Exact Solutions in the 3+1 Split

This began as some documentation Erik Schnetter wrote for the Penn State Maya code. I wanted to enter some more simple sample (3+1 splits of) exact space-times. It you see an obvious error, or have something to suggest, let me know.

There is, of course, a definite bias towards black hole space-times. I may add cosmological ones when/if I get the chance.

A few words about notation: In what follows, Greek indices such as \(\alpha\), \(\beta\), \(\mu\), \(\nu\) are four-vector indices and run from 0 to 3. Latin indices such as \(i\), \(j\), \(k\), \(l\), \(m\), \(n\) are three-vector indices and run from 1 to 3.

When using spherical polar coordinates, in general \(R\) will denote the standard “areal” radial coordinate; when dealing with a conformally flat solution, I’ll use \(r\) to denote the radial coordinate, as then it will *not* be areal. Additionally, I often use the letter \(q\) to denote the cylindrical polar quantity \(\sqrt{x^2 + y^2} = r \, \sin\theta\); most references I know use the Greek letter \(\rho\) for this purpose, but I’ve found \(\rho\) to be used for too many other purposes.

The standard reference for the 3+1 decomposition is York (1979). I’ll repeat only the most important resulting equations here.

The line element \( ds \) is given by: \[ds^2 = -\alpha^2 dt^2 + \gamma_{i j} (dx^i+\beta ^i dt)(dx^j+\beta ^j dt)\] This leads directly to the relationship between the four-metric \(g_{\mu \nu }\) on the one hand and the three-metric \(\gamma_{i j}\), the lapse \(\alpha\), and the shift \(\beta ^{i}\) on the other hand (see (Misner 1973), section 21.4): \[g_{\mu \nu} = \left( \begin{array}{cc} -\alpha ^2 + \beta_m \beta^m & \beta_j \\ \beta_i & \gamma_{i j} \end{array} \right) \Rightarrow g^{\mu \nu} = \frac{1}{\alpha^2} \left( \begin{array}{cc} - 1 & \beta^j \\ \beta^i & \gamma^{i j} \alpha^2 - \beta^i \beta^j \end{array} \right)\]

The metric connection \({\,^4\Gamma}^a_{bc}\) is defined in terms of the metric inverse, and the metric derivatives: \[{\,^4\Gamma}^{\mu}_{\nu \rho} \equiv \frac{1}{2} g^{\mu \sigma} \left[ g_{\nu \sigma, \rho} + g_{\sigma \rho, \nu} - g_{\nu \rho, \sigma} \right].\] We can write this in terms of 3+1 quantities by looking separately at the time and space components. Starting with the \(\mu=0\): \[\begin{aligned} {\,^4\Gamma}^0_{0 0} &= - \frac{1}{2\alpha^2} \left[ \left( -\alpha ^2 \right)_{,0} - \gamma_{m n, 0} \beta^m \beta^n \right] + \frac{1}{2 \alpha^2} \beta^m \left[ - \left( -\alpha ^2 + \beta_n \beta^n \right)_{,m} \right],\\ {\,^4\Gamma}^0_{0 i} &= - \frac{1}{2\alpha^2} \left[ \left( -\alpha ^2 \right)_{,i} + \beta_m \beta^m_{,i} \right] + \frac{1}{2 \alpha^2} \beta^m \left[ \gamma_{m i, 0} - \beta_{i, m} \right],\\ {\,^4\Gamma}^0_{i j} &= - \frac{1}{2\alpha^2} \left[ \beta_{i,j} + \beta_{j, i} - \gamma_{i j, 0} \right] + \frac{1}{2 \alpha^2} \beta^m \left[ \gamma_{i m, j} + \gamma_{m j,i} - \gamma_{i j, m} \right].\end{aligned}\] Now for the spatial \(\mu=i\) components: \[\begin{aligned} {\,^4\Gamma}^i_{0 0} &= \frac{1}{2 \alpha^2} \beta^i \left[ \left( -\alpha ^2 + \beta_m \beta^m \right)_{, 0} \right] + \frac{1}{2} \left( \gamma^{i m} - \frac{1}{\alpha^2} \beta^i \beta^m \right) \left[ 2 \beta_{m, 0} - \left( -\alpha ^2 + \beta_n \beta^n \right)_{, m} \right],\\ {\,^4\Gamma}^i_{0 j} &= \frac{1}{2 \alpha^2} \beta^i \left[ \left( -\alpha ^2 + \beta_m \beta^m \right)_{, j} \right] + \frac{1}{2} \left( \gamma^{i m} - \frac{1}{\alpha^2} \beta^i \beta^m \right) \left[ \beta_{m, j} + \gamma_{m j, 0} - \beta_{j, m} \right],\\ {\,^4\Gamma}^i_{j k} &= \frac{1}{2 \alpha^2} \beta^i \left[ \beta_{j, k} + \beta_{k, j} - \gamma_{j k, 0} \right] + \frac{1}{2} \left( \gamma^{i m} - \frac{1}{\alpha^2} \beta^i \beta^m \right) \left[ \gamma_{j m, k} + \gamma_{m k, j} - \gamma_{j k, m} \right].\end{aligned}\]

