>> In the last lecture, I showed that
the observed homogeneity and isotropy
of the universe implies that the metric can
be written in the Robertson-Walker form.
[ Writing Taps ]
So homogeneity and isotropy give
us the Robertson-Walker metric.
[ Writing Taps ]
And the Robertson-Walker metric is ds
squared equals minus dt squared plus a
of t squared times quantity dr squared over 1
minus kr squared plus r squared d omega squared.
Where d omega squared is shorthand for d theta
squared plus sine squared theta d fie [assumed
spelling] squared.
The part of the metric and
square brackets is just a metric
for a maximally symmetric
three-dimensional space.
There are three cases, depending on
the value of k. K is equal to plus 1,
pre-space [assumed spelling]
is a three-dimensional sphere.
And this is referred to as the closed case
when k is equal to zero, space is flat.
When k is equal to minus 1, space
is a three-dimensional hyperboloid.
And this is referred to as the open case.
Now this function a of t is
referred to as the scale factor.
The scale factor determines the size of space.
So said a little bit more precisely, remember
that on the largest length scales beyond
about 10 mega parsecs, where we're
dealing with clusters of galaxies,
matter on average is at rest in space.
So in other words, matter follows world
lines of constant, r theta and fie,
constant spatial coordinates,
and the proper distance
between those world lines is
proportional to the scale factor a.
So a determines the distance
between clusters of galaxies.
Now, one of the important quantities
that characterize the universe
is the Hubble parameter.
And it's defined as a dot that's the t
derivative a divided by a, and the current value
of the Hubble parameter, which we call H-naught.
So this is H evaluated at the
present time, which we call t-naught.
This is about 70 kilometres
per second, per mega parsec.
This number comes from observations.
And because it's positive, that
implies that a dot is positive.
So we observed that the universe
is currently expanding.
Now, theoretically, the scale factor is
determined by the Einstein equations,
but before we can write down
the Einstein equations,
we need to specify the matter
content of the universe.
So in cosmology, we usually model the matter
as composed of three types of perfect fluids.
So let's recall what a perfect fluid is?
The perfect fluid has stress-energy momentum
tensor given by the rest mass density
or rest energy density times U mu U
nu, where this is the four-velocity
of the fluid particles plus pressure
times g mu nu plus U mu U nu.
So row is the mass density, or energy density --
-- and P is the pressure.
And the relationship between mass density and
pressure is given by an equation of state --
which tells how P depends on ro.
Now the three types of perfect
fluids are referred to as matter --
-- radiation --
and dark energy.
Notice that I've used the word matter
here in the sense that it's usually meant
in gravitational physics as
non-gravitational fields.
And here I'm using the word
matter in the sense that it's used
in cosmology to refer to massive particles.
So matter in this broad sense, includes matter
in this more restrictive sense,
plus radiation and dark energy.
So let me spell this out explicitly, the three
types of perfect fluids we have are: matter --
-- that consists of massive particles, baryons,
electrons, atoms, and molecules, and so forth.
And also dark matter.
We have the second component is radiation.
So this would include electromagnetic
waves, gravitational waves.
And we usually also include
neutrinos in this list,
even though neutrinos have a very small mass,
they're usually included with the radiation.
And the third type of perfect
fluid is dark energy --
-- which is also known as vacuum energy.
So the total stress-energy-momentum
tensor for our universe consists
of these three types of perfect fluids.
So t mu nu is equal to the stress-energy
momentum tensor for the matter --
we'll call that t mu nu sub M for matter,
we have the stress, energy-momentum tensor
for radiation, we'll use the
subscript R. And for dark energy,
we'll use the subscript D. Now what about the
equations of state for these three components?
So first consider matter, the massive particles.
Now, most of the massive particles are tied
up in stars, and the stars don't interact
with one another very much except
through their gravitational interaction.
And the same is true for dark matter.
It doesn't interact with ordinary matter
except through gravitational interaction.
So in particular, for this matter component,
the pressure is pretty much negligible.
So we'll take the pressure for the
matter component of the universe to be 0.
So this is just what we've called
before a dust equation of state.
So the stress-energy-momentum tensor for the
ordinary matter is just the energy density
of the matter times the four-velocity U mu U nu.
Now for the radiation component --
-- we'll take the pressure of the radiation
to be 1/3 times the density of radiation.
And this result comes from an
analysis of black body radiation.
I don't want to go through that analysis here,
but we'll just accept this as the equation
of state that's appropriate for radiation.
So then the stress-energy-momentum
tensor for the radiation looks like rows
of R U mu U nu plus the pressure, which is
1/3 rows of R times g mu nu plus U mu U nu.
And this, of course, can be written as 4/3 rows
of R times U mu U nu plus
1/3 rows of R times g mu nu.
And by the way, that 4 velocity
here in the radiation component,
as well as the matter component
always refers to the average motion
of non-gravitational fields in the universe.
So U mu in all of these cases
is the vector 1,zero,0,0,
using the Robertson-Walker coordinate system.
And now the equation of state for the
third component, the dark energy component.
So for dark energy, the pressure
is just minus the energy density.
So the stress-energy-momentum tensor for this
component is T mu nu dark energy is equal
to row U mu U nu plus the pressure,
which is minus the energy density
times g mu nu plus U mu U nu.
And now, of course, this term
cancels this term and we're just left
with minus row sub dark energy times g mu nu.
Now the unique feature of
dark energy is, you'll notice,
the stress-energy-momentum
tensor is independent of U.
So in particular recall that the
energy density as seen by an observer
with 4 velocities say U sub O, sub script
O for observer sees an energy density --
-- which is given by T mu nu
U mu U nu for the observer.
And so now, if we evaluate
this for the dark energy,
we see that U mu times g mu nu times U nu
is minus 1, that minus 1 cancels this minus
and just leaves us with row sub D. So
this result shows that all observers,
no matter what their motion
might be relative to one another,
they all see the same energy density, namely
rows of D. Now this is the key property
of the energy density associated with vacuum
fluctuations in quantum field theories.
Namely, the energy density is
observer-independent or its Lorentz variant.
So that's why dark energy is also
referred to as vacuum energy.
So we've now specified the three components
of the matter content of the universe.
This is the dark energy component.
We have the radiation component --
-- and the dust component,
the ordinary matter component.
And we can put those together to form
the full stress-energy-momentum tensor.
Now, remember the full stress-energy-momentum
tensor must satisfy local
energy-momentum conservation.
So Del mu T mu nu must equal zero.
Now we're actually going to
assume something a bit stronger.
We're going to assume that the three
components, the matter radiation
and dark energy don't interact significantly.
So they separately the separate stress imaging
momentum tensors for those three components,
each obey the local energy-momentum
conservation.
So negligible interaction --
allows us to say that Del mu T mu nu
for the matter vanishes Del mu T mu nu,
for the radiation vanishes and Del mu T mu nu
for the dark energy component also vanishes.
Let's start by examining the energy-momentum
conservation for the matter contribution.
So this is for matter.
So the stress-energy-momentum tensor
is row for the matter times U mu U nu
where you U mu is just one 1,zero,0,0.
And now the Del U T mu nu equals 0.
And this is partial mu T mu nu plus gamma mu
mu sigma T sigma nu plus gamma nu mu sigma T
mu sigma.
And of course, this is for the matter component.
Now let's set this free index nu equal
to the coordinate T. Then we have
zero equals partial mu T mu T,
but notice the only non-zero component
of the T mu nu is that t t component.
So this first term is just partial T of
big T, t t, here we have sum of our mu,
and then the only non-sigma is when sigma is
equal to T, and this is capital T t t sub M.
And this is mu and sigma [inaudible]
both equal T we've selected nu T for T.
So this is gamma T t t t T sub M tt.
Now we can compute the Christoffel
symbols for the Robertson-Walker metric.
And what we find is that the t t t component
vanishes, and the r r t theta theta T
and fie fie T components are all equal
to a dot over a. So in this expression,
we have zero equals partial t of t
upper T T, but that's just row M.
And here we have the sum over mu.
So we ended up summing these
four Christoffel symbols.
So we have three times a dot over
a times T t t that's row sub M.
And then this term is zero
because gamma t t t is zero.
And of course, partial t
of row M is just row M dot.
