Expansion of the Universe
One of the main goals of cosmology is to figure out how the universe expands as a function of time.
1.1 Expansion and Conservation
To describe the evolution of the average universe, one needs only two kinds of equations:
1. The equation that relates the density and pressure of constituents of the universe (such as
baryons, cold dark matter, photons, neutrinos, dark energy) to the expansion of the universe,
and
2. The equation that describes the energy conservation of the constituents.
Consider a line connecting two arbitrary points in space (which is expanding), and call it L. As the
universe expands, L changes with time. As you will derive in homework using

General Relativity,
the equation of motion for L is given by

L¨(t) = −
4πG
3
L(t)
X
i
[ρi(t) + 3Pi(t)] ,

(1.1)
where ρi(t) and Pi(t) are the energy and pressure of the ith component of the universe, respectively.
Here, note that the absolute value of L does not affect the equation of motion for L. Therefore,
one may define a dimensionless “scale factor,” a(t), such that L(t) ≡ a(t)x, where x is a timeindependent
separation called a “comoving” separation, which is in units of length. In cosmology,
1
we often encounter the Hubble expansion rate, H(t), which is defined by
H(t) ≡
a˙(t)
a(t)
. (1.2)
The dimension of this quantity is 1/(time). The age of the universe can be calculated from the
above definition of H, which gives H(t)dt = da/a. Now, if we know H as a function of a instead of
t, we obtain
t =
Z
da
aH(a)
. (1.3)
Another interpretation of H is found by writing L˙(t) = H(t)L(t), which tells us that H(t)
gives a relation between the distance, L, and the recession velocity, L˙
. For this reason, it is often
convenient to write H(t) in the following peculiar units:
H(t) = 100 h(t) km/s/Mpc,
where h is a dimensionless quantity. The current observations suggest that the present-day value
of h is h(ttoday) ≈ 0.7.∗
Dividing both sides of equation (1.1) by L and using L(t) = a(t)x, we find one of the key
equations connecting the energy density and pressure to the expansion of the universe:
a¨(t)
a(t)
= −
4πG
3
X
i
[ρi(t) + 3Pi(t)] (1.4)
As expected, positive energy density and positive pressure slow down the expansion of the universe.†
This equation cannot be solved unless we know how ρi and Pi depend on time. How ρi depends
on time is given by the energy conservation equation, while how Pi depends on time is usually given
by the equation of state relating Pi to ρi and other quantities.
As you will derive in homework, the energy conservation equation is given by
X
i
ρ˙i(t) + 3a˙(t)
a(t)
X
i
[ρi(t) + Pi(t)] = 0 (1.5)
Equation (1.5) is general and does not assume presence or absence of possible interactions between
different components. If we assume that each component is conserved separately, then we have
ρ˙i(t) + 3a˙(t)
a(t)
[ρi(t) + Pi(t)] = 0, (1.6)
∗The most precise value of h(ttoday) to date from the direct measurement using low-z supernovae and Cepheid
variable stars is h(ttoday) = 0.742 ± 0.036 (Riess, Macri, et al., ApJ, 699, 539 (2009)).
†
If we ignore the effect of pressure relative to that of the energy density (which is always a good approximation
for non-relativistic matter), and write ρ(t) in terms of the total mass enclosed with a radius L,
P
i
ρi(t) = 3M
4πL3 , then
equation (1.1) becomes
L¨ = −
GM
L2
,
which is the familiar Newtonian inverse-square law. Although one must not apply the Newtonian mechanics to
describe the evolution of space (because Newtonian mechanism assumes static space), this is a convenient way to
understand equations (1.1) and (1.4)for each of the ith component. Note that the second term contains the pressure, and thus how the
energy density evolves depends on the pressure.‡
Looking at equations (1.4) and (1.5), one might think that we cannot solve for a(t) unless we
have the equation of state giving Pi(t) as a function of ρi(t) etc. While in general that would
