Q and U parameters for polarization

  • Context: Graduate 
  • Thread starter Thread starter Anne-Sylvie
  • Start date Start date
  • Tags Tags
    Parameters Polarization
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
3 replies · 2K views
Anne-Sylvie
Messages
11
Reaction score
0
Hello everyone :)

The CMB is polarized. I read in Dodelson's book "Modern Cosmology" that we can expect a signal of polarization weaker than the signal of temperature.
Is it only because the Stokes parameters obey $$I^2 = Q^2 + U^2 $$ (I drop V because Thomson scatering can't create V polarization) ? Or is there any other (more physical) reason ?

And I heard too that $$\langle Q \rangle = \langle U \rangle = 0$$ but I can't figure out why it is the case.

Thanks for your answers. :)
 

Discussion

Astronomy news on Phys.org
Anne-Sylvie said:
Hello everyone :)

The CMB is polarized. I read in Dodelson's book "Modern Cosmology" that we can expect a signal of polarization weaker than the signal of temperature.
Is it only because the Stokes parameters obey $$I^2 = Q^2 + U^2 $$ (I drop V because Thomson scatering can't create V polarization) ? Or is there any other (more physical) reason ?

And I heard too that $$\langle Q \rangle = \langle U \rangle = 0$$ but I can't figure out why it is the case.

Thanks for your answers. :)
It's because only a small fraction of the CMB photons are polarized.
 
Chalnoth said:
It's because only a small fraction of the CMB photons are polarized.

Maybe I wasn't clear enough with my question.
I am wondering why are they so few polarized photons. :)
 
Anne-Sylvie said:
Maybe I wasn't clear enough with my question.
I am wondering why are they so few polarized photons. :)
Right, it's because of the physics of how they were emitted. Most of the photons are randomized by the random scatter that occurred before the photons left the CMB. But those random motions were correlated, giving rise to some small amount of polarization.