Now, if we choose the free index nu to be
one of the other coordinate values, r theta,
or fie we just obtained the
equations zero equals zero.
So this is the full content of the local
stress, energy-momentum conservation
for the ordinary matter component.
And by the way, I should have mentioned before,
but we can assume that the density is
a function only of T by homogeneity.
It can't depend on r theta or fie.
Now, this equation is fairly easy to solve.
By separation of variables, we can write D row
M divided by row N equals minus 3 d a over a.
And now integrating both sides, we have log
of row M equals minus 3 times,
log of a plus a constant.
Now, we bring this term to the
left-hand side as log of a cubed.
Now, combining these equations, we find the
result row M times a cubed equals a constant.
So this is our result for the
matter contribution to the universe.
And it has a natural interpretation since a
is proportional to the volume of the universe.
It just tells us as a increases,
as the volume increases,
the density of ordinary matter
decreases as one over the volume.
And now we can apply local energy momentum
conservation to the radiation component.
So Del mu T mu nu for the radiation equals 0.
And the result of the analysis its
straightforward, is that row radiation times a
to the 4th is equal to a constant.
Now, this result also has
a simple interpretation.
It tells us that as the universe expands,
the density of radiation
decreases as one over the volume.
That's 3 of these factors of a,
but the energy also decreases
because the photons are being
redshifted as the universe expands.
And the redshift, as I showed
in the last lecture,
is proportional to a to the first power.
So altogether that tells us that the radiation
energy density scales as 1 over the 4th power
of the scale factor a. And finally, we can
apply local energy-momentum conservation
to the dark energy component.
So zero equals Del mu T mu
nu for the dark energy,
but T mu nu for dark energy was just minus
row for the dark energy times g mu nu.
Now the universe metric can be pulled
out from the covariant derivative.
This is minus g mu nu Del mu
row D, and now we can multiply
through by the metric, cancel
this inverse metric.
So this tells us that a covariant
derivative of row D is equal to zero.
Of course, the covariant derivative is the same
as the ordinary derivative
when it acts on a scaler.
So this just tells us that the dark
energy density is equal to a constant.
In particular, the dark energy doesn't
depend on the scale factor at all.
So as the universe expands, the
dark energy remains constant.
This is not intuitive, but it may be useful to
think of the dark energy as the vacuum energy.
So it's associated with the vacuum.
So no matter how the universe scales or
expands that energy density remains constant.
We're now ready to impose the Einstein
equation on our cosmological model.
[ Writing Taps ]
So the Einstein equation sets
the Einstein tensor g mu nu equal
to 8 pi times the total stress-energy
momentum, tensor, which has three components,
an ordinary matter component, radiation
component, and a dark energy component.
Now, if you'll recall, the
stress-energy momentum tensor
for the dark energy component is minus
the energy density times the metric.
So let's bring this term to the left-hand
side, the right, the Einstein equation,
the following by this 8 g mu nu plus 8 pi times
row sub D times little g mu nu equals 8 pi times
ordinary matter component
plus the radiation component.
Written this way, we can interpret this
term as a cosmological constant term.
We called it row sub D is a constant.
So we can define lambda, which is called the
cosmological constant as 8 pi type of row sub D.
So this is called the cosmological constant.
This terminology goes back to the original
development of general relativity by Einstein,
and Einstein went back and forth on trying
to decide whether the Einstein equation should
be written as g mu nu equals 8 pi T mu nu
or g mu nu plus a constant, called cosmological
constant, times the metric equals 8 pi T mu nu.
And originally Einstein proposed
the equations in this form.
Then he later added the cosmological constant.
Then he later took it back out, calling
that the biggest blunder of his life.
But now observations tell us that the
cosmological constant term really does need
to be there in the form of this
dark energy or vacuum energy.
Now let's bring this term back
to the right-hand side and write
out the Einstein equation in detail.
We have g mu nu equals 8 pi times
first the matter contribution
to the stress-energy momentum tensor.
That's the matter-energy
density times U mu U nu,
So it's just the stress-energy
momentum tensor for dust.
Then we have the radiation contribution
that was 4/3 row radiation times U mu U
nu plus 1/3 row radiation times g mu nu.
