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Cosmological curvature

  1. Oct 31, 2014 #1
    Good evening.

    I am a current astrophysics undergraduate who is currently having a $10 bet with my classical mechanics professor (chemistry/mathematics background) over what the official curvature of the universe is. While my academic level of understanding is still not quite high enough to fully understand the various mathematical arguments involved, I often read books on the topic such as 'Cosmology' by Edward Harrison, 'The Fabric of the Cosmos' by Brian Green, and many other related books out of sheer interest and entertainment.

    Within these sources, along with several others including NASA, I've read strong indications that for all intents and purposes, the universe's curvature on the grandest of scales is approximately flat (according to the standard Big Bang/Inflationary Cosmology model). My professor claims otherwise; citing general relativity as a reasoning.

    I've brought up these several different sources I've mentioned above, but he wants a quantitative argument, rather than qualitative. If I'm wrong, I'd love to find out why.

    Does anyone familiar with the hardcore, in-depth mathematics of this topic have any recommendations they could point me towards where I could spend some time learning this? I'd love a good challenge.
     
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  3. Oct 31, 2014 #2

    PeterDonis

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    You could both be right, because there are two possible types of curvature that are involved: spatial curvature and spacetime curvature. The current best-fit model of the universe has it spatially flat. But the spacetime of the universe is curved.
     
  4. Oct 31, 2014 #3

    marcus

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    One thing you might want to look at is on page 40, Section 6.2.3 "Curvature" of the official Planck Mission report
    http://planck.caltech.edu/pub/2013results/Planck_2013_results_16.pdf
    In particular at equations 68a and 68b which give the confidence interval on Omega_K (which gives a handle on SPATIAL curvature).

    As was pointed out universe expanding means SPACETIME curvature is negative. If space-time were flat (zero curvature) then spatial distances would not be expanding, as they seem to be. So let's not talk about spacetime curvature

    By an odd convention, probably due to some historical accident, a NEGATIVE ΩK number corresponds to a positive spatial curvature. Like a sphere surface has positive curvature, triangles add up to more than 180 degrees, lines that start parallel tend to converge, etc.

    You know the Hubble radius. roughly 14 billion LY. I'm used to 14.4 billion LY but you may have your own figure. Call it Rh

    If, instead of being zero, ΩK is in fact slightly negative, then U might be spatially a 3-sphere, hypersphere analog of ordinary 2-sphere.
    And I can tell you the RADIUS OF CURVATURE, I.E. the radius of that 3-sphere if it is considered to be the surface of a 4-ball

    Rcurv = Rh/sqrt|ΩK|

    So suppose you look at the Planck mission equation 68b and it says that the most negative ΩK can be is - 0.01
    So then the absolute value of that is 0.01
    and the square root of that is 0.1

    And suppose you think that the Hubble radius is 14.4 billion LY. then the radius of the current space hypersphere is 144 billion LY.

    But it is better to talk about its circumference, because we have no evidence that if space is a hypersphere it is embedded in a 4D space, so it doesnt really HAVE a radius . It has a "radius of curvature" which gives a handle on its curvature.

    And you can multiply the 144 Gly r.o.c. by 2π, to get the circumference. If you could freeze expansion you could circumnavigate space by going in the same average direction for that far. You'd get back to starting point.
     
  5. Oct 31, 2014 #4

    marcus

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    So that is the story about ΩK but notice that I just picked easy numbers, if you look at equation 68b it says the most negative it could be is -0.0075, so the absolute value is 0.0075 and you take the square root and divide Hubble radius by that and you get some other radius of curvature, which will be LARGER than my example.

    And the other thing to notice is that the confidence interval is roughly centered on ZERO
    which is why they say the average large-scale spatial curvature of the U is NEARLY zero. Near spatial flatness, on average, at large scale.
    Nobody knows for sure that it is perfectly flat. But because the confidence interval has shrunk down to around 1% either way, people think of
    ΩK ≈ 0, and treat it as zero for all practical purposes.

    That would make the radius of curvature either infinite or very very large. It seems like you win the bet if you bet about spatial curvature.
     
  6. Nov 1, 2014 #5
    Forgive me; I should have been more specific on what I meant. I appreciate both of your inputs so far.

    I believe our bet lies in the spatial curvature of the universe in aggregate on the largest scales. Or, as Marcus put it, our bet is basically on the value of ΩK.
     
  7. Nov 1, 2014 #6

    marcus

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    My impression is there is no "official" figure for ΩK. It's fairly common to assume it is zero when you do a calculation because it is both computationally convenient and reasonable.

