Download Undergraduate Project in Physics Alon Grubshtein Guided by Prof. Eduardo Guendelman

Undergraduate Project in Physics
Alon Grubshtein
Guided by Prof. Eduardo Guendelman
Department of Physics
Ben Gurion University
Summary:
This short paper summarizes mathematical notes covered in Joseph Silk's book: "THE
BIG BANG, The creation and evolution of the universe" (ISBN 0-7167-1085-4). It is
organized as a collection of small explanatory paragraphs, each addressing a note or
chapter in the book, and is intended to give the reader a higher level of understanding the
Big Bang mathematics, than that covered in the main sections of the book.
1. Parallax and Parsecs
If we define D to be a sector length of a circle with the radius r, and p, as the angle (in
radians) subtended at the center of the circle by the sector, then:
D rp
This relation is used in the measurement of stellar parallaxes by taking D equal to the
mean radius of the earth’s radius orbit about the sun – 1 Astronomical Unit (denoted 1
AU and equals 1.5 1013 cm ). Since D is known, and r is measured for each star, we can
determine the star’s distance by:
D
r
p
In practice, measurements are performed six month apart (the baseline then become the
diameter of earth’s orbit, and p is twice the corresponding angular shift of the star
between two observations).
1 Parsec (denoted pc.) is the distance corresponding to a parallax of 1 arc second.
1 pc 3.1 1018 cm 3.27ly
2. The Doppler Shift
We consider a star moving away from us at a velocity V, and emitting light waves at the
frequency . The interval between two such successive waves is 1/ , and during this
time the star has moved a distance of V/ Therefore the distance both waves have to
cover before they reach us is different, and hence, the stationary observer measures a
longer time interval between two successive wave crests:
1 1 V
sec.
c
Where c is the speed of light and equals ~ 3 105 Km s 1 . The ratio of the frequency
measured by the observer, to that of the star is:
1
V
observed
observed
1
c
c
1 V
emitted
emitted
c
This formula indicates the red shift result, which applies just as well to an approaching
star with a velocity of –V and a blue shift.
When V approaches c, we must make use of the special relativity formula:
1 V
V
c
observed
emitted
1 Z
, where Z
2
c
emitted
1 V
c
3. Magnitude
The unit of magnitude is a logarithmic measure of stellar brightness, where the brighter
stars have lower magnitudes. The brightness of a star in the night sky is its apparent
magnitude, m. Magnitudes satisfies the following relationship:
brightness2
m1 m2 2.5log10
brightness 1
With the naked eye, humans can usually see stars of the sixth magnitude, but strong
telescopes can detect faint galaxies almost as dim as the 30th magnitude.
Absolute magnitude, M is defined as the magnitude that a star would have at a distance of
10 Parsecs. Since the same distance is used, absolute magnitude is a measure of intrinsic
luminosity.
The absolute magnitude of our sun ( M ) is +4.8. The Milky Way has an absolute
magnitude of -20 and the most luminous quasars have an absolute magnitude of -27.
Since light received from a source decreases with the square of its distance we can write:
d
m1 m2 5log 1
d2
Where d is the distance to the star. Thus, the relationship between the apparent magnitude
and the absolute magnitude is:
m M 5log10 d 5
4. Blackbody Radiation
The spectrum, or wavelength distribution, of blackbody radiation per unit of volume is
given by the Planck distribution, which expresses the energy density of radiation between
a narrow range of wavelengths centered on :
8 hc
u
5
e
1
hc
kT
1
In this formula T is the radiation temperatures (in degrees Kelvin), h is Planck’s constant
( 6.625 10 27 erg s ), k is Boltzmann’s constant ( 1.38 10 16 erg K 1 ) and c the speed of
light. At wavelengths long compared with hc/kT this is reduced to a simpler expression
which is known as the Rayleigh-Jeans limit:
hc
kT
1
e
1
8 hc
1
hc
8 kT
1
5
5
4
kT
It is customary to express this in terms of frequencies and thus receive that the energy
density in a narrow frequency interval centered on is proportional to kT 2. This is
usually a good approximation at radio wavelength, but breaks down towards shorter
wavelengths, leading to the Ultraviolet Catastrophe.
Generally speaking, Planck’s distribution reaches a maximum at a wavelength given by
the Wien’s Displacement Law and is roughly:
u
8 hc
1
hc
kT
The total energy density of blackbody radiation is obtained by summing all contributions
of u from all wavelengths. This summation (integral) yield:
0.2
u
8
4
kT
3
aT 4
15 hc
Where T is in degrees Kelvin and the radiation density constant equals:
a 7.56 10 15 erg cm 3 k 4
Taken for 3 degrees Kelvin, the radiation amounts to:
u a (3) 4 6.1 10 13 erg cm 3 {1eV 1.6 10 12 ergs} 0.38 eV cm
3
This is roughly equal to the density of starlight in the Milky Way (~2.7k), and suggests a
peak at wavelength of 1mm (microwave).
5. The Friedmann Equation
We consider an arbitrary shell of matter expanding with the universe. The density at the
interior is uniform and equal to the mean cosmological mass density ( ). Denoting the
expansion velocity v, and present radius r, the kinetic energy per unit mass is 1/2v2. If
the mass enclosed is M, the gravitational potential energy per unit mass is GM/r.
