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PROFESSOR: OK, in
that case I can
begin by giving a quick
review of last time.
We began last time by talking
about the data of measurements
of the cosmic
microwave background,
starting with the data as
it existed in 1975 which
I advertised as being an
incredible mess, which it was.
You could easily believe that
this data fit this solid line
curve, which was what
it's supposed to fit.
But you could equally well
believe that it didn't.
Things got worse
before they got better.
There was the famous
Berkeley-Nagoya Rocket Flight
experiment of 1987 which
had a data point which
missed the theoretical curve by
16 standard deviations, which
might seem fairly disappointing.
It reminds one, by the
way, of a famous quote
of Arthur Eddington--
which you may or may not
be familiar with-- but
Eddington pointed out
that while we always say that
we should not believe theories
until the confirmed
by experiment,
it's in fact equally true that
we should not believe data
that's put forward until
it's confirmed by theory,
and that certainly
was the case here.
This data was never
confirmed by theory
and turned out to be wrong.
The beautiful data
was achieved in 1990
by this fabulous COBE
satellite experiment, which
showed-- unambiguously,
for the first time--
that the cosmic background
radiation really
does obey an essentially
perfect blackbody curve, which
is just gorgeous.
We then went on, in
our last lecture,
to begin to talk about
the cosmological constant
and its effect on the
evolution of the universe--
completely changing gear
here-- and the key issue
is the cosmological
effect of pressure.
Earlier we had
derived this equation.
The equation shows the
significant role of pressure
during the
radiation-dominated era,
but it also shows
that pressure--
if it were negative--
could perhaps
have the opposite
effect, causing
an acceleration of the universe.
Furthermore, we
learned last time
that vacuum energy-- first
thing we learned, I guess,
is just that's synonymous
with Einstein's cosmological
constant, related to Einstein's
cosmological constant lambda
by this equation.
That is, the energy
density of the vacuum
is equal to the mass
density of the vacuum times
C squared and is equal to
this expression in terms
of Einstein's original
cosmological constant.
And most important,
in terms of physics,
we learned that if we have a
non-zero vacuum energy-- vacuum
energy by definition
does not change with time
because the vacuum
is the vacuum,
it's simply the lowest
possible energy density allowed
by the laws of physics,
and the laws of physics,
as far as we know, do
not change with time,
and therefore vacuum
energy density does not
change with time-- and
that is enough to imply
that the pressure
of the vacuum has
to be equal to minus the
energy density in a vacuum,
and therefore minus
that expression in terms
of the cosmological constant--
which is exactly what will give
us a repulsive
gravitational effect,
where we put that into
the Friedmann equation.
Now I should emphasize
here that the effect
of the pressure that
we are talking about
is not the mechanical
effect of the pressure.
The mechanical effect
of the pressure
is literally zero because the
pressure that we are discussing
here is a uniform pressure,
and pressures only
cause mechanical
forces when there's
gradients in the
pressure-- more pressure
on one side than the other.
And if all this pressure
is always balanced,
the mechanical force of
the pressure is zero.
But nonetheless, that pressure
contributes-- according
to Einstein's equations--
as a contribution
to the gravitational field,
and a positive pressure
creates an attractive
gravitational field
but a negative pressure produces
repulsive gravitational fields.
And a positive vacuum
energy corresponds
to a negative pressure which,
in fact, would dominate
this equation, resulting in
a gravitational repulsion.
So it's useful to
divide the total energy
density into a normal
component plus vacuum energy,
and similarly we can
divide the pressure
into a normal component
plus the vacuum contribution
to the pressure.
The vacuum contribution
to the pressure
will instantly disappear
from all of our equations
because we know
how to express it
in terms of the
vacuum mass density.
It's just minus
Rho vac C squared.
So we can then re-write
the Friedmann equations
making those
substitutions, and we
conceive-- in the
second-order equation--
that the vacuum contribution,
negative, negative
is a positive, produces
a positive acceleration--
as we've been saying-- and a
positive vacuum energy also
contributes to the
right-hand side
of the first-order
Friedmann equation.
And in many, many situations--
although not quite all--
this vacuum energy will
dominate at late times.
It definitely falls
off more slowly
than any of the
other contributions.
The vacuum energy is a constant,
and every other contribution
to that right-hand
side falls off
with A. The only
way that Rho vac can
fail to dominate
if it's non-zero
is if you have a closed universe
that collapses before it has
a chance to dominate,
which is a possibility.