    It is reasonable because the published 95% confidence intervals typically show it within one percent of zero, or thereabouts.

    But we can't say exactly what it is, right? Every major new study that comes out gives at least one new 95% confidence interval for OmegaK, and
    typically a new "best fit" value for it.
    When people are talking carefully they say something like "flat or nearly flat". They don't claim to know that OmegaK is EXACTLY zero.

    The only reason I mentioned the Planck mission March 2013 report is because it is the most recent major report on cosmological parameters that I know.
    We already looked at their confidence intervals for OmegaK, equations 68a and 68b.
    The report also has a "best fit" value somewhere. I will try to find it and give a page reference. Of course it is not *exactly* zero, no empirical observation number ever is *exactly* zero---it's just very close. : ^)

    Yes! Look at Table 10, on page 37, of http://planck.caltech.edu/pub/2013results/Planck_2013_results_16.pdf
    The "best fit" values of OmegaK that they derive from various combinations of their Planck data with other studies are:
    -0.0105
    0.0000
    -0.0111
    0.0009

    "WP" stands for WMAP polarization (another body of data they tried combining with their own data)
    "BAO" stands for Baryon Acoustic Oscillation data, a study that is based on counts of galaxies at various redshifts to get an idea of "ripples" in the spatial distribution of primordial baryonic matter.
    This is somewhat analogous to looking at ripples in the CMB temperature map. In either case you may be looking at density waves ("sound" waves in a manner of speaking).

    Notice that if you combine Planck with WP and with BAO you get a "best fit" for OmegaK which is 0.0000, zero out to four decimal places. It's suggestive. But the published 95% confidence intervals are still allowing something like a one percent variation on either side of zero.
     
    Last edited: Nov 1, 2014
  8. Nov 2, 2014 #7

    timmdeeg

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    I think the quantitative argument follows from the CMB anisotropies. The first acoustic peak was measured at an angular scale of about one degree which confirms that the universe is spatially flat (sum of angles is 180 degrees). You should find more detailed information by searching the web.
     
  9. Nov 3, 2014 #8

    CKH

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    What about the acceleration of expansion (Dark Energy)? Is that interpreted as curvature and if so how much?
     
  10. Nov 3, 2014 #9

    bapowell

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    The curvature scalar for the FRW cosmology is [tex]R = 6\left(\frac{\ddot{a}}{a} + H^2 + \frac{k}{a^2}\right)[/tex] where the term [itex]\ddot{a}/a[/itex] gives the contribution from accelerated expansion.
     
  11. Nov 3, 2014 #10

    CKH

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    Can you explain that without the equation? Is this a spatial curvature or spacetime only? Is it a "negative" curvature (in the case of Dark Energy)?

    One more question, it the universe has uniform positive curvature, does that imply the universe is a 3-sphere with finite size, of does it only imply that inertial paths (geodesics?) are closed curves?
     
  12. Nov 3, 2014 #11

    bapowell

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    If you read earlier in the thread, we discuss that the spatial curvature is flat, whereas the Ricci curvature scalar pertains to the full spacetime geometry. Dark energy gives a positive contribution to this curvature.

    If the universe has positive *spatial* curvature, then the spatial sections are spheres.
     
  13. Nov 5, 2014 #12

    CKH

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    OK. If I understand correctly the spatial flatness is a separate issue from spatial/temporal flatness. I guess this means that at any given time space is flat? But over time it is not, due to expansion?

    In BBT there is a angular size distance and a luminosity distance, which I believe are defined as if space were Euclidean and static. However they are not actual distance measures but are related by functions involving powers of 1+z to actual distance. So, when we look into the past we don't see a flat space.

    I've read that flatness is confirmed by measuring some angular size of fluctuations of the CMB. I guess we measure the distance to the CMB by redshift then by comparing with the angular size of fluctuations when can deduce that space is flat. But we have to take into account the "distortion" of this angular measurement caused by expansion. So in this sense spacetime is not flat, but we can determine that space has been flat the whole time since the CMB?

    How would we know what the angular size of CMB fluctuations is supposed to be (independently of direct measurement)? How do we know the exact redshift of the CMB (to get the distance). Is that based on the temperature required for deionization? How do we choose the value of Ho for this measurement of flatness or doesn't it matter? Doesn't the acceleration (DE) also affect the measurement?

    By the way. Does positive curvature imply that space is finite (like a 3-sphere) or does it just mean that geodesics are closed, but space may still be infinite? Can you elaborate on what you mean by "spatial sections"?
     
