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Exact Reissner-Nordström Ricci and curvature identities

Statement

Let $M, q_{e}, q_{m} \in R$ , $Q^{2} = q_{e}^{2} + q_{m}^{2}$ , and let $J \subset (0, \infty)$ be open with $f (r) = 1 - 2 M / r + Q^{2} / r^{2} \neq 0$ on $J$ . For the dyonic RN datum of Dyonic Reissner-Nordström metric and field domains, in a spherical-polar chart its Ricci components are ${Ric}_{00} = f (\frac{f^{''}}{2} + \frac{f^{'}}{r}) = \frac{f Q^{2}}{r^{4}}, {Ric}_{11} = - \frac{1}{f} (\frac{f^{''}}{2} + \frac{f^{'}}{r}) = - \frac{Q^{2}}{f r^{4}},$ ${Ric}_{22} = 1 - f - r f^{'} = \frac{Q^{2}}{r^{2}}, {Ric}_{33} = (1 - f - r f^{'}) \sin^{2} θ = \frac{Q^{2} \sin^{2} θ}{r^{2}},$ and all mixed components vanish. The scalar curvature is zero and ${Ric}_{a b} {Ric}^{a b} = 4 Q^{4} / r^{8}$ . The Kretschmann scalar is $R_{a b c d} R^{a b c d} = f^{'' 2} + \frac{4 (f^{'})^{2}}{r^{2}} + \frac{4 (1 - f)^{2}}{r^{4}} = \frac{48 M^{2}}{r^{6}} - \frac{96 M Q^{2}}{r^{7}} + \frac{56 Q^{4}}{r^{8}} .$ All formulas hold on every component of $J$ , including $f < 0$ .

Facts & Assumptions

Given: The dyonic RN datum with length parameters $M, q_{e}, q_{m}$ , positive-radius interval $J$ , and $f \neq 0$ .

[F1]

Dyonic Reissner-Nordström metric and field domains: the base metric is $q = - f (d x^{0})^{2} + f^{- 1} d r^{2}$ , the sphere factor is $r^{2} γ_{S^{2}}$ , and $Q^{2} = q_{e}^{2} + q_{m}^{2}$ .

[F2]

The warped product of a two-dimensional Lorentzian base with a round sphere: Christoffel symbols and Ricci tensor: for $g = q + r^{2} γ_{S^{2}}$ with $\det q < 0$ , the product-chart Christoffel symbols include $Γ^{i}_{A j} = r^{- 1} (\partial_{A} r) δ^{i}_{j}$ and $Γ^{A}_{i j} = - r q^{A B} (\partial_{B} r) γ_{i j}$ ; its Ricci formulas say the mixed Ricci components vanish and ${Ric}_{A B} = Ric (q)_{A B} - 2 r^{- 1} {Hess}_{A B} r$ , ${Ric}_{i j} = (1 - z - r Δ_{q} r) γ_{i j}$ .

[F3]

The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors and Tensoriality of the curvature, its coordinate formula and its point-normal metric formula: use $R (X, Y) Z = \nabla_{X} \nabla_{Y} Z - \nabla_{Y} \nabla_{X} Z - \nabla_{[X, Y]} Z$ , $R^{a}_{b c d} = \partial_{c} Γ^{a}_{d b} - \partial_{d} Γ^{a}_{c b} + Γ^{a}_{c e} Γ^{e}_{d b} - Γ^{a}_{d e} Γ^{e}_{c b}$ , and ${Ric}_{b d} = g^{a c} R_{a b c d}$ .

[F4]

The algebraic symmetries of the curvature tensor and the symmetry of the Ricci and Einstein tensors: $R_{a b c d} = - R_{b a c d} = - R_{a b d c} = R_{c d a b}$ , so the six unordered index pairs determine the full contraction.

[F5]

For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term and 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 $g (c) \neq 0$ : for $r > 0$ and every integer $n \geq 1$ , $(r^{- n})^{'} = - n r^{- n - 1}$ ; finite sums and scalar multiples differentiate by the corresponding sums and scalar multiples.

[A1]

$r > 0$ , $f \neq 0$ , and $Q^{2} = q_{e}^{2} + q_{m}^{2}$ .

Proof

1.1givenF1F2F3F5algebra

By [F5] and the sum and scalar rules, $f^{'} = 2 M / r^{2} - 2 Q^{2} / r^{3}$ and $f^{''} = - 4 M / r^{3} + 6 Q^{2} / r^{4}$ . On the two-dimensional base, $q^{00} = - f^{- 1}$ and $q^{11} = f$ . Its only nonzero Christoffel symbols, apart from lower-index symmetry, are $Γ^{0}_{01} = f^{'} / (2 f)$ , $Γ^{1}_{00} = f f^{'} / 2$ , and $Γ^{1}_{11} = - f^{'} / (2 f)$ , by the metric formula in [F3]. Substitution gives $Ric (q)_{00} = f f^{''} / 2$ , $Ric (q)_{11} = - f^{''} / (2 f)$ , ${Hess}_{00} r = - f f^{'} / 2$ , ${Hess}_{11} r = f^{'} / (2 f)$ , $z = f$ , and $Δ_{q} r = f^{'}$ . Applying [F2] yields the four Ricci entries in the Statement; the angular factor $\sin^{2} θ$ is the corresponding component of $γ_{S^{2}}$ .