The mass of the sphere contained within the shell is:
4
M
r3
3
According to Hubble’s Law v=H(t)r, and by applying the energy conservation law:
1 2 2 4
H r
G r 2 const.
2
3
However, since the shell under consideration was an arbitrary one, it follows that this
equation must apply to all possible shells and therefore to all particles in the universe.
Suppose that at some instant the spherical shell has a radius of r0, then
r R(t )r0
Where R(t) is the scale factor that expresses the amount by which the sphere has
expanded (usually defined so that it equals 1 at present).
Defining the time-independent curvature constant for the particular sphere as -1/2kr02,
leads to the Friedmann equation:
8
k
H2
G
3
R2
Where k is the curvature parameter indicating whether the universe is opened or closed.
(Friedmann developed it as a relativistic equation, but this description is limited to a
simplified non-relativistic version).
The wavelength of radiation changes in the same way as the radius of the sphere
containing a fixed amount of matter, so we can think of both waves and particles as being
conserved quantities (assuming we neglect absorption or emission processes in the
intervening space). Thus the red shift Z:
R(tobserved )ro R(temitted )ro
1
Z
1
R(temitted )ro
R(temitted )ro
We refer to the epoch temitted as t, an arbitrary time after the big bang, and so:
1
Z
1
R(t )
This expresses the relation between Z, R, and t.
6. The Einstein – De Sitter universe
In the case of k=0, the Friedmann equation is reduced to the Einstein – De Sitter universe.
The density becomes the critical density:
3H 2
4.5 10 30 gr cm 3
c
8 G
Taking H – a measured quantity - to be 15 Km sec-1 Mly-1 and G = 6.67x10-8 cm3 gr-1 s-2.
In terms of atoms, we define the critical particle density as:
ncrit 2.7 10 6 cm 3
This corresponds to 3 atoms per cubic meter.
The reduced Friedmann equation is easy to solve, and it can be shown to be always an
accurate description of the Big Bang at early times in the matter dominated era. By
conservation of particle number, the product of with the volume of the sphere is
constant so, R3= crit, where crit is the present particle density (The scale factor is unity
at present). At the era when the dominant contribution to the density is from matter:
8
1
H2
G crit 3
3
R
Expressing H(t) in term of change of R:
v R
H (t )
r R
We receive a simple differential equation:
1
dR
8
G crit R 2
3
dt
Solving for t, by separating values yields:
3
2 3 1
2 1
t
R 2
,
3 8 G crit
3 H
Or:
2
1
1
R(t ) 6 G crit 3 t 3
and
(t )
6 Gt 2
By adopting the same value of H as before, we estimate the age of the universe to be
roughly 14 billion years. If we consider the possible uncertainty in the value of H
(ranging from 15 to 30 Km sec-1 Mly-1), the age of the Einstein – De Sitter universe could
be as short as 7 billion years.
7. The Friedmann – Lemaitre Models
We shall now commence a short study of the solution to the Friedmann equation.
First we shall consider a closed universe, with k positive. The Friedmann equation:
8
k
H2
G
3
R2
Recalling that
R 3 , we observe that by increasing R, the k term becomes more
dominant and, if R were to increase beyond a certain value, H2 will become negative.
This mathematical contradiction can be resolved if R reaches its maximum value at the
first zero of H2. That means the scale factor R reaches its maximum size when the rate of
expansion slows down to zero, and the universe subsequently recollapses (“Big Crunch”).
We consider the following two equations, with an unspecified parameter y:
4 G
R
(1 cos y )
3k
4 G
t
( y sin y )
3
3k 2
We say that these equations provide a parametric representation of the solution, and it can
be shown that they satisfy the Friedmann equation. As y varies from 0 to , R increase
from 0 to a maximum value, after which it declines to zero at 2 . In contrast, t steadily
increases with the variations of y.
This representation provides us with the periodic behavior of R(t) in a closed Friedmann
universe.
Now, let us consider an open universe. In the case, the Friedmann equation becomes:
8
k
H2
G
3
R2
2
In this model H will never become zero (provided R is not infinite), and the universe
will expand forever. The appropriate parametric solution is now:
4 G
R
(cosh y 1)
3k
4 G
t
(sinh y y )
3
3k 2
This solution is very different from the former one, and now R(t) increases without limit
as time goes on (This is a characteristic property of an open universe).
For a large enough R, and a negative k, the Friedmann equation can be approximate to:
k
H2
R2
Evaluated for the present era, we take H=H0 and R=1, we get H02=k. Since this neglects
the density term entirely, we write a more accurate statement:
k
H 0 2 (1
0)
Where 0= 0/ crit, 0 is the present matter density and crit is the matter density in the
Einstein – De Sitter universe mentioned above. Expressing R in terms of the red shift Z,
we find that the Friedmann equation for an open universe becomes:
8
H2
G 0 (1 Z )3 H 0 2 (1 Z ) 2 (1
0)
3
Whereas in the Einstein – De sitter we would have:
8
H2
G 0 (1 Z )3
3
To find out at what value of Z is it appropriate to approximate an open universe by an
Einstein – De sitter universe we equate two terms:
8
G
3
Thus:
0
(1 Z )3
H 0 2 (1 Z ) 2 (1
0
)
3H 0 2 ( 0 1)
(1
1
crit
0)
(1
1
0)
8 G 0
0
0
0
Hence we see that at a red shift much larger then 1/ 0, the Einstein – De Sitter model
provides an excellent approximation to an open universe.