  14. Nov 5, 2014 #13

    George Jones

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    I have split my response into two posts.

    As an analogy (and only an analogy), consider (part of a soap bubble) in 3-dimensional space. The soap bubble is a 2-dimensional section of 3-dimensional space. The intrinsic curvature of the soap bubble is largely independent of the properties of the 3-d space in which it is embedded. For example, a curved soap bubble could be embedded in our normal, flat 3-d space, or a flat soap bubble could be embedded in a non-Euclidean, curved 3-d space.

    A spatial section in cosmology is the the 3-d hypersurface in 4-d spacetime that results when the cosmological time coordinate is held fixed.
     
  15. Nov 5, 2014 #14

    George Jones

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    What operational definition are you using for "actual distance"?

    In cosmology, distances are defined operationally. Example include luminosity distance, angular diameter distance, proper distance, and look-back distance. It is not that one of these is the real, "actual" distance; they are all actual distances.

    At some level, it can be almost impossible to explain concepts and conceptual misconceptions using words alone; math is also a necessary component of physical explanation.
     
  16. Nov 5, 2014 #15

    CKH

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    Perhaps I'm missing your point here. Are you just saying that curvature can be measured in a coordinate-independent manner. Thus it is a property of the space (manifold?) itself without reference to an embedding?

    OK that's clear enough. Now about the curvature (of space alone, i.e. at an instant in cosmic time) and it's affects. When we talk about curvature in the context of cosmology I supposed that we assume that the curvature is constant (on large scales, from symmetry). In this context does, a positive curvature imply a finite universe, i.e. a 3-sphere? It seems strange because if that is true then we can decide something about the size of the universe by carefully measuring curvature. Positive means finite? Flat or negative means infinite?

    Thinking a little more about this, we only know about local curvature within our horizon I suppose, so we cannot really draw the conclusion that the entire universe has constant curvature. In fact, I think this is one of the problems inflation is supposed to fix (why is it locally so flat?).

    Still the question above remains in the case of global constant curvature.
     
  17. Nov 6, 2014 #16

    CKH

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    Sorry, of course distances are changing with time so there no "actual distance". Also, I believe that a Euclidean model of expansion (just imagine particles moving apart over time in any Euclidean IRF) is not a correct model of expansion because GR says that the mass/energy density causes some sort of non-Euclidean curvature.

    Getting back to my question about measuring flatness using the CMB. I am trying to understand (without the detailed math) how flatness is measured, particularly what known values are required and how they are obtained. I understand that we can measure the angular size of fluctuations in the CMB with Planck data. We also know the redshift of the CMB based on the decoupling temperature. We also have a model of how space has expanded over time. However, all of this is not enough to measure flatness. We also need to know what the expected angular scale of CMB should be. Then we can do a lot of complicated math and decide if the measured angle of fluctuations is consistent with flatness.

    My question is how do we know independently of direct measurement what the angular size of the CMB fluctuations should be?

    I understand that, but it is not practical for me to learn all the math that physicist learn over many years of study. Still I am curious about how we know what we know and that's what my question is about. A complete understanding of the calculations isn't important in that question. This is a question about the body of evidence that allows us to test flatness using the CMB.
     
  18. Nov 6, 2014 #17

    marcus

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    Here's a quick and dirty oversimple account. George Jones will I hope correct me if I am wrong. And can refine what is too simplified, if he wishes.
    We know the properties of hydrogen gas. At what temperature it begins to get so ionized it scatters light. Last scattering occurs below a certain temperature.

    So we know the temperature at the epoch of last scattering. Therefore we know the speed of sound in the hot gas.
    Because we can compare with temperature NOW we know the scale ratio (1+z). So we can calculate the nearly uniform density at last scatter time.
    And we can calculate how far the gas was back then.

    The hot gas was undulating in a curious way, ( as pockets of overdensity pull more into themselves and get denser and hotter and then push out and get rarer and cooler. ) Spherical pulsating pressure wave resonances had developed. We know the speed of sound so we can calculate the actual physical size, back then, of the resonant fluctuations.

    So we now have an isosceles triangle which existed at the moment of last scatter. The two long sides are about 42 million LY, and the base is the size of that "acoustic peak" resonant fluctuation.

    We can now calculate what the apex angle of that triangle should be (knowing the three sides) on the assumption that space was flat and check to see if it agrees with what we observe to be the angular size of predominant fluctuations. If it agrees, then space was approximately flat.