2.1step 1.1A1algebra

Contracting those diagonal entries with $g^{00} = - 1 / f$ , $g^{11} = f$ , $g^{22} = 1 / r^{2}$ , $g^{33} = 1 / (r^{2} \sin^{2} θ)$ gives $S = - Q^{2} / r^{4} - Q^{2} / r^{4} + Q^{2} / r^{4} + Q^{2} / r^{4} = 0$ and ${Ric}_{a b} {Ric}^{a b} = 4 Q^{4} / r^{8}$ . These contractions are valid for either sign of $f$ because only $f \neq 0$ is used.

2.2F2F3F4step 1.1algebra

In coordinates $(x^{0}, r, θ, ϕ)$ , the nonzero Christoffel symbols of the full metric, apart from lower-index symmetry, are $Γ^{0}_{01} = \frac{f^{'}}{2 f}, Γ^{1}_{00} = \frac{f f^{'}}{2}, Γ^{1}_{11} = - \frac{f^{'}}{2 f}, Γ^{1}_{22} = - r f, Γ^{1}_{33} = - r f \sin^{2} θ, Γ^{2}_{12} = \frac{1}{r}, Γ^{3}_{13} = \frac{1}{r}, Γ^{2}_{33} = - \sin θ \cos θ, Γ^{3}_{23} = \cot θ .$ All other entries are zero. Substitution of this complete list in the coordinate curvature formula of [F3] gives the following matrix of the 21 independent pair-pair components, with row and column labels $01, 02, 03, 12, 13, 23$ : $\begin{array}{ccccccc} 01 & 02 & 03 & 12 & 13 & 23 \\ 01 & f^{''} / 2 & 0 & 0 & 0 & 0 & 0 \\ 02 & 0 & r f f^{'} / 2 & 0 & 0 & 0 & 0 \\ 03 & 0 & 0 & r f f^{'} \sin^{2} θ / 2 & 0 & 0 & 0 \\ 12 & 0 & 0 & 0 & - r f^{'} / (2 f) & 0 & 0 \\ 13 & 0 & 0 & 0 & 0 & - r f^{'} \sin^{2} θ / (2 f) & 0 \\ 23 & 0 & 0 & 0 & 0 & 0 & r^{2} (1 - f) \sin^{2} θ \end{array} .$ For example, the entries $R_{0101}$ , $R_{0202}$ and $R_{2323}$ follow by using respectively the base-only, base-sphere and sphere Christoffel symbols in [F3]; the $03, 12, 13$ entries follow by the same substitution with the listed factors $\sin^{2} θ$ , $f^{- 1}$ and $r^{- 1}$ . The table gives all 21 pair-pair entries, not only the nonzero ones. The pair symmetries [F4] then recover every ordered component.

3.1step 2.2F4algebra

In the full contraction, each independent diagonal pair entry occurs four times. The $01$ entry contributes $4 (f^{''} / 2)^{2} = f^{'' 2}$ . Each of the four base-sphere entries $02, 03, 12, 13$ contributes $(f^{'})^{2} / r^{2}$ after raising its four indices with $g^{00} = - 1 / f$ , $g^{11} = f$ , $g^{22} = 1 / r^{2}$ and $g^{33} = 1 / (r^{2} \sin^{2} θ)$ ; their sum is $4 (f^{'})^{2} / r^{2}$ . The $23$ entry contributes $4 (1 - f)^{2} / r^{4}$ . Thus $R_{a b c d} R^{a b c d} = f^{'' 2} + 4 (f^{'})^{2} / r^{2} + 4 (1 - f)^{2} / r^{4}$ for either sign of $f$ ; only $f \neq 0$ is used.

4.1A1step 3.1algebra∎

Substituting [A1] into the three terms gives respectively $16 M^{2} / r^{6} - 48 M Q^{2} / r^{7} + 36 Q^{4} / r^{8}$ , $16 M^{2} / r^{6} - 32 M Q^{2} / r^{7} + 16 Q^{4} / r^{8}$ , and $16 M^{2} / r^{6} - 16 M Q^{2} / r^{7} + 4 Q^{4} / r^{8}$ . Adding coefficients proves the displayed polynomial.

Remarks

The calculation is a mathematical identity on $f \neq 0$ . It makes no assertion at a coordinate root, at $r = 0$ , about a distributional source there, or about a maximal extension.

Exact Reissner-Nordström Ricci and curvature identities

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources