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PROFESSOR: I'd like to
begin building up momentum
by going over what we've
already done to kind of see
how it all fits together.
So I put the last lecture
summarised on transparencies
here.
And we'll go through
them quickly,
and then we'll start
new material, which
I'll be doing on the blackboard.
So we've been studying
this mathematical model
of a universe obeying
Newton's laws of gravity.
We considered simply a
uniform distribution of mass,
initially spherical,
and uniformly expanding,
which means expanding
according to Hubble's law,
with velocities proportional to
the distance from the origin.
And Newton's laws then tell
us how that's going to evolve.
And our job was just to
execute Newton's laws
to calculate how
it would evolve.
And we did that by describing
the evolution in terms
of a function little r, which
is a function of r, i, and t.
And that's the radius at time t
of the shell that was initially
a radius r sub i.
So we're trying to track
every particle in this sphere,
not just the particles
on the surface.
And we want to verify that it
will remain uniform in time.
Matter will not collect
near the edge of the center.
And we discovered it will
remain uniform if we have a 1
over r squared force law,
but for any other force law
it will in fact
not remain uniform.
So we derived the
equations and found
that it obeyed this
scaling relationship, which
is what indicated the
maintenance of uniformity,
wholesale scale by the
same factor, which we then
called a of t.
So the physical
distance of any shell
from the origin at any time
is just equal to a of t times
the initial distance
from the origin.
And furthermore, we were able
to drive equations of motion
for the scale factor.
And it obeys the equations,
which in fact were derived
from general relativity by
Alexander Freedman in 1922,
and therefore they're called
the Freedman equations.
There's a second order
equation, a double dot,
which just tells us how
the expansion slowed down
by Newtonian gravity.
a double dot is negative,
so the expansion
is being slowed by the
gravitational attraction
of every particle in this
spherical distribution
towards every other particle.
And we also were able to
find a first order equation
by integrating the
second order equation.
And the first equation
has this form.
It could be written a
number of different ways
depending on how you arrange it.
But this is the way that
I consider most common.
And many books refer to the
second equation as the Freedman
equation.
Both of these equations were
derived by Alexander Freedman.
I think it's perfect called
both Freedman equations.
But most books do not do that.
And in addition to finding
the equations of evolution
for a of t, we also understood
how rho of t evolves.
And that really is pretty
trivial to begin with.
The Newtonian mass of this
sphere stays the same.
It just spreads out over a
larger volume as a of t grows.
So if the mass stays
the same and the volume
grows as a cubed,
then the density
has to go like one over a cubed.
And this alone could be
written more precisely
as an equation by saying the
easiest way to view the logic
is that this equation
implies that a cubed times
rho is independent of time.
And then once you know
that a cubed times
rho is independent of time,
you can write the equation
in this form,
which if I multiply
a of t cubed to the left
hand side of the equation,
we can just say that a
of t cubed times rho is
the same at time t as it
is at some other time t1.
Now in lecture last time,
we wrote this equation
where t1 was t sub i the initial
time, and a of t sub i was one.
So we didn't
include that factor.
So this is slightly
more general way
of writing it than
we did last time.
But it still has no more
content than the statement
that rho of t falls as
1 over a cubed of t.
We introduced a special set
of units to describe this.
Question, yes?
AUDIENCE: I had a question
about the Freedman equations
and what we're
calling e squared,
and the interpretation of that.
PROFESSOR: What we're
calling what squared?
AUDIENCE: h squared.
PROFESSOR: h squared, yes.
AUDIENCE: So in the
spherical universe,
we showed that if e was
positive that corresponded
to an open universe that
would expand forever.
So doing the p set for this week
for the cylindrical universe,
I think we found that that
universe would collapse.
PROFESSOR: Yes.
AUDIENCE: E was also
positive for that one.
So I was just wondering
about the interpretation
of [INAUDIBLE]
PROFESSOR: Right, well I'll
be coming to that issue later.
Let me come back to that, OK.
Since I haven't
introduced e yet.
And the slide is here.
So an interesting question.
We'll come back to that.
So to describe this
system of equations,
I like to introduce this
notion of a notch, which
is a special unit used only to
measure co-moving coordinates.
And in this case, r sub i
is our co-moving coordinate.
As our shells
move, we label them
all by where they
were a time t i.
We don't change those labels.
So those are the
co-moving coordinates,
a coordinate system that
expands with the universe.
So instead of
measuring r i in meters
or any other physical
length, I like
to measure them in a
new unit called a notch,
just to keep things separate.
And a notch is
defined so that a of t
is measured in meters per
notch, and at time t i,
one notch equals 1 meter.
But at different
times, the relationship
between notches and
meters is different
because it's given by this time
dependent scale factor of t.
If [INAUDIBLE] works to have
things depend on these units,
one find that this new
quantity that we introduced,
little k, which all we know
is that it's a constant,
has the units of 1
over a notch squared.
And that has some
relevance to us,
because it means that we can
make little k have any value we
want by choosing different
definitions for the notch.
And the notch is up for grabs.
We are just inventing
a unit to use
to measure our co-moving
coordinate system.
So we can always
adjust the meaning
of a notch so that k has
whatever value we want.
As long as we can't change its
sign by changing the units.
And if it's zero we can't
change it by changing its units.
As long as non-zero we
can make it any value
we want, and in fact that is
often used in many textbooks.
And that's I guess
what I want to talk
about next, the conventions
that are used to define a of t.
And for us, I'm going to treat
this notch as being arbitrary.
We've defined the notch
originally so that a of t i
was one meters per
notch at time t i,
and that gave the notch more
or less a specific meaning.
But the specific meaning
depends on what t i is.
One can take the
same relationship
and view it as simply
a definition of t i.
t i is the time
at which the scale
factor is one meter per notch.
We take the definition of t
i and we can let the notch
be anything we want, and
there will be some time t
i that will still make
that statement true.
So what I want to do basically
is to think of this equation,
a of t i equals 1
meter per notch,
not as a definition
of a notch, which
I want to leave arbitrary, but
rather as a definition of t sub
i.
