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PROFESSOR: A quick review of what we talked about last time.
So the first thing we did last time
was to discuss the age of universe,
considering so far only at this point,
flat matter-dominated universes where the scale factor goes
like t to the 2/3.
And we were easily able to see that the age of such a universe
was 2/3 times h inverse.
And we did discuss what happens if you plug-in numbers
into that formula using the best current value of h,
the value obtained by the Planck team last March.
The age turns out to be about 9.5 to 9.9 billion years.
And that can't be the real age of the universe, we think,
because there are stars that are older than that.
The age of the stars seems to indicate
that the university should be at least,
according to one paper I cited last time, 11.2 billion years
old, and this is younger.
So the conclusion is that our universe is not
a flat, moderate, matter-dominated universe.
We do in fact have good evidence that the universe is very
nearly flat, so it's the matter-dominated part that
has to fail, and it does fail.
We also have good evidence that the universe is dominated today
by dark energy, which we'll be talking about later.
But one of the pieces of evidence for this dark energy
is this age calculation.
The age calculation just does not
work unless you assume that the universe has
a significant component of this dark energy, which
we'll be discussing later.
We then talked about the big bang singularity,
which is an important part of understanding, when you talk
about the age, what exactly you mean by the age, age
since when.
And the point that I tried to make
there is that the big bang singularity, which gives us
mathematically the statement that the scale
factor at some time which we call zero was equal to zero,
and if you put back that formula into other formulas,
you discover that the mass density, for example,
was infinite at this magical time that we call zero.
That singularity is certainly part of our mathematical model
and doesn't go away even when we make
changes in the mathematical model.
But we don't really know if it's an actual feature
of our universe, because there's certainly no reason
to trust this mathematical model all the way back to t
equals zero, where the densities become infinite.
We know a lot about matter and we
think we can predict how matter will behave,
even the temperatures and energy densities somewhat
beyond what we measure in laboratory.
But we don't think we can't necessarily
extrapolate all the way to infinite matter density.
So these equations do break down when
you get very close to t equals zero
and nobody really knows exactly what should be said about t
equals zero.
When we talk about the age, we're
really talking about the age since the extrapolated time
at which a would have been zero in this model,
but we don't really know that it ever actually was zero.
We next discussed the concept of the horizon distance.
If the universe, at least the universe
as we know it, we want to be agnostic about what happened
before t equals zero, but certainly the universe as we
know it, really began at t equals zero in the sense
that that's when structure and complicated things
started to develop.
So since t equals zero there's been only a finite amount
of time elapsed.
And since light travels at a finite speed,
that means that light could only have traveled
some finite distance since the big bang, since t equals zero.
And that means that there's some object which
is the furthest possible object that we could see,
and any object further than that would be in a situation
where light from that object would not yet
have had time to reach us.
And that leads to this notion of our horizon distance,
where the definition of the horizon distance
is that it is defined to be the present distance of the most
distant objects which we are capable of seeing,
limited only by the speed of light.
And once able to calculate that, and in particular for the model
that we so far understood at this point,
the matter-dominated flat universe,
the horizon distance turned out to be three times c t.
Tt
Now remember, if the universe were just static
and appeared at time t ago, then the horizon distance
would just be c t, the distance that light
could travel in time t.
What makes it larger is the fact that the universe is expanding,
and that means that everything was closer together
in the early time and light could make more
progress at the early time, and then these objects have since
moved out to much larger distances.
So that allows the horizon distance to be larger than c t,
and in this particular model, it's three times c t.
Next we began a calculation, which
is what we're going to pick up to continue on now.
We were calculating how to extend our understanding of a
of t, the behavior of the scale factor,
away from the flat case, to ultimately discuss
the two other cases, the open and closed universe.
And we decided on a flip of a coin--
I promise you I flipped a coin at some point--
to start with the closed universe.
We could have done either one first.
The equation that we start with is basically the same,
it's the sine of k that makes the difference.
This is the so-called Friedmann equation,
and for a closed universe k is positive.
This is the evolution equation, but it
has to be coupled with an equation that
describes how rho behaves with time.
And for a matter-dominated universe,
rho is just representing non-relativistic matter,
which is nothing but spread as the universe expands.
And the spreading gives a factor of one
over a cubed as the volume grows as a cubed.
And that means that rho times a cubed is a constant,
and that expresses everything that there
is to know about how rho behaves with time.
OK, then after writing these equations,
we said that things will simplify a little bit, not
a lot, but a little bit if we redefine variables basically
to incorporate all the constants that appear in these equations
into one overall constant.
And we decided, or I claimed, that a good way
to do that, an economical way to do that,
is to define things so that the variables all
have units which are easily understood.
And in this case the units of length
can describe everything that we need,
so we chose to express everything
in terms of variables that have units of length.
So the scale factor itself is units of meters per notch,
and that's not a length.
And notches we'd like to get rid of because we
know they're un-physical.
That is, there's no standard for what the notch should be.
So if we divide a of t by the square root of k,
the notches disappear, and we get
something which just has the units of meters,
or units of length, and I call that a twiddle of t.
Similarly, but more obviously, t can be turned into a length
by just multiplying by c, the speed of light.
So I defined a variable t tilde which is just c times t.
So both of these new variables, with the tildes,
have units of length.
And then the Friedmann equation can
be rewritten, just reshuffling things
according to these new definitions, in this way where
all the constants are lumped into this variable alpha, where
alpha is this complicated expression, which absorbs all
the constants from the earlier equation.
Alpha also has units of length, and even
though it has a rho times a tilde cubed in it,
and both rho and a tilde each depend on time,
the product of rho times a tilde cubed is a constant because rho
times a cubed is a constant and a tilde differs only
by the square root of k, which is also a constant.
So alpha is a constant even though it
has time dependent factors.
The time dependence of those two factors
cancel each other out to give something
which is time independent and can be evaluated any old time.
So this now is our equation, and we proceeded to manipulate it.
So the first thing we did was to take it's square root,
and to rearrange it so that d t tilde appeared on one side,
and everything else was on the other side.
And everything else depends on a tilde
but not explicitly on time.
So this completely separates everything
that depends on t tilde on the left,
and everything that depends on a tilde on the right.
And now we can just integrate both sides of that equation,
and and they point to that, that there are basically
two ways of proceeding here, one of which
we already did when we did the flat case.
When we did the flat case, we integrated both sides
as indefinite integrals.
And when you carry out an indefinite integration,
you get a constant of integration,
which then becomes a constant in your solution.
And in that case we discovered that the constant really
just shifted the origin of time.
And since we had not said anything previously that
in any way determined the origin of time,
we used that constant to arrange the origin of time
so that a of zero would equal zero,
and that eliminated the constant.
Just for variety, I am going to do it another way this time.
Instead of doing an indefinite integral,
I will do a definite integral.
And if you do a definite integral,
you have to make sure you're integrating
both sides over the same range, or at least
corresponding ranges.
We have different names for the variables,
but the range of integration of the two sides
has to match in order to maintain
the equality between the two sides.
So we're going to integrate the left side from zero
to some final time.
And I'm going to call the final time t tilde variable sub
f, for f to stand for final.
And there's no real final time here.
It could be any time, it's just the final time
for the integration, and in the end
discover then how things behave at time t tilde sub f.
And then once we've figured that out we can drop the effort.
It will be a formula that will be valid for any time.
So the left hand side is integrated from zero
to time t tilde f, the right hand side
has to be integrated over the corresponding time interval.
And we would like to define a tilde and the origin of time
so that a tilde equals zero at time zero,
the same convention we used for the other case.
The standard convention in cosmology, t equals zero,
is the instance of the Big Bang.
And the instance of the Big Bang is the time
at which the scale factor vanished.
So that means when we have a lower limit of integration
of t tilde equals zero on the left hand side,
we should have a lower limit of integration
of a tilde equals zero on the right hand side.
And similarly the upper limits of integration
should correspond to each other.
So the upper limit of integration on the left hand
side was t tilde sub f, really just an arbitrary time
that we designated by the subscript f.
So the right hand side the limit should
be the value of a tilde at that time.
And I'll call that a tilde sub f.
So a tilde sub f is just defined to be
the value of a tilde at the time t tilde sub f.
And in this way the limits of integration on the two sides
correspond and now we can integrate them
and we don't need any new integration constants.
These definite integrals have definite values,
which is why they're called definite integrals, I suppose.
OK any questions about that, because that's
where we're ready to take off and start doing new material.
Yes.
AUDIENCE: I have a general question.
So regarding the issue of how we're not really sure how
to extrapolate up to t equals zero,
this is just kind of a general question.
I was wondering how we're kind of assuming
throughout that time kind of flows uniformly,
and so I've heard about something
like gravitational time dilation.
So at the beginning especially when there's
such a high density of matter or radiation,
then wouldn't that affect how, I guess, time flows?
PROFESSOR: OK good question, good question.
The question was, does things like general realistic time
dilation affect how time flows, and are we
perhaps being overly simplistic and assuming that time just
flows smoothly from time zero onward.
The answer to that is that general relativity does
predict an extra time dilation, which
in fact is built into the Doppler shift calculation
that we already did.
And there are other instances where similar things happen.
If a photon travels from the floor of this room
to the ceiling of this room, there's
a small Doppler shift, a small shift in the timing.
And you could see it in principle with clocks as well.
If you had a clock on the floor, and a clock on the ceiling,
they would not run at quite the same rate.
But to talk about time dilation, you always
have to have two clocks to compare.
In the case of the universe, we have this thing
that we call cosmic time, which can be measured on any clock.
The homogeneity assumption implies
that all clocks will do the same thing,
so so the issue of time dilation really does not arise.
Our definition of cosmic time defines
time that is the time variable that we will use.
And when we say it flows uniformly from zero up
to the present time, that word uniformly sounds sensible.
But if you think about it, I don't know what it means.
So I don't even know how to ask if it's really uniform or not.
It's certainly true that our time variable
evolves from zero up to some final time,
but smoothly or uniformly is not really a question we
can ask until one has some other clock that one can compare it
to.
PROFESSOR: Yes.
AUDIENCE: So you said at the beginning
that something like an infinite density
is just an effect of our equation.
