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PROFESSOR: OK.
Good morning, everybody.
Let's get started.
Let me just begin by asking
if there are any questions,
either about logistical issues
or about physics issues?
OK.
Today we'll be
finishing our discussion
of black-body
radiation by talking
about the actual spectrum of
the cosmic microwave background
that we find in our universe.
And then move on to talk about
the rather exciting discovery
in 1998 of the fact that
our universe today appears
to have a nonzero
cosmological constant.
So I want to begin by reviewing
what we did last time.
And one of the reasons
why I do this is I
think it's a good
opportunity for you
to ask questions that don't
occur to you the first time we
go through.
And that, from my
point of view, has
been an extraordinary success.
I think you've asked
great questions.
So we'll see what
comes up today.
We began the last
lecture by recalling,
I think from the previous
lecture actually,
the basic formulas for
black-body radiation, which
is just the
radiation of massless
particles at a
given temperature.
And we have formulas for the
energy density, the pressure,
the number density, and the
entropy density, all of which
are given in terms of
two constants, little g
and little g star, which
is the only place where
the actual nature of
the matter comes in.
G and g-star are both
equal to 2 for photons,
but these formulas
allow us to talk
about other kinds of
black-body radiation
as well, black-body radiation
of other kinds of particles.
As neutrinos are
also effectively
massless, so they contribute.
And in addition, e
plus e minus pairs,
if the temperature
gets hot enough so
that the mass of the
electrons becomes negligible
compared to kt, also contribute
to the cosmic background
radiation.
And if we want the
higher temperatures,
other particle will
start to contribute.
And at the highest
temperatures all particles
act like black-body radiation.
The general formula
for g and g star
is that there is a factor
out front that depends on
whether the particle is
a boson or a fermion,
a particle which does or does
not obey the Pauli exclusion
principle.
Fermions do not, bosons-- excuse
me, I said that backwards.
Fermions obey the Pauli
exclusion principle,
bosons do not.
G and g star are
both 1 for bosons.
But for fermions there's
a factor of 7/8 for g
and 3/4 for g star.
Yes?
AUDIENCE: Would you
mind quickly restating
why the positron-electron
pairs act
like radiation above
that temperature?
PROFESSOR: OK.
The question is, why do
electron-positron pairs
act like radiation at
these high temperatures?
And the answer is
that radiation is just
characterized by the fact that
the particles are effectively
massless.
And the effective
energy scale is
kt, that's the average thermal
energy in a thermal mix.
So as long as m e c squared
is small compared to kt,
electrons and positrons
think that they're
massless and act like
they're massless.
And as I said, if you go to
higher temperatures still,
all particles will act
like they're massless.
Coming back to the
story of g and g star,
we have the factor out
front which depends on
whether they're
bosons or fermions.
And then that just
multiplies the total number
of particle types,
whereby a particle type--
we made a complete specification
of what kind of a thing it is.
And that includes specifying
what species of particle
it is, whether it's a
particle or an anti-particle
if that distinction
exists, and what the spin
state is if the
particle has spin.
So we can try this out
now on some examples.
First example, will
be neutrinos which
play a very important role
in the early universe,
and even in the particle number
balance of today's universe.
Neutrinos actually
have a small mass,
as we talked about last time and
as I'll review again this time.
But nonetheless, as far
as cosmology is concerned,
they effectively act
like massless particles
although the story
about why they
act like massless particles
is a little complicated.
It's more than just
saying that they're
mass is small, for
reasons that we'll see.
But anyway, I'm
nonetheless going
to start by describing neutrinos
as if they were massless,
as was believed to be the
case really until 15 years
ago or so.
The massless model
of the neutrino
was a particle which
was always left-handed.
And by left-handed
what I mean is
that for neutrinos, if
you took the angular
momentum of the neutrino in
the direction of the momentum,
p hat there means dotted
with the unit vector
in the direction of
the spatial momentum,
you'd always get minus
1/2 in units of h bar.
And conversely, all new
bars are right-handed which
just means the same equation
holds with the opposite sign.
So neutrinos always
have spins that oppose
the direction of motion,
and anti-neutrinos always
have spins aligned with
the direction of motion.
Now, it's not obvious
but, if neutrinos
were massless this would be a
Lorentz invariant statement.
If neutrinos have a
mass, that statement
is obviously not learn
Lorentz invariant.
As you can see by imagining
a neutrino coming by,
and you can get
into a rocket ship,
chase it, and pass it, and
then see it going the other way
out your window because
you're going faster than it.
You would see the momentum
in the opposite direction
from the way it
looked to begin with.
But the spin would look like
it was the same direction as it
did to begin with,
and therefore the spin
would now be aligned
with the momentum
instead of opposite
the momentum.
So this could not
possibly hold universally
if the neutrino has a mass.
But for the time being our
neutrinos are massless.
So we're going to take
this as a given fact.
And it certainly is a fact
for all neutrinos that
have ever been
actually measured.
Given this model of the
neutrino, the g for neutrinos
is 7/8 because they're fermions.
Then there's a factor
of 3, because there
are three different species
of neutrinos- electron
neutrinos, muon
neutrinos, tau neutrinos.
Neutrinos come in particles
or anti-particles which
are distinct from
each other, we think.
So there's a factor of 2
associated with the particle
anti-particle duality.
And there's only one spin state.
The spin that's anti-aligned
with the momentum,
or aligned for the
anti-neutrinos.
But only one spin
state in either case.
So just a factor of
1 from spin states,
and multiplying that
through we get 21/4 for g,
and 9/2 for g star.
Yes?
AUDIENCE: If we found out
that they were Majorana,
that they were their
own anti-particles,
would that change what we expect
the temperature [INAUDIBLE]
to be?
