# A Cosmological Principle: Trying to define large-scale homogeneity

1. Aug 22, 2013

### andrewkirk

I have been trying to pin down a precise definition of large-scale homogeneity, in the context of saying, per the Cosmological Principle, that all constant-time hypersurfaces (CTHs) of a foliation are large-scale homogeneous.

Here is my attempt:

Let M represent any coordinate-independent, physical quantity that can be measured for any sphere of given centre and given finite radius in a CTH.

Let MH(P,x) be the value of M obtained from measuring M on a sphere of radius x, centred on P, in H.

Let LMH(P) = limit as x-> infinity of MH(P,x), if the limit exists. If, for given M, H and P and any e>0, there exists Ne such that x>Ne => MH(P,x)>1/e then we say that LMH(P) = infinity.

Then we say that a CTH H is “large-scale homogeneous” iff,

(1) for any M and P, LMH(P) exists (including that it may be infinity) and
(2) for any M and any points P1 and P2 in H, LMH(P1)=LMH(P2).

This definition still isn’t exactly what I’d like because it doesn’t rule out trivial, local-focused definitions of M such as MH(P,x) = 1 if there is a proton within one micron of x, otherwise zero. But I couldn’t think of an easy, concise way of ruling out such definitions (suggestions would be much appreciated!), so I’ll live with that bit of ambiguity for the time being. The sorts of M I imagine wanting to use are related to average density of mass-energy in the sphere. It could be mass-energy generally, or a particular form such as just protons. Alternatively it could be something like average pressure, average entropy or average Ricci scalar.

I would appreciate comments on this definition, especially:
- is it well-defined (unambiguous)?
- is it too narrow? (does it rule out CTHs that we would wish to regard as large-scale homogeneous?)
- is it too broad? (does it allow CTHs that we would not wish to regard as large-scale homogeneous)
- is there a reasonably concise way of ruling out annoying locally-focused measurements M like the above example with the proton?

2. Aug 26, 2013

### andrewkirk

Is there really nobody out there that has ever wondered exactly what the Cosmological Principle says? That would surprise me, because the question struck me quite forcefully the first time I read the principle.

The wikipedia article demonstrates the problem quite clearly: http://en.wikipedia.org/wiki/Cosmological_principle
Attempts it makes at stating the principle at different places in the article are:

'Viewed on a sufficiently large scale, the properties of the Universe are the same for all observers'
[But what is a sufficiently large scale? What properties are we talking about and what does it mean to say a 'property' is the same? Does it mean observers will make the same measurements? But we know they won't, because of local variation and experimental error for a start]

'the part of the Universe which we can see is a fair sample'
[what is a 'fair sample'?]

'Homogeneity means that the same observational evidence is available to observers at different locations in the universe '
[that is just wrong. Very few, if any, other parts of the universe would give exactly the same view of the surrounding observable universe as we obtain from the Earth. To make it work a clear definition of what sort of observational evidence we are talking about is needed and we need to determine what are our tolerances for two lots of evidence being 'the same']

In practice, it seems we do cosmology by literally assuming that the universe is a perfectly homogeneous, isotropic soup of mass-energy with no local variation or pattern of any kind. That's fine for imagining different kinds of hypothetical universe (flat, elliptic, hyperbolic) but we can't relate those hypothetical universes to the one we can observe unless we can clearly state in what sense ours is like the perfectly homogeneous, isotropic one. The Cosmological Principle is supposed to do that but, as stated, it's so vague than one cannot know what it is saying.

3. Aug 26, 2013

### WannabeNewton

Homogeneity and isotropy are perfectly mathematically well defined properties of space-time manifolds. See chapter 5 of Wald "General Relativity".

Let $(M,g_{ab})$ be a space-time and $\xi^a$ a smooth, future directed, normalized time-like vector field on $M$ that is twist free i.e. $\omega^a = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d} = 0 \Leftrightarrow \xi_{[a}\nabla_{b}\xi_{c]} = 0$. Then there exists a one-parameter family of space-like hypersurfaces $\Sigma_{t}$ foliating $M$ that are everywhere orthogonal to $\xi^a$. At each $t$, $\Sigma_t$ represents space relative to $\xi^a$. Then $(M,g_{ab})$ is spatially homogenous and isotropic if for all $p\in M$ and all unit space-like vectors $\sigma^a_1, \sigma^a_2 \in T_p M$ orthogonal to $\xi^a$, there exists a neighborhood $U$ of $p$ and an isometry $\varphi: U \rightarrow U$ which keeps $p$ fixed, preserves $\xi^a$, and takes $\sigma^a_1$ into $\sigma^a_2$.

