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Observer projections, pointwise Eckart vector and local C¹ Landau field

Statement

Let $V$ be a four-dimensional real vector space and let $g : V \times V \to R$ be a symmetric nondegenerate bilinear form of signature $(-, +, +, +)$ , meaning that in some basis its matrix is $diag (- 1, 1, 1, 1)$ . A vector $X$ is timelike when $g (X, X) < 0$ , and null when $X \neq 0$ and $g (X, X) = 0$ . Choose one connected component $C^{+}$ of the timelike set ${X : g (X, X) < 0}$ as the future cone; a timelike vector is future-directed when it belongs to $C^{+}$ . Components and connectedness use the basis-independent finite-dimensional real vector-space topology. Lower indices with $g$ and raise them with its inverse, summing repeated upper/lower indices over $0, 1, 2, 3$ . Let $u \in C^{+}$ be unit timelike, $g (u, u) = - 1$ , and define $Δ^{a}_{b} = δ^{a}_{b} + u^{a} u_{b}$ as in Rest fluid state, frame and geometric current definitions. For every vector $J \in V$ and symmetric contravariant rank-two tensor $T$ on $V$ , the following statements hold.

The operator $Δ$ is the $g$ -orthogonal projection onto $u^{⊥}$ , and $J^{a} = ν_{u} u^{a} + V^{a}, ν_{u} = - u_{a} J^{a}, V^{a} = Δ^{a}_{b} J^{b},$ is the unique decomposition with $V \in u^{⊥}$ .
With $ε_{u} = T^{a b} u_{a} u_{b}, Q^{a} = - Δ^{a}_{c} T^{c d} u_{d}, P^{a b} = Δ^{a}_{c} Δ^{b}_{d} T^{c d},$ one has the unique decomposition $T^{a b} = ε_{u} u^{a} u^{b} + u^{a} Q^{b} + Q^{a} u^{b} + P^{a b} .$ The scalar $p_{m e c h} = \frac{1}{3} g_{a b} P^{a b}$ and tensor $π^{a b} = P^{a b} - p_{m e c h} Δ^{a b}$ satisfy $u_{a} Q^{a} = u_{a} P^{a b} = 0$ , $u_{a} π^{a b} = 0$ , and $g_{a b} π^{a b} = 0$ . Also, $Q = 0$ if and only if $u$ is a timelike eigenvector of $T^{a}_{b}$ .
If $J$ is future timelike, then $u_{E} : = \frac{J}{\sqrt{- g (J, J)}}$ is the unique future unit timelike observer vector for which the spatial number current vanishes, $Δ_{u_{E}} J = 0$ , where $Δ_{u_{E}}^{a}_{b} : = δ^{a}_{b} + u_{E}^{a} (u_{E})_{b}$ is the clause-1 projection computed using $u_{E}$ . Call $u_{E}$ the pointwise Eckart observer vector associated with $J$ ; it is a vector, not a chart or a tetrad.
Let $(M, g, N)$ be a time-oriented smooth Lorentzian four-manifold, let $O \subset M$ be open, and let $T$ be a $C^{1}$ symmetric tensor field on $O$ . At $x_{0} \in O$ , choose an open neighborhood $O_{0} \subset O$ and a smooth future tetrad $(E_{0}, E_{1}, E_{2}, E_{3})$ on $O_{0}$ as supplied by Existence of a smooth future-directed tetrad field near every event, and put $u_{0} : = E_{0} (x_{0})$ and $Δ_{0}^{a}_{b} : = δ^{a}_{b} + u_{0}^{a} (u_{0})_{b}$ . Define $B^{3} : = {w \in R^{3} : \sum_{i = 1}^{3} (w^{i})^{2} < 1}$ and, for $x \in O_{0}$ and $w \in B^{3}$ , define $u (x, w) : = (1 - ∣ w ∣^{2})^{- 1 / 2} (E_{0} (x) + \sum_{i = 1}^{3} w^{i} E_{i} (x)) \in T_{x} M .$ The tetrad equations make $u (x, w)$ future unit timelike. If $T^{a b} (x_{0}) = ε_{0} u_{0}^{a} u_{0}^{b} + p_{0} Δ_{0}^{a b}, h_{0} : = ε_{0} + p_{0} > 0,$ then there are a neighborhood $U \subset O_{0}$ of $x_{0}$ and an open neighborhood $W \subset B^{3}$ of $0$ such that there is a unique $C^{1}$ map $w_{L} : U \to W$ for which $u_{L} (x) : = u (x, w_{L} (x))$ satisfies $Q_{u_{L} (x)} (x) = 0$ for every $x \in U$ . Thus $u_{L}$ is a $C^{1}$ observer field, meaning a $C^{1}$ vector field whose values are future unit timelike; it is distinct from the chosen tetrad field and from any coordinate chart. This is a local result; no far-from-equilibrium or global Landau-field existence is asserted.
In a Lorentz vector space, for every nonzero null vector $ℓ$ and $ρ > 0$ , the pointwise tensor $T^{a b} = ρ ℓ^{a} ℓ^{b}$ has no timelike eigenvector. Hence no future unit timelike observer vector has $Q = 0$ for this tensor, so no pointwise Landau observer vector exists.

