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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.