be true, for these equations a little mathematical trick lets us combine equations (1.4) and (1.5)
without knowing the evolution of P(t)!
First, rewrite equation (1.4) as
a¨(t)
a(t)
=
8πG
3
X
i
ρi(t) − 4πGX
i
[ρi(t) + Pi(t)] . (1.7)
Using equation (1.5) on the second term of the right hand side, we get
a¨(t)
a(t)
=
8πG
3
X
i
ρi(t) + 4πG
3
a(t)
a˙(t)
X
i
ρ˙i(t)
a˙(t)¨a(t) = 8πGa(t)˙a(t)
3
X
i
ρi(t) + 4πGa2
(t)
3
X
i
ρ˙i(t)
1
2
(˙a
2
)
· =
4πG(a
2
)
·
3
X
i
ρi(t) + 4πGa2
(t)
3
X
i
ρ˙i(t). (1.8)
As this has the form of A˙ = BC˙ + BC˙ = (BC)
·
, it is easy to integrate and obtain:
a˙
2
(t) = 8πGa2
(t)
3
X
i
ρi(t) − κ, (1.9)
where κ is an integration constant, which is in units of 1/(time)2
. (A negative sign is for a historical
reason.) Dividing both sides by a
2
(t), we finally arrive at the so-called Friedmann equation:
a˙
2
(t)
a
2(t)
=
8πG
3
X
i
ρi(t) −
κ
a
2(t)
. (1.10)
‡While it is a wrong explanation, it is useful to compare this equation to the first law of thermodynamics:
T dS = dU + P dV,
where T, S, U, and V are the temperature, entropy, internal energy, and volume, respectively. To a very good
accuracy, the entropy is conserved in the universe, dS = 0. The internal energy is U ∝ ρa3
and the volume is V ∝ a
3
,
and thus
d(ρa
3
) + P d(a
3
) = 0,
which gives
ρ˙ + 3a˙
a
(ρ + P) = 0.
This is a wrong explanation because it assumes that the pressure is doing work as a increases. However, in the
average universe, the pressure is the same everywhere, and thus there is no under-pressure region against which the
pressure can do work. Equation (1.5) must be derived using GR, which you will do in homework, but the above
thermodynamic argument is an amusing way to arrive at the same equation. Also, this gives us some confidence that
it is not crazy to think that the evolution of ρ depends on P.
3
A beauty of this equation is that it is easy to solve, once a time dependence of ρi(t) is known, which
is usually the case.
General Relativity tells us that the integration constant, κ, is equal to ±c
2/R2 where R is the
curvature radius of the universe (in units of length) and c the speed of light. When the geometry
of the universe is flat (as suggested by observations), R → ∞ (giving κ → 0), and thus one can
ignore this term. Since we have so much to learn, to save time we will not consider the curvature
of the universe throughout (most of) this lecture:
a˙
2
(t)
a
2(t)
=
8πG
3
X
i
ρi(t) (1.11)
1.2 Solutions of Friedmann Equation
In order to use solve equation (1.11) for a(t), one must know how ρi(t) depends on time.
To find solutions for a(t), let us first assume that the universe is dominated by one energy component
at a time, i.e.,
a˙
2
(t)
a
2(t)
=
8πG
3
X
i
ρi(t) ≈
8πG
3
ρi(t), (1.12)
and further assume that ρi depends on a(t) via a power-law:
ρi(t) ∝
1
a
ni (t)
. (1.13)
4
Finding the solution is straightforward:
a(t) ∝ t
2/ni
. (1.14)
This is usually an excellent approximation, except for the transition era where two energy components
are equally important. There are 3 important cases:
1. Radiation-dominated (RD) era. A radiation component (photons, massless neutrinos, or
any other massless particles) has a large pressure, PR = ρR/3,§ which gives ρR(t) ∝ 1/a4
(t),
or nR = 4. We thus obtain
aRD(t) ∝ t
1/2
. (1.15)
The expansion of the universe decelerates. With this solution, we can relate the age of the
universe to the Hubble expansion rate:
H(t) = a˙(t)
a(t)
=
1
2t
. (1.16)
2. Matter-dominated (MD) era. A matter component (baryons, cold dark matter, or any
other non-relativistic particles) has a negligible pressure compared to its energy density, PM
ρM, which gives ρM(t) ∝ 1/a3
(t), or nM = 3. We thus obtain
aMD(t) ∝ t
2/3
. (1.17)
§Again, a “wrong” derivation, but there is an intuitive way to get this result using the equation of state for
non-relativistic ideal gas (this is obviously a wrong derivation because we are about to apply non-relativistic equation
of state to relativistic gas!):
P = nkBT = ρ
kBT
hEi
,
where n is the number density, T the temperature of gas, kB the Boltzmann constant, and hEi the mean energy per