But barring that, eventually
the vacuum energy will dominate,
and once the vacuum
energy dominates,
we just have H
squared-- A dot over A
is H. H squared
equals a constant,
and that just says that
H approaches its vacuum
value, which is the square
root of 8 pi over 3 G Rho vac,
and with H being a
constant we can also
solve for A. The
scale factor itself
is just proportional to an
exponential of E to the H T,
where H is the H
associated with the vacuum.
So this will be a very easy
to obtain asymptotic solution
to the equations of the
universe, and, in fact,
we think that our real universe
is approaching an exponentially
expanding phase of
exactly this sort today.
We're not there yet, but
we are approaching it.
Yes?
AUDIENCE: So does an
exponential A of T
mean that the universe just
keeps expanding forever
and just spins out
into nothingness?
PROFESSOR: Yes, an
exponentially expanding
A means the universe will
continue expanding forever
and ordinary matter will
thin out to nothing,
but this vacuum
energy density will
remain as a constant
contribution.
So the universe would go
on expanding exponentially
forever.
Now, there is the possibility
that this vacuum that we're
living in is actually what might
be called a false vacuum-- that
is, an unstable
vacuum-- a vacuum which
behaves like a vacuum
for a long time
but is subjected to the
possibility of decay.
If that's the case,
it's still true
that most of our
future space time
will continue to
exponentially expand exactly
as this equation shows, but
a kind of a Swiss cheese
situation will develop
where decays in the vacuum
would occur in places,
producing spherical holes
in this otherwise exponentially
expanding background.
We'll be talking
a little bit more
about that later in the course.
Any other questions?
OK, well what I
want to do now is
to work on a few
calculations which
I'd like to present today.
If all goes well, we might have
as many as three calculations,
or at least two
calculations that we'll do
and one that we'll talk
about a little bit.
The first thing I
want to do-- and I
guess we just talked about
starting it last time--
is calculate the age of the
universe in this context.
How do we express the
age of the universe
in terms of measurable,
cosmological parameters,
taking into account the
fact that vacuum energy is
one of the ingredients
of our universe
along with radiation and
non-relativistic matter, which
we have already discussed.
So to start that calculation
we can write down
the first-order
Friedmann equation.
A dot over A squared is equal
to 8 pi over 3 G times Rho
and now I'm going to divide
Rho into all contributions
we know about.
Rho sub M, which represents
non-relativistic matter,
plus Rho sub radiation,
which represents radiation,
plus vacuum energy, which is
our new contribution, which
will not depend on time at all.
And then to complete
the equation
there is minus KC
squared over A squared.
And the strategy
here is really simply
that because we know the
time evolution of each
of the terms on the
right-hand side,
we will be able to start
from wherever we are today
in the universe-- which
will just take from data,
the values of these
mass densities--
and we will be able to integrate
backwards and ask how far back
do we have to go
before we find the time
when the scale factor vanished,
which is the instant of the Big
Bang.
So what we want to do is
to put into this equation
the time dependents
that we know.
So Rho sub M of T,
for example, can
be written as A of T naught
divided by A of T cubed,
times Rho sub M zero.
And all these zeroes, of
course, mean the present time.
So this formula
says, first of all,
that the mass density
falls off as 1
over the cube of
the scale factor.
A of T is the only factor
on the right-hand here
that depends on T. The
numerator depends on T sub zero,
but not T. The
other constant, T is
zero in the numerator
and Rho sub M zero.
Rho sub M zero denotes
the present value
of the mass density.
And the constants here
are just rearranged
so that when T equals
T naught, you just
get Rho is equal to
its present value.
And we can do the same
thing for radiation,
and here I won't
write everything out
because most things
are the same.
The quantity in brackets
will be the same
but this time it will
occur to the fourth power
because radiation falls
off like the fourth power
of the matter-- fourth
power of the scale factor--
and then we have
Rho radiation zero.
And then finally, for
the vacuum energy,
we will just write on the
blackboard what we already
know, which is that's
independent of time.
So this gives us the time
dependents of all three terms
here.
Now we could just go from
there, but cosmologists
like to talk about
mass densities in terms
of the fraction of the
critical density, omega.
So we're going to
change the notation just
to correspond to the
way people usually
talk about these things.
So we will recall that the
critical density-- which
is just that total
mass density that
makes little K equal
to zero and hence
the universe geometrically
flat-- so Rho sub C,
we learned, is 3 H
squared over 8 pi G
and then we will introduce
different components of omega.
So I'm going to write omega sub
X here where X really is just
a stand-in for matter or
radiation or vacuum energy.
And whichever one of
those we're talking about,
omega sub X is just a shorthand
for the corresponding mass
density, but
normalized by dividing
by this critical density.
And then I'm just going to
rewrite these three equations
in terms of omega
instead of Rho.
