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Spatial shear properties and nonnegative norm

Statement

Let $(M, g, N)$ be a smooth time-oriented Lorentzian four-manifold of signature $(-, +, +, +)$ , let $O \subseteq M$ be open, and let $u$ be a smooth vector field on $O$ satisfying $g (u, u) = - 1$ and $g (u, N) < 0$ pointwise. Use the unique torsion-free metric-compatible connection $\nabla$ of $g$ . Define

h_{a b} : = g_{a b} + u_{a} u_{b}, h_{a}^{c} : = δ_{a}^{c} + u_{a} u^{c}, θ : = \nabla_{a} u^{a},

and the shear

σ_{a b} : = h_{a}^{c} h_{b}^{d} \nabla_{(c} u_{d)} - \frac{θ}{3} h_{a b},

where parentheses denote symmetrization with factor $1 / 2$ . The field $u$ is a future unit timelike observer field in the convention of Observers, rest spaces and the spatial projection at an event. Then $σ_{a b}$ is symmetric, spatial in both indices ( $u^{a} σ_{a b} = u^{b} σ_{a b} = 0$ ), and trace-free ( $g^{a b} σ_{a b} = 0$ ). The metric restricted to $u^{⊥}$ is positive definite, so

σ^{a b} σ_{a b} \geq 0,

with equality exactly when $σ_{a b} = 0$ . More generally, every spatial vector $Y$ satisfies $g_{a b} Y^{a} Y^{b} \geq 0$ , with equality exactly when $Y = 0$ . These are pointwise geometric statements and assume no field equation.

Facts & Assumptions

Given: A smooth time-oriented Lorentzian four-manifold, an open set $O$ , a smooth future unit timelike vector field $u$ on $O$ , and the metric-selected connection $\nabla$ .

[A1]

The tangent-bundle smooth structure used by the Lorentzian geometry and vector-field definitions assumes Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )); this proof makes no further choice and does not assume unrestricted Choice.

[F1]

Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model: $g$ is a smooth nondegenerate metric of signature $(-, +, +, +)$ , and $N$ is a smooth timelike time-orientation field.

[F2]

A smooth vector field is a smooth section of the tangent bundle: a smooth vector field is a smooth section of the tangent-bundle projection; it supplies the meaning of smooth $u$ .

[F3]

Observers, rest spaces and the spatial projection at an event: a future unit timelike vector satisfies $g (u, u) = - 1$ and $g (u, N) < 0$ , and its rest space is $u^{⊥} = {Y : g (u, Y) = 0}$ .

[F4]

The observer rest-space decomposition and the spatial projection: for such $u$ , the orthogonal complement $u^{⊥}$ is three-dimensional and the restriction of $g$ to it is positive definite; $Y \mapsto Y + g (u, Y) u$ is the orthogonal projection onto it.

[F5]

Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields: $\nabla$ is torsion-free and metric-compatible, so $\nabla g = 0$ and the covariant derivative obeys the product rule.

Proof

1.1A1F1F2F3F5givenalgebra

Differentiating $g_{a b} u^{a} u^{b} = - 1$ covariantly and using [F5] gives $u^{d} \nabla_{c} u_{d} = 0$ for every $c$ . Also $u^{a} h_{a}^{c} = u^{c} + (u_{a} u^{a}) u^{c} = 0$ , and similarly $u^{b} h_{b}^{d} = 0$ . Therefore both indices of $h_{a}^{c} h_{b}^{d} \nabla_{(c} u_{d)}$ are spatial, and subtracting the spatial tensor $θ h_{a b} / 3$ preserves spatiality. Since $h_{a b}$ and $\nabla_{(c} u_{d)}$ are symmetric, their two-index projection and hence $σ_{a b}$ are symmetric.

2.1step 1.1F5algebra

Expanding $h$ and using $g (u, u) = - 1$ gives $g^{a b} h_{a}^{c} h_{b}^{d} = g^{c d} + u^{c} u^{d}$ and $g^{a b} h_{a b} = 3$ . Hence $g^{a b} σ_{a b} = (g^{c d} + u^{c} u^{d}) \nabla_{(c} u_{d)} - θ = θ + u^{c} (u^{d} \nabla_{c} u_{d}) - θ = 0$ , where the middle term vanishes by step 1.1.

3.1step 1.1F4algebra∎

By [F4], each pointwise rest space $u^{⊥}$ is a finite-dimensional positive-definite inner-product space. Choose an algebraic basis there and apply the finite Gram–Schmidt construction; each nonzero residual has positive norm and can be normalized. Since $σ$ is spatial and symmetric by step 1.1, in that basis $σ^{a b} σ_{a b} = \sum_{i, j = 1}^{3} σ_{i j}^{2} \geq 0$ , and the sum is zero exactly when every component, hence $σ$ , is zero. The same positive-definite metric gives $g (Y, Y) \geq 0$ for every spatial vector, with equality only for $Y = 0$ .

Spatial shear properties and nonnegative norm

Statement

Facts & Assumptions

Proof

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