To get the extrinsic curvature from the three-metric, we must take the future-pointing unit time-like normal to the slice, \({\hat{n}}^{\mu}\): \[\begin{aligned} {\hat{n}}_{\mu} &= \left( -\alpha, 0,0,0 \right) \\ \Rightarrow {\hat{n}}^{\mu} &= g^{\mu \alpha} {\hat{n}}_{\alpha} = \left[ \frac{1}{\alpha}, - \frac{\beta^i}{\alpha} \right].\end{aligned}\]

The “projected” four-metric is \(h_{\mu \nu} \equiv g_{\mu \nu} + {\hat{n}}_{\mu} {\hat{n}}_{\nu}\). Then we define the extrinsic curvature (York (1979) eq. (19),(35))^{1}: \[\begin{aligned}
K_{i j} & \equiv - \frac{1}{2} \mathcal{L}_{{\hat{n}}} h_{i j} \\
& = \frac{1}{2\alpha} \left( \beta_{i|j} + \beta_{j|i} - \partial _t \gamma_{i j} \right) = \frac{1}{2\alpha} \left( \mathcal{L}_{\beta} h_{i j} - \mathcal{L}_{t} h_{i j} \right)\\
& = \frac{1}{2\alpha} \left( \beta_{i,j} + \beta_{j,i} - \partial _t \gamma_{i j} - 2 \Gamma^p_{i j} \beta_p \right) \\
& = - \alpha {\,^4\Gamma}^0_{i j}.\end{aligned}\]

This can be inverted to give an expression for the three-metric’s time-derivative in terms of the extrinsic curvature: \[\gamma_{i j, 0} = \beta_{i,j} + \beta_{j,i} - 2 \alpha K_{i j} - 2 \Gamma^p_{i j} \beta_p.\]

In the special case of trivial gauge (\(\alpha = 1\), \(\beta^i = 0\)), the expressions for the four-connection simplify enormously: \[\begin{aligned} {\,^4\Gamma}^0_{0 0} &= 0, &{\,^4\Gamma}^i_{0 0} &= 0,\\ {\,^4\Gamma}^0_{0 j} &= 0, &{\,^4\Gamma}^i_{0 j} &= \frac{1}{2} \gamma^{i m} \left[ \gamma_{m j, 0} \right] = - \gamma^{i m} K_{im},\\ {\,^4\Gamma}^0_{i j} &= \frac{1}{2} \gamma_{i j, 0} = - K_{ij}, &{\,^4\Gamma}^i_{j k} &= \frac{1}{2} \gamma^{i m} \left[ \gamma_{j m, k} + \gamma_{m k, j} - \gamma_{j k, m} \right] = \Gamma^i_{jk},\end{aligned}\]

Smarr (1977) defines the *electric* and *magnetic* parts of the Weyl curvature in the 3+1 split. These are spatial tensors (that is, they are orthogonal to the unit normal to the hypersurface), with components given by \[\begin{aligned}
E_{i j} &= - R_{i j} - K K_{i j} + K_{m i} K_j^{\;m}, \nonumber \\
B_{i j} &= D_m K_{n(i} \varepsilon^{m n}_{\;\;j)}.
\label{eq:NP_elec_def}\end{aligned}\] It can be seen from this definition that the tracelessness of the electric and magnetic tensors is equivalent to the satisfying of the Hamiltonian and momentum constraints.

The Riemann and Ricci tensors (in any dimension) are given by: \[\begin{aligned} {\mbox{R}}^{a}_{\; b c d} &\equiv \partial_{c} \Gamma^{a}_{b d} - \partial_{d} \Gamma^{a} _{b c} + \Gamma^{a}_{m c} \Gamma^{m} _{b d} - \Gamma^{a}_{m d} \Gamma^{m}_{b c},\\ {\mbox{R}}_{a b} &\equiv R^{c}_{\; a c b},\end{aligned}\] the former following the Landau-Lifshitz Spacelike Convention (LLSC), as with Misner et al. (1973).

