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.

3+1 Decomposition and Conventions

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} \be