Aren't a lot of theories like inflation come out
of assuming that the universe had gotten
some infinite, dense, small area?
So what effect does this assumption that may or may not
be correct have on theories?
PROFESSOR: OK, the question is, if we are not
sure we should believe the t equals zero singularity,
how does that affect other theories like inflation, which
are based on extrapolating backwards to very early times.
And there is an answer to that.
The answer may or may not sound sensible,
and it may or may not be sensible.
It's hard to know for sure.
But one can be more detailed and ask
how far back do we think we can trust our equations?
And nobody really knows the answer
to that of course, that's part of the uncertainty here.
But a plausible answer which is kind of the working hypothesis
for many of us, is that the only obstacle to extrapolating
backwards is our lack of knowledge
of the quantum effects of gravity,
and therefore the quantum effects
of what space time looks like.
And we can estimate where that sets in.
And it's at a time called the Planck time, which is about 10
to the minus 43 seconds, and inflation,
which we'll see later as sort of a natural timescale
of about probably 10 to the minus 37 seconds.
So although that's incredibly small,
it's actually incredibly large compared
to 10 to the minus 43 seconds.
So we think there is at least a basis for believing
that things like our discussions of inflation which we'll
be talking about later, are valid even though we don't
think we can extrapolate back to t equals zero.
Yes.
AUDIENCE: I have a question about use
of the definite integrals.
PROFESSOR: Yes.
AUDIENCE: So we have a twiddle defined
as a over square root of k.
And we noticed in our equation that a
goes to zero as t goes to zero, so a twiddle also goes to zero.
How do we know then that that definite integral
is convergent, because then we have
zero over zero competing case. [INAUDIBLE]
PROFESSOR: Let me think.
Yeah, we'll see, I think is the answer.
How do we know it's convergent?
Well, we're going to actually do the integral,
and then the integral does converge.
You are right.
The integrand does become zero over zero, but that in fact
means the integrand is some finite number actually.
Both numerator and denominator go to zero,
I mean let me think about this.
I guess a tilde squared becomes negligible,
so the denominator goes like one over the square root
of a tilde.
So in fact the numerator divided by the denominator
goes like the square root of a tilde as time goes to zero.
Because you have a square root in the denominator,
the a tilde squared becomes negligible,
so it's manifestly convergent.
The integrand does not even blow up,
even though a tilde does go to zero.
It's certainly worth looking at, you're right.
One should always check to make sure integrals are convergent.
But since we will actually be explicitly doing the interval,
if it was divergent we would get a divergent answer,
and we will not as you'll see in a couple minutes.
Any other questions?
OK, so in that case, to the blackboard.
OK, so I will write on the blackboard
the same equation we have up there, so we can continue.
t twiddle f is equal to the integral from zero
to a twiddle sub f, a twiddle d a twiddle
over the square root of two alpha
a twiddle minus a twiddle squared.
OK, so what we'd like to do now is
to carry out the integral on the right hand side.
Ideally what we'd like to do is to carry out the integral
on the right hand side and get some function of a tilde f
and then we'd like to convert that function
to be able to write a tilde f as a function of time.
We actually won't quite be able to do that.
We'll end up with what's called the parametric solution,
and you'll see how that arises and what that means.
I don't need to try to describe exactly
in advance what that means.
Doing the integral can be done by some tricks,
some substitutions.
And the first substitution is based
on completing the square in the denominator,
and that motivates the substitution that we will make.
So we can rewrite this just by doing some algebra
on the denominator, which is called completing
the square for reasons that you'll
see when I write down what it is.
The numerator will stay a tilde d a tilde.
And the denominator can be written
as alpha squared minus a tilde minus alpha quantity squared.
So completing the square just means
to put the a tilde inside a perfect square.
And the nice thing about this is that now we
can shift our variable of integration,
and turn this into just a single variable instead
of the sum of the two.
And then you have an expression, which
is clearly simpler looking than this one, which
had a tildes in both places.
Now the variable integration will appear only there.
And substitution which does that obviously enough, is we're
going to let something, we can choose whatever we want,
and then once I call it x, we're going
to let x equal a tilde minus alpha.
And we're just going to rewrite that integral in terms of x.
So what we get when we do that is t
sub f tilde is equal to the rewriting of that integral,
and just substituting the a tilde becomes x plus alpha.
So we get x plus alpha where a tilde was,
and d a tilde becomes just d x.
And the denominator, which was our motivation
for making the substitution in the first place,
becomes just alpha squared minus x squared.
So this is perfectly straight forward.
The next step which is important,
is to get the limits of integration right,
because with this definite integral method
we really have to make sure that our limits of integration
are correct.
So straightforward to do that, the lower limit of integration
was a tilde equals zero, and if a tilde equals zero,
x is equal to minus alpha.
So the lower limit of integration
expressed as a value of x is minus alpha.
And the upper limit expressed as x
was a tilde f, and that means x is
equal to a tilde f minus alpha.
So the limit here is a tilde sub f minus alpha
for the upper limit of integration.
OK, now this integral is still not easy,
but it can be made easy by one more substitution.
And the one more substitution is a trigonometric substitution
to simplify the denominator.
