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Arbitrary-positive-H conformastatic Ricci and Einstein components

Statement

Let $D \subseteq R^{3}$ be open, let $H \in C^{3} (D, (0, \infty))$ , and take the C3 coordinate metric $g_{H}$ from Positive-scalar conformastatic chart data, which is Lorentzian by The positive-scalar conformastatic chart tensor is Lorentzian. All indices range over $0, 1, 2, 3$ , spatial indices $i, j, k$ range over $1, 2, 3$ , and repeated indices are summed. Use the coordinate connection $Γ^{α}_{β γ} : = \frac{1}{2} g^{α δ} (\partial_{β} g_{δ γ} + \partial_{γ} g_{β δ} - \partial_{δ} g_{β γ})$ and the explicit curvature convention $R^{α}_{β μ ν} : = \partial_{μ} Γ^{α}_{ν β} - \partial_{ν} Γ^{α}_{μ β} + Γ^{α}_{μ λ} Γ^{λ}_{ν β} - Γ^{α}_{ν λ} Γ^{λ}_{μ β}, {Ric}_{β ν} : = R^{α}_{β α ν}, S : = g^{β ν} {Ric}_{β ν}, G_{μ ν} : = {Ric}_{μ ν} - \frac{1}{2} S g_{μ ν} .$ The displayed Christoffel coefficients define the unique torsion-free, $g_{H}$ -compatible connection on the global coordinate chart, and the displayed curvature is formed from that connection. These are the standard coordinate formulas with the Riemann sign $R (X, Y) Z = \nabla_{X} \nabla_{Y} Z - \nabla_{Y} \nabla_{X} Z - \nabla_{[X, Y]} Z$ ; the C3 regularity suffices for every displayed derivative. With $H_{i}, H_{i j}, q, Δ H$ as in Positive-scalar conformastatic chart data, the only nonzero Christoffel types are $Γ^{0}_{0 i} = Γ^{0}_{i 0} = - \frac{H_{i}}{H}, Γ^{i}_{00} = - \frac{H_{i}}{H^{5}}, Γ^{i}_{j k} = \frac{δ_{i j} H_{k} + δ_{i k} H_{j} - δ_{j k} H_{i}}{H} .$ The Ricci components, scalar curvature, and orthonormal-frame Einstein components are ${Ric}_{00} = \frac{q}{H^{6}} - \frac{Δ H}{H^{5}}, {Ric}_{0 i} = 0, {Ric}_{i j} = - \frac{δ_{i j} Δ H}{H} + \frac{δ_{i j} q - 2 H_{i} H_{j}}{H^{2}},$ $S = - \frac{2 Δ H}{H^{3}}, G (n, n) = \frac{q - 2 H Δ H}{H^{4}}, G (n, e_{i}) = 0, G (e_{i}, e_{j}) = \frac{δ_{i j} q - 2 H_{i} H_{j}}{H^{4}},$ where $n = H \partial_{0}$ and $e_{i} = H^{- 1} \partial_{i}$ are the frame fields of The positive-scalar conformastatic chart tensor is Lorentzian. No harmonicity condition on $H$ is imposed.

Facts & Assumptions

Given: The open set $D$ , positive C3 function $H$ , the coordinate tensor $g_{H}$ , and the definitions of its connection, curvature, and contractions in the Statement.

[F1]

Positive-scalar conformastatic chart data supplies $g_{00} = - H^{- 2}$ , $g_{i j} = H^{2} δ_{i j}$ , $g_{0 i} = 0$ , and $H_{i}, H_{i j}, q, Δ H$ .

[F2]

The positive-scalar conformastatic chart tensor is Lorentzian supplies $g^{00} = - H^{2}$ , $g^{i j} = H^{- 2} δ_{i j}$ , $\sqrt{∣ \det g ∣} = H^{2}$ , and the frame $n = H \partial_{0}$ , $e_{i} = H^{- 1} \partial_{i}$ .

