# Where are the real definitions of homogeneity and isotropy?

1. Sep 29, 2005

### honestrosewater

What do homogeneity and isotropy mean as properties of the universe?

[skippable complaining]It seems like a pretty basic question, but I can only find analogies or short and sweet 'definitions' with no context to give them meaning (which aren't necessarily bad - just not helping me). To give you an idea of how lost and frustrated I am, here are some questions that I still can't answer. To what do homogeneity and isotropy apply? What properties are the same at every point? Direction in what - space, time? How do observers fit in? What are observers? And what exactly is observation? I've read only of qualitative observations. Don't you measure anything? Can a non-homogeneous 'thing' be isotropic, or 'appear' isotropic, at some points but not others? If so, is the thing isotropic or not? If isotropy and homogeneity apply to, eh, every 'point' or 'observer', I think homogeneity implies isotropy, though I don't really know what either of them are or what they apply to or how. And I'm just making it worse by trying to extend analogies. Is the surface of a cone isotropic from the tip? I don't even know if that question makes sense. And how does homogeneity apply to the surface of a cone? Is there supposed to be some kind of pattern of the surface? Okay, sorry, I'll stop.[/skippable complaining]

I'd love some precise definitions. Does anyone know where I can find some? Lay it on me - I'm not afraid. Please.

Oh, and I'm learning this for fun () - I don't have a textbook, but I'll pick one up if I must. google has failed me for the second time ever. And I couldn't find anything in the sticky above.

Last edited: Sep 29, 2005
2. Sep 29, 2005

### hellfire

3. Sep 29, 2005

### pervect

Staff Emeritus
In the context of General Relativity as applied to cosmology, homogeneous and isotropic basically means that the stress-energy tensor is homogeneous (the same at all points in space), and isotropic (radiation pressure is the same in all directions).

You can pretty much think of the stress-energy tensor as having three important terms - a matter density term, a pressure term which is due to energy (the cosmic microwave background), and a cosmological constant which tells us how much empty space contributes to gravity.

In the universe were not isotropic, there would be other terms of the stress-energy tensor that described the average motion of the matter in the universe - but since there isn't any preferred direction for matter to move in, these terms in the stress-energy tensor are zero.

Because the stress-energy tensor is homogeneous and isotropic, the various curvature tensors which describe the curvature of space-time are also homogeneous and isotropic.

Obviously the real universe has "lumps" - for the broad overview of cosmology, the contribution of lumps (like stars, galaxies, clusters of galaxies) are averaged out.

4. Sep 29, 2005

### honestrosewater

Okay, I think I get it. I just can't get a definition. For simplicity, let f be a function from N (points) to N2 (properties). f is homogeneous iff the range of f has exactly one member. Isotropy is then a property of the range of f? The range of f is isotropic iff for every (x, y) and (z, w) in the range of f, (x, y) = (z, w) and x = y. ?? Eh, I'm mostly going on isotropy implying homogeneity. I don't know, something is bugging me. Whatever, maybe some can see what I'm trying to do. I'll look at it again soon. I don't know what a stress-energy tensor is (I don't know what a tensor is). I'll look them up later if I can't figure this out.
Thanks. (I had read the thread with the forest analogy a while ago. It did make more sense this time, at least. )

5. Sep 30, 2005

### SpaceTiger

Staff Emeritus
To be honest, I don't think there is a universal definition of these terms, but I can share my understanding.

In perfect homogeneity and isotropy, the universe is exactly the same in every measureable way at every point and in every direction. This obviously isn't the case, so in practice it applies to the universe only on the largest scales. That is, if you average any property over, say, 100 Mpc3, you'll find that it's essentially equivalent to the property in another random 100 Mpc3. Generally, the property we're most concerned about is the density, either of ordinary matter or some other entity (radiation, vacuum energy, etc.). Applying homogeneity and isotropy to these things, in turn, leads to a universe with the general relativistic properties pervect described.

Space. There are several things in physics (like the second law of thermodynamics) that indicate a preferred direction in time. If the Big Rip or Heat Death hypotheses are correct, then cosmology also indicates a rather dramatic asymmetry in time.

This is a bit redundant, but note also that the cosmological principle does not apply to inertial frames. Although the laws of physics are the same in every inertial frame, the universe will appear different. The CMB, for example, will exhibit a dipole that depends on your velocity.

Unlike with quantum mechanics, there's nothing funny going on with the observer here. When one talks about an observer in this context, they mean only someone or something that can measure the properties of the universe from a given vantage point.

No.

Yes. Imagine a hill with the density of trees increasing as one moves down. The lack of constant density implies that it's not homogeneous, but it still appear isotropic from the top of the hill. It does not, however, appear isotropic from every point of view.

Isotropy implies homogeneity, but not the other way around. One can construct a universe for which every measurable property is the same at every point, but we see different things when we look in different directions. In other words, everything would be locally equivalent, but our view of other events in spacetime would depend on the direction in which we looked.

