Lecture 7: Origin of CMBR Flashcards

1
Q

To fully understand the origin and physical properties of the CMBR we will need to think about

A

the ratio of the number of photons and baryons (this ratio is preserved as the universe expands)

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2
Q

the present-day energy density of the CMBR is

A

ur0 = 4.17 x 10^-14 Jm^-3

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3
Q

the typical energy of a CMBR photon is

A

E__typical = 2.7kBT

=1.02 x 10^-22 J

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4
Q

dividing the CMBR energy density by the typical energy of a CMBR photon gives

A

an estimate of the present-day number density of CMBR

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5
Q

We can convert the present-day mass density of baryons to an energy density by

A

multiplying by c^2 to give ub0

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6
Q

how to find number density of baryons

A

taking the mass of a baryon to equal the proton mass

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7
Q

comparing baryon and photon number density, we see

A

that there are a few billion photons for every baryon in the universe

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8
Q

epoch of recombination

A

The CMBR photons that we observe today come to us from a time when the universe had expanded and cooled sufficiently to allow free electrons to combine with protons and form neutral hydrogen atoms. This
epoch is referred to as recombination

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9
Q

Long before the epoch of recombination, when the universe was (say) one millionth of
its present size (which means that its temperature was one million times greater than the
current temperature of the CMBR) it would have been

A

fully ionised, with photons interacting strongly with free
electrons through Thomson scattering, so the mean free path for any photon was very short

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10
Q

mean free path of a photon

A

the typical distance that a
photon could travel before being scattered by a free electron

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11
Q

As the universe expanded and the temperature dropped the interaction rate between photons and electrons also dropped and eventually the mean free path of the photons became

A

comparable to the size of the universe itself

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12
Q

mean free path of e- being approx size of universe means the the photons could

A

travel unimpeded across the universe, no longer being scattered
(since photons do not interact strongly with neutral hydrogen) for the entire duration of the universe’s subsequent evolution.

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13
Q

the point where the scattering of photons stopped occuring is know as

A

decoupling

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14
Q

decoupling marks the point when

A

the CMBR that we observe today was ‘formed’

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15
Q

CMBR is often referred to as the

A

surface of last scattering

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16
Q

The critical physics for understanding the origin of the CMBR is

A

hydrogen ionisation, related to the binding energy of hydrogen (13.6eV)

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17
Q

binding energy of hydrogen

A

energy needed to ionise from the ground state, a single eectron from the single proton in a hydrogen atom

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18
Q
A
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19
Q

how to get a crude first estimate for the temperature above which neutral hydrogen cannot form

A

equate the ionisation energy of hydrogen with the mean photon energy of a black body distribution (ie 2.7 kBT = 13.6eV)

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20
Q
A
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21
Q

equating the binding energy of hydrogen with mean photon energy does not take into account that

A

we have a thermal distribution

22
Q

why do we conclude that the universe must be cooler than 60000K before nucleosynthesis can proceed properly

A

if we have a photon-to-baryon ration of 10^9 in the universe, there must be enough high energy photons in the tail of the distribution to still ionise hydrogen (and hence prevent its formation)

23
Q

we can make a better estimate of when the CMBR formed by using

A

the Saha equation

24
Q

the Saha equation allows us to

A

compute the ionisation fraction X of a gas in thermal equilibrium

25
Q

we define X as

A

the ratio of the number density of free protons (ionised hydrogen) to the number density of baryons

26
Q

the Saha equation takes into account the

A

balance between the two relevent processes going on:

  1. the formation of hydrogen: the recombination of nuclei and electrons into atoms
  2. the ionisation of (newly-formed) hydrogen by the high-energy photons in the tail of the distribution
27
Q

for temperatures much higher than the binding energy of hydrogen, we find that X=

A

approx 1

ie the universe is fully ionised

28
Q

graph of X against z

A

starts of horizontal around 0 for small redshift then increases to then horizontal around 1 for high redshift

29
Q

why even this improved estimate of the temperature at decoupling is not quite correct

A

as the ionisation fraction drops the remaining electrons find it increasingly difficult to combine with the remaining hydrogen nuclei to form atoms because they are both becoming too rare and they do not find each other.

30
Q

the residual level of ionisation can no longer be modelled by the Saha equation since

A

the universe is no longer in thermal equilibrium (because the interactions between photons and electrons are also becoming
too rare)

31
Q

a more sophisticated treatment (beyond scope of course) predicts a residual ionisation fraction of around

A

10^-3

it also predicts the surface of last scattering to occurs at around 3000K

32
Q

a surface of last scattering temp of 3000K means that

A

corresponds to a universe one thousandth of its present size

and we observe CMBR redshift of around z=1000

33
Q

formally, the surface of last scattering is when the mean free path of the photon equals

A

the size of the universe

34
Q

from Saha equation and the solution for X as a function of z, we can see that the transition of the universe from fully ionised to fully neutral was

A

rapid but not instantaneous

(recombination and decoupling did not happen at the same time)

35
Q

difference between recombination and decoupling

A

decoupling refers to the point where photons fall out of thermal equilibrium with matter, while
recombination refers to the epoch when electrons and protons first become bound as hydrogen nuclei. Strictly speaking, then, decoupling happens after recombination.

36
Q

to work out the age of the universe at the time when the CMBR was formed we need to account for

A

the fact that the universe was radiation-dominated in its early phase, and modify our expression for E(z) accordingly.

37
Q

We can get a rough idea of the age of the universe at z ≈ 1 000, however, if we assume

A

that we live in an Einstein de Sitter and matter-dominated universe

38
Q

for z=1000, assuming that we live in an Einstein de Sitter and matter-dominated universe gives

A

CMBR emitted when universe was a fraction around 3x10^-5 of its current age

(400,000 years after Big Bang)

39
Q

omega_rel

A

general relativistic density parameter that includes the radiation energy density and the neutrinos

=omega_r + omega_v

40
Q

taking omega_m=0.3 and h=0.7m we find that z_eq=

41
Q

we find that the epoch of matter-radiation equality occurred around

A

65,000 years
(2x10^12 seconds after Big Bang)

42
Q

what does the epoch of matter-radiation equality occurring at (2x10^12 seconds after Big Bang) mean

A

the universe was already matter dominated when the CMBR was formed

43
Q

The epoch of nucleosynthesis

A

universe around 1 second old
T= around 10^10K

44
Q

period when t was between 10^-4 and 1s

A

temperature of the universe cooled to 10^10K

the universe comprised a plasma filled with free electrons, protons,neutrons, photons, neutrinos

everything strongly interacting with everything else

45
Q

period between t=10^-10 and 10^-4 s

A

temp cooled to 10^12K

universe consisted of a soup of quarks and leptons, strongly interacting

quark-hadron transition occurred and protons and neutrons formed

46
Q

the cosmic accelerator (t<10^-10s)

A

T>10^15K, close to limits of current Earth-bound particle accelerators

47
Q

the epoch of inflation (t=10^-35s)

A

T=10^27K, universe underwent a short period of accelerated expansion known as cosmological inflation

48
Q

the Planck Scale (t=10^-43s)

A

classical description of spacetime based on GR breaks down entirely and a quantum description is required

49
Q

how did we obtain 10^-43 s for Planck time

A

start with planck mass

get energy by x by c^2

get time by dividing planck length by c