The following content is provided
under a Creative Commons license.
Your support will help MIT OpenCourseWare continue
to offer high quality educational resources for free.
To make a donation or to view additional materials
from hundreds of MIT courses, visit MIT OpenCourseWare
at ocw.mit.edu.
PROFESSOR: OK, starting again.
I want to begin with a quick review
of what we said last time.
This time will be quicker than usual
because we're not really continuing from there,
we'll be starting a completely new topic today.
But I thought I'd remind you, nonetheless,
that we had a last lecture-- the universe was not
created between then and now.
And what we talked about last time
was the spacetime geodesic equation,
which is written this way.
It's a completely identical to the equation
we derived for geodesics in pure spatial situations.
The only difference really, is a difference in notation.
Instead of using i and j the tradition
is to use mu, nu, et cetera for the spacetime indices which
are sum from 0 to 3, where x super 0 is identical to t.
And in addition, the parameter which is sometimes called s
when we're talking about space, is called tau
when we're talking about time because what
the parameter refers to is the proper time
along the trajectory of the object whose geodesic we're
calculating.
We then introduced the Schwarzschild metric,
which we did not derive, but we claimed
it describes the metric for any spherically symmetric mass
distribution as long as you're talking
about the region outside where the masses are located.
And here M is the total mass of the object, which
is the only thing the metric really depends on,
G is Newton's constant, and C is of course the speed of light.
The metric has an interesting feature
that the coefficients of both the dt
squared term and to dr squared term becomes singular either 0
or infinity, depending on which one you're looking at,
a particular value of the radius called the Schwarzschild
radius, given by this formula 2 GM divided by c squared.
The bigger the mass the bigger the Schwarzschild radius,
they're proportional to each other.
The metric is singular at those points,
but I mentioned but did not prove,
that that particular singularity is
in fact what is referred to as a coordinate singularity.
It's a singularity that's there only because of the way
the coordinates were chosen.
So there are other ways of choosing the coordinates where
that singularity disappears.
There's also singularity at r equals zero,
and that singularity is real, there's
no way to remove that singularity
by a change of coordinates.
However, although r sub s is not a true singularity,
it is a horizon.
And by that we mean that if any particle or even a light beam
gets inside the Schwarzschild radius, it can never get out.
There's no geodesic which will take it out of the horizon.
And it's not even a matter of geodesics,
there are no time-like paths even
if you have a rocket which would then not follow geodesic.
There's no way to get out from inside a black hole.
We didn't show that, but that fact is claimed.
Then we calculated the geodesic for our radially falling
object.
We solve the problem of an object that
is released from rest at some initial value r sub 0,
and just let fall straight down towards the center
of the sphere.
And the the equation describing the geodesic
is just a special case of the general equation
that we had a few slides ago.
And we need only look at the radial component
if we want to track how the radius changes with time.
So there was a free index mu in the generic form
of the equation, we're setting mu equal to the r variable.
And then the equation reduces to this form
and we know what these G sub t t's and G sub r r's are,
they come from the equation for the Schwarzschild metric
on the previous slide.
And that equation can be manipulated
and eventually it simplifies to something extraordinarily
simple.
It's just the statement that d squared r d tau squared is
equal to minus GM over r squared, which looks exactly
like the Newtonian equation for something falling
in a spherically symmetric gravitational field.
But it's not really the same equation,
it just looks like it's the same equation, because the variables
both a different meanings. r and tau
both have different meanings from the r and t that
would have appeared in the Newtonian calculation.
The r variable that appears here is not really
the distance from the origin.
If you wanted to know the distance from the origin
you'd have to integrate the metric singular even--
if not even a well defined distance to the origin
because the origin singular.
And the tau that appears here is a time variable,
but it's not the time that would be read on any fixed clock,
rather it's the time that would be
read on the wristwatch of the person falling
into the spherically symmetric object, which we might consider
to be a black hole.
And we were able to solve this equation
by using essentially conservation of energy
techniques, or at least what would be called conservation
of energy if we were doing the Newtonian version
of the problem, which is the same equation even
though the variables have a different interpretation.
So we were able to calculate not r as a function of tau,
but at least tau a function of r.
And we got that equation, which is a little complicated,
but the interesting thing about it
is that it gives finite answers for every value of r going
all the way down to r equals zero.
So it means that in a finite amount of time,
as seen by the person falling into the black hole,
the person would reach r equals zero, at which point
he would disappear into the singularity.
He'd actually be ripped apart as he approached
the singularity because of tidal forces
which pull more strongly on the front part of him
than on the back part of him, stretching the object out
in the radial direction.
However, curiously, if one calculates
what this trajectory looks like as a function
of the coordinate time, t, we did actually
do that calculation but we looked
at how it would behave in the limit
as you approached-- as the particle approached
the Schwarzschild horizon.
And we discovered it would take an infinite amount of time,
as seen from the outside, for the in falling object
to reach the horizon, let go through the horizon
and get to r equals zero.
So from the outside, it looks like the object never
actually falls into the black hole,
but just gets closer and closer and closer as t approaches
infinity.
So it's an example of a very highly distorted spacetime,
where you can see very different pictures depending
on which observer you're trying to describe
the observations of.
And I think that's it.
Any questions about any of that?
I guess I'll put that back up.
OK.
On your homework you'll be applying this geodesic equation
to a model universe, to Robertson-Walker universe,
and this will serve only as an example for those calculations.
I guess there's also a homework problem about the Schwarzschild
metric, that orbits in the Schwarzschild metric
that you'll be working at.
It's all in principal straightforward
if you just look at equations and follow
what the equations tell you, thinking carefully
about what the variables mean.
OK.
In that case, let's get started on today's work,
which is a change of gear.
We're now going to be talking about black body
radiation and its effect on the universe.
I should say that my original plan was
to get into this set of lectures notes--
lecture notes six, which have not been handed out yet--
to get into those last time and to finish them today.
