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PROFESSOR: OK if there
are no questions,
we will get back to physics.
What I want to do today, as
it suggests on the slide,
is to finish the kinematics
of homogeneous expansion
that we were talking
about last time.
And the one topic in that
category that we have not
discussed yet is the
cosmological redshift
So we'll begin by
going over that.
And then we'll begin to go on
to the next topic altogether,
which is the dynamics of
homogeneous expansion --
how do we understand how
gravity affects the expansion
of the universe?
So that will be the main
subject of today's lecture,
once we've finished up the issue
of the cosmological redshift
shift.
Let me remind you that at
end of the last lecture,
we were talking about the
synchronization of clocks,
and the coordinate
system that we'll
be using to describe the
homogeneously expanding
model of the universe.
Remember, we are introducing
spatial coordinates
that grow with the
universe, so that we're
going to be assuming the fact,
literally, that the universe is
perfectly homogeneous
and isotropic, which
means that all objects will
be literally addressed,
relative to this
coordinate system.
If we're talking about
the real universe,
then there would be
some motion relative
to this coordinate system,
because the universe is not
exactly homogeneous.
But we're going to be working
for now with the approximation
that our model universe is
exactly homogeneous, which
means that all
matter is completely
at rest, relative to this
expanding coordinate system.
And now we want to talk
about how to define time,
or to review what we
said last time when
we talked about
how to define time.
What we will imagine is that in
every location in the universe
at rest, relative to
the matter, is a clock.
And each clock ticks off
time, and all those clocks
will be acceptable
as a clock which
measures the time at
relevant positions-- time
is measured locally--
but we still
have to talk about
synchronizing those clocks.
And what we said last time
is that we can synchronize
the clocks as long as there's
some cosmic phenomena that
can be seen everywhere, which
has some time evolution.
And we gave two examples-- one
is the evolution of the Hubble
expansion rate, which
can be measured locally,
and everybody can agree to set
their clocks to midnight when
the Hubble expansion
rate has a certain value.
And another cosmic
variable is the temperature
of the cosmic microwave
background radiation.
So, everybody in
this model universe
will agree that we'll set
the clocks to midnight when
the temperature of the
cosmic background radiation
goes to 5 degrees, or
any specified number.
So as long as there's a
phenomena of that sort, which
there is in our universe,
it's possible to synchronize
these clocks in a unique way.
And the important
thing to realize
is that once they're
synchronized at one time,
they will remain
synchronized as a consequence
of our assumption
of homogeneity.
That is, if everybody agrees
that the cosmic background
radiation has a temperature
of 10 degrees at midnight,
if everybody waits for 15
minutes after midnight,
everybody should see the
same fall in temperature
during that time
interval, otherwise
it would be a violation
of this hypothesis
of perfect homogeneity.
Yes, question.
AUDIENCE: Is it verified
that temperature
is invariant for all observers--
all Lorentz observers?
PROFESSOR: OK the question
is, is temperature invariant
for all observers?
And the question even included
all Lorentz observers.
It's not really invariant to
different Lorentz observers.
We're talking about a privileged
class of observers, all of whom
are at rest, relative
to the average matter.
If you move through the
cosmic background radiation,
then you don't see uniform
thermal distribution any more.
Rather what you see
is radiation that's
hotter in the forward
direction and colder
in the backward direction.
And we in fact, as I think
I have mentioned here,
see that effect in
our real universe.
We're apparently moving relative
to the cosmic background
radiation, at about 1/1000th
of the speed of light.
So it's not invariant
with respect to motion.
There's the additional
question, though is,
is it the same everywhere
in the visible universe?
As far as we can tell, it is.
There is some direct measurement
of that, that we'll probably
talk about later in
the course, by looking
at certain spectral lines
in distant galaxies.
One can effectively
measure the temperature
of the cosmic
microwave background
radiation in some
distant galaxies.
This line cannot be
seen in all galaxies,
and the extent that it's
been measured in degrees.
So certainly in our
model, we're going
to assume complete homogeneity,
so everything's the same
everywhere, and there is strong
evidence for that homogeneity.
Although it's not
exact, but there's
strong evidence for
approximate homogeneity
in the real universe.
Yes?
AUDIENCE: If you were really
close to the black hole
[INAUDIBLE].
PROFESSOR: OK.
The question is, suppose we're
a little bit more careful,
and talk about the fact that
some people might be living
near black holes, and
other people are not.
Will that affect the
synchronization of clocks
for the people who are
living near black holes?
The answer is sure, it will.
We can only synchronize
clocks cosmically
if we assume that the universe
is absolutely homogeneous.
As soon as you introduce
inhomogeneities
like black holes, or even
just stars like the sun,
they create small
perturbations, which
then make it really
impossible to expect clocks
to stay in sync with each other.
So as soon as you have
concentrations of mass,
then the fact that what
we're talking about now
is an approximation
becomes real.
But those deviations are small.
The deviations
coming from the sun
are only on the order of
a part in a million or so.
So, to a very good
approximation,
the universe obeys
what we're describing,
although if you went very
close to the surface of one
of these super-massive
black holes
in the centers of
galaxies, or something,
you would in fact
find they had a very
significant effect
on your clocks.
Any other questions?
OK.
Let me move on now.
The next topic, as
I have warned you,
is the cosmological redshift.
Now in the first lecture
beyond the overview,
which I guess was a combination
of the second and third
lectures in the course, we
talked about the Doppler shift
for sound waves, and we talked
about the relativistic Doppler
shift for light
waves-- that was all
in the context of
special relativity.
Now what we're going
to face is the fact
that cosmology is not
really governed entirely
by special relativity, although
special relativity still
holds locally in our cosmology.
But special relativity does not
include the effects of gravity,
and on a global scale,
the effects of gravity
are very important
for cosmology,
and therefore special
relativity by itself
is not enough to understand
many properties of the universe,
including the
cosmological redshift.
It turns out
though, that there's
a way of describing the
cosmological redshift which
will make it sound even simpler
than special relativity.
And I'll describe
that first, and then
afterwards, we'll
talk a little bit
about how this very
simple-looking derivation jives
with the special relativity
derivation, which must also
be correct, at least locally.
OK.
So, the question we
want to ask ourselves,
is suppose we look
at a distant galaxy,
and light is emitted
from that galaxy.
How will the frequency
of that light
shift between the frequency
it had when it was emitted,
and the frequency
that we would measure
as we received the light.
So to draw the situation on
the blackboard, let's introduce
a coordinate system, x.
And this will be our
comoving coordinate system.
X is measured in notches.
We'll put ourselves at
the origin-- there is us.
And we'll put our
galaxy out here
someplace-- there is the
distant galaxy that we
will be observing.
They galaxy will be at some
particular coordinate, which
I will call l sub c, c
for coordinate distance,
so l sub c is the coordinate
distance to the galaxy.
And then the physical
distance-- is
what we've been calling
l sub p, p for physical,
which depends on time, because
there's Hubble expansion.