One can then show that $\xi^a$ is also shear-free, meaning that $\nabla_{a}\xi_{b} = \frac{1}{3}h_{ab}\theta$, and that $\nabla_{a}\theta = (\xi^{b}\nabla_{b}\theta)\xi_{a}$ where $h_{ab}$ is, as usual, the spatial metric relative to $\xi^a$ and $\theta = \nabla_{a}\xi^{a}$ is the expansion. From this, one can derive the Friedman equations and the Friedman metric in a completely coordinate-free manner. In this context, the integral curves of $\xi^a$ represent the worldlines of observers comoving with the Hubble flow.

Last edited: Aug 26, 2013
4. Aug 26, 2013

### cristo

Staff Emeritus
The scale of homogeneity is roughly 100Mpc. That is, on scales >100Mpc the universe appears to be extremely close to homogeneous.

5. Aug 26, 2013

### andrewkirk

Thank you WannabeNewton. The definition you have quoted looks like a neat definition of homogeneity and isotropy. Unfortunately it cannot be used for the cosmological principle because the universe does not satisfy it. We know that the universe is not homogenous or isotropic. For instance if we take $p$ on the surface of the Earth, $\sigma^a_1$ pointing towards the centre of the Earth and $\sigma^a_2$ pointing circumferentially then I am pretty sure there is no neighborhood $U$ of $p$ with an isometry $\varphi: U \rightarrow U$ that keeps $p$ fixed, preserves $\xi^a$, and takes $\sigma^a_1$ into $\sigma^a_2$.

What is missing is some reference to scale, that captures the idea that, although the universe is neither homogenous nor isotropic, it in some sense starts to approximate that once we look at it on a large enough scale. I would expect the statement to contain some definitions using limits (as the scale approaches infinity), although perhaps there may be another way to do it.

6. Aug 26, 2013

### WannabeNewton

Yes certainly it is not an exact property, but rather a brilliant approximation on sufficiently large scales. You might be interested in the following: 403 Forbidden

7. Aug 26, 2013

### andrewkirk

Thanks for the link to that article WannabeNewton. It looks very relevant, and easy to read. I look forward to working my way through it.

Meanwhile, I was just wondering about the criterion for the vector field $\xi^a$ to be twist-free. It's probably not central, as it seems to be part of just describing a classic foliation like the FLRW, but nevertheless I would like to understand it. It may just be that the notation convention differs from that with which I am familiar, which is from Schutz.

The bits I don't get are

1. what is $\epsilon$?
2. what is the meaning of the square brackets in the subscripts in $\xi_{[a}\nabla_{b}\xi_{c]}$?

Last edited: Aug 26, 2013
8. Aug 26, 2013

### WannabeNewton

Hi andrew! I am familiar with Schutz text and unfortunately he never delves into the hydrodynamical formulation of GR so you might not have seen things like twist, expansion, and shear before but I can explain twist so it's no issue. $\epsilon_{abcd}$ is just the natural volume element on space-time induced by the metric tensor $g_{ab}$; it is totally antisymmetric i.e. it is antisymmetric in all of its indices. It is defined (up to a sign) by $\epsilon^{abcd}\epsilon_{abcd} = -4!$.

The square bracket in $\xi_{[a}\nabla_{b}\xi_{c]}$ picks out the antisymmetric part of $\xi_a \nabla_b \xi_c$ by permuting each index and negating after each permutation of an index: $\xi_{[a}\nabla_{b}\xi_{c]} = \frac{1}{3!}(\xi_{a}\nabla_{b}\xi_{c} - \xi_{a}\nabla_{c}\xi_{b} + \xi_{b}\nabla_{c}\xi_{a} - ...)$. $\xi_{[a}\nabla_{b}\xi_{c]} = 0$ in index-free notation is just $\xi \wedge d\xi = 0$.