We can let x equal minus alpha cosine
of theta, where theta is our new variable of integration.
And then alpha squared minus x squared
becomes alpha squared minus alpha
squared cosine squared theta.
And the alphas factor out and you have the square root of 1
minus cosine squared theta.
1 minus cosine squared theta is sine squared theta, which
is a convenient thing to take the square root of,
you just get sine theta.
And everything else also simplifies.
And the bottom line, which I will just give you,
is that now we find that t sub s tilde is just the integral of 1
minus cosine theta d theta.
That's it.
Everything simplifies to that which is easy to integrate.
We also have to keep track of our limits of integration.
The limit of integration, the lower limit of integration,
is where x equals minus alpha.
And if x equals minus alpha, that
means cosine theta equals 1.
And cosine theta equals 1 means theta equals 0.
So the limit of integration on theta is easy, it starts at 0.
And the final value is really just a value
that corresponds to the final value of x
which is a twiddle f minus alpha.
For now I'm just going to call it theta sub f.
Things that we have just determined
is the value of theta that goes with the upper limit there.
We'll figure out in a second how to write it more explicitly.
Let me first do the integral to just get that out of the way.
Doing the integral, you get alpha times 1
minus cosine theta sub f.
So this in fact becomes half of our solution.
It expresses t sub f in terms of theta sub f.
And now that we're done with the whole problem,
I'm going to drop the subscript f.
There's just some time that we're
interested in which is called t and the value of a at that time
will be called a, and the value of theta at that time
will be called theta.
So I'm just going to drop the subscript everywhere
because we are now in a situation were the subscript is
everywhere, so dropping everywhere
does not lose any information.
So one of our equations is going to be
simply t is equal to alpha times 1 minus cosine theta.
And another equation will come from figuring out
what theta sub f really is, which
I said comes from making sure that the upper limit
of integration here corresponds to the upper limit
of integration in the previous integral.
So theta sub f, I had to determine
this on another blackboard and then put the final equations
together.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yes, that's right I don't want to drop twiddles.
T twiddle, thank you, is equal to alpha times 1
minus cosine theta, and and then we have the final value of x.
Xx x is equal to a tilde minus alpha.
So the final value of x is equal to the final value
of a tilde minus alpha.
And the final value of x is also related to theta
by this equation, which is what we're trying to get,
how theta is related to the other variables.
So this is equal to minus alpha cosine of theta sub f.
And now we might want to, for example,
solve this for a tilde sub f, which just involves looking
at the right hand half of the equation,
bringing this alpha to the other side making a plus alpha.
So this implies that a tilde sub f is equal to alpha times 1
minus cosine theta sub f.
So this equation now just says that theta
sub f means what it should mean to give us
a final limit of integration that corresponds
to the final limit of integration
on our original integral, which we called a tilde sub f.
Yes.
AUDIENCE: Sorry, perhaps I missed it.
But when you are doing the integral of 1
minus cosine theta, how come--
PROFESSOR: Oh, I got it wrong.
Good point.
The integral of cosine theta d theta
is sine theta, not cosine theta.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Wait a minute.
Hold on.
Do I still have it wrong?
AUDIENCE: [INAUDIBLE]
PROFESSOR: OK hold on.
There was an alpha missing here.
That's part of the problem, coming
from the original equation here.
Let's see if I have this right now.
This is copying from a long line altogether.
So I got both factors wrong.
And then wait a minute.
I think this is right now.
There's still a wrong [INAUDIBLE]?
AUDIENCE: [INAUDIBLE]
PROFESSOR: If I differentiate sine, I get cosine.
So if I differentiate minus sine, I get minus cosine.
So differentiating this, I get this.
That should mean that integrating this I get that.
I think I have it right.
Sometimes I get things right.
It's a surprise, but--
AUDIENCE: [INAUDIBLE]
PROFESSOR: What?
AUDIENCE: The last equation you wrote.
PROFESSOR: The last equation needs to be changed, right.
This was copied from that, right.
Thanks for reminding me.
So t tilde is equal to alpha times theta minus sine theta.
OK, and now I was working out the relationship
between theta and a tilde.
And let me just remind you that all I did
was make sure that the upper limits of integration
correspond to each other.
So I'm just basically rewriting the upper limit of integration
in terms of the new variable each time
when we change variables starting from a tilde going
to a tilde to x, and from x to theta.
And then I use the equality of these two
to write a tilde in terms of theta.
And these equations now hold with the subscript f present.
When subscript f's appear everywhere we can just
drop them and say that the time that we called
t sub f now we're just going to call t.
And now we could put together our two final results
which are maybe right over here.
We have ct.
I'll eliminate my tildes altogether now.
ct, which is t tilde, is equal to alpha times theta
minus sine theta.
And a divided by the square root of k,
which is a tilde and its sub f, but we're dropping the sub f,
is equal to alpha times 1 minus cosine theta.
And this is as good as it gets.
Ideally, it would be nice if we could
solve the top equation for theta as a function of t,
and plug that into the bottom equation,
and then we would get a as a function of t.
That's what ideally we would love to have.
But there's no way to do that analytically.
In principle of course, this equation can be inverted.
You could do it numerically.
