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PROFESSOR: OK, in that case, let's get going.
I want to begin by just summarizing where we left off
last time, because we're still to some extent in the middle
of the same discussion.
We were talking about the gravitational effects
of a completely homogeneous universe that
fills all of space.
And you recall that Newton had concluded
that such a system would be stable.
But I was arguing that such a system would not be stable,
even given the laws of Newtonian mechanics.
And we discussed a few of those arguments last time.
We discussed, for example, Gauss's law formulation
of Newton's law of gravity.
And I remind you that it's a very simple derivation
to go from Newton's law of gravity,
as Newton stated it as an acceleration at a distance,
to Gauss's law.
What you do is you just show that if the acceleration
of gravity is given by Newton's law, then for any one particle
creating such a gravitational field, Gauss's law holds.
This integral is either equal to 0 or minus 4 pi GM,
depending on whether the surface that you're integrating over
encloses or does not enclose the charge or the mass.
And once you know it for one mass,
Newton tells us that for many masses
you just add the forces as vectors.
And that means that you'll be adding up
these integrals for each of the particles.
And you're lead automatically to the expression we have here.
So really, it follows very directly
from Newton's law of gravity.
On the other hand, if we apply this formulation
to an infinite distribution of mass, if Newton was right
and there were no forces, that would
mean that little g, the acceleration of gravity,
would be 0 everywhere.
And then, the integral on the left would be 0.
But the term on the right is clearly not 0,
if we have a volume with some nontrivial size that
includes some mass.
So this formulation of Newton's law of gravity
clearly shows that an infinite distribution of mass
could not be static.
Further, I showed you there's another formulation, more
modern, of Newton's law of gravity
in terms of what's called Poisson's equation.
This was really shown for the benefit of people who know it.
If you don't it, don't worry about it.
We won't need it.
But it's another way of formulating the law of gravity
by introducing a gravitational potential, phi,
and writing the acceleration of gravity
as minus the gradient of phi.
That just defines phi.
And then, you can show that phi obeys Poisson's equation,
del squared on phi is equal to 4 pi G times
rho, where rho is the mass density.
And again, one can see immediately
that this does not allow a static distribution of mass.
If the distribution of mass was static,
that would mean that g vector was 0.
That would mean that gradient phi was 0.
That would mean that phi was a constant.
And if phi is a constant, del squared phi is 0,
and that's inconsistent with the Poisson equation.
I might further add-- I don't think
I said this last time-- that from a modern perspective,
equations like Poisson's equation
are considered more fundamental than Newton's
original statement of the law of gravity
as an action at a distance.
In particular, when one wants to generalize
Newton's law, for example, to general relativity,
Einstein started with Poisson's equation,
not with the force law at a distance.
And there's nothing like the force law at a distance
and the theory of general relativity.
General relativity is formulated in a language very similar
to the language of Poisson's equation.
The key idea that underlies this distinction
is that all the laws of physics that we know of
can be expressed in a local way.
Poisson's equation is a local equation.
That's just a differential equation
that holds at each point in space
and doesn't say anything about how
something at one point in space affects something far away.
That happens as a consequence of this equation.
But it happens as a consequence of solving the equation.
It's not built into the equation to start with.
Continuing.
We then discussed what goes on if one does just
try to add up the forces using Newton's law
and action at a distance.
And what I argued is that it's a conditionally convergent
integral, which means it's the kind of integral which
has the property that it converges.
But it can converge to different things
depending on what order you add up
the different parts of the integral.
And we considered two possible orderings
for adding up the mass.
In all cases, what we're talking about here
is just a single location, P. And you
can't tell the slide is filled with mass,
but think of that light cyan as mass.
It fills the entire slide and fills the entire universe
in our toy problem here.
So we're interested in calculating
the force on some point, P, in the midst
of an infinite distribution of mass.
And the only thing that we're going to do differently
in these two calculations is we're
going to add up that mass in a different order.
And if we add up the mass ordered
by concentric shells about P, each consecutive shell clearly
contributes 0 to the force at P. And therefore, the sum
in the limit, as we go out to infinity, will still be 0.
So for this case, we get g equals 0.
We do get no acceleration at the point, P,
as long as we add up all the masses in that way.
But there's nothing in Newton's laws that tell us
what order to add up the forces.
Newton just tells you that each mass creates a 1
over r squared force, and it's a vector.
And Newton says to add the vectors.
Normally, addition of vectors commutes.
It doesn't matter what order you add them in.
But what we're going to be finding here
is that it does matter what order.
And therefore, the answer is ambiguous.
And to see this, we'll consider a different ordering.
Instead-- we'll still use spherical shells,
because that's the easiest thing to think about.
We could try to do something else,
but it's much harder to use any other shape.
But this time, we'll consider spherical shells
that are centered around a different point.
And we'll call the point that the spherical shells are
centered around, Q. We're still trying
to calculate the force at the point, P,
due to the infinite mass distribution
that fills space-- so we're doing the same problem we did
before-- but we're going to add up
the contributions in a different order.
And we discussed last time that when we do that,
it turns out that all the mass inside the sphere,
centered at Q out to the radius of P,
contributes to the acceleration at P.
And all mass outside of that, could
be divided into concentric shells, where
the point, P, is inside.
And there's no force inside a concentric shell.
So all of the rest of the mass contributes zilch.
And the answer you get is then simply the answer
that you would have for the force of a point mass located
at Q, whose mass was the total mass in that shaded region.
And clearly it's nonzero.
And furthermore, clearly, we can--
by choosing different points, Q-- make this anything we want.
We can make it bigger by putting the point, Q, further away;
and it always points in the direction of Q. Yeah, points
in the direction of Q. So we can let
it point in any direction we want by putting the point,
Q, anywhere we want.
So depending on how we add up the contributions,
we can get any answer we want.
And that's a fundamental ambiguity
in trying to apply Newton using only the original statement
of the law of gravity as Newton gave it.
OK.
So the conclusion, here, is that the action at a distance
description is simply ambiguous.
Descriptions by Gauss's law, or Poisson's law,
tell us that the system cannot be static.
