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PROFESSOR: OK, in that case, let's get started.
As usual, I like to begin by giving
a review of what we talked about last time.
This time on slides instead of on the blackboard.
We're talking mainly about relativistic energy,
or relativistic energy and momentum, and pressure,
sometimes.
The key equation is probably the most famous equation
in physics, Einstein's e equals m c squared.
And I gave some numerical examples.
I actually looked up some more numbers since then,
when I was revising the lecture notes.
So these are slightly more up to date.
But it's still true that if you could burn matter
at the rate of one kilogram per hour,
you would have about one and a half times
the total power output of the world.
And that's apparently still valid in 2011.
I only had 2008 figures, actually, at the lecture
last time.
And if you imagine the 15-gallon tank of gasoline,
and you could figure out how much that-- what it's mass is,
and convert that to energy, it turns out
that a 15-gallon tank of gasoline
could power the world for about two and half days, if you could
convert all of it into energy.
The catch, of course, is that we can't convert matter
into energy.
We can't get around the problem that, at least at the energies
that we deal with, baryon number.
And that number of protons and neutrons is conserved,
so we can't make protons and neutrons disappear.
And that means that we're limited in what we can do.
In particular, one of the most efficient things we can do
is fission uranium 235.
But when uranium 235 undergoes fission, less than 1/10
of 1% of the mass is actually converted
into energy, which is why we can't actually avail ourselves
of these fantastic numbers that would apply,
if we could literally just convert matter into energy.
We went on to talk about the relativistic definitions
of energy and momentum, and how they come together
to form a Lorentz four vector, and the underlying theme here
is that we consider ourselves users of special relativity.
Most of you I know have taken special relativity courses,
and for those of you who have, this is a review.
For those you who have not, and there are some of those
also, no need to panic.
I intend to tell you every fact that you
need to know about special relativity.
I won't tell you how to derive them all,
but I'll tell you all you'll need to know for this class.
So in particular, it's useful for this class
to recognize that energy and momentum can be put together
into a four vector, where the zeroth component
is the energy divided by the speed of light.
And the three spatial components are just the three components
of the spatial momentum, although they
have to be defined relativistically.
The relativistic definition of momentum,
at least how it relates to velocity,
is that it's equal to gamma times the rest
mass times the velocity, where gamma is the famous factor
that we've been seeing all along when
we've talked about relativity.
The Lorentz contraction factor, one
over the square root of 1 minus v squared over c squared.
The energy of a particle, relativistically,
is the same gamma, times m 0 times c squared, and it
can also be written as the square root of m 0
c squared squared, plus the momentum squared,
times c squared.
Since the momentum forms a four vector,
its Lorentz and variant square should be Lorentz and variant,
and that means that the momentum squared
minus p 0 squared should be the same in all inertial reference
frames.
And that's just equal when you put
in what p 0 means, the momentum squared
minus the energy squared, divided by c 0 squared.
And to know what value it's equal to in all frames,
it's efficient to know what it's equal to in one frame.
And the one frame where we do know
what it's equal to is the rest frame of a particle.
So in the rest frame, the momentum vanishes,
and the energy is just m 0 c squared.
So in the rest frame, we can evaluate this,
and we get minus m 0 c squared squared.
And that means that has to be the value in every frame.
And this in fact is the easy way to derive
the relationship between energy and momentum.
If we go back, the equation we had
relating energy and momentum is really exactly that equation,
rearranged.
Just to give an example of how this works, when we actually
have energy exchanges, I pointed out
that we could talk about the energy of a hydrogen atom.
And because energy and mass are equivalent,
the hydrogen atom clearly has a little bit less energy
than an isolated proton, plus an isolated electron.
Because when you bring them together
there's a binding energy, and that binding energy
is called delta e, and has a value of 13.6 electron
volts for the ground state of hydrogen.
So that tells us the mass of an hydrogen atom
is not the sum of the two masses,
but rather has this correction factor,
because we've taken out a little bit of energy for the binding.
And that means we've taken out a little bit of mass.
OK, then we talked about the mass density of radiation
and how-- building up to how that will affect the universe.
And we said that the mass density of radiation
is just the energy density divided by c squared.
And that can be taken, really, as a definition of what
we call relativistic mass, and hence
relativistic mass density.
But the important point is that this mass density actually
does gravitate the same as any other mass
density of the same value.
It really does create gravity in the same way.
Now I mentioned that things are much more complicated if you
want to talk about the gravitational field produced
by a single moving particle.
That's asymmetric, the velocity of the particle
shows up in the equations that describe the metric surrounding
a single moving particle.
But if you have a gas of particles moving
at high velocities, where the velocities nonetheless
average to zero, which tends to happen, in a gas
at least in the rest frame of the gas.
Then that gas will produce gravitational fields,
just like a static mass density.
Where the mass density is this relativistic energy divided
by c squared, the relativistic definition of the mass density.
It's also useful to know that the photon-- if we want
to describe it as a particle, is a particle of zero rest mass.
Which means that it can never be at rest,
it always moves at the speed of light.
And it also means that its energy can be arbitrarily
small, because the energy is proportional of momentum,
and the momentum of a photon can be as small as you like.
For giving frequency, of course, the energy of a photon
is fixed.
It's h times nu but if you're allowed
to vary the frequency, which you can do if you just
look at it in different frames, you
can make the energy as small as you like.
And the famous equation then, p squared minus e
squared c squared, which would have on the right,
minus m 0 squared c to the 4th m 0
squared c squared, excuse me-- has zero on the right hand
side, because m 0 is 0 And that means
that for photons, the energy is just the speed of light times
the momentum.
And that's a famous relationship that photons obey.
Now, thinking about how this gas of photons
will behave in the universe, we realized immediately
that it does not behave the same way
as a mass density of ordinary non-relativistic particles.
Which is what we have been dealing with to date.
The important difference is that in both cases,
the number density falls off like 1 over a cubed,
as the universe expands, these particles are not
created and destroyed in significant numbers,
they just persevere.
So the number density of either non-relativistic particles,
or photons, just falls off like one over the volume,
as the volume increases and the number density dilutes.
But what makes photons different from non-relativistic
particles, is that a non-relativistic particle
will maintain the energy of that particle
as the universe expands, but photons
will redshift as the universe expands.
So each photon will itself lose energy.
And it loses energy proportional to one over the scale factor.