    If the observed angle is too large, that means space was positive curved and the triangle sides are bowed outwards, a bulging triangle. If the observed angle is too small, that means space was negative curved. Picture knock-kneed rather than a bow-legged stick figure. But neither case applies. The observed angle is what it should be---so space was approximately flat.

    The reasoning here uses a type of distance called "proper distance" which is the distance you would measure by conventional means if you could pause the expansion process to give time to measure. It is the "freeze-frame" distance at a particular specified epoch. As George said, there are various types of distance used in cosmology. This one is model dependent, so checking is required to make sure that allowing for reasonable variation in the model parameters would not significantly change the conclusions---so some expertise and experience comes in there. As I say this is an oversimple account.
     
    Last edited: Nov 6, 2014
  19. Nov 7, 2014 #18

    CKH

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    That's quite helpful but I have a few more questions.

    I get the part that goes from temperature to redshift to distance. There is also volume ratio so a density ratio. But how do we know the density? If we knew what the density is now we could know the density then, but how would accurately measure current density?[/QUOTE]

    Are these the Baryonic Acoustic Oscillations? What matters in calculating their size?

    Density including DM?
    Speed of sound. I'm not sure what affect DM might have on this. This is affected by temperature and maybe by density?

    I think this may be the complicated part of the calculation. 42 billion LY is the current distance to the point from which the light was emitted but is not directly relevant to what we see now. The light we see has only traveled ~13.5 billion LY to reach us, so maybe that is the length of the sides?

    Here's what also confuses me. You can measure the apparent angular size of a galaxy and it's distance by redshift. But you cannot directly determine the size of the galaxy that way. I was lead to believe that GR/expansion requires you to adjust that apparent angular size to measure the size of the galaxy. I've read that at very high z similar galaxies actually appear larger than at lower z. There is some function of z that must be applied to the measured angle. If space is flat all along, I'm not sure what causes the need for this adjustment.
     
  20. Nov 7, 2014 #19

    timmdeeg

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    That's right, one need to measure their luminosities.
    http://m.teachastronomy.com/astropedia/article/Galaxy-Size-and-Luminosity [Broken]
     
    Last edited by a moderator: May 7, 2017
  21. Nov 7, 2014 #20

    marcus

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    Glad you get the temperature ratio implies the linear scale factor ratio. It is on the order of 1000, e.g. 3000 kelvin down to 3 kelvin.
    More precisely estimated to be 1+z = 1090

    The volume ratio is essentially the CUBE of the linear scale ratio. So density then and now are related by the cube.
    The role of DM is more subtle. It does not radiate so does not contribute to CMB. It does not have pressure so it's overdensity/underdensity oscillations occur at a different frequency and affect the alternate sizes of successive peaks. It is interesting but requires a separate discussion.
    You might ask Brian Powell, or George Jones later about the role DM plays in size of acoustic oscillations. I'm sticking to the simplified story, unrefined by additional details.

    Notice that in the following I say 42 MILLION light years (not billion) and I mention that we are using "proper distance". that is distance at a specific time, if you could pause expansion to allow measurement. It is the type of distance used to define the Hubble rate, and determine quantities like density in the Friedman equation (the basic cosmic model.)

    ==quote my post #17==

    So we now have an isosceles triangle which existed at the moment of last scatter. The two long sides are about 42 million LY, and the base is the size of that "acoustic peak" resonant fluctuation.
    ...
    The reasoning here uses a type of distance called "proper distance" which is the distance you would measure by conventional means if you could pause the expansion process to give time to measure. It is the "freeze-frame" distance at a particular specified epoch. ...
    ==endquote==

    No, not "billion". 42 million LY. If you multiply that by 1090 you get around 46 billion, the distance now.

    No 13.5 billion LY is not the length of the sides. We are not using travel time as a measure of distance. Proper distance is not travel time distance.
    And the CMB light has been traveling 13.8 billion years, not 13.5. If you take 13.8 billion and subtract 380,000 from it you still have approximately 13.8 billion.


    That is a nice effect! It is present also if space is assumed to be flat all along. Assume spatial flatness, it is the simplest and it seems approximately right. You still get that the past lightcone is tear-drop shaped. Because of expansion, beyond a certain now distance, things that are farther away now were nearer THEN, and so look bigger in the sky. Past that point (the maximum girth of the tear-drop) farther and farther things have a larger and larger angular size.

    You can see that effect clearly using Jorrie's calculator. Learn to use the calculator. Link is in my signature and in some other people's too. The calculator is the "Lightcone" link. There is a picture of the tear-drop shape light cone in the "Lineweaver...Figure 1" link in my sig.
     
    Last edited: Nov 7, 2014
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