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

[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 [itex]\mu K^2[/tex]. I think the other scaling factors are chosen to emphasize the acoustic peaks.

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'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.