And that's just because the frequency shift proportionally
to the scale factor.
And that means that the energy per photon
shifts, because quantum mechanically
we know that the energy of a photon
is proportional to its frequency.
So if the frequency redshifts so must the energy.
In exactly the same way, 1 over a f t.
Yes, question?
AUDIENCE: You said previously that neutrinos
behave like radiation in the sense
that theta energy falls is 1 over a.
What is it about them that makes this happen?
Because there are also particles with standard kinetic energy,
right?
PROFESSOR: OK, the question, in case you didn't hear it,
is why-- how did neutrinos fit in here?
I've made the claim that neutrinos act like radiation
in the early universe, but neutrinos have a non-zero mass,
so they should obey the standard formulas for particles
with nonzero masses.
The answer to that is-- there is,
I think, a simple answer, which is that as long
as the energy is large compared to the mass,
particles with masses will still act like massless particles.
It doesn't really matter if the mass is zero or not,
the key thing, really, is this equation.
So if the term on the right hand side
is small compared to either of the two on the left,
it's not much different from being zero.
And that's what happens for neutrinos
in the early universe.
And we'll see soon that if you go to early enough times,
it's true even for electron-positron pairs.
They will also act like radiation.
Any particle will act like radiation
as long as the energy is large compared to the rest energy.
So getting back to the discussion
of the early universe, if the energy of each photon
falls off like one over the scale factor,
and the number density falls off like one over the scale factor
cubed, it means that the energy density,
and hence, the mass density of radiation
will fall off like one over a to the fourth,
in contrast to the one over a cubed, that we found when
we were talking about non-relativistic matter.
And that, of course, is going to make a difference.
Because those issues play a key role in our discussions
about how the universe evolved.
An important feature, which we see immediately,
is that if we extrapolate backwards in time,
since the radiation mass density is falling off like one over a
to the fourth, and the matter density is falling off
like one over a cubed, it means that as you go back in time,
the radiation becomes more and more important
relative to the matter, by factor of the scale factor.
So if we go back far enough, we will even
find a time when the mass density in radiation
equaled the mass density in non-relativistic matter.
And we calculated about when that would be.
We said that the energy density in radiation today
is given by this number, 7 times 10 to the minus 14,
joules per meter cubed.
And I just gave you this number, I didn't derive it yet.
We will derive it, probably later today.
But for now, we're just accepting it.
And that implies we can calculate
from that the ratio of the mass densities
in radiation and ordinary matter.
Here, we use the fact that ordinary matter
can be described by having an omega
ratio to the critical density of about 0.3.
And we know how to calculate the critical density,
and that allowed us to calculate the density of ordinary matter.
And then this ratio turned out to be 3.1 times 10
to the minus four.
So radiation in today's universe is
almost negligible in its contribution
to the overall energy balance, compared
to non-relativistic matter.
But if you extrapolate backwards,
we know that this ratio will vary as one over the scale
factor.
And we could figure out what constants
to put this equation by putting in the right constant,
so that this equation gives us the right value today.
Where the right value today is 3.1 times 10 to the minus four.
And notice that this works, if we
let t be equal to t sub zero, this factors one
and we get 3.1 times 10 to the minus four.
So these two factors together they
have t zero and the 3.1 times 10 to the minus four,
are just the right factors to put in
to give us the right constant of proportionality
in that equation.
Having this equation, we can then
ask, how far back do we have to go, how much we have
to change t, for the ratio to be one?
And that's a straightforward calculation.
And the ratio of the a is then just one over 3.1 times
10 to the minus four, or 3,200.
So if we talk about it in terms of a redshift, which
is how astronomers always talk about distances, or times,
we're talking about going back to a redshift of 3,200.
We can figure out what time, then, corresponds to also,
if we know how a f t depends on time.
And we do, approximately.
For this calculation, I assume for now, we
could do better later, and we will--
but I assume for now that we could just
treat the period between matter radiation equality, so-called t
x and now as being entirely described
by a matter-dominated universe.
That's only a crude approximation,
but it will get us the right order of magnitude.
And we'll learn how to do better later in the course.
So if we assume that, then t x is just this number to the 3/2
power, cancelling the 2/3, times the age of universe, t naught.
And that turned out to be about 75,000 years.
So somewhere in the range of 100,000 years, 50,000 years,
is the time in the history of the universe when radiation
ceased to be more important than matter.
And for earlier times than that, the radiation dominated.
And that's what we refer to as the radiation dominated era.
Any questions?
OK, now I think we get on to what is really
the important subject that we want to understand,
and most of this you did yourself on the homework.
But I'll summarize the argument here.
We want to understand what this tells us
about the Friedmann Equations.
And first, we'd like to understand
what it says about pressure.
Because it turns out that pressure
is the crucial issue in determining
how fast row falls off as a expands.
So if row is proportional to one over a cubed,
we can just differentiate that, putting
in a constant proportionality temporarily, just
to keep track of what we're doing.
Since we know how to differentiate qualities
and we're less familiar with how to differentiate
proportionalities.
But what we find immediately, is that row dot is then minus 3,
where that 3 is that 3, times a dot over a times row.
On the other hand, if row of t falls like one
over a to the fourth, row dot is minus 4 times
a dot over a times row, just by differentiation.
So we get different expressions from row dot,
between radiation and matter.
And we want to explore the consequences
of that difference.
It's related to the pressure of the gas,
because we can relate the pressure to row dot.
Because we know that as a gas expands,
it loses energy, which is just equal in amount to pdV.
And we illustrated this by a piston thought experiment,
but it's true in general.
So we can apply this famous formula, dU equals minus pdV,
to a patch of the expanding universe.
And by a patch I mean some fixed region and coordinate space.
So the total energy in that region of coordinate space
will be the physical volume, which
will be a cubed times the coordinate volume.
Which is going to cancel out of this equation on both sides.
So it's a cubed times the coordinate volume
times the energy density, which is row c squared.
The rate of change of that is Du.
And then on the right hand side, we
have minus p minus b times dV, which
is the rate of change of a cubed, again
times the coordinate volume that we're talking about.
But that will cancel out on the two sides of the equation.
So this is really just a description
for the universe of the dU equals minus pdV equation.
And this can just be rearranged, expanding the time derivatives,
to give us row dot, and we get minus 3 a dot
over a, times row plus p over c squared.