PROFESSOR: No, it would not.
OK.
The question was, if
we find that they're
Majorana particles--
which I'm going
to be talking
about in a minute--
where the particles would be
their own anti-particles, which
would mean that the right-handed
anti-neutrino would really just
be the anti-particle of
the left-handed neutrino,
it would not change these
final numbers at all.
What it would do is, instead
of having the 2 for particle
anti-particle, we would
have a 2 for spin states.
So there would still be
two kinds of neutrinos,
but instead of calling them the
neutrino and the anti-neutrino,
the right words would
be right-handed neutrino
and left-handed neutrino.
But the product would
still be the same.
AUDIENCE: Wait, they have
mass and they are Majorana?
PROFESSOR: If they
have mass and Majorana,
what I just said applies.
The fact that they
have a mass would
mean at the lowest possible
temperatures they would not
act like black-body radiation.
Kt would have to be bigger than
their mass times c squared.
But that's only on the order
of electron volts at most.
So I'll talk later about why
the true model neutrinos which
have masses give the
same result as this.
OK.
Then we can also,
just as an exercise,
calculate g and g star.
It's more than an exercise.
We like to know the results.
We can calculate g and g star
for e plus e minus pairs, which
is relevant for when
kt is large compared
to the rest energy
of an electron.
And again, they
are fermions so we
get a factor of 7/8 appearing
in the expression for g,
and 3/4 appearing in the
expression for g star.
And then we just have
to multiply that times
the total number of types
of electrons that exist.
There's only one species
called an electron,
so we only get a factor
of one in the species
slot of the product.
There are both electrons
and anti-electrons
where the anti-electrons are
usually called positrons.
So we get a factor of 2
in particle anti-particle.
Two spin states because an
electron can be spin up or spin
down, and that
gives us 7/2 and 3.
Given that, we can go
ahead and calculate
what the energy density
and radiation should
be for the present
universe given
the temperature of the
photons, the temperature
of the cosmic
microwave background.
And in doing that there's
an important catch which
is something which is the
subject of a homework problem
that you'll be doing
on problem set seven.
When the electron-positron
pairs disappear
from the thermal
equilibrium mix,
if everything were still
in thermal contact,
its heat would be shared between
the photons and the neutrinos
in a way that would keep
a common temperature.
But in fact, when the e plus
e minus pairs disappear,
things are not in
thermal contact anymore.
And in particular, the
neutrinos have decoupled.
They're effectively not
interacting with anything
anymore.
So the neutrinos keep
their own entropy
and do not absorb any entropy
coming from the e plus
e minus pairs.
So all the entropy of the
e plus e minus pairs gets
transferred only to the photons.
And that heats the photons
relative to the neutrinos
in a calculable
amount, which you
will calculate on
the homework problem.
And the answer is that the
temperature of the neutrinos
ends up being only
4/11 to the 1/3 power,
times the temperature
of the photons.
And that's important
for understanding
what's been happening in the
universe since this time.
That ratio is maintained
forever from that time onward.
So if we want to write down
the formula for energy density
and radiation today it
would have two terms.
The 2 here is the g for the
photons, and this, times
that expression is the
energy density in photons.
The second term is the
energy density in neutrinos.
And it has the
factor of 21/4 which
was the g factor for neutrinos.
But then there's also
a correction factor
for the temperature, because
on the right hand side here
I put t gamma to the fourth.
So this factor corrects it
to make it into t neutrino
to the fourth, which
is what we need there
to give the right energy
density for the neutrinos today.
And this is just that
ratio to the fourth power.
And once you plug in
numbers there it's
7.01 times 10 to the minus 14th
joules per meter cubed, which
is, from the
beginning, what we said
was the energy density in
radiation of the universe
today.
OK.
Finally I'd like to come back
to this real story of neutrinos
and their masses and why, even
though they have small masses,
the answers that we gave for the
massless model of the neutrino
are completely
accurate for cosmology.
We've never actually measured
the mass of a neutrino.
But what we have seen is
that neutrinos of one species
can oscillate into neutrinos
of the other species.
And it turns out,
theoretically, that that
requires them to have a mass.
And by seeing how
fast they oscillate
you can actually
measure the difference
in the mass squareds
between the two species.
So it's still possible
actually, in principle,
that one of the species
could have zero mass.
But they can't
all have zero mass
because we know the differences
in the squares of their masses.
So in particular, delta
m squared 2 1 times c
to the fourth, meaning the
mass expressed as an energy,
is 7.5 times 10 to the
minus 5 eV squared.
And larger values
obtained for 2 3,
which is 2.3 times 10 to
the minus 3 eV squared.
We're still talking
about fractions of an eV.
The other of the three
possible combinations
here are just not known yet.
Now, if neutrinos have a mass,
that does actually change
things rather dramatically
because of what we said about--
the statement that the neutrinos
always align their spins with
their motion just cannot be
true if neutrinos have a mass.
And more generally,
for any particle
with a mass of arbitrary
spin j, the statement
is that, the component of j
along any particular axis--
and we'll call it
the z-axis-- always
takes on the possible
values in terms of h bar
going from minus j up
to j with no emissions.
It's different for
massless particles.
For massless particles
every one of these elements
on the right hand side is
independent and, by itself,
a Lorentz invariant possibility.
But, coming back to
neutrinos-- if the neutrinos
have a mass, in addition to
the left-handed neutrinos there
has to be a
right-handed partner.
And the question then is,
what's the story behind that?
And it turns out we
don't know the story.
But we know two
possible stories.
And one of the
possible stories is
that could be what's
called a Dirac mass.