[F3]

Positive-base powers have derivative $(u^{α})^{'} = α u^{α - 1}$ for $u > 0$ , and the ordinary one-variable chain and product rules apply. (Continuity and derivatives of positive-base real powers, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , 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 $g (c) \neq 0$ )

[F4]

Since $H$ is C3, its second coordinate partials commute: $\partial_{i} H_{j} = \partial_{j} H_{i}$ . (Clairaut--Schwarz theorem for continuous second partial derivatives)

[A1]

$H$ is C3, positive everywhere on $D$ , and independent of $x^{0}$ . No condition $Δ H = 0$ is assumed.

Proof

1.1givenF1F2F3algebra

Because $M$ has one global coordinate chart, the displayed $Γ^{α}_{β γ}$ define a connection by $\nabla_{\partial_{β}} \partial_{γ} = Γ^{α}_{β γ} \partial_{α}$ and the connection Leibniz rule. The formula is symmetric in $β, γ$ , so the connection is torsion-free. Lowering its upper index gives $g_{ρ α} Γ^{α}_{β γ} = \frac{1}{2} (\partial_{β} g_{ρ γ} + \partial_{γ} g_{ρ β} - \partial_{ρ} g_{β γ})$ ; adding the expressions with $(ρ, γ)$ interchanged cancels opposite derivative terms and yields $\partial_{β} g_{ρ γ}$ , which is metric compatibility. Applying the formula to $g_{00} = - H^{- 2}$ and $g_{i j} = H^{2} δ_{i j}$ gives $Γ^{0}_{0 i} = \frac{1}{2} (- H^{2}) (2 H^{- 3} H_{i}) = - H_{i} / H$ , $Γ^{i}_{00} = - \frac{1}{2} H^{- 2} (2 H^{- 3} H_{i}) = - H_{i} / H^{5}$ , and $Γ^{i}_{j k} = (δ_{i j} H_{k} + δ_{i k} H_{j} - δ_{j k} H_{i}) / H$ . Every other type vanishes because $g$ is diagonal and time-independent. The derivatives use [F3] and the positivity of $H$ .

2.1F2step 1.1algebra

Any other torsion-free, $g_{H}$ -compatible connection $\hat{\nabla}$ differs from $\nabla$ by the tensor $C (X, Y) : = {\hat{\nabla}}_{X} Y - \nabla_{X} Y$ . Torsion-freeness gives $C (X, Y) = C (Y, X)$ , while subtracting the two compatibility identities gives $C_{X Y Z} : = g_{H} (C (X, Y), Z) = - C_{X Z Y}$ . Thus $C_{X Y Z} = C_{Y X Z} = - C_{Y Z X} = - C_{Z Y X} = C_{Z X Y} = C_{X Z Y} = - C_{X Y Z}$ , so $C = 0$ by nondegeneracy of $g_{H}$ . Therefore the displayed connection is unique. Its trace is $Γ^{α}_{α i} = Γ^{0}_{0 i} + \sum_{k = 1}^{3} Γ^{k}_{k i} = - H_{i} / H + 3 H_{i} / H = 2 H_{i} / H$ , while $Γ^{α}_{α 0} = 0$ .

3.1givenF1F2step 1.1step 2.1algebra

In the Ricci formula, $\partial_{i} Γ^{i}_{00} = - Δ H / H^{5} + 5 q / H^{6}$ by differentiating $- H_{i} H^{- 5}$ . The trace-product term is $Γ^{i}_{00} Γ^{α}_{α i} = - 2 q / H^{6}$ . The final product contraction $Γ^{α}_{0 β} Γ^{β}_{0 α}$ has two nonzero orders for each spatial $i$ , namely $Γ^{0}_{0 i} Γ^{i}_{00}$ and $Γ^{i}_{00} Γ^{0}_{i 0}$ ; each is $H_{i}^{2} / H^{6}$ , so its subtraction contributes $- 2 q / H^{6}$ . Thus ${Ric}_{00} = q / H^{6} - Δ H / H^{5}$ .