1 Z
8. The Radiation Era
At very early times, the radiation density made a larger contribution to the total density
then did the matter. We have shown (4) that the radiation density:
u aT 4 erg cm 3
By applying Einstein’s relation of matter and energy we find the corresponding massenergy density is:
u aT 4
radiation
c2
c2
We denote by u the Planck distribution energy density per unit volume per unit
wavelength interval. The number density of photons is obtained by dividing u by the
energy of a photon of wavelength and summing over all wavelengths. This equals the
number of photons per unit volume, and is:
3
kT
20.3T 3 photons cm 3
hc
In the 3 degrees blackbody radiation there are about 550 photons per cubic centimeter.
The effects of expansion leaves the blackbody spectrum unchanged. If the wavelength is
red shifted by a factor of R, then the energy distribution decreases by a factor of R4, due
to both energy loss (factor of R) and the expansion of volume (factor of R3). However,
blackbody radiation maintains its spectral distribution and remains blackbody radiation at
lower temperatures as the universe expands.
The radiation temperature at present T(t0)=3 K, leads to a present mass density of
radiation which is only:
6.8 10 34 gr cm 3 .
radiation (t0 )
However:
4
radiation (t )
radiation (t0 ) R
Whereas:
3
matter (t )
matter (t0 ) R
We infer that at early ages, radiation becomes dominant over matter in its gravitational
influence. This dominance first occurred at a time when the scale factor was at a red shift:
4.5 10 30
matter
Z eq
6000
7 10 34
radiation
At earlier eras the Friedmann equation is well approximated by:
8 G radiation (t0 )
H2
R4
n
60.4
And hence:
8 G
R
radiation
(t0 )
Solving for the scale factor:
R(t )
1
R
1
32 G
3
radiation (t0 )
4
t
1
2
We receive:
32 G
2
radiation (t0 )t
3
In terms of radiation temperature:
R(t )4
radiation
(t )
radiation
R(t )
(t0 )
4
3
32 Gt 2
1
3c 2 4 1 1010
T
deg k
32 Ga
t
t
(t is measured in seconds from the big bang.)
At t=1 seconds, the average energy of a blackbody radiation photon is equal to
2.3x106eV. This exceeds the rest mass energy of an electron (5x105eV).
Somewhat later, the photon energy drops below the rest mass of an electron, and
consequently particle pairs creation occurs.
At later eras the general result of the radiation temperature is given by:
T 3(1 Z ) deg k
It implies that the number density of both photons and particles vary in a similar way (T3)
and their ratio is a constant of the expansion. This statement holds true only after the
complete annihilation of electron – positron pairs (at times later then 1 second).
Subsequently there are approximately 108 photons per hydrogen nucleus. Prior to that, the
number of particles pairs comparable with the number of photons was present. This is the
basis of the statement that the early universe was asymmetric.
9. Nucleosynthesis
a. Neutron production. When electron – positron pairs are in thermal equilibrium with the
radiation field, at temperatures exceeding 1010 degrees, some of the reactions that occur
are:
p e
n
p
n e
When the average energy of a blackbody photon drops below the creation threshold, the
electron – positron pairs annihilate rather abruptly. As a consequence these reactions
stops and neutron are no longer created or destroyed.
While thermal equilibrium is maintained, the neutrons abundance can be expressed as:
(m
m ) c2
n
p
nneutron
kT
e
n proton
Where mn and mp are the masses of the neutron and proton, and e is the natural
logarithmic base. The mass difference, determines the relative proportions of neutrons.
The residual neutron abundance can be obtained to a rather good approximation by
choosing T=1010 degrees, since the reactions cease and the neutron abundance freezes at
this temperature. The result is that one neutron remains for every five protons.
b. Helium synthesis. The fate of a neutron might be to decay into a proton and an
electron. However, this decay might take as long as 15 minutes for a free neutron.
Instead, nuclear reactions occur, and neutrons tend to react with protons to make
deuterium. If the temperature is too high, the deuterium does not survive. In fact, there is
a narrow range of temperatures over which reactions are possible (around 109 degrees).
At one hand, high temperatures are needed for a nucleus to come close enough for the
nuclear forces to take over, and on the other hand, if the particles are moving too fast they
will not stay near each other long enough for the forces to work at all.
Once the universe cools down to 109 degrees (at the age of approximately 90 seconds) the
reactions begin. First we have:
n p
d
1
Which produces Deuterium (or 1 H - heavy hydrogen), and is followed rapidly by a
sequence of reactions:
d d
Helium3 n
d
d
tritium
p
d tritium
Helium 4 n
(Tritium is an unstable, ultra heavy hydrogen, consisting of one proton and two neutrons).
It so happens, that there are no stable nuclei of masses 5 and 8, and as a result nuclear
reactions rapidly diminish after making helium. Small amounts of other light elements
are made, but it is not possible to make significant amounts of heavier nuclei such as
carbon (mass 12). These require higher densities such as those found in stellar interiors,
to allow three helium nuclei to fuse in a carbon nucleus (known as Triple-Alpha).