Facts & Assumptions

Given: The Lorentz vector space and future unit timelike vector in clauses 1–3, and for clause 4 the time-oriented smooth Lorentzian four-manifold, open sets, $C^{1}$ symmetric tensor field, chosen smooth future tetrad, and pointwise perfect-fluid value stated there.

[F1]

Rest fluid state, frame and geometric current definitions: $Δ^{a}_{b} = δ^{a}_{b} + u^{a} u_{b}$ ; for a vector current, $ν_{u} = - u_{a} J^{a}$ and $V^{a} = Δ^{a}_{b} J^{b}$ ; for a symmetric tensor, $Q^{a} = - Δ^{a}_{c} T^{c d} u_{d}$ and $P^{a b} = Δ^{a}_{c} Δ^{b}_{d} T^{c d}$ .

[F2]

Local graph form at a nonsingular derivative minor: a $C^{1}$ map on an open Euclidean domain with an invertible derivative minor has a unique local $C^{1}$ graph of its zero set.

[F3]

Existence of a smooth future-directed tetrad field near every event: every point of a time-oriented smooth Lorentzian four-manifold has a neighborhood carrying a smooth future-directed tetrad field.

[F4]

The observer rest-space decomposition and the spatial projection: the orthogonal complement of a unit timelike observer is positive definite, and its spatial projection is the orthogonal projection onto that complement.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): the local tetrad supplier uses only the library's ${A C}_{ω}$ tangent-bundle smooth-structure convention; no full axiom of choice is assumed.

Proof

1.1givenF1algebra

For every vector $X$ , $u_{a} (Δ X)^{a} = u_{a} X^{a} + (u_{a} u^{a}) (u_{b} X^{b}) = 0$ and $Δ u = 0$ . If $X \in u^{⊥}$ , then $Δ X = X$ . Hence $im Δ = u^{⊥}$ , $\ker Δ = R u$ , and $Δ^{2} = Δ$ . Also $g (Δ X, Y) = g (X, Y) + g (X, u) g (u, Y) = g (X, Δ Y)$ for all $X, Y$ , so $Δ$ is self-adjoint and is the orthogonal projection onto $u^{⊥}$ .

1.2F1algebra

Since $δ^{a}_{c} = Δ^{a}_{c} - u^{a} u_{c}$ , expansion in both indices gives $T^{a b} = Δ^{a}_{c} Δ^{b}_{d} T^{c d} - u^{a} u_{c} Δ^{b}_{d} T^{c d} - Δ^{a}_{c} T^{c d} u_{d} u^{b} + u^{a} u^{b} u_{c} u_{d} T^{c d} .$ By the definitions this is the stated decomposition. Orthogonality of $Q$ and $P$ follows by $u_{a} Δ^{a}_{c} = 0$ , so $π$ is spatial as well. Since $g_{a b} Δ^{a b} = 3$ , the definition of $p_{m e c h}$ gives $g_{a b} π^{a b} = 0$ . The component formulas recover $ε_{u}, Q, P$ from any such decomposition, so it is unique.

