Inhomogeneous and anisotropic dust models

[hellfire]

The cosmological models of Lemaitre-Tolman-Bondi describe spherically symmetric universes with isotropic but inhomogeneous space, i.e. there are concentric shells with different mean mass densities. The LTB line element is:

$$ds^2 = -dt^2 + \frac{(R'(t, r))^2}{1+2E(r) r^2}dr^2 + R^2(t, r)d\Omega^2$$

R' denotes derivate of R wrt r. I had two questions. First, how can this metric be derived? (e.g. it would be nice to have similar arguments as used to derive the homogeneous and isotropic FRW metric). Second, how would the metric look like if the observer is not located at the centre of the spheres but at a slight distance of it (leading to anisotropy)?

 

[Berislav]

First, how can this metric be derived? (e.g. it would be nice to have similar arguments as used to derive the homogeneous and isotropic FRW metric)

I don't know anything about this solution, but it seems to me that the authors derived the metric by imposing spherical symmetry on the metric:

$$ds^2=-f(r,t)dt^2+h(r,t)dr^2+r^2(d\theta^2+\sin^2 \theta d\phi^2)$$

The unknown functions were probably derived by imposing that the universe was spatially isotropic at the origin (through the rotational isometry of the metric at that point).

What is $R(t,r)$?

Second, how would the metric look like if the observer is not located at the centre of the spheres but at a slight distance of it (leading to anisotropy)?

An universe which can be called isotropic need not be isotropic at every point (which would mean that it was homogeneous, as well).
My guess is that metric should remain the same, otherwise it wouldn't be much of a metric.

 

[hellfire]

Thank you for your answer.

I don't know anything about this solution, but it seems to me that the authors derived the metric by imposing spherical symmetry on the metric:

$$ds^2=-f(r,t)dt^2+h(r,t)dr^2+r^2(d\theta^2+\sin^2 \theta d\phi^2)$$

The unknown functions were probably derived by imposing that the universe was spatially isotropic at the origin (through the rotational isometry of the metric at that point). [/tex]

I will think about this, but at first sight I am not able to figure out how to get the LTB line element from the expression you wrote above.

What is $R(t,r)$?

I guess this is something similar to the scale factor in the FRW metric.

An universe which can be called isotropic need not be isotropic at every point (which would mean that it was homogeneous, as well).
My guess is that metric should remain the same, otherwise it wouldn't be much of a metric.

You are right; the metric tensor would be the same. I mean, how would the line element look like in the coordinates of such an observer? (As there would be no spherical symmetry).

 

[Berislav]

I will think about this, but at first sight I am not able to figure out how to get the LTB line element from the expression you wrote above.

The rotational isometry of the metric at the origin, together with the stress-energy for those concetric shells should provide the necessary constraints to derive the metric exactly. This is easier said than done.

I guess this is something similar to the scale factor in the FRW metric.

In my opinion the solutions aren't complete without the expression for $R(t,r)$. This function should differ drastically from the scale factor in FRW cosmologies, since it must express the non-homgeneous nature of the space-time

You are right; the metric tensor would be the same. I mean, how would the line element look like in the coordinates of such an observer? (As there would be no spherical symmetry).

When I read your post I assumed that by line element you meant the notation used to represent the metric:

$$ds^2= \sum_{\mu, \nu} g_{\mu \nu}dx^\mu dx^\nu$$

This will not change, unless you change the coordinate basis.

 

[hellfire]

Altough I have not derived it yet, I think I can imagine now how to do it. Starting from the metric you wrote, scale t such that $$\inline g_{tt} = -1$$:

$$ds^2= - dt^2 + F(r,t)dr^2+ R(r, t)^2 d\Omega^2$$

Then, imposing Einstein’s equations for an isotropic pressureless dust with $$\inline T_{tt} = \rho(r, t)$$, should lead to the correct values of F and R. Actually, this is the same procedure used to fix the coefficients in the Schwarzschild metric, but I had wrongly expected that it can be derived making use of geometrical considerations only, as it is done with the FRW metric. (May be this was already clear to you).

When I read your post I assumed that by line element you meant the notation used to represent the metric:

$$ds^2= \sum_{\mu, \nu} g_{\mu \nu}dx^\mu dx^\nu$$

This will not change, unless you change the coordinate basis.

An observer which is not located at the center will make use of a different coordinate system (isotropy is not longer valid), therefore it will change.

 

[Berislav]

Then, imposing Einstein’s equations for an isotropic pressureless dust with LaTeX graphic is being generated. Reload this page in a moment., should lead to the correct values of F and R.

Yes, I believe that that would be sufficient to derive the metric.

I had wrongly expected that it can be derived making use of geometrical considerations only, as it is done with the FRW metric. (May be this was already clear to you).

It was clear to me that a dust-filled cosmology can not be solved without the stress-energy tensor, if that's what you mean. Also note that the scale factor in FRW cosmology was defined through the use of Einstein field equations.

An observer which is not located at the center will make use of a different coordinate system (isotropy is not longer valid), therefore it will change.

It is still unclear to me what you mean. Any difference in the line element between the two points will be defined by $F(r,t)$ and $R(r,t)$. If you want to change the coordinates you can to that simply if you can express a mapping between the old and new coordinates.

 

[hellfire]

It is still unclear to me what you mean. Any difference in the line element between the two points will be defined by $F(r,t)$ and $R(r,t)$. If you want to change the coordinates you can to that simply if you can express a mapping between the old and new coordinates.

I thought (intuitively) that the loss of isotropy would give a different form to that expression, but what you write seams correct. I will think about it. Thanks again.