3.2givenF1step 1.1step 2.1algebra

For ${Ric}_{0 i}$ , the derivative terms vanish because all coefficients are time-independent and $Γ^{α}_{α 0} = 0$ . In the trace-product term, a nonzero $Γ^{α}_{0 i}$ requires $α = 0$ , but then $Γ^{β}_{β 0} = 0$ . In the last contraction, the only potentially nonzero first factors are $Γ^{0}_{0 j}$ or $Γ^{j}_{00}$ ; their paired factors have a lower time-space connection slot that vanishes. Hence ${Ric}_{0 i} = 0$ .

3.3givenF1F2F3F4step 1.1step 2.1algebra

For spatial $i, j$ , $Γ^{k}_{i j} = (δ_{k i} H_{j} + δ_{k j} H_{i} - δ_{i j} H_{k}) / H$ . Differentiating and summing $k$ gives $\partial_{k} Γ^{k}_{i j} = 2 H_{i j} / H - 2 H_{i} H_{j} / H^{2} - δ_{i j} Δ H / H + δ_{i j} q / H^{2}$ . The trace term is $- \partial_{j} Γ^{α}_{α i} = - 2 H_{i j} / H + 2 H_{i} H_{j} / H^{2}$ , using the commuting mixed partials of [F4]. The other trace product is $Γ^{k}_{i j} Γ^{α}_{α k} = (4 H_{i} H_{j} - 2 δ_{i j} q) / H^{2}$ .

4.1step 1.1step 3.3algebra

In the last Ricci contraction, the time-index product is $Γ^{0}_{i 0} Γ^{0}_{j 0} = H_{i} H_{j} / H^{2}$ . The spatial numerator is $(δ_{a i} H_{b} + δ_{a b} H_{i} - δ_{i b} H_{a}) (δ_{b j} H_{a} + δ_{b a} H_{j} - δ_{j a} H_{b})$ . Summing over $a, b$ gives the nine terms $H_{i} H_{j}, H_{i} H_{j}, - δ_{i j} q, H_{i} H_{j}, 3 H_{i} H_{j}, - H_{i} H_{j}, - δ_{i j} q, - H_{i} H_{j}, H_{i} H_{j}$ , hence the spatial product is $(5 H_{i} H_{j} - 2 δ_{i j} q) / H^{2}$ . Including the time product, the full product before the Ricci minus sign is $(6 H_{i} H_{j} - 2 δ_{i j} q) / H^{2}$ . Combining this with step 3.3 gives ${Ric}_{i j} = - δ_{i j} Δ H / H + (δ_{i j} q - 2 H_{i} H_{j}) / H^{2}$ .

4.2step 3.1step 3.3F2algebra

Contracting the Ricci table with $g^{00} = - H^{2}$ and $g^{i j} = H^{- 2} δ_{i j}$ gives $g^{00} {Ric}_{00} = - q / H^{4} + Δ H / H^{3}$ and $\sum_{i j} g^{i j} {Ric}_{i j} = q / H^{4} - 3 Δ H / H^{3}$ . Their sum is $S = - 2 Δ H / H^{3}$ .

5.1step 3.2step 3.1step 3.3step 4.1step 4.2F2algebra∎

The coordinate Einstein component is $G_{00} = {Ric}_{00} - \frac{1}{2} S g_{00} = q / H^{6} - 2 Δ H / H^{5}$ , so $G (n, n) = H^{2} G_{00} = (q - 2 H Δ H) / H^{4}$ . For spatial indices, $G_{i j} = {Ric}_{i j} - \frac{1}{2} S g_{i j} = (δ_{i j} q - 2 H_{i} H_{j}) / H^{2}$ because the $δ_{i j} Δ H / H$ terms cancel; therefore $G (e_{i}, e_{j}) = H^{- 2} G_{i j} = (δ_{i j} q - 2 H_{i} H_{j}) / H^{4}$ . The mixed components vanish by step 3.2.

Remarks

For constant $H$ , all displayed curvature components vanish. For nonconstant harmonic $H$ , the scalar curvature is zero but the Einstein tensor need not vanish. The table is not restricted to electrovacuum, and it is not a physical field equation.

Arbitrary-positive-H conformastatic Ricci and Einstein components

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

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