For any particular value of t you
could figure out what value of theta makes this work,
and then plug that value of theta into here.
But there's no analytic way that you
can write theta as a function of t.
It's not a soluble equation.
So this is called a parametric solution in the sense
that theta is a parameter.
And as theta varies, both t and a vary in just the right way
so that a is always related to t in the correct way
to solve our original differential equation.
That's what's meant by a parametric solution.
We can also see from this equation,
or maybe from thinking back about how things are defined
in the first place, how theta varies
over the lifetime of our model universe.
Theta we discovered starts at 0.
We discovered that when we wrote our integral over there.
And we could also probably see it from here.
We start our universe at a equals 0.
And at a equals 0, we want 1 minus cosine theta to be 0
and theta equals 0 does that.
So theta starts at 0, which corresponds to a equals 0,
and it also corresponds putting theta equals 0 here
to t equals 0.
So we have arranged things the way
we intended so that a equals 0 happens
the same time t equals 0 happens.
And then theta starts to grow.
As theta grows, a increases, the universe
gets bigger until theta reaches pi.
And when theta reaches pi, cosine of pi is minus 1,
you get a factor of 2 here, 2 alpha.
That's as big as our universe gets.
And then as theta continues beyond that,
1 minus cosine theta starts getting smaller again.
So our universe reaches a maximum size
when theta equals pi, and then starts to contract.
And then by the time theta equals 2 pi,
you are back to where you started from,
a is again equal to 0.
We have universe which starts at 0 size, goes to a maximum size,
goes back to 0 size, giving a Big Crunch at the end.
And that's the way this closed universe behaves.
It turns out that these equations actually
correspond to some simple geometry.
It corresponds to a cycloid.
And you may or may not remember what a cycloid is.
There are the equations written on the screen, which
are hopefully the same equations that I have on the blackboard.
Can't always count on these things it turns out.
But yeah, they are the same equations, that's healthy.
And I have a diagram here which explains
this cycloid correspondence.
A cycloid is defined as what happens in this picture.
Let me explain the picture.
We have a disk shown in the upper left
in its original position with a dot on the disk, which
is initially at the bottom.
And initially we're going to put this disk
at the origin of our coordinate system
to make things as simple as possible.
And then we're going to imagine that this disk rolls
without slipping to the right.
And the path that this dot traces out,
which is shown along that line, is a cycloid by definition.
That's what defines a cycloid.
It's the path that a point on a rolling disk evolved through.
So what I would like to do is convince you
that this geometric picture corresponds to those equations.
Let me put the equations higher to make sure everybody
can see them.
And it's actually not so complicated
once you figure out how to parse the pictures.
So what I've drawn in the upper right is a blow up of this disk
after it's rolled through some angle theta.
And I even made it the same angle theta in the two cases.
But this is just a bigger version
of what's in the corner there showing the disk after it
rolled a little bit.
So after it's rolled through an angle theta, what
we want to verify is that the horizontal and vertical
coordinates here correspond to these two equations.
And if they do, it means that we're tracing out
the behavior of those two equations.
So look first at the horizontal component.
The horizontal component is the ct axis.
So that should correspond to alpha times theta
minus alpha times sine theta.
And you can see that in the picture.
We're talking about the horizontal component
of the coordinates of this point P.
And the point is that you can get
to P starting at the origin.
Now remember we're only looking at horizontal motion
so we could start anywhere on that line.
We could start at the origin, go to the right by alpha theta,
and to the left by alpha times sine theta, and we get there.
And that's exactly what this formula says.
So if we just understand the alpha theta and the alpha sine
theta of those two lines we have it made.
So let's look first at what happens where
this first arrow comes from, the alpha theta line.
That's just the total distance that the point of contact
has moved during the rolling process.
And the claim is that as something rolls,
really the definition of rolling without slipping,
is that the arc length that is swept out by the rolling
is the same as the length along the surface on which it's
rolling.
You could imagine, if you like, that as it
rolls there's a tape measure that's
wrapped around it that gets left on the ground as it rolls.
And if you could picture that happening,
the existence of that movie really
guarantees that the length on the ground
is the same as what the length was when the tape measure was
wrapped around the cylinder.
So the length that the point of contact has moved
is just alpha times the angle through which
the disk is rolled.
So that explains the alpha theta label on that line.
To get the alpha sine theta on the line above,
that's the distance between the point P
and a vertical line that goes through the center of the disk.
And that's just simple trigonometry on this triangle.
The radius of our circle is alpha.
And then by simple trigonometry, this length
is alpha times sine theta, which is what the label says.
So to summarize what we've got here
we can find the x-coordinate of the horizontal
according to the point P by going
to the right a distance alpha theta,
and then back to the left the distance of minus alpha theta.
And that gives us an x-coordinate,
which is exactly the formula that appears in the ct formula.
So ct works.
The horizontal component of that dot
is just where it should be to trace out the equations that
describe the evolution of a closed universe.
Similarly we can now look at the vertical components
of that dot.
Again it's most easily seen as the difference
of two contributions.
We're trying to reproduce this formula that says that's alpha
minus alpha cosine theta.
