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Cosmic Microwave Background
2/5
Paolo de BernardisDipartimento di Fisica, Universita’ La Sapienza, Roma
Sao Jose dos Campos12/Sep/2011
IV INPE Advanced School on AstrophysicsRadio Astronomy for the 21st Century

What is the CMB According to moderncosmology:An abundant background of photons filling the Universe.
• Generated in the very earlyuniverse, less than 4 μs after the Big Bang (109γ for each baryon) from a small asymmetry
• Thermalized in the primevalfireball (in the first 380000 years after the big bang) byrepeated scattering against freeelectrons
• Redshifted to microwavefrequencies (zCMB=1100) and diluted in the subsequent 14 Gyrs of expansion of the Universe
γ2→+ bb
t
10−6s
1013s
1017s
visibleNIRMW
visibleNIRMW
T=3000K
T=3K
em
now
em
now
rrz ==+
λλ1
T >1GeV
B(ν)
B(ν)
Today 400γ/cm3
bb −

σ (cm1) wavenumber (= 1/λ(cm))

Environment ! Space : to remove atmospheric emissionCryogenics : to limit instrumental emission

COBEFIRAS• COBEFIRAS was a
cryogenic MartinPuplett FourierTransformSpectrometer withcomposite bolometers. It wasplaced in a 400 km orbit.
• A null instrumentcomparing the specificsky brightness to the brightness of a cryogenic Blackbody

All this is cooled at 2K (271oC)

Fourier TransformSpectrometers (FTS)
• To measure spectra, you useinterference (prism, grating, FP … )
• In the case of the FTS only 2 light beams interfere: this is the simplestexperimental configuration, but resultsin a complex encoding of the spectrum.

• Take the beam to beanalyzed (A), transform itinto a quasiparallel beam(C), and split it in two beams(D).
• Delay one of the two beams(E), driving it along a longeroptical path (x).
• Recombine the beams on the detector (H and J), and record the detected power vs. the optical pathdifference (this is called the interferogram).
• The interferogram is the Fourier transform of the spectrum of the incomingradiation (as shown below).
Recipe for a FTS
ZPD
OPD

( )( ) )42cos()(
)2cos()()(xctRTE
ctRTEtE
o
o
πσπσσσπσσσ
+++=
• The OPD (optical pathdifference) is 2x.
• For a perfectlymonochromatic radiation withwavenumber σ (=1/λ) the resulting field on the detector will be
• Here RT is the efficiency of the beamsplitter (frequencydependent, in general)
ZPD
OPD
Elementary theory of the FTS

Elementary theory of the FTS( ) ( ) )42cos()()2cos()()( xctRTEctRTEtE oo πσπσσσπσσσ ++=
( ) [ ][ ]( ) [ ] ( ) )4cos1(2)(11)(
)(
)()()()(
2442
4224222
*2
xrtEeertE
eeeeeertE
tEtEtExI
oxixi
o
xictictixictictio
πσσσσσ
σσπσπσ
πσπσπσπσπσπσ
+=+++=
=++=
==∝
−
−−−
)4cos()()( xrtIxI πσσ=−
( ) σπσσσ dxrtSIxI )4cos()()(0∫∞
=−
• The power on the detector I(x) will beproportional to the mean square electrical field:
• So the interferogram is• If the input radiation is not monochromatic, and
each wavenumber has amplitude S(σ):The specificBrightness and the interferogram are related by a FourierTransform

Moving mirror displacement x (μm)
Det
ecte
dP
ower
I(x)
Det
ecte
dP
ower
I(x)
( ) σπσσσ dxrtSIxI )4cos()()(0∫∞
=−

Elementary theory of the FTS
• For obvious reasons we cannot extend x toinfinity ! If the maximum displacement of the moving mirror is xmax , all we can do is tocompute
• S’ is an approximation of the real spectrum S • The main difference is in the effective
spectral resolution of the spectrometer, which for S’ is limited to approx. 1/(2xmax).
( ) ( ) dxxIxIrtS )4cos()()( πσσσ ∫∞
∞−
−=
( ) ( ) dxxIxIrtSx
x
)4cos()()(max
max
' πσσσ ∫−
−=