6. Oct 1, 2005

### honestrosewater

Okay, I don't think I know enough to understand this in the same way that you guys understand it. I don't know half of the terms used to define tensor; I don't know half of the terms used to define those terms, and so on - I don't know a whole heap of terms. I had to look up Mpc. I might learn all of that eventually, but if anyone could step down to my level for a moment and help me understand it in my own way, well, I would be very happy and grateful. That's all I can think of to offer now.

The most useful statement I've seen is isotropy implies homogeneity (I like those kind of statements). So taking it, SpaceTiger's forest analogy, and the view that everything is just some kind of set...

Let g be a function from Sm to Tn, where m and n are integers > 0 (I suppose the members of S and T are numbers, but they could possibly be other things); Sm is a set of points, each point being specified by m coordinates (or perhaps something to do with m dimensions?); Tn is a set of values of n properties (i.e., you measure something and get a number; the order of the members of the n-tuples tells you which properties they are. For instance, if it makes sense for a point to have density, you could assign density the first position; 'the presence of a tree' the second position, maybe a 1 if a tree is present, 0 otherwise - whatever). I hope a finite number of properties is enough? If not, I suppose you could take a finite number at a time and just have an infinite number of functions. I hope assigning each point exactly one value for each property isn't a problem either. I imagine Sm is usually R3, but I'm not used to working in it, so I'll use something simpler.

So g assigns values of properties to all of your points. You then do everything else - measure in different directions, change scales, etc. - by specifying and working with subsets of g.
For example, I could interpret SpaceTiger's three forests, minus the trees, as assigning a (real) height to each point in a plane.

i) flat. f: R2 --> R, f(x, y) = c, where c is a constant in R.

ii) inclined plane. i: R2 --> R, i(x, y) = x (right?)

iii) hill. h: R2 --> R, h(x, y) = -|x| (or maybe h(x, y) = -(x2)? I don't know how to make a hill - it would slope in all directions?)

For, say, f, measuring in the positive direction along the x-axis from a point (a, b), or 'looking', say, North along a line, could be represented by (the restriction of f to N)

N = {((x, b), f(x, b)) : ((x, b), f(x, b)) in f and x > a}

or maybe you wouldn't want to include a - whatever. Looking in the opposite direction, South, would be

S = {((x, b), f(x, b)) : ((x, b), f(x, b)) in f and x < a}

And similarly for other directions. You then determine whether f is isotropic or homogeneous by comparing these subsets. I don't actually know yet what conditions must be met for either. It may be, in this case, that a constant function is the only isotropic kind. Is it? Hint?
i is homogeneous and non-isotropic, right? Is j: R2 --> R, j(x, y) = x3 also homogeneous and non-isotropic? I'll guess that j is non-homogeneous, but it's not a very confident guess.

Alright, I should probably stop now. Just one last thing: I don't know what a 'higher dimension' interval would be, but if you wanted to use a different scale (averaging the properties) for, say,

k: R --> Rn

I imagine you would just define the subsets with intervals:

Irp = {((s, (t1, ..., tn)), k(s, (t1, ..., tn))) : ((s, (t1, ..., tn)), k(s, (t1, ..., tn))) in k and s in [r, (r + p)]}

p, r, s, t are reals. Maybe you want open or half-open intervals - whatever.

Well, bless you for reading this. Does it make sense to anyone?

__
Just remembered: For the forests, I was thinking that homogeneity just requires 'straightness', with or without an 'incline'. A constant function is a special case of this (no incline). So this works for isotropy implying homogeniety. But the hill (esp. the top) and j: R2 --> R, j(x, y) = x3 confuse me.

Last edited: Oct 1, 2005
7. Oct 1, 2005

### SpaceTiger

Staff Emeritus
Alright, let me try this from a different point of view. Homogeneity basically implies:

$$F(\vec{x})=F(\vec{x'})$$

for all $\vec{x}, \vec{x'}$ in the universe. The function F is any measurable property of a local region of spacetime and x is the spatial position. Note that

$$F(\vec{x},t)\ne F(\vec{x'},t')~if~t \ne t'$$

where t is proper time since the Big Bang.

Isotropy is a bit more complicated, but boils down to:

$$f(\vec{r},\vec{\theta})|_{\vec{x}}=f(\vec{r},\vec{\theta'})|_{\vec{x}}$$

for all $\vec{r}, \vec{x}, \vec{\theta}, \vec{\theta'}$ in the universe. In this context, $\vec{\theta}$ is the angle in which the observer is looking, $\vec{r}$ is the distance along that sightline, and f is any measurable property of those points in spacetime.

Several things. Firstly, notice that the definition of homogeneity includes equating measurables in the primed and unprimed coordinate systems, while that of isotropy only states that the condition is true from all vantage points, but does not compare them in any way. This would seem to make the second condition weaker, but in actuality, it ends up comparing different points in space via the function, f.

Second, the isotropy definition includes looking at different times because the finite speed of light prevents us from viewing the universe as it is at one point in time. However, since only the radial coordinate is dependent on time and I did not specify that the functions, f, are equal for all $\vec{r}, \vec{r'}$, this does not imply homogeneity in time.

There's more I can say, if you like, but I wanna check to make sure that this is making sense before I do.