Practically all neutrons end up forming helium nuclei. This is easily explained if we
consider the following: The probability that a neutron will collide and react with a proton
is roughly given by its geometrical size (or cross section), which we will take as r2 (r is
the nuclear radius). If there are n protons per cubic cm moving at velocity v, then the
number of collisions a neutron will undergo per second is simply ( r2)nv. We require
each neutron to undergo at least one collision within the age (y) of the universe at 109
degrees. This condition requires that the following quantity must be equal to unity for
most of the neutrons:
Q ( r 2 )nvt
We know that t is approximately 100 seconds; v is the velocity of a neutron at 109
degrees, and equals:
kT
v
3 108 cm s 1
mn
-13
The nuclear radius is 10 cm; and the particle density:
n
Resulting in:
ncrit
T
3
3
Q
10
26
2.7 10
6
3 108
3
3 108
102
8 104
Even when we consider a more accurate rate than the geometrical rate the result is Q=103,
which is still a large number. We conclude that a neutron is likely to react with a proton.
Since each helium nucleus contains two neutrons, we predict that Big Bang produces one
helium nucleus for every ten protons. This prediction is essentially independent of the
highly uncertain value of the present density, since the value of Q is so large (even had
we overestimated n by a 100 our result would not change).
c. Deuterium. Deuterium is an important by-product of the nuclear reactions that
synthesis helium. In a closed universe essentially no deuterium survives, because, if we
evaluate the likely fate of a deuterium nucleus by estimating Q for it to react, a far
smaller value is obtained then in the previous case. Very roughly, when the deuterium
density has dropped to one thousandth of the proton density, if the particle density is at
one tenth of its critical value, Q becomes less then unity, and the decay of deuterium can
be halted. However, if n is close to its critical value, Q is larger then 10, and almost all
deuterium is destroyed. In an open universe, a significant fraction of the Big Bang
deuterium would be preserved.
10. Decoupling
A critical factor in determining the state of ionization of matter in the early universe is the
ionization of newly formed atoms by blackbody radiation photons. High energy photons
(in the short wavelength limit, with wavelengths small compared to hc/kT) are most
effective at ionizing. The number of ionizing photons (relative to the particle density) is
proportional to:
I
ni
ni 3 107 157000 T
e kT
e
n
n
Here, I denote the energy required to ionize an unexcited hydrogen atom, and is
equivalent to a temperature of 157000 deg K. Though we may expect hydrogen atom
formation and decoupling to occur as soon as there are insufficient ionizing photons to
keep the atoms ionized, the temperature has to drop well below that for the exponential
factor to become sufficiently small, because of the large number of blackbody photons
(about 108) for every atom. In fact, ni/n becomes less than unity, only when T has fallen
bellow ~8000 degrees.
The Saha Equation describes the equilibrium between ionized and neutral phases in terms
of the ionized fraction x:
3
2 me kT 2 I kT
e
1 x
nh3
This equilibrium involves a balance between ionizations by the remaining photons and
electron recombination.
Solution of this equation as T drops below several thousand degrees reveals a very abrupt
change in the state of the matter from being ionized to being neutral at about 4000
degrees.
x2
11. Interaction of Radiation with Density Fluctuations
We consider density fluctuations prior to the Decoupling Era (on scales where
gravitational effects are important), separately.
a. Isothermal fluctuations. We consider an electron that is incorporated in the density
fluctuation that is moving at a small velocity v, relative to the background radiation. An
observer in the rest frame of the electron would see a net flux f of radiation, amounting to
4/3 vaT4, in the direction of motion. The fact that there is a flux of radiation means that a
drag force is exerted on the electron. If the mean free path of a photon is then the force
exerted on the electron is due to the excess radiation pressure over a distance . The
excess pressure is just f/c, and the drag force is equal to f/ c, or:
4 v aT 4
Fdrag
3 c
(The drag force is always in the opposite direction to that of the motion of electron and
hence the negative sign). The photon mean free path is:
1
n
Where is the electron scattering cross section, and is equal to 0.7x10-24cm2. This gives:
4 v
Fdrag
aT 4 n
3 c
The gravitational force is:
GM
Fgrav
G 2L
L2
Taking the mass of the fluctuation to be roughly equal to the amount of matter at the
cosmological density within a sphere of radius L. The ratio of these two forces:
Fdrag
aT 4 v
Fgrav L Gcm p
Assuming an Einstein – De Sitter cosmology this ratio equals 10-8(1+Z)5/2.
At Z=1000, prior to decoupling the radiation drag force will always exceed the
gravitational force.
The radiation drag force therefore has the effect of inhibiting any motion of fluctuations.
It also helps us to understand why the matter and radiation temperatures are so well
coupled prior to decoupling, even though processes that are ordinarily characteristic of
thermal equilibrium may not be occurring.
Any excess energy of the electrons (if the electron temperature exceeds the radiation
temperature) is lost. The process also works in reverse: if the electron temperature is
lower than the radiation temperature. In this case, the electrons will heat up until the
matter and radiation temperatures are equal.
b. Adiabatic fluctuations. The fact that radiation has a finite mean free path means that it
can eventually escape by diffusion from fluctuations of any given size. For sufficiently
small scale fluctuations, this diffusion time is rapid and will be less then the age of the
universe. The radiation will diffuse away and smooth (or damp) the fluctuations. Because
of its short mean free path, the diffusing radiation homogenizes the fluctuation as it
escapes.
Let us attempt to estimate a critical size for fluctuations to be damped by this process. If
the fluctuation size is L, and we apply a random walk with a step length of for the
photon, we get a solution of N2 scatterings to diffuse a distance equal to N scattering
lengths (“Drunkard problem”).