I don't think that's going to be possible
because I didn't get into them last time,
and I don't think I'll be able to finish them today.
But I would like today to be sort of the closing for what's
needed for the problem set due Monday and for the quiz
next week.
So, shortly after today's lecture
I will send you an email telling you
where the cutoff is as far as the reading
and the lecture notes.
And I also hope to post the lecture notes by tomorrow.
And I also will be posting a set of review problems
as we had for quiz one.
And I hope to get that done by tomorrow.
You may have noticed that not all of my hopes are filled,
but I do my best, and I'll try.
OK are there any logistic questions or anything
before we go on?
Yes.
AUDIENCE: Can you post the solutions
to the previous problem set, the one
that we turned in on Friday?
PROFESSOR: Oh, um, yes I can, I should.
OK I will.
OK, that's a third item I should try to get done today.
Thanks for reminding me.
And the solutions to the problem set
that you'll be handing in Monday will
be posted very shortly after you hand them in
so that people can start talking about them
and prepare for the quiz.
Yes.
AUDIENCE: Do we have any day at which the videos might be up?
PROFESSOR: Ah, I've request about that and all I was told
was that they're doing their best.
So, I did look into it, but I don't know the answer.
I hope that you'll have all the videos available to study
for the quiz, but I don't know if that's going to happen
or not.
OK.
in that case, the new topic is Blackbody Radiation
and the Early History of the Universe.
So far we've dealt with a universe which
contains only non-relativistic matter
and that, as we said from the beginning,
describes our universe for the bulk of its history.
But in the early period, the universe
was in fact dominated by radiation,
as we will now be calculating.
And in the more recent period, the universe
is dominated by dark energy, which
we'll be talking about immediately after we
finish talking about radiation.
So the important point here is that even though we don't think
of light as having mass, light certainly
has energy and relativistically we
know that energy and mass are equivalent.
The key equation that actually dominates today's lecture
is perhaps the most famous equation physics,
E equals MC squared, energy and mass are equivalent.
And the numbers-- I'll just give you
some numbers for this equation.
You've probably already aware that the numbers
are kind of out of sight.
One kilogram, a point to that equation,
is equivalent to-- I don't have any figures to give you--
8.9876, in case you really want to know it accurately, times 10
to the 16th, most important to know the exponent there,
joules.
And it's also perhaps interesting to translate this
into the kind of energy units that
are use when we talk about power consumption
in practical situations.
It corresponds to 2.497 times 10 to the 10th kilowatt hours,
which is a lot of kilowatt hours.
An interesting comparison, is the total power consumption
of the world, which turns out to be
comparable to a kilograms worth of things.
I looked things up in the Wikipedia,
and it told me that in 2008-- which
is the most recent year it had numbers for,
which is a little surprising, that's so far in the past--
the total world power consumption
for the year equaled about, I'm rounding off here,
about 150 petawatt hours.
Now, I always have to look up peta when I see it,
I haven't quite learned what peta means yet.
But they translated this, this means 150 trillion kilowatt
hours.
So 150 times 10 to the 12th kilowatt hours.
And if we divide this by the number of hours in the year--
to ask how much is per hour, which
is I think a natural thing to think about if you're using
kilowatt hours to measure the power, especially the energy--
the power that goes with this is about 17 times 10
to the ninth kilowatts.
And therefore 17 times 10 to the 9th kilowatt hours per hour.
And if you compare these two numbers,
it means that if you could convert one kilogram per hour
into pure energy, that would be about
equal to 1.5 times the world's power usage in 2008.
So if you could convert matter completely to energy,
as they do on Star Trek, it would
mean that you could fill up your tank
of your typical American car and power
the world for about two days.
But of course we can't undo this,
that's the important fact concerning power.
Only a small fraction of the mass
of the uranium in a nuclear reactor
is actually converted as power, a fraction of a percent.
So you don't get nearly as much power
as this calculation would indicate,
but in principle this much energy
is contained in the matter that we have around us all the time.
OK.
I want to introduce a few formulas which
we'll be using sooner or later concerning
the relativistic treatment of momentum and energy,
which is what we're getting into here.
And we're not trying to derive relativity in this course,
so I'm just trying to quote the results that we'll be needing.
So, it is useful to introduce an energy momentum four-vector,
which has a 0-th component and an i-th component,
where i refers to the spatial indices 1, 2, and 3.
And sometimes I might write this as p zero p
with a vector sign over it, where the vector sign indicates
the three components 1, 2, 3.
The momentum here is the momentum.
It differs in its relationship to velocity
from what we have from Newton, and I'll
write that in a minute, but this momentum
is the conserved physical momentum of an object.
And p0 is also conserved.
p0 is just an abbreviation for the energy divided by C.
And this quantity forms a four dimensional
vector in special relativity.
And when we say that it's a four-vector,
we're actually making a definite statement
about how it transforms from one Lorentz frame to another.
a four-vector is something which transforms
in exactly the same way as x super mu, four spatial and time
coordinates transform.
And in particular, we learned that there
was an invariant associated with spacetime transformations this
s squared, which was x squared plus y squared plus z squared
minus C squared t squared.
And the same thing will happen here.
p squared, which means the Lorentz invariant
square of the four-vector is the Lorentz [INAUDIBLE] again here,
and it's again equal to the sum of the squares of the 3
spatial components minus the square of the time component.
And that can be written out as the square of the momentum--
spatial momentum-- minus E squared C squared.
And the claim is that this is also Lorentz invariant.
And we could figure out what Lorentz invariant quantity
it's equal to by having knowledge
that this is the same in all frames.
We can evaluate it in the simplest frame.
And the simplest frame would be the rest frame of the object.
In the rest frame of the object the p would be zero
and this would then just be minus E squared over C squared.
E squared would be M squared C to the fourth,
so that implies that this is equal to minus M0 C squared
squared where M0 is often called the rest mass.