And we'll soon try to figure out exactly how we
do expect it to behave.
Now, I still want come back to one argument, which was really
the argument that persuaded Newton in the first place.
Newton said that if we want to calculate
the acceleration of a certain point in this infinite mass
distribution, we have a symmetry problem.
All directions from that point looking outward look identical.
If there's going to be an acceleration acting
on any point, what could possibly
determine the direction that the acceleration will have.
So that's the symmetry argument, which is a very sticky one.
It sounds very convincing, in Newton's reasoning.
There could be no acceleration, simply because there's
no preferred direction for the acceleration to point.
To convince Newton that that's not a valid argument
would probably be hard.
And I don't know if we could succeed in convincing him
or not.
Don't get the chance to try.
But if we did have a chance to try,
what we would try to explain to him
is that the acceleration is usually measured relative
to an inertial frame.
That's how Newton always described it.
And to Newton, there was a unique inertial frame-- unique
up to changes in velocity-- determined
by the frame of the fixed stars.
That was the language that Newton used.
And that defined his inertial frame.
And all of his laws of physics were
claimed to hold in this inertial frame.
On the other hand, if all of space is filled with matter
and it's all going to collapse as we're claiming,
there isn't any place to have any fixed stars.
So the whole idea of an inertial frame really disappears.
There's no object, which one can think of as being at rest
or being non-accelerating with respect
to any would-be inertial frame.
So in the absence of an inertial frame,
one really has to admit that all accelerations, just
like all velocities, have to be measured
as relative accelerations.
We could talk about the acceleration
of one mass relative to another.
But we can't talk about what the absolute acceleration is
of a given mass, because we don't
have an inertial frame with which
to compare the acceleration of that object.
So when all accelerations are relative,
then there are more options here.
And it turns out that the right option-- the one that we'll
eventually deduce-- is an option that
looks similar to Hubble's law.
Hubble's law is a law about velocities.
And it says that from the point of view of any observer,
all the other objects will look like they're
moving radially outward from that observer.
And in spite of the fact that that description makes
it sound, emphatically, like the observer you're talking about
is special, you can transform to the frame of any other observer
and thus seeing exactly the same thing.
So having one observer seeing all the other velocities
moving outward from him does not violate homogeneity.
It does not violate any of the symmetries
that we're trying to incorporate in the system.
And the same thing is true for acceleration.
So I'm not going to try to show it now.
We will be showing it in the course of our upcoming
calculations.
But in the kind of collapsing universe
that we're going to be describing,
any observer can consider himself or herself
to be non-accelerating.
And then, that observer would see all the other particles
accelerating directly towards her.
And although that description makes
it sound like the person at the center is special,
it's not true.
You can transform to any other observer's frame.
And each observer can regard himself
as being non-accelerating and would see all the other objects
accelerating radially towards him.
OK.
So now we are ready to go on and try
to build a mathematical model, which will tell us
how a uniform distribution of mass will behave.
Now in doing this, we first would
like to tame the issue of infinities.
And we are going to do that by starting out
with a finite sphere.
And then at the very end, we'll let
the size of that sphere go to infinity.
So our goal is to build a mathematical model of our toy
universe.
And what we want to do is to incorporate
the three features that we discussed,
isotropy, homogeneity, and Hubble's law.
And we're going to build this as a mechanical system,
using the laws of mechanics as we know them.
We're, in fact, going to be using a Newtonian description.
But I will assure you that although we
are using a Newtonian description,
the answer that we'll get will in fact be the exact answer
that we would have gotten with general relativity.
And we'll talk later about why that's the case.
But we're not wasting our time doing
only an approximate calculation.
This actually is a completely valid calculation,
which gives us, in the end, entirely the correct answer.
So we're going to model our universe
by introducing a coordinate system.
I'm now just going to really re-draw
the picture that's up there.
But if I draw it down here, I can point to it better.
We're going to imagine starting at our toy universe
as a finite sized sphere of matter.
And we're going to let t sub i be
equal to the time of the initial picture.
t sub i need not be particularly special in any way,
in terms of the life of our universe.
Once we construct the picture, we'll
be able to calculate how it would behave--
at times later than ti and at times earlier than ti.
ti is just where we are starting.
At time, ti, we will give our sphere a maximum size.
I'm calling it maximum, because I'm
thinking of this as filled with particles.
It's, therefore, the maximum radius for any particle.
It's just the radius of the sphere.
So R sub max i is just the initial radius.
An initial means at time, t sub i.
We're going to fill the sphere with matter.
And we'll think of this matter as being
a kind of a uniform fluid, or a dust of very small particles,
which we can also think of as a fluid.
And it will have a mass density, rho sub i.
OK.
So it's already homogeneous and isotropic, at least isotropic
about the center.
And now, we want to incorporate Hubble's law.
So we're going to start all of this matter, in our toy system
here, expanding and expanding in precisely the pattern
that Hubble's law requires-- namely, all velocities will
be moving out from the center with a magnitude proportional
to the distance.
So if I label a particle by-- I'm sorry,
v sub i is just for initial.
For any particle-- there's no way
we're indicating a particle-- at the initial time,
the velocity will just obey Hubble's law.
It will be equal to some constant, which I'll call H sub
i-- the initial value of the Hubble expansion
rate-- times the vector r, which is just
a vector from the origin to the particle.
That tells us where the particle is that we're talking about.
So v sub i is the initial velocity of any particle.
H sub i is the initial Hubble expansion rate.
And r is the position of the particle.
OK.
As I said, we're starting with a finite system,
which is completely under control.
We know, unambiguously, how to calculate-- in principle
at least-- how that system will evolve,
once we set up these initial conditions.
And at the end of the calculation,
we'll take the limit as R max i goes to infinity.
And that, we hope, will capture the idea
that this model universe is going
to fill the infinite space.
Now, I did want to say a little bit here--
only because it's something which
has entered my scientific life recently
and in interesting ways.
I would say a few things about infinities.
Now, this is an aside, which means
that if you're struggling to understand what's
going in the course, you can ignore this
and don't worry about it.