So this tells us how to relate row dot to the pressure.
And it tells us that the formula that we started
with a long time ago, which just said that row fell off
like 1 over a cubed, was synonymous with saying
the pressure is zero, for a gas of non-relativistic particles,
the pressure is negligible.
But for radiation, clearly if we're
going to get a four instead of a three,
the pressure will be non-negligible.
And in fact, it implies that the pressure
is exactly equal to one third of the energy
density for a gas of radiation.
OK, knowing that, we can now look back at the Friedmann
equations and ask, how do they stand up?
Are they still consistent, or do we have to modify something?
And this is really the crucial point.
What we know are these three equations,
which are the two Friedmann equations
and the equation for row dot.
And we could see immediately that those equations are not
independent of each other.
The easiest thing to see is that if we
start with the top equation, we could differentiate it.
And since the top equation has a dot in it,
when we differentiate, we'll get an equation for a double dot.
But the equation will also involve row dot,
if we take the time derivative of that top equation.
But if we know what row dot is, we
could put that in, and in the end
we'll get an equation for a double dot by itself.
And it will in fact agree with the equation
on the middle line.
Again, things would be inconsistent.
Things are consistent, we didn't make any mistakes.
If we derive the equation for a double dot
by using the first and third of those equations,
then we'll get the second of those equations.
But the catch is that now we want
to modify the third of those equations,
the equations for row dot.
And then the Friedmann equations as we've
written them will not be consistent anymore.
Because we'll have a different equation for row dot,
we'll get a different equation for a double dot.
So we have to decide what gives.
What can we change to make everything consistent?
And here, the rigorous way of proceeding
is to look at general relativity and see what it says,
and the answer we're going to write down
is exactly what general relativity says.
But we can motivate the answer in, I think,
a pretty sensible way, by noticing
that as the universe expands, we'd
expect the energy density to vary continuously,
because energies are conserved.
And we also expect a dot to vary continuously,
because basically, the mechanics of the universe
are like ton's laws.
And velocities don't change discontinuously.
You can apply a force, and that causes
velocities to have a rate of change.
But velocities don't change instantaneously,
unless you somehow apply an infinite force.
And the same thing will be true with the universe.
On the other hand, accelerations can change instantaneously.
You could change the force acting on a particle,
in principle, as fast as you want,
and the acceleration of the particle
will change at that same rate.
So if we look at these equations,
we would expect that the first equation and the third equation
would not be allowed to involve the pressure.
Because the pressure basically is a measure of a force.
Pressures can change instantaneously.
So what you need to do, if we're going
to make these equations consistent in the presence
of pressure-- which changes the row dot equation,
the only equation we can change is the second one.
And then we can ask ourselves, what
do we have to change it to make the three equations consistent?
And this is what you looked at on your homework.
And the answer is that the a double dot equation
has to be modified to give the equation
at the bottom of the screen here.
And this is the correct form of the a double dot equation
in cosmology.
And this is what we'll be using for the rest of the term,
this is exactly what you would get from general relativity.
As long as we're talking about homogeneous
and isotropic universes, this formula as exact
as far as we know.
OK, any questions about that?
Yes.
AUDIENCE: Why when we derive that equation do we use--
PROFESSOR: This equation?
AUDIENCE: Or the one above that.
PROFESSOR: Yeah.
AUDIENCE: We use dU equals minus pdV?
I mean, I agree with that, but couldn't we
use a more complete version?
Like, the complete version of the first law
of thermodynamics, that dU equals TDS minus pdV.
PROFESSOR: OK, yeah.
The question was when we wrote down dU equals minus pdV,
why did we not include a plus TDS term here,
which could also be relevant.
The answer is that for the applications
we're interested in-- you're quite right,
it could be important, but for the applications
that we're interested in, which is the expanding
gas in the universe, the expanding gas in the universe
will be making use of this fact.
It really does expand adiabatically, that is,
there's nothing putting heat in or out,
and everything is remaining very close to thermal equilibrium,
which means that entropy does not spontaneously change.
So the TDS term we will be assuming is very, very small,
and that's accurate.
And you're right, if that were not the case,
there would be further complications in terms
of figuring out what row dot is.
Let me point out here that this equation actually
does contain a somewhat startling perhaps fact
about gravity it says that in the context
of general relativity.
And that's really the context that we're in,
even though we haven't learned a lot of general relativity.
But it says that in the context of general relativity,
pressures, as well as mass densities,
contribute to the gravitational field.
A double dot is basically a measure
of how fast gravity is slowing down the universe.
And this says that there's a pressure.
It can also help to slow down the universe.
Meaning that pressure itself can create a gravitational field.
In the early universe, where we go back
to this radiation dominated period,
we know that the pressure is one third of the energy density.
That says that this pressure term
is the same size exactly as the mass density term.
So in the radiation dominated phase,
the pressure is just as important in effect
for slowing down the universe as is the mass density.
In today's universe it's negligible.
Well, we'll come back that.
The dark energy has a non-trivial pressure,
but the pressure of ordinary matter in today's universe
is negligible.
The other important fact about this equation
is that energy densities, so far as we know,
are always positive.
We don't know for sure what the ultimate laws of physics
are, but for all the laws of physics that we know,
energy densities are positive.
On the other hand, pressures actually
can be negative for some kinds of material.
And we'll talk a little bit more about how
to get a negative pressure later.
But this formula tells us that positive pressures
act the same way as positive mass densities,
creating an attractive gravitational field which
slows down the expansion of the universe.
But if there could be a material with a negative pressure,
this same equation, which would presumably still hold,
and believe it does, would tell us
that that negative pressure would actually
cause the universe to accelerate, because
of its gravitational effects.
Now, we're not talking about the mechanical effects
of the pressure.
Mechanical effects of pressure only
show up when there are pressure gradients, when
the pressure is uneven.
So the very large air pressure in this room,
which really is quite large, we don't feel all,
because it's acting equally in all directions.
Uniform pressures do not produce forces.
So the mechanical effects of the pressure in the early universe,
which we're assuming is completely homogeneous,
is zilch.
There is no mechanical effect.
But what we're seeing in this equation
is a gravitational effect caused by the pressure.
And it's obviously gravitational,
business the effect is proportional to Newton's
capital G, a constant determining
the strength of gravity.