So l sub p of t, as we've said
a number of times already,
is a of t times l sub c.
The scale factor, which
depends on time, times
the coordinate distance,
which does not depend on time.
So everything just expands
with the scale factor a of t.
So this describes the
situation, and now
what we want to
ask ourselves, is
suppose a wave is being
emitted by the galaxy--
and we'll be trying to track
the distance between wave
crests, which determines
what the wavelength is.
Since we'll only be
interested in wave crests,
we will talk in
language where we just
imagine there's a
pulse at each crest,
and what happens
in between doesn't
matter for what
we're talking about.
So we want to track successive
pulses emitted by the galaxy.
Now the important
feature of our system
is that we have
argued that we know
how to track light waves through
this kind of coordinate system.
If x is our cosmic coordinate,
dx dt, the coordinate velocity
of light, is just equal to the
ordinary velocity of light,
c, but rescaled by
the scale factor.
And the scale factor
here is playing
the role of converting
meters to notches.
So c is measured in
meters per second.
By dividing by a of
t, we get the speed
in notches per second,
which is what we want,
because x is measured, not
in meters, but in notches.
A notch being the arbitrary
coordinate-- the arbitrary unit
that we adopt to describe our
comoving coordinate system.
Now the important
feature of this equation,
for our current purpose, is that
the speed of light, as we're
going to follow
these light pulses
through our coordinate
system, depends on time,
but it does not depend on x.
Our universe is homogeneous,
so all points x are the same.
So two pulses will travel at
the same speed at the same time,
no matter where they are.
And that's all we really
need, to understand
the fact that if one
pulse leaves our galaxy
and is coming towards
us-- I should do that
with my right hand, because
the second pulse is going
to be my other hand-- as
that second pulse follows it,
the second pulse,
at any given time--
even though the speed
will change with time,
but at any given
time-- the second pulse
will be traveling at the same
speed as the first pulse.
And that means that it'll
look something like this.
The speed might
change with time,
but as long as they both travel
at exactly the same speed
at any given time, they will
stay exactly the same distance
apart in our comoving
coordinate system.
Delta x, the x distance
between the two pulses,
will not change with time.
And if the coordinate distance
does not change with time--
the physical distance is
always the scale factor
times the coordinate
distance-- it
means that the physical
wavelength of the light pulse
will simply be stretched
with the scale factor, which
means you'll be stretched with
the expansion of the universe,
in exactly the same way as any
other distance in this model
universe will be stretched
as the universe expands.
So that's the key idea,
and it's very simple,
and those words
really say it all.
Delta x equals constant
implies delta l
physical is
proportional to a of t,
and that implies that the
wavelength of the light,
as a function of t, is
proportional to a of t.
Wavelength is actually what I
was calling delta l physical,
the distance between
these two pulses,
where each pulse represents
a crest of the wave.
And lambda is the standard
letter of the wavelength.
Now the wavelength is related
to the period of a wave
simply by the
relationship that lambda
is equal to c times delta t.
Wavelength is just the distance
the wave travels in one period.
So if lambda is
proportional to a of t,
so is the time interval, delta
t, the period of the wave,
going to be
proportional to delta t.
So we have been
defining the redshift
in terms of the period.
So delta t observed over
delta t at the source
is equal to lambda observed
over lambda at the source.
Lambda and delta t are
proportional to each other.
And-- let me finish and
I'll get to you, OK?
AUDIENCE: Yes.
PROFESSOR: This then,
the ratio of the lengths,
is just the amount
by which universe
has stretched over that time.
So just the ratio of the scale
factors at the two times.
So this is equal to just a of
the time of observation, which
I'll call t sub o, over a of
the time of the source, t sub s.
So this is the scale
factor at source,
and the numerator here is the
scale factor at observation.
And this ratio of times,
or ratio of wavelengths,
or ratio of scale factors,
is defined to be 1 plus z,
as we have always done.
The ratio of the time intervals
we had defined originally
as 1 plus z, we'll
keep that definition,
and that defines the
redshift shift, z.
Question now?
Yes.
AUDIENCE: Is that
definition of lambda,
does that have anything to do
with the Lorentz invariant?
Like, it just kind of
struck me as the first term?
PROFESSOR: Not
sure what you mean?
What-- Lorentz invariant what?
AUDIENCE: Like the c delta
tau squared equals c delta t--
PROFESSOR: Oh.
Well, the delta t could
be put into that formula,
but that's formula could
measure any delta t.
AUDIENCE: Yeah
PROFESSOR: So of course
Lorentz is a special case,
but any delta t would be a
special case of that formula,
so I don't think
there's a lot to say
about it being a special case.
AUDIENCE: All right, cool.
PROFESSOR: Any other questions?
Yes?
AUDIENCE: Is this like
fundamentally different?
Or is it similar [INAUDIBLE]?
PROFESSOR: [INAUDIBLE] I
was going to come to that.
That's the question of how the
cosmological redshift relates
to the special
relativity redshift
that we derived earlier, and
I'm coming to that immediately.
Good question,
we're getting there.
Any other questions,
though, before I go there?
In my point of view,
that's the next topic.
OK.
OK, so let me move on to
exactly that question.
How does this relate
to what we already
said about the redshift?
This answer-- I would like
to quantify things and say
that it differs in two
ways from the calculation
that we've done previously.
And the first is-- the
reason why it's important
to us-- is that this actually
takes into account, effects
which were not taken into
account by our earlier
calculation.
In particular, even
though we derived this
by a very simple
kinematic argument, which
didn't seem to involve
much math at all,
it actually is
incredibly strong,
in that it encompasses not only
special relativity, but also
general relativity.
It includes all the
effects of gravity.
If you think about
what gravity might
do to what we're
talking about, gravity
doesn't change the fact
that the speed of light
is going to be c over a of t.
That really is just
a unit conversion,
combined with the fundamental
physics assumption
that the speed of
light is always
measured at c, relative
to any observer.
So when we put in
gravity, this relationship
continues to hold--
that was really
all we used to drive
this-- so gravity is not
going to affect the answer.
If you think about
special relativity,
is there something left out?
Everything I said
here, Newton would
have understood perfectly.
I didn't have to mention
time dilation, which
was crucial to our special
relativity calculation
of the redshift shift.
Did I make a mistake?
Is there some place where time
dilation should come in here?
The answer, really, is
no, if you think about it.
We had two clocks
involved in our system,
a clock on the galaxy,
and a clock at us,
which we used to measure
the period of emission,
and the period of reception, but
those clocks are each at rest,
relative to matter
in the region--
even though they're moving
with respect to each other--
so by definition, they
do measure cosmic time.
Cosmic time is a very
peculiar kind of time,
it's not the time in
any inertial frame.
These clocks are moving
with respect to each other,
so if you were defining
inertial frame time,
their clocks could
never be synchronized
and would never agree
with each other.
But in this concept
of cosmic time,
they do agree with each
other, by construction.
And since each clock is at rest,
relative to its local matter,
it measures this t that
we're talking about,
this cosmic time variable.