But if you're interested in thinking about these concepts,
the concept of infinity has sort of struck cosmology in the nose
in the context of the multiverse,
which I spoke of a little bit about in the overview lecture,
and which we'll get back to at the end of the course.
The multiverse has forced us to think much more
about infinities than we had previously.
And in the course of that, I learned some things
about infinity that in fact surprised me.
For the most part, in physics, we
think of infinities as the limit of finite things,
as we're doing over here.
So if we want to discuss the behavior of an infinite space,
we very frequently in physics start
by discussing a finite space, where things are much easier
to control mathematically.
And then we take the limit as the space
gets bigger and bigger.
That works for almost everything we do in physics.
And I would say that the reason why it works
is because we are assuming fundamentally
that physical interactions are local.
Things that are vastly faraway don't affect what happens here.
So as we make this sphere larger and larger,
we'll be adding matter at larger and larger radii.
That new matter that we're adding
is not going to have much affect on what happens inside.
And in fact, for this problem, we'll
soon see that the extra matter we had on the outside
has no effect whatever what happens inside,
related to this fact that the gravitational field
inside a shell of matter is 0.
So that's a typical situation and gives physicist
a very strong motivation to always try
to think of infinities as the limit of finite systems.
What I want to point out, however,
is that that's not always the right thing to do.
And there are cases where it's, in fact, emphatically
the wrong thing to do.
Mathematicians know about this, but physicists tend not to.
So I want to point out that not all infinities are
well-described as limits of finite systems.
I don't want to call this into question.
This, I think, is absolutely solid.
And we will continue with it after I go off
on this side discussion.
But to give you an example of a system which
is infinite and not well-described as a limit
of finite systems-- this is a mathematical example-- we could
just think of the set of positive integers, also
called the natural numbers.
And I'll use the letter that's often used
for it, N with an extra line through it to make
it super bold or something like that.
So that's the set of integers.
And the question is, suppose we consider
trying to describe the set of positive integers,
as a limit-- a finite set.
So we could think about the limit from N
goes to infinity-- as N goes to infinity--
of the natural numbers up to N. So we're
taking sets of integers and taking bigger and bigger sets
and trying to take the limit, and asking,
does that give us the set of integers?
One might think the answer is yes.
What I will claim is that this is definitely not
equal to the set of integers.
And in fact, I'll claim that the limit does not
exist at all, which is why it couldn't possibly
be equal to the set integers.
And to drive this home, I just need
to remind us what a limit means.
And since this is not a math course,
I won't give a rigorous definition.
But I'll just give you an example
that will strike bells and things that you
learned in math classes.
Suppose we want to talk about the limit
as x goes to 0 of sine x over x.
But we all know how to do that.
It's usually called L'Hopital's rule or something.
But you could probably just use the definition of a limit
and get it directly.
That limit is 1.
And what we mean when we say that,
is that as x gets closer and closer to 0,
we could evaluate this expression for any value of x
not equal to 0-- for 0, itself, it's ambiguous.
But for any value of x not equal to 0, we can evaluate this.
And as x gets closer and closer to 0,
that evaluation gives us numbers that
are closer and closer to 1.
And we can get as close to 1 as we
want by choosing x's as close to 0 as necessary.
And that's usually phrased in terms of epsilons and deltas.
But I don't think I need that here.
The point is that, the limit is simply the statement
that this can be made as close to 1
as you like by choosing x as close as possible to the limit
point, 0.
Now, if you imagine applying the same concept
to this set-- the set of integers from 1 to N--
the question is, as you make N large,
does it get closer to the set of all integers?
Are the numbers from 1 to 10 close to the set
of all integers?
No.
What about 1 to a million?
Still infinitely far away.
1 to a billion?
1 to a google?
No matter what number you pick as that upper limit,
you're still infinitely far away from the set of all integers.
You're not coming close.
So there's no sense in which this
converges to the set of all integers.
It's just a different animal.
Now does this make a difference?
Are there any questions where it matters
whether you think of the integers as being defined
in some other way, or by this limit?
Or maybe first say how it is defined.
If you ask mathematicians how you define the set of integers,
I think all of them will tell you, well, we
used the Peano axioms.
And the key thing in the Peano axioms--
if you look at them-- that controls the fact that there's
an infinite number of integers, is the successor axiom.
That is, built into these Peano axioms that
describe the integers mathematically
is a statement that every integer has a successor.
And then, there are other statements
that guarantee that the successor is not
one of the ones that's already on the list.
So you always have a new element as the successor
to your highest element so far.
And that guarantees from the beginning, the set of axioms
is infinite.
It's not thought of as the limit of finite sets and cannot be
thought of as the limit of finite sets.
Because, no finite set is, in any way,
resembling an infinite set.
So does it matter?
Are there questions where we care
whether we could describe the integers this way or not?
And the only questions I know sound kind of contrived,
I'll admit.
But I also point out that in mathematics, the word,
contrived doesn't cut much ice.
If you discover a contradiction in some system,
nobody's going to tell you whether you should ignore
that contradiction because it's contrived.
If it's really a contradiction, it counts.
So a question where it makes a difference,
whether you think of the integers as being defined
as infinite from the start, or whether you think of it--
or try to think of it-- as a limit of this sort,
is a question such as, what fraction of the integers are
so large that they cannot be doubled and still be
an integer?
And notice that, if we consider this set, for any N,
no matter how large N is, we would conclude that half
of the integers are so large that they cannot be doubled.
And that would hold no matter how large we made N.
On the other hand, if we look at the actual set of integers,
we know that any integer can be doubled
and you just get another integer.
So that's an example of a property of the integers, which
you would get wrong if you thought that you could think
of the integers as this kind of a limit.
So you really just can't.
Any questions about that?
Subtle point, I think.
AUDIENCE: How does that relate to--
PROFESSOR: To this?
It's just a warning that you should
be careful about treating infinity
as limits of finite things.
That's all it is.
It does not relate to this.
That's why I said it was an aside when I started.
Aside means something worth knowing but not
directly related to what we're talking about.
OK.
So we're going to proceed with this model.
Let me mention one other feature of the model
to discuss a little bit, the shape that we started with.