Note that Maple’s Tensor package follows the conventions of Misner et al. (1973) for Riemann, but *not* for Ricci.

This agrees with Misner et al. (1973) and Cook. Wald (1984) uses the opposite sign (eq (10.2.13),(E.2.30)) but is self-consistent.↩

\label{sec:ks_general}

An important sub-class of black-hole space-times can be written in *Kerr-Schild* form. The four-metric is written \[g_{\mu \nu} = \eta_{\mu \nu} + 2 H \ell_{\mu}\ell_{\nu} \Rightarrow g^{\mu \nu} = \eta^{\mu \nu} - 2 H \ell^{\mu} \ell^{\nu},\] where \(\ell_{\mu}\) is a flat-space null vector: \[\ell_{\mu} \ell^{\mu} \equiv \eta^{\mu \nu} \ell_{\mu} \ell_{\nu} = 0.\] From this generic form, we can deduce something of the 3+1 decomposition: \[\begin{aligned}
\alpha &= \frac{1}{\sqrt{1 + 2 H \ell_0^2}},\\
\beta_i &= 2 H \ell_0 \ell_i,\\
\beta^i &= \frac{2 H \ell_0 \ell^i}{1 + 2 H \ell_0^2},\\
\gamma_{i j} &= \eta_{i j} + 2H \ell_i \ell_j,\\
\gamma^{i j} &= \eta^{i j} - \frac{2H}{1 + 2 H \ell_0^2} \ell^i \ell^j,\\
\Rightarrow \gamma_{i j,k} &= 2 \left[ H_{,k} \ell_i \ell_j + H \ell_{i,k} \ell_j + H \ell_i \ell_{j,k} \right], \\
K_{i j} &= \alpha \left[ \ell_i H_{,j} + \ell_j H_{,i} + H \ell_{i,j} + H \ell_{j,i} + 2 H ^2 \left( \ell_i \ell_m \ell_{j,m} + \ell_j \ell_m \ell_{i,m} \right) + 2 H \ell_i \ell_j \ell_m H_{,m} \right].\end{aligned}\]

This agrees with the formula presented in eqn (35) of Yo et al. (2002).

\label{sec:by_general}

Bowen et al. (1980) introduced a set of partial solutions to the constraint equations, where the momentum constraint is solved automatically by extrinsic curvature that can incorporate bulk ADM linear and/or angular momentum. The data is not completely determined – there is a conformal factor \(\psi\) that must be found by solving the Hamiltonian constraint elliptic equation. The physical metric quantities are: \[\label{eq:by_general} \gamma_{i j} = \psi^4 \delta_{i j} \;\; , \;\; K_{i j} = \psi^{-2} \hat{K}_{i j},\] where the conformal extrinsic curvature contains angular- and/ linear-momentum terms: \[\begin{aligned} \hat{K}_{i j} &= \frac{3}{r^3}\left[ \epsilon_{k i m} \, S^m \, n^k \, n_j + \epsilon_{k j m} \, S^m \, n^k \, n_i \right] \nonumber \\ & + \frac{3}{2 r^2}\left[ P_i \, n_j + P_j \, n_i - (\delta_{i j} - n_i \, n_j ) P^k \, n_k \right] \label{eq:by_K_P_S} \\ & \mp \frac{3 a^2}{2 r^4}\left[ P_i \, n_j + P_j \, n_i + (\delta_{i j} - 5 n_i \, n_j) P^k \, n_k \right] \nonumber,\end{aligned}\] where \(r\) is a conformal radial coordinate, centred on the singularity. The factor \(\psi\) then must be found by solving the Poisson-like vacuum Hamiltonian constraint: \[\label{eq:by_ham_const} \Delta \psi + \frac{1}{8} \psi^{-7} \hat{K}_{i j}\hat{K}^{i j} = 0.\]

Bowen-York data relates very neatly to the ADM quantities, but has two main disadvantages: (a) the Hamiltonian constraint (\ref{eq:by_ham_const}) must be solved explicitly to obtain a solution of the Einstein equations, and (b) even once solved, the data does *not* describe a “clean” black hole – there’s always some radiation on top of the hole, which will radiate away to infinity, or into the hole, over time. In contrast, Kerr-Schild data (subsection \ref{sec:ks_general}) represents both sp

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