What are the latest exciting results from WMAP's three-year data release?

[oldman]

What is it about the WMAP results that tells us the universe is flat?

I've been browsing back in this thread and the "What is CMB" thread to answer this question, and gather (mainly from Space Tiger and Garth) that a flat geometry for the universe is deduced from a best fit of theory (treating many factors?) to the high frequency peaks (of an assumed power-law spectrum of evolved primordial density fluctuations?).

Is this anywhere near correct?

I'd like to understand just what is it about the flat geometry that produces the good fit to the data. Perhaps the sensitivity of the Sachs-Wolfe effect to geometry, or something more subtle or quite different?

I realize from the threads that analysing the spectrum is a highly technical matter. But I'd love a simple explanation.

 

[hellfire]

As fas as I know, the main data to infer about geometry is the angular size of the first peak. The first peak is a primary anisotropy, the greatest of acoustic nature. Its physical size is determined by the size of the particle horizon at decoupling. The relation between the observed angular size of the first peak and its physical size depends on the geometry of space and on the distance to the last scattering surface. Thus, to infer about geometry from the observed angular scale of the firs peak it is needed an assumption about the size of the particle horizon at decoupling as well as about the distance to the last scattering surface. For example, a positive curvature implies that the physical scale of the first peak should be smaller than in a flat space.

 

[marcus]

hellfire, oldman,
maybe we could do a calculator experiment about this!

we know, because of the temperature, that the CMB is at z = 1100.

we could convert that to distance using two assumptions, with Ned Wright's calculator: we could assume flat and we could assume some nonflat case.

then we would get two different figures for the area of the lastscatter sphere---two different ideas of the actual physical size of the universe at the time of decoupling.

this seems doable (using Wright's calculator) with some simple arithmetic

from the temperature we could deduce the speed of sound in that medium, and we would have two separate cases of what the size of the medium is---maybe we could get some intuition about what hellfire says about the size of the first acoustic peak.

then, we would expect that the angular size we calculate in the FLAT case would match the observed CMB power spectrum---i.e. fit the mottled way it looks. and the angular size we calculate from the NONFLAT case would NOT match the observed CMB picture. so we would be doing a crude imitation of the professional CMB analyst routine.

hands dirty CMB interpretation you can do in your own kitchen. I would like to see it, if anyone's game.

 

[hellfire]

The impact of the curvature in the relation between actual size and measured angular size of the first peak is given through the definition of the angular diameter distance d_A (for which you have started a thread recently).

In general, one has that the actual size l relates to the angular size \theta:

l = d_A \theta

In the cosmological calculator you get, for example for z = 1100:

d^{SCDM}_A = 24.08 Mly
d^{OCDM}_A = 74.26 Mly

For:
- The flat model SCDM (standard cold dark matter model): \Omega = 1, with \Omega_m = 1
- The open model OCDM (open cold dark matter model): \Omega = 0.3, with \Omega_m = 0.3

If you assume a measured angular size of about 1° for both models, you see immediately that the actual size of the first peak would be smaller in the flat model than in the open model.

This may help to illustrate the angular diameter distance in different models.

But the problem is that both models produce a different angular size for the first peak (it cannot be assumed that both are 1° today). To calculate the actual size of the first peak one should proceed as you propose, i.e. the size of the sound horizon should be calculated. How to do this I don't know.

Afterwards, aplying then the formula \theta = l / d_A, you could calculate the expected angular size in sky of the first peak for both models.

 

[oldman]

The impact of the curvature in the relation between actual size and measured angular size of the first peak ...

Thanks, Marcus and Hellfire, for your answers and calcs. It seems that finding out from the WMAP results that the universe's geometry is flat is much easier than I thought. Viva Euclid!

 

[hellfire]

I have read that it is usually assumed that the speed of sound during recombination is equal to c_s = c/ \sqrt{3} (would be nice if someone could check this). The sound horizon is:

s = \int_0^{t_{rec}} dt \frac{c_s}{a}

this would mean that it is 1 / \sqrt{3} times the size of the particle horizon at recombination. In my calculator I have an output of the particle horizon for a specific redshift, but I see now that I have made a silly mistake there and that this output field is incorrect. I will see if I can correct this today. Then we would have the values s^{OCDM}, s^{SCDM}.