And when the pulse
arrives at us,
when we measure
delta t on our clock,
that's exactly the
quantity that, in the end,
we want to talk about--
delta t sub observer.
The quantity measured
on our clock,
which is a clock which
also measures cosmic time.
So there's no place for
any time dilation to enter.
It's not that we forgot
it, it's not there.
It's not part of
this calculation.
So this result, as
simple as it looks,
actually fully
encompasses the effects
of both special
relativity and gravity.
Now let me just mention,
it's not obvious
how gravity came in here.
I'm telling you it
satisfies-- includes
all the effects of gravity.
Where is gravity hidden?
Let me throw that
out as a question.
How does gravity affect
this calculation,
even though I didn't
have to mention gravity
when I described
the calculation?
Yeah, in back.
AUDIENCE: The scale factor?
PROFESSOR: That's
right, the scale factor.
We have not yet talked
about how a of t evolves.
And the evolution of the
a of t will explicitly
involve the effects of gravity.
And that's why this
result depends on gravity,
even though we didn't
need to use gravity,
or say anything about
gravity to get the results.
So this is the first difference.
This calculation includes
the effect of gravity.
Which is through a of t.
Now, because this calculation
seems to include everything
that the first calculation
included and more,
you'd expect to be
more complicated,
but it's less complicated.
Could we have saved
ourselves a lot
of time last week by just
giving this calculation,
and deriving the
other answer from it.
The answer is, not easily,
it would not have saved time,
one can't, in principle,
do it that way.
But the other
important difference
between these two
calculations is the variables
that you're using to
express your answer.
Once you ask a question, if
you ask the question vaguely,
there could be many different
answers to that question,
depending on what variables
are used to express the answer.
So what we're doing here is
we're expressing the redshift z
for objects which are in fact at
rest in the comoving coordinate
system.
The special relativity
calculation-- I think I'm
going to need
another blackboard.
The special relativity
calculation, on the other hand
gives z as a function
of the velocity,
as measured in an inertial
coordinate system.
So the answers are
just being expressed
in terms of totally
different things,
and the answer is so
simple here because a of t
already incorporates
a lot of information,
and we've just taken
advantage of that
to be able to give a very simple
answer in terms of a of t,
without yet saying how we're
going to calculate a of t.
Yes.
AUDIENCE: [INAUDIBLE]
two questions.
One is about that constant time.
PROFESSOR: Yes.
AUDIENCE: How is that different
than the Newton or Galilean
idea of absolute time?
PROFESSOR: OK.
The question was how does
the notion of cosmic time
differ from Newton's
or Galileo's
notion of absolute time?
And the answer is
perhaps not much.
Operationally, I think it
is pretty much the same,
but the real point is that
Newton and Galileo did not
know anything about relative
effects like time dilation.
So for them, it was just
obvious that all clocks ran
at the same speed,
and time was naturally
universal-- naturally absolute.
In this case, we're
aware of the fact
that moving clocks run
at different speeds.
So if we were to take
these clocks between us
and the galaxy, and
transport one to the other,
depending on what path we
used to transport them on,
in the end, they would probably
not agree with each other.
So, we're setting
up a definition
of what we're going to define
locally as time, recognizing
that what time means here,
versus what time means there,
is a consequence of our
assumptions about how we define
things.
It is not given
automatically by the fact
that all clocks will
run at the same speed.
Follow up?
AUDIENCE: Yeah.
An addition, this is
slightly different.
So, in the special
relativity calculations,
[INAUDIBLE] z could
be [INAUDIBLE]
PROFESSOR: Absolutely.
AUDIENCE: So here we're
only seeing a red shift,
but we would obtain a blue
shift if we allowed a of t
to be decreasing, right?
PROFESSOR: That's right.
If the universe contracted,
we would get a blue shift.
AUDIENCE: [INAUDIBLE]
PROFESSOR: That's right.
It would correspond to the
special relativity case.
I was going to say a few words
about the correspondence,
but I'll answer questions first.
Yes.
AUDIENCE: I'd kind
of like to add on
to that question
regarding the causal time.
PROFESSOR: Yes.
AUDIENCE: Isn't the
fact that you've
scaled the speed
of light, that's
what takes care of this
discrepancy between the clocks
themselves?
PROFESSOR: The question is,
does the fact that we've
rescaled the speed
of light take care
of the discrepancy of times?
Well, partially, but
it doesn't say anything
about what moving
clocks will do.
If you had a clock moving
through this universe,
you would have to calculate a
time dilation for that clock,
just as in any other case.
AUDIENCE: What about the two
end points, say, of the path.
Is that why you're scaling
the speed of light?
PROFESSOR: Not really.
The scaling of
the speed of light
really comes about through
the scaling of space.
This in fact is just the scale
factor that we scale space.
Time is measured
locally on every clock,
and we don't think of
it as being rescaled.
The speed of light
looks different,
simply because a notch
is changing with time.
And that formula tells you how
to convert meters per second,
which will always be the
same to the speed of light,
to notches per second,
which will change
as the size of a notch changes
AUDIENCE: Right,yeah.
OK.
I understand.
PROFESSOR: Yes.
AUDIENCE: Further in
the line of questioning
about cosmological
time-- so we expect
that us and that
other galaxy have
simultaneous clocks
relative to the cosmic time,
and also we expect our own
clocks to be simultaneous
with our cosmological
clocks, I assume.
So if we-- Is that true?
PROFESSOR: That's right, yes.
Our own clock is just
an example of one
of the clocks sitting
on the place called us,
and all clocks sitting on that
place will behave the same way.
And they define the local
definition of cosmic time.
AUDIENCE: So if we take those
clocks and move very slowly
across to the other galaxy,
in cosmological-- in comoving
coordinates, we wouldn't expect
there to be any time dilation,
in the respect that
clocks stay simultaneous.
Safe to say, that
we would think it
would be simultaneous
with us the whole time,
until we got to
the other galaxy.
And then it would
still be simultaneous.
But, they're moving at
a speed relative to us,
so we wouldn't
expect [INAUDIBLE].
PROFESSOR: Right.
OK.
You raise a good
question, which I
would have to think
about the answer.
If we brought-- if we
carried our clock very slowly
to this galaxy, and the
limit was infinitely slow,
would it agree
when it got there?
Let me think about that,
and answer it next time.
I'm not altogether sure.
Any other questions?
OK.
I want to say something
about the relationship
between these two calculations.
What would happen if we
tried to actually compare
the answers that we got for
the relativistic Doppler shift,
and for this answer, for
the cosmological redshift.
There's really
only one case where
it would be legitimate
to compare them.
Since the calculation
we just did
was supposed to include
the effects of gravity,
and special relativity
calculation does not
include the effects of
gravity, the only way
we should be able to compare
them, and see that they agree,
would be the case where
gravity is negligible.
And one can talk about a
cosmological model where
gravity is negligible,
there's nothing
inconsistent about that.
If gravity were
negligible, what would we
expect for the behavior of
a of t for this [INAUDIBLE]
question.