And when I say it's often called the the rest mass, what I mean
is that nobody ever mistakes the word "rest
mass" to mean anything else, if anybody says "rest mass,"
and he knows what he's talking about, he means this.
But this is often sometimes just called the mass
because sometimes people only talk about masses
as being rest masses.
But I'll try to call this the rest
mass because I will use the word mass in other ways.
We could also relate this momentum to the velocity,
and in doing that we will again encounter this factor of gamma
that we found kinematically earlier.
AUDIENCE: Yes, question.
Did you forget to divide by C squared there?
PROFESSOR: Um, yes.
There's too many C's here.
Absolutely.
It should have units of momentum.
So it should have units of mass times velocity squared.
Thank you.
So, the quantity gamma, is the same quantity
we encounter at the beginning of the course
when we talked about time dilation and Lorentz
contraction, it just depends on the velocity and approaches
infinity as the velocity approaches the speed of light.
And the physical momentum of a particle, relativistically
is equal to gamma times M sub 0 times V.
Where V is the ordinary velocity-- this
is special relativity, we're not concerned
with coordinate velocity versus physical velocity yet--
and gamma is that factor.
So the momentum is larger than what you would get a la Newton.
The energy can also be written down.
And this formula is one expression
we can use to find the energy in terms of the momentum.
The energy in terms of the momentum
is M0 C squared squared plus p squared C squared.
And it also can be written in terms of the velocity
as just gamma times M0 C squared where the velocity appears
in the gamma.
And a special case of this is when the particle is at rest.
Might as well write this, the energy of a particle at rest,
which we might call E0, is just M0 C squared, which gets us
back to where we started with E equals MC squared.
Now, I might just say a quick word
about where these formulas come from, what idea underlies them.
I'm not going to make any attempt
to derive them because we just don't have time.
It'd be easy to drive them, but we want to do other things.
But logically, where they come from
is simply the observation, by Einstein originally,
that if one has the Lorentz transformations relating
what one inertial observer sees to another inertial observer,
if one used those transformations but used
the Newtonian definitions of energy and momentum,
then you would find immediately that if energy and momentum
were conserved in one frame you then know how to calculate what
happens in other frames by using the transformations,
you'd find it would not be conserved other frames.
So if the conservation of energy and momentum
are to be a universal principle of physics, which Einstein
wanted to maintain, it would be necessary to redefine
energy and momentum.
Now they're defined in ways so that they approach
the Newtonian values for small velocities,
but for velocities of the order of the speed of light
they're different.
And they have the property-- not completely obvious from what
we wrote, well it is actually completely obvious from what
we wrote-- they have the property
that if it's conserved in on frame,
it's conserved in all frames.
And what makes it obvious-- maybe
the connections between these different formulas
are not obvious-- but I did tell you
that p super mu transforms as a four-vector,
meaning it transforms the same way as x super mu transforms,
which are linear transformations.
And that's enough to guarantee that if p mu is conserved
in one frame, it has to be conserved in all frames
because delta p mu, the change in p mu,
would also be a four-vector, and if a four-vector vanishes
in one frame, it vanishes in all frames.
OK.
Well I wanted to give you sort of a quick example of how
this works in practice.
So I to just talk about the energetics of a hydrogen atom.
A hydrogen atom consists of a proton,
with a mass that we'll call M sub p, and an electron,
with the mass that we'll call M sub
E. And, as, if you imagine starting with the electron
and proton arbitrarily far apart and bring them together,
what you can discover experimentally-- of if you know
quantum mechanics you can calculate
theoretically-- energy is released,
because you're releasing potential energy as you
bring the electron into the atom.
And the amount of energy released
is 13.6 electron volts.
And the important E equals MC squared
implication which I want to point out here
is that loss of energy.
This energy would be extracted from the system
as you made the hydrogen atom.
The fact that you've extracted energy from the system
means that now the system should have less energy
than it had to start with.
Initially it had the rest energy of the proton
and the rest of electronic, M sub p C squared and M sub E C
squared.
Now it has less energy by delta E,
and that means it also has to have less mass.
So the mass of a hydrogen atom is not
the sum of the mass of the proton and electron,
as it would be in Newtonian mechanics,
but is less by an amount proportional
to this energy given off, delta E, 13.6 electron volts.
And just putting in the C squares in I
hope the right places, the mass of a hydrogen atom
will be equal to the mass of a proton
plus the mass of an electron, but then
minus the binding energy expressed in mass units-- delta
E over C squared.
OK.
So I guess, just probably one more topic
I want to talk about in terms of just basic special relativity,
and this actually get's into general relativity.
I wanted to find the relativistic mass
of any system just being its energy divided by C squared.
And this means that the relativistic mass of a particle
increases with its velocity.
The energy of a single moving particle
is gamma times M0 C squared.
That would say the by this definition
the relativistic mass of that particle is gamma times M0.
And I might mention that this concept of relativistic mass
is disparaged in many books on special relativity.
It's certainly a concept that you can do without,
so people who emotionally are bothered by it
can get along without it because it is just
the energy divided by C squared.
And in fact a lot of work in special relativity
is done in units where C is equal to 1
and then it is just the energy.
Especially if you use C equals 1 you
could dispense with this concept of relativistic mass.
We're not going to be using C equals 1,
so the phrase relativistic mass will
allow us to abbreviate E divided by E
squared in a convenient way.
But the important thing is not the definitions,
the important thing is what properties
does this relativistic mass have,
whether or not one chooses to call it relativistic mass
or E divided by C squared.
And it as an important property concerning
the gravitational field created by matter.
Now the gravitational field of a single moving object
is complicated.
If we were talking about, say, a moving star
that was moving at a velocity large enough
so we care about relativity, the way we calculate that actually
as we start with the Schwarzschild metric, which
will describe the metric of the star outside of the matter
if it were stationary.