The scattering length will be L/ and the time per scattering would be /c, giving a total
diffusion time of: L2/ c. Diffusion time is less then the age of the universe if:
L2
!t
L ! ct
c
(We note that ct is essentially the radius of the observable universe).
After atomic hydrogen has formed, scattering becomes unimportant and the diffusion
becomes ineffective. Maximum scale of fluctuations that been damped, expressed in
terms of their mass:
3
4
4
ct 2
Md
L3
n mp
3
3
Adopting values for n and t appropriate for decoupling this expressions is roughly
Md=1012 -5/4. This is actually an underestimate of Md for the following reason: During
the process of decoupling, the photon scattering distance is rapidly increasing. Initially it
is very small compared with the size of any fluctuations of interest, but it ends up being
comparable with the size of the observable universe. This is because the decoupling
process decouples radiation from matter by removing almost all the free electrons that are
responsible for the scattering. While the photon mean free path is comparable with the
size of the fluctuation, considerable additional damping can occur.
12. Gravitational Instability
Fluctuations that survive decoupling are subject to gravitational instability, if they are of
sufficiently large scale. Sir James Jeans criterion for gravitational instability, states that
fluctuations of scale greater than a critical length (denoted RJeans), will be unstable and
grow larger. This is basically due to the competition between pressure and gravitational
forces.
To examine this criterion we consider a medium of density , a sound speed of vs and a
fluctuation of diameter R, and compare the typical kinetic energy with the gravitational
potential energy of a particle. Evidently the thermal energy dominates for small R, but
gravity takes over for large R. The thermal and gravitational energies are in balance at
R=RJeans, where:
vs
2
2
vs 2 G 2 RJeans
RJeans
G
While a more rigorous derivation yields:
RJeans
vs
G
Study of the behavior of fluctuations in an expanding universe leads to the conclusion
that, on scale smaller then the critical length, the fluctuations oscillate like pressure
waves. On larger scales, the density fluctuations are weakly unstable. This is manifested
in a slow growth in amplitude of the density fluctuations.
We call the mass contained within a sphere of diameter RJeans the Jeans mass:
M Jeans
vs
3
2
3
G
While the universe is radiation dominated, the mixture of matter and radiation maintains
a high sound speed: vs=c/(3)1/2. This means that the Jeans mass encompasses all the mass
within the observable horizon. Immediately before decoupling this mass exceeds that of
the largest clusters of galaxies. After decoupling, the sound velocity drops abruptly to a
value characteristic of hydrogen at 2000 degrees Kelvin, which is roughly 4 Km s-1. To
computer the Jeans mass now, we set the following values:
vs 4 105 cm s 1
6 10 30 (1 Z )3 gr cm 3
In the Jeans mass equation, and get:
1
M Jeans 109
solar masses 105 M
3
1
2
2
(1 Z )
The growth and eventual collapse of the gravitationally unstable fluctuations must lead to
the formation of the galaxies.
To estimate that rate at which fluctuations larger than the Jeans length grow, we examine
the parametric representation of the solution to the Friedmann equations derived earlier:
R A(1 cos y ) and t B ( y sin y )
We consider a region that is slightly denser than the background. It will tend to lag
behind the expansion of the rest of the universe. We associate a deviation R and a
density enhancement
with respect to the scale factor R of the fluctuation, and the
background density , respectively. At early times, the parameter y can be treated as
small, and we can write the Taylor expansion for both Cosine and Sine function to obtain:
1 2 1 4
y3
R " R A( y
y ) and t B
2
24
6
Since we are considering small perturbations we must have:
2
4
R y 2 t 3 and " R y 4 t 3
Furthermore, since the multiple (R+ R)( + ) is a constant by virtue of mass
conservation in the disturbed volume, the fractional change in density must be equal to
3 R/R. It follows that:
"
y2
t
2
3
R
Or, the density contrast of fluctuations increases proportionately to the expansion scale
factor.
For example, if during decoupling, fluctuations were present with amplitude of 1 percent,
they would have become large by a red shift of 10 when the universe has expanded 100
times (this result is only true if the curvature term is still negligible at a red shift of 10).
In a closed universe, the rate of growth of density perturbations tends to be enhanced over
the rate in an Einstein – De Sitter universe. In an open universe, this growth is inhibited,
because a low density universe (with an important curvature term), is essentially coasting
along on the kinetic energy of the expansion. The gravitational energy term in the
Friedmann equation is relatively small, making the self gravitational attraction of
fluctuations become unimportant. Once the curvature term become significant, density
fluctuations which are not large enough to re-collapse, cease to grow. Such an effect may
be important on very large scales, corresponding to super clusters of galaxies.
13. Collapse of Density Fluctuations: Galaxy Formation
Applying the theorem stating that in a sphere with symmetric matter distribution only the
matter contained within a spherical shell contribute to the gravitational forces acting on
that shell, we decompose the fluctuation into a series of spherical shells. For a given shell
with radius r, we write:
r A(1 cos y ) and t B( y sin y )
We note that though both A and B are still constants, they may change from shell to shell
(they depend on the mass within any given shell). By specifying the mass M within a
shell, and the total energy (1/2 E) per unit mass of the shell:
1
1 2 GM
E
V
2
2
r
We receive the relation A=GM/E and B=GM/E3/2 (which will not be derived here). If we
identify E with kr2 we receive a somewhat similar equation to that derived previously,
only now, k is no longer a constant , rather, it varies from shell to shell.