We're starting with a sphere.
You might ask, why a sphere?
Well, a sphere is certainly the simplest thing
we could start with.
And a sphere also guarantees isotropy-- or at least
isotropy about the origin.
Isotropy about one point.
We could, by doing significantly more work,
have started, for example, with a cube
and let the cube get bigger and bigger.
And as the cube got bigger and bigger,
it would also fill all space.
And we might think that would be another way
of getting the same answer.
And that would be right.
If we did it with a cube, it would be a lot more work.
But we would, in fact, get the same answer.
The cube has enough symmetry, So it, in fact,
will be the same as sphere in this case.
I'm not going to try to tell you how to calculate it
with an arbitrary shape.
But I will guarantee you that a cube
will get you the same answer.
On the other hand, if we started with a rectangular solid,
where the three sides were different-- or at least not all
equal to each other-- then would be
starting with something which is asymmetric in the first place.
Some direction would be privileged.
And then, if we continued-- as we're
going to be doing for our sphere--
starting with this irregular rectangular solid,
we would build in an anisotropy from the very beginning.
We end up with an anisotropic model of the universe.
So since we're trying to model the real universe, which
is to a high degree isotropic, we
will start with something which guarantees the isotropy.
And the sphere does that.
And it's the simplest shape which does that.
OK.
Now, we're ready to start putting in some dynamics
into this model.
And the dynamics that we're going to put in for now
will just be purely Newtonian dynamics.
And in fact, we'll be modeling the matter
that makes up this sphere as just
a dust of Newtonian particles-- or if you like,
a gas of Newtonian particles.
These particles will be nonrelativistic-- as implied
by the word Newtonian.
And that describes our real universe
for a good chunk of its evolution
but not for all of it.
So let me-- before we proceed-- say a few words
about the real universe and the kind of matter that
has dominated it during different eras of evolution.
In the earliest time, our universe, we believe,
was radiation dominated.
And that means that, if you follow
the evolution of our universe backwards in time,
as one goes to earlier and earlier times, the photons that
make up the cosmic background radiation blue shift.
We've learned that they red shift as universe expands.
That means if we extrapolate backwards, they blue shift.
They get more energetic for every photon.
But the number of photons remains constant, essentially,
as one goes backwards in time.
They just get squeezed into a small volume.
And they become more energetic.
Now, meanwhile, the particles of ordinary matter-- and dark
matter as well, the protons and whatever the dark matter is
made of-- also gets squeezed, as you go backwards in time.
But they don't become more energetic.
They remain just-- a proton remains
a particle whose mass is the mass of a proton times c
squared.
So as you go backwards, the energy density
in the radiation-- in the cosmic microwave background
radiation-- gets to be larger and larger compared
to the energy density in the matter.
And later, we'll learn how to calculate this.
But the two cross at about 50,000 years.
Yes.
AUDIENCE: If we think of particles as waves,
then how does that make sense that they wouldn't change?
PROFESSOR: Right.
If we think of particles like waves,
how does it make sense that they don't change?
The answer is, they do a little bit.
But we're going to be assuming that these particles have
negligible velocities.
What happens is their momentum actually does blue shift.
But a blue shift is proportional to the initial value.
And if the initial value is very small,
even when it blue shifts, the momentum
could still be negligible.
So for the real universe, between time, 0,
and about 50,000 years, the universe
was radiation dominated.
And we'll be talking about that in a few weeks.
But today, we're just ignoring that.
Then, from this time of about 50,000 years
until about 9 billion years-- so a good chunk
of the history of the universe-- the universe
was what is called matter dominated.
And matter means nonrelativistic matter.
That's standard jargon in cosmology.
When we talk about a matter-dominated universe,
and even though we don't use the word nonrelativistic,
that's what everybody means by a matter dominated universe.
And that's the case we're going to be discussing,
just ordinary nonrelativistic matter filling space.
And then for our real universe, something else
happened at about 9 billion years in its history
up to the present-- and presumably in the future
as well-- is the universe became dark energy dominated.
And dark energy is the stuff that
causes the universe to accelerate.
So the universe has been accelerating
since about 9 billion years from the Big Bang.
Now, I think we mentioned-- I'll mention it again quickly--
there's no conversion of ordinary matter to dark energy
here, which you might guess from this change in domination.
It's just a question of how they behave as the universe expands.
Ordinary matter has a density, which decreases as 1
over the cube of the scale factor.
You just have a fixed number of particles scattering out
over a larger and larger volume.
The dark energy-- for reasons that we
will come to near the end of the course,
in fact, does not change its energy
density at all as the universe expands.
So what happened 9 billion years ago
was simply that the density of ordinary matter
fell below the density of dark energy.
And then, the dark energy started to dominate
and the universe started to accelerate.
Today, the dark energy, by the way, is about 60% or 70%
of the total.
So it doesn't dominate completely.
But it's the largest component.
OK.
For today's calculation, we're going
to be focusing on this middle period
and pretending it's the whole story.
And we will come back and discuss these other eras.
We're not going to ignore them.
But we will not discuss them today.
So we will be discussing this case
of a matter-dominated universe.
And we're going to be discussing it using Newtonian mechanics.
And in spite of the fact that we're
using Newtonian mechanics, I will assure you--
and try to give you some arguments later--
it gives exactly the same answer that general relativity gives.
OK.
So how do we proceed?
How do we write down the equations
that describe how this sphere is going to evolve?
We're going to be using a shell picture.
That is, we will describe that sphere
in terms of shells that make it up.
So in other words, we will divide the matter
at the initial time.
And then, we'll just imagine labeling those particles
and following them.
So at the initial time, we're going
to divide the matter into concentric shells.
So each shell extends from some value of r sub i--
the initial radius-- to r sub i plus dr sub i.
Our shells will be infinitesimal in their thickness.
And each shell will have a different r sub i.
And we could let all the dr sub i's be the same, if we want.
We can let each shell be the same thickness.
We can let them be different.
It doesn't matter.
Now a reason why we can get away with thinking of the matter
purely has being described by shells
is that we know that we started with all
of the velocities radial.