$\omega^a = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d}$ is the definition of the twist 4-vector $\omega^a$ yes. Notice that it is a purely space-like vector. The twist has a nice physical interpretation. Choose any particle in the time-like congruence and imagine an observer comoving with this particle. The observer carries with him a clock and three mutually perpendicular gyroscopes that are non-rotating (i.e. Fermi-Walker transported, if you've seen that before). If $\omega^{a} \neq 0$ then the observer comoving with the chosen particle will see infinitesimally nearby particles in the congruence rotate relative to his gyroscopes. So the twist codifies a local "curling" of neighboring particles around a given particle. The reason I use the word "curl" is due to the following: consider a locally inertial frame momentarily comoving with the particle. In such a frame, the components of the twist become $\omega^0 = 0$ and $\omega^{i} = \epsilon^{i0nm}\partial_{n}\xi_{m} = \partial_{j}\xi_k - \partial_k\xi_j$ where $j\neq k\neq i$ are all fixed spatial indices. So in a momentarily comoving locally inertial frame, the twist is just the usual curl from vector calculus.

A congruence is (at least locally) hypersurface orthogonal if and only if $\xi_{[a}\nabla_{b}\xi_{c]} = 0$. It can be shown that $\xi_{[a}\nabla_{b}\xi_{c]} = 0$ if and only if $\omega^{a} = 0$ i.e. if the congruence is irrotational. So a twist free time-like congruence is one for which we can find a one-parameter family of space-like hypersurfaces foliating the space-time such that the congruence is (at least locally) orthogonal to all the hypersurfaces.

9. Aug 28, 2013

### andrewkirk

Thanks for your notes. I've been reading up on Grassman algebras and wedge products to understand the twist. It's a bit of a diversion from homogeneity but it's good to take the opportunity to understand wedge products, which have been a known hole in my understanding for a long while.

Can you please explain the following?
By 'purely space-like' I assume you mean that $\omega$ is orthogonal to $\xi$. It is not obvious to me why that is so. I assume it's something to do with the fact that $\xi$ is time-like and that there are spacelike hypersurfaces orthogonal to $\xi$, but I'm a bit rusty and can't see the steps to demonstrate the link. Can you show why the orthogonality follows?

Thank you.

10. Aug 28, 2013

### WannabeNewton

Sure! $\epsilon^{abcd}$ is totally antisymmetric so, in particular, it is antisymmetric in $a$ and $b$. Hence $\omega^a \xi_a = \epsilon^{abcd}\xi_a \xi_b \nabla_c \xi_d = 0$ because we are contracting two antisymmetric indices with two symmetric indices, the latter coming from the fact that $\xi_a \xi_b = \xi_b \xi_a$.

11. Sep 3, 2013

### andrewkirk

I've been thinking some more about capturing the 'large scale' aspect of the Cosmological Principle. I tried for a while to adapt the Wald definitions, but gave up in the end because they are all defined pointwise, and so do not seem capable of being extended so as to identify a manifold that fails the definition of strict homogeneity but is homogeneous 'at the large scale'. I feel that we are going to have to use an integral in the definition, to average over large enough volumes.

The Wald definition clarifies one important idea, which is that statements of homogeneity are only about the shape of the manifold. Other non-gravitational aspects such as whether the mass in a volume is principally neutrons, protons or their anti-matter counterparts are invisible to the definition. It seems to be implying that interactions based on fundamental forces (EM, strong, electroweak) are ignored in cosmology. If that's not right I would appreciate correction.

This encourages me to make the definition in the OP more specific. Rather than requiring any physical quantity whatsoever (represented by M in the OP) to be almost invariant between large enough sample volumes, we can define M to be the density of mass-energy, if that's the only thing that matters to cosmology. Again I would appreciate correction if that's wrong.

At first I thought we couldn't measure the average mass-energy in a volume because mass-energy is represented by a $(^2_0)$ tensor rather than a scalar, and hence any integration of just the energy density component of that tensor will be coordinate dependent. But I think the foliation saves us from that. We choose a coordinate system for which the time basis vector is parallel to the foliation time vector and the three spatial basis vectors are orthogonal to that. That makes $T^{\alpha\beta}$ take the form of a diagonal block arrangement of a (1,1) and a (3,3) matrix. The (1,1) part is the energy density that we want to measure, and it is invariant between different choices of basis for the (3,3) spatial part.

That enables us to define large-scale homogeneity as follows:

First define:
$H_t$ as the hypersurface of constant time t in the foliation.

$d(t,P,Q)$ is the length of the shortest path in $H_t$ from point $P$ to point $Q$, both of which are in $H_t$. The length of path is of course determined by integration using the metric tensor.

$S(t,P,r)$ is the 'spherical' set of all points $Q\in H_t$ such that $d(t,P,Q)<r$.

