 with B(p,δ)⊆U, the bound remaining valid on the smaller ball.

3.1step 2.1F1

Apply [F1] with Ω:=B(p,δ) and K:={p}: there is χ∈Cc∞(B(p,δ)) with 0≤χ≤1 and χ=1 on a neighborhood of p. Extended by zero, χ lies in Cc∞(U;R) and satisfies supp⁡χ⊆B(p,δ).

4.1step 2.1step 3.1F1F4algebra

Put φ:=uχ, extended by zero to Rn. Then φ is continuous and compactly supported in U by [F4] and [F1], satisfies φ≥0 on U because φ=0 where χ=0 and φ(x)=u(x)χ(x)≥(u(p)/2)χ(x)≥0 for x∈B(p,δ) by step 2.1, and has φ(p)=u(p)>0; hence φ is not the zero function.

5.1step 3.1step 4.1givenF5F6

The hypothesis at the admissible test function χ gives ∫Uuχ dλ=0, that is ∫Uφ dλ=0: since φ≥0, the signed integral equals the nonnegative integral of φ over the measurable set U by [F5], and that integral is a finite real number by [F6].

5.2step 4.1F2F3

By the contrapositive localization [F2] applied to the continuous function φ with φ(p)≠0, there is a ball V compactly contained in U with ∫Vφ dλ≠0; as φ≥0, monotonicity in [F3] applied to the zero function below φχV gives ∫Vφ dλ≥0, so this integral is strictly positive.

6.1step 5.1step 5.2F3discharge-contradiction∎

Since V⊆U, the nonnegative measurable functions φχV and φχU satisfy φχV≤φχU pointwise, so monotonicity in [F3] gives ∫Vφ dλ≤∫Uφ dλ=0, contradicting step 5.2; therefore no point p∈U with g(p)≠0 exists and g=0 on U.

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Compact variations of the quadratic field functional: first variation and the coefficient equation

Statement

Assume the Axiom of Countable Choice (ACω). Let U⊆R4 be a bounded open set, let N⊆R4 be an open neighbourhood of its closure, let κ>0, and let A:N→V=R4 and J:N→V be smooth (Ck Euclidean maps and diffeomorphisms). Use the Minkowski form η and the index conventions of The Minkowski affine four-space: the form η, causal cones, frames and Lorentz maps: Aμ denotes the components of A in the standard basis, Aν:=ηνμAμ, Fμν:=∂μAν−∂νAμ, Fμν:=ημαηνβFαβ, and JνAν is the indicated contraction of the components of J with the covector Aν. Put L:=−14κFμνFμν+JνAν and S[A]:=∫UL dλ, the 4-dimensional Lebesgue integral of L over U. Then S[A] is a finite real number, and for every h∈Cc∞(U;V), with hν:=ηνμhμ and Gμν:=∂μhν−∂νhμ: (a) the map s↦S[A+sh] is the quadratic polynomial S[A]+s∫Uℓ1 dλ+s2∫Uℓ2 dλ, where ℓ1:=−κ−1Fμν∂μhν+Jνhν and ℓ2:=−14κGμνGμν, so that its derivative at s=0 is ∫U(−κ−1Fμν∂μhν+Jνhν)dλ; (b) that first variation also equals ∫U(κ−1∂μFμν+Jν)hν dλ; (c) A is stationary for S, meaning that the derivative in (a) vanishes for every h∈Cc∞(U;V), if and only if ∂μFμν=−κJν for every ν on U, all derivatives being classical partial derivatives of the smooth field A. This is a pure mathematical statement about the functional S: existence of a stationary field A is not asserted, and identifying κ with the vacuum permeability, A with a physical potential or S with a physical action is a later physical model choice.