That Hubble's law that we assumed
says that the velocities were proportional
to the radius vector, which is measured from the origin.
So all our initial velocities were radial outward.
And furthermore, if we think about how
the Newtonian force of gravity will come about
for this spherical object, they will all be radial as well.
So all the motion will remain radial.
There will never be any forces acting on any of our particles
that will push them in a tangential direction.
So all motion will be radial.
And as long as we keep track of the radius of each particle,
its angular variables, theta and phi--
which I will never mention again--
will just be constant in time.
So we will not need to mention them.
So all motion is radial, because the v's start radial.
And there are no tangential forces,
where tangential means any direction other than radial.
So we're just saying that all the forces are radial.
And that means the motion will stay radial.
To describe the motion, I want to give a little bit more teeth
to this statement that we're going
to be describing it in terms of shells.
We'll going to describe the motion by a function, r, which
will be a function of two variables, r sub i and t.
And this is just the radius at time, t,
of the shell that was at radius, r sub i, at time, t sub i.
So each shell is labeled by where it is at time, t sub i.
And once we label it, it keeps that label.
But then, we'll talk about this function, r, which tells us
where it is at any later time.
And we could also think about earlier times if we want.
So r of r sub i, t is the radius of the shell that started at r
sub i, where we're looking at the radius at time, t.
Now, I should warn you that if you look in any textbook,
you will see a simpler derivation
than what I'm about to show you.
So you might wonder why am I going to so much trouble
when there's an easier way to do it.
And the answer is that the calculation I
show you will, in fact, show you more than what's
in the textbooks-- and Ryden's textbook, for example.
What most textbooks do-- including Ryden,
I believe-- is to assume that this motion will continue
to obey Hubble's law and continue
to be completely uniform in the density.
And then, you could just calculate
what happens to the outside of the sphere.
And that governs everything, if you
assume that will stay uniform.
We are not going to assume that it will stay uniform.
We will show that it will stay uniform.
And from my point of view, it's a lot better
to actually show something than to just guess
it and we write about, without showing it.
So that's why we're going to do some extra work.
We will actually show that the uniformity, once you put in,
is preserved under time, under the laws of Newtonian
evolution.
I guess I'll leave the picture.
Maybe, I'll leave all of that.
OK.
Now, there's an issue that's a little bit complicated
that I'll try to describe.
And again, this is a subtlety that's
probably not mentioned in the books.
We have these different shells that are evolving.
And we can calculate the force on any given shell
if we know the matter that's inside that shell.
That's what it tells us.
If everything is in shells, the shells
have no effect on the matter that is inside the shell.
So if we want to know the effect on a given shell,
we only need to know the matter that's inside that shell.
The matter outside has no force.
So it's very important that we know
the ordering of these shells.
Now, initially, we certainly do.
They're just ordered according to r sub i.
But once they start to move around,
there's, in principle, the possibility
that shells could cross.
And if shells crossed, our equations of emotion
would have to change, because then different amounts of mass
would be acting on different shells.
And we'd have to take that into account.
Fortunately, that turns out not to be a problem.
It just sounds like it could be a problem.
And the way we're going to treat it
is to recognize that, initially, all these shells are moving
away from each other, because of the Hubble expansion.
If Hubble's law holds, any two particles
are moving away from each other with a relative velocity
proportional to their distance.
So that holds for any two shells, as well.
So if shells are going cross, they're
certainly not going to cross immediately.
There are no two shells that are approaching
each other initially.
All shells are moving apart initially.
This could be turned around by the forces--
and we'll have to see.
But what we can do is, we can write down equations
that we know will hold, at least until the time where there
might be some shell crossings.
That is, we'll write down equations
that will hold as long as there are no shell crossings.
Then, if there was going to be a shell crossing,
the equations we write down would
have to be valid right up until the time
of that shell crossing.
And, therefore, the equations that we're writing down
would have to show us that the shells are going to cross.
The shells can't just decide to cross independently
of our equations of motion.
And what we'll find is that our equations will tell us
that there will be no shell crossings.
And the equations are valid as long
as there are no shell crossings.
And I think if you think about that, that's
an airtight argument, even though we're never
going to write down equations that will tell us
what would happen if shells did cross.
OK.
So we're going to write equations
that hold as long as there are no shell crossings.
OK.
As long as there's no shell crossings,
then the total mass inside of any shell
is independent of time.
It's just the shells that are inside it.
So the shell at initial radius, r sub i, even at later times,
feels the force of the mass inside.
And we can write down a formula for that mass inside.
The mass inside the shell, whose initial radius is r sub i,
is just equal to the initial volume of that sphere, which
is 4 pi over 3 r sub i cubed, times the initial mass
density, rho sub i.
So that's how much mass there is inside
a given shell when the system starts.
And that will continue to be exactly how much mass is inside
that shell as the system evolves,
unless there are shell crossings.
And our goal is simply to write down equations that are valid
until there might be a shell crossing.
And then, we can ask whether or not
there will be any shell crossings.
OK, now, Newton's law tells us how
to write down the acceleration of an arbitrary
particle in this system.
Newton's law tells us the acceleration
will be proportional to negative-- I'll
put a radial vector at the end in r hat.
So it's direction, minus r hat, will
be the direction enforced on any particle.
And the magnitude, it will be Newton's constant times
the mass enclosed in the sphere of radius, r sub i,
divided by the square of the distance of that shell
from the origin.
And that's exactly what we called this function, r of ri,
t.
It's the radius of the shell at any given time.
So it's r squared of r sub i and t.
And then as I promised you, there's a unit vector, r hat,
at the end.
So the force is-- and the acceleration, therefore,--
is in the negative r hat direction.
So this holds for any shell, whether which
shell we're talking about is indicated by this variable,
r sub i.
r sub i tells us the initial position of that shell.
So is everybody happy with this equation?
This is really the crucial thing.
Once you write this down, everything else
is really just chugging along.
So everybody happy with that?
It's just the statement from Newton,
that if all the masses are ranged spherically
symmetrically, the mass inside any shell-- excuse me,
the force due to the mass that's on a shell that's at a larger
radius than you are, produces no acceleration for you.
