# Difference between spatially homogeneous and isotropic?

1. Oct 23, 2006

### pivoxa15

What are their differences?

Spatially homogeneous is when there is uniform composition of space

Spatially isotropic is when you look anywhere, they look the same

Is it the case that one is visit anywhere, it is the same and the other is look anywhere they look the same?

They seem very similar but what are their differences?

2. Oct 23, 2006

### Meir Achuz

Isotropic means there is no angular dependence to what you see.

3. Oct 23, 2006

### lightarrow

If the composition of space is uniform in every region, then, uniformity clearly implies isotropy as well; but this situation is not necessary to apply Noether's Theorem.

When you write the lagrangian of the system, you have to choose the coordinates q_i, so you can have a traslational symmetry of the system with respect of one of these coordinates, but not with respect another.

Think, for instance, to a region or space made of two parallel different regions, each of which is uniform. If you choose one of the coordinates parallel to the separation between the two regions, when you change the value of that coordinate, you remain in one of the two regions, so the lagrangian doesn't change; it would change, instead, with respect to a coordinate which is perpendicular to the first, or with respect to the direction, that is, an angular coordinate, because in that case, you are forced to go from one region to another.

4. Oct 23, 2006

### pmb_phy

Consider an example, that of the dielectric constant of a crystal. In general this constant is a tensor quantity. In one direction the constant has one value and in another direction it has the same values. However these values will depend on the direction you're looking at and not on your location.

Best wishes

Pete

5. Nov 21, 2006

### Chris Hillman

Homogeneous vs. isotropic

Hi, pivoxa15,

The other respondents apparently assumed you were asking about the difference between homogeneous and isotropic -materials-, but is it possible that you were actually asking about homogeneous versus isotropic cosmological models?

If so, it might be helpful to see some simple cosmological models which lack one or both of these features.

Here is a simple exact dust solution in gtr, the plane symmetric case of the Kasner dust (1921):
$ds^2 = -dt^2 + \frac{(t+K)^2}{(t-K)^{2/3}} \, dx^2 + (t-K)^{4/3} \, \left( dx^2 + dy^2 \right),$
$K < t < \infty, -\infty < x, \, y, \, z < \infty$
This is a dust solution because the Einstein tensor matches the matter tensor of a pressureless perfect fluid ("dust") in which the world lines of the dust particles are timelike geodesics whose tangent vectors are given by the timelike unit vector field $$\partial_t$$. Thus, the lapse of coordinate time along one of these world lines agrees with the lapse of proper time as measured by an observer riding on the dust particle. (Cosmologists would say that we are using a "comoving chart".)

The self-isometries of this spacetime (considered as a Lorentzian manifold) form a four dimensional Lie group. These isometries, restricted to the hyperslice $$t=t_0$$, consist of combinations of translations in any direction and rotations about a vertical axis (parallel to $$\partial_z$$) around any point. That is, the z direction is "distinguished", so it is not isotropic. But the isometry group is transitive on the hyperslices, i.e. all points within each slice are equivalent, so it is spatially homogeneous.

Lorentzian manifolds which have one "slicing" have infinitely many. But the one we are using is preferred because each hyperslice is everywhere orthogonal to the world lines of the dust particles, which in turn form a distinguished timelike (geodesic) congruence because these dust particles are the source of the gravitational field (also, the only matter in sight).

There are many other well known cosmological models which are spatially homogeneous but not isotropic, such as the LTB dust models, the Kantowski-Sachs dusts, the Ellis-MacCallum dust, and so on.

Here is a simple exact dust solution which is not homogeneous, the Senovilla-Vera dust (1999):
$ds^2 = -dt^2 + \left(1 - \frac{t^2+r^2}{k^2} \right) \, dz^2 + dr^2 + r^2 \, d\phi^2,$
$t^2 + r^2 < k^2, \; -\infty < z < \infty, \; -\pi < \phi < \pi$
This model once again features dust particles whose world lines have tangent vectors of form $$\partial_t[/itex], so that again the lapse of coordinate time along one of these world lines agrees with the elapsed proper time as measured by an observer riding on the dust particle. These dust particles are once again the sole source of the gravitational field, and their contribution to the matter tensor matches the Einstein tensor, so we have an exact dust solution. The self-isometries of this spacetime form a two dimensional Lie group. These isometries, restricted to a hyperslice [tex]t=t_0$$, consist of combinations of translations along z and rotations about the vertical axis through any point. This group cannot possibly be transitive on a three dimensional submanifold (the hyperslice), however, since it is only two dimensional, so this model is not spatially homogeneous. Indeed, various curvature invariants show that the curvature is larger or smaller on average depending on how far you are from $$r=0$$.

There are other known dust solutions which can be written down in closed form which are inhomogeneous, including the Szekeres dust (1975). As you can see from the date, the Senovilla-Vera dust is a very recent discovery, suggesting that not all simple solutions have already been found.

Many but not all of the known dust solutions are generalizations of the best known cosmological models, the FRW dusts, which are both spatially homogeneous and isotropic. In fact, some of these give "nonlinear perturbations" of the FRW dusts and can serve as realistic models of the universe in which we live.

Chris Hillman

Last edited: Nov 21, 2006