So the equation is telling us that a positive pressure
creates a gravitational attraction, which
would cause the universe to slow down in its expansion.
But a negative pressure would produce
a gravitational repulsion, which would cause universe
to speed up.
And we now know that, today-- in fact,
for the last five billion years or so-- our universe
itself is actually accelerating under the influence
of something.
And the only explanation we have is
that the something that's causing the universe
to accelerate is the repulsive gravity caused
by some kind of a negative pressure material.
And that negative pressure material
is what we call the dark energy.
And we'll talk a little more later about what it is.
It's very likely just vacuum energy.
But we'll come back to that later in the course.
Yes?
AUDIENCE: When we looked at the toy example of the piston
in the cavity, the pressure of the gas
was pushing outwards against the wall of the container.
But we can't have that view of the universe, really,
because there's no exteriors in the universe.
PROFESSOR: There's no walls, right.
AUDIENCE: So how should we view pressure in the sense that--
PROFESSOR: OK.
Yeah.
OK, the question is, in our toy problem involving the piston,
we had walls for the pressure to push against.
And that was where the energy went.
It went into pushing the walls.
When we're talking about the universe, there are no walls.
How does that analogy work?
What plays the role of the walls?
And the answer, I think, is that the role of the walls, when
we're talking about the universe, first of all, you
can ignore it if you just took in a small region.
You could still just say, the small region
is pushing out on the regions around it.
And I think that's enough to make the logic clear.
But it still leaves open the question of, ultimately,
where does this energy go?
So saying it goes from here to there
doesn't help you unless you know where
it goes after it goes from and there and there.
So you might want to ask the question more generally,
where does the energy ultimately end up?
And then I think the answer is that it ends up
in gravitational potential energy.
You could certainly build a toy model, where you just
have a gas in a finite region, self-contained under gravity.
And then you'd have to make up some kind of a mechanism
to cause it to expand.
But when you cause it to expand, you'll
be pulling particles apart, which
are attracting each other gravitationally.
And that means you'll be increasing
the gravitational potential energy
as you pull the gas apart.
So I think, ultimately, the answer is the energy imbalance
that we seem to be seeing here is taken up
by the gravitational field so that, all in all, energy
still conserved.
Any other questions?
OK, in that case, let us continue on the blackboard.
OK, first thing I want to look at it
is just the behavior of a radiation-dominated flat
universe.
So a flat universe is going to obey
H squared equals 8 pi over 3 G rho,
and then the potential minus kc squared over a squared.
Hard to write dotted lines on the blackboard.
But this potential term is not there, because k equals 0.
We're talking about the flat case.
So for a flat case, we could just express H in terms of rho.
And we know how rho behaves for radiation.
Rho falls off as 1/a to the fourth.
So H squared is proportional to 1/a to the fourth.
That means that H itself is proportional to 1/a squared.
So we can do that.
a-dot over a is equal to some constant over a squared,
a-dot over a being H. And now we can multiply both sides
by a, of course.
And we get a-dot is equal to a constant over a.
And this we can integrate.
The way to integrate is to put all the a's on one
side and all of the dt's on the other side.
So we get ada, writing this as da/dt.
So ada is equal to the constant times dt.
And then, as we've done before, when
we're talking about matter, it's the same calculation there.
I just did different power of a appears
so we'd know how to do it.
Integrating, we get 1/2a squared is equal to the constant times
t, and then plus a new constant of integration, constant prime.
Now we make the same argument as we've made in the past.
We have not yet said anything that
determines how our clocks are going to be set.
So we can choose to set our clocks in the standard way,
which is to set our clocks so that t equals
0 corresponds to the moment where a is equal to 0.
And if a is going to be equal to 0 when t is equal to 0,
it means constant prime is going to be equal to 0.
So by choosing the value of constant prime,
we really are just determining how
we're going to set our clocks, how
we're going to choose the 0 of time.
And we'll do that by setting constant prime equal to 0.
And then we get the famous formula
for a radiation-dominated universe, a of t
is just proportional to the square root of t,
or t to the 1/2 power.
And this is for a radiation-dominated flat
universe, replacing the t to the 2/3
that we have for the matter-dominated flat universe.
Once we know that a is proportional
to the square root of t-- and for the flat universe,
the constant proportionality mean nothing, by the way.
It's not that we haven't been smart enough
to figure out what it is.
It really has no meaning whatever.
You could set it equal to whatever you want,
and it just determines your definition
of the notch, your definition of how you're
going to measure the comoving coordinate system.
Once we know this, we should know pretty much everything.
So in particular, we can calculate
h, which a-dot over a.
And the constant proportionality drops
when we compute a-dot over a.
As we expect, it has no meaning.
So it should not appear in the equation for anything
that does have physical meaning.
So H is just 1/2t, the 1/2 here coming
from differentiating the t the 1/2 power.
We can also compute the horizon distance.
So the physical horizon distance, l sub p horizon,
where p stands for physical, is equal to the scale factor
times the coordinate horizon distance.
And the coordinate horizon distance
is just the total coordinate distance
that light could travel from the beginning of the universe.
And we know the coordinate velocity of light
is c divided by a.
So we just integrate that to get the total coordinate distance.
So it's the integral from 0 to t of c
over a of t prime dt prime.
And since a of t is just t to the 1/2,
this is a trivial integral to do.
And the answer is 2 times c times t.
So in a radiation-dominated universe,
the horizon's distance is twice c times t.
For a static universe, the horizon distance
would just be c times t.
It would just be the distance light can travel in time t,
but more complicated in an expanding universe.
For the matter-dominated case, we
discovered that the horizon distance was 3ct,
if you remember.
For the radiation-dominated case, it's 2ct.
And finally, an important equation
is that, going back to here, where we started,
this equation relates H to rho.
We found out that, merely by knowing
the universe is radiation-dominated,
without even caring about what kind of radiation
it is, how much neutrinos, how much photons, whatever--
doesn't matter-- merely by knowing the universe
is radiation-dominated, we were able to tell that H is 1/2t.
And if we know what H is, that formula tells us
we also know what rho is.
So without even knowing what kind of radiation
is contributing, we know that, for a radiation-dominated
universe, rho is just equal to 3 over 32
pi Newton's constant G times time, little t, squared.
It's rather amazing that we can write down that formula
without even knowing what kind of radiation is contributing.