The only acceleration you feel is
due to the masses at smaller radii.
And that's what this formula says.
OK, good.
Now, what I want to do is-- this is a vector equation--
but we know that we're just taking these spherical motions.
And all we really have to do is keep
track of how r changes with time, little r.
So we can turn this into an ordinary differential equation
with no vectors for this function, little r.
The acceleration has a magnitude, which is just,
r double dot.
And that is then equal to minus--
and we write again what M of ri is from this formula.
So I'm going to write this as minus 4 pi over 3 G times r sub
i cubed rho sub i-- coming from that formula-- divided
by r squared.
And I'm going to stop writing the arguments.
But this r is the function of r sub i and t.
But I will stop writing its arguments.
And the two dots means derivative with respect
to time.
Yes.
AUDIENCE: I guess I don't understand how come we would,
calculating the mass, we use r sub i, which
is the radius at some initial time, ti.
But then, when we're saying the distance
from that mass to our point, we use this function, r,
instead of r sub i.
PROFESSOR: Right.
An important question.
The reason is that we're using the initial radius,
but we're also using the initial density.
So this formula certainly gives us
the mass that's inside that circle at the beginning.
And then, if there's no shell crossings,
all these circles move together.
The mass inside never changes.
So it's the same as it was then.
While, the distance from the center of that mass
does change, as the radius changes.
So the denominator changes, the numerator does not.
Yes.
AUDIENCE: So, the first equation, g, that's
the gravity--
PROFESSOR: g is the acceleration of gravity.
AUDIENCE: So, why does it have-- it has units of-- no,
it doesn't.
Sorry.
PROFESSOR: Yes, it should have units of acceleration,
as our double dot should have units of acceleration.
Hopefully, they do.
OK.
Now, as the system evolves, r sub
i is just a constant-- different for each shell,
but a constant in time.
And imagine solving this one shell at a time.
Rho sub i is, again, a constant.
It's the value of the mass density at the initial time.
So it's going to keep its value.
So the differential equation involves a differential
equation in which little r changes with time, and nothing
else here does.
So this is just a second order differential equation
for little r.
So, we're well on the way to having a solution,
because second order differential equations
are, in principle, solvable.
But the one thing that you should all
remember about second order differential equations
is that in order to have a unique solution,
you need initial conditions.
And if it's a second order equation--
as Newton's equations always are-- you
need to be able to specify the initial position
and the initial velocity before the second order
equation leads to a unique answer.
So that's what we want to do next.
We want to write down the initial value
of r and the initial value of r dot.
And then, we'll have a system, which
is just something we can turn over to a mathematician.
And if the mathematician is at all smart,
he'll be able to solve it.
So initial conditions.
So we want the initial value of r sub i--
and initial means a time, t sub i.
That's where we're setting our initial conditions.
So we need to know what r of r sub i,
t sub i is, which is a trivial thing
if you keep track of what this notation means.
What is that?
r sub i, good.
The meaning of this is just the radius at time, t sub i,
of the particle, whose radius at time, t sub i, was r sub i.
So to put together all those tautologies,
the answer is just, obviously, r sub i.
OK.
And then we also want to be able to solve this equation
and have a unique solution-- we want the initial value of r
sub i dot.
And initial means, again, at time, t sub i.
And that's specified by our Hubble expansion.
Every initial velocity is just H sub i times the radius.
So if the radius is r sub i, this
is just H sub i times r sub i.
That's just the Hubble velocity that we put in to the system
when we started it.
OK.
So now, we have a system, which is purely mathematical.
We have a second order differential equation
and initial conditions on r and r dot.
That leads to a unique solution.
Now it's pure math.
No more physics, at least at this stage of the game.
However, there are interesting things
mathematically that one can notice
about this system of equations.
And now what we're going to see is the magic
of these equations for preserving
the uniformity of this system.
It's all built into these equations.
And let's see how.
The key feature that's somewhat miraculous
that these equations have is that r
sub i can be made to disappear by a change of variables.
I'll tell you what that is in a second.
It also helps if you know how to spell "disappear."
We're going to define a new function, u.
I just made up a letter there.
Could have called it anything.
But u of r sub i and t is just going
to be defined to be r of r sub i and t divided by r sub i.
Given any function, r of r sub i and t,
I can always define a new function,
which is just the same function divided by r sub i.
Now, let's look at what this does to our equations.
My claim is that r sub i was going to disappear.
And now, we'll see how that happens.
OK, if u is defined this way, we can write down the differential
equation that it will obey, by writing down
an equation for u double dot.
And u double dot will just be r double dot divided by r sub i.
So we take the equation for r double dot over here and divide
it by r sub i .
So we can write that as minus 4 pi over 3 G-- for some reason,
I decided to leave the r sub i cubed
in the numerator for now-- and just put an extra r sub
i in the denominator-- the one that we divided by-- times r
squared.
And now, what I'm going to do is just replace r .
We're trying to write an equation for u.
r is related to u by this equation. r is r sub i times u.
So I can write this as minus 4 pi over 3 G r sub i
cubed rho sub i over-- now we have-- u
squared times r sub i cubed.
I left the numerator alone.
I just rewrote the denominator by placing
r by r sub i times u.
And we get this formula.
And now, of course, the r sub i cubes cancel.
And I think the reason why I kept it separated this way
when I wrote my notes is at this shows very
explicitly that what you have is a cancellation between an r
sub i here, which was the power of r that appears in a volume--
r sub i cubed was just proportional to the volume
of the sphere-- and r sub i cubed down here,
where one of them came from just a change of variables,
and the other was the r squared that appeared in the force law.
So this cancellation depends crucially on the force law,
being a 1 over r squared force law.
If we had 1 over some other power of r--
even if it differed by just a little bit-- then
the r sub i would not drop out of this formula.
And we'll see in a minute that it's the dropping out of r sub
i which is crucial to the maintenance of homogeneity
as the system evolves.
As it evolves, Newton tells us what happens.
We don't get to make any further choices.
And if Newton is telling us there's a 1
over r squared force law, then it remains homogeneous.
