# [Users] Isotropic Initial Data Evolving Towards Anisotropy

Peter Diener diener at cct.lsu.edu
Fri May 27 14:53:22 CDT 2022

Hi Nick,

If I understand your notation g_yy(0,1,0) correctly to mean g_yy along the
y axis and g_yy(1,0,0) to g_yy along the x-axis, the statement you make is
not correct. What you would expect in a spherically symmetric spacetime is
that g_xx(1,0,0)=g_yy(0,1,0)=g_zz(0,0,1), i.e. gxx along x should be the
same as g_yy along y and g_zz along z.

Cheers,

Peter

On Fri, 27 May 2022, Nick Olsen wrote:

> Hello Erik
> I see what you're getting at here, but as I understand it the metric should
> also be invariant under rotations and so for example gyy should not change
> depending on the axis you look at. More explicitly, isotropy should have
> that g_yy(0,1,0)=g_yy(1,0,0) or any similar combination, but the data I have
> is showing otherwise.
>
> Nick
>
> On Fri., May 27, 2022, 2:33 p.m. Erik Schnetter, <schnetter at gmail.com>
> wrote:
>       Nick
> Here is an example:
>
> Take the 3-metric ds^2 = a dr^2 + dθ^2 + (sin θ)^2 dϕ . It is
> spherically symmetric.
>
> Along the z axis, you have gxx = gyy = 1, but there is gzz = a. The
> metric tensor itself (as object in tangent space) is not spherically
> symmetric. It is only spherically symmetric as object on the manifold.
>
> -erik
>
>
> On Fri, May 27, 2022 at 10:56 AM Nick Olsen <n.olsen.3.711 at gmail.com>
> wrote:
>       Hello Erik
> Forgive the late reply, it's been a busy few days. As I
> understand things isotropic and spherically symmetric should be
> the same thing in this case, with an isotropic metric taking the
> form -a(r)^2 dt^2+b(r)^2 ds^2, so the fact that g_xx=/=g_yy and
> its value depends on direction is what has me worried.
>
> Nicholas Olsen
>
> On Fri., May 13, 2022, 2:28 p.m. Erik Schnetter,
> <schnetter at gmail.com> wrote:
>       On Thu, May 12, 2022 at 3:28 AM Nick Olsen
>       <n.olsen.3.711 at gmail.com> wrote:
>       Hello Everyone
> I am running into a problem where I evolve a
> Gaussian shell scalar field alongside the BSSN
> equations using the Scalar/ScalarInit/ScalarBase
> thorns, where the initial data is isotropic but
> evolves to an anisotropic solution. More
> specifically, along the x axis I have g_yy=g_zz and
> along the z axis I have g_xx=g_yy, with g_xx along
> the x axis equal to g_zz along the z axis, despite
> having isotropic initial conditions. The point is
> illustrated by the first image being the plot of
> g_xx and g_zz along their respective axes at a later
> time, and the rest of the diagonal metric values
> being shown in the second image. Additionally, T_ij
> shows a similar problem, where If so, T_xx along the
> x axis and T_zz along the z axis are equal to
> eachother, but not the rest of the diagonal entries
> of T_ij, which are all equal.
>
>
> Nils
>
> What you describe sounds isotropic.
>
> I assume that by saying "isotropic" you mean "spherically
> symmetric", i.e. the solution only depends on the radius r
> and not on the angles \theta or \phi.
>
> If so, then scalars should be the same in every direction,
> vectors should point in the radial direction, and tensors
> will look a bit more complicated. but "g_xx in the x
> direction is the same as g_zz in the z direction" sounds
> correct: If you rotate this tensor from the x to the z
> axis, then you're essentially exchanging x and z
> directions.
>
> The tensor itself does not need to remain spherically
> symmetric. (From your description above it sounds as if
> you assumed this was the case.)
>
> -erik
>
>
>
>       gxxx.PNG gxxz.PNG
>       I have attached the parameter file used to get
>       these results, which is a modified version of
>       the test parameter file found in the Scalar
>       thorn bundle.
>
> Thanks,
> Nicholas Olsen
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>
>
> --
> Erik Schnetter <schnetter at gmail.com>
> http://www.perimeterinstitute.ca/personal/eschnetter/
>
>
>
> --
> Erik Schnetter <schnetter at gmail.com>
> http://www.perimeterinstitute.ca/personal/eschnetter/
>
>
>