particle. For relativistic particles in thermal equilibrium, hEi ≈ 3kBT, which gives P ≈ ρ/3. Now, actually, it turns
out that the error we are making by using non-relativistic equation of state for relativistic gas cancels out precisely
the error we are making by using an approximate relation hEi ≈ 3kBT. This gives us the exact relation, P = ρ/3
for relativistic particles. More precisely, the equation of state for relativistic gas takes on the form P = (1 + )ρ
kBT
hEi
with hEi = 3(1 + )kBT, giving P = ρ/3. Here, ' 0.05 and −0.10 for Fermions and Bosons, respectively.
5
The expansion of the universe decelerates. With this solution, we can relate the age of the
universe to the Hubble expansion rate:
H(t) = a˙(t)
a(t)
=
2
3t
. (1.18)
3. Constant-energy-density-dominated (ΛD) era. A hypothetical energy component (let’s
call it Λ) whose energy density is a constant over time, nΛ = 0. In this case we cannot use
equation (1.14). Going back to equation (1.12) and setting ρΛ = constant, we get ˙a/a =
constant, whose solution is
aΛD(t) ∝ e
Ht
, (1.19)
where an integration constant, H, is the same as the Hubble expansion rate (which is a
constant for this model). The expansion of the universe accelerates, which must mean that,
according to the acceleration equation (1.4), the pressure of this energy component is negative.
The conservation equation (1.5) tells us that such a component indeed has an enormous
negative pressure given by
PΛ = −ρΛ. (1.20)
While this looks quite strange, we now know that something like this may actually exist
in our universe, as the current observations suggest that the present-day universe is indeed
accelerating.
6
1.3 Equation of State of “Dark Energy” and Density Parameters
The matter has PM ρM; the radiation has PR = ρR/3; and Λ has PΛ = −ρΛ. This motivates
our writing the equation of state of the ith component in the following simple form:
Pi = wiρi
. (1.21)
Here, wi
is called the “equation of state parameter,” and can depend on time (although it is usually
taken to be constant).
Why this form? It is important to keep in mind that there is no fundamental reason why we
should use this form. This form is often used either just for convenience, or simply for parametrizing
something we do not know. At the very least, this form is exact for radiation, wR = 1/3, and for
Λ, wΛ = −1. For matter, since wM 1, the exact value does not affect the results very much.
The equation of state parameter is almost exclusively used for parametrizing “dark energy,”
which is supposed to cause the observed acceleration of the universe. If we assume that w for dark
energy, wDE, is constant, then the current observations suggest that (Komatsu, et al., ApJS, 192,
18 (2011))
wDE = −0.98 ± 0.05 (68% CL). (1.22)
In other words, the energy density of dark energy is consistent with being a constant (wDE = wΛ =
−1).
Determining wDE with better accuracy may tell us something about the nature of dark energy,
especially if wDE 6= 1 is found with high statistical significance, as it would tell us that dark energy
is something dynamical (time-dependent).
Ignoring a potential interaction between dark energy and other components in the universe
(e.g., dark matter), the energy density of dark energy obeys (see equation (1.6))
ρ˙DE(t) + 3a˙(t)
a(t)
(1 + wDE) ρDE(t) = 0, (1.23)
whose solution is ρDE(t) ∝ [a(t)]−3(1+wDE)
. On the other hand, if we do not assume that wDE is a
constant, then the energy density of dark energy obeys
ρ˙DE(t) + 3a˙(t)
a(t)
[1 + wDE(t)] ρDE(t) = 0, (1.24)
whose solution is
ρDE(t) ∝ e
−3
R
d ln a[1+wDE(a)]
. (1.25)
Putting these results together, we obtain the Friedmann equation for our Universe containing
radiation, matter, and dark energy (but not curvature) as
a˙
2
(t)
a
2(t)
= H2
(t) = 8πG
3
ρM(t0)
a
3
(t0)
a
3(t)
+ ρR(t0)
a
4
(t0)
a
4(t)
+ ρDE(t0)e
−3
R a(t)
a(t0)
d ln a[1+wDE(a)]
, (1.26)
where t0 is some epoch, which is usually taken to be the present epoch.