Spectral Resolution• Consider a monochromatic line with
wavenumebr σo: the interferogram is
• The Fourier integral, limited to +xmax, is:
• This is an approximation of the real S(σ) which would be a delta function centeredin σo : in place of a delta, we get a sinc, with a halfwidth approx. 1.23/(2xmax).
( )
( ) ( )( ) maxmax
max'
'
44sin
)4cos()4cos(max
max
xxxIS
dxxxIS
o
oo
x
xoo
σσπσσπσ
πσπσσ
−−
=
⇒= ∫−
)4cos()( xIIxI oo πσ=−

σ (cm1) wavenumber (= 1/λ(cm))
Working with continuous radiation, lowspectral resolution is not a real problem: with a maximum displacement of the moving mirror of 1 cm we get a resolution of about 0.5 cm1 (15 GHz), perfect to describe the blackbody curve.Working with spectral lines, one wouldhave to increase the displacement. 1m is surely feasable.

Beamsplitter problems
• What we get is the input spectrum times the efficiency of the beamsplitter.
• If the latter goes to zero, we cannot retrievethe spectrum.
• So we need good beamplitters, ideally withrt=0.25, independent on frequency.
( ) ( ) dxxIxIrtSx
x
)4cos()()(max
max
' πσσσ ∫−
−=

• The simplestbeamsplitter is a dielectric slab, withrefaction index n and thickness t.
• Due to multiple reflections inside the slab, the transmittedand reflected fieldscan be computed asthe sum of an infinite number of components withdecreasing amplitude(a converging series) and increasing phasedelay.
the beamsplitter
rrt2
r3t2r5t2
r7t2
t rtr2t
t2r2t2
r4t2r6t2
nσθπδ 'cos4 nd=
( ) ( )( ) ...32cos
22cos2cos2cos(25
232
+++
+++++−=
δπσ
δπσδπσπσ
cttr
cttrctrtctrEE oFrom this, the efficiency rt(σ) is computed

'cosθσ
ndm
m =
the beamsplitter• The efficiency is a
periodic functionwith zeros at wavenumbers
• Whateverthickness and refraction indexyou select, this isnot efficient at lowfrequencies.
PolyethyleneTerephthalate(mylar or melinex)n=1.7

Linear Polarizers• Linear polarizers can be used as high efficiency
achromatic beamsplitters at long wavelengths.• A linear polarizer is an optical device transmitting
only the projection of the E field of the EM waveparallel to a given direction, which is called the principal axis of the polarizer.
• Unpolarized radiation (where the E field direction in the wavefront is random) is transformed intolinearly polarized radiation (where the E fielddirection is constant) when crossing a polarizer.
x
y
k
x
y
k
x
y
PA

• The component of the incoming field orthogonal tothe PA is either absorbed or reflected, dependingon the perticular polarizer used.
• Note that
• So, for θ=45° incidence, half of the intensity istransmitted (and half is reflected or absorbed).
Linear Polarizers• The transmitted field E’ can
be computed projecting the incoming field E along the Principal Axis:
Ein
Eout, T
PA
x
y
φEout, Rφφ sincos yxpar EEE +=( )( )⎪⎩
⎪⎨⎧
+==+==
φφφφφφ
sinsincoscossincos
,'
,'
yxypary
yxxparx
EEEEEEEE
θθ 22222 coscos'' IEEEI par ==== (Malus law)
θ

• At long wavelengths metallic wire gridsact as ideal polarizers:
• Radiation with E parallel to the wiresinduces a current in the wires, so the polarizer acts as a metallic mirror: radiation is fully reflected and is nottransmitted.
• Radiation with E orthogonal to the wires cannot induce a current in the wires, so it is transmitted.
• Radiation at a generic angle from the wires is partially transmitted(orthogonal component) and partiallyreflected (parallel component).
• If the spacing of the wires a and theirdiameter d are much less than the wavelength, the wire grid is very closeto an ideal polarizer, with its principalaxis orthogonal to the wires.
• Wire grid polarizers can be used as ideal beamsplitters for radiation at 45° wrt the wires: half of the intensity is transmitted, and half is reflected, without any wavelength dependance.
• Wire grids can be machined easilyusing a lathe and tungsten wire, whichis available in long coils.
E
E
Principal axis
Principal axis

• In the MartinPuplettconfiguration, radiation isprepared by the first polarizer, then is split by the second, and is recombined bythe third.
• There are two sources. The beam from source 1’ isreflected one time more thanthe beam from source 1.
• For each metallic reflectionthere is a 180° phase changeof the electric field, so the detector will measure the difference in spectralbrightness between source 1 and source 1’.
• The instrument becomes a zero instrument, comparingthe brightness of two sources(see later).
• In the case of FIRAS one source was the sky, the otherone was an internalblackbody.
Martin PuplettInterferometer

Martin Puplett Interferometer• Two input ports
and two output ports.
• Uses twounpolarizedsources, and twodetectorssensitive to the power.
• Let’s study the operationfollowing the beams.

Martin Puplett Interferometer• The input polarizer
A reflects radiationfrom source S1’ and transmits radiationfrom S1
• Assume that the polarizer wires are horizontal (mainaxis of the polarizervertical)
• The beam at position 3 has a vertical componentfrom S1 and a horizontalcomponent from S1’.

Martin Puplett Interferometer• The input polarizer
A reflects radiationfrom source S1’ and transmits radiationfrom S1
• Assume that the polarizer wires are horizontal (mainaxis of the polarizervertical)
• The beam at position 3 has a vertical componentfrom S1 and a horizontalcomponent from S1’.

Martin Puplett Interferometer• The wires of the
beamsplitter polarizerB are oriented to beseen by incomingradiation at 45o fromthe drawing plane.
• In this way, B reflectsa fraction of the vertical componentfrom S1 and a fractionof the horizontalcomponent from S1’ .
• So beam 4 will bepolarized at 45° and will consist of equalcontributions fromboth S1 e da S1.

The beamsplitterpolarizer B reflects radiationfrom both sources
3
4

Martin Puplett Interferometer
• In addition, B transmits a fractionof the verticalcomponent from S1and a fraction of the horizontalpolarization from S1’.
• So beam 4’ ispolarized at 45° and consists of equalcontributions fromboth S1 and S1’

The beamsplitter B transmits and reflects radiationfrom both sources
3
4
4’

Martin Puplett Interferometer• The roof mirror CD
reflects beam 4 intobeam 6 back to the beamsplitter B.
• A roof mirrror isused in place of a normal mirrorbecause it rotatesthe polarizationplane by 90o.
• In this way beam 6, which had beenreflected by B, nowis transmittedtowards the detectors.

A roof mirrorrotates the polarizationplane by 90o.
46
5

Martin Puplett Interferometer
• In the same way the roof mirrorC’D’ reflects beam4’ back towardsthe beamsplitter(as 6’)
• The polarizationplane is rotatedby 90°, so thatbeam 6’, whichhad beentransmitted (as4’) now isreflected towardsthe detectors.

Martin Puplett Interferometer• The output polarizer G
has wires parallel to the drwaing plane.
• The rays coming from the beamsplitter, which are at 45°, have both horizontaland vertical components(coming from bothsources) so theycontribute to both the beams towards the detectors (bothtransmitted and reflected)
• In this way both detectorsreceive radiation fromboth sources, whichpassed through both the arms of the interferometer.

Martin Puplett Interferometer
• The fundamentaldifference is thatradiation from source S1’underwent 4 reflections, while radiation from S1 underwent only 3 reflections. Since eachreflection produces a 180°phase shift, the instrument measures the diffence betweeninterferograms producedby S1’ and S1

Quantitative treatment: uses Jones Calculus
• Jonex matrices are used to describe linearlypolarized radiation (Jones 1941)
• The interaction of the E field of the EM wave withan optical component is described by a 2x2 matrix:
• They work only for fully polarized radiation. Forpartially polarized radiation one can use Mullercalculus (loosing any phase information).
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
INy
INx
OUTy
OUTx
EE
JJJJ
EE
,
,
2221
1211
,
,

• If E is the amplitude of the electric field of the EMW, it is represented as in the following examples:
• Linear polarizationaligned along x axis
• Linear polarizationaligned along y axis
• 45° from x axis
• 45° from x axis
• Circular polarization(right)
• Circular polarization(left)
⎟⎟⎠
⎞⎜⎜⎝
⎛01
E
⎟⎟⎠
⎞⎜⎜⎝
⎛10
E
⎟⎟⎠
⎞⎜⎜⎝
⎛11
2E
⎟⎟⎠
⎞⎜⎜⎝
⎛−11
2E
⎟⎟⎠
⎞⎜⎜⎝
⎛− i
E 12
⎟⎟⎠
⎞⎜⎜⎝
⎛i
E 12
E field

Reference system
• Ideal single mirror, orthogonal to xz plane:
• Ideal roof mirror, orthogonalto xz plane:
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=10
01M
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟
⎟⎠
⎞⎜⎜⎝
⎛−
=1001
1001
1001
RM
Mirrors
• Comoving withthe light beam :
x
y
z
y
x
z
mirror
k
k

Linear polarizers• Transmission :
Polarizer with horizontalprincipal axis:
• Transmission : Polarizer with verticalprincipal axis:
• Transmission : Polarizer with principalaxis at angle φ from x axis
• Reflection : Polarizerwith principal axis at angle φ from x axis
⎟⎟⎠
⎞⎜⎜⎝
⎛0001
⎟⎟⎠
⎞⎜⎜⎝
⎛1000
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛=
ϕϕϕϕϕϕ
ϕ 22
sinsincossincoscos
tP
Ein
Eout, T
PA
x
y
φ
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛
−−
=ϕϕϕ
ϕϕϕϕ 2
2
cossincossincossin
rP
Eout, R

Delay
• Introduced by an optical path differenceδ=4πσx : this is common for bothpolarizations, so
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛=
δ
δ
δ ii
ee
D0
0

• The two sourcesS1 and S’1 , are described by Jonesvectors
• beam 3, after the input polarizer(with horizontalprincipal axis), is
⎟⎟⎠
⎞⎜⎜⎝
⎛=
y
x
AA
S1 ⎟⎟⎠
⎞⎜⎜⎝
⎛=
y
x
BB
S '1
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=+=y
xrt B
ASPSPS '113 )0()0(

• beam 4 (reflectedby the beamsplitter) and beam 4’(transmitted by the beamsplitter) willbe:
⎟⎟⎠
⎞⎜⎜⎝
⎛++
=⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟
⎟⎠
⎞⎜⎜⎝
⎛−−
==yx
yx
y
xr BA
BAB
ASPS
21
1111
21)4/( 34 π
⎟⎟⎠
⎞⎜⎜⎝
⎛−−
=⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟
⎟⎠
⎞⎜⎜⎝
⎛==
yx
yx
y
xt BA
BAB
ASPS
21
1111
21)4/( 3
'4 π

• Since roof mirrorsare represented byunity matrix, wehave also
⎟⎟⎠
⎞⎜⎜⎝
⎛++
==yx
yx
BABA
SS21
46
⎟⎟⎠
⎞⎜⎜⎝
⎛
−−
=⎟⎟⎠
⎞⎜⎜⎝
⎛−−
== δδ
iyx
iyx
yx
yx
eBAeBA
BABA
DDSS)()(
21
21'
4'6
• S6 is transmitted by the beamsplitter, while S6’ isreflected, so
'66
'668 11
1121
1111
21)4/3()4/( SSSPSPS rt ⎟⎟
⎠
⎞⎜⎜⎝
⎛−−
+⎟⎟⎠
⎞⎜⎜⎝
⎛=+= ππ

• So
⎟⎟⎠
⎞⎜⎜⎝
⎛
++−−−−+
=)1()1(
)1()1(21
8 δδ
δδ
iy
ix
iy
ix
eBeAeBeA
S
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−+==
0)1()1(
21)0( 89
δδ iy
ix
teBeA
SPS
• After the interaction with the output beamsplitter the beams tothe detectors are
'66
'668 11
1121
1111
21)4/3()4/( SSSPSPS rt ⎟⎟
⎠
⎞⎜⎜⎝
⎛−−
+⎟⎟⎠
⎞⎜⎜⎝
⎛=+= ππ
⎟⎟⎠
⎞⎜⎜⎝
⎛+−−
==)1()1(
021)0( 8
'9 δδ i
yi
xr eBeA
SPS

• Now we can compute the power on the detectors:
• Both detectors measure a constant intensity (equal tohalf of the sum of the intensities from the twosources), plus a modulated intensity (modulated bythe optical path difference), whose amplitude is the difference of the spectra from the two sources.
• The interferogram is zero if the two sources havethe same spectrum.
δδδ cos22
)]cos1()cos1([21 222222
9yxyx
yx
BABABAI
−+
+=−++=
δδδ cos22
)]cos1()cos1([21 222222'
9yxyx
yx
BABABAI
−−
+=++−=
→+= ** yyxx EEEEI

( ) ( )[ ] [ ]{ } σπσσσσ dxrtSSCxI REFSKYSKY 4cos1)()(0
+−= ∫∞
( ) ( )[ ] [ ]{ } σπσσσσ dxrtSSCxI REFCALCAL 4cos1)()(0
+−= ∫∞

• To measure a few K blackbody, you need a cryogenic referenceblackbody, withvariable temperature. Otherwise you do notnull the signal.
• Practical design of a Blackbody cavity: seee.g. Quinn T.J., Martin J.E., (1985), Phil. Trans. R. Soc. Lond., A316, 85: who made a radiometric measurement of the Boltzmann constant (precise to 5 significant digits !)

FIRAS• The FIRAS guys were able to change the temperature of
the internal blackbody until the interferograms were flatzero.
• This is a null measurement, which is much more sensitive than an absolute one: (one can boost the gain without saturating !).
• This means that the difference between the spectrum of the sky and the spectrum of a blackbody is zero, i.e. the spectrum of the sky is a blackbody with the sametemperature as the internal reference blackbody.
• This also means that the internal blackbody is a realblackbody: it is unlikely that the sky can have the samedeviation from the Planck law as the source built in the lab.

σ (cm1) wavenumber

• Isotropic expansionor contraction:For every observer :r = physical distanceχ = comoving distancea(t) = scale factor
• FRW metric: the most generalhomogenous isotropic metric
• 1/k2 = curvature of space• If we want to know how the universe expands, we need to
integrate the Einstein Field Equations
( ) ( ) ( )⎥⎥
⎦
⎤
⎢⎢
⎣
⎡−−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−= 22
2
2
2222 sin1
)( ϕθχθχχ
χ ddk
dtadtcds
Tc
GG 48π
−=Curvature Tensor(derived from the metric of the universe)
Stressenergy tensor(describes the energycontent of the universe)
r1 (t) = χ1 a(t)
r2 (t) = χ2 a(t)
r3 (t) = χ3 a(t)
1
2
3
MW

Evolution of the scale factor• a(t) is the solution of the Friedmann equation, which is found from
Einstein’s field equations for a homogeneous isotropic medium, and can be rewritten:
•
• The solution a(t) depends on the different kinds of energy densitiesrelevant at the considered epoch. To integrate, we need the Ωs:
• The very first result is that the universe cannot be static: a=a(t). • From observations we know that the universe expands today, so a(t)
is growing today.• To say more on the previous and future behaviour of the scale factor
we need to estimate the cosmological parameters Ωi and Ho.• Cosmology recently entered in a “precision” phase, were the
cosmological parameters have been estimated with good precision. • Precision measurements of the CMB played a key role in this
process.
{ }Λ−−− Ω+Ω+Ω+Ω=⎟⎠⎞
⎜⎝⎛ 2342
2
aaaHaa
KoMoRoo&
Radiation Matter Cosm. constantCurvature
GH o
ococ
oioi π
ρρρ
83;
2
,,
==Ω

Radiation Phase• From Friedmann equation is evident that at early epochs (a small)
the expansion is driven by radiation:
• Note that the expansion rate tends to infinity at the beginning(near the Big Bang), and then decreases with time.
• In this phase the solution a (t) can be found analytically:
• We know Ho from Hubble’s law. ΩR has contributions from CMB photons, but also from all other relativistic particles present at earlyepochs. So the extrapolation using only the energy density of the CMB would not be precise.
{ } 4223422
−Λ
−−− Ω≈Ω+Ω+Ω+Ω=⎟⎠⎞
⎜⎝⎛ aHaaaH
aa
RooKoMoRoo&
{ } ( ) 2/12/12 2)(2
tHtatHa oRoRoo Ω=⇒Ω=
aa /&

log a(t)
log ρ(t)
3−= amom ρρ
4−= aror ρρThe radiation phase continues until the energydensity of radiation becomes comparable tothe energy density of non relativistic matter.
34 −− = eqMoRo aaeq ρρ
como
CMB
Mo
Ro cTaeqρ
σρρ
Ω≈=
34 /
5104 −×≈eqayt eq 2000≈
veryrough
log aeq(t)
Note: at the end of the radiation phase the temperature T=TCMB/a was still > 105 KThe universe was still ionized and opaque.

Matter Phase• When the energy of non.relativistic matter becomes
dominant
• In this phase the solution a (t) can be found analytically:
{ } 4223422
−Λ
−−− Ω≈Ω+Ω+Ω+Ω=⎟⎠⎞
⎜⎝⎛ aHaaaH
aa
MooKoMoRoo&
3/22/32/3 )(
23)()(
32
⎟⎠⎞
⎜⎝⎛ +−Ω=⇒−Ω= oooMoomoo
a
a
attHtattHao
a(t)
Hot
t1/2t2/3 From this equation we can estimate how
long it took to go from a=105 (end of radiation phase) to a=103(recombination).
The result is 380000 years.This number is important for the following.

time
Distance between
two positions
At the very beginning, the physical distancebetween two positions r(t)=χa(t) increasedat a rate larger than the speed of light. Afterwards, the expansion rate decreased.
The result is the presence of causal horizons in the Universe
{ } ( ) 2/12/12)( tHta oRoΩ=r(t)=χa(t)
422
−Ω≈⎟⎠⎞
⎜⎝⎛ aH
aa
Roo& 42
2−Ω≈⎟
⎠⎞
⎜⎝⎛ aH
aa
Moo&
3/22/3)(
23)( ⎟
⎠⎞
⎜⎝⎛ +−Ω= oooMo attHta

Expansion vs Horizon
time
Distance travelledby light since the big bang
Distance betweentwo observersr(t)=χ12 a(t)
r(t)=ct
At this time the two observers seeeachother
At this time the two observers do not see eachother: they are separated by a causal horizon.

Horizons at recombination
time
Distance travelledby light since the big bang Distance between
two observersr(t)=χ12 a(t)
r(t)=ct
Conversely, at any given time, there is a physical separationbetween observers which marks the causal horizon: observers more distant than this separation did not have enough time to exchangelight signals. They are causally disconnected.
A generic time tx

Horizons at recombination
time
Distance travelledby light since the big bang Distance between
two observersr(t)=χ12 a(t)
r(t)=ct
Observers more distant than this separation did not have enoughtime to exchange light signals. They are causally disconnected. Observers closer than this separation have been in causal contact.
Distance between twoobservers entering the horizon at txr(t)=χ34 a(t)
A generic time tx

Horizon• The physical phenomena happening within
the causal horizon ar different from the phenomena at scales outside the horizon.
• Forces are transmitted at most at the speedof light, so phenomena outside the causalhorizon are frozen until they enter the horizon.
• We should be able to see the effects of causal horizons, impressed in the image of the CMB.

Horizons• At recombination (t=380000 years), only regions of the
Universe closer than 380000 light years have had the possibility (enough time) to interact.
• They might be very different, since they could notinteract during all the previous history of the Universe, from the Big Bang to recombination !
• However, measurements show that this is not the case.• CMB anisotropy measurements.
T1
T2
1°
1°
3800
00 ly
3800
00 ly
14 000 000 000 ly
10°
• That length, as seen from a distance of the order of c/Ho = 14 Glyrs, has an angularsize of about 1 degree.

Measuringanisotropy
• Anisotropy means thatbrightness is a function of the observed direction.
• Large brightness variationswould be evident in a map.
• However, since its discovery, it was evident that the CMB isa very isotropic background.
• At a few mm of wavelength, the brightness of the sky isdominated by the CMB, and isisotropic to at least 1 partover 100.
• Only if we remove the average brightness and increase the contrast of the image, we start to seestructures.
• But this requires special experimental configurations, removing a large background an enhancing smallfluctuations.

Measuring anisotropy• A way to remove the
commonmode signalis to alternate on the detector twocontiguous regions of the sky, and measurethe AC signalremoving the DC signal electronically.
• A small AC signalsynchronous to the modulation can beextracted fromdetector noise using a demodulationtechnique (a lockinamplifier).
Lockin
In ref.
In S.
Out
Towa
rdssky
region
A
Toward
s sky re
gion B
detector
wobblingmirror

• Using a modulation/demodulationtechnique has 3 purposes:
• Remove the common mode signal (instrumentaland atmospheric emission– to first order)
• Remove noise at frequencies different fromthe modulation frequency
• Produce a signalproportional to the Difference of Brightnessbetween the two observedregions.
Lockin
Lockin
In ref.
In S.
Out
Towa
rdssky
region
A
Toward
s sky re
gion B
detector
wobblingmirror
( ))()1( ABtBtBERAP satmatmosphatminstrA +−+Ω=( ))()1( BBtBtBERAP satmatmosphatminstrB +−+Ω=
{ })()( BBABERtAP ssatm −Ω=Δ

t
ASBS
Lockin
t
1+
1−
Signal from detector
Reference signal from modulator (wobbling mirror)
BA SS −Signal times Reference
t
Lockin
In ref.
In S.
Out
[ ]BABAout SSSStRtSS −=−×+×== 212121 11)()(In ref.
In S.
X Out∫T
dttVT 0
)(1
Offset due to commonmode emission
A:
B:

Lockin
In presence of noise :
Lockin
In ref.
In S.
Out
[ ]
[ ] '1)(1)(11
)()()(
21
21
21
21
21
nSStntnSS
tRtntSS
BA
BA
out
+−=
=−×+×+−×+×=
=+=
In ref.
In S.
X Out∫T
dttVT 0
)(1
n’ is a zeroaverage noise: due to the lack of correlationbetween n(t) and R(t), n’ tends to 0 if the average period islong enough (many modulation cycles). To quantify, we need to specify the noise better.
The combination multiplier + integrator is equivalent to a bandpass filter centered on the reference frequency.

• Let’s consider for simplicity a sinewave signal :
• The signal from the sky will be at the same frequency of the reference, while the noise will have contributions at allfrequencies.
• The reference also can be a sinewave at the same frequency asthe signal:
• At the output of the multiplier we get
• If the lowpass filter has a cutoff frequency lower thanfc=1/T

)( RSSout HVV ωω −=
Sω
)( RSH ωω −)( RSH ωω +−
Rω
outV
The sky signal is exactly at the samefrequency as the reference, with no phase delay, so it is transferred tothe output as it is.
Noise contributes to all frequencies, but only those within the bandpassΔω=2(2πfc)=4π/T will contribute tothe output signal.
In case of wideband noise the improvement in S/N is verysignificant !
0
Δω
sV

max
max
2//
2;
TfNSNS
TwNfwN
in
out
VoutVin
=
==
If wV is the spectral density of the noise (in V2/Hz) :

Early measurements of CMB anisotropy
at scales of the order of the horizon. So regions which have neverbeen in causal contact before produce the same CMB brightness, withoutstanding precision. How is this possible ? This is the paradox of horizons.
410−