Next step would be to know how to go from the value of s to the size of the first peak...?

 

[hellfire]

For the \LambdaCDM model we might try to do the calculation backwards. First, the angular diameter distance for z = 1100 is:

d^{\Lambda CDM}_A = 41.34 Mly

Now, if I assume that the size of the first peak is equal to the sound horizon size l = s (?), then, applying l = d_A \theta, with \theta = 1°, I would get:

s^{\Lambda CDM} = 0.7 Mly

This value seams meaningful to me. But the point is this value for s should arise from the horizon formula I put above. Then, inserting in l = d_A \theta it should lead to the angular size of the first peak. I have been trying to modify my calculator to give such an output but I did not succeed; the value for s I am getting with the calculator is about 833 Mly.

Could anyone find out the value of the particle horizon and sound horizon at z = 1100 for the \LambdaCDM model?

 

[marcus]

hellfire, I want to echo the appreciation expressed by oldman.
Your demonstration of the down-and-dirty-nitty-gritty of the first acoustic
seems to satisfy oldman (at least for now) and although I have not
followed all your steps I feel generally better about it too.
sometime I hope we locate an online tutorial article about this,
but for now we seem to have gotten a better handle on it.

 

[oldman]

Your demonstration ... of the first acoustic
seems to satisfy oldman (at least for now) and ...I feel generally better about it too.

Marcus, in this thread Hellfire and yourself seem to agree that the main flat-geometry-indicator, if I may call it that, is the angular width of the first acoustic peak. Fine -- I think I grasp the explanations you both so kindly gave.

Yet in your most recent post in the thread "WMAP 3 and spatial closure" (#108) you quoted a statement: "...However, altering the geometry of universe mainly affects the positions of the CMB acoustic peaks,..." This seems to imply that it is the position of the peaks rather than a width that is the main flat-geometry-indicator. This confuses me again.


The WMAP results are rich and important to grasp. A list of the main conclusions, each with a statement of which feature/s of the results they are attributed to would be very illuminating for the uninformed. Tabulated along lines like:

Geometry is flat..... deduced from first peak angular width

Baryonic matter is 4%...deduced from fit to peaks l > 150

Dark energy is 75% ... deduced from distance between 3rd and 4th peaks

and so on, perhaps. (The entries above are of course a fiction ... I have no idea of how to draw such a table up).

The tutorial article you mentioned sounds like a good idea!

 

[Garth]

Geometry is flat..... deduced from first peak angular width

This is a statement agreed by all in the community.

My personal beef - this statement more generally should be:

"Geometry is conformally flat...deduced from first peak angular width"
as the WMAP data is angular in nature and conformal transformations are angle preserving.

Garth

 

[hellfire]

Marcus, in this thread Hellfire and yourself seem to agree that the main flat-geometry-indicator, if I may call it that, is the angular width of the first acoustic peak. Fine -- I think I grasp the explanations you both so kindly gave.

Yet in your most recent post in the thread "WMAP 3 and spatial closure" (#108) you quoted a statement: "...However, altering the geometry of universe mainly affects the positions of the CMB acoustic peaks,..." This seems to imply that it is the position of the peaks rather than a width that is the main flat-geometry-indicator. This confuses me again.

Both are basically the same. You can convert from the position of any multipole \ell to its angular scale \theta with:

\ell \sim \frac{180^{\circ}}{\theta}

 

[oldman]

Both are basically the same. You can convert from the position of any multipole \ell to its angular scale \theta with:

\ell \sim \frac{180^{\circ}}{\theta}

Thanks, Hellfire. But how does this change of label along the x-axis change "width" into "position" ? The two are qualitatively different. I'm stupid today.

 

[hellfire]

Take a look to the picture in wikipedia. You can see that the x-axis is the number of the multipole moment \ell. The peak at \ell ~ 200 tells you that the power is strongest at that value. This is the first peak. When we talk about the angular width of the first peak we are not talking about the width of the peak in this picture, but about the conversion of \ell to \theta I gave you before. E.g. a smaller angular scale \theta of the first peak would just mean that it would be located more to the right, at higher values of \ell. This means that curvature shifts the position of the first peak to the left or to the right in this picture.

 

[oldman]

Thanks again. I understand now.