And after defining
t sub i, I want
to just forget about t sub i.
t sub i in fact will not
enter our equations anywhere.
So we don't need to
remember its definition.
I decided like the cubit.
All of us know that the
cubit is some unit distance,
but we don't care what it is
because we never use cubits.
Same thing here.
We'll just never use t sub i.
And since we're never
going to use it.
We don't need to remember
how it was initially defined.
It's only of
historical interest.
So the bottom line then is
simply that for us the notch
is just an undefined
unit of distance
in the co-moving
coordinate system.
Other people use
different definitions.
Ryden, for example
uses the definition
where a of t sub, a of the
present time is equal to 1.
And we would interpret that
as meaning one meter per notch
today.
And that's a perfectly
good definition.
And we can use it
whenever we want,
because our notch is
initially undefined.
So that allows us the
freedom to define it
in any particular problem
in any way that we want.
Many other books take
advantage of the fact
that this quantity k has units
of inverse notch squared,
even though they don't say that.
But that means you could rescale
the co-moving coordinate system
to make k equal to
whatever value you want.
So in many books k is always
equal to plus or minus 1
if it's non-zero.
The co-moving coordinate
systems is just scaled.
We would do is we scaling of the
notch to make k have magnitude
1.
OK, having derived
these equations,
the next step is
to go about asking
what do the solutions to
the equations look like.
And that's where things
start getting interestingly
when we start getting some
real nontrivial results.
The Freedman equation,
the first order one,
could be rewritten this way.
It's just a rewriting
of rearranging things.
And I used here the fact
that rho times a cubed
was a constant.
If we took our original form
of the Freedman equation,
this would be rho
sub i times a cubed
of t sub i, which would be 1.
But knowing that rho times
a cubed is a constant,
we can let t argument
here be anything we want,
which is what I'll do, just
to emphasize that we don't
care anymore what t sub i was.
It really has disappeared
from our problem.
It was just our way
of getting started.
So this equation holds.
And we can use it to discuss
how the different classes
of solutions will behave.
And what we can see very
quickly from this equation
is that the behavior
of the solutions
will depend crucially
on the sign of k.
And it's useful
here to remember,
although we don't
need to know anything
more than this
proportionality, that k
is proportional to the negative
of something that we call e.
And that the thing
that we call e
is related to the overall
energy of this thing.
So we'll keep that in
the back of our minds.
But it's really
only for intuition.
Everything that
we're going to say
follows directly
from this equation,
where we don't know
anything about e.
There are three
types of solutions,
depending on whether k is
positive, negative, or zero,
and those are the options
for what k might be.
It's a real number.
So first we consider
with the solutions
where k is less than zero, which
means e is greater than zero.
And e being greater
than zero means
the system has more
energy than zero.
And in this case,
zero energy would
correspond to having all the
particles infinitely far away.
So the potential
energies would be zero.
And all particles the rest,
so the kinetic energies
would be zero.
So in particular, zero energy
would correspond to the system
being completely dispersed,
no longer compact.
And in this case, our system
has more energy than that.
And more energy
than that means it
can blow outward without limit.
And we see that directly
from the equation.
If the second term has
a negative value of k,
then the second term
itself is positive.
And the first term is
also always positive.
And that means
that a dot squared,
no matter what happens
to the first term,
is always at least bigger
than the second term,
which is a constant.
And if a dot is always
bigger than some constant
means that a grows indefinitely.
And that's called
an open universe.
And it goes on
expanding forever.
Second case we'll
consider is k greater
than zero, which corresponds
to e less than zero.
And that means since 0
corresponds to the system
being completely dispersed,
e less than 0 means
the system does not
have enough energy
to ever become
completely dispersed.
So we'll have some maximum size.
And the maximum size follows
immediately from this equation.
a dot squared has
to be positive.
It can ever get negative.
It can become zero, but
it can never get negative.
In this case, the minus k
c squared term is negative.
And that means that if this
term gets to be too small,
the sum will be negative,
which is not possible.
So this term cannot
get to be too small.
And since a of t is
in the denominator,
that means a of t cannot
get to be too big.
And you can easily
derive the inequality
that a has to obey for
the right hand side
to always be positive.
And in that case you
[INAUDIBLE] a max,
which is what you would get
if you just set the right hand
side equal to 0, given
by that expression.
a can never get bigger than
that, because if it did,
the right hand side
of the equation
become negative,
which is not possible.
So this universe will
reach a maximum size,
which we just calculated,
and then [INAUDIBLE]
will come back.
So we already have a very
nontrivial result here.
Given a description of
a universe of this type,
we can calculate
how big it will get
before it turns
around and collapses.
And this of universe
ultimately undergoes
a big crunch when it collapses,
where the word big crunch
was made as an analogy
to the phrase big bang.
It's called a closed universe.
And then finally, we've now
considered the k less than zero
and k greater than zero.
There are many cases
when k equals zero.
And that's called
the critical mass
density or critical universe.
And we can figure
out what it means
in terms of the mass density.
This again is our
Freedman equation.
If k is zero, this
last term is absent.
So we just have a relationship
between rho and h, the Hubble
expansion rate.
And we can solve that.
And the value of rho which
satisfies that equation
is called the critical density.
As the density is equal
to the critical density,
it means that k is zero.
And that's called a flat case.
We'll figure out in a
minute how it evolves.
It's not clear if it will be
collapsed or stop or what.
But we'll find out soon.
It's really on the
borderline between something
which we know expands
and something which
we know collapses.
And it's called a flat universe.
The word flat suggests
geometry, and we'll
be learning about that later.
General relativity tells
us a little bit more
than we learn here.
These equations are
all exactly true
in the context of
general relativity.
But general relativity
also tells us
that these equations
are connected
to the geometry of space.
And only for this critical mass
density is the space Euclidean.
The word flat here is used
in the sense of Euclidean.
So to summarize what we've said,
if the mass density is bigger
than this critical value,
we get a closed universe,
which reaches a maximum
size and then collapses.
If the mass density is less
than the critical density,
we get an open universe, which
goes on expanding forever.