And for Dirac mass
what it means is that,
the right-handed neutrino is
simply a new type of particle
which just happens to be a
particle that we've never
seen, but a particle
which would have
a perfectly real existence.
It would however, to fit
into theory and observation,
be an extremely weakly
interacting particle.
The interactions of
the right-handed one
do not have to be the
same as the interactions
of the left-handed one.
That is, the interactions
can depend explicitly
on p hat dot j.
So depending its
value, you could
affect what the
interactions are, again,
in a Lorentz invariant way.
And in practice, the
right-handed neutrinos
would indirect so
weakly that we would not
expect to see them
in the laboratory.
And we would not expect
even in the early universe
that they would
have been produced
in any significant number.
So even though it would
be a particle that,
in principle, exists, we
would not expect to see it.
And we would not expected it
to affect the early universe.
Alternatively-- and in some
ways a more subtle idea--
is that the mass of
the neutrino could
be what's called
a Majorana mass.
Where Majorana, like Dirac,
is the name of a person--
perhaps less well
known than Dirac,
but made important contributions
in this context nonetheless.
In this case, it can only
occur if lepton number is not
conserved.
And if lepton number
is not conserved then
there are really
no quantum numbers
that separate the
particle that we
call a neutrino
from the particle
that we call an anti-neutrino.
And if that's the
case, the particle
that we call the anti-neutrino
could, in fact, just
be the right-handed
partner of the neutrino.
So for the Majorana
mass case we don't
need to introduce any new things
that we haven't already seen.
We just have to rename
the thing that we've
been calling the anti-neutrino
the neutrino with helicity
plus 1 instead of minus 1-- it's
with j hat dot p with j dot p
hat, equal to plus 1/2
instead of minus 1/2.
So these would just
be the two spin states
of the neutrino instead of the
neutrino and the anti-neutrino.
And that's a possibility.
And this would also change
nothing as far as the counting
that we did.
It would just change where
the factors go instead
of having a factor
of 2 for particle
anti-particle and
this type counting.
We'd have a factor of 2
in the spin state factor,
and a factor of 1 in the
particle anti-particle
[INAUDIBLE].
The particle and
the anti-particle
would be the same thing.
OK.
Any questions about that?
OK.
Finally-- and I think this is
my last slide of the summary.
At the end of the lecture
we just pointed out
a number of tidbits
of information.
We can calculate the temperature
of the early universe
at any time from the
formulas that we already
had on the slide.
We know how to calculate the
energy density at any time.
And by knowing about
black-body radiation
we can convert that
into a temperature.
And for an important
interval of time,
which is when kt is small enough
so that you don't make muon
anti-muon pairs, but
large enough so that
electron-positron pairs
act like they are massless,
and this very large range
kt is equal to 0.860 m e v,
divided by the square root of
time where time is measured
in seconds.
So in particular, at 1
second kt is 0.86 m e
v. And it does
apply at 1 second.
Because 0.86 m e v
is in this range.
We also then talked
about the implications
of the conservation of entropy.
If total entropy is
conserved, the entropy density
has to just fall off like 1
over the cube of the volume.
Total entropy is conserved
for almost all processes
in the early universe.
So the entropy falls
off like 1 over a cubed.
And that means that,
as long as we're
talking about a period of time
during which little g does not
change-- and little g only
changes when particles freeze
out, like when the
electron-positron pairs
disappear-- but as long as
little g doesn't change,
s [INAUDIBLE] 1 over
a cubed means simply
that the temperature
falls like 1 over a.
And when little g changes you
can even calculate corrections
to this as, effectively
you're doing
when you calculate
this relationship
between the neutrino temperature
and the photon temperature.
And finally, we talked
about the behavior
of the atoms in the universe
as the universe cools.
For temperatures above
about 4,000 degrees
the universe, which
is mainly hydrogen,
is mainly a hydrogen plasma.
Isolated protons and electrons
zipping through space
independently.
At about 4,000 Kelvin-- and
this is a stat [? mac ?]
calculation, which we're not
doing-- but using the answer.
At about 4,000 Kelvin--
which is a number which
depends on the density of
hydrogen in the universe,
it's not a universal
property of hydrogen--
but for the density of
hydrogen in the universe,
at about 4,000 Kelvin
hydrogen recombines.
It becomes neutral atoms.
And slightly colder, at about
3,000, the degree of ionization
becomes small enough
so that the photons
become effectively free.
The photons decouple.
In between 4,000 and 3,000 the
hydrogen is mostly neutral,
but they're still enough ionized
so that the photons are still
interacting.
So the most important
temperatures--
the 3,000 Kelvin,
when the photons
are released, when the
photons are no longer trapped
with the matter of the universe.
And last time we estimated the
time at which that happens.
That should be a small t, sorry.
The time of decoupling
is about 380,000 years.
And that number is
actually very accurate,
even though we didn't
calculate it very accurately.
And that's the
end of my summary.
Any questions about the summary?
OK.
In that case, let's
go on to talk first
about the spectrum of the
cosmic background radiation.
And then we'll move
on to talk about
the cosmological constant.
CMB is cosmic
microwave background.
And that's a very, very standard
abbreviation these days.
So when the cosmic microwave
background was first
discovered by Penzias and
Wilson in 1965-- which,
I might point out,
is going to have
its 50th anniversary
in the coming year--
they only measured
it at one frequency.
It was a real tour de force to
measure it at the one frequency
and to convince themselves that
the buzz that they were hearing
in their detector was
not just some kind
of random electrical noise,
but really was some signal
coming from outer space.
And the main clue that
it was some signal coming
from outer space
was that they were
able to compare it
with a cold load,
a liquid helium-cooled
source, and find
that that comparison worked
the way they expected.