And then you can just make a transformation
to a moving frame.
You're allowed to use any coordinates
you want in general relativity, so transforming
to coordinates that describe the moving frame is no problem.
But it distorts the field in a complicated way, nonetheless,
a way that you can deal with.
And what you find, of course, is that what
you get would be asymmetric once you transform to the moving
frame, it would show the signs of the velocity
that you used to transform from the original spherically
symmetric Schwarzschild metric to the new frame.
So the bottom line is that the gravitational field of a moving
object is not isotropic, it's more complicated than that,
just as the electric field would be.
But if we have a gas of particles,
which is pretty much what we have in the early universe.
If we have a gas of particles in a box moving every which way,
then if we thought of this box as being an object
that we're only going to look at from the outside--
a black box in the classic use of the phrase
black box-- the mass of the black box
really would just be the sum of the relativistic masses
of the particles.
And the isotropy of the metric that any one particle would
generate would be canceled by averaging or summing
over all the particles going every which direction, because
on average the velocity of particles inside the box is 0.
So this relativistic mass, when you're talking about a gas,
really is the mass per particle.
And if you divide that by the volume
really you do get the mass density,
which is a relevant mass density in terms
of talking about how this matter would
generate gravitational fields.
Yes.
AUDIENCE: So why do some people have
emotional problems with it?
PROFESSOR: I think some people have emotional problems with it
because when one thinks about pedagogy in a course,
for example, one worries about people confusing it
with the rest mass.
I think that's the reason.
And I guess there are other possible sources of confusion,
so your question is a good one.
Another source of confusion is that this mass does not
fit into an F equals ma equation.
So in calling this the mass you might
suggest to students who aren't paying attention to every word
that you say, you might go ahead and put F equals
ma for this mass, that does not work.
So it has some of the properties of a mass,
but not-- by no means all of them.
But in particular for us is important
because we're going to be interested in the gravitational
field of a gas and then it really
is the mass that determines that.
In the more formal language in general relativity,
it's the mass-- it's the energy density, that
appears in the equations that produce gravitational fields,
and then this really is the energy density
except for a factor of C squared.
OK, any questions?
Because now I'm going to leave this formalism
and get into what role this radiation could
play in the early universe.
So now I'd like to talk about radiation in particular,
and for now I mean electromagnetic radiation
just ordinary photons.
And we're not accustomed to thinking of light
as having mass, but we know light
has energy and energy is related to mass by a factor of C
squared.
So we can write down the formula that
says that rho is equal to u divided by C squared.
Where u is the energy density, which
we know electromagnetic fields have.
And rho will be the mass density of radiation.
Now, photons have zero rest mass,
so if we apply for example this formula for a photon
we would set M sub 0 equal to 0.
And we said that photons have zero rest mass what
we mean is that there's no lower limit
to the energy a photon can have.
In general, the rest mass determines
the lowest possible energy a particle could have,
which is when it's at rest.
Photon can never be rest and there's
no lower limit to what its energy can be.
So for photons, M0 is equal to 0 and that
implies that the energy of the photon
is just C times the magnitude of its momentum.
And this formula one can derive just for electromagnetic waves
is purely classical EM-- you don't
have to be talking about photons,
but since it's true for a classical electromagnetic wave
it had better also be true for photons because we think
of this classical electromagnetic waves
as really being made out of photons.
So the energy that exists in the universe
in the form of electromagnetic radiation
will have an energy density, which we know how to calculate.
And we know to calculate the momentum of any given photon.
Now that will average to zero, but nonetheless
if we imagine talking about a box of photons, which
are bouncing off the walls, the fact
that each photon carries momentum
means that there will be a pressure on the walls.
And we will be interested in that pressure,
we'll be calculating a formula for it in a minute.
So, what we now want to talk about
is what happens when we put a photon gas into the universe
and allow the universe to expand.
What happens to the radiation energy density,
or equivalently mass density, as the universe expands?
This turns out to be a very easy question
to answer if we think of the energy
as being made out of photons.
We would get an equivalent identical answer
if we use classical Maxwell's equations
and talked about how energy density is
a [INAUDIBLE] of electromagnetic fields behaved.
It would be more work actually do it that way,
but we would get the same answer.
In terms of photos, we could simply
notice that the number density of photons-- photons
are not going to disappear as the universe expands,
the number density will just keep
the same number of photons, but as the universe expands
those photons will occupy a larger volume.
So it's exactly the same as what we
said about non-relativistic matter,
the number density of photons will fall like 1
over the cube of the scale factor, which just
says that photons are conserved.
And the volume of any region grows like a cubed.
I should also mention that I'm using gamma here.
Gamma of course also sometimes meant 1
over the square-root of 1 minus V squared over C squared,
but besides that use, gamma is also
just a label that means photons.
It comes from the idea of gamma rays,
but it's actually used in this context
for any kind of a photon no matter
what its frequency is, whether it's
a gamma ray or an x-ray or visible light, or infrared.
Yes.
AUDIENCE: Is this assumed for time average?
Because photons can be absorbed, right?
I mean, at least like in a small [? slice ?]
can't there be a decided non-relativistic matter?
PROFESSOR: Is this true on the--?
Well, the validity of this formula-- you're
right this formula's not exact, photons can be absorbed.
But in terms of what happens as the universe evolves,
that's a very, very, very minor process,
especially when we're talking about the early universe
when there isn't really anything around to absorb them.
So, especially for the early universe and even pretty well
today, if we're talking about the cosmic background
radiation, which is the bulk of the photons,
this formula's a very good approximation.
Yes.
AUDIENCE: Even though the photons in the early universe
created a lot of massive particles,
[INAUDIBLE] didn't that affect the expansion?
PROFESSOR: Are you're asking, will the photons
be important if there's a lot of mass of particles?
Is that your question?
AUDIENCE: Would the photons, won't they
decay into matter antimatter?