So Rho sub M of T becomes then
3 H naught squared over 8 pi G
times the same A of T
zero over A of T cubed,
but not I'm multiplying,
omega sub M zero.
And from the definitions
we've just written,
this equation is just a
rewriting of that equation.
And we can do the same thing,
of course, for radiation.
Rho radiation of T is
equal to the same factor
out front, the same
quantity in brackets
but this time to
the fourth power,
and then times omega
radiation at the present time.
And finally, Rho vac doesn't
really depend on time--
but we'll write it as if
it was a function of time--
it consists of the same factor
of 3 H naught squared over 8 pi
G, and no powers of the
quantity in brackets,
but then just multiplying
omega sub vac zero.
[ELECTRONIC RINGING]
Everybody should turn off
their cell phone, by the way.
[LAUGHING]
OK, sorry about that.
OK, now to make the
equation look prettier,
I'm going to rewrite
even this last term
as if it has something
to do with an omega.
And we can do that
by defining omega sub
K zero to be equal to minus
KC squared over A squared of T
naught times H naught
squared, which is just
the last term that appears
there [INAUDIBLE] of factor
of H squared, which we'll
be able to factor out.
And doing all that, the original
Friedmann equation can now
be rewritten as H squared-- also
known as A dot over A squared--
can be written as H naught
squared-- oh, I'm sorry,
one other definition I
want to introduce here.
This ratio-- A of T naught
over A-- keeps recurring,
so it's nice to give
it a name, and I'm
going to give 1
over that a name.
I'm going to let X equal
the scale factor normalized
by the scale factor today.
And I might point out that
in Barbara Ryden's book, what
I'm calling X is just what
she calls the scale factor,
because she chooses to normalize
the scale factor so that it's
equal to 1 today.
So we haven't done that yet
but we are effectively doing it
here by redefining a
new thing X. Having done
that, the right-hand side
of the Friedmann equation
can now be written
in a simple way.
It's H naught squared
over X squared
times a function F of X--
which is just an abbreviation
to not have to write something
many times-- this is not,
by any means, a
standard definition.
It really is just for today.
It allows us to save some
writing on the blackboard.
So I'm going to, for today,
be using the abbreviation
F of X is equal to omega sub
M zero times X plus omega sub
radiation zero times no
powers of X plus omega
sub vac zero times X to
the fourth, and finally
plus omega sub K
zero times X squared.
And this just lists
all the things
that would occur in parentheses
here if we factored everything
out.
Notice I factored out
some powers of X squared,
so the powers of X that appear
here do not look familiar,
but the relative powers do.
That is, omega should fall
like-- omega matter should
fall like 1 over X cubed.
Omega radiation should fall
off like one power faster
than that, and it does.
This is one less power
of X there than there,
and omega vacuum should fall
off like four powers of X
different from radiation,
and it does, et cetera.
But there's no real
offset here that makes
the factors there
not look familiar.
OK, all of this was just to
put things in a simple form,
but there's one other very
useful fact to look at.
Suppose we now apply
this for T equals
T zero, which means
for X equals 1.
OK, it's true at any
time, but in particular
we can look at what
it says for X equals 1
and it tells us something
about our definitions
that we could have
noticed in other ways--
but didn't notice yet-- which is
that we set X equal to 1 here.
These just becomes
the sum of omegas.
The powers of X's
all become just ones.
And the left-hand side
is just H zero squared,
which matches the
H zero squared here
because at T equals T naught
H squared is H zero squared,
so these H zeros
squareds cancel.
So applying it to
T equals T naught X
equal to 1, what
you get is simply
1 is equal to omega sum M zero
plus omega sub radiation zero
plus omega sub vac
zero plus omega sub K
zero, which gives
us the simplest
way of thinking about what this
omega sub K zero really means.
We defined it initially up
there in terms of little K,
but for this equation, we
can see that omega sub K
zero really is just
another way of writing
1 minus all the other omegas.
So you could think of this as
being the definition of what
I'm calling omega sub k zero.
So it's a language in
which you essentially
think that the total
omega has to equal 1,
and whatever is not contained
in real matter becomes
a piece of omega sub k, the
curvature or contribution
to omega.
OK, now it's really
just a matter
of simple manipulations
and I-- the main purpose
of defining F of x is
to be able to write
these simple manipulations
simpler than they would be
if you had to write out
what F of x was every time.
We're first just going
to take the square root
of the key equation up there--
the Friedmann equation-- and we
get a dot over a is also
equal to x dot over x.
Note that the constant of
proportionality there, a
of t naught-- which
is a constant--
cancels when you
take a dot over a.
So a dot over a is the
same as x dot over x.
And that-- just taking the
square root of that equation--
is H naught over x.
Hold on a second.
Yeah, we're taking the
square root of the equation,
so yeah, we had H naught
squared over x squared.
Here we have H naught
over x and then just times
the square root of f.
And these x's cancel each other.
Wait a minute.
Oh, I'm sorry.
They're not supposed
to cancel because I
didn't write this
quite correctly.
That should have
been x to the fourth.
Apologies.
And now here we have x squared.
And this can just
be-- by manipulating
where the x's go--
rewritten as x times dx dt.
So I multiply the
whole equation by x
squared to get rid of
that factor on the right,
and now on the
right-hand side we just
have H zero times
square root of F.
And now I just want to do
the usual trick of separating
the dx pieces from dt
pieces in this equation.
And we can rewrite
that equation as dt
is equal to 1 over
H naught times
x dx over the square root of
F. And maybe I'll rewrite it
as the square root of F
of x to make it explicit
that F depends on x.
So this is just the
rewriting of this equation,
moving factors around,
and in this form
we can integrate
it and determine
the age of the universe.
The present age of the
universe can be obtained just
by integrating this
expression from the big bang
up to the present.
And that will be the integral
dt from the big bang up
to present, the
sum of all the time
intervals from the
big bang until now.
And it's just equal
to 1 over H zero times
the integral of x dx over
the square root of F of x.
And now-- just to think about
the limits of integration--
what should limits
of integration be?
AUDIENCE: Zero to one.
PROFESSOR: I hear zero to
one, and that's correct.
We're integrating from the
big bang up to the present.
At the big bang,
a is equal to zero
and therefore x
is equal to zero.
And, at the present time,
t is equal to t naught
and therefore x is equal to 1.
So we integrate up
to one if we want
the present age of the universe.
We could also integrate
it up to any other value
of x that we want, and it will
tell us the age of universe
when the scale factor
had that value.
So this is the final,
state of the art
formula for the age
of the universe,
expressed in terms of
the matter contribution
to omega, the radiation
contribution to omega,
and the vacuum
contribution to omega,
and the value of H naught.
Those are the only ingredients
on the right-hand side there.
And then you can
calculate the age.
And it's the completely
state of the art formula.
It's exactly what the Planck
people did when they told you
that the age of the universe
was 13.9 billion years, using
that formula.
Now as far as actually doing the
integral, in the general case,
the only way to do
it is numerically.
That's how it's usually done.
Special cases can be
done analytically.
We've already talked about
the case where there's
no cosmological constants, no
vacuum energy, but just matter
and curvature-- omega,
in this language.
There's another special
case which can be done,
which is the case
that involves vacuum
energy and
nonrelativistic matter.
And this is the flat case, only,
that can be done analytically.
So it corresponds
to omega radiation
equals omega k
equals zero, and that
means that omega matter plus
omega vac is equal to 1,
because the sum of all the
omegas is always equal to 1.
So in this case I can
write an answer for you.
And I don't intend to try
to derive this answer,
but it's worth knowing that
can be written analytically.
That's the main point, I guess.
So it does get divided
into three cases
depending on whether omega
matter is larger than, smaller
than, or equal to 1.
So the first case will
be if omega matter
zero is greater than 1.
And if omega matter
is greater than 1,
that corresponds to
omega vac less than 1
because the sum of the two is
equal to 1 in all cases, here.
So omega vac has
to be less than 0.
So this is not our real
universe but it's a calculation
that you can do, and
it's 2 over 3 H naught
times the inverse tangent of
the square root of omega sub
m zero 1 over the square root
of omega sub m zero minus 1.
So if you plug this
integral into mathematica,
you should get that answer or
something equivalent to it.
For the case-- the
borderline case, here--
where omega matter
zero equals 1,
that's the special case
where omega vac is equal to 0
because the sum of
these is always one.
So this special
case in the middle
is the case we
already knew, it's
just the matter-dominated
flat universe.
So that's two thirds H inverse.
So it's 2 over 3 H naught.
And then, finally, if omega
matter zero is less than 1
and then omega vac is positive.
And this [INAUDIBLE]
approximation
is our universe, that is, that
our universe has possibly zero
curvature-- in any case,
unmeasureably small curvature--
and very, very small radiation
for most of its evolution.
So this last case
is our universe
except for cases that are near
the radiation-dominated era,
and the formula here is
2 over 3 H naught times
the inverse hyperbolic
tangent of 1
minus-- square root, excuse
me-- of 1 minus omega sub m
zero over the square root
of 1 minus omega sub m zero.
OK, so this is just
a result obtained
by doing that integral
for the special case
that we're talking about.
Now, I don't know
any simpler way
to write it except
as these three cases.