I hear a constant.
Constant is certainly
a possibility,
but it's not the
only possibility,
so try to think a
little harder, and ask
if there are other
possibilities.
Yes.
AUDIENCE: I'm sorry, could
you rephrase the question?
PROFESSOR: Rephrase
the question.
The question is, if
gravity were negligible,
what would we expect for the
behavior of the scale factor
a of t?
And so far, it's been suggested
that it could be a constant,
and that's true, but that's
not the most general answer.
Yes.
AUDIENCE: It could be negative.
PROFESSOR: Could be negative?
I don't know what
would mean, actually.
AUDIENCE: What do you--
PROFESSOR: It would mean
the universe was inside out.
AUDIENCE: Oh.
PROFESSOR: It would
really Just mean
that you've reversed
your coordinates.
I don't think it would
have any significance.
AUDIENCE: Oh, the
expansion would actually
be a contraction?
PROFESSOR: Oh, well
it could decrease
with time, that's not the
same as being negative.
AUDIENCE: Oh, I'm sorry
PROFESSOR: It could always
increase or decrease with time,
whether gravity
is present or not.
For our universe it's
increasing with time,
but one could imagine
a contracting universe.
Yes, Aviv.
AUDIENCE: Linear?
PROFESSOR: Linear.
That's right.
If there's no gravity, a of t
should be a constant times t.
The constant could be
zero, and then a of t
is-- and maybe I
should say it should
be a constant plus
a constant times t,
and then in a special case
it could just be a constant.
But it should vary
linearly with time.
And that simply means that
all velocities are constant.
If all velocities are constant,
then a of t is varied linearly
with time, so that
the distance --
the famous relationship is the
distance of a of t times l sub
c.
If this distance were
growing linearly with time,
it could just be a constant
velocity, which is certainly
allowed in the
absence of gravity.
It would mean that a of t
was growing linearly in time.
So that would be the special
case of absence of gravity, a
of t growing linearly in time.
And one can always
set the constant
that would be added to
the linear to be zero,
just by choosing zero of time
to be the time at which a of t
is zero.
So, in the absence of gravity,
one can say that a of t
should just be
proportional to t.
So for that special case,
these two calculations
should really agree.
And it will be, I'm pretty
sure, an extra-credit homework
problem coming up soon, in
which you'll get a chance
to calculate that.
It's not easy,
which is why it will
be an extra-credit problem,
probably, not required problem,
because it involves
understanding the relationship
between these two
coordinate systems.
The special relativity
answer is given
in inertial coordinate system
which, when gravity is present,
doesn't exist at all.
In the presence
of gravity, there
is no global inertial
coordinate system.
But without its
action, there is.
But it's related to this
coordinate system, where
everything's expending
in a complicated way,
because of the various
time dilations and Lorentz
contractions associated
with the motions that
are taking place in
our expanding universe.
So what you'll need
to do is to figure out
the relationship between
these two coordinate systems.
And when you do, and
actually compare the answers,
is you find that they
actually do agree exactly.
This is all perfectly consistent
with special relativity,
but the special case
where there is no gravity.
OK.
Ready to leave cosmological
redshift altogether,
unless there are any
further questions?
OK.
In that case, Onward to
the next major topic.
We've now finished
what I wanted to say
about the kinematics of
homogeneously expanding
universes, and now we're ready
to talk about the dynamics.
What happens when we try
to think about what gravity
is going to do to
this universe, to be
able to calculate how a of t
is going to vary with time.
That will be the only
goal, to understand
the behavior of a of t.
Now this problem, in a way,
goes back to Isaac Newton.
And I might just give
a little aside here,
and mention that one of the
fun things about cosmology,
actually, is that
if one looks back
at the history of cosmology,
many great physicists have made
great blunders in trying to
analyze cosmological questions.
And in the discussion
today, we'll
be discussing one of
Newton's blunders.
And to me, it's very
consoling to know that even
physicists as great as Newton
can make stupid mistakes.
And he actually did
make a stupid mistake,
in terms of analyzing
the cosmological effect
of his own theory of gravity.
At issue was Newton's
view of the universe,
and Newton, like everybody,
really, until Hubble,
believed that the
universe was static.
He imagined the universe
as a static distribution
of stars scattered
through space.
And early in his career,
from what I understand
of the history, he assumed
that this distribution of stars
was finite, and an
infinite background space.
But he realized at
some point that if you
had a finite distribution of
mass, in otherwise empty space,
that everything would
attract everything else,
with his one over r squared
force of gravity-- which
he knew about, he invented
it-- and the result would
be everything would
collapse to a point.
So he decided that
would not work,
but he was still sure
everything was static.
Because everything
looked static,
stars don't seem
to move very much.
So he asked what
he could change,
and decided that instead of
assuming that the stars made up
a finite distribution,
he could assume
that they were an infinite
distribution, sharing
all of space.
And he reasoned--
and this is really
where the fallacy
showed up-- but he
reasoned that if the stars
filled the infinite space
that, even though they would
all be tugging on each other
through the force
of gravity, they
wouldn't know which way to go.
And since they wouldn't
know which way to go,
because they'd be tugged
in all directions,
they would stand still.
So he believed that an infinite,
uniform, distribution of mass
would be stable-- that there'd
be no gravitational forces
resulting from the masses in
this infinite distribution.
And I have some
quotes here, which
I think are kind of cute,
so I'll show them to you.
Newton had a long discussion
about these issues
with Richard Bentley,
the theologian.
And we get to read about it,
because all these letters have
been preserved.
In fact, I'm told that the
original letters are actually
still in existence at
Trinity College in Cambridge
University.
And you can find
them on the web even,
I'll give you a web reference
for the text of these letters,
and they're in books
and various places.
So let me read this to you.
I think it's a cute quotation.
"As to your first
query"-- by the way,
I think we don't have the
letters that Richard Bentley
sent to Newton,
only the responses.
But Newton fortunately
responded in a way
that made the
questions pretty clear,
so it's not an important
problem in understanding what's
going on.
Newton says, "It seems to me
that if the matter of our sun
and planets and all the matter
of the universe were evenly
scattered throughout
all the heavens,
and every particle had an innate
gravity toward all the rest,
and the whole space throughout
which this matter was scattered
was but finite, the matter
on the outside space would,
by its gravity, tend toward
all the matter on the inside"--
this is a finite universe he's
talking about -- and he says,
"that by its consequence,
everything would fall down
into the middle of
the whole space,
and there compose one
great spherical mass."
So, there he's describing
how it would not
work if you had a finite
collection of matter.
But, he says, "If
the matter was evenly
disposed throughout
an infinite space,
it could never convene into
one mass, but some of it
would convene into
one mass, and some
into another, to as to
make an infinite number
of great masses, scattered
at great distances
from one to another, throughout
all that infinite space."
So he thought there'd be local
coagulation, which of course is
what we see in our real world.
We see stars that have
formed, and now we
know about galaxies,
which Newton
had no way of knowing about.
That's the kind of coagulation
process that he's discussing.