But as long as that radiation falls
off as 1 over the scale factor to the fourth,
and as long as we know the universe is flat,
then we know what that energy density has to be.
This is crucial here, by the way.
The energy could be anything if we did not
assume that the universe was flat.
OK, any questions about this?
Yes?
AUDIENCE: If we assumed that it was almost flat,
would we be able to have any bounds on it?
PROFESSOR: OK, question is, if we assumed
that it was almost flat, would we
be able to have any bounds on it?
The answer is, yeah, if you were quantitative
about what you meant by "almost flat,"
you could know how almost true that formula would have to be.
OK, if there are no other questions,
I want to switch gears slightly now
and go back to talk about some of the basic underlying physics
that we are going to need, and in particular, the physics
of black-body radiation.
So this is really just a little chapter of a stat mech course
that we're inserting here, because we need it.
And because it comes from another course,
we're not going try to do it in complete detail.
But I'll try to write down formulas that make sense.
And that will give us what we need to know to proceed.
So that will be the goal.
So what is black-body radiation?
The physical phenomenon is that, if one imagines
a box with a cavity in it-- that's
supposed to be a box with a cavity in it,
in case you can't recognize the picture-- if the box is held
at some uniform temperature t-- t is temperature--
then is claimed and verified experimentally
that the cavity will fill up with radiation-- in this case,
we're really just talking about electromagnetic radiation--
the calving will fill up with electromagnetic radiation
whose characteristics would be determined solely
by that temperature t and will therefore
be totally independent of the material that makes up the box.
Roughly speaking, I think the way
to think about it is to say that the box will fill up
with radiation at temperature t.
And saying that the radiation has temperature t
is enough to completely describe the radiation.
It doesn't matter what kind of a box
that radiation is sitting in.
So the box will fill with radiation at temperature t.
And that radiation is called black-body radiation.
Like many things in physics, it has a variety of names,
just to confuse us all.
So it's also called cavity radiation,
which makes a lot of sense, given
the description we just gave.
And it's also sometimes called just thermal equilibrium
radiation.
This is radiation at temperature t.
I haven't really justified the word "black-body"
radiation yet, so let me try to do that quickly.
The reason why it can be called black-body radiation-- and this
will be important for some things in cosmology;
I'm not sure if it will be important to us or not,
but certainly important to know-- the reason why it's
called black-body radiation is because we imagine inserting
into this cavity a black body, in the literal sense.
What is the literal sense of a black body?
It's a body which is black in the sense
that all radiation that hits it is absorbed.
Now, this black body is still going to glow.
If you heat a piece of iron or something
to very high temperatures, you see it glow.
That glow is not reflection.
That glow is emission by the hot atoms
in the piece of iron or whatever.
Emission is different from reflection.
When we say it absorbs everything,
we mean it does not reflect anything.
But it will still admit by thermal de-excitation.
The crucial distinction between reflection and thermal emission
is that reflection is instantaneous.
When a light beam comes in, if it's reflected,
it just goes back out instantaneously.
Emission, thermal emission, is a slower process.
Atoms get excited, and eventually, they
de-excite and emit radiation.
So it takes time.
And that's the distinction.
We're going to assume that this body is black in the sense
that there's no reflection.
OK, now we're going to make use of the fact
that we know that thermal equilibrium works.
That is, if you let any isolated system sit long enough,
it will approach a unique state of thermal equilibrium
determined by its constituents, which you've put in
to begin with, but otherwise independent of how exactly
you arrange those constituents.
So if we put in, for example, a cold black body,
it will start to get harder, warming up
to the same temperature as everything else.
If we put in an extra hot black body,
it would emit energy and start to cool down
to the temperature of everything else.
But eventually, this black body will
be at the same temperature as everything else.
And we're going to be assuming here that the box itself
is being held at some fixed temperature t.
So wherever energy exchange occurs
because of this black body, it will
be absorbed by whatever is holding
the outer box at the fixed temperature.
So in the end, if we wait long enough,
this black body is going to acquire the same temperature
as everything else and hold that temperature.
Now, if it's holding that temperature,
it means that the energy input to the box, to the black body,
will have to be the same as the energy
output of the black body.
Now, the black body is going to be absorbing radiation,
because we have radiation here, and we
said that any radiation that hits it is absorbed.
That was the definition of "black."
So it's clearly absorbing energy.
If it's not going to be heating up--
and we know that it's not, because it's
in thermal equilibrium; the temperature will remain fixed--
in order for it to not heat up, it
has to radiate energy, as well.
And the energy it radiates has to be exactly the same
as the energy it's absorbing once it
reaches thermal equilibrium.
So in equilibrium, the black body, BB, radiates at same rate
that it absorbs energy.
This radiation process is this slow process
of thermal emission.
There are atoms inside this black body that are excited.
Those atoms will de-excite over time,
releasing photons that will go off.
And the important thing about that slow mechanism
is that, if we imagine taking this black body out
of its cavity, but not waiting long enough for its temperature
to change-- so we'll assume its temperature
is still the same, t.
So this is a picture of the same black body at temperature t,
but now outside the cavity.
Its radiation rate is not going to change when we take it
outside the cavity, because the radiation was caused by things
happening inside the black body, which are not changed when we
put it in or out of the cavity.
So it will continue to radiate at exactly the same rate
that it was radiating when it was in the cavity.
And that means it's going to radiate
at exactly the same rate as the energy
that it would have absorbed if it were bathed
by this black-body radiation.
So essentially, it means it will emit black-body radiation
with exactly the intensity that the black-body radiation would
have on the outside if the black body were still
inside the cavity.
So it radiates with exactly the same intensity as the energy
that it would receive if it were inside the cavity.
And furthermore, you could even elaborate a bit
on his argument to show that the radiation that it radiates
has exactly the same spectrum, exactly the same decomposition
into wavelengths, as the black-body radiation
inside the cavity.
And the way to see that is to imagine
surrounding this black body by absorption filters that only
let through certain frequencies.
And the point is that, no matter what frequencies
you limit going through this filter,
you have to stay in equilibrium.
It will never get hotter or colder.
So that means that each frequency by itself
has to balance, has to have exactly the same emission
as it would have abortion if the black body were just
exposed to black-body radiation surrounding it.
So it radiates black-body radiation.
And the intensity and spectrum must
match what we call black-body or cavity radiation.
So the cavity radiation has to exactly mimic
the radiation emitted by this black body.
