2
$\begingroup$

I understand that Cosmic Microwave Background Radiation Map provided by WMAP survey in HEALPix pixelization format is nothing but an array of temperatures associated with the cmb radiation coming from each line of sight i.e. we have a temperature value for each pixel of the map. The map array values are something like 0.047.. so they can't be abosolute temperature T. Either these values are (T-Tmean)/Tmean or they are (T-Tmean). Which of the two they are?

I am referring to the CMB map available as FITS file- "wmap_band_iqumap_r9_7yr_W_v4.fits" at this NASA/Goddard data page. The product description contains no answer to my question.

$\endgroup$
7
  • 1
    $\begingroup$ The CMB is uniform to one part in 100,000, I believe, so these values can not be absolute temperature differences. They are too large by three order of magnitude or so. Unless, of course, the dipole hasn't been subtracted, yet, in which case it might be the right order of magnitude, but I am not an expert on the CMB, so I might be talking total nonsense here. But just from looking at the magnitude the interpretation doesn't seem right. $\endgroup$ Commented Apr 23, 2023 at 19:31
  • 1
    $\begingroup$ According to lambda.gsfc.nasa.gov/product/suborbit/POLAR/… : "First of all, the temperature variations in the CMB are very, very small, and the CMB is uniform up to about 1 part in 100,000. So the variances in temperature have a range of 2.7K ± 0.00003.". So whatever units the data product you have in mind uses, it's not calibrated in absolute temperature differences. $\endgroup$ Commented Apr 23, 2023 at 19:36
  • 2
    $\begingroup$ Which data product are you looking at? Is there an associated paper? Can you link those? When I had a question like this fifteen years ago, I think I was looking at some kind of radio intensity data. $\endgroup$
    – rob
    Commented Apr 23, 2023 at 20:21
  • 1
    $\begingroup$ At a high level you are correct as to what a CMB temperature map represents. However, to be precise requires knowing exactly what data you are referring to. There should be documentation associated with the data release that tells you exactly what the data are. $\endgroup$
    – Andrew
    Commented Apr 24, 2023 at 1:55
  • $\begingroup$ @rob I am referring to the CMB map available as FITS file- "wmap_band_iqumap_r9_7yr_W_v4.fits" at lambda.gsfc.nasa.gov/product/wmap/dr4/… $\endgroup$ Commented Apr 30, 2023 at 22:58

1 Answer 1

2
$\begingroup$

Your main link contains the sentence

A detailed description of how these files are formatted is available here.

The linked page says

As previously stated, the first FITS extension contains the maps. Each row in the table represents a single pixel. The number of columns depends upon which maps are included; these columns may be:

  • TEMPERATURE: The Stokes I, or temperature, measurement in mK (thermodynamic).
  • Q_POLARISATION: The Stokes Q polarization measurement in mK (thermodynamic).
  • U_POLARISATION: The Stokes U polarization measurement in mK (thermodynamic).
  • SPUR_SIGNAL: The bandpass mismatch component.
  • N_OBS: The effective number of observations.

The data-download page you link also includes just the file headers for the W-band data you’re looking at, which also say the temperatures are in millikelvin.

The first paper in the bibliography, Jarosik et al. 2011, includes several sky maps whose temperature scales are positive and negative tens of microkelvin, which is consistent with the numbers you describe in your question. That suggests the temperatures are in fact stores as differences from the monopole (mean) temperature, as you have guessed.

The “calibration” section of the 2011 paper says that the absolute gain and baseline of the detectors were determined separately for each hour’s worth of data, following the same procedures as Henshaw et al. 2009. The 2009 paper cites a monopole temperature of 2.725 K (from Mather et al. 1999) in their discussion of the dipole subtraction.

$\endgroup$
2
  • $\begingroup$ My question has been now been answered. I sincerely thank you for your effort @rob. $\endgroup$ Commented May 1, 2023 at 18:33
  • 1
    $\begingroup$ You are welcome! Consider clicking the green checkmark so that other users will know that your problem is solved. $\endgroup$
    – rob
    Commented May 1, 2023 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.