And if the mass
density is exactly
equal to the critical density,
that's called a flat k.
So we'll explore a little
bit more in a minute.
It's interesting to know what
this critical density is.
It depends on the
expansion rate.
But the expansion rate has now
been measured quite accurately.
So if I take the
value of 67.3, which
is the value that comes from
the Planck satellite combining
their results with
several other experiments,
they get a value of 67.3.
And when we'll put
that into this formula,
the number one gets
is 8.4 times 10
to the minus 30 grams
per centimeter cubed,
which is only about 5 proton
masses per cubic meter.
It's an unbelievably empty
universe that we live in.
I say the universe
that we live in
because in fact the mass
density of our universe
is very close to
this critical value.
It's equal to it to within about
half of a percent we now know.
An important
definition, which we'll
be continuing to use
through the course
and which cosmologists
always use,
is omega, where omega
means capital Greek omega.
And that's just defined
to be the actual density
of the universe,
whatever it is, divided
by this critical density.
OK, the one remaining thing
that we did last time,
and we'll summarize
this and go on,
is we figured out
what the evolution is
for a flat universe.
And we can do that just by
solving the differential
equation, which is a fairly
simple differential equation.
If we leave out the k
term, the Freedman equation
becomes a dot over a
squared is equal to 8 pi
g over 3 times rho And we
know how rho depends on a.
It's proportional
to 1 over a cubed.
So the right hand side
here is some constant
divided by a cubed.
And by just
rearranging things, we
can rewrite that
as da over dt is
equal to some constant
over a to the 1/2.
In this slide I use this
symbol const a number of times.
Those constants are not
all equal to each other.
But they're all
constants, which you
can keep track of
if you wanted to.
But there's no need
to keep track of them
because they have no bearing
on the answer anyway.
So I just called
these constants const.
So again, da over dt
equals const over a
to the one half, which is an
easy differential equation
to solve.
We just multiply through by
dt and a to the one half,
and write it in as a
to the one half times
da equals constant times
dt, which can easily
be integrated both
sides of the equation
as indefinite integrals.
And then we get 2/3
times a to the 3/2
is equal to a constant times
t, where this constant happens
to be the same as that constant,
for whatever that's worth,
plus a new constant of
integration, c prime.
Then we argue that
the value of c prime
depends on how we
synchronize our clocks.
If we reset our clock by
changing t by a constant
that would change
the value of c prime.
And since we haven't
said anything yet
about how we're
going set our clock,
we're perfectly
free at this point
to just say that we're going to
set our clock so that t equals
0 corresponds to the same
time that a is equal to zero.
The initial singularity
of the universe
starts from zero
size with a as zero.
So if we do that when
a is zero, t is zero.
That means that c prime is zero.
So setting c prime
equal to zero is just
a choice of how
to set our clocks.
So we do that.
And then we can take the
2/3 power of this equation.
And since constants are
just constants that we don't
care about, we
end up with a of t
is proportional to t to the 2/3.
And proportional is
all we need to know,
because the constant
of proportionality
would depend on the
definition of the notch.
And we haven't
defined the notch.
And we don't need
to define the notch.
And so anything that depends on
the constant of proportionality
will never enter
any physical answer.
It will be relevant to
questions like how many notches
are a certain
distance on your map.
But for any physical
answer, we don't care.
If we want to talk
about our map,
we could just define a notch to
be whatever we want it to be.
OK, that's the
end of my summary.
Any questions about any of that?
I'm sorry.
I didn't come back to
answer your question
about the cylinder.
The cylinder problem does
end up always giving you
a closed universe no matter
what the parameters are.
It always collapses.
And even though its
energy as you compute it
would turn out to be positive,
the difference though
is that for the case
of the cylinder,
the potential energy
does not go to 0
as the thing becomes
infinitely big.
The potential energy has a
logarithmic diversion in it.
So the zero is just
placed differently
for the case of the
cylinder problem.
So it ends up being closed no
matter how fast it's expanding.
Yes?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Well that is true.
Certainly the
differential equations
break down when a
is equal to zero.
The mass density
goes to infinity.
But we're still certainly
free to set our clocks
so that the equations
themselves, when extrapolated
to 0, would have the
property that a equals zero
when t is equal to zero.
It's certainly
correct, and I was
going to be talking
about this in a minute,
that you should not trust
these equations back
to t equals zero.
But that doesn't
stop you for choosing
whatever you want
as your origin of t.
And if these are the
equations we have,
the simplest way to deal
with these equations
is to use the zero of t when
equations say that a was zero.
Any other questions?
OK, in that case, we will
leave the slides for a bit
and proceed on the blackboard.
So so far I think
we have learned
two varied nontrivial things
from this calculation.
We learned how to calculate
the critical density.
We learned how to calculate what
density the universe would have
to have so it would re-collapse.
And we've also learned that
if the universe is closed,
we can calculate how large it
will get before it collapses.
So those are two very
nontrivial results coming out
of this Newtonian calculation.
The next question I
want to ask is still
about the flat universe.
It's a fairly trivial
extension of what we have.
Given this formula
for a of t, I would
like to calculate the
age of a flat universe.
If you were living in a flat
universe that was matter
dominated like the
one we're describing,
how would you determine
how old it was?
And the answer is that
it's immediately related
to the Hubble expansion
rate, and the age
can be expressed in terms of
the Hubble expansion rate.
To see that-- so we're
calculating the age of a matter
dominated flat universe.
And these age
calculations will be
extending as we go
through the course.
So in the end, we'll
have the full calculation
for the real model that
we have of our universe.
But you have to start somewhere.
So we're starting
with just a flat,
matter dominated universe.
We know that a of t is equal
to some constant, which
I will call little v
times t to the 2/3 power.
Previously I just
used proportional to,
but now it's just more
convenient to give a name
to the constant of
proportionality.
We'll never need
to know what it is.
But v is some constant
of proportionality.
This by the way already
tells us something else,
which wasn't obvious
from the beginning, which
is our flat universe does go
on expanding forever, somewhat
like an open universe.
An important difference
is that if you calculate
da dt for the open
universe, that approaches
a constant as time
goes to infinity.