And the main reason
for believing
it was cosmological
rather than local
is that they got the
same reading no matter
what direction they
pointed their antenna.
This just took a lot of
radio technique skill
to convince themselves
that it wasn't just
some radio tube that was
malfunctioning or something.
They even worried
that it may have
been caused by pigeon
droppings in their antenna,
I actually read about
in Weinberg's book.
But they finally convinced
themselves that it was real.
They were still not
convinced really
that it was a sign
for the big bang
and-- you may recall,
again, from reading Weinberg
that there were two papers
published back-to-back.
The experimental paper
by Penzias and Wilson,
which really just
described the experiment,
mentioning that a
possible explanation
was in this other paper
by Dickie, Peebles, Roll,
and Wilkinson which described
the theory that this
was radiation that
originated with the big bang.
But it's all based on one
point at one frequency.
Shortly afterwards, I
guess within the same year,
Roll and Wilkinson
were able to measure it
at a slightly
different frequency.
And when I wrote my
popular-level book
I tabulated all of the data
that was known in 1975.
And this mess is the graph.
This shows sort
of the full range
of interesting frequencies.
The solid line here is the
expected theoretical curve
corresponding to a
modern measurement
of the temperature
2.726 degrees Kelvin.
All of the interesting
historical points
are in this tiny little
corner on the left, which
is magnified above.
The original Penzias
and Wilson point
is way down here at
very low frequencies
by the standards of radiation
at 2.726 degrees Kelvin.
The Roll and Wilkinson
point is there.
These blobs indicate error bars.
The [? cyanogen ?] points
that you read about
in Weinberg are shown
there and there.
The first measurement that
showed that, it didn't only
go up but started to go down
like black-body radiation
should, was a balloon flight--
this 1971 balloon flight
which produce that
blob and that bound.
This was an experiment
by MIT's own Ray Weiss.
And it was very
important in the history
because it was the first
evidence that we weren't just
seeing some straight
line, but we
were seeing something which
did indeed rise and fall
the way black-body
radiation should.
A later balloon flight in
1974 produced error bars
that are shown by
this gray area.
Incredibly broad.
So the bottom line
that this graph
was intended by me to
illustrate is that, in 1975 you
could believe that this was
black-body radiation if you so
wished.
But there was not
really a lot of evidence
that it was
black-body radiation.
The situation did not
get better quickly.
The next significant
measurement came
in 1987 which was a
rocket flight, which
was a collaboration
between a group at Berkeley
and a group at Nagoya, Japan.
I believe it was the
Japanese group that
supplied the rocket and
the American group that
supplied the instrumentation.
And they measured the
radiation at three points.
I can give you the
number that goes
with those graphical points.
I guess what I
have tabulated here
is the effect of temperature
that those points correspond
to.
As you can see from
the graph, those points
are all well above
the black-body curve.
Significantly more
radiation than what
was expected by
people who thought
it should be black-body.
And 0.2 up there
would correspond
to a temperature of 2.955 plus
or minus 0.017 K. The size
of the vertical bars
there are the error
bars that the
experimenters found.
And 0.3 was t equals 3.175
plus or minus 0.027 K.
So these were higher
temperatures then
the 2.7 that fit the lower
part of the spectrum.
And very, very small error bars.
So this data came out in 1987.
And, in truth, nobody knew
what to make out of it.
The experimental group were
well aware that this was not
what people wanted them to find.
And they certainly
examined their data
very carefully to figure at
what could have conceivably gone
wrong.
And they were going
around the country
giving talks about this.
And I heard one or
two of them in which
they described how surprised
they were by the results,
but emphasized that they
analyzed the experiment very,
very carefully and couldn't
find anything wrong with it.
And this was the
situation for awhile.
I should point out that I think
this point number three is
something like 16 standard
deviations off of the theory.
And usually when somebody
makes a measurement that's
three or four standard
deviations off of your theory,
you really start to worry.
16 standard deviations is
certainly a bit extreme.
Nonetheless, nobody had any
good explanation for this.
So, well, different people
had different attitudes.
There were some people who
tried to construct theories
that would account for this.
And there were others who
waited for it to go away.
I'm pretty sure
I was among those
who waited for it to go
away, and we were right.
So the next important
piece of data
came from the first
satellite dedicated
to measuring the cosmic
background radiation.
The famous COBE Satellite--
Cosmic Background Explorer-- I
guess I didn't write
down the name here.
Oh, it's in the title.
Preliminary measurement of the
cosmic microwave background
spectrum by the Cosmic
Background Explorer,
COBE Satellite.
So COBE was the first
satellite dedicated
to measuring the cosmic
background radiation.
It was launched in 1989,
I guess, and released
its first data in
January of 1990.
Back in those days there
was no internet or archive.
So you may or may not know that
the way physics results were
first announced
to the world were
in the form of what were
called pre-prints, which
were essentially xeroxed copies
of the paper that were sent out
to a mailing list.
Typically, I think,
institutions had mailing lists
of maybe 100 other institutions.
And every physics department
had a pre-print library
that people can go to and
find these pre-prints.
So this is the COBE pre-print.
90-01, the first
pre-print from 1990.
And this is the data.
So it is kind of
breathtaking, I think.
It suddenly changed the entire
field, and in some sense
really change cosmology
for the field.
Where we only had
approximate ideas of the way
things worked, to suddenly
having a really precise science
in which precise
measurements could be made,
and cleared up the
issue of the radiation.
It wasn't just a mess like this,
or a terrible fit like that,
but a fantastically good fit.
Really nailing the radiation
as having a thermal spectrum.
So the history is that John
Mather presented this data
at the January 1990 American
Physical Society meeting,
and was given a
standing ovation.
