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Meaning of a Flat, Open, and Closed Universe 
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#1
Jan2813, 06:06 PM

P: 10

Hello all,
I'm reading up on cosmology and the potential shapes of the universe based on its density in relation to the critical density. When one says that a universe is "flat," what precisely does this mean? Does this mean that the spacetime of the universe is described by Euclidean geometry? Similarly, what does it mean to say the universe is "open" or "closed"? Thanks in advance! 


#2
Jan2813, 08:42 PM

P: 1,857

Open and closed is simply infinite or finite.
For the geometry its a factor of its critical density. If the density parameter is greater than 1 then the universe is closed (finite) and positively curved like the surface of a ball. If the density parameter is less than 1 then its open with negative curvature like a saddle. (open equals infinite) If its exactly equals 1 then its flat and infinite WMAP results show that its flat. In the flat geometry this expansion is suppose to be at a steady rate. Dark energy is the reason its accelerating. 


#3
Jan2913, 12:56 AM

P: 10

Thanks, that's very helpful.
When speaking of flatness or curvature, I presume you mean flatness/curvature of spacetime, not just space? Also, what sort of physical effects would you expect to see from an open or closed universe due to its curvature? For example, if you created a huge triangle across a very large segment of space, would you expect the sum of its angles to be more or less than 180 degrees? 


#4
Jan2913, 03:26 AM

P: 887

Meaning of a Flat, Open, and Closed Universe
i am not an astrophysicist, but here's what i believe.
when we talk about flat or curved, we are actually talking about the metric of spacetime, which is related to the metric tensor G. the metric tensor is a (covariant) rank 2 tensor whose elements are made up of the dot products of the direct basis vectors: g[itex]_{uv}[/itex] = e[itex]_{u}[/itex][itex]\bullet[/itex]e[itex]_{v}[/itex] the direct basis vectors are defined by e[itex]_{u}[/itex] = ds/dq[itex]^{u}[/itex] where q[itex]^{u}[/itex] is a variable in your coordinate system and s is the expression for a length element vector in your coordinate system. for example: lets say we have a Cartesian coordinate system in 2D. s = x x + y y. the 2 basis vectors would be then e[itex]_{x}[/itex] = d(x,y)/dx = dx/dx x + dy/dx y = x e[itex]_{y}[/itex] = d(x,y)/dy = dx/dy x + dy/dy y = y as a rank 2 tensor we can conveniently represent it as a matrix. the Euclidian metric tensor that we are familiar with represents the graphs we are familiar with: g[itex]_{xx}[/itex] = e[itex]_{x}[/itex][itex]\bullet[/itex]e[itex]_{x}[/itex] = x[itex]\bullet[/itex]x = 1 g[itex]_{yy}[/itex] = e[itex]_{y}[/itex][itex]\bullet[/itex]e[itex]_{y}[/itex] = y[itex]\bullet[/itex]y = 1 g[itex]_{xy}[/itex] = g[itex]_{yx}[/itex] = e[itex]_{x}[/itex][itex]\bullet[/itex]e[itex]_{y}[/itex] = x[itex]\bullet[/itex]y = 0 thus the entire metric tensor G is the following matrix: [1 0] [0 1] note that it is a symmetric matrix. that means the basis vectors are orthogonal. also note that the magnitudes of the basis vectors are one. thus this is a coordinate system made from an orthonormal basis; just as we expect. that seems too easy. lets do a nontrivial example: polar coordinate systems in 2D. x = rcosθ, y = rsinθ s = rcosθ x + rsinθ y e[itex]_{r}[/itex] = d(rcosθ x + rsinθ y)/dr = cosθ x + sinθ y e[itex]_{θ}[/itex] = d(rcosθ x + rsinθ y)/dθ = r(sinθ x + cosθ y) g[itex]_{rr}[/itex] = e[itex]_{r}[/itex][itex]\bullet[/itex]e[itex]_{r}[/itex] = cosθ x * cosθ x + 2 cosθ x * sinθ y + sinθ y * sinθ y = cos[itex]^{2}[/itex]θ + 0 (because x and y are orthogonal) + sin[itex]^{2}[/itex]θ = cos[itex]^{2}[/itex]θ + sin[itex]^{2}[/itex]θ = 1 g[itex]_{rθ}[/itex] = g[itex]_{θr}[/itex] = e[itex]_{r}[/itex][itex]\bullet[/itex]e[itex]_{θ}[/itex] = 0 g[itex]_{θθ}[/itex] = e[itex]_{r}[/itex][itex]\bullet[/itex]e[itex]_{θ}[/itex] = r()sinθ x * r()sinθ x 2 rsinθ x * cosθ y + rcosθ y * rcosθ y = r[itex]^{2}[/itex]cos[itex]^{2}[/itex]θ + r[itex]^{2}[/itex]sin[itex]^{2}[/itex]θ = r[itex]^{2}[/itex] so G = [1 0] [0 r[itex]^{2}[/itex]] this is obviously not a flat surface. this gets more complicated for other shapes. now we have set the stage to talk about curved space  as we see, space CAN be curved! 