 

[hellfire]

I have a question about that picture in wikipedia, may be someone can answer.

The y-axis corresponds to the power of the multipole. It is written as a function of C_{\ell} or as \mu K^2. I do not see how both magnitudes are equivalent:

The \mu K^2 means that the y-axis gives the deviation from the mean temperature for a given multipole.

On the other hand, the C_{\ell} indicates that the power is calculated making use of the two-point correlation function. To my understanding this would mean that the y-axis gives the deviation from a Poisson distribution of anisotropies for a given multipole.

This understanding seams to be incorrect, because this seams not to be equivalent to a temperature deviation.

 

[SpaceTiger]

The y-axis corresponds to the power of the multipole. It is written as a function of C_{\ell} or as \mu K^2. I do not see how both magnitudes are equivalent:

The former is the quantity being plotted and the latter are its units. In general, the power spectrum is just the Fourier transform (in the spherical expansion) of the two-point correlation function. The latter, I believe, is just:

C(\theta)=<\delta T (\vec{e_1}) \delta T (\vec{e_2})>

where the \delta T are the deviations from the mean temperature at a given point in the sky. This, of course, has units of temperature squared. Since the anisotropies are on micro-Kelvin scales, the units of the angular power spectrum are also given in \mu K^2[/tex]. I think the other scaling factors are chosen to emphasize the acoustic peaks.<br /> <br /> If the anisotropies are Gaussian (that is, described by a Gaussian random field), then the power spectrum is a complete description of them. As best we can measure, the anisotropies are indeed Gaussian, as predicted by inflation. Inflation also predicts small deviations from Gaussianity, but we&#039;re not yet at the level where we can detect that.

 

[hellfire]

Thanks for your answer, but I still don't get it. According to my knowledge the two-point correlation function measures the deviation from an homogeneous distribution of anisotropies. If the distribution of l = 200 or 1° anisotropies is homogeneous through the sky, wouldn't this mean that the correlation should vanish, independently from the fact that these have a higher temperature than the average? What is C(\theta) telling us exactly?

 

[SpaceTiger]

Thanks for your answer, but I still don't get it. According to my knowledge the two-point correlation function measures the deviation from an homogeneous distribution of anisotropies. If the distribution of l = 200 or 1° anisotropies is homogeneous through the sky, wouldn't this mean that the correlation should vanish, independently from the fact that these have a higher temperature than the average?

Suppose I filled the sky with fluctuations (both hot and cold spots) that had typical sizes of order 1°. What would we expect from the correlation function at angular scales of 0.5°? If I look at just one point in, say, a cold spot, then most of the points at distances 0.5° away should also be cold (the size of the fluctuation is larger than the angular scale we're probing). This means that, for this one point, the quantity I quoted above should be positive (the product of two negative temperature fluctuations). In the hot spots, both temperature fluctuations will be above the mean, so the correlation function will again be positive. Thus, averaged over the whole sky, we expect the correlation function at 0.5° to be positive.

This is not the case at much larger angular scales, however. If I look again at a point in a hot spot and compare it to a point 5° away, I will be just as likely to run into a hot spot as a cold spot. Thus, the correlation function at 5° should come out to zero (or nearly zero) when averaged over the whole sky.

What is C(\theta) telling us exactly?

It's telling us about the relative amplitudes of fluctuations of different angular sizes. C(\theta) is somewhat more difficult to interpret than the power spectrum because, as you can imagine, a sky full of 1° fluctuations will produce correlations at all scales less than about a degree. When you combine this with fluctuations at smaller scales, it becomes difficult to distinguish fluctuations of different sizes. The power spectrum, however, tells you directly about the relative contributions of fluctuations at various scales (in this case, expressed in terms of the spherical wavenumber, l). The more power there is at a given l, the larger the amplitude of fluctuations at that scale.

 

[hellfire]

It's clear now, thanks!

 

[Harry Costas]

Hello All

Have a look at
http://astro.uwaterloo.ca/~mjhudson/research/threed/

Loacation of our super clusters of galaxies

 

[marcus]

Hello All

Have a look at
http://astro.uwaterloo.ca/~mjhudson/research/threed/

Loacation of our super clusters of galaxies

these look like nice pictures, I have not watched the animations yet (haven't checked that they are online)

thanks.