Professor Charles
Bailyn: Okay,
welcome to the second part of
Astro 160.
This is going to be about black
holes and relativity.
And just to give you kind of a
preview, the whole point of
black holes is that,
of course, they emit no light,
so you can't see them directly.
And so, the question arises,
"How do you know that they're
there?"
And the reason you can
demonstrate that black holes
exist is because they're in
orbit around other things and
you can see the motion of the
other things that interact
gravitationally with the black
hole.
This concept should be
familiar, to a certain extent,
because it's exactly the same
thing we've been doing for
discovering exoplanets.
You don't see the exoplanet
directly.
What happens is that there's
something else that you can see
that's affected by the presence
of the exoplanet.
So, exactly the same thing
happens with black holes.
And so, we're going to use the
same equations,
the same concepts,
to explore this very different
context.
So, black holes can't be seen
directly.
And so, instead of detecting
them directly,
you use this combination of
orbital dynamics and things like
the Doppler shift to infer their
presence,
and more than just inferring
their presence,
to infer their properties.
Now, the context is more
complicated.
And, in particular,
we're no longer going to be
using Newton's laws – Newton's
Law of Gravity,
Newton's Laws of
Motion--because there is a more
comprehensive theory that
replaced Newton,
which is necessary to
understand these things.
That more complex theory is
Einstein's Theory of Relativity.
So, we're going to be using
some relativity rather than
Newtonian physics.
This gets weird very fast, okay?
And so, I'm not going to start
there.
I'm going to start with a kind
of Newtonian explanation for
what black holes are,
we'll do that this time,
and then the weirdness will
start on Thursday.
So, the first concept and the
easiest way, I think,
to understand black holes is
the concept of the escape
velocity.
This is a piece of high school
physics.
Some of you may have
encountered it before.
And it just means how fast you
have to go to escape from the
gravitational field of a given
object.
If you go outside and you shoot
up a rocket ship or something
like that, how fast do you have
to shoot it up so that it
doesn't fall back to the Earth?
And so, you can define an
escape velocity for the Earth,
or for any other object for
that matter,
which is just how fast you have
to go to escape its
gravitational field.
There is an equation associated
with this.
It looks like this,
V_escape,
that's the escape velocity,
2GM / R all to
the 1/2 power.
And this is the speed required
to escape the gravitational
field of an object;
supposing that the object has
mass equal to M and
radius equal to R.
Oh, one other assumption,
I'm assuming here that you're
standing on the--that you start
from standing on the surface of
the object.
If you are on the surface.
Okay.
This equation should look
vaguely, but not a hundred
percent familiar to you,
because you derived something
that looked a lot like it on the
second problem set,
where you worked out the
relationship between the
semi-major axis of an orbit and
the speed and object had to go
in to be in that orbit,
and what--the way that
calculation worked out it was
the velocity equals GM
over the semi-major axis,
a.
So that 2 wasn't there in that
derivation, but otherwise,
the form of this is actually
quite similar to that.
And so, let me explain why that
is.
Here's some object.
It's got a radius of R
and a mass of M.
And imagine that you've got
something that's in orbit around
this object, but it's in orbit
right above the surface.
It's just skimming the surface
of the thing.
That would be impossible in the
case of Earth,
because the friction from the
atmosphere would slow you down,
but in a planet without an
atmosphere, this would be
possible.
Imagine you're just sort of
skimming over the tops of the
mountains, or whatever.
So here you are in orbit,
this is something in a nice
circular orbit right above the
surface of this object,
and it's going around.
And it's got some velocity,
which I'm going to call the
circular velocity.
And as we calculated in the
previous section of the course,
that's GM / a,
which is, in this case,
GM / R, because
you're skimming the surface,
to the 1/2 power.
So that's how fast you have to
go to stay in orbit.
If you go slower than
that--let's take a different
color, here.
If you go slower than that,
you crash into the surface
right away, because you
don't--you're not going fast
enough to stay in orbit.
So, that's what happens in the
case--if you have a velocity
less than the circular velocity.
What happens if you're going
faster than the circular
velocity?
Well, you're going so fast that
you move further away from this
object.
You don't stay skimming the
surface.
So, here you are,
going a little bit faster.
And now you're not in a
circular orbit and your orbit
ends up looking something like
this--nice elliptical orbit.
And this is a velocity greater
than the circular velocity,
but still less than the escape
velocity because you haven't
escaped the gravitational field,
because you're still in an
orbit.
It's an elliptical orbit now,
but you're still in an orbit,
and it'll come back to the same
place.
Now, if you're going at
somewhat faster than that at the
escape velocity,
then you never come back.
You continue all the way out to
infinity.
Your orbit continues to change
direction a little bit due to
the gravitational force of this
thing.
You continue to slow down,
but in the end you never come
back.
