# Q and U parameters for polarization

1. May 17, 2012

### Anne-Sylvie

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.

2. May 18, 2012

### Chalnoth

It's because only a small fraction of the CMB photons are polarized.

3. May 18, 2012

### Anne-Sylvie

Maybe I wasn't clear enough with my question.
I am wondering why are they so few polarized photons. :)

4. May 18, 2012

### Chalnoth

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.