Total Conserved Charges of Kerr Spacetime with One Rotation Parameter in 5-Dimensions Using Poincaré Gauge Theory

Abstract

A pentad field, which creates Kerr spacetime, with one rotation parameter in 5-dimensions is provided. We calculate the total conserved charges of this pentad using the approach of “invariant conserved currents”. Regularized expression through relocalization is used to get the known form of conserved charges of Kerr 5-dimensions spacetime. In contrast, the covariant calculation of conserved charge led to non-convergent results.

Appendices

Appendix A: Notation

In this study, Latin indices i,j,⋯ refer to local holonomic space-time coordinates while Greek indices α, β, ⋯ label (co)frame components. Particular frame components are denoted by hats, $\hat {0},\hat {1}$, etc. As usual, the exterior product is denoted by ∧, while the interior product of a vector ξ and a p-form Ψ is denoted by ξ⌋Ψ. The vector basis dual to the frame 1-forms 𝜗 ^α is denoted by e _α and they satisfy e _α ⌋𝜗 ^β=δ _α ^β. Using local coordinates x ⁱ, we have $\vartheta ^{\alpha }=h^{\alpha }_{i} dx^{i}$ and $e_{\alpha }=h^{i}_{\alpha } \partial _{i}$ where $h^{\alpha }_{i}$ and $h^{i}_{\alpha }$ are the covariant and contravariant components of the tetrad field. We define the volume 4-form by

$$ \eta \overset{\text{def.}}{=} \vartheta^{\hat{0}}\wedge \vartheta^{\hat{1}}\wedge \vartheta^{\hat{2}}\wedge\vartheta^{\hat{3}}. $$

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Furthermore, with the help of the interior product we define

$$\eta_{\alpha} \overset{\text{def.}}{=} e_{\alpha} \rfloor \eta = \ \frac{1}{3!} \ \epsilon_{\alpha \beta \gamma \delta} \ \vartheta^{\beta} \wedge \vartheta^{\gamma} \wedge \vartheta^{\delta}=^{*}\vartheta_{\alpha}, \ \ \text{ \ \ refers \ \ to \ \ Hodge \ \ dual \ \ operator} $$

where 𝜖 _{α β γ δ} completely antisymmetric with 𝜖 ₀₁₂₃=1.

$$\eta_{\alpha \beta} \overset{\text{def.}}{=} e_{\beta} \rfloor \eta_{\alpha} = \frac{1}{2!}\epsilon_{\alpha \beta \gamma \delta} \ \vartheta^{\gamma} \wedge \vartheta^{\delta}=^{*}\left(\vartheta_{\alpha} \wedge \vartheta_{\beta}\right), $$

$$\eta_{\alpha \beta \gamma} \overset{\text{def.}}{=} e_{\gamma} \rfloor \eta_{\alpha \beta}= \frac{1}{1!} \epsilon_{\alpha \beta \gamma \delta} \ \vartheta^{\delta}=^{*}\left(\vartheta_{\alpha} \wedge \vartheta_{\beta}\wedge \vartheta_{\gamma}\right), $$

which are bases for 3-, 2- and 1-forms respectively. Finally,

$$\eta_{\alpha \beta \mu \nu} \overset{\text{def.}}{=} e_{\nu} \rfloor \eta_{\alpha \beta \mu}= e_{\nu} \rfloor e_{\mu} \rfloor e_{\beta} \rfloor e_{\alpha} \rfloor \eta=^{*}\left(\vartheta_{\alpha} \wedge \vartheta_{\beta} \wedge \vartheta_{\mu} \wedge \vartheta_{\nu}\right)=\epsilon_{\alpha \beta \gamma \delta}, $$

is the Levi-Civita tensor density. The η-forms satisfy the useful identities:

$$\begin{array}{@{}rcl@{}} \vartheta^{\beta} \wedge \eta_{\alpha} & \overset{\text{def.}}{=} & \delta^{\beta}_{\alpha} \eta, \qquad \vartheta^{\beta} \wedge \eta_{\mu \nu} \overset{\text{def.}}{=} \delta^{\beta}_{\nu} \eta_{\mu}-\delta^{\beta}_{\mu} \eta_{\nu}, \\ \vartheta^{\beta} \wedge \eta_{\alpha \mu \nu} &\overset{\text{def.}}{=}& \delta^{\beta}_{\alpha} \eta_{\mu \nu}+\delta^{\beta}_{\mu} \eta_{\nu \alpha}+\delta^{\beta}_{\nu} \eta_{\alpha \mu}, \\ \vartheta^{\beta} \wedge \eta_{\alpha \gamma \mu \nu} & \overset{\text{def.}}{=} & \delta^{\beta}_{\nu} \eta_{\alpha \gamma \mu}-\delta^{\beta}_{\mu} \eta_{\alpha \gamma \nu }+\delta^{\beta}_{\gamma} \eta_{\alpha \mu \nu}-\delta^{\beta}_{\alpha} \eta_{\gamma \mu \nu}. \end{array} $$

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The line element $ds^{2} \overset {\text {def.}}{=} g_{\alpha \beta } \vartheta ^{\alpha } \bigotimes \vartheta ^{\beta }$ is defined by the space-time metric g _{α β} .