So if we start at vertical coordinate 0, which
means on the x-axis, we can begin
by going up to the center of the disk.
And the disk has radius alpha, so that's
going up the distance alpha.
And then we go down the distance of this piece of the triangle,
going from the center of the disk
to the point which is parallel to the point P.
And that again is just trigonometry on that triangle,
and it's alpha cosine theta by simple trigonometry.
So we can get to the elevation of point P
by going up by alpha and down by alpha times cosine theta.
And that's exactly that formula.
So it works.
The x and y components, the horizontal
and vertical components of that dot,
are exactly the two formulas here.
So one of them can be thought of as the x-axis, and one of them
can be thought of as the y-axis.
And the rolling of the disk just traces out
the evolution of our closed universe.
So closed universes evolve like a cycloid.
Any questions about that?
OK, great.
OK, Let me just mention that this angle theta is sometimes
given a name.
It's called the development angle of the universe
or of the solution.
And that is just intended to have the connotation that theta
describes how developed the universe is
and theta has a fixed scale.
It always goes from 0 to 2 pi over the lifetime
of this closed universe, no matter
how big the closed universe might be.
That brings me to my next question I want to mention.
How many parameters do we have in this solution?
The way we've written it, it looks
like it can depend on both alpha and k because both of them
appear in the answer.
And k is positive for our closed universes.
These formulas will not make sense if k were negative,
square root of k appears there.
We don't want anything to be imaginary.
But k can have any value in principle
and these equations would still be valid.
So on the surface it appears like there's a two parameter
class of closed universe solutions.
But that's actually not true.
Can somebody tell me why it's not true?
Yes.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Exactly.
Yes, since k has units of 1 over notch squared,
you can change k to anything you want
by changing your definition of a notch.
And there's nothing fixed about the definition of a notch.
So k is an irrelevant parameter.
If we change k we're just rescaling the same solution,
and not actually creating a new solution.
So there's really only one parameter class of solutions.
One could, for example, fix k to always be 1,
and then we'd have a one parameter
class of solutions indicated by alpha.
Alpha, unlike k, really does have a clear, physical meaning
related to the behavior of the universe.
And we could see what that is if we
ask, what is the total lifetime of this universe from beginning
to end, from Big Bang to Big Crunch.
We can answer that by just looking at the ct equation.
From Big Bang to Big Crunch, we know that theta evolves from 0
to pi back to 2 pi, which is the same as 0.
So theta goes through one cycle of 2 pi
during a lifetime of our model universe.
As theta goes from 0 to 2 pi, sine theta starts at 0
and eventually comes back to 0 when theta equals 2 pi.
But theta increases from 0 to 2 pi over one cycle.
So over one cycle of our universe,
ct increases by alpha times 2 pi.
So that tells us what the total lifetime of the universe
is, I'll call it t total.
And we get it by just writing c times t total equals 2 pi
times alpha.
And I think I can do this one without making a mistake. t
total is then 2 pi alpha divided by c.
And we can even check our units there.
Alpha has units of length, so length divided by c
becomes a time, c being a velocity.
So it has the right units.
And that's the total lifetime of the universe.
It's just determined by alpha.
So alpha can be viewed as just the measure
of the total lifetime of the universe, which can be anything
for different sized closed universes.
Alpha is also related to the maximum value of a over root k.
And a has no fixed meaning, this is meters per notch.
But a over root k does have units of meters.
We haven't yet really seen what it means physically,
because that's related to the geometry of the closed universe
which we'll be discussing later.
But in any case as a mathematical fact,
we could always say that the maximum value of a over root k
is determined by alpha.
So a max over root k is equal to the maximum value
of this expression.
And this expression has a maximum value
when theta equals pi, which gives cosine theta
equals minus 1, which gives a 2 here.
And that's as big as it ever gets,
so that's just equal to 2 alpha.
So alpha is related in a very clear way to the total lifetime
of our universe, and is also related to a over root
k, although we haven't really given
a physical meaning to a over root k yet.
But we know it has dimensions of meters.
OK, the next calculation I want to do
is to calculate the age of the universe
as a function of measurable things.
We learned for the flat matter dominated case
that there was a simple answer to that.
The age was just 2/3 times the inverse Hubble parameter.
So what you do now is get the analogous formula here,
it follows in principle immediately
from our description of the evolution.
But we have to do a fair number of substitutions
before we can really see how to express the age in terms
of things that we're interested in.
The formula here tells us directly
how to express the age of the universe. t
is the age of the universe as a function of alpha and theta.
But if you tell an astronomer to go out and measure alpha
and theta so I could calculate the age,
he says what in the world are alpha and theta.
So what we'd like to do is to express the age
in terms of things that astronomers know about.
And the characterizations of the universe
like this that an astronomer would know about
would be the Hubble expansion rate,
and some notion of the mass density.
And the easiest way to talk about the mass density
is in terms of omega, the fraction
of the critical density that the actual mass density has.
So our goal is going to be to manipulate these equations.
All the information is already there.
But our goal will be to manipulate these equations
to be able to express the age t in terms of h and omega.
OK, so first we need to remind ourselves what omega is.
The critical density is defined as that density
which makes the universe flat.