It is, however, a single
analytic function,
and when you graph it--
and I'll show you a graph--
it is one smooth function right
across the range of these three
cases, which is similar
to what we found
the earlier for the flat,
matter-dominated case.
So let's see.
This is not that yet.
This is the case that
we did a long time ago,
actually, the case of a
matter-dominated universe
with nothing but
nonrelativistic matter
and possibly with curvature.
And I can remind
you, here, that what
we found for that model is
that we tended to get ages
there were too young.
So if we take a
reasonable value for H
of 67 to 70 kilometers per
second per megaparsec--
which is in this range--
and take a reasonable value
for omega-- which is somewhere
between, say, 0.2 and 1
depending on what you
consider reasonable,
this model doesn't work
anyway-- but if you
take omega anywhere
between 0.1 and 1,
you get numbers
for the age which
are in the order of 10 billion
years, which is not old enough
to be consistent
with what we know
about the ages of
the older stars.
And especially if you think
that omega should be one,
you get a very young age of
more like 9 billion years,
which is what we found earlier.
This is just a graph of
those same equations.
But, if we include
vacuum energy,
it makes all the difference.
So this now is a graph
of those equations.
What's shown is the age, T
naught, as a function of H
and for various values of
omega sub m, the same omega sub
m that's called omega sub
m zero on the blackboard.
And shown here are the Barbara
Ryden a benchmark point,
which is the left-most of these
two almost overlapping points.
And also shown here
is the favored point
from the WMAP satellite
seven-year data.
They lie almost on
top of each other.
I didn't get a chance to
plot the Planck point, which
is the one that we
would consider the most
authoritative these
days, but I'll
add that before I
post the lectures.
It lies almost on top of
these, and it corresponds
to a Hubble expansion
rate of a little under 70,
and a vacuum energy
contribution of about 0.7,
and therefore a matter
contribution of about 0.3.
This curve.
And it gives an age of 13.7,
13.8 billion years-- perfectly
consistent with estimates of
the age of the oldest stars.
So this age problem
which had been,
until the discovery of the
dark energy, a serious problem
in cosmology for
many, many years
goes away completely once
one adds in the dark energy.
So that's it for
the age calculation.
Are there any questions about
the age of the universe?
Yes?
AUDIENCE: So when
you say dark energy,
are you using that synonymously
with vacuum energy?
PROFESSOR: Sorry, yes.
I used the word
dark energy there
and I've been talking
about vacuum energy,
and what's the relationship?
When I said dark energy I
really meant vacuum energy.
In general, the way
these words are used
is that vacuum energy has
a very specific meaning.
It really does mean the
energy of the vacuum,
and by definition,
therefore, it does not
change with time, period.
We don't know for
sure what this stuff
is that's driving the
acceleration of the universe,
and hence the name dark energy,
which is more ambiguous.
I think the technical
definition of dark energy
is it's whatever the
stuff is that's driving
the acceleration
of the universe.
And the other conceivable
possibility-- and observers
are hard at work trying to
distinguish, experimentally,
between these two options-- the
other possibility is that it
could be a very
slowly evolving scalar
field of the same type
that drives inflation
that we'll be
talking about later.
But this would be a
much lower energy scale
than the inflation of
the early universe,
and much more slowly evolving.
So far, we have not yet
found any time variation
in the dark energy.
So, so far, everything we' have
learned about the dark energy
is consistent with
the possibility
that it is simply vacuum energy.
Question.
AUDIENCE: Is the amount
of dark energy related
to the amount of dark matter?
PROFESSOR: Is the amount
of dark energy related to.
the amount of dark matter?
No.
They're both numbers
and they differ
by a factor of 2 and
1/2 or so, but there's
no particular relationship
between them that we know of.
AUDIENCE: But
doesn't dark matter
imply that they have a
certain attraction to bodies
around it, which is
a form of energy?
PROFESSOR: Yeah, well let's
talk about this later.
AUDIENCE: Do we have any idea
what dark energy is at all?
PROFESSOR: OK, the
question is, do we
have any idea what
dark energy is at all?
And the answer is probably yes.
That is, I think there's a good
chance it is vacuum energy.
Now if you ask what is
vacuum energy, what is it
about the vacuum that gives
it this nonzero energy,
there we're pretty
much clueless.
I was going to talk
about that a little more
at the end of today,
if we get there.
But whatever property
of the vacuum
it is that gives it its
energy-- we know of many,
it's just a matter
of what dominates
and how they add up--
the end result is pretty
much the same as far as the
phenomenology of vacuum energy.
So we understand the
phenomenology of vacuum energy,
I would say, completely.
