366 Cosmic microwave background anisotropies
inhomogeneities. The root-mean-square difference between a local measurement
and the cosmic mean is known ascosmic variance. This difference is due to the
poorer statistics of a single observer and depends on the number of appropriate
representatives of the random inhomogeneities within an horizon. The variance is
tiny at small angular scales but substantial for angular separations of 10 degrees or
more.
Cosmic variance is an unavoidable uncertainty, but experiments usually introduce
additional uncertainty by measuring only a finite fraction of the full sky. The total
difference from the cosmic mean, calledsample variance, is inversely proportional
to the square root of the fractional area measured, approaching cosmic variance as
the covered-by-measurements area approaches the full sky.
Because of the homogeneous isotropic nature of the random fluctuations one
can calculate the cosmic mean of the angular correlation functionC(θ)by simply
averaging over the observer positionx 0 and keeping the directionsl 1 andl 2 fixed.
For the Gaussian field, this is equivalent to an ensemble average of the appropriate
random Fourier components,〈 (^) k (^) k′〉=| (^) k|^2 δ
(
k+k′)
, etc. Keeping this remark
in mind and substituting (9.32) into (9.33), after integration over the angular part
ofk,thecosmic meanof temperature autocorrelation function can be written as
C(θ)=∫(
(^) k+
δk
4
−
3 δk′
4 k^2∂
∂η 1)(
(^) k+
δk
4
−
3 δk′
4 k^2∂
∂η 2)∗
×
sin(k|l 1 η 1 −l 2 η 2 |)
k|l 1 η 1 −l 2 η 2 |k^2 dk
2 π^2, (9.35)
where after differentiation with respect toη 1 andη 2 , we setη 1 =η 2 =η 0 .Using
the well known expansion
sin(k|l 1 η 1 −l 2 η 2 |)
k|l 1 η 1 −l 2 η 2 |=
∑∞
l= 0( 2 l+ 1 )jl(kη 1 )jl(kη 2 )Pl(cosθ), (9.36)wherePl(cosθ)and jl(kη)are the Legendre polynomials and spherical Bessel
functions of orderl,respectively, we can rewrite the expression forC(θ)asa
discrete sum overmultipole moments Cl:
C(θ)=1
4 π∑∞
l= 2( 2 l+ 1 )ClPl(cosθ). (9.37)The monopole and dipole components (l= 0 ,1) have been excluded here and
Cl=2
π∫∣∣
∣
∣
(
(^) k(ηr)+
δk(ηr)
4
)
jl(kη 0 )−3 δ′k(ηr)
4 kdjl(kη 0 )
d(kη 0 )∣
∣∣
∣
2
k^2 dk. (9.38)