LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Perfect-fluid divergence identity and its equivalence with the energy-balance and Euler equations

Statement

Let $Ω \subseteq A$ be open, let $ε, P : Ω \to R$ be $C^{1}$ , let $n : Ω \to V$ be a $C^{1}$ future unit timelike field ( $η (n, n) = - 1$ ) and set $T^{a b} : = (ε + P) n^{a} n^{b} + P η^{a b}$ . Write $D ε : = n^{a} \partial_{a} ε$ , $ϑ : = \partial_{a} n^{a}$ , $a^{b} : = n^{a} \partial_{a} n^{b}$ and $h^{b c} : = η^{b c} + n^{b} n^{c}$ (these formulas define the first-order expansion scalar, acceleration and raised rest projection for the stipulated $C^{1}$ field; they agree with Kinematics of an observer congruence: acceleration, expansion, shear, vorticity and Born rigidity and The flat covariant derivative on affine Minkowski space and fields along curves when $Ω$ is nonempty and the field is smooth). Then the identity $\partial_{a} T^{a b} = n^{b} (D ε + (ε + P) ϑ) + ((ε + P) a^{b} + h^{b c} \partial_{c} P)$ holds on $Ω$ , and consequently $\partial_{a} T^{a b} = 0$ on $Ω$ if and only if $D ε + (ε + P) ϑ = 0$ and $(ε + P) a^{b} + h^{b c} \partial_{c} P = 0$ on $Ω$ . The two equations are the energy-balance and Euler equations of the model; the lemma asserts no equation of state, no thermodynamic relation and no causality condition. No choice principle is used.

Facts & Assumptions

Given: The affine Minkowski model with inertial chart $x$ , constant $η_{a b} = η^{a b} = diag (- 1, 1, 1, 1)$ and coordinate partials $\partial_{a}$ (Oriented time-oriented affine Minkowski space (the spacetime primitives), Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ); open $Ω$ , $C^{1}$ functions $ε, P$ , a $C^{1}$ future unit timelike field $n$ (Norms of four-velocity and four-acceleration, and the observer rest-space decomposition,(c), C^k tensor fields on affine Minkowski space and their inertial components); $T^{a b} = (ε + P) n^{a} n^{b} + P η^{a b}$ is a $C^{1}$ symmetric tensor field. On nonempty $Ω$ its force density is $f^{b} = \partial_{a} T^{a b}$ and conservation means $f = 0$ (Stress-energy tensor, force density and local balance on flat spacetime). The scalar $ϑ = \partial_{a} n^{a}$ and vector $a^{b} = n^{a} \partial_{a} n^{b}$ are defined by these first-order formulas; on nonempty $Ω$ with smooth $n$ they agree with the expansion and acceleration of Kinematics of an observer congruence: acceleration, expansion, shear, vorticity and Born rigidity and The flat covariant derivative on affine Minkowski space and fields along curves.

[F1]

Product rule and constancy: $\partial_{a} (Φ n^{a}) = (\partial_{a} Φ) n^{a} + Φ \partial_{a} n^{a}$ for $C^{1}$ functions and fields, and $\partial_{a} η^{b c} = 0$ . (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ )

[F2]

Since $n_{b} n^{b} = η (n, n) = - 1$ is constant, $n_{b} n^{a} \partial_{a} n^{b} = \frac{1}{2} n^{a} \partial_{a} (n_{b} n^{b}) = 0$ ; hence $a^{b} = n^{a} \partial_{a} n^{b}$ lies in $n^{⊥}$ , as does $h^{b c} \partial_{c} P$ because $n_{b} h^{b c} = η^{b c} n_{b} + n^{b} n^{c} n_{b} = 0$ . (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space)

Proof

technique · direct. If $\Omega=\varnothing$, the tensor-field definition supplies the unique empty fields; all displayed component derivatives and products are empty maps, so the identity and both directions of the pointwise equivalence are vacuous. No nonempty-domain stress-energy or congruence definition is invoked in this branch. In steps 1.1–4.1 assume $\Omega\ne\varnothing$

1.1F1given

First expansion. By [F1], using that $η^{a b}$ is constant, $\partial_{a} T^{a b} = \partial_{a} [(ε + P) n^{a} n^{b}] + η^{a b} \partial_{a} P = n^{b} \partial_{a} [(ε + P) n^{a}] + (ε + P) n^{a} \partial_{a} n^{b} + \partial^{b} P,$ where $\partial_{a} [(ε + P) n^{a}] = (\partial_{a} ε + \partial_{a} P) n^{a} + (ε + P) \partial_{a} n^{a} = D ε + D P + (ε + P) ϑ$ with $D P : = n^{a} \partial_{a} P$ , while $n^{a} \partial_{a} n^{b} = a^{b}$ and $\partial_{a} n^{a} = ϑ$ .

2.1F2step 1.1

Reconstruction of the identity. Since $h^{b c} = η^{b c} + n^{b} n^{c}$ , one has $\partial^{b} P = h^{b c} \partial_{c} P - n^{b} D P$ . Substituting into step 1.1 cancels $n^{b} D P$ and gives $\partial_{a} T^{a b} = n^{b} (D ε + (ε + P) ϑ) + (ε + P) a^{b} + h^{b c} \partial_{c} P$ . By [F2] the vector $Y^{b} : = (ε + P) a^{b} + h^{b c} \partial_{c} P$ is in $n^{⊥}$ .

3.1step 2.1

Equivalence with the two equations. Contracting the identity of step 2.1 with $n_{b}$ gives $n_{b} \partial_{a} T^{a b} = n_{b} n^{b} (D ε + (ε + P) ϑ) + n_{b} Y^{b} = - (D ε + (ε + P) ϑ)$ because $n_{b} n^{b} = - 1$ and $Y \in n^{⊥}$ . Hence conservation $\partial_{a} T^{a b} = 0$ implies $D ε + (ε + P) ϑ = 0$ , and substituting this back into the identity of step 2.1 gives $Y^{b} = 0$ ; conversely, if both $D ε + (ε + P) ϑ = 0$ and $Y = 0$ hold, then the identity of step 2.1 gives $\partial_{a} T^{a b} = 0$ . The two equations displayed are exactly the energy-balance equation and the Euler equation.

4.1step 2.1step 3.1∎

The divergence identity and the equivalence with the two conservation equations are proved in steps 2.1 and 3.1. The argument holds for any $C^{1}$ data with $η (n, n) = - 1$ and asserts no equation of state, no thermodynamic relation, no positivity of $ε$ or $P$ , no energy condition and no causality condition; all operations are finite component computations and no choice principle is used.

Perfect-fluid divergence identity and its equivalence with the energy-balance and Euler equations

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources