On 5 Nov 2017, at 09:28, Nisa Amir <nisaa...@math.qau.edu.pk> wrote:
> Actually, when in the einsteinexact thorn there is a mathematica package
> named EinsteinExact. The documentation said that if you want to add some new
> space time add the space time in that mathematica package. I want to add non
> kerr which is in polar coordinates to that package, so that it generates a
> new thorn but I dont know how to do that.
> Kindly guide me.
Hi,
Yes, you can add a new spacetime to the Metrics package (in
Cactus/arrangements/EinsteinExact/m/Metrics/metrics). See the examples in
there. For example, Schwarzschild.m:
(* ::Package:: *)
{
"Name" -> "Schwarzschild",
"Description" -> "Schwarzschild spacetime",
"Dimensions" -> 4,
"Coordinates" -> {t, r, \[Theta], \[Phi]},
"Parameters" -> {M},
"Metric" -> {{-1 + 2 M / r, 0, 0, 0},
{0, 1/(1 - 2 M / r), 0, 0},
{0, 0, r^2, 0},
{0, 0, 0, r^2 Sin[\[Theta]]^2}},
"SignDet" -> -1
}
The EinsteinExact package then uses this information to generate an exact
solution thorn, which can be used for initial data in a numerical relativity
evolution. However, it only supports metrics which are explicitly given in
Cartesian coordinates, so for example, the Schwarzschild example won't work
with it (the Metrics package is generic, and is used also in other contexts,
outside EinsteinExact).
There is a check in arrangements/EinsteinExact/m/EinsteinExact.m that the
metric is in Cartesian coordinates. I suspect that the only reason for this
check is that it wouldn't know what Cactus gridfunctions to use for anything
else.
Since you want to do the evolution in polar coordinates, I assume that you are
going to have some mapping between x, y and z, and r, th and ph? I must admit
I have never tried to do something like this.
There are people on this list who have done evolutions in polar coordinates;
maybe one of them could give some advice about whether what you are trying to
do is feasible? Maybe you could give more details about what you are trying to
do?
PS: *please* include users@einsteintoolkit.org in the CC when you reply. If
you reply just to me, then nobody else benefits from the discussion, and nobody
else has the opportunity to help you.
--
Ian Hinder
http://members.aei.mpg.de/ianhin
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