And he later won the
Nobel Prize for this work.
He was the head of the team
that brought this data.
He won the Nobel Prize in
2006 along with George Smoot,
who was responsible for one
of the other experiments
on the COBE satellite.
Yes?
AUDIENCE: So do we
know what happened
with the other measurements?
PROFESSOR: To tell
you the truth,
I don't think the
other measurements
ever-- the other people
ever really published
what they think happened.
But the widespread rumor,
which I imagine is true,
is that they were seeing
their own rocket exhaust.
And there were, I think,
some arguments going on
between the Americans
and the Japanese,
with the Americans more or less
accusing the Japanese of not
really telling them how
the rocket was set up.
Yes?
AUDIENCE: Are the error bars
plugged on those points,
or is it just that good?
PROFESSOR: Those
are the error bars?
AUDIENCE: OK.
PROFESSOR: And even
more spectacular,
a couple years later,
I guess it was-- this
was actually just based on nine
minutes of data or something
like that.
But a couple years later they
published their full data set,
where the size of the error bars
were reduced by a factor of 10.
And still a perfect fit.
They didn't even
know how to plot it,
so I think they
plotted the same graph,
and said the error
bars are a factor of 10
smalled than what's shown.
It was gorgeous.
So I think I forgot to tell you
what the spectrum is supposed
to look like exactly.
And this is just
a formula that I
want you to understand
the meaning of, but not
the derivation of.
We-- as with the other stat mech
results that we're relying on,
we're going to relegate their
derivation to the stat mech
course that you either
have taken or will take.
But the spectrum is
completely determined
because the principle
of thermal equilibrium
is sort of absolute in
statistical mechanics.
And in order for a
black-body radiating
object to be in
thermal equilibrium
with an environment
at that temperature,
it has to have not only the
right emission rate but also
the right spectrum.
If the spectrum
weren't right you
could imagine putting
in filters that
would trap in some frequencies
and let out others.
And then you would move away
from thermal equilibrium
if the spectrum were right
or wrong because you'd
be trapping in more
radiation-- you could arrange
for the filters to trap in more
radiation than they are letting
out.
So the spectrum is calculable.
And in terms of-- I guess
this is energy density.
I have to admit, I usually
call energy density u
and in these notes
here it's called rho.
We'll figure out the units after
I write it down and make sure
that it is energy density.
Rho sub nu of nu d nu,
means-- with this product
it means the total energy
density, energy per volume,
per frequency interval, d
nu-- well, it's times d nu,
so if you multiply
by times nu, this
is the total energy for
frequencies between nu and nu
plus d nu.
And the formula is 16 pi
squared h bar nu cubed, divided
by c cubed times 1 over e to
the 2 pi h bar nu over kt,
minus 1 d nu.
OK.
And actually, the unit's
not that transparent.
I believe this is energy
density and not mass density.
But maybe I'll make sure of
that and let you know next time.
And this is what
produces that curve
that you saw on the slides.
I've included the
subscript nu here
to indicate that it's the number
which, when you multiply it
by d nu, gives you the energy
density between nu and nu
plus d nu.
If instead you wanted
to know the energy
density between lambda
and lambda plus d lambda,
there'd be a kinematic
factor that you'd
have to put in here-- the factor
that relates d lambda to d nu.
And you could imagine
working that out.
I might add that,
in Weinberg's book,
he actually plots
both sub lambda of nu.
So his curve looks somewhat
different than the curves
that I showed you.
This is not exactly
the same thing.
Now, what this extremely
accurately black-body curve
proves is that the
early universe really
was very accurately in
thermal equilibrium.
And that can only happen if the
early universe was very dense.
And of course, our
model of the universe
goes back to infinite density.
So the model predicts
that it should
be in thermal equilibrium.
But in particular, the
numbers that we have here,
if you ask how much could
you change the model
and still expect these
curves the answer
is roughly that, all of the
important energy-releasing
processes have to have
happened before about one
year after the Big Bang.
Anything that happened
after one year
would still show
up as some glitch
in the black-body spectrum.
So the big bag model really
is confirmed back to about one
year on the basis of
this precise measurement
of the spectrum of the
cosmic background radiation.
And the COBE measurement
is still, by the way,
the best measurement
of the spectrum.
We've had other very
important experiments,
that we'll talk
about later, which
measure the non-uniformity
of the black-body radiation.
Which is very small,
but nonetheless very,
very important [? effect. ?]
So we've had WMAP
and now Planck which
have been dedicated to
measuring the anisotropies
of the radiation.
COBE also made
initial measurements
of the anisotropies.
And we'll be talking
about anisotropies later
in the course.
Yes?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Sorry?
AUDIENCE: The units
of the right-hand side
are energy density.
PROFESSOR: Energy density.
OK.
Thanks.
OK.
Good.
So my words were right.
I should have called
it u, I think,
to be consistent with
my usual notation.
Thanks.
OK.
Any other questions
about the CMB?
Because if not, we're going
to change gears completely
and start talking about one of
the other crucially important
observational discoveries in
cosmology in the last 20 years.
OK.
So what I want to
talk about next
is the very important
discovery originally made
in 1998-- also resulting
in a Nobel Prize--
that the universe
is accelerating.
And this was a
discovery that involved
two experimental groups,
and a total of something
like 52 astronomers
between the two groups.
Which actually meant that--
I'm exaggerating slightly,
I suppose.
But it really
involved the majority
of the astronomers of the
world, and therefore there
weren't a lot of astronomers
to argue with them about
whether or not the
result was right.
But there still
was some argument.
The announcement was initially
made at a AAS meeting
in January of 1998 by--
which group was first?
I think that was the High-Z
Supernova-- where are they?
Yeah.