PROFESSOR: Will the photon decay into matter antimatter?
No, not really.
It is, in principle, possible for two photons
to collide and produce an electron positron pair.
It's actually a rather small cross section for that.
And all of these processes in the early universe
will rapidly reach an equilibrium,
which we'll be talking about more a little later, where
there'll be just as many photons converting into E plus E
minus pairs as there will be E plus E minus pairs colliding
and making photons.
So the early universe is assumed to reach equilibrium
very quickly, and all the description we'll be giving
will be a description of the universe that's
in thermal equilibrium with these processes will
tend to iron out.
We will learn that they don't always
cancel each other because the universe is cooling,
and that means it can't be exactly in thermal equilibrium.
And there are some cases where the effect of that cooling
is significant and we'll be talking about those.
But for the most part, if the cooling is slow,
which it is for the most part compared to other processes,
everything stays in thermal equilibrium.
OK.
Those are good questions.
We've gotten a little ahead of what I wanted to talk about.
So for now I'm just imagining a free photon gas,
which is an excellent approximation
for the early universe.
And those photons just continue to exist
as the universe expands, so their number density
falls off as 1 over a cubed.
But there's another affect that goes
on which is that the photons are redshifting.
And we already know about that, but now we're
going to take it into account in terms of the energy balances.
So we know that the frequency of a photon at some time t2
divide by its frequency at some time t1,
and here nu equals frequency, is just
diluted by the expansion of the universe.
This ratio is 1 over 1 plus z, where
z is the redshift between these two times.
But written out in more detail which
is a formula we'll actually be using,
is just a of t1 divided by a of t2.
When the scale factor doubles, all the frequency is half.
And that means that all the photons are lowering
in frequency, and we also know that photons are quantized.
The energy of a photon can't be any old thing,
but in fact, the energy of a photon is equal to h times nu.
Where little h is what's called Planck's constant.
And numerically there are various units
you could express it in, but it's 4.136 times
10 to the minus 15th electron volt seconds.
So if you measure a frequency an inverse seconds,
you get an energy in electron volts from that formula.
The important thing for now though is
that this says that the energy of each photon-- being
proportional to its frequency, and the frequency being
proportional to 1 over the scale factor--
the energy of each photons is proportional to 1 over a of t.
And then the total energy density
of photons in photon on gas which
I'll call u sub gamma, the energy density of the gas,
can be thought of as the number density of photons
times the energy of each photon.
And the number density is falling off
like 1 over a cubed, the energy is falling off like 1 over a.
And therefore, this is proportional to 1
over a to the fourth.
So as the universe expands, the density
of non-relativistic matter falls off like 1 over a cubed--
as we've been talking about some time now--
but the energy of radiation falls off faster,
like 1 over a to the fourth because
of the red shifting of the photons.
OK, now once we know this, we can ask ourselves
what happens if we look at our universe going backwards,
knowing where we are now where we come from?
And if the energy density of photons
is falling off faster than the energy density of matter,
it would mean that the ratio is getting smaller
as we go forward in time.
But that of course implies that the ratio
gets larger as we go backwards in time.
So as we go backwards in time, the radiation
becomes more and more important, and there actually
is going to be a time when the radiation will equal the matter
and at earlier times the radiation will dominate.
Today I'll just give you a number for now,
we'll learn later how to calculate it,
but for today the total radiation energy
density in the universe is equal to 7.01 times 10
to the minus 14 joules per meter cubed.
And this actually includes two kinds of radiation,
it includes photons and also neutrinos,
which at least in the early universe
behaved just like radiation.
And we'll be talking more about neutrinos later so don't worry
if you don't have any idea what a neutrino is.
But for now it's just another contribution to the radiation,
and we can measure basically this is all
based on measuring the temperature
of the cosmic microwave background radiation--
and we'll learn later how to make that conversion.
But once you measure the temperature
of the cosmic background radiation
and have a theory about how many neutrinos there should be,
that's actually all theoretical and we'll
talk about that later as well.
One can determine what the energy
density of that radiation is.
And it corresponds to a mass density just dividing by C
squared of 7.80 times 10 to the minus 31 kilograms per meter
cubed.
And when I think of mass densities
I always like to think of in centimeters per-- excuse me,
grams per centimeter cubed because I'm
used to the density of water being one gram per centimeter
cubed and I like to be able to make that comparison.
So just making that conversion, usually I use SI units,
but some things just seem to make more sense in other units.
So it's 10 to the minus 34 grams per centimeter cubed,
so 10 to the minus 34, or maybe 10 to the minus 33,
times the density of water.
And this is incredibly low even compared
to the critical density of our universe,
and the actual density we know is
very near this critical density.
Let me remind you that the critical density
we derived a formula for, and when we put numbers
into that formula we found that was
equal to 1.88 times little h0 squared,
which is Hubble's units-- Hubble's constant in units
of 100 kilometers per second per megaparsec--
I'll write that in a second-- times 10
to the minus 29 grams per centimeter cubed.
So just writing down the equation for little h sub 0
is where capital H sub 0, Hubble's expansion rate,
is equal to 100 times little h sub 0
kilometers per second per megaparsec.
Yep.
AUDIENCE: What's the, I guess, what's
the motivation of normalizing the Hubble
constant in this way?
PROFESSOR: Well, I think the real motivation is
that the astronomers like these peculiar units of kilometers
per second per megaparsec, but if your favorite unit is called
a kilometer per second per megaparsec
you don't want to have to say those units very often.
So H0 is dimensionless, so it's a dimensionless way
of talking about the Hubble constant.
But that's the only importance, there's
no real-- no deep significance to it.
But it's a standard notation, so it's worth knowing.
And then finally, we can write down
how much the radiation contributes to omega-- omega
sub r, r is going to indicate radiation.
The notation, by the way, will be--
and I realized I've already violated that notation.
This really should have been r.