And he-- oops, sorry.
"And thus might the
sun and the fixed stars
be formed, supposing the
matter were of a lucid nature."
That's a cute phrase.
I can tell you what it
means, it may not be obvious.
But at this point,
nobody had any idea
what the sun was
made out of, and why
the sun was different
from the earth.
In fact nobody really had
much of a real idea what
the earth was made
out of either, here.
Chemistry wasn't
really invented yet.
So the assumption
was that there were
two kinds of matter, lucid
matter and opaque matter.
Where lucid matter is the
stuff the sun is made out of,
and the stars, that glows,
and is fundamentally different
in some way, that was of
course not understood at all,
from opaque matter, which is
what the Earth is made out of.
You can't see through it, and
it doesn't, obviously, glow.
So here, when he's
talking about matter
forming the stars
and the sun, he
says if the matter
was lucid, if it
was the kind of
matter that glows.
Going on-- so far, what he
said sounds pretty good.
Going on, he goes
on now to talk more
about this lucid
versus opaque business.
And I think it's cute.
I don't know where
exactly it's going,
but it shows something about
Newton's personality, which
one might not have
known otherwise.
"But how the matter
should divide itself"--
I should also warn you, all
of this is one sentence.
If you think sometimes my
sentences sound convoluted,
just think how lucky you are
that you don't have Newton here
as your lecturer.
This is just impossible.
So, "But how the matter should
divide itself into two sorts"--
how we'd have lucid and opaque
matter in the right places--
"and that part of it which is to
comprise a shining body should
fall down into one mass, and
make a sun, and the rest,
of which is fit to
compose an opaque body,
should coalesce not into one
great body, like shiny matter,
but into many little ones"--
somehow he's forgotten about
the stars here, when he's
talking about the sun
and the planets, many
planets and one sun.
So he says that, "how the opaque
matter should fall instead
into many little
masses"-- and then
he talks about
other possibilities.
It's wonderful the way he
lists all the possibilities.
"Or," he says, "if the sun
were at first an opaque body,
like the planets, or if the
planets were lucid bodies like
the sun, how he alone"--
he being the sun,
if you track everything back,
"how the sun alone should be
changed into a shiny body,
while all the"-- lost track --
"where all they"--
of the planets--
"continue to be opaque,
or"-- he's considering all
possibilities-- "or they all
be changed to opaque ones,
while he,"-- the sun-- "remains
unchanged as a lucid one."
He does not know how to explain
all that, is what he's saying.
Bottom line of the
sentence is, I don't know,
I don't have a clue.
And he says, "I don't think
it's explicable by mere natural
causes, but am
forced," Newton says,
"to ascribe it to the
council and contrivance
of a voluntary agent."
So the theory of
intelligence design,
as well the theory
of gravity, actually
both go back to
Newton, it turns out.
Newton was a very
religious person,
and in certain
aspects of physics,
he was happy to ascribe
to a voluntary agent,
as he calls it.
I have some references
here, and I'll
be posting this so
you'll be able to read
those references and type
them in, if you want.
Now Newton decided
that you could not
have a finite distribution,
because it would collapse.
If you had an
infinite distribution,
he thought it would be
stable, but he apparently
had heard different arguments
to that same conclusion.
And one argument
that you might give
for saying that the infinite
distribution would be stable
would be the argument that
if you look at the force
one any one particle, there is
an infinite force pulling it
to the right-- my right,
your left-- and an infinite
force pulling it to
my left, your right,
and since they're both infinite,
they would cancel each other.
Newton did not
accept that argument.
He was sophisticated enough
to realize that infinity minus
infinity isn't necessarily zero.
And he has a bit of
a tirade on that,
that I thought
was worth quoting.
And this is a second letter
to the same Richard Bentley.
I guess it was Bentley
who made this argument,
and Newton rejected it.
Infinity minus infinity,
Newton realized, is ambiguous.
It's not something that
we should necessarily
think of zero.
"But you argue in the next
paragraph of your letter that
every particle of the matter
in the infinite space has
an infinite quantity
of matter on all sides,
and by consequence, an
infinite attraction every way,
and therefore must
rest in equilibrio,
because all infinities
are equal:"--
he's summarizing Richard
Bentley's argument--
"yet you suspect a parologism"--
that means logical error,
I think-- "in this argument: and
I can see the paralogism lies
in the position that all
infinities are equal.
The generality of mankind
consider infinities no other
ways than indefinitely"-- and
in this sentence they said all
infinities are equal-- "though
they would speak more truly
if they should say that they
are neither equal nor unequal,
nor have any certain difference
or proportion, one to another."
So he realizes that the ratio
of infinity could be anything,
and infinity minus infinity
could be anything, all of which
is consistent with
our modern view of how
to do the mathematics.
"In this sense,
therefore, no conclusions
can be drawn from them about
the equality, proportions
or differences of things, and
they that attempt to do so
usually fall into paralogisms."
He goes on, now I just have
one more Newton quote--
I like Newton quotes--
I have one more
Newton quote, again
from the same series of letters.
These are all from
1692 and 1693,
I believe, where he
gives an example--
I think this follows the
quotes of the previous slide
immediately-- where he gives
an example of a false argument
that you get into-- and
apparently it's an argument
that he had heard
from other people--
if you think all
infinities are equal.
What he says is, "So when men"--
he doesn't say who men are,
and I don't really
know the history.
He may referring to some
particular philosophers
at the time-- "when men argue
that the infinite divisibility
of magnitude by saying
that an inch may be divided
into an infinite number of
parts, the sum of those parts
would be an inch--
and a foot can
be divided into an
infinite number of parts,
the sum of those
parts must be a foot--
and therefore, since all
infinities are equal,
these sums must be equal."
Understand the argument here.
He's saying that if
you divide and inch
into an infinite
number of parts--
this is all you've
been given as a foil.
He's not claiming the
argument is right,
he's claiming it's wrong-- that
argument is that if you divide
an inch into an infinite
number of parts,
you get an infinite
number of points,
if you put them together,
you get an inch.
If you divide a foot into
an infinite number of parts,
you get an infinite
number of points,
and if you put them together,
you should get a foot.
But they're both
an infinite number
of points in the description.
So if you think all
infinities are equal,
the infinite number of
points that make an inch
should be the same as the
infinite number of points that
make a foot, therefore a
foot should equal an inch,
obviously.
Right.
Not right, he know.
So he says that the
falseness of the conclusion
shows an error in the
premises, and the error lies
in the position that all
infinities are equal.
So Newton has given
us a very nice example
of how you can convince
yourself that you
get into logical
paradoxes if you pretend
that all infinities are equal.
But, this does not
change the fact
that Newton was still convinced
that an infinite distribution
of mass would be stable.
The argument that convinced
him was not the infinity
on each side, but
rather the symmetry.
Newton's argument,
the one he believed,
was that if you
look at any point
in this infinite distribution,
if you look around that point,
all directions would look
exactly the same, with matter
extending off to infinity,
and therefore there'd
be no direction that
the force should
point on any given particle.