Conventions. Repeated Greek indices are summed over {0,1,2,3}. With ημν:=η(eμ,eν) the matrix entries in the ordered basis (e0,e1,e2,e3) one has ημν=diag⁡(−1,1,1,1)μν, and ημν denotes the entries of the inverse matrix; since diag⁡(−1,1,1,1) is its own inverse, ημν=ημν and ημαηαν=δνμ (The Minkowski affine four-space: the form η, causal cones, frames and Lorentz maps). Thus Aν=ηνμAμ=(−A0,A1,A2,A3)ν, the lowering map m↦(mν)ν is the componentwise involution m↦(−m0,m1,m2,m3) with inverse mμ=ημνkν=(−k0,k1,k2,k3)μ, and likewise for h. For h∈Cc∞(U;V) the zero extension to N is smooth; A+sh denotes the map x↦A(x)+sh(x) on N, and ∫Uf dλ denotes the Lebesgue integral of the zero extension of f∣U to R4 (Integrable real and complex functions, and their integrals, Integral over a measurable subset).

Facts & Assumptions

Given: The Axiom of Countable Choice ACω; a bounded open set U⊆R4; an open N⊇U‾; a real κ>0; smooth maps A,J:N→V=R4; the Minkowski form η and conventions of [F1]; Fμν:=∂μAν−∂νAμ, Fμν:=ημαηνβFαβ, L:=−14κFμνFμν+JνAν and S[A]:=∫UL dλ; and, for parts (a)-(c), a fixed h∈Cc∞(U;V) with hν:=ηνμhμ and Gμν:=∂μhν−∂νhμ.

[F1]

Minkowski form and index conventions: V=R4 has the ordered basis e0,…,e3, the form η(v,w)=−v0w0+v1w1+v2w2+v3w3 is symmetric bilinear with matrix diag⁡(−1,1,1,1), that matrix is its own inverse, and nondegeneracy and the tensor-component transformation are as stated there (The Minkowski affine four-space: the form η, causal cones, frames and Lorentz maps).

[F2]

Componentwise smoothness, compact support and zero extension: a map is smooth exactly when all its components are, a member of Cc∞(U;V) has compact support in the open U and extends by zero to a smooth compactly supported map on N, and constant-coefficient linear combinations of components preserve smoothness and compact support (Ck Euclidean maps and diffeomorphisms, Test function space d of an open set). Hence h↦(hν)ν is a bijection of Cc∞(U;V) onto Cc∞(U;R4) with inverse k↦mμ:=ημνkν by [F1].

[F3]

Cutoff functions, in ZF: for compact K⊆Ω⊆R4 with Ω open there is χ∈Cc∞(Ω) with 0≤χ≤1 and χ=1 on a neighborhood of K (Test function cutoffs and euclidean localization).

[F4]

Compactness of the closure: U‾ is closed and, since U is bounded, bounded; a closed bounded subset of R4 is compact, with no choice principle (Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[F5]

Integrability of continuous compactly supported functions: let F:R4→R be continuous with compact support K. If K=∅ then F=0 and ∫F dλ=0; otherwise M:=max⁡K∣F∣ is a finite real number (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value), K is bounded (A compact subset of a metric space is closed and bounded) and hence contained in a closed box Q with λ(Q)<+∞ (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included), and ∣F∣≤MχQ gives ∫∣F∣ dλ≤∫MχQ dλ=Mλ(Q)<+∞ (Monotonicity and nonnegative homogeneity of the nonnegative integral, The nonnegative integral agrees with the simple integral on simple functions, The integral of a nonnegative simple function), so F is integrable (Integrable real and complex functions, and their integrals). Under ACω every continuous map between Euclidean spaces is Borel measurable (Continuous functions on Euclidean spaces are Borel measurable), the Borel sigma-algebra is generated by the open sets (The Borel sigma-algebra of a topological space) and every Borel set is Lebesgue measurable (Assuming countable choice, every Borel subset of Rn is Lebesgue measurable); consequently, for a continuous f on an open G⊆R4, the zero extension fχG is Borel measurable, because for a≥0 one has {fχG>a}=G∩{f>a}, open as f is continuous on the open G, while for a<0 it is the union of the closed set R4∖G and that open set, and the rays (a,+∞) generate the Borel sigma-algebra.

