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Finite spacelike slab and normal flux charge

1 · Prerequisites

2 · Summary

This page develops the finite-box geometry and integration inputs before stating the balance. A slab has a compact product-box closure in a positively oriented Lorentzian chart; its spatial region is the open rectangle Ω and its slices are the spacelike level sets x0=t. The current is C1 on a neighborhood of the closure, and the slice charge is the signed flux against the future unit normal and induced three-volume: Finite spacelike slab and normal flux charge. The coordinate identity converting that normal integral to wJ0 is proved in Normal slice charge equals its coordinate density integral.

The balance proof uses the locally extended C1 density formula Metric divergence density identity for a C¹ current and one-dimensional FTC on each coordinate direction. It gives the exact equation QJ(t1)−QJ(t0)+ΦJ(B)=∫K∇aJa volg. Writing P:=∫KΠ volg and ΔQ:=QJ(t1)−QJ(t0), the balance gives ΔQ=P−ΦJ(B). Thus the endpoint charge is nondecreasing across this slab exactly when P≥ΦJ(B). Under the theorem's hypothesis Π≥0, zero net outward side flux is sufficient, but not necessary: production can outweigh positive outward flux, and negative net flux contributes positively to ΔQ. The independent flat-current calculations in proof steps 1.4, 2.2 and 4.1 of Complete finite-slab production and outward-flux balance verify both nonzero-flux possibilities directly. The side flux is a three-form integral, so the side faces need not admit unit normals.

The diagram is a 1+1 dimensional cross-section of the four-dimensional coordinate box. Two spatial coordinates are suppressed. It marks the two spacelike slice caps and the side boundary only; it makes no assertion about the causal character of the side faces and is not used in any proof.

\begin{tikzpicture}[>=stealth]
  \draw[->] (0,0) -- (0,3.7) node[above] {$x^0$};
  \draw[->] (0,0) -- (5.2,0) node[right] {$x^1$};
  \draw[thick] (1,0.8) -- (4.2,0.8);
  \draw[thick] (1,2.8) -- (4.2,2.8);
  \draw[thick] (1,0.8) -- (1,2.8);
  \draw[thick] (4.2,0.8) -- (4.2,2.8);
  \node[left] at (1,0.8) {$\Sigma_{t_0}$};
  \node[left] at (1,2.8) {$\Sigma_{t_1}$};
  \node[right] at (4.2,1.8) {$B$};
  \node[below] at (2.6,0.4) {spatial interval};
\end{tikzpicture}

The companion page gives an explicit flat example in which current production vanishes but the charge in an open box decreases because current leaves through its side boundary: Zero divergence with decreasing slice charge in an open region.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

Finite coordinate boxes and their faces are compact Jordan sets

Statement

Let m≥1 be an integer. For a canonical coordinate vector x:{0,…,m−1}→R, write xj:=x(j−1) for 1≤j≤m. Accordingly, ∏j=1mIj denotes {x∈Rm:x(j−1)∈Ij for 1≤j≤m}; this is only a relabelling of the coordinates in the cited suppliers. Let Q=∏j=1m[aj,bj]⊆Rm,aj<bj. Then Q is compact and Jordan measurable, and its boundary has Jordan content zero. For any product R=∏j=1mIj where each Ij is one of [aj,bj], (aj,bj), [aj,bj) or (aj,bj], the set R is a bounded Jordan set with closure contained in Q. Every continuous f:Q→R is Riemann integrable over R. When m≥2, each closed coordinate face of Q is compact and Jordan measurable in its own (m−1)-dimensional coordinate space; the same endpoint-variant and integrability conclusions hold for products in that face.

Facts & Assumptions

Given: A positive integer m and finite real endpoints aj<bj for j=1,…,m.

[F1]

An axis-parallel rectangle has volume equal to the product of its side lengths (Axis-parallel rectangles in Rm and their volume).

[F2]

Jordan outer content is the infimum of total volumes of finite axis-parallel rectangle covers; Jordan measurability is equality of inner and outer content (Jordan inner and outer content and Jordan measurable bounded sets in Rm). Content zero implies nullity in the covering sense used in Lebesgue's criterion, and content zero passes to subsets (Measure zero and content zero in Rm by countable and finite cube covers).