And if there's no direction
in that force at the point,
it must be zero.
That was the argument
Newton believed.
OK.
What I want to do now is to
talk about this in a little bit
more detail, and
try to understand
how modern folks would
look at the argument.
And by the way, I might
just add a little bit more
about the history first.
Newton's argument,
as far as I know,
was not questioned by anybody
for hundreds of years,
until the time of
Albert Einstein.
Albert Einstein, in trying
to describe cosmology using
his new theory of
general relativity,
was the first
person, as far as I
know, to realize
that even if you
had an infinite distribution
of mass, it would collapse--
and we'll talk about why.
And Einstein did realize that
the same thing would happen
with Newtonian physics,
it's not really
a special feature of
general relativity,
it just somehow
historically took
the invention of
general relativity
to cause people to
rethink these ideas
and realize that
Newton had been wrong.
So, what's going on.
The difficulty in trying
to analyze things the way
in which Newton did is that
Newton was thinking of gravity,
in the language that he
first proposed it, as a force
at a distance.
If you have two objects in
space, the distance r apart,
they will exert a force on
each other proportional to one
over r squared.
Since the time of
Newton, other ways
of describing Newtonian
gravity itself
have been invented, which
make it much more clear what's
going on.
The difficulty in
using Newton's method--
we'll talk about in more
detail in a few minutes--
but it's simply that we try to
add up all of these one over r
squared forces, you
get divergent sums
that you have to figure
out how to interpret.
But to understand that
Newton couldn't possibly
have been right, the
easiest thing to do
is to look at other formulations
of Newton's gravity.
And I'll describe two
of them, both of which
will probably have some
familiarity to you.
The first one I'm
quite sure will.
And I'm going to describe it
by analogy with Coulomb's law,
because 802 goes a little
further with Coulomb's law
than any course you are
likely to have taken
has gone with gravity.
But Coulomb's law is really
the same as the force
law of gravity.
So Coulomb's law
says that any charged
particle will create an
electric field, which
is the charge divided by the
distance squared times the unit
vector pointing
radially outward.
That's Coulomb's law.
People can-- sometimes
there's constants in here,
depending on what
units you measure q in,
but that won't be
important for us.
So I'm going to
assume we're using
this where that constant is one.
You know that Coulomb's
law can be reformulated
in terms of what we
call Gauss's law.
If Coulomb's law is true, you
can make a definite statement
about what happens
when you integrate
the flux of the electric
field over any surface.
It's proportionate to the
total amount of charge inside.
So Coulomb's law
implies Gauss's law,
which says that the integral
over any closed surface of E
dotted into da is equal to 4
pi times the total enclosed
charge.
q encloses the total amount
of charge inside that volume.
And what constants appear
depends on what constants
appear here, which depends
on what units you're using,
but these equations
are consistent.
Those are the correct
constants, if you
measure charge in a way which
makes the electric field be
given by that simple formula.
OK, so I'm going to assume you
know this, that you learned it
in 802 or elsewhere.
If this is true,
then, since this
is the same inverse square
law, if we write down
Newton's law of gravity,
almost as Newton
would have written
it, we can express it
as the acceleration
of gravity at a given
distance from an object.
So we could write
Newton's law of gravity
by saying the
acceleration of gravity
is equal to minus Newton's
constant times the mass
of the object, the analog
of the charge up there,
divided by r
squared times r hat.
Again, it's the
inverse r squared law,
and the point radiating outward
is just like Coulomb's law,
except for the
constant out front.
The constant actually
has the opposite sign,
which is important
for some issues,
but not for what
we're saying now.
The important point
is that this can also
be recast as a
Gauss's law, and it's
called Gauss's law of gravity.
And the only thing that differs
is a constant out front,
so it's a trivial
transformation.
The integral over
any closed surface
of the gravitational
acceleration vector, little g
dotted into da is
equal to minus 4 pi
g times the total mass enclosed.
The only difference is the
minus sign, and the factor
of g, which follow from the
difference of the minus sign
and the factor of g in
the formula on the left.
OK, does everybody believe that?
OK, now let's think about
this homogeneous distribution
of mass that Newton was
trying to think about.
Newton's claim
was that you could
have a homogeneous
distribution of mass filling
all of the infinite space, and
that would be static, that is,
there would be no acceleration.
No acceleration means Newton
is claiming in this language
that little g could
be zero everywhere.
But if you look at this formula,
if little g is zero everywhere,
then the integral of g over
any surface is going to zero,
and therefore the total mass
enclosed had better be zero.
But if we have a uniform
distribution of mass,
the total mass enclosed
will certainly not
be zero for anything
with non-zero volume.
So clearly this assertion that
the system would be static
was in direct contradiction
with the Gauss's law formulation
of Newton's law of gravity.
Just for the fun of
it, I'll give you
another similar argument using
another more modern formulation
of Newtonian gravity.
Another way of formulating
Newtonian gravity, which
you may or may not have seen--
and if you haven't seen it,
don't understand what I'm
saying, don't worry about it,
it's not that important.
But for those of you
who have seen it,
I'll give you this argument.
Another way of formulating
Newtonian gravity
is to introduce the
gravitational potential.
So I'm going to
use the letter phi
for the gravitational potential.
I'll tell you in a second how
that relates to gravity-- well,
I guess I'll tell you now.
It's related to the
gravitational acceleration
by g is equal to minus
the gradient of phi,
and gradient of phi is
something that you probably all
learned in 802, but I'll
write down the formula anyway.
It's equal to i hat, a unit
vector in the x direction,
times the derivative
of phi with respect
to x, plus j hat, a unit
vector in the y direction,
times the partial of
phi with respect to y,
plus k hat times the partial
of phi with respect to z.
And once one defines this
gravitational potential,
one can write down
the differential form
of the Gauss's
law, which becomes
what's called
Laplace's equation.
And it says the
del squared phi is
equal to 4 pi times
Newton's constant times rho,
where rho is the mass density.
And this is called
Laplace's equation,
and if you're given
the mass density,
it allows you to find the
gravitational potential,
and then you can
take its gradient,
and that determines what g is.
And it's equivalent to the
other formulations of gravity.
But it gives us another
test of Newton's claim
that you could have a
homogeneous distribution
of matter, and no
gravitational forces.
If there are no
gravitational forces,
then g would have to be zero,
as we said a minute ago,
and this formulation
of g is zero,
that implies the
gradient of phi is zero.
If we look at the formula for
the gradient, it's a vector.
For the vector to be zero,
each of the three components
has to be zero, and
therefore the derivative
of phi with respect
to x has to vanish,
the derivative of phi with
respect to y has to vanish,
the derivative of phi with
respect to z has to vanish,
that means phi has to
be constant everywhere,
it has no derivative
with respect
to any spacial coordinate.
So if g vanishes,
the gradient of phi
vanishes, and phi is a
constant throughout space.
And if phi is a constant
throughout space,
now we can look
at this formula--
and I forgot to write down
the definition of del squared.