But otherwise, it would not.
And that's, I think, a very interesting fact.
Continuing, we've now crossed off these r sub i's.
And we get now a simple equation,
u double dot is equal to minus 4 pi over 3 G rho sub i
over u squared with no r sub i's in the equation anymore.
And that means that this u tells us
the solution for every r sub i.
We don't have different solutions
for every value of r sub i anymore.
r sub i has dropped out of the problem.
We have a unique solution independent of r sub i.
And that means it holds for all r sub i.
I'm sorry.
I jumped the gun.
The conclusions I just told you about are true,
but part of the logic I haven't done yet.
What part of the logic did I leave out?
Initial conditions, I heard somebody mumble.
Yes.
Exactly.
To get the same solution, we have
to not only have a differential equation that's
independent of u sub i, but we don't have a unique solution
unless we look at the initial conditions.
We better see if the initial conditions depend on u sub i.
And they don't either.
That's the beauty of it all.
Starting with the r initial condition, the initial value
of u of r sub i, t sub i, will just
be equal to the initial value of r divided by r sub i.
But the initial value of r is r sub i.
So here, we get r sub i divided by r sub i, which
is 1, independent of r sub i.
The initial value of u is 1, for any r sub i.
And similarly, we now want to look at u dot of r sub i and t.
And that will just be r dot divided by r sub i, where
we take the initial value of r dot.
The initial value of r dot is Hi times r sub i.
The r sub i's cancel.
And u sub i dot is just H sub i.
So now, we have justified the claim
that I falsely made a little bit prematurely a minute ago.
We have a system of equations, which
can be solved to give us a solution for u, which
is independent of r sub i, which means every value of r sub i
has the same equation for u.
That-- if you stare at it a little bit--
gives us a physical interpretation
for this quantity, u.
u, in fact, is nothing more nor less than the scale factor
that we spoke about earlier.
We have found that we do have a homogeneously expanding system.
We started it homogeneously expanding,
but we didn't know until we looked at equation of motion
that it would continue to homogeneously expand.
But it does.
And that means it can be described by a scale factor.
We have u of r sub i, t.
In principle, it was defined so it
might have depended on r sub i.
That's why I've been writing these r sub i's all along.
But now, we discovered that the u's are completely
determined by these equations, which
have no r sub i's in them-- at least none that survived.
We had an r sub i divided by r sub it,
but that's not really an r sub i, as you know.
It's 1.
So u is independent of r sub i and can
be thought of as just a function of t.
And we can also change its name to a of t
to make contact with the notion of a scale factor
that we discussed last time.
And we can see that the way u arises,
in terms of how it describes the motion, r of r sub i
and t is just equal to-- from this equation-- r sub
i times u, which is now called a.
So r, of r sub i and t, is equal to a of t times r sub i.
That's just another way of rewriting our definition
that we started with of u.
So what does that mean?
These r sub i's are comoving coordinates.
We've labeled each shell by its initial position, r sub i.
And we've let that shell keep its label, r sub i,
as it evolved.
That's a comoving coordinate.
It labels the particles independent of where they move.
And what this is telling us is that the physical distance--
from the origin in this case-- is
equal to the scale factor times the coordinate distance.
That is the coordinate distance between the shell
and the center of the system, the origin.
OK.
Any questions about that?
OK.
It's now useful to rewrite these equations
in a few different ways.
Let's see.
First of all, we have written our differential equation,
up here, in terms of rho sub i.
And that was very useful because rho sub i is a constant.
It doesn't change with time.
Still, it's sometimes useful to write the differential equation
in terms of the temporary value of rho, which
changes with time, to see what the relationships are
between physical quantities at a given time.
And we could certainly do that, because we
know what the density will be at any given time.
For any shell, we can calculate the density
as the total mass divided by the volume.
We know this is going to stay uniform, because everything
is moving together, when you just have motion, which
is an overall scale factor that multiplies all distances.
So the density will be constant.
And we can calculate the density inside a shell
by taking M of r sub i-- which we already have formulas
for and is independent of time-- and dividing it
by the volume inside the shell, which is 4 pi r cubed.
And doing some substitutions, M of r sub i
is just the initial volume times the initial density.
So that is-- as we've written before-- 4
pi over 3 r sub i cubed times rho sub i.
And then, we can write the denominator.
r is equal to a times r sub i.
The physical radius is the scale factor
times the coordinate radius.
So here, we have a cubed times r sub i cubed.
And now, notice that almost everything cancels.
And what we're left with is just rho sub i divided by a cubed.
So that's certainly what we would've guess, I think.
The density is just what it started but then
divided by the cube of the scale factor.
And our scale factor is defined as 1
at the initial time-- the way we've set up these conventions.
So that is just the ratio of the scale factors cubed
that appears in that equation.
As the universe expands, the density
falls off by 1 over the scale factor cubed.
We can also now rewrite the equation for a double dot.
a is u, so we have the equation up there.
But we could write it in terms of the current mass density.
Starting with what we have up there,
it's minus 4 pi over 3 G rho sub i over a squared.
Notice, that I can multiply numerator and denominator by a.
And then, we have rho sub i over a cubed
here, which is just the mass density in any given time.
So I can make that substitution.
And I get the meaningful equation
that a double dot is equal to minus 4 pi over 3 G
rho of t times a.
So this equation gives the deceleration
of our model universe in terms of the current mass density.
And notice that it does, in fact, depend only
on the current mass density.
It predicts the ratio of a double dot over a.
And that, you would expect to be what should be predicted,
because remember, a is still measured in notches per meter.
So the only way the notches can cancel out
is if there is an a on both sides.
Or if you could bring them all to one side
and have a double dot divided by a.
And then, the notches go away.
And you have something which has physical units being related
to something with physical units.
OK.
We said at the beginning that when we were finished,
we were going to take the limit as R max initial goes
to infinity.
And lots of times when I present this,
I forget to talk about that.
And the reason I forget to talk about that is, if you notice,
R max sub i doesn't appear in any of these equations.
So taking the limit as R max sub i goes to infinity
doesn't actually involve doing anything.