And we've calculated that the critical density is
equal to 3 h squared over 8 pi times newtons
constant, capital G. We can then write the mass density rho
as omega times the critical density,
which is just the definition of omega.
Omega is rho divided by rho c, the actual mass density divided
by the critical density.
And putting what rho c is, we can express rho
as 3h squared omega divided by 8 pi G. And being very pedantic,
I'm just going to rewrite that in the form
that we're actually going to use it by multiplying through.
8 pi over 3 G rho, taking these factors
and bringing them to the other side,
becomes just equal to h squared times omega.
And you might recognize this particular combination
as what appears in the Friedmann equation.
The Friedmann equation told us that a dot over a squared
is equal to 8 pi over 3 G rho, minus kc
squared over a squared.
And in order to get the substitutions that I want,
I'm going to just rewrite this putting
h squared for a dot over a squared.
8 pi over 3 G rho we said we could
write as 8 squared times omega.
And then we have minus kc squared over a squared.
Note that we really have here a tilde squared.
This is a squared divided by k if I put them together.
So this term can be written as minus c squared
over a tilde squared.
And this accomplishes one of our goals.
It allows us to express a tilde in terms of the quantities
that we want in our answers, h and omega.
And if we can do the same for theta,
we have everything we need to express the age.
So the implication here is that a tilde squared
is equal to c squared divided by h squared times omega minus 1.
To take the square root of that equation,
to find out what a tilde is, we need
to think a little bit about signs and things like that.
A tilde is always positive.
This is the scale factor divided by the square root of k,
square root of k is positive, scale factors
are always positive the way we defined it.
So we can take the square root of that
taking the positive square root of the right hand side.
Omega is bigger than 1 for our case, so omega minus 1
is a positive number, 8 squared is a positive number.
So taking the square root there offers no real problem.
We can write a tilde is equal to c over,
I guess this point I might not notice until later.
h h can be positive or negative over the course
of our calculation.
We're going to talk about an expanding
phase and a contracting phase.
So when we take the square root of h squared,
we want the positive number to give us a positive a tilde.
So it's the positive square root that we want,
which is the magnitude of h, not necessarily h.
When h is positive, the magnitude of h is h.
h could be negative though, and the magnitude of h
is still positive, and then the square root of omega
minus 1, which is always positive.
So that's our formula for a tilde in terms of h and omega.
OK, now we want to evaluate alpha,
and I guess I did not keep the formula for alpha quite as
long as we needed it.
When we defined alpha in the first place,
let me remind you how it was defined, 4 pi over 3 G
rho a tilde cubed over c squared.
And that can be evaluated using our formula for rho
and we'll put that in for rho, and what we get
is c over 2 times the magnitude of h
using this formula for a tilde as well.
And then omega over omega minus 1 to the 3/2 power.
Just using this formula, and we know
how to express rho from the right hand side,
and we know how to express a tilde from this formula here.
So everything's straightforward, and this is what we get.
And now I want to use these to rewrite this equation,
a over the square root of k is equal to alpha times 1
minus cosine theta.
I'm going to replace a over root k by this formula.
I'm going to replace alpha by that formula.
So this implies, rewriting it, that c
over the magnitude of h times the square root of omega
minus 1 is equal to c over twice the magnitude of h times omega
minus 1 to the 3/2 power times 1 minus cosine theta.
And now we've had to survive some boring algebra.
But notice that now most things cancel away here.
We get a very simple relationship between theta
and h and omega, actually just omega .
In particular, when we solve that,
we get simply that cosine theta is
equal to two minus omega over omega.
So theta is directly linked to omega.
If you know omega, you know theta, If you know theta,
you know omega, by that formula.
And we can rewrite this the other way
around by solving for omega if we want.
Omega is equal to 2 over 1 plus cosine theta.
Now we can look at this qualitatively
to understand how omega is going to behave.
At the very beginning cosine theta is equal to 1,
theta is equal to 0, so omega is 2 over 1 plus 1, which is 1.
So at very early times omega is driven to 1 even
in a closed universe.
As theta gets larger, cosine theta gets less than 1.
This then becomes more than 1.
So omega starts to grow as the universe starts to evolve.
At the turning point, when the universe
has reached its maximum size, theta is pi,
cosine theta is minus 1, omega is
infinite at the turning point.
That may or may not be a surprise.
But it you think about it, it's obvious.
At the turning point h is 0, therefore the critical density
is 0, but the actual density is not 0.
And the only way the actual density can be 0
while the critical density is 0 is for omega
to be infinite, so we should have expected that.
And then the return trip, the collapsing phases,
a mirror image of the expanding phase,
omega goes from infinity at the turning point back to 1
at the moment of the Big Crunch.
Yes.
AUDIENCE: I'm confused with what the universe would
look like when it gets to infinity.
PROFESSOR: It would look static.
It's temporarily static.
It's reached a maximum size and is
about to turn around and collapse, so h is 0.
AUDIENCE: OK, but like with the density.
PROFESSOR: Well we could calculate the density.
It's some number which depends on alpha.
AUDIENCE: OK, but that doesn't diverge or anything.
PROFESSOR: It doesn't diverge or anything, no.
It's just some finite density.