That was the High-Z
Supernova Search Team.
And then there was also a group
largely based at Berkeley.
The High-Z Supernova Search Team
was actually fairly diffused,
although based to some
extent at Harvard.
And the Supernova
Cosmology Project
was based rather
squarely in Berkeley,
headed by Saul Perlmutter.
And they both agreed.
And what they found
was, by looking
at distant supernovae of a
particular type-- type 1a--
they were able to use these
supernovae as standard candles.
And because supernovae
are brighter
than other standard candles
that had been studied earlier
in history, they were able to go
out to much greater distances.
And that means to look
much further back in time
than previous studies.
And what they discovered
was that the expansion rate
of the universe
today was actually
faster, and not slower,
than the expansion rate
about five billion years ago.
And that was a big shock because
until then everybody expected
that gravity would
be slowing down
the expansion of the universe.
And when these guys started
to make these measurements
they were just simply
trying to figure out
how fast the universe
was slowing down.
And they were shocked to
discover that it was not
slowing down, but
instead speeding up.
Initially there was some
controversy about it.
People did try to invent other
explanations for this data.
But the data has, in fact, held
up for the period from 1998
to the present.
And in fact, it has been
strongly supported by evidence
from these anisotropies in the
cosmic microwave background
radiation, which we'll
be talking about later.
But it turns out, you can
get a lot of information
from these anisotropies in the
cosmic background radiation.
So the picture now is
really quite secure,
that the acceleration-- the
expansion of the universe
is actually accelerating,
and not decelerating.
And the simplest
explanation for that,
which is the one that--
well, certainly because it's
the most plausible, and the one
that most of us take seriously,
and it's the only one that fits
the data extraordinarily well.
So we've not seen any reason
not to use this explanation.
The simplest explanation is
that there's a nonzero energy
density to the
vacuum, which is also
what Einstein called the
cosmological constant.
So we should begin
by writing down
the equations that
describe this issue.
So we've learned how to write
down the second order Friedmann
equation, which
describes how the scale
factor of the
universe accelerates.
And on the right-hand side,
once we included materials
with nonzero
pressures, we discover
that we need on the
right-hand side, rho plus 3 p,
over c squared--
excuse me-- times a.
Now when the cosmological
constant was born,
was when Einstein
first turned his theory
of general relativity
to cosmology.
Einstein invented the theory
of general relativity in 1916.
And just one year
later, in 1917,
he was applying it to
the universe as a whole
to see if he could get
a cosmological model
consistent with
general relativity.
Einstein at that point was
under the misconception
that the universe was static,
as Newton had also thought,
and as far as we
know, as everybody
between Newton and
Einstein thought.
If you look up at the stars, the
universe looks pretty static.
And people took
this very seriously.
In hindsight, it's a little
hard to know why they took it
so seriously, but they did.
So when Einstein
discovered this equation
he was assuming that
the universe consisted
of basically
non-relativistic stuff.
Stars are essentially
non-relativistic hunks
of matter.
So he thought that
rho would be positive,
the effective pressure
would be zero.
And he immediately
noticed that this equation
would imply that the
scale factor would
have a negative acceleration.
So that if you tried to
set up a static universe
it would instantly collapse.
And as we talked
about earlier, Newton
had talked himself out
of that conclusion.
And I think the real
difference, as I think we also
talked about earlier, was
that Newton was thinking
of the law of gravity as
an action at a distance,
where you determine the
total force on something
by integrating the forces
caused by all other masses.
And then things get complicated
and divergent, actually,
for an infinite,
static universe.
And Newton managed
to convince himself
that you could have a static
universe of that type,
a statement that we now
consider to be incorrect
even in the context of
Newtonian mechanics.
But this fact that
it's incorrect even
in the context of
Newtonian mechanics
was really not discovered
until Einstein wrote down
this equation.
And then Einstein himself
also gave a Newtonian argument
showing that, at least with
a modern interpretation
of Newtonian mechanics.
It doesn't work in
Newtonian gravity
either to have a
static universe.
But Einstein was still convinced
that the universe was static.
And he realized that he could
modify his field equations--
the equations that we
have not written down
in this course, the equations
that describe how matter create
gravitational fields--
by adding a new term
with a new coefficient in front
of it which he called lambda.
And this extra
term, lambda, could
produce a kind of a universal
gravitational repulsion.
And he realised he had to adjust
the constant to be just right
to balance the amount of
matter in the universe.
But he didn't let
that bother him.
And if he adjusted
it to be just right,
and the universe was
perfectly homogeneous,
he could arrange
for it to balance
the standard force of gravity.
We can understand what
lambda does to the equations
because it does, in fact,
have a simple description
in terms of things that we have
discussed and do understand.
That is, you could
think of lambda
as simply corresponding to
a vacuum energy density.
Einstein did not
make that connection.
And not being an
historian of science,
I can speculate
as much as I want.
So my speculation is that,
the reason this did not
occur to Einstein
is that Einstein
was a fully classical
physicist who was not
at this time or
maybe never accepting
the notions of quantum theory.
And in any case,
quantum field theory
was still far in the future.
So in classical physics the
vacuum is just plain empty.
And if the vacuum
is just plain empty
it shouldn't have
any energy density.
The quantum field theory
picture of the vacuum,
however, is vastly more complex.
So to a modern quantum
field theory-oriented
theoretical physicist
the vacuum has
particle, anti-particle pairs
appearing and disappearing
all the time.
We are now convinced
that there's also
this Higgs field that has
even a nonzero mean value
in the vacuum.
So the vacuum is a
very complicated state
which, if anything
characterizes it,
it's simply the state of
lowest possible energy density.
