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