The big issue, which
I'll talk about either
at the end of
today or next time,
is trying to estimate the
magnitude of the vacuum energy,
and there we're really
totally clueless,
as I will try to describe.
OK, that's it for my slides.
OK, I wanted to now talk
about another very important
calculation, which is
basically the calculation which
led to the original evidence
that the universe is
accelerating to begin with.
OK, discovery that the
universe was accelerating
was made, as I said earlier,
by two groups of astronomers
in 1998, and the key observation
was using a type 1a supernovae
as standard candles to
measure the expansion
rate of the universe
versus time,
looking back into the past.
And basically what they found
is that when they look back
about 5, 6 billion years,
the expansion rate then
was actually slower
than expansion rate now,
meaning that the
universe has accelerated.
And that was the
key observation.
So the question
for us to calculate
is, what do we expect, as a
function of these parameters,
for redshift versus luminosity?
These astronomers, by
using type 1a supernovae
as standard candles,
are basically
using the luminosity
measurements of these type
1a supernovae as estimates
of their distance.
So what they actually measured
was simply luminosity verses
redshift, and that's what we
will learn how to calculate,
and the formula
that will get will
be, again, exactly
the formula that they
used when they were trying to
fit their data-- to understand
what their data
was telling them--
about possible acceleration
of the universe.
So the calculation
we're about to do
is really nothing
new to you folks
because we have
calculated luminosities
in another contexts.
Now we will just write down the
equations in their full glory,
including the contribution
due to vacuum energy.
So we'd like to do
these calculations
in a way that allows
for curvature, even
though-- in the end--
we're going to discover
that the curvature
of our universe
is-- as far as anybody
can tell-- negligible.
But people still look for
it and it still very well
could be there at the
level of one part in 1,000
or something like that.
But at the level of 1 part
in 100, it's not there.
So we begin by writing down
the Robertson-Walker metric,
ds squared is equal
to minus c squared
dt squared plus a
squared of t times dr
squared over 1 minus little k
times r squared plus r squared
d theta squared plus sine
squared theta d phi squared,
end curly brackets.
OK, so this is the metric
that we're familiar with.
We're going to be interested,
mainly, in radial motion,
and if you're interested
mainly in radial motion,
it helps to simplify the
radial part of this metric
by using a different
radial variable.
And we've done
this before, also.
At this point, we
really need to pick
whether we're talking
about open or closed.
If we're talking
about flat, we don't
need to do anything, really.
If you eliminate k,
here, the radial part
is as simple as it gets.
But if we want talk
about open or closed,
it pays to use
different variables,
and the variable
that we'd use would
be different in two cases.
So I'm going to consider
the closed-universe case.
And I'm going to introduce
an angle, sine of psi
being equal to the
square root of k-- which
is positive in this
case-- times little r.
And this psi is, in fact, if
you trace everything back,
the angle from the
w-axis that we originally
used when we
constructed the closed
Robertson-Walker universe
in the first place.
But now we're essentially
working backwards.
We've learned to know
and love this expression,
so we're going to
just rewrite it
in terms of the new
variable, sin of psi
equals the square
root of k times r.
And from this, by
just differentiation,
you discover that deep psi
is equal to the square root
of k times dr over cosine psi.
And that is equal
to the square root
of k times dr over the square
root of 1 minus kr squared.
So this, then, fits in very
nicely with the metric itself.
The metric is just the
square of this factor,
and therefore it is just
proportional to deep psi
squared all by itself.
And rewriting the whole
metric, we can write it
as ds squared is equal to
minus c squared dt squared
plus a new scale factor--
which I'll define
in a second in terms of the old
one-- times deep psi squared
plus-- now, the
angular term becomes
nonstandard instead of just
having an r squared here,
we have sine squared of psi.
Which is, of course,
proportional to r squared.
And that multiplies d theta
squared plus sine squared
theta d phi squared,
end curly brackets.
And a tilde is just
equal to our original
a divided by the
square root of k.
So we scaled it.
And I should mention that
I'm putting a tilde here
because we've already written
an a without a tilde there,
and they're not equal to other.
If you want to just
start here, you can,
and then there's no
need for the tilde.
You could just call
this the scale factor
and it doesn't need a tilde.
The tilde is only to distinguish
the two cases from each other.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Sorry?
AUDIENCE: Do a and a tilde
have different units?
PROFESSOR: Didn't hear you?
AUDIENCE: Do a and a tilde
have different units?
PROFESSOR: They do have
different-- yes, a and a tilde
do have different units.
That's right, and
that's because in what
one might call
conventional units here,
r is some kind of a
coordinate distance.
So in my language I'd
measure it in notches,
and then a has units
of meters per notch.