But because of, basically,
the uncertainty principles
of quantum mechanics, the
lowest possible energy density
does not mean that all
the fields are just zero
and stay zero.
They're constantly
fluctuating as they must
according to the
uncertainty principle, which
applies to fields
as well particles.
So we have no reason
anymore to expect the energy
density of the
vacuum to be zero.
So from a modern perspective
it's very natural
to simply equate the idea
of the cosmological constant
to the idea of a nonzero
vacuum energy density.
And there are some
unit differences-- just
constants related to
the historical way
that Einstein added
this term his equations.
So the energy density of
the vacuum-- which is also
the mass density of
the vacuum times c
squared-- is equal to Einstein's
lambda times c to the fourth,
over 8 pi G.
And this is really just
an historical accident
that it's defined this way.
But this is the way
Einstein defined lambda.
Now, if the vacuum
has an energy density,
as the universe
expands the space
is still filled with vacuum.
At least, if it was
filled with vacuum.
If it was matter
it would thin out.
But we can imagine a region
of space that was just vacuum,
and as it expands it would
have to just stay vacuum.
What else could it become?
And that means that we
know that, for a vacuum,
rho dot should equal zero.
Now we've also learned earlier,
by applying conservation
of energy to the
expanding universe,
that rho dot in an expanding
universe, is equal to minus 3
a dot over a.
Or we could write
this as h times
rho plus p over c squared.
This is basically
a rewriting of d u
equals minus p d v, applying
it to the expanding universe.
So I won't re-derive it.
We already derived it.
Actually, I think you
derived it on the homework,
was the way it actually worked.
But in any case,
this immediately
tells us that if
rho dot is going
to be 0 for vacuum
energy, this has to be 0.
And therefore p vacuum has
to be equal to minus rho
vacuum times c squared.
And if we know the energy
density in the pressure
of this stuff called
vacuum, that's
all we need to know to put it
into the Friedmann equations
and find out how things behave.
Otherwise this vacuum energy
behaves no differently
from anything else.
It just has a
particular relationship
between the pressure
and the energy density,
with a very peculiar feature-
that the pressure is negative.
And that's an important
feature because we
had commented earlier that
a negative pressure can
drive acceleration.
And now we're in a good position
to see exactly how that works.
To sort of keep
things straight I'm
going to divide the mass
density of the universe
into a vacuum piece
and a normal piece,
where normal represents matter,
or radiation, or anything else,
if we ever discover
something else.
But in fact it will just
be matter or radiation
for anything that we'll
be doing in this course,
or anything that's really
done in current cosmology.
And similarly, I'm going to
write pressure as p vac plus p
normal.
"N" is for normal.
But p vac I don't
really need to use,
because p vac I can rewrite
in terms of rho vac.
So in the end I can
express everything just
in terms of rho vac.
And I can write down the second
order Friedmann equation.
And it's just a matter of
substituting in that rho
and that p into the
Friedmann equation
that we've already written.
And we get minus 4 pi
over 3 G, times rho
normal plus 3 p
normal, over c squared.
And the vacuum pieces-- have two
pieces because there's a vacuum
piece there and a
vacuum piece there. it
can all be expressed in terms
of rho vac and collected.
And what you get is
minus 2 rho vac times a.
Showing just what we
were talking about.
That because of that minus
sign, multiplies that
minus sign, vacuum energy drives
acceleration, not deceleration.
And that's why vacuum
energy can explain
these famous results of 1998.
And we'll see later that,
for the same reason vacuum
energy or things
like vacuum energy
can actually drive the
expansion of the universe
in the first place in
what we call inflation.
Yes?
AUDIENCE: So for the equation
without the cosmological
constant it's, let's say, rho
and p are about the constant,
then wouldn't that
be the equation
for a simple harmonic
function [INAUDIBLE]
or the oscillation
of a [INAUDIBLE]
is some negative
constant times a?
PROFESSOR: That's right, except
that you would probably not
believe the equations
with the bounds.
AUDIENCE: OK.
PROFESSOR: And when a went
negative you wouldn't really
have a cosmological
interpretation anymore,
I don't think.
But it is, in fact, true that
if rho and p were constants--
I'm not sure of any model that
actually does that-- this would
give you sinusoidal behavior
during the expanding
and contracting phase.
Yes?
AUDIENCE: [INAUDIBLE] the vacuum
energy is constant over time,
is it also makes
sense [INAUDIBLE]?
AUDIENCE: Are you asking,
does it make sense
for maybe the vacuum
energy to change with time?
I think, if it
changed with time,
you wouldn't call
it vacuum energy.
Because the vacuum
is more or less
defined as the lowest
possible energy
state allowed by
the laws of physics.
And the laws of physics,
as far as we know,
do not change with time.
It's certainly true that, in a
completely different context,
you might imagine the laws of
physics might change with time.
And then thing would
get more complicated.
But that would really take you
somewhat outside the sphere
of physics as we know it.
You could always explore
things like that,
and it may turn out to be right.
But at least within
the context of physics
as we currently envision
it, vacuum energies
are constant, pretty
much by definition.
Now I should maybe qualify
that within the context of what
we understand, there may,
in fact, be multiple vacua.
For example, if you
have a field theory
one can have a potential
energy function
for one or more fields.
And that potential
energy function
could have more than
one local minimum.
And then any one of those local
minima is effectively a vacuum.
And that could very
likely be the situation
that describes the real world.
And then you could
tunnel from one vacuum
to another, changing
the vacuum energy.
But that would not be
a smooth evolution.
That would be a
sudden tunneling.
OK.
So this is what happens to
the second order Friedmann
equation.
It is also very useful to look
at the first order Friedmann
equation, which is a dot over
a squared, 8 pi over 3 G.