That is, the universe
keeps on expanding
at some-- minimal rate forever.
In this case, if you calculate
the adt, it goes to 0 at times.
So the flat universe
expands forever,
but at an ever, ever,
ever decreasing rate.
We know how to
relate a of t to h.
The Hubble expansion
rate is a dot over a,
we learned a long time ago.
And if we know what a
is, we know what this is.
So this is just 2 over 3t.
The 2/3 coming from
differentiating the 2/3.
And that gives you a
t to the minus 1/3,
but then you're
also dividing by a,
which turns that t to the minus
1/3 to a t to the minus 1.
So you get 2 over 3t
is the final answer.
So this is the relationship
now between h and t,
and the question we asked
is how to calculate the age.
The age is t.
This is all defined as where t
is equal to 0 at the big bang.
So t really is the time
elapsed since the big bang.
So we're left immediately
with a simple result,
that t is equal to
2/3 times h inverse.
Now this result immediately
makes rigorous contact
with something that we talked
about in vague terms earlier.
If you are so unfortunate
as to badly mis-measure h,
you can get a pretty wild
answer for the h of your model
universe.
And Hubble mis-measured
h by about a factor
of seven comparative to
present modern values.
He got h to be too high.
His value of h was too high
by a factor of about seven,
and that meant that when big
bang theorists calculated
the age of the
universe were getting
ages that were too low
by a factor of seven.
And in particular
that meant they
were getting ages of the
order of 2 billion years.
And even back in
the 1920s and '30s,
there was sufficient
geological evidence
that the Earth was older
than 2 billion years.
There was also
significant understanding
of stellar evolution,
that starts
took longer to evolve
than 2 billion years.
So the big bang model was
in trouble from the start,
largely because of this
very serious mis-measurement
in the early days of the
Hubble expansion rate.
If we put in some
numbers, this of course
is not an accurate model for
our universe, we now know.
Our universe is now currently
dark energy dominated.
But nonetheless, just
to see how this works,
we can put in numbers.
So h, using this Planck value
that I quoted earlier 67.3
plus or minus 1.2 kilometers
per second per megaparsec.
To be able to get
an answer in years,
one has to be able to convert
this into inverse years.
h is actually an inverse time.
And a useful conversion
number is 1 over 10
to the 10 years is
equal to almost 100,
but not quite, 97.8 kilometers
per second per megaparsec,
which allows you to convert
these Hubble expansion rate
units into inverse years.
And using that one finds that
the age of the universe using
the 2/3 h inverse formula
is 9.7 plus or minus
0.2 billion years.
Now this number played a role
in the fairly recent history
of cosmology.
Before 1998, when
the dark energy
was discovered, which kind of
settled all these questions,
but before 1998, we thought the
universe what matter dominated.
It might have been open.
It didn't have to be flat.
That was debated.
It looked more open and flat.
But some of us wanted to
hold out for a flat universe
because we were
fans of inflation
and admired inflation's
other successes,
which we'll learn about
later in the course,
and thought therefore that
the universe should be flat,
and wanted to try to
reconcile all this.
And the problem was that like
cosmology in the '20s and '30s
when the age of the
universe that you calculate
was too young, the same thing
was happening here before 1998,
when we thought the universe
was matter dominated.
This is the age that we got,
modified a little bit by having
different values of h,
but pretty close to this.
At the same time,
there were calculations
about how old the
universe had to be
to accommodate the oldest stars.
And in the lecture notes
I quote a particular paper
by Krauss and Chaboyer.
Lawrence Krauss is an
MIT PhD by the way.
And what they decided by
studying globular clusters,
which are supposed
to contain the oldest
stars that astronomers know
about, that the oldest stars,
they said, had an age of
12.6 plus 3.4 minus 2.2
billion years.
And this is a 95%
confidence number.
That is, instead of using
one sigma, which are often
used to quote errors,
these are two sigma errors,
which have probabilities
of being wrong by only 5%
if things work properly
according to the statistics.
So then we're going to
think about 95% limits.
So they were willing to
take the lower limit here,
which was 10.4.
So they got a minimum
age of 10.4 billion years
for the oldest stars.
But they also argued that the
stars really could not possibly
start to form until
about 0.8 billion years
into the history
of the universe.
And doing a little bit of
simple arithmetic there,
they decided that the
minimum possible age
for the universe at the
95% confidence level,
would be 10.4 plus 0.8 or 11.2.
And 11.2 is older than
9.7, and by a fair number
of standard deviations,
although [INAUDIBLE]
were somewhat bigger in
1998 than they are now.
But in any case, this led to I
think what people at the time
regarded as a tension between
the age of the universe
question and the possibility
of having a flat universe.
A flat universe
seemed to produce
ages that were too young
to be consistent with what
we knew about stars.
Yet there was still
evidence in terms
of the desired to make
inflationary models
work to indicate that
the universe was flat.
And actually it's also
true that by 1998, there
was evidence from the
Kobe satellite measuring
fluctuation in the cosmic
background radiation, which
also suggested
that omega was one,
that the universe with flat.
So things didn't fit together
very well before 1998.
And this was at the
crux of the argument.
It all got settled with the
discovery of the dark energy,
which we'll learn how to
account for in a few weeks.
When we includes dark energy
in these calculations,
the ages go up,
and everything does
come into accord
with the idea now
that the age of the
universe is estimated
at 13.8 billion years.
And that's consistent with the
Hubble expansion rate given
here, as long as one
has dark energy and not
just relativistic matter.
OK any question about that?
OK next thing I wanted
to say a little bit about
is what exactly we
mean here by age.
And the question of what we
mean by age does of course
connect to the question of how
do we think it actually began.
Because age means time since
the beginning presumably.
And the answer really
is that we don't
know how the universe began.
The big bang is often said to be
the beginning of the universe.
But I would argue that
we don't know that,
and I think most
cosmologists would
agree that we don't know that.
