Viewing a single comment thread. View all comments

amaurea t1_j2kth2z wrote

Reply to comment by Aseyhe in Is any "movement" visible in the fluctuations of the CMB over time, or does it appear static? by JarasM

What's 0.07° (4.2 arcmin) based on? It's somewhat close to Planck's FWHM at its highest frequencies (the main CMB frequencies are more like 5-7 arcmin), but Planck isn't the state of the art at small scales - that's the South Pole Telescope (SPT) and Atacama Cosmology Telescope (ACT) with resolutions of about 1 arcmin.

Though as you say in another comment, there's hardly any CMB left at those scales. ACT has published a foreground-cleaned CMB temperature power spectrum with significant signal detection up to l = 3700 which corresponds to roughly 0.10° = 5.4 arcmin 0.05° = 2.9 arcmin.

I think I read a paper at some point about the feasibility of detecting the time-derivative of the CMB. If I remember correctly, it was actually the largest scales that were considered the most promising there, not the smallest. Those scales change extremely slowly, but the signal is also much brighter there, and from what I remember that won out.

6

Aseyhe t1_j2kty6j wrote

I just lazily took pi/l for l=2500, the largest l that Planck papers plot. Indeed, the ground-based telescopes push to somewhat higher l.

Hmm, around 1 degree (where the fluctuation power peaks), the time scale for the CMB to change would be of order a billion years, or one part in ~10^(9) per year. I wonder how far off that kind of sensitivity is.

5

amaurea t1_j2kuo6l wrote

Yes, they typically plot out to l=10000, but all the power out there comes from point sources (if one doesn't clean those away). I used 360°/l, but if I use 180°/l the ACT foreground cleaned spectrum becomes too faint to be detectable at 2.9 arcmin.

5

amaurea t1_j2m8d9y wrote

(Actually, come to think of it, shouldn't the formula be sqrt(4pi)/(lmax+1)? Alms from 0 to lmax have (lmax+1)² degrees of freedom (sum_0^lmax (2l+1)). This is enough information to split the full sky into (lmax+1)² pixels, which would then have a side length of sqrt(4pi)/(lmax+1). This works out to 1.13*pi/l, so very close to your formula and far from the 2pi/l I had been using.)

3