#5
Jan2913, 03:39 AM

P: 887

now we have the metric tensor. what do we do with it? well first we note that coefficients of the elements of the matrix are not necessarily 1. these coefficients represent magnitude. we can now introduce a concept called the metric coefficient, or in physics language, the scale factor.
the scale factor for an element of the matrix h[itex]_{i}[/itex] = [itex]\sqrt{g_{i}}[/itex] note that for the polar coordinate, we have a scale factor with r. That means as r increases, the other basis vector increases! could spacetime behave the same way? sure! we can also note that certain coordinates do not lead to a symmetric metric tensor. this would result in a nonorthogonal space. Sure, that sounds really complicated, why is it useful? It turns out that in spacetime, things travel along their geodesics if you give no force input. A geodesic is the closest coordinate distance between 2 points; going against the geodesic requires a force. If we have a 4D spacetime that's Euclidian, the geodesic would be a straight line. But if it was not Euclidian, but rather defined by some other metric tensor, the geodesic might not be a straight line. Indeed, it may lead towards something. Such as a gravity source. This is why we talk about spacetime being bent by mass  the metric tensor in presence of mass changes. This actually leads to a scary thought: if it were not for the electromagnetic force, the degeneracy force... where does the geodesics lead? all particles are trying to follow the geodesics, to one point: their center of mass. all particles are trying to become black holes. Sleep on that. However these problems are questions of local scale. The universe as a whole may also be defined by a metric tensor that is not Euclidian. That would mean geodesics would not be straight lines, which would mean triangles would not have their angles add up to 180 degrees. 


#6
Jan2913, 04:55 AM

P: 13

without getting all heavy on the math and terminology, EVERYTHING will naturally try to follow its geodesic path...
...you don't see an ocean going ship try to jump out of the water...it just follows a path around a curved surface...local direction changes notwithstanding... ...a force on that very ship MAY make it jump out of the water, but that is not its natural predilection ....and so the very same with all objects containing mass, whereby anything massive will ''fall'' and it is only because an initial force has pushed the massive object outward, its tendency will be to gradually fall back in to its more stable, if not placid, state.. ...our physics on earth are just a mirror for the more volatile physics in the wider universe ... 


#7
Jan2913, 07:31 AM

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#9
Jan2913, 09:18 AM

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P: 4,782

There's been a lot of noise in this thread, so I thought I'd try to clarify somewhat. First, take the first Friedmann equation:
[tex]H^2 = \rho(a)  {k \over a^2}[/tex] (Pedantic note: I have neglected the constants in an attempt to improve clarity) Here we have four things that need to be considered. The first is [itex]H[/itex], the rate of expansion. The second is the total mass/energy density [itex]\rho[/itex]. This is a function of the scale factor [itex]a[/itex] because the density changes as the universe expands (and how it changes depends upon the sort of matter/energy). The scale factor is the amount by which things have expanded. We typically define the scale factor as being equal to one today, which means that when the scale factor was 0.5, things were, on average, half as far apart. The final parameter, [itex]k[/itex], is the spatial curvature: [itex]k=0[/itex] is flat, [itex]k > 0[/itex] is closed, [itex]k < 0[/itex] is open. What does this mean? Well, to see that, look at the other terms in the equation: they are the expansion rate and the density of the universe. If you have a high energy/mass density, but a low expansion rate, then you have a large positive curvature [itex]k[/itex], and the universe is closed. As long as you just have normal matter, this expansion will be too slow for gravitational attraction of all that dense matter and it will collapse in on itself. Similarly, if the expansion is much faster than the energy density, then this is balanced by having a large negative curvature, which is an open universe that tends to expand forever. There is a middle point where the expansion rate exactly balances the energy density, and that is what is called a flat universe. Because of the way General Relativity operates, this has geometric effects so that if you have a significant amount of spatial curvature, triangles drawn across large regions of the universe, e.g. by sending light across billions of light years, will tend to have angles which don't add up to 180 degrees. I'll end on a pedantic point: the open/flat/closed bit doesn't actually say what the eventual fate of the universe is. The issue is with regard to how the effect of the curvature scales with the expansion: [tex]H^2 = \rho(a)  {k \over a^2}[/tex] If we have an expanding universe, and some of the stuff that makes up [itex]\rho(a)[/itex] dilutes more slowly than [itex]1/a^2[/itex], then if the universe is allowed to expand enough, that form of slowlydiluting energy will dominate, and the fact that there is curvature becomes less and less important. This is the case with dark energy, for example. So if this extra stuff that dilutes more slowly tends to make the universe expand faster, as in the case of dark energy, then a closed universe will expand forever anyway. Similarly, one can have a negative dark energy that causes an open universe to recollapse. In reality, our universe appears to be very close to flat, with [itex]k < 0.01[/itex] in some units, meaning that the expansion rate [itex]H^2[/itex] differs from the density [itex]\rho[/itex] by no more than one part in a hundred. 