As we extrapolate
backwards, we're
using our knowledge of physics
that we measure in laboratories
and physics that we confirm
with other astrophysical type
observations, but nonetheless,
as we get closer and closer
to t equals zero, the mass
density in this approximation
grows like 1 over the
scale factor cubed, which
means it goes up
arbitrarily large.
Later we'll learn that when the
universe gets to be very young,
we have to include radiation and
not just relativistic matter.
The dark energy is actually
totally unimportant
when we go backwards in time.
It becomes important when
we go forwards in time.
But when we put in radiation,
it does not solve this problem.
The universe still
requires a mass density
that goes to infinity as
we approach t equals zero.
People had wondered
whether maybe that's
an idealization associated
with our approximation
of exact homogeneity
and isotropy
which after all do break
down at some level.
Maybe if we put in a slightly
inhomogeneous and slightly
anisotropic universe
and ran it backwards,
maybe the mass density
would not climb to infinity.
Hawking proved that
that was not a way out.
The universe would
become singular.
He didn't really prove the
mass density went to infinity,
but he proved it becomes
singular in other ways
as t went to zero no matter
what geometry you put in.
So the bottom line is that
classical general relativity
does predict a
singularity of some sort
as we extrapolate
backwards in time.
But the important
qualification is
that once the mass density goes
far above any mass densities
that we've had any
experience with,
we really don't know how
things are going to behave.
And we don't really know how
classical general relativity
holds in that regime.
And in fact, we have
very strong ways
is to believe that classical
general relativity will not
hold in that regime.
Because classical
general relativity
is after all a classical
theory, a theory in which one
talks about fields that
have definite values
at definite times
without incorporating
the ideas of the uncertainty
principle of quantum theory.
So nobody in fact knows how to
build a theory in which matter
is quantized and gravity
is not quantized.
So all the smart money bets on
the fact that gravity is really
a quantum theory, even though
we don't yet quite understand it
as a quantum theory.
And that as we go back in
time, the quantum effects
become more and more important.
So there's no reason to trust
classical general relativity
as we approach t equals
zero, and therefore no reason
to really take this
singularity seriously.
Furthermore, we'll even see
at the end of the course
that most inflationary
scenarios imply
that what we call
the big bang is not
a unique beginning
of the universe.
But rather it now
seems pretty likely,
although we sure don't
know, that our universe
is part of a multiverse,
where we are just
one universe in the multiverse,
and that the big bang, what
we call the big bang, is really
our big bang, the beginning
of our pocket universe.
But before that the space
of time already existed.
The big bang is
just a nucleation
of a phase transition.
It's not really a beginning.
And that there was
other stuff that
existed before what
we call the big bang.
I should add though that the
inflationary scenario does not
provide any answer whatever
to the question of how did it
all ultimately begin.
That's still very
much an open question.
And it's clear that
inflation by itself
does not even offer an
answer to that question.
So when we talk about
the age of the universe,
what are we talking about?
What we're talking about is
the age, the amount of time
that has elapsed, since
this event that we
call the big bang.
The big bang might not have been
the beginning of everything,
but certainly the evidence
is overwhelmingly strong
that it happened.
And we could talk
about how much time
has elapsed since it happened.
And that's the t that we're
trying to calculate here.
And it will be offset
by a tiny amount
by changing the history in
the very, very early stages,
but only by a tiny
fraction of a second.
So the uncertainties
of quantum gravity
are not important in calculating
the age of the universe.
Although they are important
in interpreting what
you mean by it.
I think we don't really mean
the origin of space and time,
but rather simply
the time has elapsed
since the event
called the big bang.
OK any questions about that?
All right, next event
I want to talk about
is that if the
universe as we know
it began some 13.8
billion years ago to use
the actual current number, that
would mean that light could
only have traveled some finite
distance since the beginning
of the universe as
we know it, meaning
the universe since the big
bang, and that would mean there
would be some maximum distance
that we could see things.
And beyond that there
might be more things,
but they'd be things for
which the light has not yet
had time to reach us.
So that's an important
concept in cosmology,
the maximum distance
that you could see.
It goes by the name
the horizon distance.
If you're sailing on
the ocean, the horizon
is the furthest
thing you can see.
So what we want to do now
is to calculate this horizon
distance in the model
that we now understand,
the flat matter
dominated universe.
And this of course
is also a calculation
that we will be
generalizing as we
go through the
course learning how
to treat more and more
complicated cases and more
and more realistic cases.
So this horizon distance, I
should define it more exactly.
It's the present
distance, and maybe I
should even stick
the word proper here.
I've been usually using
the word physical distance
to refer to the
distance to an object
as it would be measured
by rulers, which
are each along the way
moving with the velocity
of the average matter
at those locations.
That is also called
the proper distance,
which is [INAUDIBLE] calls it.
And this horizon
distance is defined
as the present proper distance
of the most distant objects
that can be seen, limited
only by the speed of light.
So we pretend we have telescopes
that are incredibly powerful
and could see anything,
any light that
could have reached us.
But we know the light has
a finite propagation time,
so take that into account in
talking about this horizon
distance.
OK so what is the horizon
distance going to be?
Well remember that the
coordinate velocity of light
is equal to c divided by a of t.
I should maybe start by
saying before we get down
to details, that you might think
naively that the answer should
be the speed of light times
the age of the universe.
That's how far light can travel.
And so if the universe was
static and just appeared
a certain time in the past,
that would be the right answer.
I would just start
off at the beginning
and travel at speed c.
But it's more complicated
because the universe
has been expanding all along.
And it started out with
a scale factor of 0.
And furthermore, what
we're looking for
is present distance
in these objects,
and the objects of
course have continued
to move after the light that
we're now seeing has left them.
So it's a little
more complicated
than just c times
the speed of light.
And we'll see what
it is by tracking it
through very carefully.
We'll imagine a light
beam that leaves
from some distant object.
And the light beam will get the
furthest if it leaves earliest.
So we want the earliest
possible light beam
that could have left
this distant object.
And that would be a light
beam that left at literally t
equals zero.
So the light beam
leaves the object at t
equals zero and
reaches us today.
And we want to know how
far away is that object?
That's the furthest
object that we could see,
objects for which we can
only see the light that
was emitted from the
object at t equals zero.
