Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Q and U parameters for polarization

  1. May 17, 2012 #1
    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. :)
     
  2. jcsd
  3. May 18, 2012 #2

    Chalnoth

    User Avatar
    Science Advisor

    It's because only a small fraction of the CMB photons are polarized.
     
  4. May 18, 2012 #3
    Maybe I wasn't clear enough with my question.
    I am wondering why are they so few polarized photons. :)
     
  5. May 18, 2012 #4

    Chalnoth

    User Avatar
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Q and U parameters for polarization
  1. Hubble parameter (Replies: 30)

  2. Hubble parameter (Replies: 2)

  3. The density parameter (Replies: 9)

Loading...