And that's the motivation for calling it
black-body radiation.
Now, if this black body absorbed some radiation
and reflected some, then it would emit different radiation.
So it is important that this body
be black, in the sense that it doesn't reflect anything.
All radiation hitting it is absorbed.
And only under that assumption do
we know exactly what it's going to emit.
Yes?
AUDIENCE: So is this only true for right
after you take it out?
PROFESSOR: Well, it will start to cool after you take it out.
And as it cools, its temperature will change.
But if you account for the changing temperature,
it will be true at anytime, actually.
But the temperature will change.
AUDIENCE: Because, in the black cavity,
it has things exciting it.
And when you take it out, there's
no photons, no constant radiation to excite it.
So it can radiate--
PROFESSOR: Yes.
Once you take it out, it's no longer being excited.
And I think, technically, you're right.
Once you take it out, it will not only cool,
but it will cease to be at a uniform temperature.
And that's basically what we're saying
if we're saying that the atoms that are excited
won't necessarily be in the right thermal distribution
as they would be if it was on the inside.
That would be a statement that is not
any longer in thermal equilibrium.
But as long as the radiation is slow,
you could just account for the changing temperature.
You would know how it radiates.
And I think that's a very good approximation.
Although in principle, it will cease
to be in thermal equilibrium, as soon as you take it out,
the edges will be cool, and the center will be hot.
And you'd have to take into account all of those things
to be able to understand how it radiates.
Any other questions?
OK, next, I want to talk a little bit
about what this black-body radiation is.
And one can begin by trying to understand
it purely classically, which, of course,
is what happened historically.
In the 1800s, people tried to understand cavity radiation
or black-body radiation using classical physics, Maxwell's
equations, to describe the radiation.
And then, in a nutshell-- we're just
trying to establish basic ideas here--
one can try to treat a field statistical-mechanically
by imagining not fields in empty space,
but fields in some kind of a box.
In this case, it doesn't necessarily
have to be the cavity that we're talking about.
It could be a big box that just enclosed the system somehow
to make it easier to talk about.
And in the end, you could imagine
taking that box to infinity, this theoretical box
that you use to simplify the problem.
But once you put the system in a box,
then a field, like the electromagnetic field,
can always be broken up into normal modes,
standing wave patterns that have an integer or a half integer
number of wavelengths inside the box.
And no matter how complicated the field is inside the box,
you could always describe it as a superposition
of some set of standing waves.
In general, it takes an infinite number
of standing wave components to describe an arbitrary
field-- that is, with shorter and shorter wavelengths.
But you can always-- and this is Fourier's theorem--
you can always describe an arbitrary
field in terms of the standing waves.
And that's good for the point of view
of thinking about statistical mechanics,
because you could think about each standing wave
almost as if it were a particle.
It really is a harmonic oscillator.
So if you think you know the statistical mechanics
of harmonic oscillators, each standing wave in the box
is just a harmonic oscillator, so simple.
We now try to ask what is the thermodynamics
of this system of harmonic oscillators.
And the rule for harmonic oscillator is simple.
Stat mech tells you that you have
1/2 kT per degree of freedom in thermal equilibrium.
The energy of a system should just
be 1/2 kT per degree of freedom.
Having said that, all the complicate questions
come about by asking ourselves what
is meant by degree of freedom.
But for the harmonic oscillator, that has a simple answer.
A harmonic oscillator has two degrees
of freedom-- the kinetic energy and the potential energy.
So the energy of a harmonic oscillator
should just be kT per degree of freedom.
And we could apply that to our gas in the box and we could,
ask how much energy should the gas absorb
at a given temperature?
What should be the energy density of the gas at a given
temperature-- this gas of photons.
But it was noticed in the 1900-- the 1800s
that this doesn't work because there's no limit to how short
the wavelengths can be.
And therefore, there's not just some finite set
of harmonic oscillators.
There's an infinite set of harmonic
oscillators where you have more and more harmonic
oscillators at shorter and shorter wavelengths
ad infinitum, no limit.
And that came to be known as Jean's Paradox.
So what it suggests is that if this classical stat mech
worked-- which obviously it's not working.
But if it did work, it would mean
that as you tried to put a gas in just an empty box in contact
with something at a fixed temperature,
the box would absorb more and more energy without limit.
And ultimately, it would presumably
cause the temperature of the whatever
is trying to maintain the temperature
to go to 0 as energy gets siphoned off
to shorter and shorter wavelengths of excitations.
That obviously isn't the way the world behaves.
We'd all freeze to death if it did.
So something has to happen to save it.
And it wasn't at all obvious for many years
what it was that saves it.
But this Jean's Paradox turns out
to be saved by quantum mechanics.
And the important implication of quantum mechanics
is that the energy of a harmonic oscillator
is no longer allowed to have any possible value,
but is now quantized as some integer times
h times nu, the frequency-- h being
Planck's constant, nu being frequency, n
being some integer where this integer might be called
the excitation level of the harmonic oscillator.
Depending how you choose your 0, you might have an n plus 1/2
there.
But that's not important for us right now.
It will be important later actually.
But for now, we'll just allow ourselves to readjust the 0
and just think of it as n times h nu, or h bar omega.
Now, this makes all the difference,
statistical mechanically.
One can apply statistical mechanics
using basically the same principles
to the quantum mechanical system.
And the key thing now is that for the very short wavelengths,
which are the ones that were giving us trouble--
the infinities came to short wavelengths.
For the short wavelengths where nu is high, h nu is high.
And it means that there's a minimum ante
that you could put in to excite those short wavelength
harmonic oscillators.
And it's a large number.
You either put in a large amount of energy or none at all.
Quantum mechanics doesn't let you do anything in between.
Now remember, the classical mechanics answer
was that you have kT in each harmonic oscillator.
And kT would be small compared to h nu,
if we're talking about a very short wavelength.
So the classic answer is just not
allowed by quantum mechanics.
You either have to put in nothing or an amount much, much
larger than the classical answer.
And when you do the statistical mechanics quantum mechanically,
which is not a big deal really, you
find that when you're confronted with that choice,
the most likely answer is to put in no energy at all.
So quantum mechanics freezes out these short wavelength modes.
And then the n produces a finite energy density
for a gas of photons.
Yes?