At the turning point, it's just a finite density
that can be expressed in terms of alpha.
Sounds like a good homework problem.
I think maybe I'll do that.
OK, so this now allows us to express theta
as a function of omega, which is what we wanted.
If we express theta as a function of omega,
and alpha as a function of omega and h, we have our answer.
We have t expressed as a function of h and theta.
So there are choices about how exactly to express omega
in terms of theta that will involve
inverse trigonometric functions.
And anything that can be expressed
in terms of an inverse cosine, can also
be expressed as an inverse sine by doing
a little bit of manipulations.
We have our choice here about what we want to do.
But in any case, the answer already
has a factor of sine theta in it.
So it's most useful to express theta
as the inverse sine of what we get from that formula,
to express the answer in what at least to me
is the simplest form.
So I'm going to manipulate this a little bit
to find out what sine theta is in terms of omega,
and then invert that to express theta in terms of omega.
So sine theta is of course plus or minus
the square root of 1 minus cosine squared theta.
And cosine theta we know in terms of omega,
so we can express this in terms of omega.
And if you do that it's straightforward enough algebra.
It's plus or minus, depending on which sign of the square root
is relevant, and we'll talk about that in a minute.
It's plus or minus the square root of 2 times omega
minus 1 over omega.
So we can express theta as the inverse sine
of this expression.
And now what I want to do is to make use of this
to put into this expression using the value for alpha
that we've already calculated and wrote over there.
So we get our final answer, which
is that t is equal to omega over twice
the magnitude of the Hubble expansion rate times omega
minus 1 to the 3/2 power times the arc sine or inverse
sine of twice the square root of omega minus 1 over omega.
That's just the theta that appears in this formula written
in terms of sine theta or the inverse sine of the quantity
that we determined was the sine of theta.
And then I see here I should have a plus or minus
because we haven't figured out our signs yet.
Either is actually possible, depending
on where we are in the evolution.
And then minus or plus twice the square root of omega
minus 1 over omega.
That's a plus or minus, this is a minus or plus.
And the reason I wrote one upstairs and one downstairs
is that we don't yet really know how to evaluate theta,
but the sign of this term is always
going to be the negative of the sign of that term, this
or that minus sign.
So theta can be the inverse sine of this expression
with either sign of the sine.
Sorry for the puns.
But whichever it is, it's the same on there
as it is here but with a minus sign in front, that minus sign.
OK, so this is our final formula for the age.
But we still need to think a little bit about the s-i-g-n
signs of these inverse functions that appear here,
that and that, and straightforward if you just
take it case by case.
And in the notes we have a table which
I'll put shortly on the screen.
But let's start by just talking about the earliest phase where
the universe is shortly after the Big Bang,
so that the development angle is a small angle.
We know that theta is going to go
from 0 to 2 pi over the lifetime of our universe.
So I now want to think of theta being small.
And small means small compared to any number you think about.
So when theta is small the sine theta is nearly theta.
Both are positive.
And in that case the sine theta being positive
means this is the positive root in this equation,
and therefore the positive root in that equation.
So for early times this would be the plus sign,
and that would mean that this would be the minus sign.
Again the minus sign just coming from there
and theta being positive.
So for early times it's a plus and minus.
And the arc sine is itself ambiguous.
For early times the angle we know
is going to be just a little bit bigger than 0.
So that's the evaluation that we make of the arc sine function.
Pi plus that would also give us the same sine.
It would be another possible value for the arc sine.
And of course 2 pi plus that would also
be another possible root.
So you have to know which root to take
to know the right answer here because as an angle,
0 is equivalent to 2 pi, but as a time, 0 is not
at all equivalent to 2 pi.
So you do have to know the right one to take.
We'll continue doing that on a case by case basis.
Those are the equations, that's the formula for the age,
and that's the formula for the age with a description of which
roots to take for each case, which just comes out
by following the evolution, we know
that theta is going from 0 to 2 pi,
and this last column, the inverse sine of the expression
which means the expression that appears here.
For the smallest angle is 0 to pi over 2.
We can think of this actually as defining our columns.
Theta starts at 0 so the time lengths between 0
and pi over 2, a time length between pi over 2,
a time length between pi and 3 pi over 2,
and a final time length between 3 pi over 2 and 2 pi.
And the first two correspond to the expanding phase, second two
correspond to the contracting phase.
We can easily see what values of omega
are relevant in those cases.
Omega we said starts at 1 and gets
larger, the borderline where the angle is pi over 2 one
could just plug into these formulas
and see amounts to omega equals 2.
So that is a division line between these first two
quadrants just calculated from the value of theta.
Omega then goes to infinity as we
said, comes backwards and back to 1 in the end.
And in this column we just figure
out which sign choice corresponds
to getting the right value for omega
and the angle that appears in the arc
sine of our formula for the age, the formula there.
So any one of these I claim is very obvious.
Seeing the whole picture takes time
because I think you really have to look at each case one
at a time to make sure you understand it in detail.
But if you understand the initial expanding phase
that's what corresponds to our universe
if our universe were closed.
And the others are just as easy.
You just have to take them one at a time I think.
OK, we're going to end there.
We will continue on Thursday.