Let $M(t,P,r)=\frac{\int_{S(t,P,r)}T^{00}dV}{ \int_{S(t,P,r)}dV}$ where $dV$ is the volume element in $H_t$. $M(t,P,r)$ is the average mass-energy density in the 'sphere' $S(t,P,r)$. As discussed above, this is well-defined because $T^{00}$ is fully determined by the foliation, and invariant to the spatial coordinate system.

Let $LM(t,P) = lim_{r\rightarrow\infty} M(t,P,r)$, if the limit exists. This is the average 'large-scale' density of mass-energy around point $P$.

Then we say that $H_t$ is “large-scale homogeneous” iff:

(1) for every $P\in H_t$, $LM(t,P)$ exists and
(2) for any points $P$ and $Q$ in $H_t$, $LM(P)=LM(Q)$.

The 'homogeneity' part of the Cosmological Principle then states that there is a foliation with the required properties (eg twist-free time-like vector field $\xi$), under which, for all times $t$, the hypersurface $H_t$ is "large-scale homogeneous".

What do you think?

Last edited: Sep 3, 2013
12. Sep 8, 2013

### andrewkirk

It occurs to me that the preceding definition is not strong enough, as it would characterise the following spacetime as large-scale homogeneous, and I don't think we'd want to do that.

The spacetime AltH is foliated into infinite, flat, constant-time hypersurfaces (CTHs), each of which is made up of alternating, parallel, flat slabs of emptiness and constant-density dust, where the slabs get thicker as they move away from an origin point. Adopting a Cartesian coordinate system for each CTH, we describe this formally as follows:

$T^{00}(t,x,y,z) = \rho$ if $int(\frac{log(|x|+1)}{a(t)})\$mod$\ 2 = 0$, otherwise $0$.​

Here $int(x$) is the largest integer less than or equal to $x$ and $a(t)$ is a time-based scale parameter, which we do not need to specify, as it doesn't affect our results.

This formula makes the width of the slabs in $H_t$ increase exponentially as we move away from the origin in either the positive or negative $x$ direction. The consequence of this is that the scale needed to approximate homogeneity becomes increasingly large as we move away from the origin. Certainly, for every point $P\in H_t$ the limit $LM(t,P)$ exists and is equal to $\rho/2$, but the size of 'sphere' $S(t,P,r)$ needed to achieve an average density around $P$ that is sufficiently close to $\rho/2$ increases exponentially as $P$ moves away from the origin.

It's a little like a function that is continuous but not uniformly continuous. What we want is 'uniform homogeneity' so that not only is the limit criterion in the previous post met, but also the rate of convergence to homogeneity as we increase scale is the same throughout the universe.

With that in mind, here is my revised definition of large-scale homogeneity that achieves that:

The first few definitions are identical to previously:

First define $H_t$ as the hypersurface of constant time t in the foliation.

$d(t,P,Q)$ is the length of the shortest path in $H_t$ from point $P$ to point $Q$, both of which are in $H_t$. The length of path is of course determined by integration using the metric tensor.

$S(t,P,r)$ is the 'spherical' set of all points $Q\in H_t$ such that $d(t,P,Q)<r$.

Let $M(t,P,r)=\frac{\int_{S(t,P,r)}T^{00}dV}{ \int_{S(t,P,r)}dV}$ where $dV$ is the volume element in $H_t$.
$M(t,P,r)$ is the average mass-energy density in the 'sphere' $S(t,P,r)$. As discussed above (in the first part of post #11), this is well-defined because $T^{00}$ is fully determined by the foliation, and invariant to the choice of spatial coordinate system.

Now this next paragraph is the only different bit. Mercifully , it's shorter than the previous definition.

Then we say that $H_t$ is “large-scale homogeneous” iff $\exists \rho\geq 0$ such that, for every $\epsilon>0$ there exists $R_\epsilon>0$ such that, for every $P\in H_t$, $r\geq R_\epsilon\Rightarrow |M(t,P,r)-\rho|<\epsilon$.

This definition will not class the above example AltH as homogeneous because, given an $\epsilon<\rho/2$, no matter what value of $R_\epsilon$ we choose, there will be points $P$ sufficiently far away from the origin that the surrounding 'sphere' $S(t,P,R_\epsilon)$ is either entirely within a dust slab or entirely within an empty slab, giving $M(t,P,R_\epsilon) = \rho$ or $0$, so that $|M(t,P,r)-\rho|>\epsilon$.