And then we have the dark energy contribution,
which is minus row D times g mu nu.
And now we can combine these
terms on the right-hand side.
And when I write this down again, I
want to also lower one of the indices.
It turns out that some of the
formulas are a little bit nicer
with a mixture of one up and one down index.
So this is equal to 8 pi times row
M plus 4/3 row R times U mu U nu,
and the remaining terms are 8 pi times 1/3
rows of R minus rows of D times Delta mu nu.
And now we just need to compute the Einstein
tensor for the Robertson-Walker metric.
And we can set u upper mu equal to
1,zero,0,0, and you lower nu equal to minus 1,
ero,0,0 and here are the results.
The t t component is minus 3
times k plus a dot squared divided
by a squared that's G upper T lower t
equals minus 8 pi, the minus sign comes
from this minus sign times row M plus 4/3 row
R plus 8 pi times 1/3 row R minus row D. We can
simplify the right-hand side.
This is minus 8 pi times row M plus these
two terms combined to just give row R,
and from here we have row D. So this
is just the total energy density
in all three components of
the non-gravitational fields.
And now if we solve this equation for a dot,
we find a dot squared equals minus k plus 8 pi
over 3 times row M plus row
R plus row D times a squared.
So this is the result of the t t
component of the Einstein equation.
What about the other components?
Well, the off-diagonal components --
-- just give zero equals zero.
And the other diagonal components.
So the r r component, theta theta component,
and pi pi component, I'll give the same thing,
namely minus 2 a a double dot
plus a dot squared plus k divided
by a squared equals eight pi times 1/3
row R minus row D. Now this equation can
be simplified.
We have 2 a a double dot plus a dot
squared equals minus k plus 8 pi
over 3 times 3 row D minus
row R times a squared.
Now it turns out that this equation
is not independent from this equation.
In fact, we can use this equation to derive
this equation and let me sketch that derivation.
So we'll take this equation and first
differentiate with respect to time t,
the t coordinate, we have 2 a dot, a double dot
equals the [inaudible] of k is, of course, zero.
Then we have plus 8 pi over 3 times
the derivative with respect to T
of row M times a squared plus row R times
a squared plus row D times a squared.
And now let's recall that row M
is a constant divided by a cubed.
So that means row M times a squared is a
constant over a, and we can differentiate.
Now, if we differentiate this with
respect to T, we have minus the constant
over a squared times a dot, but the constant
over a cube is row M. So this is equal
to minus row M times a times a dot.
Now recall that row R is proportional
to 1 over a to the fourth power.
So when we differentiate row R times a squared,
we obtain minus 2 row R times a times a dot.
And finally for row D, row D is just a constant.
So the derivative of row D times a squared
is just equal to 2 row D times a a dot.
Now, if we put all that together, this equation
becomes 2 a dot times a double dot equals 8 pi
over 3 times minus row M minus 2 row
R plus 2 row D times a times a dot.
Now multiply by a and divided by a dot to
obtain 2 a times a double dot equals 8 pi
over 3 times minus row M minus 2
row R plus 2 row D times a squared.
The next step is to add the original equation,
which was a dot squared equals minus k plus 8 pi
over three times row M plus row
R plus row D times a squared.
So when we add these, we have 2 a a double
dot plus a dot squared on the left-hand side,
the right-hand side is minus k, plus 8
pi over three times the row Ms cancel.
We have a minus row R and a
plus 3 row D times a squared.
Now, if I haven't made any mistakes,
this equation should be precisely --
This equation here, and it is.
So I've shown that this equation
can be used to derive this equation.
So there's no new content here.
The only independent component of the
Einstein equation is this t t component,
and this is referred to as
the Friedman equation.
Of course, we shouldn't be
surprised that the various components
of the Einstein equations are
not independent from one another.
You will recall that the components of
the Einstein equations are linked together
by equation Del mu g mu nu
equals 8 pi Del mu T mu nu.
The left-hand side here is zero, always
0 by the contracted Bianchi identity.
And the right-hand side is equal to
zero by energy-momentum conservation.
And that's always a consequence
of the matter equations of motion.
So now let's examine in some
detail the Friedman equation.