On the other hand,
here psi is an angle.
It is naturally dimensionless.
So one doesn't introduce
notches in this case,
and therefore a just
has units of length--
a tilde, rather-- just
has units of length.
OK, now we want to imagine
that some distant galaxy is
radiating-- or a distant
supernova, perhaps--
and we want to ask,
what is the intensity
of the radiation that
we receive on earth?
And we'll draw the
same picture that we've
drawn at least twice
before, if not more.
We'll put the source
in the middle.
We'll imagine a sphere
surrounding the source,
with the source of
the center, and we'll
imagine that the sphere has been
drawn so that our detector is
on the surface of the sphere.
This will be the detector.
And we'll give a symbol for
the area of the detector.
It will be a.
And we'll imagine drawing this
in our co-moving coordinate
system where psi is
our radial variable.
So the sphere here
will be at some value,
psi equals psi sub d-- where d
stands for detector-- and psi
equals zero at the center.
OK, I'm going to make the
same kind of arguments
we've made in the past.
We say that the
fraction of light
that hits the sphere--
which hits the detector--
is just equal to the
area of the detector
over the area of the sphere.
Now, the area of detector
is, by definition, a.
The area of the
sphere we have to be
a little bit careful
about because we
have to calculate the area of
the sphere using the metric.
Now, the metric is
slightly nontrivial,
but the sphere is just described
by varying theta and phi.
And if we just vary theta and
phi, this piece of the metric
is what we're used to-- it's
the standard Euclidean spherical
element-- and the coefficient
that multiplies is just
the square of the
radius of that sphere.
So the radius of our sphere
is a tilde times sine psi.
That's the important thing
that we get from the metric.
The thing that multiplies
d theta squared and d phi
squared, et cetera,
is just the square
of the radius of the sphere
that determines distances
on the surface of the sphere.
So what goes here is 4 pi
times the radius squared.
So it's 4 pi times a tilde
squared of t naught times
sine squared of psi sub d.
It's t naught because
we're interested in what
happens when we detect
this radiation today.
Our detector is detecting it
today and has area a today,
and we want to compare it with
entire sphere that surrounds
this distant source as
that sphere appears today,
so that all of the distances
are measured today,
and therefore can be
properly compared.
The other thing we
have to remember
is the effect of the redshift.
The redshift,
we've said earlier,
and it's just a repetition,
it reduces the energy
of each photon by a factor
of 1 plus z, the redshift,
and similarly it reduces
the rate at which photons
are arriving at the sphere by
that same factor-- 1 plus z.
It basically says that
any clock slows down
by a factor of 1 plus
z, and that clock
could be the frequency
of the photon-- which
affects its energy
proportionally--
or the arrival rate
of the photons.
That's also a clock that get
time dilated in the same way.
So we get two factors of 1
plus z sub s, I'll call it.
s for z of the source,
the z between the time
of emission at the
source to a time
where it arrives at us today.
So 1 plus z is equal to
a of t naught divided
by a of t emission.
I'll just put it
here to remind us.
One from redshift of photons
and one from arrival rate.
OK, putting that
together we can now
say that the total
power received is
equal to the power originally
emitted by the source-- p
will just be the power emitted
by the source-- divided
by 1 plus the redshift
z of the source squared,
and then just
times the fraction.
A over 4 pi a twiddle
squared sine squared psi d.
And then, finally, what
we're really interested in
is J-- the intensity
of the source
as we measure it-- which
is just the power received
by our detector
divided by its area.
So from this formula we
just get rid of the A there.
We can write it as the power
emitted by the source, capital
P, divided by 4
pi 1 plus z sub s
squared a twiddle squared
of t naught times sine
squared psi sub d.
Now that effectively
is the answer
to this question except
that we prefer to rewrite it
in terms of things
that are more directly
meaningful to astronomers.
a twiddle is not particularly
meaningful to the astronomer.
The redshift is, that's OK.
But a twiddle is
not particularly
meaningful to an astronomer, nor
is sine squared of psi sub d.
Now, many astronomers who
know general relativity
can figure this out,
of course, but it's
our job to figure it out.
We would like to
express this in terms
of things that are directly
measured by astronomers.
So to do that, first
of all, a tilde--
to get a tilde related
to other things,
it really just goes
back to the definition
that we gave for
omega sub k sub 0.
And if you look back
at that definition,
you'll find that a
of t naught tilde
is just equal to c times the
inverse of the present Hubble
expansion rate times the
square root of minus omega
sub k comma zero.
And this is for the
close-universe case.
The closed-universe case,
little k is positive.
But if you remember the
definition of omega sub k
naught-- maybe I should write
it back on the blackboard,
or is it findable?