AUDIENCE: Seems like if you were to sum over
like all the possible wave numbers, that-- well,
so the energy is inversely related to wavelength, right?
So even if you quantize it, like for large wavelengths,
isn't the sum still like a sum of one over lambda wavelength,
with its derivative?
PROFESSOR: You're saying, isn't there also
a divergence at the large wavelength n?
AUDIENCE: Because that sum doesn't
seem like it would work.
PROFESSOR: Right.
No, that's important.
The reason that's not a problem is
that if you're talking about the energy in a box,
the wavelength can't be bigger than the box.
The largest possible wavelength is twice the box so
that half a wavelength fits in the box.
If you're talking about the energy in the whole infinite
universe, then we expect the answer to infinite.
And it is.
There's no problem with having infinite total energy
if you want to have a finite energy
density throughout an infinite universe.
So the size of the box cuts off the large wavelengths.
And quantum mechanics cuts off the small wavelengths.
So in the end, one does get a finite answer
for the energy density of black-body radiation.
And that's crucial for our survival,
crucial for the existence of the universe
as we know it, and also crucial for the calculations
that we're about to do.
OK, so when one does these calculations initially
for photons only, what we'd find is that the energy density
is equal to a fudge factor, which I'm going to call g.
And you'll see later why I'm introducing a fudge factor.
For now, g is just 2.
But later, we'll generalize the application of this formula,
and g will have different values.
But for now, we're dealing with photons.
There's a factor of 2 there, but I'm going to write 2 as g,
writing g equals 2 underneath.
And then the pi squared over 30--
you can really calculate this-- times
kT to the fourth power divided by h bar c cubed, h bar
being Plank's constant divided by 2 pi and little k
being Boltzmann constant.
So this is calculated just by thinking of the gas in a box
as a lot of harmonic oscillators and applying standard stat
mech to each harmonic oscillator,
but you apply the quantum mechanical version of the stat
mech to each harmonic oscillator.
And you can also find, by doing the same kind of analysis,
that the pressure is 1/3 the energy density,
which we also derived earlier by different means.
And it's all consistent so you get the same answer every time,
even if you think about it differently.
So here, I have mine deriving it directly from the stat mech.
You can also, from the stat mech,
calculate the number density of photons in thermal equilibrium.
And that will be equal to -- again, there's a factor of 2.
But this time, I'll call the factor
of 2 g star, where g star also equals 2 for photons.
But when we generalize these formulas,
g will not necessarily equal g star,
which is why I'm giving it two names.
And this g star multiplies zeta of 3,
where zeta refers to the Riemann zeta function, which
I'll define in a second, divided by pi squared times
kT cubed divided by h bar c cubed.
OK.
OK, so I need to define this zeta of 3.
It's 1 over 1 cubed plus 1 over 2 cubed plus 1 over 3 cubed
plus dot dot dot.
It's an infinite series.
And if you sum up that infinite series,
at least to three decimal places, it's 1.202.
OK, then there's one other formula
that will be of interest to us.
And that'll be a formula for the entropy density.
Now, if you've had a stat mech course,
you have some idea of what entropy density means.
If you have not, suffice it to say for this class
that it is some measure of the disorder in the sense
of the total number of different quantum states
that contribute to a given macroscopic description.
The more different microstates there
are that contribute to a macroscopic description,
the higher the entropy.
And the other important thing about entropy to us
besides that vague definition-- which will be enough--
but the important thing for us is
that under most circumstances, entropy will be conserved.
It's conserved as long as things stay
at or near thermal equilibrium.
And in the early universe as the universe expands,
that's the case.
So for us, the entropy of our gas
will simply be a conserved quantity
that we can make use of.
And we will make use of it in some important ways.
And we could write down a formula
for the entropy density of photons.
And it's g, where this g in fact the same g as over there.
It is related to the energy.
So it's the same g that appears in two cases, 2
in both cases for protons by themselves.
And then there are factors that you can calculate--
2 pi squared over 45 times k to the fourth T cubed over h bar
c cubed.
OK, this time, the number of k's and T's do not match.
That's mainly due to the conventions about
how entropy is defined.
It's not really anything deep.
I might mention at this point that the 2's
that I've been writing for everything-- g equals 2,
g star equals 2-- the reason those 2's are written
explicitly rather than just absorbing the factor of 2
into the other factors is that photons are characterized
by the fact that there are two polarizations of photons.
So if I have a beam of photons, they
could be right-handed or left-handed.
And anything else could be considered
a superposition of those two.
So there are two independent polarizations.
And it's useful to keep track of these formulas
as the amount of energy density per polarization.
Thus, different kinds of particles
will have different numbers of polarizations.
So if we know the amount per polarization,
we'll be able to more easily apply it to other particles.
Yes?
AUDIENCE: Sorry, just to kind of bring up the same question,
if we-- I read that in the early universe,
the temperature is constantly changing or it's cooling.
PROFESSOR: Right.
AUDIENCE: So then if the temperature is changing,
then how can we say that the entropy is constant?
PROFESSOR: OK, important question--
we'll be getting to it very soon.
But since you asked the question, I'll ask it now.
The question was if the universe is expanding,
and the entropy density is going down
because it thins, how can that happen--
I guess it was asked the other way around.
If the temperature is falling, how
can entropy be conserved if this is
the formula for entropy density?
And the answer-- when I tell you, you'll see it's obvious.
We don't expect the entropy density
to be conserved if the entropy is conserved.
The entropy thins out as the universe expands.
So if we just had a gas with nothing else changing,
we expect the entropy density to go down like 1 over the scale
factor cubed, just like the number density of particles.
So if s is going to go down like 1 over the scale factor cubed,
that would be consistent with this formula
if the temperature also fell as 1 over the scale factor.
So that to cubing it made things match.
And that's what we'll find.
The temperature falls off, like 1 over the scale factor.
And that's consistent with everything
that we said about energy densities and so on.
OK.
Next thing I want to talk about is
neutrinos, which I told you earlier
contributes in a significant way to the radiation energy
density in the universe today.
Neutrinos are particles which for a long time,
were thought to be massless.
Until around 2000 or so, neutrinos
were thought to be massless.
Now we know that in fact, they have a very small mass, which
complicates the description here.
It turns out that cosmologically, neutrinos
still act as if they were massless
for almost all purposes and for really all purposes
that we'll be dealing with in this class, although if we were
interested in the effects of neutrinos on structure
formation, we'd be interested in whether or not
the neutrinos have a small mass or whether it's
smaller than that.