[F4]

A bounded set is Jordan measurable exactly when its boundary has Jordan content zero (A bounded set in Rm is Jordan measurable iff its boundary is null, equivalently of content zero).

[F5]

A continuous real function on a compact Jordan set is Riemann integrable (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set).

[F6]

A bounded function on a rectangle whose discontinuity set is null in the covering sense is Riemann integrable (Lebesgue's criterion in Rm: a bounded function on a closed nondegenerate rectangle is Riemann integrable iff its discontinuity set is null).

[F7]

A bounded function on a bounded Jordan set is Riemann integrable when its zero extension to a containing rectangle is Riemann integrable (The Riemann integral of a bounded function over a bounded Jordan measurable set).

Proof

technique · direct finite-cube covering
1.1givenF3algebra

The box Q is closed and bounded because it is a finite product of closed bounded intervals; [F3] therefore makes Q compact. Its boundary is the union of the 2m coordinate faces Fr,−:={x∈Q:xr=ar} and Fr,+:={x∈Q:xr=br}, since a point of Q is interior exactly when all its coordinates lie strictly between the endpoints.

2.1F1F2step 1.1algebra

Put C:=max⁡(1,max⁡1≤r≤m∏j≠r(bj−aj+1)), with empty product 1. Given ε>0, choose 0<δ<min⁡(1,ε/(2mC)). For each face xr=c, where c=ar or br, cover every other interval by the Nj:=⌈(bj−aj)/δ⌉ intervals [aj+kδ,aj+(k+1)δ], 0≤k<Nj, and use [c,c+δ] in coordinate r. Their products are finitely many closed cubes of side δ covering that face. By [F1], their total volume is δm∏j≠rNj≤δ∏j≠r(bj−aj+δ)≤δC. For m=1 this is one interval of length δ per endpoint. Joining the covers of all 2m faces gives total cube volume at most 2mδC<ε. Hence ∂Q has content zero in the finite-cube sense of [F2], as well as zero Jordan outer content.

3.1F2F4step 1.1step 2.1

The box is bounded and has boundary of content zero by step 2.1, so [F4] makes Q Jordan measurable. For any stated endpoint variant R, its boundary is contained in the same coordinate faces from step 1.1, hence has content zero by step 2.1; since R is bounded, [F4] makes it Jordan measurable as well.

4.1F2F5F6F7step 3.1

Let f:Q→R be continuous. By [F5], f is Riemann integrable on Q and therefore bounded there. For any endpoint variant R, extend f∣R by zero to Q. At every point outside ∂R this extension is either f on a neighborhood in R or zero on a neighborhood in Q∖R‾; thus its discontinuities lie in ∂R. The boundary has content zero by step 3.1, hence is null by [F2]. Then [F6] makes this zero extension integrable on Q, and [F7] gives the Riemann integrability of f∣R over R.

5.1F3step 1.1step 2.1step 3.1step 4.1∎

For m≥2, a closed coordinate face of Q is a closed rectangle in Rm−1, so it is compact by [F3]. Applying steps 1.1–4.1 in dimension m−1 proves its Jordan measurability and the stated integrability of every endpoint variant on that face.

Remarks

This lemma proves only finite-dimensional box and face regularity. It asserts no integrability or convergence result for unbounded sets or exhaustions.

LemmaStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

Metric divergence density identity for a C¹ current

Statement

Assume the Countable Choice convention ACω of The Axiom of Countable Choice (ACω) used by the smooth tangent-bundle structure in Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model. Let (M,g,N) be a time-oriented smooth Lorentzian four-manifold of signature (−,+,+,+) with its metric-selected connection ∇, and let (U,x) be a smooth chart. Write w=∣det⁡(gab)∣. Let J be a C1 vector field on U; the connection derivative on such a field is defined by its usual chart formula ∇bJa:=∂bJa+ΓacbJc, whose coefficients are continuous. Then, throughout U, ∇aJa=w−1∂a(wJa). The right-hand side is the chart expression of the scalar divergence ∇aJa; no higher regularity of J than C1 is required.

Facts & Assumptions

Given: The ACω background, a smooth Lorentzian metric and its metric-selected connection, a smooth chart (U,x), and a C1 vector field J on U.