And in its native
way of being written
we would just have
8 pi over 3 G rho,
minus k over-- kc
squared over a squared.
And all I want to do now
is replace rho by rho
vac plus rho n.
And this is a first
order Friedmann equation.
And we can expand rho n
if we want more details,
as rho matter plus
rho radiation.
And rho matter, we know,
varies with time proportional
to 1 over a cubed.
Rho radiation behaves with
time as 1 over a to the fourth.
So all of the terms
here, except for rho vac,
fall off as a grows.
And that implies that
if you're not somehow
turned around firsts,
which you can be--
you could have a
closed universe that
collapses before vacuum
energy can take over.
But as the universe gets larger,
if it doesn't turn around,
eventually rho vac will win.
It will become larger
than anything else
because everything else is just
getting smaller and smaller.
And once that starts to
happen everything else
will get smaller and
smaller, faster and faster,
because a will start
to grow exponentially.
If rho vac dominates--
which it will, as I said,
unless the universe
re-collapses first--
so for a large class
of solutions rho vac
will dominate-- then you
can solve that equation.
And you have h, which is
a dot over a, approaches,
as a goes to infinity, the
square root of 8 pi over 3 G
rho vac.
So h will approach a fixed
value for a universe which
is ultimately dominated
by rho vacuum.
And if a dot over
a is a constant,
that means that a
grows exponentially.
So we could maybe give
this a name-- h vac.
The value h has
when it's completely
dominated by the vacuum energy.
And then we can
write that a of t
is ultimately going to be
proportional to e to the h vac
times t.
Which is what you get when
you solve the equation,
a dot over a equals
this constant.
OK.
Now one thing which you
can see very quickly--
let's see how far I
should plan to get today.
OK.
I'll probably make one
qualitative argument
and then start a calculation
that won't get very far.
I will continue next time.
One qualitative
point which you can
see from just glancing
at these equations
is that the cosmological
constant, when
added to the other
ingredients that we've already
put into our model
universe, will
have the effect of
increasing the age
of the universe for
a given value of h.
And that's something that we
said earlier in the course,
we're looking forward to.
Because the model of
the universe that we're
been constructing
so far have always
turned out to be too young
for the measured value of h.
That is, the oldest
stars look like they're
older than the universe.
And that's not good.
So we'd like to make
universe look older.
And one of the beauties of
having this vacuum energy,
as far as making
things fit together,
is that it does make
the universe older.
And the easiest way to
see that-- at least a way
to see that-- is to imagine
drawing a graph of h versus t.
Hubble expansion rate versus t.
And if we look at
the formula for h
here we see that the
rho vac piece just
puts in a floor as
h evolves with time,
instead of going to 0 as it
wood in most models-- at least,
as it would in
open-universe model.
It stops at some floor.
And certainly for the models
that we've been dealing with,
h just decreases to
some-- this is supposed
to represent the present time.
So this is previous models.
Now as you might say, that what
I'm trying to describe here
is not quite a theorem if you
considered closed universes
where this k piece
could be causing
a positive-- a negative
contribution to h, which
is then decreasing with time.
Things can get complicated.
But for the models that
we've been considering which
are nearly flat, that
k piece is absent.
And then we just have pieces
that go like, 1 over a cubed,
1 over a to the
fourth, and constant.
All of which are positive.
Then in the absence of vacuum
energy we would have h falling.
And with the presence
of vacuum energy
it would not fall
as fast because we
have this constant piece
that would not be decreasing.
So this is previous models.
This is with rho vac.
And I'm always talking
about positive rho vac
because that is what
our universe has.
So this would be the
two different behaviors
of h for the model
without vacuum energy
and the model with
vacuum energy.
And if we're trying to calculate
the age of the universe
we would basically be
extrapolating this curve back
to the point where h was
infinite, at the big bang.
And we could see that,
since this curve is always
below this curve,
it will take longer
before it turns up
and becomes infinite.
So the age will always increase
by adding vacuum energy.
With rho vac h equals infinity
is further to the left.
And notice that I'm comparing
two different theories,
both of which are
the same age today.
Because that's what
we're interested in.
We've measured the value of h.
We're trying to infer
the age of the universe.
OK.
Maybe I'll just say a couple
words about the calculation
that we'll be starting
with next time.
We want to be able to
precisely calculate
things like the age
of the universe,
including the effect
of this vacuum energy.
And we'll be able to do that
in a very straightforward way
by using this first
order Friedmann equation.
We know how each term in
this Friedmann equation
varies with a.
And we can measure
the amount of matter,
and the amount of radiation,
and in principal the amount
of curvature-- it's
negligibly small--
in our current universe.
And once you have
those parameters
you can use that
equation to extrapolate,
to know what h was at
any time in the past.
And that tells you
how the derivative--
it tells you the value of a
dot at any time in the past.
And if you know the value of
a dot at any time in the past,
it's a principle just
a matter of integration
to figure out when a was 0.
And that's the calculation that
we'll begin by doing next time.
And we'll be able to get
an integral expression
for the age of the universe
for an arbitrary value.
We'll, at the end,
express the matter density
and the radiation
density as fractions
of omega, fractions of
the critical density.
And for any value
of omega matter,
omega radiation, and we'll
even express the curvature
as an omega curvature.
The effective fraction
of the critical density
that this term represents.
And in terms of those
different omegas,
we'll be able to
write down an integral
for the total age
of the universe.
And that really is going
to be state of the art.
That is what the
Planck team uses
when they're analyzing their
data to try to understand what
the age of universe is
according to the measurements
that they're making.
So we will finally come
up to the present as far
as the actual understanding
of cosmology by the experts.
So that's all for today.
see you all next Tuesday.
