Physical Foundations of Cosmology

9.10 Polarization of the cosmic microwave background 397

characterized entirely by the two-dimensional, second rank symmetric polarization
tensor

Pab≡

1

I

(

〈EaEb〉−

1

2

〈EcEc〉gab

)

, (9.134)

where the metric tensorgab=ea·eband its inverse are used to raise and lower the
indices, e.g.Ea=gacEc. The brackets represent an average over a time interval
much exceeding the typical inverse frequencies of the wave. The scalar product of
two three-dimensional vectorseais defined, as usual, with respect to the spatial Eu-
clidian metric. The overall intensity of the radiation is proportional toI≡〈EcEc〉.
If the light is polarized, then the brightness temperature of the radiation after it
passes through the polarizer depends on its orientationm=maeaand the temper-
ature variations areδT(m)∝Pabmamb.Thus, by measuring this dependence, one
can determine the polarization tensorPab.

Problem 9.18Calculate the polarization tensor and the fraction of polarization

P≡−4 det

∣

∣Pba

∣

∣ (9.135)

in two extreme cases: completely polarized and completely unpolarized radiation.

Let us assume thatP=0 and consider the eigenvalue problem for the matrix
Pba:

Pbapa=λpb (9.136)

with positive eigenvalueλ.Normalizing an eigenvectorpain such a way thatp^2 ≡
pcpc= 2 λ,we can express the polarization tensorPabthrough thepolarization
vector paas

Pab=papb−^12 p^2 gab. (9.137)

In fact, one can see that forPabgiven by (9.137), the vectorpais the solution
of (9.136) withλ=p^2 / 2 .The other independent solution of (9.136) is a vector
fαperpendicular topa, that is, fapa= 0 .Using the orthogonality condition, we
immediately obtain from (9.137) that the appropriate eigenvalue is negative and
equal to−p^2 / 2 ,in complete agreement with the fact that the polarization tensor
is traceless:Paa= 0 .The fraction of polarization can be expressed through the
magnitude of polarization vector as

P≡−4 det

∣

∣Pba

∣

∣=p^4. (9.138)