[A1]

The tangent-bundle smooth structure used by the cited Lorentzian supplier records the Axiom of Countable Choice ACω; this is the only choice principle inherited here (The Axiom of Countable Choice (ACω)).

[F1]

For w=∣det⁡(gab)∣, the metric-density derivative and the Christoffel trace satisfy ∂aw=12wgbc∂agbc and Γbba=w−1∂aw (The derivative of the metric density and the chart formula for the divergence of a vector field clauses (a)–(b)). These clauses concern the smooth metric and do not require a smooth current.

[F2]

The metric-selected connection has smooth Christoffel coefficients Γabc=12gad(∂bgdc+∂cgbd−∂dgbc); its derivative on a vector field has components ∂bJa+ΓacbJc (Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields clauses (a), (c)). The same coordinate rule is defined for C1 components by the C1 chain rule, so the resulting first derivative is continuous.

Proof

technique · direct
1.1A1F1

By [F1], w is smooth and positive and the Christoffel trace obeys Γaab=w−1∂bw after renaming the dummy index. The identity follows from the metric-density derivative and its trace formula, both valid for the given smooth metric.

2.1step 1.1F2algebra∎

Since J is C1, [F2] gives ∇aJa=∂aJa+ΓaabJb. Multiplying by w and substituting step 1.1 yields w∇aJa=w∂aJa+(∂aw)Ja=∂a(wJa), by the product rule for C1 functions. Dividing by w>0 proves the stated identity.

Remarks

The published density lemma states its divergence formula for smooth vector fields. This local extension uses exactly the first-derivative coordinate formula and therefore covers the C1 currents required by the finite-slab balance; it does not change the published supplier.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Finite spacelike slab and normal flux charge

Definition

Use a time-oriented smooth Lorentzian four-manifold (M,g,N) of signature (−,+,+,+) and an independent orientation, with the exact meanings in Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model and Oriented smooth manifolds and oriented charts. The tangent-bundle smooth structure follows the stated Countable Choice convention The Axiom of Countable Choice (ACω). Let (U,x) be a positively oriented chart with x=(x0,x1,x2,x3) and x(U)=V⊆R4. Coordinates here are mathematical real coordinates; no physical units are assigned.

Choose real numbers t0<t1 and ai<bi for i=1,2,3, and let Ω=∏i=13(ai,bi),Ω‾=∏i=13[ai,bi],K=[t0,t1]×Ω,K‾=[t0,t1]×Ω‾⊂V. The finite integration slab K is a bounded Jordan set and its closure K‾ is a compact coordinate box. Assume K‾ has an open neighborhood in V on which the spatial metric block hij:=gij is positive definite and the inverse-metric component g00 is negative. Thus each slice Σt={x−1(t,x1,x2,x3):(x1,x2,x3)∈Ω} is spacelike. Assume also that the future unit normal defined below is future-directed in the given time orientation. The closure K‾, its closed coordinate faces, the open spatial rectangle Ω, the mixed-endpoint slab K, and the corresponding mixed-endpoint side-face rectangles are Jordan sets by Finite coordinate boxes and their faces are compact Jordan sets.

Write gab for the inverse matrix to gab and put w:=∣det⁡(gab)∣,α:=(−g00)−1/2,na:=−αga0,na:=−α δa0. Then n is the future unit normal to every Σt; its unit and normal properties follow from gabnanb=−1 and gabnb=na. Let dVh:=det⁡(hij) dx1dx2dx3 be the Riemannian volume element of the induced positive-definite slice metric. In this positively oriented chart the metric volume form is volg=w dx0∧dx1∧dx2∧dx3 by The metric volume form, current flux through 3-faces and finite-chain charges clause (a).

A current here is a C1 vector field J=Ja∂a on an open neighborhood of x−1(K‾); equivalently, its coordinate components Ja are C1. Its current three-form is the pointwise contraction ωJ:=ιJvolg, where (ιJvolg)(X1,X2,X3):=volg(J,X1,X2,X3). In coordinates, ωJ=w(J0dx1∧dx2∧dx3−J1dx0∧dx2∧dx3+J2dx0∧dx1∧dx3−J3dx0∧dx1∧dx2). The slice orientation is chosen so that the future unit normal precedes a positive slice basis. Since n0=1/α>0 and the spacetime chart is positive, this slice orientation is dx1∧dx2∧dx3; the pullback of ωJ to Σt is wJ0dx1∧dx2∧dx3. Define its signed slice charge (the normal flux through the spacelike slice) by QJ(t):=∫Ω−g(J,n) dVh. Every displayed integral is the Riemann integral over the indicated bounded Jordan set, as defined in The Riemann integral of a bounded function over a bounded Jordan measurable set. The current and metric extend continuously to the compact coordinate closure, and Finite coordinate boxes and their faces are compact Jordan sets proves that the open spatial and mixed-endpoint rectangles are Jordan measurable and that these continuous densities are Riemann integrable there.