An interesting result of these equations, is that as time increases with the value of y, r
reaches a maximum value (at y= ) and then decreases. In other words, at time t* = B the
shell has expanded to the greatest radius it will attain. As the shell turns around near time
t* its velocity lags more and more behind the Hubble expansion, until if finally drops
down to zero, and the shell collapses. By the time 2t* the shell completely re-collapses.
At t* the density of matter within the shell is not much greater than the density of the
unperturbed background. At 2t*, density is formally infinity, though what really happens
is that the pressure will intervene to halt the collapse. This will lead to galaxy formation
at a finite (nonzero) value of the radius.
14. Characteristic Masses and Dimensions of Galaxies
We consider a density fluctuation that begins to collapse. The balance between the
pressure which will halt the collapse and the gravity forces is determined by the Jeans
criterion. If the cloud is much more massive than the Jeans mass appropriate to the
density and temperature, it may keep collapsing. The cloud heats up as it collapses, and
the Jeans mass will increase. The collapse will stop, when the forces are in balance, or
when the Jeans mass first becomes comparable with the cloud mass.
The systematic rise of temperature is crucial for this process. As the density increases the
gas may begin to radiate. If the gas cloud can radiate away the compression energy
acquired at its collapse, it will continue to collapse.
Possible fragmentation may eventually result in stars. These fragmentations occur
whenever the Jeans mass is less that that of the cloud, in which case fluctuation will
develop and collapse on scales comparable to the Jeans mass. Cooling will cause the
Jeans mass to decrease (proportional to T3/2 -1/2), since the density increase with the
collapse, and the cloud will fragment into smaller and smaller sub fragments.
We require that the cooling time be comparable with the characteristic collapse time
scale, to derive a simple condition for a hydrogen gas cloud to cool. Both time scales can
be expressed in terms of temperature and density.
At temperatures below ~ 3x105 degrees, cooling is mainly due to the combination of
electrons and protons. Cooling time for a gas at temperature T and density n is roughly:
3
T 2
sec
tcool 3 10
n
For the gravitational collapse time we consider a gas cloud of mass M, and radius R that
is allowed to cool. Collapse velocity will be:
GM
V
R
And hence the time for it to collapse appreciably will be:
5
3
3 1015
n
R
R 2
3 1
t grav
v
4 G
GM
(By converting mass density to particle density)
The ratio of these time scales:
tcool
t grav
10
10
T
3
sec
2
n
However:
M Jeans
10
35
T
3
2
n
gr
And therefore:
tcool
M Jeans
M Jeans
45
t grav 10 gr 5 1011 M
This suggests that only when the Jeans mass had decrease enough will a collapsing cloud
be able to cool, fragment, and create stars. Early collapse will not be associated with star
formation, since the gas remains hot.
This argument, however, must be modified if considering higher initial temperatures,
where the gas is too hot to decouple and radiates mostly by a process known as free-free
radiation. Cooling time for this process is approximately:
T
tcool 3 1011
sec
n
Leading to:
tcool
RJeans
T
10 4
t grav
n 3 105 ly
This indicates that star formation in hot collapsing cloud will only occur once the Jeans
length has fallen to 300000 light years.
15. When Did Galaxies Form?
Knowing both the age of the universe and of our galaxy can help us infer the red shift at
which our galaxy had formed. This however, is not possible, because knowledge of both
of these ages is a mere approximation, and most galaxies could have formed either a large
red shift (Z=100) or relatively recent (Z=3-4).
We rely on the nature of the orbits of the oldest stars when attempting to infer the time of
our galaxy formation. These stars have a highly elongated orbit for a very long time,
bringing them out to the radius at which they formed. We deduce that the galaxy as a
whole must have collapsed from a region extending beyond the orbits of the oldest stars.
We estimate the density our galaxy would have if it were to fill such a volume of 200000
light years. Adopting a mass of 1012 solar masses, leads to a mean initial density of 10-25
gram per cubic centimeter. The density of the universe at any red shift:
#$%
(1 Z )3 gr cm 3
(Where =1 for a universe of critical density, and less then 1 for an open universe)
Consequently, the red shift of our initial galactic density:
25
1 Z
1
3
At larger red shifts, the universe was denser than our galaxy could have initially been.
Our galaxy couldn’t have formed by collapsing at an earlier era than given by this
condition. If our galaxy was initially much larger, then its density would have been rather
low and it could have formed relatively recent. Unfortunately, even the extent of our
galaxy is poorly known, bringing us back to estimates as low as 3 or 4 for galaxy
formation.
16. Formation Of Galaxy Clusters
A galaxy cluster may initially have been a density fluctuation that was small in amplitude
compared to the background matter, which grew by gravitational instability to form
galaxies. These galaxies collapsed and fell toward one another creating the clusters. This
process is quite similar to the galaxy creation process, only that the concept of pressure
has not meaning when describing the evolution of cluster galaxies. Estimation of the
cluster final radius is rather easy, because no energy is lost from the systems by collisions
(galaxies don’t often collide like atoms).
When the galaxies just begin to fall into the cluster, the energy of the cluster is entirely in
the form of gravitational energy:
GM
E
R*
Where M is the cluster’s mass, and R* is its radius.