So we're going to use our
co-moving coordinate system
to trace things.
All calculations are done
most straightforwardly
in the co-moving
coordinate system.
And we know that light travels
in the co-moving coordinate
system at the rate of dx dt is
equal to c divided by a of t.
And this really just says
that as the light passes
any observer in this
co-moving coordinate system,
the observer sees speed c, as
special relativity tells us
he must.
But we need to convert it
into notches per second
to be able to trace it through
the co-moving coordinate
system.
And the relationship between
notches and meters is a of t.
So a of t is just a
conversion factor here
that converts the local speed
of this light pulse from meters
per second to notches
per second, which
is what dx dt has
to be measured in.
So this will be the speed.
That means that the
maximum distance that light
will travel, still
measured in notches
in co-moving
coordinates, will just
be the integral of the speed.
The integral of dx
dt is just delta x.
So it would be the
integral of dx dt dt,
integrating from 0 up to
t zero, the present time.
Now this is not the final
answer that we're interested in.
We want to know the
present physical distance
or the present proper
distance of this object
that's the furthest
object that we can see.
And the way to go from
co-moving distances
to physical businesses is to
multiply by the scale factor.
And we are interested
in the distance
today, so we multiply
by the scale factor
today, the present value
of the scale factor.
So the answers to
our problem, which
I will call l sub p or
sub [INAUDIBLE] of t.
I want to get the word horizon
into the subscript someplace.
So I will call it l sub
p comma horizon, which
means the physical distance
to the horizon at time t
is just equal to x
max that we have here
times the present value
of the scale factor.
So it's a of 2
[INAUDIBLE] times x max.
Or the final formula, just
substituting in x max,
will be of a of t naught
times the integral from 0
to t naught of dx dt.
I'm going to substitute
c over a of t.
Let me just remind
you that [INAUDIBLE]
variable integration
should never
have the same symbol as
the limits of integration,
because that just
causes confusion.
They're never really
the same thing.
So I called the
limits t sub zero.
So it's perfectly OK to call
the variable of integration t.
In the notes, I call
the value of the time
that we want to
calculate this as t,
and then I use t prime for
the variable of integration.
Whatever you do,
you should make sure
that those are not the same.
There's one variable
that corresponds
to the variable that varies
from the initial time
to the final time.
And then there's
also another value
that represents the final time.
OK, so now all we have
to do is plug in a of t
is a constant times t to
the 2/3 into this formula
and we have our answer.
Notice that it does obey
an important property.
There's an a in the numerator
and an a in the denominator.
And that means that when we
put in the formula for a of t,
the constant of proportionality
b will cancel out.
And it must, as the
constants proportionality is
measuring the notches or notches
per seconds to the 2/3 power
but proportional to the notch.
And the answer can't
depend on notches
because notches are not
really a physical unit.
But it works.
That's an important check.
So-- now it's just a
matter of plugging in here,
and maybe I'll do it explicitly.
I'll leave out the
b's [INAUDIBLE]
so you see the cancel.
We have b times t zero to the
2/3 times the integral from 0
to t zero, times c over b
times t to the 2/3 times dt.
The b's cancel as I claimed.
The integral of t to the minus
2/3 is 3 times t to the 1/3.
We then subtract the t to
the 1/3 giving it the value t
zero on the positive
side, and then
we subtract the
value [INAUDIBLE]
same expression at zero.
But t to the 2/3 when
t is zero vanishes.
So we just get t zero to
the 2/3 from the upper limit
of integration.
I'm sorry, to the 1/3 power.
We integrate minus 2/3.
We get t to the 1/3.
The t to the 1/3
multiplies the t
to the 2/3 giving us
a full one power of t.
So what we're left with is
just 3 times c times t zero,
which has the right units.
It should have units
of physical distance.
Speed times time
is the distance.
And it has a surprising
factor of 3 in it.
The naive answer would have
just been c times t zero,
saying that the light
at time t travels
so it travels a
distance c times t zero.
That would be true, as I said,
in a stationary universe.
But the universe
is not stationary.
It's expanding.
And the fact it's
expanding means
that you expect this to be
bigger than c times t zero.
It means that at earlier
times, things were closer.
So the light can save time by
leaving early and traveling
a good part of the
distance while distances
are smaller than they are now.
And it's a full factor of 3.
So that is the horizon
distance for a flat,
[INAUDIBLE] universe.
We can also, since we
know how to relate t
sub zero to the
Hubble expansion rate,
we can express the
horizon distance
if we want in terms of
the Hubble expansion rate
by just doing that substitution.
So this then becomes 2 times
c times the current Hubble
expansion rate inverse.
So these are both
valid expressions
for the horizon distance
in this particular model
of the universe.
Any questions about the
meaning of horizon distance?
There is actually a
subtlety about the meaning
of horizon, which I
should talk about.
The initial value of the scale
factor in our model is zero.
It's t to the 2/3, and t goes to
0, the scale factor goes to 0.
Things are of course
singular there,
where you don't really trust
exactly what the equations are
telling us at t equals zero.
But that's certainly
how they behave.
a of t goes to 0 as t goes to 0.
That means initially everything
was on top of everything else.
So if everything was on
top of everything else,
why is there any
horizon distance?
Couldn't anything have
communicated with anything
at t equals zero, when the
distance between anything
and anything was zero?
The answer to that is
perhaps somewhat ambiguous.
We of course don't really
understand the singularity.
We don't claim to
understand the singularity.
And therefore anything
you want to believe
about the singularity
at t equals
zero you are welcome to believe.
And nobody intelligent is
going to contradict you.
They might not reinforce
you, but they're not
going to contradict you
either, because nobody knows.
So it is conceivable that
everything had a chance
to communicate with
everything else at t
equals zero at the singularity.
And it's conceivable that when
we understand quantum gravity,
it may even tell us that.
We don't know.
What is still the case is
that if you strike out t
equals zero exactly, then
everything is well defined.
You can ask what happens if a
photon is sent from one object
to another leaving that
object at time epsilon,
when epsilon is slightly
later than zero.