Let B denote the union of the six side faces of K, with outward boundary orientation; each face is integrated over its relative interior, a bounded Jordan rectangle whose closure is compact. The upper xi face has induced orientation (−1)idx0∧⋯dxi^⋯∧dx3; the lower face has its opposite. Thus the ith current-three-form pullback integrates to the upper-minus-lower wJi coordinate expression below. Define the net outward side flux ΦJ(B) as the sum over the three spatial coordinate directions of upper-face flux minus lower-face flux. More explicitly, for i∈{1,2,3} let Ωi be the product Jordan rectangle in the three coordinates other than xi, with endpoint convention x0∈[t0,t1] and xj∈(aj,bj) for j≠i; then ΦJ(B):=∑i=13∫ΩiwJi∣xi=bi−wJi∣xi=ai d3xi^. The integrals use the standard increasing-coordinate order on each face. For a C1 current the three-form coefficients are continuous, so the displayed rectangular integrals exist. This extends the smooth-current flux convention of The metric volume form, current flux through 3-faces and finite-chain charges clause (b) to the present finite coordinate faces. No unit normal to a side face is part of this definition; in particular the flux form remains defined if a side face is null. The term “charge” names this mathematical flux integral only; it does not assert a physical conservation law or an entropy interpretation.

Remarks

  • The definition is local to the supplied chart and finite product slab with compact closure. It makes no claim about arbitrary regions, rough boundaries, noncompact exhaustions, or limits at infinity.
  • If a later physical model assigns units and identifies J with an entropy current, those units and that interpretation are additional premises. This mathematical definition has no unit convention.
LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

Normal slice charge equals its coordinate density integral

Statement

Assume the ACω convention and the spacelike slab data of Finite spacelike slab and normal flux charge. For every slice Σt and every C1 current J in that definition, its normal flux charge is QJ(t)=∫ΩwJ0 d3x, where w=∣det⁡(gab)∣ is the metric density in the supplied chart.

Facts & Assumptions

Given: The coordinate box K, positive-definite spatial block hij, g00<0, scalar α=(−g00)−1/2, future unit normal na=−αδa0, and charge QJ(t)=∫Ω−g(J,n)det⁡h d3x.

[A1]

The smooth Lorentzian geometry uses the ACω convention recorded in The Axiom of Countable Choice (ACω) and Finite spacelike slab and normal flux charge.

[F1]

The supplied definition gives h=(gij) positive definite, g00<0, na=−αδa0 and the normal-flux definition of QJ(t) (Finite spacelike slab and normal flux charge).

Proof

technique · direct block determinant and inverse-matrix calculation
1.1A1F1algebra

In the coordinate basis write the metric matrix as G=(g00bTbh), where bi=gi0. Since h is positive definite it is invertible. Subtract bTh−1 times the lower block rows from the first row; this determinant-preserving operation gives det⁡G=det⁡(h) S,S:=g00−bTh−1b. To identify the inverse entry, solve Gz=e0: the lower equations give zsp=−h−1bz0, and the first equation then gives Sz0=1. Therefore g00=z0=S−1, so S<0 by [F1].

2.1step 1.1F1algebra

From S=(g00)−1<0 and the definition of α, α=−S. As det⁡h>0, step 1.1 yields w=∣det⁡G∣=det⁡h−S=αdet⁡h.

3.1step 2.1F1algebra∎

Lowering the index of n gives na=−αδa0 by [F1], hence −g(J,n)=−Jana=αJ0. Substitute step 2.1 into the defining integrand of QJ(t) to obtain −g(J,n)det⁡h=αJ0det⁡h=wJ0. Integrating over Ω proves the statement.