We follow the galaxies through the re-collapse. At first, their motion is probably inward,
and after they have been passed by and been deflected by other galaxies, they acquire non
radial velocities. Since the energy is conserved the mean radial velocity is lowered, and
the galaxies bounce back to smaller radius than that from which they fell.
If the final radius of the cluster is R0, then the energy per unit mass:
GM 1 2
E
v
R0
2
Assuming equilibrium, we invoke the Virial Theorem which requires:
1
GM
GM
2 v2
0
v2
2
R0
R0
This leads to:
1 GM
2 R0
E
But since the energy is conserved:
1
R*
2
The Cluster collapses by a factor of two in radius.
A similar argument may also apply to the formation of an elliptical galaxy, if
fragmentation into stars occurs before the protogalaxy has re-collapsed.
R0
17. Star Formation
The above mentioned fragmentation of a collapsing cloud goes on until star formation.
This happens when the role of cooling diminishes due to fragments that are sufficiently
opaque to inhibit radiation from escaping. Once these heat up, pressure forces can
balance gravity and a protostar is formed.
It is possible to estimate the Jeans mass for a fragment when opacity first becomes
significant. We consider such an opaque fragment, which satisfies the Jeans criterion.
The rate at which a particle in this fragment will cool is on the average:
T4
erg cm 3 s 1
RJ
Where is the Stephan Boltzman constant, and equals 4.6x10-5 erg cm-2 K-4. This
follows when we regard the fragment as radiating like a blackbody. Radiation escapes
from a depth in the fragment roughly equal the Jeans length.
The cooling time scale is the ratio of the thermal energy divided by the rate at which
cooling occurs by radiation:
3nkT RJeans
tcool
2
T4
Equating this with the gravitational collapse time:
1
1
t grav
G
Gnm
(Where m is the average mass of a particle, roughly 1.3 times the mass of a proton if we
allow for the presence of helium), we obtain:
3 nkT
2 T4
kT
mG
5
1
G
n
The fragment mass:
2T 2
3 k 32
m
3
kT 2
M Jeans
R
6
mG
6
Assigning the appropriate values, we find that the critical fragment mass is equal to:
3
Jeans
1
kT 4
M
mc 2
The constant above varies from 0.001 to 0.01 for T spanning the range from 10 to 104
degrees. Clouds will cool and adjust to a temperature within this range.
Because this expression for MJeans is rather insensitive to T we have demonstrated that the
minimum fragment size will be rather less than that of the smallest known stars, with
mass of 0.01 solar mass.
18. Lifetimes of Stars of Differing Masses
Massive stars have high luminosities and are wasteful of their nuclear fuel. As such they
are short lived stars. In contrast, lower mass stars tend to radiate in a relatively feeble
manner - they use their nuclear fuel thriftily and, as a result, are long lived.
We demonstrate this by considering a star of mass M, radius R and central temperature T,
in hydrostatic equilibrium. The balance between pressure and gravity is expressed in
terms of the resulting energies:
kT GM
mp
R
Luminosity is approximately:
4 R2 T 4
L
R
Where the numerator expresses the amount of energy radiated by a blackbody of radius
R, and the denominator is the total mass of material in a column of unit area. Radiation
must overcome the opacity of this material to escape, and hence, the higher the column
density, the lower the emergent flux.
The opacity is modifying the escape radiation from the center of the star. At the surface,
the luminosity can be expressed in terms of a much lower surface temperature Ts by:
L 4 R 2 Ts 4
The opacity is proportional to its density, and the emergent flux is attenuated to the total
opacity of a column of material, and therefore to the term R. We have:
T 4 R T 4 R4
L
L M3
equilibrium:
M
T M /R
The luminosity of a star increases roughly as the third power of its mass. This can be
expressed as:
L
L
M
M
3
The nuclear fuel supply of a star can be easily calculated by Einstein’s E=mc2. Taking
into account that when hydrogen is fused into helium, only 0.7 percent of the original rest
mass is liberated as energy, we can infer the lifetime of a star. Since the total nuclear
energy supply is available in the star’s core, which constitutes about 10 percent of the
star’s mass, we get:
M
Enuclear 0.007 0.1 Mc 2 1051
erg
M
(Taking 2x1033 grams for 1 solar mass)
Therefore, the lifetime on the hydrogen burning main sequence is given by:
t
Enuclear
L
1051. M
3.8 10
33
M
10
3
M
10
M
M
2
years
M
(Using a solar luminosity of 3.8x10 erg s )
We can see that the lifetime of a star is inversely proportional to the square of its mass.
The sun for example, which has only been burning hydrogen for about 4 billion years,
has some 1- billion years to live before its fuel is exhausted.
After exhausting its hydrogen fuel supply, a star may proceed to the burning of heavier
elements until iron is synthesized. It may then end its life and wither quietly as a white
dwarf, or explosively as a neutron star or black hole. Its end depends on whether its final
mass is less or greater than ~3 solar masses.
33
-1
19. Mass of a White Dwarf
As a star evolves, its core collapses to higher and higher density and temperature as
heavier nuclei are fused. This process halts with the fusion of iron, after which, any
further collapse will only absorb energy. Since the star is continuously losing energy by
radiation, it loses pressure support and collapses to increasingly higher densities.
This collapse will be halted, for stars below a certain mass, when a critical density is
reached. The atoms are so tightly packed, resulting in a new form of pressure –
degeneracy pressure. This pressure is best understood by the Pauli Excursion Law, which
forbids two electrons to be in the same place (and state) at the same time.