And you can ask how
long does that photon
take to go from
object a to object b?
And that's really exactly
the calculation we did,
except instead of going
down to t equals zero,
you go down to t
equals epsilon, where
epsilon is the
earliest time that you
trust your classical
calculations.
Then you'd be asking how far
is the furthest object that we
could see during this classical
era, the era that starts from t
equals epsilon?
The only difference would
be to put an epsilon there
instead of 0.
And the answer-- you
can go through it--
differs only by some
small multiple of epsilon.
And if epsilon is small, it
doesn't change the answer
at all.
Physically what is
going on is that if you
go to very, very early times
and look at two objects a and b
and trace them back, the
distance between them
does get smaller and smaller
as epsilon goes to 0.
So you might think that
communication would be trivial.
But at the same time as the
distance is going to zero,
you can calculate
the velocities.
h remember is going
like 2 over 3t.
h is blowing up.
The velocities between
these two objects
a and b are going to go to
infinity at the same time
that the distance between
them goes to zero.
So even though they
will become very close,
if one sends a light
beam to the other,
the other is actually
moving away faster
than the light would be moving.
The light would
eventually catch up,
but the amount of time it would
take for the light to catch up
is exactly what this
integral is telling us.
So there is no
widespread communication
that is possible once
these equations are valid.
They really do say that
in the early universe,
things just cannot talk to other
things because the universe is
expanding so fast.
And the maximum distance
that you can see
is more than c times t zero,
but less than infinity.
Yes?
AUDIENCE: Why does it
matter how far away
the furthest galaxies
we can see are now?
Because we're
seeing them as they
were a long time ago when
they were closer to us.
PROFESSOR: Yea, now we certainly
are seeing them a long time ago
when they were closer to us.
That's right.
I would just say that this
is sort of a figure of merit.
If you want to describe what
you think the universe looks
like now, you would assume
that those galaxies that you're
seeing billions of light years
in the past are still there,
and you'd extrapolate
to the present.
So it's relevant to
the picture that you
would draw in your mind
of what the universe looks
like at this instant, although
that picture would be based
on things you haven't
actually seen.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yes?
AUDIENCE: Is the
fact that this number
is greater than c t zero
proof that the universe is
the actual space of the
universe is expanding?
PROFESSOR: I don't think
so because if you just
have the objects moving
and a space that you regard
as absolutely fixed
and then you ask
the present distance of that
object given that you can see
a light pulse that was
emitted by that object,
it will still be bigger than
ct, because the object continues
to move after it
emits the light pulse.
So I don't think this is
proof of anything like that.
I think I should add
that certainly we
think of this as
the space expanding.
That is certainly
the easiest picture.
But you can't have any
absolute definition
of what it means for
the space to expand.
AUDIENCE: So you're
saying it has
moved three times the
original distance it
was when it sent out an impulse?
PROFESSOR: No it actually was
zero distance when it sent out
the light pulse, because the
first object we in principle
could see is an object for which
the light pulse left it at t
equals zero when it literally
was zero distance from us.
And the light pulse actually
then gets further away from us
and comes back and
eventually reaches us.
We have enough information
here to trace it.
And that's how it behaves.
OK any other questions?
OK, next thing I
want to do, and I
don't know if we'll
finish this today or not.
But we'll get ourselves started.
We'd like to now, having
solved the equation
for the flat universe
for determining a of t,
we would now like
to do the same thing
for the open and
closed universes.
And we'll do the
closed universe first
because it's a
little bit simpler.
I mean they're about
equal, but we're
going to do the
closed universe first
because it's first in the
notes that I've written.
So we know what equation
we're trying to solve.
This is really now
just an exercise
in solving
differential equations.
So one has to be clever
to solve this equation.
So we will go
through it together
and see how one can be clever
and find the solution to it.
The equation is a
dot over a squared
is equal to 8 pi
over 3 g rho minus k
c squared over a
squared, with rho of t
being equal to rho i times a
cubed of t i over a cubed of t.
And I'm writing i here.
I could just as well write 1.
It could be any time.
Rho times a cubed isn't
dependent of time.
So you can put any
time you want there.
And the numerator is
just a fixed number.
OK, so our goal is to
solve this equation.
The first thing I want
to do is something
which is usually
a good thing to do
when you are given some
physical differential equation.
Physical differential equations
usually have constants in them
like g and c squared, which
have different units which
complicate things.
And they don't complicate
things in any intrinsic way.
They just give you extra
factors to carry around.
So it's usually a little
cleaner to eliminate
them to begin with by redefining
variables that absorb them.
One can do that by
defining variables
that have the simplest
possible units that
are available for
the problem at hand.
And in our case, we
have these complications
that k is measuring in
inverse notch squared,
and a is meters per notch.
We could simplify
all that by defining
some auxiliary quantities.
So in particular, I'm
going to define an a
with a tilde above it, which is
sort of like the scale factor,
but redefined by the
square root of k.
And this is the case
when k is positive.
I said we're going to do
the closed universe first.
I would not want to divide
by the square root of k
if k were negative.
That would be confusing.
Now the nice thing
about this is remember
k has units of
inverse notch squared.
a has units of meters per notch.
That means the
notches just go away.
So this has units
of length only.
And that was the
motivation for dividing
by the square root of k,
to get rid of the notch.
Similarly, time has units
of meters per second.
I'm trying to
minimize the number
of distinct physical quantities
that we have in our problem.
So I'm going to define
a t twiddle which
is just equal to c times t.
This is of course no difference
in just saying the working
units is c is equal to
1, which a lot of times
is something people say.
This is the same
thing except I'm
a little bit more
explicit about.
So t twiddle now is
units of length also.
So the idea is to
translate everything
so that everything is
the same units, which
would be meters or
whatever physical unit
of length you want to use.
OK, now I'm just going
to rewrite the Freedman
equation using
these substitutions.
And to make way for
the substitutions,
I'm first just going to
divide the Freedman equation
by kc squared.
This is k to the 3/2.
OK now when I divided
by kc squared,
the kc squared over a squared
there became-- I'm sorry.