It really just involves pointing out that the answers we got
are independent of how big the sphere is,
as long as everything we want to talk
about fits inside the sphere.
Adding extra matter on the outside
doesn't change anything at all.
So taking the limit as you add an infinite amount of matter
on the outside-- as long as you imagine doing it
in these spherical shells-- is a trivial matter.
So the limit as R max sub i goes to infinity
is done without any work.
OK.
I would like to go ahead now, and in the end,
we're going to want to think about different kinds
of solutions to this equation and what they look like.
For today, I want to take one more step, which
is to rewrite this equation in a slightly different way, which
will help us to see what the solutions look what.
What I want to do is to find a first integral
of this equation.
OK.
To find a first integral, I'm going to go back to the form
that we had on the top there, where everything
is expressed in terms of rho sub i rather than rho.
And the advantage of that for current purposes
is that I really want to look at the time dependence of things.
And rho has its own time dependence,
which I don't want to worry about.
So if I look at the formula in terms of rho sub i,
all time dependence is explicit.
So I'm going to write the differential equation.
It's the top equation in the box there,
but I'm replacing u by a, because we renamed u as a.
And I'm going to put everything on one side.
So I'm going to write a double dot plus 4 pi
over 3 G rho sub i divided by a squared equals 0.
OK, that's our differential equation.
Now, it's a second order differential equation,
like we're very much accustomed to from Newtonian mechanics--
as this is an equation which determines
a double dot, the acceleration of a
in terms of the value of a.
A common thing to make use of in Newtonian mechanics
is conservation of energy.
In this case, I don't know if we should
call this conservation of energy or not.
We'll talk later about what physical significance
the quantities that we're dealing with have.
But certainly, as a mathematical technique,
we can do the same thing that would
have been done if this were a Newtonian mechanics problem,
and somebody asked you to derive the conserved energy.
Now, you might have forgotten how to do that.
But I'll remind you.
To get the conserved energy that goes with this equation,
you put brackets around it.
You could choose whether you want
curly brackets or square brackets or just
ordinary parentheses.
But then, the important thing is that it's
useful to multiply the entire equation
by an integrating factor, a dot.
And once you do that, this entire expression
is a total derivative.
This equation is equivalent to dE-- for some E
that I'll define in a second-- dt
equals zero, where E equals 1/2 a dot squared minus 4 pi over 3
G rho sub i over a.
And you can easily check.
If I differentiate this, I get exactly that equation.
So they're equivalent.
That is, this is equivalent to that.
So E is a conserved quantity.
Now if one wants to relate this to the energy of something,
there are different ways you can do it.
One way to do it is to multiply by an expression, which I'll
write down in a second, and to think of it
as the energy of a test particle at the surface of this sphere.
I'll show you how that works.
I'm going to find something, which
I'll call E sub phys-- or physical-- meaning,
it's the physical energy of this hypothetical test particle.
It's not all that physical.
But it will just be defined to be m r sub i squared times E.
m is the mass of my test particle.
r sub i is the radius of that test particle expressed
in terms of its initial value.
And then, if we write down what E phys looks like,
absorbing these factors, it can be written as 1/2 m times a dot
r sub i squared minus GmM of r sub i divided by a r sub i.
And that's just some algebra-- absorbed these extra factors
that I put into the definition.
And now, if we think of this as describing a test
particle, where r sub i is capital R sub i max-- so we're
talking about the boundary of our sphere-- then,
we can identify what's being conserved here.
What's being conserved here is 1/2 m v squared.
a dot times r sub i would just be the velocity
of the particle of the boundary of the sphere.
And then, minus G times the product of the masses
divided by the distance between the particle in the center.
And that would just be the Newtonian energy-- kinetic
energy plus potential energy, where the potential energy
is negative-- of a point particle on the boundary, where
we let r sub i be capital R sub i
max-- the boundary of the sphere.
Now, if we want to apply this to a particle inside the sphere,
it's a little trickier to get the words right.
If the particle is inside the sphere--
if r sub i is not equal to the max--
this is not really the potential energy of the particle.
Can somebody tell me why not?
Well, maybe the question is a little vague.
But if I did want to calculate the potential energy
of a particle inside the sphere-- that's
meant to be in the interior.
You can't really tell unless it's the actual diagram.
But that's deep inside sphere.
I would do it by integrating from infinity G da,
and ask how much work do I have to do
to bring in a particle from infinity and put it there.
And in doing this line integral, I get a contribution
from the mass that's inside this dot, which
is what determines the force on that dot.
But I also get a contribution from what's outside the dot.
So I don't get-- if I wanted to calculate
the actual potential energy of that point--
I don't get simply Gm times the mass
inside divided by the distance from the center.
It's more complicated what I get.
And in fact, what I get is not conserved.
Why is it not conserved?
I could ask you, but I'll tell you.
We're running out of time.
It's not conserved, because if you
ask for the potential energy of something
in the presence of moving masses,
there's no reason for it to be conserved.
The potential energy for a point particle
moving in the field of static masses is conserved.
That's what you've learned in [? AO1 ?] or whatever.
But if other particles are moving,
the total energy of the full system will be conserved.
But the potential energy of a single particle-- just thought
of as a particle moving in the potential
of the other particles-- will not be conserved.
OK.
What is conserved, besides the energy
of this test particle on the boundary,
is the total energy of this system,
which can also be calculated.
And that, in fact, will be one of your homework problems
for the coming problem set-- to calculate
the total energy of that sphere.
And that will be related to this quantity
with a different constant of proportionality
and will be conserved for the obvious physical reason.
For the particles inside, what one can imagine--
and I'll just say this and then I'll stop--
you can imagine that you know that the motion
of this particle is uninfluenced by these particles outside.
And therefore, you could pretend that they're not there
and think of it as an analog problem, where
the particles outside of the radius of the particle you're
focusing on simply do not exist in this analog problem.
For that analog problem, this would be the potential energy.
And it would be conserved.
And you could argue that way, that you
expect this to be a constant.
And you'd be correct.
But it's a little subtle to understand
exactly what's conserved and why and how to use it.
OK, that's all for today.