1.3F3givenA1algebra

In the chosen tetrad of [F3], identify symmetric contravariant tensors with the ten-dimensional vector space ${Sym}^{2} (R^{4})$ and write $T^{A}_{B} = T^{A C} η_{C B}$ for their mixed components. For $w \in B^{3}$ , put $γ (w) = (1 - ∣ w ∣^{2})^{- 1 / 2}$ and ${\hat{u}}^{A} (w) = γ (w) (1, w^{1}, w^{2}, w^{3})$ , the tetrad components of $u (x, w)$ from clause 4, and define $Φ_{i} (T, w) : = T^{i}_{A} {\hat{u}}^{A} (w) - w^{i} T^{0}_{A} {\hat{u}}^{A} (w) (i = 1, 2, 3) .$ This is a smooth map on the open subset ${Sym}^{2} (R^{4}) \times B^{3} ≅ R^{10} \times B^{3}$ of $R^{13}$ , and $Φ (T_{0}, 0) = 0$ . At $T_{0} = diag (ε_{0}, p_{0}, p_{0}, p_{0})$ as a contravariant matrix, the mixed matrix is $diag (- ε_{0}, p_{0}, p_{0}, p_{0})$ . Since $\partial_{w^{j}} {\hat{u}}^{0} (0) = 0$ and $\partial_{w^{j}} {\hat{u}}^{k} (0) = δ_{j}^{k}$ , direct differentiation gives $\partial_{w^{j}} Φ_{i} (T_{0}, 0) = p_{0} δ_{i j} + ε_{0} δ_{i j} = h_{0} δ_{i j}$ . Thus $D_{w} Φ (T_{0}, 0) = h_{0} I_{3}$ is invertible by $h_{0} > 0$ .

2.1step 1.1F1algebra

Put $ν_{u} = - u \cdot J$ . Then $J - ν_{u} u = J + u (u \cdot J) = Δ J = V$ , and [step 1.1] shows $V \in u^{⊥}$ . If $J = ν u + W$ with $W ⊥ u$ , contracting with $u$ forces $ν = ν_{u}$ and applying $Δ$ forces $W = V$ .

2.2step 1.1F1algebra

$Q = 0$ is equivalent to $Δ (T^{♯} u) = 0$ , where $(T^{♯} u)^{a} : = T^{a b} u_{b}$ . By [step 1.1], $\ker Δ = R u$ , so this holds exactly when $T^{a}_{b} u^{b}$ is a scalar multiple of $u^{a}$ , the timelike-eigenvector condition.

3.1step 2.2step 1.3F2algebraconstruct

Apply [F2] with $q = 3$ to $Φ (T, w) = 0$ . The $3 \times 3$ minor in the $w$ variables is invertible by step 1.3, so near $(T_{0}, 0)$ the zero set is a unique $C^{1}$ graph $w = φ (T)$ . Choose an open ball $W \subset B^{3}$ about $0$ within the graph neighborhood and shrink the tensor neighborhood so that $φ (T) \in W$ . In the chosen tetrad, the component map $x \mapsto T^{A}_{B} (x)$ is $C^{1}$ , because $T$ is $C^{1}$ and the tetrad is smooth. Shrink to a neighborhood $U \subset O_{0}$ on which these components remain in that tensor neighborhood; then $w_{L} (x) : = φ (T (x))$ is $C^{1}$ , and $u_{L} (x) : = u (x, w_{L} (x))$ is a $C^{1}$ future unit timelike field. The equations $Φ_{i} (T (x), w_{L} (x)) = 0$ say that the spatial components of $T^{A}_{B} (x) u_{L}^{B} (x)$ equal $w_{L}^{i} (x)$ times its time component, so $T^{A}_{B} (x) u_{L}^{B} (x) = λ (x) u_{L}^{A} (x)$ for a scalar $λ (x)$ . By step 2.2 this is equivalent to $Q_{u_{L} (x)} (x) = 0$ , and uniqueness of the local graph gives the claimed uniqueness of $w_{L}$ .

3.2givenstep 2.1F1algebra

Let $r : = \sqrt{- g (J, J)} > 0$ . Then $u_{E} = J / r$ is future unit timelike and $Δ_{u_{E}} J = J + u_{E} (u_{E} \cdot J) = J + (J / r) (- r) = 0$ . Conversely, if a future unit timelike $u$ has $Δ_{u} J = 0$ , step 2.1 gives $J = λ u$ for a scalar $λ$ . Since both $J$ and $u$ are future timelike, $λ > 0$ ; hence $- g (J, J) = λ^{2}$ forces $λ = r$ and $u = u_{E}$ .

4.1F4givenalgebra∎

If $g (ℓ, u) = 0$ , then $ℓ \in u^{⊥}$ , and [F4] makes the restricted metric positive definite; since $g (ℓ, ℓ) = 0$ , this would force $ℓ = 0$ . Therefore $g (ℓ, u) \neq 0$ , so $T^{♯} u = ρ ℓ g (ℓ, u)$ is a nonzero null vector. It cannot equal $λ u$ for any scalar $λ$ : if $λ = 0$ it is nonzero, and if $λ \neq 0$ the right side is timelike while the left side is null. Thus no timelike eigenvector exists.

Observer projections, pointwise Eckart vector and local C¹ Landau field

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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