Remarks

The calculation is restricted to the supplied spacelike coordinate slices. It makes no assertion that the coordinate component J0 alone is invariant or that charges from different slicings agree.

TheoremStatement: Literature-sourcedProof: AI-generatedOpen item page →

Complete finite-slab production and outward-flux balance

Statement

Assume the ACω background of Finite spacelike slab and normal flux charge. Let K, Σt, the C1 current J, its slice charge QJ(t) and its net outward side flux ΦJ(B) be as in Finite spacelike slab and normal flux charge. Use the oriented metric volume form of The metric volume form, current flux through 3-faces and finite-chain charges clause (a). Put Π:=∇aJa, using the C1 divergence of Metric divergence density identity for a C¹ current. Then the exact finite-slab balance is QJ(t1)−QJ(t0)+ΦJ(B)=∫KΠ volg. Here the four-volume integral is the Riemann integral on the bounded Jordan slab K, whose compact closure is the supplied coordinate box. If Π≥0 everywhere on K, then QJ(t1)+ΦJ(B)≥QJ(t0). In particular, nonnegative local production implies QJ(t1)≥QJ(t0) when the net outward side flux is zero. If Π=0, the identity instead reads QJ(t1)−QJ(t0)=−ΦJ(B), so zero production alone does not imply constant slice charge.

Facts & Assumptions

Given: The ACω convention, a positively oriented Lorentzian coordinate slab K=[t0,t1]×Ω with nondegenerate rectangular factors and compact closure K‾, spacelike slices, its future unit normal and induced volume, and a C1 current on a neighborhood of K‾.

[A1]

The Countable Choice assumption is the tangent-bundle background recorded in The Axiom of Countable Choice (ACω) and Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model; no additional choice is used.

[F1]

The definition gives the coordinate slice charge QJ(t)=∫Ω−g(J,n)det⁡h d3x and net side flux as the sum of upper-minus-lower wJi face integrals (Finite spacelike slab and normal flux charge).

[F2]

On each slice, the normal charge equals its coordinate density integral QJ(t)=∫ΩwJ0 d3x (Normal slice charge equals its coordinate density integral).

[F3]

For a C1 current, ∂a(wJa)=w∇aJa (Metric divergence density identity for a C¹ current).

[F4]

The sets K, Ω and each Ωi are bounded Jordan products, and every continuous density on their compact coordinate closures is Riemann integrable on them. In particular, the two endpoints of each closed coordinate interval have content zero, by the one-dimensional case of Finite coordinate boxes and their faces are compact Jordan sets.

[F5]

For an integrable function on a bounded Jordan set whose coordinate sections are integrable outside a content-zero set, its integral equals the iterated section integral; in particular this applies to continuous functions on a product rectangle (Fubini over a bounded Jordan set when all but a content-zero family of sections are integrable).

[F6]

If G is differentiable on a closed interval and G′ is integrable, then its integral is G(b)−G(a), including the one-sided endpoint values (The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a)).

[F7]

The integral over a Jordan set is the Riemann integral of its zero extension, and on a rectangle an integrable function's integral is the supremum of lower Darboux sums; lower sums use the cell infima times nonnegative rectangle volumes (The Riemann integral of a bounded function over a bounded Jordan measurable set, The lower and upper Darboux integrals over a nondegenerate rectangle in Rm, Lower and upper Darboux sums over a grid partition in Rm).

[F8]

In a positively oriented chart, volg=w dx0∧dx1∧dx2∧dx3 (The metric volume form, current flux through 3-faces and finite-chain charges clause (a)).

Proof

technique · direct; apply one-dimensional FTC on each coordinate and sum the six oriented side-face terms
1.1A1F3F4

By [F3], wΠ=∂0(wJ0)+∑i=13∂i(wJi) on the coordinate box. Since g and w are smooth and J is C1, this integrand and all face integrands are continuous; [F4] makes their compact-box integrals finite.

1.2F2F5F6

By [F2], QJ(t)=∫ΩwJ0 d3x. Fix (x1,x2,x3)∈Ω. Apply [F6] to the function x0↦wJ0(x0,x1,x2,x3) on [t0,t1], then integrate over Ω using [F5]. This gives ∫K∂0(wJ0) d4x=∫Ω(wJ0(t1,x)−wJ0(t0,x)) d3x=QJ(t1)−QJ(t0).