#10
Jan2913, 11:48 AM

P: 234

I wonder, why the 1° angle corresponding to this certain peak in the CMBR, yields the spatial geometry. Because to my understanding the triangle measured by analysing the direction of incoming photons can't be a triangle "at a given instant in time". Any comment appreciated. 


#11
Jan2913, 09:09 PM

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P: 4,782

http://arxiv.org/abs/astroph/9905116 This can be understood geometrically as the curvature affecting the angular size of objects we see which, in turn, impacts our estimate of their distance. The curvature estimate then comes from comparing the estimated distances between things that are very far and things that are closer. One of the most effective is comparing the typical separation between hot/cold spots on the CMB and the typical separation of galaxies in the relatively nearby universe (this is known as the Baryon Acoustic Oscillation measurement). The separation in either case is viewed as an angular diameter distance. 


#12
Jan3013, 11:58 AM

P: 234

If I understand your explanation correctly, the density spot on the CMB (angular size 1 degree, z = 1100) develops into a BAO (certain measurable angular size, small value of z). Now, from page 5 "The angular diameter distance Da is defined as the ratio of an object's physical transverse size to it's angular size." I guess "physical transverse size" means a distance measured with a ruler. I have in mind that the size of the density spot on the CMB is around 150 Mpc (ruler!). So, with the known increase of z one should be able to calculate the size of the corresponding BAO. I hope this is correct so far. But I still have no idea how to construct this very triangle "at a given instant in time" (which's sum of angles is 180°) by the comparison of the angular diameter distances of said objects. 


#13
Jan3013, 04:30 PM

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#14
Jan3113, 11:12 AM

P: 234

You mentioned the comparison of two "objects", the spot on the CMB and a characteristic galaxy separation. Does this mean, as in either case the true size is known, that one obtaines two values of the spatial geometry independent of each other? 


#15
Jan3113, 12:46 PM

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#16
Jan3113, 12:53 PM

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#17
Feb113, 03:50 AM

P: 234

Kindly comment and correct what's wrong. Having not studied physics, I appreciate any help very much. 


#18
Feb113, 05:56 AM

P: 234

The spatial geometry refers to the sum of angles = 180° of a triangle at a certain instant of time. Or assuming the expansion is "frozen". The triangle shall be formed by an object of a known size (from one edge to the other) and the distance between this object and our worldline. Both distances are true, meaning measured with rulers. At t = t1 the angular size of said object is β1. Together with it't true size the sum of the angles is 180°, the geometry of the space is euclidean. Then the universe expands (whereby the object participates) for x billion years and stops to be "frozen" again. At t = t2 "watching" the frozen state nothing remarkable has changed, except that the triangle is much larger. But the sum of angles is still 180° and β2 = β1. The reason is that in contrast to the spacetime curvature the spatial curvature doesn't change over time. If I am right, the spacetime curvature is constant only in case the expansion is linear over time. At t = t2 we see  while the universe expands  the light of the object, which was emitted at t1 x billion years back in time and measure it's angular size βexp. My question is how is βexp related to β1 and to β2 resp.? It seems the only given quantities are the true size of the object at t = t1, it's angular size βexp at t = t2 and and it's redshift, which is related to the lookback time (t2  t1). Perhaps I missed something. How do we calculate the spatial geometry from this (whereby I suspect that the measured angle βexp does not coincide with β1, β2 resp.)? I have the notion that an angle measured by light rays says because of the dynamics involved something about the spacetime curvature, but not about the spatial curvature. I might be wrong, but am much interested to learn why. Any help appreciated. 


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