Last edited: Sep 8, 2013
13. Sep 10, 2013

### andrewkirk

Whoops! I just realised that I have used $H_t$ instead of $\Sigma_t$ in my above posts for the constant-time hypersurface (CTH) at time $t$. this is different from the terminology introduced by WannabeNewton in post #3 and may cause needless confusion, and now it's too late to edit my posts and change $H_t$ to $\Sigma_t$. Never mind. They mean the same thing. I'll use $\Sigma_t$ from now on, and I'll use WannabeNewton's trerminology for the other things as well.

We can use a variant of the AltH manifold defined in the previous post, to demonstrate a spacetime that is large-scale homogeneous but not large-scale isotropic. Let's call it AltI. It's like AltH except that all slabs have the same width $a(t)$. That is:

$T^{00}(t,x,y,z) = \rho$ if $int(\frac{x}{a(t)})\$mod$\ 2 = 0$, otherwise $0$.​

Consider the point $P=(t,\frac{3a(t)}{2},0,0)\in \Sigma_t$. This is in the middle of an 'empty slab'. A straight line through $P$ parallel to the $y$ axis will have $T^{00}=0$ (zero mass-energy density) all the way along its infinite length, because it remains in the same empty slab all the way. However, a straight line through $P$ parallel to the $x$ axis will have $average(T^{00})=\frac{\rho}{2}$, because it passes through an infinite number of empty and non-empty slabs. So $\Sigma_t$ is not isotropic at any scale because the average densities along the directions of the $x$ and $y$ axes are different at $P$, no matter how far we pursue those directions.

But the CTHs of AltI meet the definition of large-scale homogeneity because all slabs are the same width, so the average density in a large enough sphere will approach $\frac{\rho}{2}$. So AltI is large-scale homogeneous but we would not wish to say it is large-scale isotropic.

Here is a definition of large-scale isotropy that will exclude AltI, and any other spacetime I can think of that we would wish to exclude.

Let $\gamma_{\vec{v}}:[0,\infty)\rightarrow \Sigma_t$ be the unique spatial geodesic (meaning a geodesic of the sub-manifold $\Sigma_t$, not a geodesic of $M$) through $P$ with initial velocity $\vec{v}$ at $P$.

Define $\mu:[0,\infty) \rightarrow R$ by $\mu(s) = \sqrt{g_{ij}(\dot{\gamma}_\vec{v})^i(\dot{\gamma}_\vec{v})^j}|_{\gamma_\vec{v}(s)}$

then define $K(t,P,\vec{v},r)=\frac{\int_0^r T^{00}(\gamma(s))\mu(s)ds}{\int_0^r\mu(s)ds}$. This is the average density along the spatial geodesic from $P$ to $\gamma(r)$.

Then we say that $\Sigma_t$ is “large-scale isotropic” iff $\exists \rho\geq 0$ such that, for every $\epsilon>0$ there exists $R_\epsilon>0$ such that, for every $P\in \Sigma_t$ and every $\vec{v}\in T_PM$ orthogonal to $\vec{\xi}$ (recall that $\vec{\xi}$ is the time-like vector field that determines the foliation), $r\geq R_\epsilon\Rightarrow |K(t,P,\vec{v},r)-\rho|<\epsilon$.

That is, for any point $P$ in $\Sigma_t$, the average density along a geodesic curve of $\Sigma_t$ in any direction away from $P$ approaches a limit of $\rho$ as the curve length increases, and a uniform lower bound on the rate of convergence can be achieved for all points in $\Sigma_t$.

Some observations

The definition of perfect homogeneity and isotropy in WannabeNewton's post #3 above looks like a definition of just perfect isotropy, as it is about directions. My guess is that that's all that's necessary because perfect homogeneity would follow from perfect isotropy. A simple definition of perfect homogeneity would be that $T^{00}(P)$ is the same for every point $P\in\Sigma_t$

So it seems that perfect isotropy entails perfect homogeneity. I suspect that perfect homogeneity also entails perfect isotropy, so that the two conditions are equivalent, but I don't have a proof ready to hand. Does anybody have a proof or counter-example?

With large-scale homogeneity and isotropy, the spacetime AltI demonstrates that we can have large-scale homogeneity without large-scale isotropy. So unlike in the 'perfect' case, large-scale isotropy and large-scale homogeneity are not equivalent. However I feel (but again don't yet have a proof) that large-scale isotropy probably entails large-scale homogeneity. Does anybody have a proof or counter-example?

14. Sep 10, 2013

### Chronos

At some scale, it is difficult to avoid the conclusion the universe is homogenous and isotropic. Is it true on the scale of the observable universe? Unknown. So far it looks that way.