Equation of State of “Dark Energy” and Density Parameters
The matter has PM ρM; the radiation has PR = ρR/3; and Λ has PΛ = −ρΛ. This motivates
our writing the equation of state of the ith component in the following simple form:
Pi = wiρi
. (1.21)
Here, wi
is called the “equation of state parameter,” and can depend on time (although it is usually
taken to be constant).
Why this form? It is important to keep in mind that there is no fundamental reason why we
should use this form. This form is often used either just for convenience, or simply for parametrizing
something we do not know. At the very least, this form is exact for radiation, wR = 1/3, and for
Λ, wΛ = −1. For matter, since wM 1, the exact value does not affect the results very much.
The equation of state parameter is almost exclusively used for parametrizing “dark energy,”
which is supposed to cause the observed acceleration of the universe. If we assume that w for dark
energy, wDE, is constant, then the current observations suggest that (Komatsu, et al., ApJS, 192,
18 (2011))
wDE = −0.98 ± 0.05 (68% CL). (1.22)
In other words, the energy density of dark energy is consistent with being a constant (wDE = wΛ =
−1).
Determining wDE with better accuracy may tell us something about the nature of dark energy,
especially if wDE 6= 1 is found with high statistical significance, as it would tell us that dark energy
is something dynamical (time-dependent).
Ignoring a potential interaction between dark energy and other components in the universe
(e.g., dark matter), the energy density of dark energy obeys (see equation (1.6))
ρ˙DE(t) + 3a˙(t)
a(t)
(1 + wDE) ρDE(t) = 0, (1.23)
whose solution is ρDE(t) ∝ [a(t)]−3(1+wDE)
. On the other hand, if we do not assume that wDE is a
constant, then the energy density of dark energy obeys
ρ˙DE(t) + 3a˙(t)
a(t)
[1 + wDE(t)] ρDE(t) = 0, (1.24)
whose solution is
ρDE(t) ∝ e
−3
R
d ln a[1+wDE(a)]
. (1.25)
Putting these results together, we obtain the Friedmann equation for our Universe containing
radiation, matter, and dark energy (but not curvature) as
a˙
2
(t)
a
2(t)
= H2
(t) = 8πG
3
ρM(t0)
a
3
(t0)
a
3(t)
+ ρR(t0)
a
4
(t0)
a
4(t)
+ ρDE(t0)e
−3
R a(t)
a(t0)
d ln a[1+wDE(a)]
, (1.26)
where t0 is some epoch, which is usually taken to be the present epoch.
7
Now, taking t → t0, we find the present-day expansion rate
H2
0 ≡ H2
(t0) = 8πG
3
[ρM(t0) + ρR(t0) + ρDE(t0)] ≡
8πG
3
ρc(t0), (1.27)
which has been determined to be H0 ≈ 70 km/s/Mpc. Here, ρc(t0) is the so-called “critical density”
of the universe, which is equal to the total energy density of the universe when the universe is flat.
The numerical value of the critical density is
ρc(t0) ≡
3H2
0
8πG = 2.775 × 1011 h
2 M Mpc−3
. (1.28)
The critical density provides a natural unit for the energy density of the universe, and thus it is
convenient to measure all the energy densities in units of ρc(t0). Defining the so-called density
parameters, Ωi
, as
Ωi ≡
ρi(t0)
ρc(t0)
, (1.29)
one can rewrite the Friedmann equation (1.26) in a compact form:
H2
(t)
H2
0
= ΩM
a
3
(t0)
a
3(t)
+ ΩR
a
4
(t0)
a
4(t)
+ ΩDEe
−3
R a(t)
a(t0)
d ln a[1+wDE(a)] (1.30)
Basically, most of the literature on cosmology (within the context of General Relativity) use this
equation as the starting point.¶ Taking z = 0, one finds that all the density parameters must sum
to unity: P
i Ωi = 1.
In summary, the Friedmann equation is a combination of two key equations: (1) the equation
describing how the universe decelerates/accelerates depending on the energy density and pressure of
the constituents, and (2) the equation describing the energy conservation of the constituents. Once
the Friedmann equation is given with the proper right hand side containing the energy densities of
the relevant constituents of the universe, we can find a(t) as a function of time easily.
¶An interesting possibility is that General Relativity may not be valid on cosmological scales. There are scenarios
in which the form of the Friedmann equation is modified. One widely-explored example is the so-called DvaliGabadadze-Porrati
(DGP) model (Dvali, Gabadadze & Porrati, Phys. Lett. B485, 208 (2000)). In this scenario, the
Friedmann equation is modified to:
H
2
(t) −
H(t)
rc
=
8πG
3
X
i
ρi(t),
where rc is some length scale below which General Relativity is restored. (For r rc, the potential is given by
−GN m/r where GN is the ordinary Newtonian gravitational constant. For r rc, the potential is modified to
−G5m/r2
and decays faster. G5 is the gravitational strength in the 5th dimension.) This model has attracted a huge
attention of the cosmology community, as it was shown that this modified Friedmann equation gives an accelerating
expansion without dark energy. Namely, even when the right hand side contains only matter, the solution for this
equation can still exhibit an accelerating expansion. As this is a quadratic equation for H(t), we can solve it and find
H(t) = 1
2
1
rc
±
r
1
r
2
c
+
32πG
3
ρM(t)
.
At late times when ρ(t) becomes negligible compared to the other term, one of the solutions is given by a(t) ∝ e
t/rc
,
i.e., an exponential, accelerated expansion.
8
At present, the radiation is totally negligible compared to matter, ΩR/ΩM ' 1/3250, and the
dark energy density is about 3 times as large as the matter density, ΩDE/ΩM ' 2.7 (with ΩM ' 0.27
and ΩDE ' 0.73).
1.4 Redshift
As the universe expands, the wavelength of light, λ, is stretched linearly:
λ(t) ∝ a(t), (1.31)
which implies that photons lose energy as E(t) ∝ 1/a(t).
This is something one can observe, by comparing, for example, the observed wavelength of a
hydrogen line to the rest-frame wavelength that we know from the laboratory experiment. We often
use the redshift, z, to quantify the stretching of the wavelength:
1 + z ≡
λ(t0)
λ(temitted)
. (1.32)
The present-day corresponds to z = 0.
Using equation (1.31), we can relate the observed redshift to the ratio of the scale factors:
1 + z =
a(t0)
a(temitted)
. (1.33)
Using this result in the Friedmann equation (1.30), we obtain the most-widely-used form of the
Friedmann equation:

