Abstract
Calculate quasi-local measures such as masses, momenta, or angular momenta and related quantities on closed two-dimentional surfaces, including on horizons.
[NOTE: Ignore the stuﬀ below. You can do that much easier.]
The papers about dynamical horizons contain integrals over the 3D horizon world tube, expressed e.g. as
$$\begin{array}{rcll}\int X\phantom{\rule{2.77695pt}{0ex}}{d}^{3}V& & & \text{(1)}\text{}\text{}\end{array}$$
where $X$ is some quantity that lives on the horizon. These integrals have to be transformed into a $2+1$ form so that they can be conveniently evaluated, e.g. as
$$\begin{array}{rcll}\int X\phantom{\rule{2.77695pt}{0ex}}A\phantom{\rule{0.3em}{0ex}}{d}^{2}S\phantom{\rule{0.3em}{0ex}}dt& & & \text{(2)}\text{}\text{}\end{array}$$
where ${d}^{2}S$ is the area element on the horizon cross section contained in $\Sigma $, and $dt$ is the coordinate time diﬀerential. The factor $A$ should contain the extra terms due to this coordinate transformation.
Starting from the $3$-volume element ${d}^{3}V$, let us ﬁrst decompose it into the $2$-volume element ${d}^{2}S$ and a “time” coordinate on the horizon, which we call $\sigma $. Note that $\sigma $ will generally be a spacelike coordinate for dynamical horizons. Let $Q$ be the induced $3$-metric on the horizon, and $q$ be the induced $2$-metric on the cross section. Then it is
$$\begin{array}{rcll}{d}^{3}V& =& \sqrt{detQ}\phantom{\rule{2.77695pt}{0ex}}d\mathit{\theta}\phantom{\rule{0.3em}{0ex}}d\varphi \phantom{\rule{0.3em}{0ex}}d\sigma & \text{(3)}\text{}\text{}\\ & =& \frac{\sqrt{detQ}}{\sqrt{detq}}\phantom{\rule{2.77695pt}{0ex}}{d}^{2}S\phantom{\rule{0.3em}{0ex}}d\sigma & \text{(4)}\text{}\text{}\end{array}$$
because ${d}^{2}S=\sqrt{detq}\phantom{\rule{0.3em}{0ex}}d\mathit{\theta}\phantom{\rule{0.3em}{0ex}}d\varphi $.
The coordinate time diﬀerential $dt$ and the diﬀerential $d\sigma $ will in general not be aligned because the horizon world tube will in general not have a static coordinate shape. It is
$$\begin{array}{rcll}d\tau & =& \left(cosh\alpha \right)\phantom{\rule{0.3em}{0ex}}dt+\left(sinh\alpha \right)\phantom{\rule{0.3em}{0ex}}ds& \text{(5)}\text{}\text{}\\ d\sigma & =& \left(cosh\alpha \right)\phantom{\rule{0.3em}{0ex}}ds+\left(sinh\alpha \right)\phantom{\rule{0.3em}{0ex}}dt& \text{(6)}\text{}\text{}\end{array}$$
where $s$ is a radial coordinate perpendicular to the horizon and also perpendicular to $t$, and $\tau $ is perpendicular to $\sigma $ and lies in the plan spanned by $t$ and $s$. $\tau $ and $\sigma $ are depend on $t$ and $s$ via a Lorentz boost. Thus we have
$$\begin{array}{rcll}\frac{d\sigma}{dt}& =& \left(cosh\alpha \right)\phantom{\rule{0.3em}{0ex}}\frac{ds}{dt}+\left(sinh\alpha \right)\phantom{\rule{0.3em}{0ex}}\frac{dt}{dt}& \text{(7)}\text{}\text{}\\ & =& sinh\alpha \phantom{\rule{1em}{0ex}}\mathrm{\text{.}}& \text{(8)}\text{}\text{}\end{array}$$
Putting everything together we arrive at
$$\begin{array}{rcll}\int X\phantom{\rule{2.77695pt}{0ex}}\frac{\sqrt{detQ}}{\sqrt{detq}}\phantom{\rule{2.77695pt}{0ex}}\left(sinh\alpha \right)\phantom{\rule{0.3em}{0ex}}{d}^{2}S\phantom{\rule{0.3em}{0ex}}dt\phantom{\rule{1em}{0ex}}\mathrm{\text{.}}& & & \text{(9)}\text{}\text{}\end{array}$$
Starting from
$$\begin{array}{rcll}{N}_{R}& =& \left|\partial R\right|& \text{(10)}\text{}\text{}\end{array}$$
we ﬁnd, since the radius $R$ changes only in the $\sigma $ direction,
$$\begin{array}{rcll}{N}_{R}^{2}& =& {g}^{\sigma \sigma}\phantom{\rule{0.3em}{0ex}}\left({\partial}_{\sigma}R\right)\phantom{\rule{0.3em}{0ex}}\left({\partial}_{\sigma}R\right)\phantom{\rule{1em}{0ex}}\mathrm{\text{.}}& \text{(11)}\text{}\text{}\end{array}$$
If we assume ${\partial}_{\tau}R=0$ and write ${\partial}_{t}R=\u1e58$, and use the relations between $\sigma $ and $t$ from above, we get
$$\begin{array}{rcll}\u1e58& =& {\partial}_{t}R& \text{(12)}\text{}\text{}\\ & =& \frac{\partial \tau}{\partial t}{\partial}_{\tau}R+\frac{\partial \sigma}{\partial t}{\partial}_{\sigma}R& \text{(13)}\text{}\text{}\\ & =& sinh\alpha \phantom{\rule{0.3em}{0ex}}{\partial}_{\sigma}R& \text{(14)}\text{}\text{}\end{array}$$
[NOTE: but ${\partial}_{t}\alpha \ne 0$.] and therefore
$$\begin{array}{rcll}{\partial}_{\sigma}R& =& \frac{1}{sinh\alpha}\phantom{\rule{2.77695pt}{0ex}}\u1e58\phantom{\rule{1em}{0ex}}\mathrm{\text{.}}& \text{(15)}\text{}\text{}\end{array}$$
Additionally we have ${g}^{\sigma \sigma}={g}^{ab}{\sigma}_{a}{\sigma}_{b}={g}_{ab}{\sigma}^{a}{\sigma}^{b}$ where ${\sigma}^{a}$ is the unit vector in the $\sigma $ direction, i.e.
$$\begin{array}{rcll}{\tau}^{a}& =& \left(cosh\alpha \right)\phantom{\rule{0.3em}{0ex}}{t}^{a}+\left(sinh\alpha \right)\phantom{\rule{0.3em}{0ex}}{s}^{a}& \text{(16)}\text{}\text{}\\ {\sigma}^{a}& =& \left(cosh\alpha \right)\phantom{\rule{0.3em}{0ex}}{s}^{a}+\left(sinh\alpha \right)\phantom{\rule{0.3em}{0ex}}{t}^{a}& \text{(17)}\text{}\text{}\end{array}$$
In order to use the IsolatedHorizon thorn on existing data (postprocessing), the following procedure is necessary.
Computing time-independent quantities.
The 3-metric and the extrinsic curvature must be available in HDF5 ﬁles.
This gives you the time-independent variables on the horizon, i.e., mostly the spin. It also allows you to look for apparent horizons if you don’t know where they are.
If data for the extrinsic curvature is not available, but those for the 3-metric, lapse, and shift for consecutive time steps are (that is, if you have data suitable for ﬁnding event horizons), then one needs to reconstruct the extrinsic curvature ﬁrst. There is a thorn AEIThorns/CalcK that helps with that. It reads the data for the 4-metric timestep after timestep, calculates the time derivative of the 3-metric through ﬁnite diﬀerencing in time, and then determines the extrinsic curvature from that, and writes it to a ﬁle. Once you have it, you can go on as above. CalcK has a small shell script that tells you what to do.
In general, things become more interesting if a static conformal factor is involved (since more variables are present), especially if it is output only once (since it is static), which means that one has to mix variables from diﬀerent time steps.
The thorns involved in this procedure have some examples. In general, this is NOT a “just do it” action; you have to know what you are doing, since you have to put the pieces together in your parameter ﬁle and make sure that everything is consistent. We may have a vision that you just call a script in a directory that contains output ﬁles and the script ﬁgures out everything else, but we’re not there yet. All the ingredients are there, but you’ll have to put them together in the right way. Think Lego.
2D output is given on a rectangular grid. This grid has coordinates which are regular and have a constant spacing in the $\mathit{\theta}$ and $\varphi $ directions. Cactus output has only grid point indices, but does not contain the coordinates $\mathit{\theta}$ and $\varphi $ themselves.
In gnuplot, one can deﬁne functions to convert indices to coordinates:
$$\begin{array}{rcll}\mathit{\theta}\left(i\right)& =& \left(i-g\mathit{\theta}+0.5\right)\ast \pi \u2215n\mathit{\theta}& \text{(18)}\text{}\text{}\\ \varphi \left(j\right)& =& \left(j-g\varphi \right)\ast 2\ast \pi \u2215n\varphi & \text{(19)}\text{}\text{}\end{array}$$
where $g\mathit{\theta}$ and $g\varphi $ is the number of ghost points in the corresponding direction, and $n\mathit{\theta}$ and $n\varphi $ the number of interior points. Here are the same equations in gnuplot syntax:
Usually, nghosts=2, ntheta=35, and nphi=72. i and j are is the integer grid point indices. Note that ntheta and nphi in the parameter ﬁle include ghost zones, while their deﬁnitions here do not include them. In general, nphi is even and ntheta is odd, because the points are staggered about the poles.
A test plot shows whether the plot is symmetric about $\pi \u22152$ in the $\mathit{\theta}$ and $\pi $ in the $\varphi $ direction. Also, plotting something axisymmetric with bitant symmetry vs. $\mathit{\theta}$ and vs. $\pi -\mathit{\theta}$, and vs. $\varphi $ and $2\pi -\varphi $, should lie exactly on top of each other.
There are also scalars origin/delta_theta/phi which one can use in the above equations. Then the equations read