So, it's a non-repeating orbit.
And, if you're going even
faster than the escape velocity,
then you get to infinity even
faster.
And, more importantly,
when you get there,
you're still moving pretty
fast.
This, you gradually slow down.
So, all of these different
kinds of orbits--the one that
crashes into the planet,
the one that skims the planet,
the one that's an elliptical
orbit, the one that escapes
altogether--those are all within
a factor of the square root of 2
of each other.
Because, remember,
the escape velocity here is
equal to (2GM /
R)^(1/2).
So, just by increasing your
orbital speed by a factor of the
square root of 2,
you go from being in a nice
circular orbit to escaping the
gravitational field of the
object altogether.
So, you can calculate escape
velocities of things.
Let's do that once.
The escape velocity of
Earth--that would be--the escape
is (2GM /
R)^(1/2).
2G = 7 x 10^(-11)
M is, for the Earth,
is 6 x 10^(24) R for the
Earth is something like 7 x
10^(6).
This all has to be taken to the
½ power.
The 7s cancel.
What do we have here?
Let's see, 12 x 10^(13).
24 - 11 over 10^(6) 1/2--let's
see, that's 1.2.
14 - 6 = 8, times 10^(8),
to the 1/2.
That's something like 10^(4)
meters per second,
because I've used the value of
G that's appropriate for
meters per second.
So, that's about 10 kilometers
a second.
So, that's the escape velocity
of the Earth.
If you go outside,
you throw a football up into
the air at 10 kilometers per
second, it's not going to come
back down.
Try it--good exercise.
Okay, you could calculate the
escape velocity of any
particular object.
You could calculate the escape
velocity of the human being--of
a human.
A human has a mass of 100
kilograms, a radius of about 1
meter;
this is kind of a big human,
but we'll go with it.
Let's see,
V_escape is
equal to 2GM over--a
half--there's a very famous book
about physics with the title,
"Assume A Spherical Cow," and
this--so, this is the kind of
thing that physicists like to
do.
It's an idealized situation.
We have this perfectly
spherical human being and--okay
so let's do the calculation,
2 x 7 x 10^(-11) x 100,
that's 10^(2),
over 1, to the square root.
That's 14 x 10^(-9).
Call it (1.4 x 10^(-8))^(1/2).
That's something like 1.2,
let's call it 1.
1 x 10^(-4) meters per second.
Or--let's see,
that's 1/10 of a millimeter per
second.
That's less than 1 meter per
hour.
And that's why we don't go into
orbit around each other.
Okay?
Because you're always moving
way, way, way faster than the
escape velocity of the people
you're interacting with.
The escape velocity of the
Earth is 10 kilometers a second.
So, you're not moving that fast.
So, you're bound to the Earth.
But when you're hanging around
with your friends,
you're probably going faster
than 1 meter per hour,
and therefore you're moving
faster than the escape velocity.
You don't feel a great effect
of the gravitational force of
other people.
This is kind of
anti-Valentine's Day
calculation, right?
Because it proves that human
beings are not attracted to each
other;
at least not by the force of
gravity.
And so, you can do this
calculation for any given
object.
Fine, so what's a black hole?
Black hole has,
actually, a very simple
definition.
A black hole is simply
something in which the escape
velocity is greater than or
equal to the speed of light.
c is the expression for
the speed of light.
This is 3 x 10^(8) meters per
second.
And, if you've got something
where the escape velocity is
greater than the speed of light,
that's what a black hole is.
And it makes a certain amount
of sense.
If the escape velocity's
greater than the speed of light,
then light won't escape this
object, so you won't be able to
see it, right?
Because, the way you see
something is you see the photons
that come off it.
And, there's nothing
particularly extraordinary about
this, as far as it goes.
In fact, the middle of--this
was already talked about and
worked out in some detail in the
middle of the eighteenth century
by an otherwise obscure English
clergyman named,
I think, John Michell,
and he did the following
calculation.
He said, okay now,
if this is going to be true,
how big is such an object?
How dense does it have to be?
And, you can work this out.
Supposing the escape velocity
is equal to the speed of
light--so c = (2GM
/ R) ^(1/2).
You can square both sides and
regroup, and you have R =
2GM / c^(2).
This is now--wasn't called this
in the eighteenth century,
but it's now called the
Schwarzschild radius.
Schwarzschild was a
contemporary of Einstein.
And, this is the size something
has to be in order for its
escape velocity to be equal to
the speed of light.
And a black hole--another
definition of a black hole is
something in which the radius of
the object is less than the
Schwarzschild radius--because if
the radius is less,
then the escape velocity will
be even greater.
So as I say,
this was worked out in
obscurity in the middle of the
eighteenth century,
and nobody thought anything
much about it.