This is sometimes called the
Friedman Robertson-Walker equation.
And it's usually written as a dot
squared equals minus k plus 8 pi
over 3 times the total energy
density times a squared.
So this equation determines the
scale factor of the universe.
This a of t, for a given choice of matter
content and a choice for this constant k
which is either plus or minus 1 or zero.
Now for our basic cosmological model,
row consists of an ordinary baryonic matter
plus dark matter component row sub m,
or radiation component row sub R, and a dark
energy component row sub D. And recall rows
of m satisfies rows sub m
times a cubed is a constant.
Now we can evaluate this concept by
evaluating row sub m and a at the present time.
So if we use a subscript zero to
denote quantities at the current time,
this is row sub m zero times a sub 0 cubed.
And likewise, the radiation component satisfies
row r times a to the fourth is a constant.
And that's just equal to the present
value of row sub R times the present value
of the scale factor to the fourth power.
And of course row sub D is already a constant so
we can write that as the value of the constant
at the present time row sub
D zero just for emphasis.
And now we can write the total energy
density as row sub m zero times a 0 cubed
over a cubed plus row sub R zero
times a zero to the fourth power
over a to the fourth plus row D 0.
And the Friedman equation becomes a
dot squared equals minus k plus 8 pi
over 3 times row sub m zero times a zero cubed
over a plus row R zero a 0 to the fourth power
over a squared plus row D 0 times a squared.
Written this way, we see more explicitly
that this is just a differential equation
for the scale factor a as a function of time T.
My next task will be to rewrite this equation,
the Friedman equation, in terms of
dimensionless variables, so let me remind you
that the subscript zero he
refers to the present time.
[ Writing Taps ]
We can evaluate this equation
at the present time.
We'll have a zero dot squared
equals minus k plus 8 pi
over 3 times row m zero times a 0 squared plus
row r 0 times a 0 squared plus row D 0 times a
0 squared.
And now divide through by a 0 dot squared.
We have 1 equals minus k over a 0 dot squared
plus 8 pi over 3 times row m 0 a 0 squared
over a 0 dot squared plus 8 pi over
3 times row R 0 times a 0 squared
over a 0 dot squared plus 8 pi over 3 times
row D 0 a 0 squared over a 0 dot squared.
Now each one of the terms in this
equation is a dimensionless constant.
They're constants meaning
they're independent of time T,
and we can also check that
they're dimensionless.
For example, if you look at this term, this
is really 8 pi over three times g row m 0
over C squared, times a 0
squared over a 0 dot squared.
And you can check that this is dimensionless.
So let's define this quantity to be
omega m, this constant, and likewise,
we'll define this constant to be omega
r and this constant is omega d. So each
of these omegas is a constant proportional to
the present-day value of the energy density
of one of the three components of the universe.
So omega m is proportional to the
present-day value of the energy density
in ordinary baryonic matter and dark matter.
omega r is proportion to the present
value of radiation energy density
and omega d is proportional to the dark energy
density, which is, of course, constant in time.
And for convenience, let's also define this
term as omega k. Unlike the other omegas,
omega k is not proportional
to an energy density.
Its main significance is its sign.
So when omega k is positive, that
means little k must be negative,
which means that space is a
three-dimensional hyperboloid.
If omega k is zero, that means space
is flat, and if omega k is negative,
that means little k is positive
in spaces in 3-dimensional sphere.
Now from this equation, we can see that
all of these constants are linked together
by the equation 1 equals omega k plus omega m
plus omega r plus omega d. So in particular,
these four constants are not
independent of one another.
Now let's return to the Friedman equation,
and let's divide through by the
present value of a dot squared.
So this will be a dot squared
over a dot zero squared.
This term will be omega k. This term will
be proportional to omega m and so forth.
And the result is a dot squared over a zero dot
squared equals omega k plus omega m times a 0
over a plus omega r times a 0 squared
over a squared plus omega d
times a squared over a 0 squared.
Now this differential equation, the
independent variable is the cosmic time t.
And the dependent variable
is the scale factor a. So I'd
like to introduce dimensionless versions of
these independent and dependent variables.