[F6]

Signed integrals, monotonicity and linearity: for measurable real f with at most one of ∫f+, ∫f− infinite one sets ∫f dλ:=∫f+dλ−∫f−dλ, and f is integrable when both are finite; for nonnegative measurable f the integral over a measurable set E is ∫Ef dλ=∫fχE dλ; if 0≤f≤h are nonnegative measurable then ∫f dλ≤∫h dλ, and if c>0 then ∫cf dλ=c∫f dλ, while the zero function has integral 0; on L1 the integral is linear, ∫(αf+βg) dλ=α∫f dλ+β∫g dλ (Integrable real and complex functions, and their integrals, Integral over a measurable subset, Monotonicity and nonnegative homogeneity of the nonnegative integral, The Lebesgue integral is linear on L1(μ)).

[F7]

Partial derivatives and their linearity: for a map on an open subset of R4, ∂μ is the directional derivative in the standard direction eμ (Directional derivatives and partial derivatives of a map U⊆Rm→Rn), and differentiation is linear, D(f+cg)=Df+cDg for a real constant c (Sums and scalar multiples of totally differentiable maps are totally differentiable with the expected derivatives), so ∂μ(f+cg)=∂μf+c ∂μg componentwise.

[F8]

Weak derivatives and the test identity, under ACω: a real function u of class C1 on an open Ω⊆Rn has each classical partial derivative ∂μu locally integrable and equal to its weak derivative (Classical derivatives agree with weak derivatives), so for every test function ψ∈Cc∞(Ω;R) the defining test identity of the weak derivative gives ∫Ωu ∂μψ dλ=−∫Ω(∂μu)ψ dλ (Weak derivative of a locally integrable function).

[F9]

Polynomial derivative: for real a0,a1,a2 the polynomial function P(s)=a0+a1s+a2s2 is differentiable at every real c, and P′(c)=a1+2a2c; in particular P′(0)=a1, the term a0 contributing 0 and the term a2s2 contributing 2a2c (For a natural n≥1 the function x↦xn is differentiable everywhere with derivative ι(n) x n−1; for n=0 it is the constant 1, with derivative 0; for a natural n≥1 the function x↦x−n is differentiable at every x≠0 with derivative −ι(n) x−n−1; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, clause 4).

[F10]

Fundamental variation lemma, under ACω: if a continuous g:U→Rm on an open U⊆Rn satisfies ∫U⟨g,k⟩ dλ=0 for every k∈Cc∞(U;Rm), then g=0 on U (The fundamental lemma of the calculus of variations for continuous coefficients).

[F11]

Euclidean inner product: ⟨x,y⟩=∑i<mxiyi on Rm, so ∑νCνhν=⟨C,k⟩ for the arrays C=(C0,…,C3) and k=(h0,…,h3) (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn).

Proof

technique · direct; expand the functional along a compactly supported variation, move the derivative onto the coefficient with the weak-derivative test identity, and read off the coefficient equation with the fundamental variation lemma
1.1givenF1F3F4

The conventions of [F1] fix the index conventions of the Statement, and x↦Fμν(x) and x↦Fμν(x) are smooth on N because A is smooth and ημαηνβ is constant; hence L is smooth on N. The closure U‾ is a compact subset of N by [F4], so [F3] with Ω:=N and K:=U‾ yields χ∈Cc∞(N) with 0≤χ≤1 and χ=1 on U‾.

2.1step 1.1F5F6

The product Lχ, defined to be 0 outside supp⁡χ, is continuous with compact support, so ∫R4∣Lχ∣ dλ<+∞ by [F5]; since χ=1 on U, the nonnegative measurable functions ∣L∣χU and ∣L∣χ satisfy ∣L∣χU≤∣L∣χ pointwise on R4, so [F6] gives ∫U∣L∣ dλ≤∫R4∣Lχ∣ dλ<+∞; hence L is integrable over U and S[A]=∫UL dλ is a finite real number.

2.2step 1.1F1F2F5F7

The zero extension of h to N is smooth with compact support in U by [F2], and for every real s the map A+sh:N→V is smooth with components (A+sh)ν=Aν+shν by [F1] and [F7]; every product below vanishes outside the compact set supp⁡h⊆U, so the corresponding integrands are continuous with compact support and integrable by [F5].