Michell sort of pondered for a
little bit what such an object
would look like,
and he decided it would look
dark.
And true enough, but who cares.
And nobody thought anything
further about it for 150 years
until Einstein came along at the
start of the twentieth century,
and came up with the Theory of
Relativity.
And one of the pieces,
important pieces of relativity
is that the speed of light,
c is a very,
very special velocity,
and has rather bizarre and
extremely profound properties.
And it was only at that point
that the concept of the black
hole was recognized as something
that was in any way out of the
ordinary.
And so, for 150 years this idea
kind of lay dormant until it was
resurrected by a profound change
in the thinking about the
underlying physics.
So, this is one of these kinds
of historical fables that we've
hit a couple of times.
This is the fable of Michell's
discovery of black holes.
And the moral of this little
fable might be that the
importance of a result changes
depending on the context;
result changes,
sometimes dramatically,
with context.
So, something that looks quite
unimportant for a long time can
all of a sudden become a very
big deal.
And so, it happened in this
case.
Okay, so, one question you can
ask is that's all fine,
how big is the Schwarzschild
radius for some bunch of
objects?
Let's try the Sun;
here, let me start a new piece
of paper.
So, how big is the
Schwarzschild radius of the Sun?
Schwarzschild radius,
again, 2GM /
c^(2) = 2 x 7 x 10^(-11).
Mass of the Sun,
now, 2 x 10^(30).
Divided by c^(2).
c is 3 x 10^(8) and
we'll square it.
Okay, so let's see,
2 x 2 = 4, times 7 is 30.
30 x 10^(19) 30 - 11 = 19
Divided by--3 x 3 = 10.
(10^(8))^(2) = 10^(16).
So, that's 3 x 10^(3),
because 19 - 16 = 3--oh,
in meters, because we're still
in MKS units,
because that what we're using.
That's the units of G
that we've chosen.
3 x 10^(3);
that's 3 kilometers.
That's pretty small for the Sun.
And so, you might imagine that
such objects are rare,
or perhaps even non-existent,
because it would have to be
incredibly small,
and therefore,
incredibly dense,
to have a strong enough
gravitational field to prevent
light from escaping.
And so, you might have thought
that this is an entirely
theoretical concept,
and that no matter how
interesting it is,
there's no point in really
studying it further,
because you're unlikely to ever
encounter one of these things in
real life.
But one of the strange things
that was--has been known for
quite some time is that black
holes really should exist,
and that they are predicted to
exist.
This has been known for quite a
while--known for at 70 years,
that black holes should exist.
And the reason they should is
that they are one of the
possible end points of stellar
evolution--the evolution of
stars.
And so, now I want to summarize
how stars evolve.
I should say,
this is the subject of whole
courses.
You can take Astro 350 and they
will talk about this for the
entire semester.
You can take Astro 110 and then
they'll talk about it for a
month.
But we're going to do it in
about twenty minutes,
save you time and effort,
here.
So, a star's life is determined
by the competition between two
forces;
so, star's lifetime and
evolution is determined by two
forces.
One is gravity,
which has the tendency to hold
the star together.
And now, the thing about
gravity is, it ought to work.
But stars have no solid surface.
So, if you can imagine gravity
pulling on an atom on the
surface of the star,
why shouldn't that atom just
fall all the way down to the
bottom of the star?
Because there's no solid
surface to prevent it from doing
so.
And the answer is that gravity
isn't the only force operating,
there's also pressure.
So, there's gravity,
which pulls stuff in,
and pressure,
which has the tendency to push
out.
And these things balance in
most stars.
In the Sun, for example,
these two forces are in balance
at all points in the Sun,
and this balance goes by the
technical name hydrostatic
equilibrium.
Hydro, because it's a fluid,
it's not a solid surface.
Static--nothing's moving,
and equilibrium is just
balance.
And, to be a little more
precise, the way this
works--here's the surface of
some star.
Here's some point within the
star, and there are two kinds of
forces acting on this point.
There's gravity,
which is pulling the thing
toward the center of the star,
and then there's pressure
forces in two different ways.
The outer regions of the star
exert a pressure inward.
So, there's an inward pressure.
And the inner regions of the
star exert a pressure outward.
And the outward pressure has to
be greater than the inward
pressure by exactly the right
amount to counteract gravity.
So, it basically looks like
P_out -
P_in = gravity.
And this holds true at all
points.
Now, in order--it's the other
way around right?
Thank you.
P_in -
P_out--yeah,
right.
P_in minus
P_out has to
equal gravity.
And that requires that the
pressure on the inside has to be
bigger than the pressure on the
outside, because you don't have
negative gravity;
at least, not until the third
part of this course.
But, at the moment,
you don't have any negative
gravity, and so the pressure on
the inside has to be greater
than the pressure on the
outside.