Del squared phi is defined to
be the second derivative of phi
with respect to x squared, plus
the second derivative of phi
with respect to y squared, plus
the second derivative of phi
with respect to z squared.
So if phi is a
constant everywhere,
as it would have to be if there
were no gravitational forces,
then one can see immediately
from this equation
that del squared phi
would have to be zero,
and one can see
from this equation
that rho would have
to be zero, there
would have to be
no mass density.
But Newton wanted to
have a non-zero mass
density, the matter
of the universe spread
out uniformly over
an infinite space.
So this is another demonstration
that Newton's argument
was inconsistent.
Yes.
AUDIENCE: I'm sorry,
what does phi represent?
PROFESSOR: Phi is really
defined by these equations,
it's defined, really,
by this equation.
The name is that it's the
gravitational potential.
AUDIENCE: Potential.
PROFESSOR: And its
physical meaning
is simply that it
gives you another way
of writing what g is.
AUDIENCE: Yeah.
PROFESSOR: Any other questions?
OK, so the conclusion seems
to be that Newton has not
gotten the right answer,
here, but we still
have to analyze Newton's
argument a little bi
more carefully, to see
exactly where he went wrong.
So, the next thing
I want to talk about
is the ambiguity associated with
trying to add up the Newtonian
gravitational
forces, as Newton was
thinking, for an
infinite universe.
I mentioned that the real
problem with Newton's
calculation is that the quantum
he was calculating actually
diverges, and you have to
be more careful about trying
to calculate it
in a reliable way.
So to make this
clear, I want to begin
by giving an example of this
general notion of integrals
that give ambiguous values.
And I want to define just a
couple of mathematical terms.
I want to consider
just-- again, starting
talking about general functions,
and when integrals are well
defined and when they're not.
I want to imagine that we just
have some arbitrary function
f of x where x would not
just be one variable.
We'll generalize this to
three dimensions, which
is the case that we'll
be interested in,
but we'll start by talking
in terms of one variable.
If we have a function
f of x, we can
discuss what I'll
call I sub 1, which
is the integral, from
minus infinity to infinity,
of f of x dx.
This is exactly the
kind of integral
that you're thinking
of when we wanted
to-- thinking about adding up
all the gravitational forces
acting on a given body.
Now I want to consider the
case where I1 is finite.
I'm sorry.
I need to first define more
carefully what I mean by I1.
OK, to even define what you mean
by this minus v to infinity,
you should say something
a little bit more precise.
So we could define I1 a
little bit more precisely,
and I'll call this I1
prime, for clarity.
This will really
just be a clearer way
of describing what
one probably meant
when one wrote the first line.
We can define the integral
from minus infinity
to infinity as the limit,
as some quantity L goes
to infinity, of the integral
from minus L to L of f of x dx.
So this says to do the
integral from minus L to L,
and if we assume f of
x is itself finite,
this is always finite.
I will assume f of
x itself is finite,
we'll only worry about the
convergence of the integral.
So for any given L, this is
a number, then you can ask,
does this number approach a
limit as L goes to infinity?
And if it does, you say that's
the value of this integral.
That just defines what
we mean by the integral
from minus infinity to infinity.
I want to now consider the
case where that exists.
So consider the case where
I1 prime is-- I'll write
is less than infinity, meaning
it has some finite value.
The limit as L goes
to infinity exists.
But now, I want
to also consider--
and I'll move on to
the next blackboard--
to consider this-- consider
an integral that I'll call I2,
for future reference,
which is just
defined to be the integral from
minus infinity to infinity.
Defined as the same kind
of limit that we used here,
but I won't rewrite it.
I'll just assume that the
integral from minus infinity
to infinity means that limit.
But I want to consider the
integral from minus infinity
to infinity of the absolute
value of f of x dx.
And now I want to
introduce some terminology.
If I2 is less than
infinity, if it converges,
then I1 is called
absolutely convergent.
So absolutely convergent
means that it would converge,
even if you had
absolute value signs.
Conversely, this
I2 is divergent--
and I'll just write that
as I2 equals infinity,
if that limit does not exist,
if its a divergent integral.
But remember, we
assumed I1 did exist,
so I1 still converges, but
it's called conditionally
convergent.
So if an integral
converges, but the integral
of the absolute value of that
same client does not converge,
that's the case that's called
conditional convergence.
And the moral of
the story, that I'll
be beginning to tell you now,
is that conditionally convergent
integrals are very dangerous.
What makes them dangerous
is that they're not really
well defined.
You can get any value
you want by adding up
the integrand in
different orders.
As long as you stick to
a particular order, which
is how we define the symbol,
you will get a unique answer,
but if, for example, you
just shift your origin,
you can get a
different answer, which
is something you
don't usually expect.
You usually think of
just integrating over
the whole real line,
it doesn't matter
what you took to be
the center of the line.
So things become much
less well defined
when one is discussing
conditionally
convergent integrals.
And before we get to
the particular integral
that we're really
interested in, which
is trying to add up the
gravitational forces of
and infinite distribution of
matter, which I'll get to,
I'm going to give you an example
of a very simple function that
just illustrates this ambiguity,
that the integral converges,
but is not absolutely
convergent.
You can get any answer
you want by adding it
in different orders-- adding
up the pieces of the integral
in different orders.
So let me consider an
example-- and this is just
to illustrate the ambiguity--
the example I'll consider
will be a function f of x, which
is defined to equal plus 1 if x
is greater than zero, and minus
one if x is less than zero.
And I have neglected
to specify what
happens if x is
exactly equal to zero,
but when you integrate,
that doesn't matter.
A single point never matters.
So you could measure
it's anything
you want at x equals zero,
it won't change anything
you're going to be saying.
Let me draw a graph of this.
f of x versus x.
I'll put plus 1 there,
and minus 1 there.
The function is plus 1.
Maybe I have a little
bit of colored chalk
here to draw the function.
The function is plus 1 for
all positive values of x,
and minus one for all
negative values of x.
And there's the function.
And if we integrate
it symmetrically,
following this
definition of what
we mean by integrating from
minus infinity to infinity,
we do get a perfect
cancellation.
When you integrate
from minus L to L,
we get zero, because you
get a perfect cancellation
between the negative parts
and the positive parts.
And then if you take the
limit as L goes to infinity,
the limit of zero is zero.
There's not really any
ambiguity to that statement.
So in the order specified,
this has unique integral,
which is zero.
But, it depends on how you've
chosen to add things up.
In particular, if you
just change your origin,
and integrate starting moving
outward from the new origin,
you'll get a different
answer, and that's
what I want to illustrate next.
Suppose-- suppose we
consider the limit
as L goes to infinity, we'll
pick the limit the same way,
but instead of integrating
from minus L to L,
we can integrate from a minus
L to a plus L of f of x dx.
Now this is really
the same integral,
we've just basically
changed our origin
by integrating from a outwards.
In the special
case a equals zero,
it's exactly the same
as what we did before,
but if a is non-zero, it
means that our integral
is centered about x equals
a, instead of centered
about x equals zero.