Well, I'm not going to change it, it'll get too messy.
But I'm going to start using the notation where gamma indicates
photons and little r indicates radiation.
And the difference is that there are other kinds of radiation
besides photons, in particular we've already
added in neutrinos in part of what we're calling radiation.
So omega sub r, which now includes photons and neutrinos,
is just defined to be the mass density radiation divided
by the critical density.
And that turns out, when you combine these numbers,
to be 4.15 times 10 to the minus 5 little h0 to the minus 2.
And then for h0 equals 0.67, which is the Planck satellite
value for little h0, we finally get
omega sub r is equal to 9.2 times 10 to the minus 5.
So roughly 10 to the minus 4.
The fraction of the mass density, or energy density
today is about 10 to the minus 4 fraction and radiation.
And actually as I write this, this actually
calls to mind another reason for defining little h sub 0, which
is that if you write formulas in terms of little h sub 0,
they remain valid between one year and next year.
The observational value of the Hubble parameter
is still floating around and differs,
for example, every time I teach this course.
So the formulas in terms of little h sub 0
stay, and then you plug in the current value of little h sub 0
to get the best value that one can currently write down.
Now we know the Hubble constant to within a few percent, which
is much better than it used to be, but it's still floating.
The Planck value was somewhat lower than the previous except
a value, which was about 0.70.
OK.
So we know enough information to extrapolate backwards
and calculate when this radiation would've
equaled the energy density of matter.
Because we know how the ratio changes.
It changes by a factor of a, the scale factor,
because the energy density of non-relativistic matter
is falling off like 1 over a cubed.
The energy density of radiation is flowing off like 1
over a to the fourth.
So the ratio between them just changes
by a factor of a, decreasing as a gets larger.
So we can write RHO radiation of t divided by RHO matter of t.
The m there means non-relativistic matter.
This is just equal to the current value.
And the current value is gotten by taking that number
and, where I forgot to write down is that omega matter today
is about 0.30.
So the ratio of the two, omega radiation over omega matter,
which is the same as RHO radiation over RHO matter,
is about 3.1 times 10 to the minus 4.
And that number is about to appear in this equation.
If we want the ratio as a function of time,
we start with this value today, 3.1 times 10 to the minus 4,
and then we can just multiply that by the scale factor
today divided by the scale factor at the time
that we want to know it, because we know it falls off as 1
over the scale factor.
And by putting an a of t0 here, this just
guarantees that if we let t equal t0,
we get this number, which is the right ratio for today.
OK.
Now it's just a matter of arithmetic.
We also know that for a matter-dominated universe.
And for now, we're going to estimate when the radiation
energy will equal the matter density.
This will only be an estimate.
We're going to estimate it by assuming
that we can approximate the universe as matter dominated
between now, all the way back until that time.
There are two errors in that calculation.
We're not taking into account here the era of acceleration
where dark energy is playing a significant role.
We'll learn later how to do that.
And also, as we approach this time when they're
making equal contributions, we will run into a regime
where the contribution of the radiation
itself will be relevant.
So this is only an estimate.
But what we're going to do is we're
going to assume that we can treat this
as a matter-dominated universe with a of t proportional
to the 2/3.
And then we can plug numbers into this formula
and ask, when was the ratio 1?
And when the ratio was 1, we call
that the time of equality, using the subscript EQ for equality
to indicate anything having to do with that crossing point.
And what we find is that the z of equality,
and z is just the ratio of the a's, is-- according
to this calculation, it would be 3.1 times 10
to the minus 4 minus 1.
I fibbed when I said that z is the ratio of the a's.
It's offset by a little bit.
It's 1 plus z that's the ratio of the a's.
And that's why there's a minus 1 there.
And numerically, this is about equal to 32,000.
So if we look back in the history of the universe,
we can define looking back in terms of the redshift.
If you look back to a redshift of 3,200, we get to the time
when matter and radiation had the same energy density.
And we can know what time that is if we assume t
to the 2/3, which again is only a crude approximation.
We don't necessarily expect to get the right answer here.
But we expect to get the right order of magnitude.
So t-equality, according to this situation,
would be about 75,000 years after the Big Bang-- just
converting the scale factor that we just
calculated to a time using that formula, treating this as t
to the 2/3.
So this says that about 75,000 years after the Big
Bang, the energy densities of matter and radiation
were equal.
In the earlier times, the universe
was radiation dominated.
The radiation exceeded the matter in its energy density.
Yes?
AUDIENCE: It seems like zEQ is relatively about 31
[INAUDIBLE].
PROFESSOR: You might be right if we would just
look to these formulas.
When I calculated this at home, I
kept more decimal places all the way
and rounded off each answer to one significant figure.
And that's not the same as taking the answer to one
significant figure and calculating
and then rounding off to ones in every figure.
So I think there's always an ambiguity of, roughly speaking,
1 in the last decimal place whenever
you're rounding numbers off.
AUDIENCE: So we just plug that [INAUDIBLE]
PROFESSOR: 3,220-- starting with this
or starting with a more accurate number than the 3.1.
AUDIENCE: Just 1/3.1.
PROFESSOR: 1/3.1.
OK.
So, OK?
AUDIENCE: Sorry.
AUDIENCE: Wait no, [INAUDIBLE] like we have 1
over 3 times 10 negative 4.
That's 3 times 10--
PROFESSOR: It's 3.1.
AUDIENCE: You have 1 over 3.1.
PROFESSOR: 1 over 3.1.
So you have to divide.
AUDIENCE:I do that all the time.
AUDIENCE: [INAUDIBLE]
PROFESSOR: So apparently it's even right
if you just calculate with that.
But there is actually some ambiguity.
The numbers I'm giving you probably
have some uncertainty of 1 in the last digit,
depending on how you calculate.
But the number they give you, I think
they're the ones that you get if you start
using this and the 0.67 and from then on do everything
to large numbers in decimal places
and round off at each stage.