I'm doing more than
dividing by kc squared.
I'm dividing by kc squared
and multiplying by a squared.
So I have turned
that last term into 1
by multiplying by its inverse.
On the left hand side,
the 1 over a squared
went away when I
multiplied by a squared.
So I just have a dot squared
divided by kc squared so far.
We'll simplify that shortly.
And the middle term,
I took the liberty
of multiplying by
the square root of k,
and then we have a
k to the 3/2 here.
If we absorb this into
that, it would just
be the kc squared factor
that we divided by.
And the a squared
has been also divided
into two pieces, a cubed and a.
So together these
two factors make up
the factor of a squared that
we multiplied that term by.
So it's the same
thing we had, just
multiplied by the common factor.
And now the nice thing
is that this is, in fact,
our definition of
da tilde over dt.
The c turns the dt into a
dt tilde, and the 1 over k
turns the da into da tilde.
So the left hand
side now is simply
da tilde over dt tilde squared.
And the right hand side,
rho times a cubed remember
is a constant.
So the only thing that depends
on time on the right hand side
is the a divided by mu k,
which I've isolated here.
Earth That's a tilde.
So I'm going to take all of
this, which is a constant,
and just give it
a name, constant.
So this term becomes a constant,
we'll call 2 alpha divided
by a tilde.
And then we still have minus 1.
And this constant,
which I'm calling alpha,
is just all of this factor
except for a factor of 2.
So it's 4 pi over 3
g rho a tilde cubed
divided by c squared.
a tilde because we had a
cubed divided by k to the 3/2,
and that's a tilde cubed.
So I've just rearranged things.
But now everything has
the units of length.
a tilde has units of length.
t tilde has units of length.
This is dimensionless,
which is good
because that's
also dimensionless.
Alpha, if you work
out all this stuff,
has units of length. a
tilde has units of length.
This is length divided
by length, dimensionless.
We haven't really changed
anything significant.
But at least as far as
keeping track of factors,
that equation is
the one we'll solve.
And the factors are now absorbed
into the constant alpha, which
is the only thing we
have to worry about.
And we don't need to worry
about that until the end.
Yes?
AUDIENCE: [INAUDIBLE]
PROFESSOR: As long as these two
are evaluated at the same time,
it doesn't matter.
AUDIENCE: [INAUDIBLE]
PROFESSOR: That's right.
One does need to remember
that these two are
to be evaluated at the
same time, whatever it is.
And the product does
not depend on time.
So I didn't write the arguments.
I could have, and I could have
just put in an arbitrary time.
But it would be the same time
for the rho and the tilde.
I will do that.
It will be rho of
some t one times
a tilde cubed, evaluated the
same t one for any t one.
This product is
independent of time.
OK, now that is the
kind of equation
where we could move things
from one side of the equation
to the other to reduce it
to doing ordinary intervals.
So I can multiply by
d tilde and divide
by the expression on
the right hand side
and get an expression
that says that dt tilde is
equal to da tilde divided
by the square root of 2
alpha over a tilde minus 1.
And since this has an a
tilde in the denominator
of the denominator, I'm
going to multiply through
by a tilde to
rationalize things.
So I'm going to rewrite
this as a tilde da tilde
over the square root of 2 alpha
a tilde minus a tilde squared.
OK this looks better
than we've had so far.
Now in principle, if we imagine
we can do that integral,
we can just
integrate both sides.
We get an equation
that says t tilde is
equal to some expression
involving the final value of a.
So we're going to
imagine doing that.
And we will actually be
able to carry it out.
When we integrate, there are
two ways you can proceed.
When I solve the
analogous problem
for the flat case
with the t to the 2/3
it was a much simpler equation.
But you might remember
that at one point,
I had an equation
where I calculated
the indefinite
integral of both sides,
and then I got a constant
of integration, which I then
argued should be set
equal to zero if we wanted
to find the zero of time
to be the zero of a.
The situation is really
exactly the same,
but to show you
an alternative way
of thinking about
it, this time I'm
going to apply definite
integrals to both sides.
And if you apply a definite
integral to both sides,
the thing to keep in mind is
that you should be integrating
over the same physical
interval on both sides.
So on the left hand
side, I'm going
to integrate from zero
up to some t tilde final.
t tilde final is just
some arbitrary time,
but I'm going to give it
the name t tilde final.
And that integral
of course we can do.
It's just t tilde final.
And that should be equal to the
integral of the right hand side
over the same period of time.
But the right hand
side is not expressed
as an integral over time.
It's expressed as an
integral over a tilde.
So we have to ask
ourselves, what
do we call the integral
over a tilde that
corresponds to the integral over
time from 0 to t tilde sub f.
And the lower
limits should match.
And we know what we want a tilde
to be at t tilde equals zero.
We're going to use
the same dimensions
we had in the other case.
We're going to define
the zero of time
to be the time when the
scale factor is zero.
So t tilde equals
zero should correspond
to a tilde equals zero.
So to get the lower
limits of integration
to correspond to the same
time, we just put zero her.
And zero here doesn't
mean time zero.
It means a tilde equals zero.
But that's what we want.
And for the upper limit of
integration, that should just
be the value of a tilde
at the time t tilde sub f.
So I will call
that a tilde sub f,
where I might make
a note on the side
here that a tilde sub f is
e tilde of t tilde sub f.
It's just the final
value of a tilde, where
final means anytime I choose
to stop this integration.
The interval is valid
over any time period.
So these are moments
of integration
where the only thing
that's new on this line--
now I just copy from the
line above, 2 alpha times
a tilde minus a tilde squared.
So t tilde f is equal
to that integral.
And our goal now would
be to do that integral.
And if possible-- it
won't quite be possible.
We can see how
close we can come.
But if possible we would like
to invert that relationship
that we get from
doing this integral,
to determine a tilde as
a function of t tilde.
That's what we would love to do.
It turns out that's
not quite possible.
But we will nonetheless be
able to obtain something called
a parametric solution
to this problem.
And we'll stop now.
But I will tell you next
Tuesday after the quiz
how we proceed here to get
a solution to this problem.
