<div dir="ltr">Hello everyone<div><br></div><div>Working from the general spherically symmetric form of a metric I verified the data is in fact isotropic. I appreciate both of your inputs on this, it helped a lot.</div><div><br></div><div>Nick</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Fri, May 27, 2022 at 3:53 PM Peter Diener <<a href="mailto:diener@cct.lsu.edu">diener@cct.lsu.edu</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">Hi Nick,<br>
<br>
If I understand your notation g_yy(0,1,0) correctly to mean g_yy along the <br>
y axis and g_yy(1,0,0) to g_yy along the x-axis, the statement you make is <br>
not correct. What you would expect in a spherically symmetric spacetime is <br>
that g_xx(1,0,0)=g_yy(0,1,0)=g_zz(0,0,1), i.e. gxx along x should be the <br>
same as g_yy along y and g_zz along z.<br>
<br>
Cheers,<br>
<br>
Peter<br>
<br>
<br>
On Fri, 27 May 2022, Nick Olsen wrote:<br>
<br>
> Hello Erik <br>
> I see what you're getting at here, but as I understand it the metric should<br>
> also be invariant under rotations and so for example gyy should not change<br>
> depending on the axis you look at. More explicitly, isotropy should have<br>
> that g_yy(0,1,0)=g_yy(1,0,0) or any similar combination, but the data I have<br>
> is showing otherwise. <br>
> <br>
> Nick<br>
> <br>
> On Fri., May 27, 2022, 2:33 p.m. Erik Schnetter, <<a href="mailto:schnetter@gmail.com" target="_blank">schnetter@gmail.com</a>><br>
> wrote:<br>
> Nick<br>
> Here is an example:<br>
> <br>
> Take the 3-metric ds^2 = a dr^2 + dθ^2 + (sin θ)^2 dϕ . It is<br>
> spherically symmetric.<br>
> <br>
> Along the z axis, you have gxx = gyy = 1, but there is gzz = a. The<br>
> metric tensor itself (as object in tangent space) is not spherically<br>
> symmetric. It is only spherically symmetric as object on the manifold.<br>
> <br>
> -erik<br>
> <br>
> <br>
> On Fri, May 27, 2022 at 10:56 AM Nick Olsen <<a href="mailto:n.olsen.3.711@gmail.com" target="_blank">n.olsen.3.711@gmail.com</a>><br>
> wrote:<br>
> Hello Erik <br>
> Forgive the late reply, it's been a busy few days. As I<br>
> understand things isotropic and spherically symmetric should be<br>
> the same thing in this case, with an isotropic metric taking the<br>
> form -a(r)^2 dt^2+b(r)^2 ds^2, so the fact that g_xx=/=g_yy and<br>
> its value depends on direction is what has me worried. <br>
> <br>
> Nicholas Olsen <br>
> <br>
> On Fri., May 13, 2022, 2:28 p.m. Erik Schnetter,<br>
> <<a href="mailto:schnetter@gmail.com" target="_blank">schnetter@gmail.com</a>> wrote:<br>
> On Thu, May 12, 2022 at 3:28 AM Nick Olsen<br>
> <<a href="mailto:n.olsen.3.711@gmail.com" target="_blank">n.olsen.3.711@gmail.com</a>> wrote:<br>
> Hello Everyone<br>
> I am running into a problem where I evolve a<br>
> Gaussian shell scalar field alongside the BSSN<br>
> equations using the Scalar/ScalarInit/ScalarBase<br>
> thorns, where the initial data is isotropic but<br>
> evolves to an anisotropic solution. More<br>
> specifically, along the x axis I have g_yy=g_zz and<br>
> along the z axis I have g_xx=g_yy, with g_xx along<br>
> the x axis equal to g_zz along the z axis, despite<br>
> having isotropic initial conditions. The point is<br>
> illustrated by the first image being the plot of<br>
> g_xx and g_zz along their respective axes at a later<br>
> time, and the rest of the diagonal metric values<br>
> being shown in the second image. Additionally, T_ij<br>
> shows a similar problem, where If so, T_xx along the<br>
> x axis and T_zz along the z axis are equal to<br>
> eachother, but not the rest of the diagonal entries<br>
> of T_ij, which are all equal.<br>
> <br>
> <br>
> Nils<br>
> <br>
> What you describe sounds isotropic.<br>
> <br>
> I assume that by saying "isotropic" you mean "spherically<br>
> symmetric", i.e. the solution only depends on the radius r<br>
> and not on the angles \theta or \phi.<br>
> <br>
> If so, then scalars should be the same in every direction,<br>
> vectors should point in the radial direction, and tensors<br>
> will look a bit more complicated. but "g_xx in the x<br>
> direction is the same as g_zz in the z direction" sounds<br>
> correct: If you rotate this tensor from the x to the z<br>
> axis, then you're essentially exchanging x and z<br>
> directions.<br>
> <br>
> The tensor itself does not need to remain spherically<br>
> symmetric. (From your description above it sounds as if<br>
> you assumed this was the case.)<br>
> <br>
> -erik<br>
> <br>
> <br>
> <br>
> gxxx.PNG gxxz.PNG<br>
> I have attached the parameter file used to get<br>
> these results, which is a modified version of<br>
> the test parameter file found in the Scalar<br>
> thorn bundle.<br>
> <br>
> Thanks,<br>
> Nicholas Olsen<br>
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> <br>
> <br>
> <br>
> --<br>
> Erik Schnetter <<a href="mailto:schnetter@gmail.com" target="_blank">schnetter@gmail.com</a>><br>
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> <br>
> <br>
></blockquote></div>