We know it's non-zero.
We don't know what the neutrino mass is, by the way.
What we actually know from observations
is that there are three types of neutrinos.
And those are called flavors.
And I'll use the letter nu for the word neutrino.
And those three types of neutrinos
are called nu sub e, called the electron neutrino;
nu subbed mu, called the muon neutrino, and nu sub tau,
called the tau neutrino.
And these letters e, mu, and tau link to the names of particles.
This is the electron neutrino.
This is the muon neutrino connected
to a particle called the muon, which
is like the electron but heavier and different.
And this is linked to a particle called the tau, which is also
like the electron but much more heavier but otherwise similar
in its properties.
And the neutrinos are linked in the sense
that when a neutrino is produced, depending
on how you start, it is very typically produced
in conjunction with one of these other particles.
So an electron neutrino is typically
produced in conjunction with an electron.
And similarly, a muon neutrino is typically
produced in conjunction with a muon.
And a tau neutrino is typically produced in conjunction
with a tau.
Now, what does this have to do with neutrino masses?
We've never actually measured the mass of neutrino.
So we only know that they have mass indirectly.
What we have seen is one flavor of neutrino turn
into another flavor.
And it turns out, in the context of quantum field theory
and I think this does make a certain amount of sense
just by intuition, if a particle is massless,
it can never change into anything.
The process by which one changes into another
is pretty quantum mechanical and a little hard
to understand anyway.
But if the particles were really massless,
they would move at the speed of light.
And if the particles were moving at the speed of light,
if the particle had any kind of a clock on the particle,
that clock would literally stop with the particle
moving at the speed of light.
So if particles are massless, any internal workings
that that particle might have to be frozen.
That is, if it's a clock, it has to be
a clock that's stopped completely.
And for reasons that are essentially that,
although they can be made more formal and more rigorous,
a truly massless particle could never
undergo any kind of change whatever.
It would have to stay exactly like it
looks like to start with.
Because it just has no time.
So the fact that these neutrinos turn into each other
implies that they must have a nonzero mass.
It must not really be moving at quite the speed of light.
And that's the way the formalism works.
And we could set limits on the masses
based on what we know about the transitions between one
kind of neutrino and another.
Yes?
AUDIENCE: How do we explain photon decaying
to an electron-positron pair then?
PROFESSOR: A photon decaying to an electron-positron pair?
AUDIENCE: Yeah.
PROFESSOR: The answer is a free photon never does decay
to an electron-positron pair.
Photons can collide with something
and produce an electron-positron pair.
But that collision, that's a more complicated process.
What I'm saying is-- I'm sorry.
This process of conversion-- maybe I should have clarified--
happens just as the neutrinos travel.
It's not due to collisions.
Due to collsions, complicated things
can happen whether the particle is massless or not.
But a massless particle simply in transit cannot undergo any
kind of transition.
And these neutrinos are seen to undergo transitions simply
being in transit without any collisions.
In terms of my clock analogy and a stopped clock,
I think the reason the photon can convert
into electron-positron pairs if it collides with something
is that when it collides, it essentially breaks the clock.
You don't have a photon that's just
moving along without time anymore.
OK, so these neutrinos have masses.
And maybe I should, at this point,
write down some bounds on these masses.
m squared 21 times c to the fourth is equal to 7.50 plus
or minus 0.2 times 10 to the minus 5 electron volts squared.
These numbers come from the latest particle data tables,
which I gave references for in the notes.
So this is the difference of the mass squared.
And delta m 23 squared times c to the fourth
to turn it into the square of an energy
is 2.32 plus 0.12 minus 0.08 times 10
to the minus third electron volts squared.
So one thing you notice immediately
is that these are incredibly small masses.
Remember, the proton weighs 938 MeV,
three million electron volts.
And these are fractions of one electron volt.
So by the standards of particle physics,
these are unbelievably small energies,
unbelievably small mass differences.
But they're there.
They have to be there for the physics we know to make sense.
The other thing that you may notice
about this notation-- which I don't want to elaborate
on but I'll just mention-- this is called 21.
This is called 23.
There's no 1, 2, or 3 there.
There's an e and a mu and a tau.
The complication here is something very quantum
mechanical.
The e, mu and tau labels are labels
which basically label the neutrino according
to how the neutrino is created.
It turns out that their mass eigenstates-- states which
actually have a definite mass-- are not
the e, the mu or the tau.
In fact, if the e had a definite mass,
that would be saying that an e would just
propagate as a particle with a certain mass.
It would not convert into anything else.
The fact that an e converts into other particles--
a nu sub e converts into other particles
is really the statement that nu sub e is not
a state with a definite mass.
But there are states with definite masses
which could be expressed quantum mechanically as superpositions
of these flavor eigenstates.
So nu sub 1, nu sub 2 and nu sub 3
are states of neutrinos that have definite masses.
And each one of them is a superposition of nu sub
e, nu sub mu and nu sub tau.
Yes?
AUDIENCE: How come we don't have a delta n from 31?
PROFESSOR: Good question.
I think it's just the lack of knowledge.
I don't think there's any reason it's not defined.
I'm sure it is defined.
I think it's just lack of knowledge.
And if I knew more details about how these things were measured,
I could give you a better story about that.
But I don't, frankly.
So in the end, it's a rather complicated quantum mechanical
system which we're not going to go into any details about.
What more should I say today?
OK, let me just mention for today
and we'll continue next time after the quiz,
for our purposes, we're going to treat these neutrinos
as if they're massless.
And it turns out that that's actually extraordinarily
accurate from the point of view of cosmology,
at least for the kind of cosmology
that we're doing where we're just
interested in the effect of these neutrinos
on the expansion rate of the universe.
And treating them as massless particles,
I will shortly give you the formulas
for how they contribute to the black-body radiation.
But it I think there's no point in my writing them now.
I'll just write them again in the beginning of next period.
But they do contribute to the black-body radiation
and in a way that we actually know how to calculate.
And they have a noticeable effect
on the evolution of our universe.
OK, that's all for today.
Good luck on the quiz on Thursday.
I'll be here to help proctor.
I think Tim [INAUDIBLE] will be here too.
And I'll see you more intimately either at my office hour
tomorrow or at lecture a week from now.