You'll get the numbers I've given you.
In any case, all these are really, at best,
order of magnitude estimates.
So worrying about whether or not the last figure is accurate
is not a big deal.
Yes?
AUDIENCE: I don't really understand
how you got t [INAUDIBLE] without telling was a of t dot
is [INAUDIBLE].
PROFESSOR: I'm sorry.
There is actually a piece of information
I used I forgot to write here.
You're absolutely right.
I used t0 is equal to 13.8 times 10 to the nine years.
And then everything can be related to t0
if you know how things are proportional to t.
You're absolute right.
I did not give you all the information
necessary for that calculation.
Now I think I have.
I haven't done the arithmetic for you.
But otherwise it's all there.
OK.
Now I might mention that in Ryden's book,
she does the calculation taking into account everything--
matter, radiation, cosmological constants.
And her number for t-equality is 47,000 years,
which verifies that we have the right order of magnitude.
And actually, the biggest difference
between her number and my number is not
that she's taken into account these more sophisticated
things, but rather that she used a different value
for the Hubble expansion rate than I'm using.
She's using a value that was current at the time she wrote
her book, which was like '72, I think.
h0 equals 0.72 instead of [INAUDIBLE] 0.67.
And that does make a significant difference here.
But either of these numbers are, I think, probably
within the range of uncertainty of when it really happened.
But it's on that scale, on the scale of 50,000 years, 100,000
years, something of that order.
So there was a significantly long period
compared to human lifetime when the universe was
radiation dominated.
But it's a very small fraction of the overall history
of the universe, but nonetheless does
have important features that happened during that time
period.
Now, if we want to understand those features,
we have to understand how a radiation-dominated universe
evolves, which is what we're going to get to next.
OK.
The next little chapter than I'm going
to be talking about-- the dynamics
of a radiation-dominated universe.
This is a chapter that you more or less
get to work out the equations for yourself
on one of the homework problems.
That's part of this week's set.
So I will try here to outline the logic.
But because all the calculations are in the homework,
I will basically skip the calculations themselves,
and let you do them for yourselves
as part of the homework.
But where we start is we have written down Friedman equations
for the matter-dominated case.
And I'll start by reminding us what those were.
And then in addition to these two equations, which
describe our expanding universe, which we've derived sometime
ago for-- and to remind us here, this
is for a matter-dominated universe.
And matter-dominated means non-relativistic matter
dominated.
And going along with these equations,
we also know that RHO of t for non-relativistic matter
falls off like one over a cubed of t.
This can be converted into a differential equation for RHO.
That is, we can calculate Rho dot from this equation.
And the way to see that is probably most easily
to start on a new blackboard and write that equation not as
a proportionality, since it's hard to differentiate
a proportionality, but we can write it
in an arbitrary constant of proportionality.
And then it becomes an equality.
So I'm going to write the equation as RHO of t
is equal to some constant, b, divided by a cubed of t.
And this we know how to differentiate.
We can write Rho dot is equal to minus b over a
to the fourth of t times a dot.
And that is equal to minus 3.
I'm sorry--there's a 3 here.
Minus 3 times a dot over a times the original RHO.
So we can forget the intermediate steps,
and we just arrived at the equation
that RHO dot is equal to minus 3 a dot over a times RHO.
And we can think of that as going along
with equations one and two.
Maybe I'll even give that a number.
Equation three will be RHO dot is
equal to minus 3 a dot over a times RHO.
Now for radiation, there will be a 4 here.
The 4 will arrive the same way as the 3 arrives there.
It's just the power that appeared in the factor of a.
So for radiation, this last formula we know
is going to be modified, which is the key point.
Note that these three formulas are not
independent of each other.
If we know, for example, equation one, which
is an equation for a dot, we could
differentiate that with respect to time
and get an equation for a double dot.
When we do that, everything has to be differentiated.
So it involves differentiating a,
but that just expresses things in terms of derivatives of a.
But the new quantity that gets introduced is RHO.
If we wanted to differentiate this equation with respect
to time, we have to know what RHO dot is.
But we do.
That's what equation three tells us.
So we can differentiate equation one, use equation three,
and we can derive an equation for a double dot.
And if these equations are consistent,
it'd better be equation two.
And it will be.
You can check it.
And actually, I think any two of these equations
can be used to derive the third.
Those equations just are-- really
a set of two independent equations
and one dependent equation.
You can shuffle it any way you want.
But, now what we want to do is to consider
a different kind of matter.
Instead of non-relativistic matter,
we're considering photon matter.
And in particular, we know that it's
going to change equation three.
So for radiation, 3 gets modified
into 3 prime, which is the equation that
says that RHO dot is equal to minus 4 a dot over a times RHO.
So how are we going to fix these equations?
Now they're inconsistent.
If we change three and don't change either one or two,
we know that we're inconsistent, because any two
of those equations can be used to derive the third.
So we're in trouble.
Either equations one or two will also
have to be modified if we're going to modify equation three.
OK.
Before we go on, I'd like to say a little more about why
this equation is different from that equation.
One might think that it should just
be governed by the conservation of energy.
After all, we just write down an equation
for RHO dot-- how energy density changes with time.
Shouldn't conservation of energy determine that?
It does.
But there is an extra element to conservation of energy
that we need to take into account,
and that is the pressure of the gas affects
what happens to its energy as it expands.
So before we get back to the early universe,
I just want to consider a gas in a piston chamber.
And I'm going to let the piston have an area a, and inside
we're going to have a volume v. Just to define our notation.
If we have a gas inside a piston chamber
and let the piston chamber enlarge by pulling out
on the piston, the gas has a pressure, in general,
and that pressure will exert a force on the piston.
So if I allow the piston to move to the right,
that gas will be exerting a force
on the piston in the direction that it's moving.
And that means the gas will be doing work on the piston.