We explore this pressure by using the Heisenberg Uncertainty Principle:
&p &x ' h
(Where p and x are uncertainty in the momentum of an electron and its position
respectively)
We imagine squeezing the electrons until they are x apart, leading to a density of ( x)-3.
We write the momentum as:
1
h
&p
h n 3
&x
The pressure P exerted by an electron:
P
F
A
1
N &p
&t
1
N &p &v
&x
N
&p
2
n
5
3
h2
me
2
2
3
me
&x
&x
&x
We note that this pressure is not dependent of temperature.
Hydrostatic equilibrium requires a balance between gravity and pressure, so for a star of
radius R we infer:
GM M (
GM P
P R2
n mp 2
2
R
R
R
By the previous calculation:
2
2
1
M P
M 3
n 3
R M 3
3
R
n
R
The more massive the star, the smaller it must become. This suggests a limiting mass.
Another physical effect intervenes to modify the expression we derived for pressure. As
the electrons become tightly squeezed their random motions approach c. The pressure:
4
P nc &p hc n 3
This relativistic degenerate gas requires a modification of the previous expression:
n mp
GM
R2
hc n
R
4
3
M
R
hc
n
Gm p
1
3
hc
3M
Gm p 4 R 3m p
1
3
M
hc
Gm p 2
3
2
mp
This is the limiting mass of a white dwarf, known as Chandrasekhar mass, and is equal
~1.4 solar masses. Stars less massive are supported by the degeneracy pressure, whereas
more massive stars continue collapsing until the atomic nuclei themselves become tightly
squeezed, to form neutrons (neutron star). Very massive stars will not be held even by the
neutron degeneracy and collapse furthermore, to form a black hole.
20. Bode’s Law
This is a prescription for calculating planetary distance, and played a role in the discovery
of the planet Neptune in 1846. The distance D to the nth planet:
Dn 0.4 (0.3) n AU
The method gives accurate distances for the inner planets but begins to break down for
Saturn, Uranus and Neptune. It fails to predict Pluto’s orbit.
21. Cosmological Tests
a. Density. Evaluating the Friedmann equation for the present era results in an equation
for the curvature constant:
8 G
k
( 0
H0 (
1)
crit )
3
(Where crit is the value of density appropriate to the Einstein – De Sitter model)
It is immediately clear that if >1, or if 0 exceed crit we have a positive k, and a closed
universe, and if the situation is just the opposite we have an open universe. So,
measurement of 0 would provide a conclusive cosmological test for the type of our
universe. Unfortunately, we can only have a lower limit on this value because of the
many types of non luminous matter. This limit is roughly equivalent to =0.1.
b. Red Shift Magnitude Test. The luminosity distance is a concept introduced for
measuring distances to galaxies. The flux of radiation received from a source is
diminished by the red shift. The energy of each photon is decrease by 1+Z, and so does
the rate at which photons arrive. The net result is that we receive from the galaxy:
L
flux
erg cm s 1
2
4 r (1 Z ) 2
This provides us with an effective luminosity distance of rlum=r(1+Z). This can be
computed by measuring the flux, if we know L.
To proceed further, we evaluate r along an actual light ray, yielding (general relativity):
2c
1
r
1
H0
1 Z
Using the definition of magnitude: m=M+5(log r -1), where r is expressed in parsecs:
2c
1
1
5
H0
1 Z
This is the required relation for a flat universe. Curved models yield different functions of
red shift. All models however, look alike at small red shifts and show that m and log Z
are linear for these shifts. This translates to a straight line intercepting with either axes at
a point yielding H0. At small red shifts, Z is a direct measure of distance, but for more
remote object a nonlinear term introduces a curvature that depends on whether the
universe is open or closed. (We can think of space in an open universe as being stretched)
m
M
5log
c. Angular Diameter. If we consider a galaxy of actual diameter d and apparent angular
diameter A, at distance r, the relation between d and A can be shown to be:
A
d r
(1 Z )
At small red shifts, this reduces to the formula obtained in a local region of space: d=rA.
The angular size is obtained by substituting the formula for r in terms of Z, valid for an
Einstein – De Sitter universe, yielding:
d
1 Z
A (1 Z )
dH 0
r
2c 1 1 Z
An interesting aspect of this formula is that at small red shift, A varies as Z-1 whereas at
large Z, A must increase as 1+Z. This means that the angular diameter goes through a
minimum, and hence at larger red shifts (1.25 for Einstein – De Sitter) galaxies appear to
grow larger! This surprising result is due to light focusing by the gravitational field of the
universe, and may occur even in the open model, for larger Z. It offers a way to
distinguishing between models by measuring angular diameters of very distant galaxies.
This is exceedingly difficult test to perform in practice, because the images are not sharp
enough, and edges are not well defined.
d. Number Counts. We consider a static universe with a uniform distribution of sources of
luminosity L. If f is the flux from a source at distance r:
L
L
f
r
2
4 r
4 f
We denote by n0 the local number density of sources, and write the total number N of
sources brighter than f as:
N
4
4
3
r n0
n0
3
2
L
1
3
4 f
f 2
Adopting this result to the Einstein - De Sitter model is done by replacing r with the
luminosity distance rlum=r(1+Z):
N
4
n0
L
4 f
3
2
1
1 Z
3