It's not findable.
The original definition big
A for this omega sub k naught
was just minus Kc squared over
a squared of t naught H naught
squared.
So this is just
rewriting of that,
and for our closed-universe
case, k is positive,
omega sub k naught is
negative, this is then
the square root of a positive
number with that extra minus
sign, so everything
fits together.
So that takes care
of expressing a tilde
in terms of measurable things.
We use this formula.
Expansion rate is measurable,
omega sub k comma zero
is measurable.
And then, in terms of
sine squared psi sub d,
we obtain that by
reminding ourselves
that we know how to trace light
rays through this universe.
Light waves just travel
locally at the speed c,
they travel locally
on null geodesics.
So if we're looking
at a radial light ray,
this metric tells us-- if we
apply it to a radial light ray
where ds squared equals minus dc
squared-- where ds squared has
to be zero-- that says
that minus c squared dt
squared plus a twiddle squared
of t d psi squared equals zero.
This is just the
equation that says
we have a null line,
a null radial line.
That implies that
deep psi dt has
to equals c over a
twiddle of t, which
is a formula that,
in other cases,
we try to motivate just
by using intuition.
But, in that case,
we were probably not
talking about curved
universe where
the intuition is a
little bit less strong.
But you see it does
follow immediately
from assuming that we're
talking about a null geodesic
in the Robertson-Walker metric.
Now, the point is that the
Friedmann equation, which we've
been writing and rewriting,
tells us what to do with that.
The Friedmann equation
basically allows
us to integrate that
because it allows
us to express a in
terms of x, and we
know some things about x.
So let me try to get that
on the blackboard, here.
We know that H squared-- which
is x dot over x squared--
can be written as H zero squared
over x to the fourth times
this famous function F of x.
And psi of a given redshift,
according to this equation,
could just be obtained by
integration of the time
that the source emits
the radiation up
to the present time of c
over a twiddle of t times dt.
And now to rewrite this
in terms of redshift,
we can use the fact that
1 plus z is equal to 1
over x because we know
how to relate 1 plus z
to the scale factor.
1 plus z is just the
ratio of scale factors
and it's precisely the ratio
that we called 1 over x.
And we can then
differentiate this equation
and find that dz is equal to
minus a twiddle of t naught
over a twiddle of t squared
times a twiddle dot times
dt, rewriting x in terms of
a of t naught over a of t.
And this, then, is equal to
minus a twiddle of t naught
times H of t times, oops,
times dt over a tilde of t.
And this allows us to replace
the dt that appears there
and the final relationship
is that psi of s
is equal to 1 over a tilde of
t naught times the integral
from zero up to z sub
s of c over H of z dz.
Yeah, I think that
looks like it works.
So it really is just a
matter of changing variables
to express things in terms
of H and integrating over z
instead of integrating over t.
And the usefulness
of that is simply
that z is the variable
that astronomers
use to measure time.
And this then can be
written in more detail,
and it really finishes
the answer more or less.
Psi of z sub s can be
written just-- writing
in what a tilde is according
to our definition here--
square root of the
magnitude of omega comma k
zero-- this could also have
been written as minus omega of k
comma 0 because we know
it's a negative quantity--
and then times the integral from
0 to z sub s and integral dz.
Now I'm just writing
H as a function of z.
Earlier we had written H--
it's no longer on the screen,
I guess-- earlier we had
written H in terms of F of z,
oh, excuse me, F of x.
x is related to z
simply by this formula.
So since the
integral was written
with z as the variable
of integration,
I'm going to rewrite the
integrand in terms of z,
but it really is just
our old friend F of x.
So it would be the square
root of omega sub m zero 1
plus z cubed plus omega sub
radiation zero times 1 plus z
to the fourth, all
inside the square root,
here, plus omega sub vac
zero plus omega sub k
zero times 1 plus z squared.
And this, then, is the
answer for psi of z.
And then we put that into here
and replace a twiddle by this,
and we get a formula for
what we're looking for,
an expression for the
actual measured intensity
of the source at the Earth in
terms of the parameters chosen
here-- the current
values of omega
and the redshift of the source.
And that's all that goes
into this final formula.
So if you know the
current values of omega
and the redshift
of the source, you
can calculate what you expect
the measure intensity to be
in terms of the
intrinsic intensity.
And that's exactly
what the supernova
people did in 1998 using
exactly this formula--
nothing different--
and discovered that,
in order to fit their data,
they needed a very significant
contribution from this vacuum
energy, namely a contribution
in the order of 60 or 70%.
So we will stop there for today.
We will continue on Thursday
to talk a little bit more
about the physics
of vacuum energy.
