Where are the real definitions of homogeneity and isotropy?by honestrosewater Tags: definitions, homogeneity, isotropy, real 

#1
Sep2905, 01:37 AM

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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 nonhomogeneous '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. 



#2
Sep2905, 02:28 AM

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May be this will be helpful:
http://www.physicsforums.com/showthread.php?t=86948 



#3
Sep2905, 04:28 AM

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You can pretty much think of the stressenergy 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 stressenergy 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 stressenergy tensor are zero. Because the stressenergy tensor is homogeneous and isotropic, the various curvature tensors which describe the curvature of spacetime 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
Sep2905, 06:13 AM

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Where are the real definitions of homogeneity and isotropy?
Okay, I think I get it. I just can't get a definition. For simplicity, let f be a function from N (points) to N^{2} (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 stressenergy 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
Sep3005, 10:45 AM

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



#6
Oct105, 03:07 AM

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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 S^{m} to T^{n}, where m and n are integers > 0 (I suppose the members of S and T are numbers, but they could possibly be other things); S^{m} is a set of points, each point being specified by m coordinates (or perhaps something to do with m dimensions?); T^{n} is a set of values of n properties (i.e., you measure something and get a number; the order of the members of the ntuples 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 S^{m} is usually R^{3}, 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: R^{2} > R, f(x, y) = c, where c is a constant in R. ii) inclined plane. i: R^{2} > R, i(x, y) = x (right?) iii) hill. h: R^{2} > R, h(x, y) = x (or maybe h(x, y) = (x^{2})? 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 xaxis 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 nonisotropic, right? Is j: R^{2} > R, j(x, y) = x^{3} also homogeneous and nonisotropic? I'll guess that j is nonhomogeneous, 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 > R^{n} I imagine you would just define the subsets with intervals: I_{rp} = {((s, (t_{1}, ..., t_{n})), k(s, (t_{1}, ..., t_{n}))) : ((s, (t_{1}, ..., t_{n})), k(s, (t_{1}, ..., t_{n}))) in k and s in [r, (r + p)]} p, r, s, t are reals. Maybe you want open or halfopen 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: R^{2} > R, j(x, y) = x^{3} confuse me. 



#7
Oct105, 11:57 AM

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Alright, let me try this from a different point of view. Homogeneity basically implies:
[tex]F(\vec{x})=F(\vec{x'})[/tex] for all [itex]\vec{x}, \vec{x'}[/itex] in the universe. The function F is any measurable property of a local region of spacetime and x is the spatial position. Note that [tex]F(\vec{x},t)\ne F(\vec{x'},t')~if~t \ne t'[/tex] where t is proper time since the Big Bang. Isotropy is a bit more complicated, but boils down to: [tex]f(\vec{r},\vec{\theta})_{\vec{x}}=f(\vec{r},\vec{\theta'})_{\vec{x}}[/tex] for all [itex]\vec{r}, \vec{x}, \vec{\theta}, \vec{\theta'}[/itex] in the universe. In this context, [itex]\vec{\theta}[/itex] is the angle in which the observer is looking, [itex]\vec{r}[/itex] 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 [itex]\vec{r}, \vec{r'}[/itex], 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
Oct205, 04:30 AM

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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 [itex]_{\vec{x}}[/itex] mean? I can look it up if you can state it in words. 



#9
Oct205, 04:57 AM

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Is this correct:
Sorry, I'm not trying to be a pain, and I do really appreciate you help. 



#10
Oct205, 01:05 PM

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#11
Oct205, 01:09 PM

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#12
Oct205, 02:59 PM

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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.
In 3D, do you want two angular coordinates (in [itex]f(\vec{r},\vec{\theta})= f(\vec{r},\vec{\theta'})[/itex])  to rotate in two directions? 



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Oct205, 03:44 PM

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#14
Oct205, 04:41 PM

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#15
Oct905, 05:02 AM

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#16
Oct1105, 02:18 AM

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#17
Oct1105, 10:14 AM

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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 moreorless correct, and finally the CMB showed that at z ~ 1000 the universe isotropic and, therefore most probably homogeneous, to one part in 10^{5}. Garth 


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