Lecture 3: Gravitational Effects on Temperature Anisotropy
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Part I: Sachs-Wolfe Effect(s)
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Evolution of photon’s energy
Sachs & Wolfe (1967)
• Let’s find a (formal) solution for p by integrating this equation over time.
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γ i is a unit vector of the direction of photon’s momentum:
Newtonian
gravitational potential
Scalar curvature
perturbation Tensor perturbation
= Gravitational wave
Evolution of photon’s energy
Sachs & Wolfe (1967)
• Let’s find a (formal) solution for p by integrating this equation over time.
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γ i is a unit vector of the direction of photon’s momentum:
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1 ap
d(ap)
dt =
Evolution of photon’s energy
Sachs & Wolfe (1967)
• Let’s find a (formal) solution for p by integrating this equation over time.
5
γ i is a unit vector of the direction of photon’s momentum:
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1 ap
d(ap)
dt =
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d
dt + ˙
because
Formal Solution (Scalar)
or
Line-of-sight direction
“L” for “Last scattering surface”
Sachs & Wolfe (1967)
Present-day time
Comoving distance (r)
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Formal Solution (Scalar)
Line-of-sight direction Initial Condition
Sachs & Wolfe (1967)
Comoving distance (r)
Formal Solution (Scalar)
Line-of-sight direction
Comoving distance (r) Gravitational Redshit
Sachs & Wolfe (1967)
Formal Solution (Scalar)
Line-of-sight direction
“integrated Sachs-Wolfe” (ISW) effect
Sachs & Wolfe (1967)
Comoving distance (r)
Part II: Initial Condition
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Initial Condition
Only the data can tell us!
• Were photons hot, or cold, at the bottom of the potential well at the last scattering surface?
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“Adiabatic Initial Condition”
The initial condition that fits the current data best
• Definition: “Ratios of the number densities of all species are equal everywhere initially”
• For i th and j th species, n i (x)/n j (x) = constant
• For a quantity X(t,x), let us define the fluctuation, δX, as
• Then, the adiabatic initial condition is
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n i (t initial , x)
¯
n i (t initial ) = n j (t initial , x)
¯
n j (t initial )
Example of the adiabatic initial condition
Thermal equilibrium
• When photons and baryons were in thermal equilibrium in the past, then
• n photon ~ T 3 and n baryon ~ T 3
• That is to say, thermal equilibrium naturally gives rise to the adiabatic initial condition, because n photon / n baryon = constant
• This gives
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• “B” for “Baryons”
• ρ is the mass density
A Big Question
• How about dark matter?
• If dark matter and photons were in thermal equilibrium in the past, then they should also obey the adiabatic initial condition
• If not, there is no a priori reason to expect the adiabatic initial condition!
• The current data are consistent with the adiabatic initial condition. This means something important for the nature of dark matter!
We shall assume the adiabatic initial condition throughout the lectures
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Adiabatic solution
Was the temperature hot or cold at the bottom of potential?
• At the last scattering surface, the temperature
fluctuation is given by the matter density fluctuation as
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T (t L , x)
T ¯ (t L ) = 1 3
⇢ M (t L , x)
¯
⇢ M (t L )
Adiabatic solution
Was the temperature hot or cold at the bottom of potential?
• On large scales, the matter density fluctuation during the matter-dominated era is given by
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T (t L , x)
T ¯ (t L ) = 1 3
⇢ M (t L , x)
¯
⇢ M (t L ) = 2
3 (t L , x)
Hot at the bottom of the potential well, but…
⇢ M / ⇢ ¯ M = 2
Adiabatic solution
Was the temperature hot or cold at the bottom of potential?
• Therefore,
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T (ˆ n)
T 0 = 1
3 (t L , r ˆ L )
This is negative in an over-density region!
Part III: Gravitational Lensing
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Equation of motion for photons
Evolution of the direction of photon’s momentum
• Instead of the magnitude of photon’s momentum, write the equation of motion for photon’s momentum
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in terms of the unit vector of the direction of photon’s momentum, γ i :
y
x
“u” labels photon’s path
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d i
dt = 1 a
X 3
j =1
( i j ij ) @
@ x j ( + )
The sum of two potentials!
Einstein’s what could-have-been the biggest blunder
Φ or Φ+Ψ?
• In 1911, Einstein calculated the deflection of light by Sun, and concluded that it would be 0.87 arcsec.
• At that time, Einstein had not realised yet the role of spatial curvature (Ψ).
Thus, his metric was still ds 42 = –(1+2Φ)dt 2 + dx 2 . As a result, his prediction was a factor of two too small: the correct value is 1.75 arcsec.
• In 1914, the expedition organised by Erwin Freundlich (Berliner Sternwarte) to detect the deflection of light by Sun during the total solar eclipse failed.
• In 1916, Einstein predicted 1.75 arcsec by incorporating Ψ, which is equal to Φ.
• In 1919, the expedition organised by Arthur Eddington (Cambridge Observatory) confirmed Einstein’s prediction.
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What if Freundlich’s
expedition was successful?
Getting 1.75 arcsec
Let’s calculate!
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d i
dt = 1 a
X 3
j =1
( i j ij ) @
@ x j ( + )
Sun
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d 2
dt = 2 2 X
j
j @
@ x j 2 @
@ x 2
Look at i=2:
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= (2nd order) + 2GM b
[(x 1 ) 2 + b 2 ] 3/2
Integrating over dt = dx 1 , we obtain
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2 = 4GM b
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= 8.49 ⇥ 10 6 rad = 1.75 arcsec Yay!
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R = 6.96 ⇥ 10 8 m M = 1.99 ⇥ 10 30 kg
(
Gravitational lensing effect on the CMB
What does it do to CMB?
• The important fact: the gravitational lensing effect does not change the surface brightness.
• This means that the value of CMB temperature does not change by lensing;
only the directions change.
• You might be asked during your PhD exam: “Is the uniform CMB temperature affected by lensing?” The answer is no.
• Only the anisotropy (and polarisation; Lecture 8) is affected:
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T lensed (ˆ n) = T unlensed (ˆ n + d)
Basak, Prunet & Benumbed (2008)
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T unlensed (ˆ n) 24
Basak, Prunet & Benumbed (2008)
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T lensed (ˆ n) = T unlensed (ˆ n + d)
Gravitational lensing effect on the CMB
Deflection angle and the “lens potential”
• The vector “d” is called the deflection angle. For the scalar perturbation, we can write d as a gradient of a scalar potential (like the electric field):
with
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T lensed (ˆ n) = T unlensed (ˆ n + d)
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d = @
@ n ˆ
r L : the comoving distance from the observer to the last scattering surface
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