8. Oct 2, 2005

### honestrosewater

I don't understand what you're saying yet, but it sounds like what I was looking for. So I'm working on it - it'll just take me some time.
Is it crucial that I understand proper time? I think it would take me at least a day to learn enough about relativity to get any kind of grasp on it. I think I've got everything else covered except for
- primed and unprimed coordinate systems; I couldn't find any definitions, but it sounds like a transformation - an isometry?
- what does $|_{\vec{x}}$ mean? I can look it up if you can state it in words.

9. Oct 2, 2005

### honestrosewater

Is this correct:
? Is the 'closed system' part significant (because I don't know what it means in this context)?
Sorry, I'm not trying to be a pain, and I do really appreciate you help.

Last edited: Oct 2, 2005
10. Oct 2, 2005

### SpaceTiger

Staff Emeritus
No, all that's important is that you understand that our observations of distant objects can only be done in the past; that is, there is no way for us to see how they are now. As a result, when we talk about how an observer "sees" the universe from their vantage point, we must take into account the fact that they're looking back in time.

The "coordinate systems" refer only to the location in space and time from which the measurement is being taken. Between the primed and unprimed systems, the times are assumed to be the same, but the spatial coordinates differ.

That was poor notation on my part. It technically means that the function is evaluated at position $\vec{x}[/tex], but I didn't include position in the arguments of the function, so it's somewhat nonsensical. All I'm saying is if you look at the universe from some spatial position (denoted by [itex]\vec{x}$), it will look the same in all directions. I then go on to say that this equality is true for all such spatial positions.

11. Oct 2, 2005

### SpaceTiger

Staff Emeritus
The universe is generally assumed to be a closed system, so I wouldn't worry about it. That definition is addressing the homogeneity and isotropy of the laws of physics, something which is necessary for the homogeneity and isotropy of the universe, but not sufficient.

12. Oct 2, 2005

### honestrosewater

Oh, you just made me so happy. Not only do I think I'm finally getting closer to really understanding homogeneity and isotropy, their consequences, etc., but in order to read your post, I had to learn about vector spaces this morning and got interested in it and other stuff - it's actually pretty easy and cool, so I'm finally going to learn linear algebra and ...whatever else.

Why is only the radial coordinate dependent on time? Eh, I'll read it again, but I'm not sure I'll figure it out on my own.

In 3D, do you want two angular coordinates (in $f(\vec{r},\vec{\theta})= f(\vec{r},\vec{\theta'})$) - to rotate in two directions?

13. Oct 2, 2005

### SpaceTiger

Staff Emeritus
Because the effective time delay (and corresponding time dependence) comes from the finite speed of light. Since the light moves along this radius vector to reach our detectors, the time dependence will be tied to it. As for the angular coordinates, there are indeed two of them in three spatial dimensions, but the equation is the same in every other respect, so it won't alter the concept.

14. Oct 2, 2005

### honestrosewater

Okay, I get the basic idea, and I think I understand everything else you've said. If you still want to say more, I'm all ears. If you don't, that's fine too.

15. Oct 9, 2005

### SpaceTiger

Staff Emeritus
I was just going to discuss the actual application of this principle in a little more detail. The way it was described is extremely idealistic and does not apply to the real universe. Otherwise, to disprove homogeneity or isotropy, one would need only spin themselves around. In actuality, it's more of a limiting hypothesis; that is, as one approaches infinitely large scales, the universe approaches homogeneity and isotropy. So far, this has been supported by the power spectra of both the cosmic microwave background and the overall structure of galaxies in the universe, but it's possible that, one day, large-scale structures will enter our horizon that exhibit large inhomogeneities or anisotropies. We can never say for sure that this won't happen. Really, the principle is just a formalized application of Occam's Razor. It will likely only be abandoned if observational data gives us a strong reason to do so.

16. Oct 11, 2005

### honestrosewater

Yeah, I was wondering about the motivation behind the assumption, but I don't know enough to ask more questions - it just struck me as being, I don't know, 'fishy' for some reason I still can't put my finger on. Anyway, thanks a lot for helping.

17. Oct 11, 2005

### Garth

The motivation behind the assumption was to simplify the Einstein Field Equation by introducing such symmetry that its cosmological case could be solved!

Observations subsequently showed that it was more-or-less correct, and finally the CMB showed that at z ~ 1000 the universe isotropic and, therefore most probably homogeneous, to one part in 105.

Garth

18. Oct 11, 2005

### honestrosewater

Practically or theoretically solved?

19. Oct 11, 2005

### Garth

Theoretically. The solution of Einstein's FE in the cosmological case, i.e. homogeneous and isotropic, gave the Friedmann family of models on which our present understanding of mainstream cosmology still depends.

Garth

20. Oct 11, 2005

### SpaceTiger

Staff Emeritus
Be extremely dubious about anything Garth has to say about theoretical cosmology -- he has a theory to peddle. In fact, it was not introduced to make the equations easier to solve, it was introduced, as I said, as a form of Occam's Razor, and has so far stood up to observational tests. In modern astrophysics, it's also further justified by inflationary theory, for which we have several forms of indirect evidence.