ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck pass

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

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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 components in its rest frame and in a moving frame

Example

Let $T^{a b} = (ε + P) n^{a} n^{b} + P η^{a b}$ be the perfect-fluid tensor of Perfect-fluid divergence identity and its equivalence with the energy-balance and Euler equations, with $n$ future unit. In an inertial frame in which $n = e_{0}$ (fluid at rest) the components are $T^{00} = ε$ , $T^{11} = T^{22} = T^{33} = P$ and all other components zero. Under the passive boost along $x^{1}$ with parameter $β$ , $∣ β ∣ < 1$ , $γ = (1 - β^{2})^{- 1 / 2}$ , the primed components become $T^{' 00} = γ^{2} (ε + P) - P, T^{' 01} = T^{' 10} = - γ^{2} (ε + P) β, T^{' 11} = γ^{2} (ε + P) β^{2} + P, T^{' 22} = T^{' 33} = P,$ all remaining components zero, and the observer contraction $T^{a b} (e_{0}^{'})_{a} (e_{0}^{'})_{b} = T^{' 00}$ differs from $ε$ by $(γ^{2} - 1) (ε + P)$ .

Facts & Assumptions

Given: The affine Minkowski model with inertial frame and chart $x$ , constant $η = diag (- 1, 1, 1, 1)$ (Oriented time-oriented affine Minkowski space (the spacetime primitives), Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space, Inertial frames and inertial charts of the affine Minkowski model); $C^{1}$ data $ε, P$ , a future unit field $n$ (Norms of four-velocity and four-acceleration, and the observer rest-space decomposition); the perfect-fluid field $T$ of Perfect-fluid divergence identity and its equivalence with the energy-balance and Euler equations; the passive boost along $e_{1}$ with $γ = (1 - β^{2})^{- 1 / 2}$ and $∣ β ∣ < 1$ (The passive Lorentz boost with parameter beta, The passive boost is a proper orthochronous Lorentz map with inverse B(-beta), Rational powers $a^{r}$ of a positive base).

[F1]

Rank-two component law $T^{' a b} = Λ^{a}_{c} Λ^{b}_{d} T^{c d}$ under the inertial transition. (Transformation of tensor, covector, derivative and current components under affine Lorentz maps,(iii))

[F2]

Observer decomposition: for a continuous symmetric field $T$ and future unit observer $n$ the contraction $ε = T^{a b} n_{a} n_{b}$ is the energy density component of the unique observer decomposition, and in an adapted frame with $n = e_{0}$ it equals $T^{00}$ . (Observer decomposition of a symmetric stress-energy tensor)

Verification

technique · direct

1.1given

Rest-frame components. With $n = e_{0}$ , $n^{a} = δ_{0}^{a}$ and $η = diag (- 1, 1, 1, 1)$ : $T^{00} = (ε + P) (n^{0})^{2} + P η^{00} = (ε + P) - P = ε$ ; $T^{i i} = (ε + P) (n^{i})^{2} + P η^{i i} = 0 + P = P$ for $i = 1, 2, 3$ ; and $T^{0 i} = (ε + P) n^{0} n^{i} + P η^{0 i} = 0$ .

2.1F2step 1.1

Observer contraction in the rest frame. By [F2] applied to the observer $e_{0}$ the contraction is $ε = T^{a b} (e_{0})_{a} (e_{0})_{b} = T^{00}$ , consistent with step 1.1.

2.2F1step 1.1

Boosted components. The boost matrix has $Λ^{0}_{0} = Λ^{1}_{1} = γ$ , $Λ^{0}_{1} = Λ^{1}_{0} = - γ β$ and preserves the other two spatial directions. Substituting the rest-frame components of step 1.1 into [F1] and summing the nonzero terms gives $T^{' 00} = γ^{2} ε + γ^{2} β^{2} P = γ^{2} (ε + P) - P$ , $T^{' 01} = γ (- γ β) ε + (- γ β) γ P = - γ^{2} (ε + P) β$ , $T^{' 11} = γ^{2} β^{2} ε + γ^{2} P = γ^{2} (ε + P) β^{2} + P$ , $T^{' 22} = T^{' 33} = P$ , $T^{' 23} = 0$ and $T^{' 02} = T^{' 03} = T^{' 12} = T^{' 13} = 0$ .

3.1F2step 2.2given

The boosted observer contraction. Because the passive boost $B (β)$ is a proper orthochronous Lorentz map (The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)), the primed basis is an orthonormal future-oriented frame, so the vector $e_{0}^{'}$ with primed components $e_{0}^{' a} = δ_{0}^{a}$ is future unit; by [F2] its energy contraction is $ε^{'} : = T^{a b} (e_{0}^{'})_{a} (e_{0}^{'})_{b} = T^{' 00} = γ^{2} (ε + P) - P$ , so $ε^{'} - ε = (γ^{2} - 1) (ε + P)$ : this difference vanishes exactly when $β = 0$ (i.e. $γ = 1$ ) or $ε + P = 0$ , and $ε^{'} \geq ε$ whenever $ε + P \geq 0$ ; no positivity of $ε + P$ is assumed. The transverse diagonal components remain $P$ .

4.1step 2.2step 3.1∎

The computation uses the observer decomposition to identify the numerical contraction $T^{' 00}$ as the energy density of the boosted observer; it asserts no equation of state, no positivity of $ε$ or $P$ and no measurement.

Perfect fluid components in its rest frame and in a moving frame

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources