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Cosmic Microwave Background
2/4
Paolo de Bernardis and Silvia MasiDipartimento di Fisica, Universita’ La Sapienza, Roma
Nizza, 12/Sep/2012
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 (cm-1) wavenumber (= 1/λ(cm))
Environment ! Space : to remove atmospheric emissionCryogenics : to limit instrumental emission
COBE-FIRAS• COBE-FIRAS was a
cryogenic Martin-Puplett Fourier-TransformSpectrometer 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 quasi-parallel 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
• 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):
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()()(2)( 2 xrtEIxI o πσσσ=−
( ) σπσσσ dxrtSIxI )4cos()()(0∫∞
=−
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
wavenumber 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 half-width approx. 1.23/(2xmax).
( )
( ) ( )( ) max
maxmax
'
'
44sin
)4cos()4cos(max
max
xxxIS
dxxxIS
o
oo
x
xoo
σσπσσπσ
πσπσσ
−−
=
⇒= ∫−
)4cos()( xIIxI oo πσ=−
σ (cm-1) 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 cm-1 (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 d.
• 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
r3t2
r5t2
r7t2
t rtr2t
t2
r2t2
r4t2
r6t2
nσθπδ 'cos4 nd=
( ) ( )( ) ...32cos
22cos2cos2cos(25
232
+++
+++++−=
δπσ
δπσδπσπσ
cttr
cttrctrtctrEE o
From 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 particular 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 Martin-Puplettconfiguration, 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 drawing 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 difference 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 2
2
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 i
i
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
⎟⎟⎠
⎞⎜⎜⎝
⎛
−−
=⎟⎟⎠
⎞⎜⎜⎝
⎛−−
== δ
δ
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 constant (vs OPD) if the twosources have the same spectrum.
δδδ cos22
)]cos1()cos1([21 2222
229
yxyxyx
BABABAI
−+
+=−++=
δδδ cos22
)]cos1()cos1([21 2222
22'9
yxyxyx
BABABAI
−−
+=++−=
→+= **yyxx EEEEI
( ) ( )[ ] [ ]{ } σπσσσσ dxrtSSCxI REFSKYSKY 4cos1)()(0
+−= ∫∞
( ) ( )[ ] [ ]{ } σπσσσσ dxrtSSCxI REFCALCAL 4cos1)()(0
+−= ∫∞
COBEFIRAS
• 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 (cm-1) wavenumber