So, by our ordinary notions of Newtonian conservation
of energy, we would know that the gas would lose energy,
and we can even calculate how much energy it loses.
And the formula is easy enough to get in a Newtonian context.
It's just du is equal to minus the pressure
of the gas times the change in volume.
A famous formula.
And this just comes about by saying
that the work that's done is the force times the distance.
The force is the pressure times the area.
And the volume is the area times the distance.
And putting those things together,
you get this formula immediately.
Now this formula is actually much more general
than the quasi derivation that I just showed.
It works no matter what the shape of the gas is.
If you put a gas in any kind of a container
and let that container enlarge, even in an irregular way,
the work that the gas will do will always
be equal to minus the pressure of the gas
times the change in the volume.
We can apply this to the early universe.
It actually works.
The difference between our two cases
is that our non-relativistic matter has no pressure at all.
We're just talking about particles sitting at rest.
They're not bouncing off of any walls.
They're not creating any pressure,
while the photons are moving around all the time.
And if you imagine a box of them,
they'd be hitting against the walls of that box,
exerting a pressure.
And we're now in a position to relate the pressure
to the difference between the 3's and the 4's that
appear in those two equations for RHO dot in the two cases.
To apply this naive idea to a piece of the universe,
we can imagine choosing-- we're going to choose some fixed
volume in our co-moving coordinate system.
So our box, the volume that we're talking about
will actually be expanding with the universe
but be fixed in co-moving coordinates.
And the physical volume therefore of our box
will be a cubed of t times the coordinate volume
of the box-- the volume and not just cubed.
And this volume will be independent of time.
There's time dependents there.
The physical volume of our box will be enlarging.
The total energy in our box, the total gas energy,
which I'll call capital U, will just
be the physical volume times the energy density.
The energy density is energy per physical volume.
And we can now apply this formula
using this U and this v.
And here again is one of these cases where
I'm going to be skipping steps because you're
going to be doing it in detail on the homework.
But by putting these equations together,
what you'll find is that d dt of a cubed times RHO times c
squared-- this is just d dt of a cubed times the energy
density-- basically, the left hand side
of that equation divided by dt and divided by v coordinates--
is equal to minus p times d dt of a cubed.
And this is just the PDV term from the right hand side
of that equation rewritten in terms of the variables.
And you'll be doing this for homework.
I'm just getting straight the factors
to make sure I have them right.
OK.
Reshuffling that equation-- and again, this
is a homework problem-- you can turn that
into an equation for RHO dot.
And what you'll get is minus 3 a dot
over a times RHO plus p over c squared.
And now we can see how our two cases emerge.
If the pressure is 0, we get minus 3 a dot
over 8 times RHO, which is what we
had for non-relativistic matter.
And the photon gas is going to have a pressure,
and we could read off from this formula
to know what the pressure has to be to turn the 3 into a 4.
The pressure has to be a factor of a third.
So you determine for this that the pressure for light
is 1/3 of the energy density, or 1/3 times RHO c squared.
And that's what you need to turn the 3 into a 4.
So we now have indirectly calculated the pressure
of light, and this agrees with any other calculation
for the pressure of light that you might do.
It's by no means the only way to calculate it.
And now finally, we're in a position--
and we'll just do this quickly to decide
how to modify these equations.
Now, we're not in a position to determine that rigorously.
It can be determined rigorously by doing general relativity,
which we're not doing at that level.
But we can still motivate the answer.
One of these two equations is going
to have to be changed to accommodate
a more general expression for RHO dot.
The top equation we know is really
an equation for conservation of energy.
That's how we got it in the Newtonian case
where little k ended up being partial to the energy
in the Newtonian case.
But this is basically a conservation
of energy equation.
And that's what you expect, just given
your general notion of mechanics as well.
If you have a second order equation,
a second order differential is with respect to time.
That's the force equation.
And if you have a first order differential
equation with respect to time, 1/2 mv squared plus v of r
equals constant.
That's energy conservation.
Same thing here.
And we know that energy cannot suddenly change.
If we imagine-- I guess the first experiment I want to do
is imagining somehow there's an explosion
throughout all of space.
I imagine putting pieces of TNT throughout space
and arranging for, at the same cosmic time, for all of them
to be ignited.
And that would suddenly change the pressure of the universe,
but it would not change the energy density.
The energy density would be conserved.
So the bottom line is that pressures can change
discontinuously, but energy densities cannot.
And since this equation is the conservation of energy
equation, we'd expected that nothing can change suddenly
here, that the pressure term cannot contribute here,
because if it did, the pressure term would changed suddenly.
Nothing else in this equation would change suddenly.
There would be no way the equation could be satisfied.
But if we added a pressure term to the second equation,
that would allow the pressure to change discontinuously
as the TNT went off.
And that would change a double dot discontinuously.
And there's nothing wrong with a double dot changing
discontinuously.
If you suddenly the apply a new force to a particle,
you suddenly change its second derivative of its motion.
You suddenly change its acceleration.
So that's OK.
So any pressure to this term make sense.
Adding pressure to this equation does not make sense.
And then we can just ask, what do you
have to do to this equation if we're going to add a pressure
term to make all three equations now
consistent with the new equation for RHO dot?
It's your homework problem to answer that question,
but the homework tells you the answer,
and I'll write the answer on the board right now,
and then we'll consider today's lecture over.
The bottom line is that equation number one
has to be modified into one prime, which
says that equation number two has to be modified.
What am I talking about?
And the new equation is a double dot
is equal to minus 4 pi over 3 G times RHO
plus 3p over c squared times a.
And now we have a consistent set of Friedman equations,
and these are the Friedman equations
that we would have gotten if we had done everything
using general relativity from the beginning.
And we'll stop there.
And we will meet again next Tuesday.
And I'll send you an email about-- there
will be at least one homework problem on the problems set
that will have to be held over to the following problems set.
I'll send you an email about that
and post it on the website.