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Perfect stress and particle balances with entropy identity — Examples

1 · Prerequisites

2 · Summary

An explicit null-dust tensor illustrates that a timelike Landau eigenframe need not exist for a symmetric stress tensor. The calculation uses the stated $(-, +, +, +)$ convention and the Landau-frame definition cited in the example.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-generatedOpen item page →

An explicit null-dust tensor without a timelike Landau frame

Example

Take $M = R^{4}$ with Minkowski metric $g = diag (- 1, 1, 1, 1)$ and choose the time orientation $N = (1, 0, 0, 0)$ . By Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model, a nonzero causal vector $v$ is future-directed exactly when $g (v, N) < 0$ , which here is equivalent to $v^{0} > 0$ . Let $ℓ = (1, 1, 0, 0)$ , so $g (ℓ, ℓ) = 0$ and $g (ℓ, N) = - 1 < 0$ ; thus $ℓ$ is future-directed null. Let $ρ > 0$ and define the symmetric tensor $T^{a b} = ρ ℓ^{a} ℓ^{b}$ . A Landau frame for this tensor means a future unit timelike vector $u$ for which the projected energy flux $Q^{a} = - Δ^{a}_{c} T^{c d} u_{d}$ vanishes. The current frame convention of Observer projections, pointwise Eckart vector and local C¹ Landau field gives the unique stress decomposition; when $Q = 0$ , contracting it with $u_{b}$ gives $T^{a}_{b} u^{b} = - ϵ_{u} u^{a}, ϵ_{u} = T^{a b} u_{a} u_{b} .$ This explicit tensor has no Landau frame.

Facts & Assumptions

Given: Minkowski space with signature $(-, +, +, +)$ ; the future-directed null vector $ℓ = (1, 1, 0, 0)$ ; $ρ > 0$ ; and an arbitrary candidate future unit timelike vector $u$ .

[A1]

The geometric supplier carries only Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Observer projections, pointwise Eckart vector and local C¹ Landau field (current batch-53 draft): its clause 2 states the unique decomposition $T^{a b} = ε_{u} u^{a} u^{b} + u^{a} Q^{b} + Q^{a} u^{b} + P^{a b}$ , with $Q^{a} = - Δ^{a}_{c} T^{c d} u_{d}$ and $u_{a} Q^{a} = u_{a} P^{a b} = 0$ . Contracting the decomposition on its first index with $u_{a}$ gives $u_{a} T^{a b} = - ε_{u} u^{b} - Q^{b}$ ; symmetry of $T$ then gives $T^{a}_{b} u^{b} = - ε_{u} u^{a}$ when $Q = 0$ , where $ε_{u} = T^{a b} u_{a} u_{b}$ . This example proves its null-tensor case directly; the supplier's separate clause 5 is not used. Its current input hash is stale, so this use remains conditional.

[F2]

Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model: for a chosen time orientation $N$ , a nonzero causal vector $v$ is future-directed exactly when $g (v, N) < 0$ . Here $N = (1, 0, 0, 0)$ , so any future-directed timelike candidate $u$ has $u^{0} > 0$ ; $ℓ = (1, 1, 0, 0)$ is future-directed null because $g (ℓ, N) = - 1 < 0$ and $g (ℓ, ℓ) = 0$ .

Verification

1.1A1F2givenalgebra

Write $u = (u^{0}, u^{1}, u^{2}, u^{3})$ . Future timelikeness gives $u^{0} > 0$ and $- (u^{0})^{2} + (u^{1})^{2} + (u^{2})^{2} + (u^{3})^{2} = - 1$ , so $∣ u ∣^{2} = (u^{0})^{2} - 1 < (u^{0})^{2}$ . Consequently $ℓ_{b} u^{b} = - u^{0} + u^{1} < 0$ , since $u^{1} \leq ∣ u ∣ < u^{0}$ . Hence $ϵ_{u} = ρ (ℓ_{b} u^{b})^{2} > 0$ .

2.1F1F2step 1.1algebra∎

Contract the stress decomposition in [F1] on its first index with $u_{a}$ . Using $u_{a} u^{a} = - 1$ , $u_{a} Q^{a} = 0$ , and $u_{a} P^{a b} = 0$ gives $u_{a} T^{a b} = - ϵ_{u} u^{b} - Q^{b}$ . Since $T$ is symmetric, this means $T^{a}_{b} u^{b} = - ϵ_{u} u^{a}$ if $Q = 0$ . By the definition of $T^{a}_{b}$ and the metric, the left side is $ρ ℓ^{a} (ℓ_{b} u^{b})$ , a nonzero null vector by step 1.1. The right side is nonzero timelike because $ϵ_{u} > 0$ and $u$ is timelike. Equality is impossible because equal vectors have equal quadratic norm, while these norms are respectively zero and negative. Hence $Q \neq 0$ for every candidate future unit timelike $u$ , and the tensor has no Landau frame.