H2
(z)
H2
0
= ΩM(1 + z)
3 + ΩR(1 + z)
4 + ΩDEe
3
R z
0
d ln(1+z)[1+wDE(z)From this result, it follows that the best way to determine the equation of state of dark
energy is to measure H(z) over a wide range of z. If we can only measure the expansion
rates at z 1, then Taylor expansion of equation (1.34) with ΩR ΩM and ΩDE ' 1 − ΩM gives
H2
(z 1)
H2
0
≈ 1 + 3ΩMz + 3(1 + wDE)(1 − ΩM)z. (1.35)
As we know from observations that |1 + wDE| is small (of order 10−1 or less), the third term is
tiny compared to other terms, making it difficult to measure wDE. This is why we need to measure
H(z) over a wide redshift range

## joi, 31 martie 2016

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Anomalies, however, are abundant. With a discrepancy

RăspundețiȘtergereof 5200 km/s “Stephan’s Quintet, a cluster of galaxies

(NGC 7317-20), gives evidence that some redshifts may

not be directly related to distance. The galaxies are still

believed to lie at the same distance, which can be

estimated from various indicators.” 8

Another group,

NGC 2903, similarly has an anomaly of 6000 km/s.

A useful expansion parameter may be calculated by

RăspundețiȘtergerefirst selecting the length that light can travel in one of our

time intervals. For convenience, this would be a length of

1 million lightyears (the distance a light pulse travels in

the time of one million years).

We know that this distance (sometimes called a

comoving coordinate distance) increases by a fractional

amount i every million years. The increase can be

expressed as,

∆ distance = 1MLY × i ,

then divide by our chosen time interval,

∆ distance / ∆t = (1MLY / 1MY) × i .

The left side is simple the definition of average speed,

and on the right side 1MLY ÷1MY is, by definition, the

speed of light c, and can be replaced by 300,000 km/s.

Thus,

Speed of expansion = c × i .

Finally we divide both sides by the length of 1MLY

(alternately 1 mega-parsec favored by astronomers),

which is the coordinate length that references the

expansion:

v /1MLY = (c i) /1MLY.

The left side defines the expansion parameter. The right

side is easily evaluated.

Space expansion parameter = (c i) /1MLY (6-6)

= 300,000 km/s × 0.00006226 /1MLY

Space expansion parameter ≅ 18.7 km/s per MLY

We have, in effect, determined the value of the space

expansion parameter —known as Hubble’s ‘constant’ in

conventional cosmology— by using the cellular structure

of the Universe and the associated galaxy motions

induced by aether dynamics.