1.3F1F4F5F6F7

For each fixed i∈{1,2,3} and the other three coordinates, the section of K in the xi coordinate is (ai,bi). By [F7], its integral is defined using the zero extension to [ai,bi]. This extension differs from the continuous derivative xi↦∂i(wJi) only at the two endpoints, which have content zero by [F4]; since the functions are bounded, covers of those points by intervals of arbitrarily small total length show from the Darboux definition that the two integrals agree. Apply [F6] on [ai,bi], then integrate the endpoint difference over the remaining face coordinates by [F5]. This gives ∫K∂i(wJi) d4x=∫ΩiwJi∣xi=bi−wJi∣xi=ai d3xi^. By [F1], summing these three terms gives exactly ΦJ(B), with outward orientation and no choice of a unit normal on any side face.

1.4A1F1F2F3F5F6constructalgebra

To verify the nonimplication by a current independent of any consuming counterexample, use M=R4 with g=diag⁡(−1,1,1,1), future direction ∂0, the standard positive coordinate orientation, and K=[0,1]×(0,1)3. Then w=1, n=∂0 and dVh=d3x, so these are the slab data of [F1]. For real constants a,b∈[−1/4,1/4], set J=(2+ax0)∂0+bx1∂1. This is smooth on all of R4; on the compact slab closure, J0≥7/4>∣J1∣ and J0>0, so it is even future timelike there and on a neighborhood of the closure. Directly from [F1]–[F3] and the elementary rectangular integrals [F5]–[F6], Π=a+b,QJ(t)=2+at,QJ(1)−QJ(0)=a,ΦJ(B)=b,∫KΠ volg=a+b. Indeed only the x1 side pair contributes, and its upper-minus-lower density is b on a face parameter rectangle of volume one. These evaluations do not use the balance conclusion or any later example.

2.1step 1.1step 1.2step 1.3F5F8algebra

Add the temporal identity of step 1.2 and the three spatial identities of step 1.3. Additivity of the finite Riemann integral and [F5] give ∫K∑a=03∂a(wJa) d4x=QJ(t1)−QJ(t0)+ΦJ(B). By step 1.1 and the coordinate expression volg=w dx0∧dx1∧dx2∧dx3 in [F8], the left side is ∫KΠ volg, proving the equality.

2.2step 1.4algebra

In the family of step 1.4 take a=−1/4 and b=1/4. Then Π=0, ΦJ(B)=1/4 and QJ(1)−QJ(0)=−1/4. This admissible smooth current has zero production and strictly decreasing slice charge, proving that zero production alone does not imply charge constancy.

3.1step 2.1F7algebra

If Π≥0, then Πw≥0 because w>0. The zero extension of Πw from K to a containing rectangle is nonnegative; each of its lower Darboux sums is nonnegative by [F7], so ∫KΠ volg≥0. The equality in step 2.1 yields QJ(t1)+ΦJ(B)≥QJ(t0); setting the net side flux to zero gives the stated slice-charge inequality. If Π=0, step 2.1 gives QJ(t1)−QJ(t0)=−ΦJ(B).

4.1step 1.4algebra∎

Taking instead a=b=1/4 gives positive production Π=1/2 whose integral 1/2 exceeds the positive outward flux 1/4, with positive charge change 1/4. Thus zero side flux is sufficient under nonnegative production but is not necessary for nondecreasing charge. The choice a=1/4, b=−1/4 also gives zero production, negative net side flux and positive charge change. These explicit cases establish both possibilities without treating a later consuming counterexample as a premise.

Remarks

  • This is an equality on the stated finite rectangular coordinate slab. It asserts no arbitrary-domain Stokes theorem, infinite exhaustion, boundary at infinity, or convergence result.
  • For nonnegative production, zero net outward side flux is sufficient for endpoint charge monotonicity. The exact criterion is that integrated production is at least the net outward side flux, as illustrated in proof steps 1.4, 2.2 and 4.1. The theorem places no causal-character restriction on the side faces; their flux is defined by the three-form/coordinate face integral, so null side faces need no unit normal.
  • This is a pure mathematical identity. A physical entropy interpretation, units, and a local second-law premise must be supplied separately.

5 · Examples, experiments and counterexamples

None yet.

Sources