3.1step 2.2F1F7

For every real s and all μ,ν one has F(s)μν=Fμν+sGμν and F(s)μν=Fμν+sGμν on U, where F(s)μν:=∂μ(A+sh)ν−∂ν(A+sh)μ and F(s)μν:=ημαηνβF(s)αβ: the first identity is linearity of the partial derivative, ∂μ(Aν+shν)=∂μAν+s∂μhν and ∂ν(Aμ+shμ)=∂νAμ+s∂νhμ [F7], and the second follows by multiplying by the constant coefficients ημαηνβ and collecting terms [F1].

4.1step 3.1algebra

Multiplying out with step 3.1 gives F(s)μνF(s)μν=FμνFμν+2sFμνGμν+s2GμνGμν, and FμνGμν=FμνGμν=2Fμν∂μhν: the first equality renames the summation indices, and in the second one substitutes Gμν=∂μhν−∂νhμ, relabels μ↔ν in the second sum and uses Fνμ=−Fμν, so the two sums coincide. Since (A+sh)ν=Aν+shν, the source term is Jν(A+sh)ν=JνAν+sJνhν, and therefore L(A+sh)=L+sℓ1+s2ℓ2 pointwise on U with ℓ1=−κ−1Fμν∂μhν+Jνhν and ℓ2=−14κGμνGμν.

5.1step 2.1step 2.2step 4.1F6F9

The functions L, ℓ1, ℓ2 are integrable over U by steps 2.1 and 2.2, so linearity of the integral [F6] applied to step 4.1 gives S[A+sh]=∫U(L+sℓ1+s2ℓ2) dλ=S[A]+s∫Uℓ1 dλ+s2∫Uℓ2 dλ for every real s; thus s↦S[A+sh] is the polynomial P(s)=a0+a1s+a2s2 with the finite real coefficients a0=S[A], a1=∫Uℓ1 dλ and a2=∫Uℓ2 dλ, and [F9] gives P′(0)=a1=∫U(−κ−1Fμν∂μhν+Jνhν)dλ. This is (a).

6.1step 5.1F2F6F8

For each fixed ν the functions x↦Fμν(x), μ=0,1,2,3, are of class C1 on the open U by step 1.1, and hν∈Cc∞(U;R) by [F2]; applying [F8] to each of the four functions and summing over μ gives ∫UFμν∂μhν dλ=−∫U(∂μFμν)hν dλ. Substituting this into step 5.1 and using linearity [F6] yields ∫Uℓ1 dλ=∫U(κ−1∂μFμν+Jν)hν dλ, which is (b).

7.1step 5.1step 6.1F2F10F11

Put Cν:=κ−1∂μFμν+Jν; each Cν is continuous on U by step 1.1, and by steps 5.1 and 6.1 the first variation of S at A in the direction h equals ∫UCνhν dλ=∫U⟨C,k⟩ dλ, where k:=(hν)ν∈Cc∞(U;R4) and the pairing is that of [F11]. Assume first that A is stationary, so that this first variation vanishes for every h∈Cc∞(U;V). Given any k∈Cc∞(U;R4), [F2] provides h∈Cc∞(U;V) with hν=kν, so ∫U⟨C,k⟩ dλ=0; the fundamental variation lemma [F10] applied to the continuous array C gives C=0 on U, that is ∂μFμν=−κJν on U for every ν.

8.1step 5.1step 6.1step 7.1given∎

Conversely, if ∂μFμν=−κJν on U for every ν, then C=0 and steps 5.1 and 6.1 give vanishing first variation for every h∈Cc∞(U;V), so A is stationary; hence stationarity is equivalent to the coefficient equation, which is (c). The point h∈Cc∞(U;V) was arbitrary throughout, so (a)-(c) hold as stated; existence of a stationary field A and the identifications of κ, A and J with physical quantities are not asserted, and the Axiom of Countable Choice entered only through [F5], [F8] and [F10], the expansions of steps 3.1-5.1 being choice-free.

5 · Examples, experiments and counterexamples

None yet.

Sources