So we can draw that
on the blackboard.
If we let a be over
here, our integral
will go from a minus
L, and that will
be to the left of
distance L, you
will extend to a plus L,
which will be to the right
by distance L. The
integral defined
by the equation on the
blackboard at the left
will correspond to that
region of integration.
And the specification is that we
should do that interval first,
and then take the limit
as L goes to infinity,
and see what we get.
It's easy to see
what we will get.
Once L is bigger
than a, you can see
that the answer won't change
any more, as we make L bigger.
As you make L bigger,
we will always
be adding a certain amount
of minus 1 on the left,
and certain amount-- the same
amount of plus 1 on the right,
and they will cancel each
other once L is bigger than a.
And we don't care about
small l, because we're
only interested in taking
the limit of large L,
but we should look
at what happens
when L equals a, and then
from any bigger value of L
will give us exactly
the same number.
And when L equals
a, the integral
will go from 0 up to
2a-- a plus L which
is a equals L, so that's 2a.
So the integral will be
only on the positive side,
and we'll have a
length of 2a, and that
means the integral will
be 2a, because we're just
integrating one from 0 to 2a.
And that will be what we get
for any bigger value of L
also, because as we
increase L, as I said,
we just get a cancellation
between adding more plus 1
on the right, and adding
more minus 1 on the left.
So this limit has a
perfectly well defined value,
which is 2a.
And a is just where we
chose to start integrating,
so a could be anything.
We could choose a
to be anything we
want if we're free to
integrate in any order.
So we can get any
answer we want,
if we're free to
integrate in any order,
to add up the pieces of
this integral in the order
that we choose.
And that is a
fundamental ambiguity
of conditionally
convergent integrals.
And what we'll
see is that trying
to add up the force on a
particle in an infinite mass
distribution is
exactly this kind
of conditionally
convergent integral.
And that's why you get
any answer you want,
and it doesn't
really mean anything
unless you do things
very carefully.
OK.
Let's move on.
We only have a few
minutes left, which
I guess means I will
set up this calculation,
but not quite get the answer,
and we'll continue next time.
I actually have some
diagrams here on my slides.
What I want to do now,
is calculate the force
on some particle in an
infinite mass distribution,
and show you that I can
get different answers,
depending on what
order I add things up.
I will add things up in a
definite order at each stage,
so I will get a definite answer
at each stage, though I'll
get different answers, depending
on what ordering I choose.
So, we're going
to start by trying
to calculate-- and the only
thing [? of interest, ?]
actually, in calculating
the gravitational force
on some point, p in an
infinite distribution of mass.
Mass fills the slide, and
everything, out to infinity.
And we're going to add up
that mass in contributions
that are specified.
And for our first
calculation, we're
going to add up the
forces for masses
that are defined in
concentric shells, where we're
going to take the
innermost shell first,
then the second shell,
then the third shell,
going outward from the center.
In that case, it's easy to see
that the force on p calculated
in that order of
integration is 0,
because every shell has
p exactly at the center,
and by symmetry, it
has to cancel exactly.
In fact, we know-- and we'll
use this fact shortly--
that the gravitational field
of a shell, inside the shell,
is exactly zero-- Newton
figured this out--
and outside the shell, the
gravitational field of a shell
looks exactly the same as the
gravitational field of a point
mass located at the
center of the shell
with the same total mass.
So we're going to be
using those facts.
And clearly those facts
indicate that, for this case,
the answer is 0.
P equals 0.
Now we're going to consider a
more complicated case-- going
too far, here, don't want
to tell you the answer yet--
this more complicated case,
we're going to still calculate
the force at the
point p, but we're
going to choose concentric
spherical shells
which are centered around
a different point, q.
So q just defines
the shells that we're
going to use for
adding things up,
and we're still going to add up
all the shells out to infinity,
so we're going to be
adding up the force on p
due to the entire infinite
mass distribution,
but we'll be taking
those contributions
in a different
order, because we're
going to be ordering it
according to shells that
are all centered on q, starting
with the innermost, and then
the second, and then
the third, and so on.
Now in this case, we can first
talk about the contribution
of the shaded region, which are
all the shells around q which
have radii which are less
than the distance to p.
For all of these shells,
p lies outside the shell.
And therefore all
of those shells
act just like a point mass,
with the same total mass
concentrated at q, the
center of all those spheres.
So the mass that's
in the shaded region
will give a contribution
to the force
at p, which is just equal
to the force of the mass
given by the same total mass
the point q, located at q.
On the other hand,
all the shells outside
will be shells for
which p is inside.
P is no longer at the
center of those shells,
but Newton figured out, and
I'll assume that we all believe,
it doesn't matter.
Inside the spherical shell,
the gravitational force
is zero anywhere,
no matter how close
you are to the boundaries.
It just cancels out perfectly.
As you get closer
to one boundary,
you might think you'd be
pulled toward that boundary,
but-- let me just tell you
what's happening here--
as you get closer
to one boundary,
it is true that the
force pulling you
towards the particular particles
at the boundary get stronger,
because it's 1 over r
squared, but as you get close
to this boundary, there's
more mass on this side,
because all the mass
except for a little sliver
is on the other side.
And those two effects
cancel out exactly.
So the force on a
particle inside a shell
is exactly zero, as you can
prove very easily by the way,
from the Gauss's law of
formulation of gravity.
And therefore, the outer
shells give no contribution.
So we've completely
calculated now the force at p
is just equal to the force
due to the shaded mass.
It's just given by
that simple formula,
it's g times the total
mass, divided by b squared,
that would be it's
distance between q and p.
And it's non-zero.
So you get 0 or non-zero
answer, depending
on what ordering you
chose for adding up
the pieces of the mass
that are going to make up
this infinite distribution.
And furthermore, this answer
could be anything you want,
because I could let
b be anything I want.
And this answer depends
on b, and becomes
arbitrarily large in
magnitude as b gets bigger.
The mass grows like b cubed.
It might look like
it falls with b,
but actually it grows with b.
And we could get a
point in any direction,
by choosing q on any side
we want of p, so we can get,
really, any answer
we want by using
this particular way of
adding up the masses.
Yes.
AUDIENCE: Well, although we
can get any answer we want,
every answer [INAUDIBLE]
PROFESSOR: Every
answer, say again?
AUDIENCE: Like every
single one of those
answers corresponds to a setup.
I mean like the g
equals 0, [INAUDIBLE]
PROFESSOR: Well the
reason it's a problem
is that these shells
don't really exist.
We're just thinking
about these shells.
The shells only
determine what order
we are going to use for adding
up the different contributions.
The matter is just
uniformly distributed
and there's no shells present.
The shells are purely
a mental construct,
which should not
affect the answer.
This is not part of the
physical system at all.
The shells only
reflect the order
that we have used to
add up the masses.
So we'll stop there.
If anybody has questions,
we can talk after class,
and we can talk more about
the question at the beginning
of the next class, but
class is over for now.