So first for the independent variable,
let's define t tilde to be t divided
by the Hubble time t sub H. So recall
the Hubble time t sub H is just one
over the present value of the Hubble parameter.
So that's a zero over a 0 dot.
And for the dependent variable,
let's define a tilde as a function
of t tilde to be a of t divided by a zero.
And we can also write this as a of
t tilde times T H divided by a zero.
And now if we differentiate this with respect
to t tilde, what we find is a tilde dot,
and by dot here we mean differentiation with
respect to his argument t tilde is equal to --
now differentiating this with respect to t
tilde, we first differentiate a with respect
to his argument, that gives us a dot.
And then we differentiate the argument with
respect to t tilde, which gives us a factor
of t sub H, and then that's
divided by a sub zero.
And now using this expression for T sub H,
we see that a tilde dot is equal
to a dot divided by a dot 0.
Using these results, we can
rewrite the Friedman equation.
So the left-hand side, this
is just a tilde dot squared.
This factor a 0 over a is just
1 over a tilde, and so forth.
So what we obtain is a tilde dot squared
equals omega k plus omega m divided
by a tilde plus omega r divided by a tilde
squared plus omega d times a tilde squared.
Now, one of the nice features of these
rescale variables is that a tilde
at the present time is just equal to 1.
And you can see that from the definition.
If we evaluate this at the present
time, we just have a 0 over a 0 is 1.
And similarly, if we evaluate
this equation at the present time,
we just have a dot 0 over a dot 0 which is 1.
So the rate of change of a
tilde at the present time is 1.
Now let me put these key results together
with our previous equation
relating to your omega constants .
So 1 is equal to omega k plus
omega m plus omega r plus omega d.
We can now use these equations to build model
universes, and we just need to pick values
for the omegas, and then solve
this differential equation,
which is of course just the
Friedman equation written in terms
of these dimensionless variables.
And when we pick the omegas,
we should keep in mind
that on physical grounds we
expect omega m to be positive.
This is proportional to the present
value of the ordinary matter density.
And likewise, the radiation should be positive.
On the other hand, omega d which comes from
vacuum fluctuations, could be either positive
or negative, and then also omega k
could be either positive or negative.
A positive value of omega k corresponds to
an open universe or spaces a hyperboloid.
In a negative value of omega k corresponds
to a closed universe or spaces a sphere.
We can deduce the qualitative
behavior of different types
of solutions by rewriting this equation.
Let's multiply by one half --
-- and let's bring these
terms, these three terms
over to the left-hand side,
and call these u effective.
And we'll leave this term on the right-hand
side and remember we've multiplied 2 by half.
Or u effective is minus 1/2 times omega
m divided by a tilde plus omega r divided
by a tilde squared plus omega
d times a tilde squared.
Written in this way, the Friedman equation looks
like the equation for a point particle moving
in one dimension with coordinate a
tilde in the potential u effective.
So the particle has a mass equal 1
and energy equal 2 omega k over 2.
Now there are two main cases
to consider depending
on whether omega d is positive or negative.
So let's first consider the case
in which omega d as negative.
And now let's draw an energy diagram.
So this is the energy axis,
and this is the a tilde axis.
Now when a tilde is small,
these two terms dominate.
And remember omega a m and omega r are positive.
So along with this negative sign, that means
the behavior looks like this for small a tilde.
But as a tilde becomes larger.
This term becomes more important.
And since omega d is negative
along with this minus sign,
this has the behavior of an upturned parabola.
So the effective potential looks like this.
This is u effective.
And remember the total energy is 1 half omega
k. So for example, if 1/2 omega k is positive,
then the behavior is this
effective particle moves from zero,
reflects off the potential,
and then comes back to 0.
So this corresponds to a universe that starts
with a scale factor of zero, has a big bang,
expands to a maximum scale factor,
and then returns in a big crunch.
And similarly, if so this is the
case, when 1/2 omega k is positive.
If 1 half omega k is negative --
-- then we have similar behavior.
The universe begins in a big bang,
expands, then recontract to a big crunch.
And this is the case in which omega k is 0.
Again, same, qualitatively the same behavior,
big bang, maximum expansion, then a big crunch.
And these correspond to the cases in which
the universe is open, flat or closed.