Appendix B: Necessary Calculations of Conserved Quantities

The non-vanishing components of the Riemannian connection, of the first coframe, have the form:

$$\begin{array}{@{}rcl@{}} &&{} {\widetilde{\Gamma}_{\hat{1}}}^{\hat{0}}= -\frac{[(r^{2}+a^{2})dt+a\sin^{2}\theta[a^{2}(1+\cos^{2}\theta)+2r^{2}] d\phi]}{ {{\mathcal{ B}}_{3}}^{4}\sqrt{r^{4}+a^{4}+a^{2}(r^{2}+2M)(1+\cos^{2}\theta)}}, \\ &&{} {\widetilde{\Gamma}_{\hat{2}}}^{\hat{0}}= -\frac{Ma^{2}\sin2\theta\sqrt{r^{2}+a^{2}-2M}[a\sin^{2}\theta d\phi+dt]}{ {{\mathcal{B}}_{3}}^{4}\sqrt{r^{4}+a^{4}+a^{2}(r^{2}+2M)(1+\cos^{2}\theta)}},\\ &&{} {\widetilde{\Gamma}_{\hat{3}}}^{\hat{0}}= \frac{Ma\sin\theta[\{2r^{3}+ra^{2}(1+\cos^{2}\theta)\}dr-a^{2}\sin\theta\cos\theta(r^{2}+a^{2}-2M) d\theta]}{\sqrt{r^{2}+a^{2}-2M}[r^{4}+a^{4}+a^{2}(r^{2}+2M)(1+\cos^{2}\theta)]},\\ &&{} {\widetilde{\Gamma}_{\hat{1}}}^{\hat{2}}= \frac{[a^{2}\sin\theta\cos\theta dr+r(r^{2}+a^{2}-2M) d\theta]}{{{\mathcal{B}}_{3}}^{2}\sqrt{r^{2}+a^{2}-2M}},\\ &&{} {\widetilde{\Gamma}_{\hat{1}}}^{\hat{3}}= \frac{r\sin\theta\sqrt{r^{2}+a^{2}-2M}[(2Ma^{2}\sin^{2}\theta-{{\mathcal{B}}_{3}}^{4}) d\phi+aM dt]}{ {{\mathcal{B}}_{3}}^{4}\sqrt{r^{4}+a^{4}+a^{2}(r^{2}+2M)(1+\cos^{2}\theta)}},\\ &&{} {\widetilde{\Gamma}_{\hat{1}}}^{\hat{4}}= \frac{-\cos\theta \sqrt{r^{2}+a^{2}-2M}d\psi}{{{{\mathcal{B}}}_{2}} },\\ &&{} {\widetilde{\Gamma}_{\hat{2}}}^{\hat{3}}=- \frac{\cos\theta\{[r^{6}+a^{2}\cos^{2}\theta(r^{2}+a^{2}-2M)(a^{2}\cos^{2}\theta+2r^{2})+r^{2}a^{2}(r^{2}-2M)] d\phi+2aM(r^{2}+a^{2}) dt\}} {{{\mathcal{B}}_{3}}^{4}\sqrt{r^{4}+a^{4}+a^{2}(r^{2}+2M)(1+\cos^{2}\theta)}}, \\ &&{} {\widetilde{\Gamma}_{\hat{2}}}^{\hat{4}}= \frac{-r\sin\theta d\psi}{{{{\mathcal{B}}}_{2}} }. \\ \end{array} $$

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Using (22), the non-vanishing components of the vector ξ ^α have the form

$$ \xi^{0}= {{\mathcal{B}}_{1}}\zeta^{0}, \qquad \qquad\xi^{3}= -{\mathcal{B}}_{4}\zeta^{0}+{\mathcal{B}_{5}}\zeta^{3},\qquad \qquad\xi^{4}= {\mathcal{B}}_{6}\zeta^{4} . $$

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If we take pentad (17), as well as the trivial Weitzenböck connection, Γ^α _β =0, we finally get the non-vanishing components of the translation momentum, $\widetilde {H}_{\hat {\alpha }}$

$$\begin{array}{@{}rcl@{}} &&{\widetilde{H}_{\hat{0}}}=\frac{2\sqrt{r^{2}+a^{2}-2M}\sin2\theta(2a^{2}[r^{2}+M]+a^{2}[2r^{2}+a^{2}-2M)\cos^{2}\theta+3r^{4})(d\theta \wedge d\phi \wedge d\psi)} {{{\mathcal{B}}_{3}}\sqrt{r^{4}+a^{4}+a^{2}(r^{2}+2M)(1+\cos^{2}\theta)}},\\ &&{\widetilde{H}_{\hat{3}}}=\frac{4Mar^{2}\cos\theta\sin^{2}\theta(a^{2}(1+\cos^{2}\theta)+2r^{2})(d\theta \wedge d\phi \wedge d\psi)} {{{\mathcal{B}}_{3}}^{3}\sqrt{r^{4}+a^{4}+a^{2}(r^{2}+2M)(1+\cos^{2}\theta)}}. \end{array} $$

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