The Exponential Equations for Space Expansion

Equation (6-1) above provides a simple, intuitive,

approach to space expansion. The increment factor (1+ i)

is applied repeatedly to the growing coordinate length in

the same way that an interest factor is applied repeatedly

to a growing monetary investment. The formal method is

to use the expression for the relative rate of change of a

co-ordinate length r with respect to time:

dr/dt ÷ r = v ⁄ r = k , (6-7)

where k is constant when space is expanding uniformly.

The expansion is described by the ratio of the rate of

change of a length divided by that length. Note that the

value of k depends not on the length units, but only on the

time units chosen. Constant k is simply our space

expansion parameter with its length units cancelled out.

Let us, then, replace k with the space expansion

parameter (which we symbolize as H) and write the

relative rate of change equation as,

dr/dt ÷ r = v ⁄ r = H . (6-8)

Now if we choose our units so that v is in km/s and r is

in MLYs, then the expression could easily be mistaken for

the Hubble term used in conventional cosmology. The

identity confusion is but momentary; only until one

realizes that ‘their’ Hubble expansion is applied to the

entire visible universe, while our space expansion H is

applied only within the confines of cosmic cells —the

cosmic cells of a non-expanding universe

that it takes

RăspundețiȘtergereabout 300 Gigayears (GY) of comoving expansion to

convey a point, or test particle, starting one lightyear from

the center of a void and ending at the interface

150,000,000 lightyears from the center of the void. How

long does it take to reach the halfway point at 75,000,000

lightyears? Remarkably, it takes 290 GY for expansion to

reach one half the radius of a full-size cosmic bubble.

This leaves only 10 GY in which to expand the balance of

the distance to the interface; and is achieved by a

relentless increase in both the speed and acceleration of

the outward space flow (caused by expansion). Obviously

comoving expansion takes a very long time, both in

relative and absolute terms. The prolonged slow

expansion and almost negligible space flow in the central

portion of a void, leads to an interesting possibility.

Part of the DSSU theory of galaxy formation is

described as follows: As space expands in three spatial

dimensions and flows radially outward from the cosmic

bubble’s central void, space accumulates matter by a

formation process in which primitive matter emerges

from the aether, from the fundamental fluctuators that

constitute aether. The primitive matter grows and evolves

—manifesting as conventional energy and mass particles.

The important point here is that matter accumulation

within the void depends primarily on time and

consequently on radial position.

Now if we divide the total expansion-flow time of

300 GY (Graph 3) into two equal time periods along the

full nominal radius: then 150 GY is spent along the first

million lightyears of length (actually considerably less

than one million lightyears, only 11,400 LY, using

equation (6-1a)); and 150 GY along the much longer 149

million lightyears, of the latter portion of the radius. In

descriptive terms, it is as if space sits leisurely at the core

of the void for 150 Gigayears and then spends another

150 Gigayears expanding completely across the void (to

the interface boundary). This is a disproportionate

consequence of the ‘miracle’ of compounding or

exponential growth!

Back to the galaxy formation process. A vital quantity

for determining the rate of galaxy formation is missing.

What is the rate of matter formation and accumulation per

unit of volume? Equivalently one may ask, how long does

it take for a galaxy to form from pure vacuum energy and

its derivatives? It could not possibly be a short time span

—otherwise the voids would not be voids and would be

filled with proto-galaxies and mature galaxies. It would

have to be as long as possible. A reasonable assumption is

that the time span of formation is not more than 150 GY.

By the time a region of matter and energy accumulation

reaches the interface it will have evolved into a full grown

elliptical. This result is predictable and observable (only

the evolution time is contentious but seems reasonable).

If we accept this conservative time frame for the

formation of galaxies in expanding and flowing space,

and we recognize that the same time span (about 150 GY)

and the same rate of space expansion occurs in the central

core (approx. one MLY radius) of the void, we can

reasonably surmise that galaxies also form, and even

mature, in this region. It is possible that galaxy formation

is great enough to sustain a small cluster of galaxies. The

result would be a void core-region where expansion

space-flow is actually radially inward. The geometric

center of each void may actually be a region of net spacecontraction.

Without knowing the rate of matter formation

per unit of volume, the size and degree of contraction

remains speculative.

Full grown galaxies arriving at the interface is an

observable fact; the existence of galaxies in the center of a

cosmic-bubble void is an interesting idea and actually has

been reported but not verified.