Open, and closed referring to
the hyperboloid or the 3-sphere.
And the common feature of all these universes
is that they all begin in a big bang --
-- and end in a big crunch.
Now let's consider the case
in which omega d is zero.
So this term is absent from
the effect of potential.
So omega d equals zero and this
was the standard assumption
in cosmology until relatively recently.
The energy diagram looks like this.
A tilde axis, energy access and the effect
of the potential curve looks like this.
So this is u effective.
Now in this region where omega k is positive,
this is the case of an open universe.
We see the universe starts with a
big bang and then expands forever.
So there's a big bang but no big crunch.
Of course, it could do the opposite,
it could start with the infinite scale
factor and come in with the big crunch.
So we could have a big crunch, now big bang.
And if omega k is negative.
This is the case of a closed universe.
You see that the universe starts with a
big bang, reaches a maximum scale factor,
and then returns in a big crunch.
So we have a big bang and a big crunch.
And in the case when omega k is equal to 0.
That's the flat case.
Qualitatively, it's the same as
the case when the universe is open.
And now let's take a look at the case
in which the dark energy is positive.
So omega d is greater than zero.
So here's the energy diagram in this case.
The effect of potential begins with a
negative 1 over a tilde squared behavior.
Then as a tilde becomes large, it
becomes a minus a tilde squared behavior.
In this case, we see that all open
universes and flat universes as well
as some closed universes have the same
qualitative behavior where they begin
with a big bang and expand forever.
Or they could do the opposite.
They could start from infinite scale
factor and just end in a big crunch.
So these cases have either a big
bang or a big crunch, but not both.
And this includes the flat case, which
is right along a tilde axis here.
So this is open, flat and closed.
But in the closed case, we have
some other behaviors that can occur.
We can have the universe start with a big
bang, expand to a maximum scale factor,
and then return into a big crunch.
So this is a case of a closed universe
with the big bang and a big crunch both.
And we can also have the type of
behavior in which the universe starts
with an infinite scale factor, comes in
and reflects off of the potential here,
and then expands back out
to an infinite scale factor.
So this would be a case of the closed
universe with no big bang and no big crunch.
Just for fun, let me give you some
examples of this type of behavior.
If you take omega d equal to 0.3, so
if the positive, omega m equals 0.6
and omega r equal 2.5 -- and remember the
omegas are not independent from one another.
So these imply that omega
k is equal to minus 2.4.
So if you plot the effective
potential using these parameters,
you'll find a curve that looks like this.
And the energy of the effect of particle
remembers 1/2 omega k. So it's minus 1.2.
And the horizontal line representing
the energy of the particle occurs
down here below the peak of the potential.
So this value of energy right here is minus 1.2.
And for an example of this type of
behavior, we can take omega d equal
to 3.2 omega m equals 0.2 omega r
equals zero, and then these imply
that omega k is equal to minus 2.4.
So again, the energy of the
effective particles minus 1.2.
And what distinguishes this case from
this case, the reason we choose the branch
of this horizontal line that's to the right side
of the peak is because of the value of a tilde.
Remember, a tilde equal 1 corresponds
to the scale factor at the present time.
And when you plot this graph
using this set of parameters,
you find that a tilde equal
1 occurs right about here.
But when you plot this, the graph
with this set of parameters,
you find that a tilde equal 1 is over here.
Of course, our primary interest
is our own universe --
-- and observations tell us that for
our universe, omega d is about 0.68.
Omega m is about 0.32, omega r is small.
It's on the order of 10 to the
minus 5, and omega k is also small.
Its absolute value is less than about 0.01.
So our universe is very nearly
flat and it has omega d positive.
So it's represented in this
diagram by the a tilde axis.
So this is our universe, and we know
it's expanding, not contracting.
So this is us, our universe.
So if you plot this graph using
these parameters, what you'll find is
that a tilde equal 1 corresponds
to a point about right here.
Just beyond the peak of the potential.
So you can see our universe
began with a big bang --
-- and it expanded, but the expansion slowed
down until we passed the peak of the potential.
And then the expansion begins to speed up.
So at this point, the expansion is accelerating.
And if the simple model holds, then our
universe will continue to expand forever.