Okay, so what is pressure?
Cast your mind back to high
school chemistry.
Remember high school chemistry?
It doesn't matter if you don't.
I'll tell you everything you
need to know.
There's something called an
ideal gas, and there's something
that the ideal gas does,
which is to exert gas pressure.
Your high school chemistry
teacher probably wrote down
something that looks like this:
PV = nRT. And
here's the thing about this.
V is volume in this case;
P is the pressure;
n is the number of
particles per volume.
And so, the key thing here is
that n divided by
V is equal to the
density, basically,
times a constant.
Because, remember,
density is mass per volume.
If you take the number of
particles and you multiply by
the mass of each particle,
that'll give you the total
amount of mass in a given
region.
You divide by the volume,
that equals the density.
And so, this also is a
constant, this R thing.
And so, what you get is
P is equal to a constant,
times the density,
times the temperature.
So, this is how physicists
think of the ideal gas law,
because we prefer to work in
terms of the density.
Okay, so here's the pressure.
And the pressure on the inside
had better be bigger than the
pressure on the outside,
or this isn't going to balance,
which means,
either the density or the
temperature,
or both, had better be larger
in the middle of the star than
it is in the outer parts of the
star.
So, inside of the star,
the T and/or the ρ has
to be bigger than it is on the
outside.
Now, it turns out that if you
just keep the temperature
constant all the way through the
star, you never achieve this
balance.
So, if only the density varies,
then inner regions do have
higher pressure but the increase
in density also increases the
force of gravity,
because gravity is dependent on
how much mass there is.
And, if you increase the
density, you also increase the
amount of mass.
So, you have higher pressure,
but you also have higher
gravity.
And it turns out that you can
prove, mathematically speaking,
that no balance is possible,
because it always ends up being
the case, for gas pressure,
that the amount you have to
increase the density by,
if you're only increasing the
density, will also increase the
gravity, and you'll never get a
balance.
So, the consequence of that is
that the inner parts of a star
must be hotter than the outer
parts.
Otherwise, the star wouldn't
exist;
it would collapse.
And this is true.
The inside of the Sun turns out
to be something like 10^(7)
degrees.
The surface of the Sun turns
out to be something like 6 x
10^(3).
So yes, indeed,
the inside very much hotter
than the Sun,
that is very much hotter than
the outside,
in the Sun, and that's what
keeps the Sun in balance.
And there's a problem with this.
And the problem is that there
is something called
thermodynamics.
And one of the laws of
thermodynamics is that heat
tends to flow from places where
it's hot to places where it's
cool.
This is evident in everyday
life.
If you take a little piece of,
I don't know,
molten lead or something,
and you drop it in a bucket of
water,
the heat from the lead will
spread into the water.
The water will increase in
temperature very slightly.
The heat will come out of that
piece of lead.
The lead will solidify,
and everything will come into a
kind of temperature balance.
Similarly, if you put a
snowball in some hot place,
it'll melt.
Why?
Because the heat from around it
will go into the snowball.
The temperatures will try and
equalize each other and
they'll--it'll come out even.
So, this law of thermodynamics
is why the snowball has no
chance in hell.
And so, this happens in stars
too, right?
So, the heat in the center of
the star flows out.
And when it gets to the
surface, it's radiated.
At the surface,
it radiates,
and that's the energy that we
see coming from the star.
It's this heat that was in the
center.
It's gotten to the surface.
It's now radiating away out
into the cold depths of space,
and that's what we see.
But, that means that the
temperature in the center of the
star, which is holding the star
up, decreases,
and then the star wouldn't hold
itself up.
So you require--in order for
the star to hold itself out--an
energy source at the center of
the star.
And this does two things.
It replaces all that lost heat,
and it preserves the
equilibrium of the star so it
doesn't collapse.
Okay, so this is all pretty
abstract.
And it was known that this had
to be true before they figured
out what the energy source was.
It was known for people just
sort of thinking in very general
terms about how the Sun could
exist--understood that there had
to be some kind of large source
of energy down in the middle of
the star.
And, notice it has to be in the
middle of the star.
It does no good for the energy
to be created all the way
through the star,
because if it's created all the
way through the star,
then the temperature is
distributed throughout the star
and you don't get a situation
where it's much hotter in the
middle than it is on the
outside.
So, everybody knew for quite a
while that there was energy
being created in the center of
the star.
They just didn't understand how
that was done.
And then when people invented
nuclear physics in the 1930s and
'40s and '50s,
it was understood that this
comes from nuclear
reactions--nuclear fusion,
in particular.
In the case of the Sun,
it's the fusion of hydrogen,
atoms together to make helium,
that does this.
And that releases energy in the
same way that a nuclear bomb
does.
The problem with this is that
eventually you run out of
hydrogen, or whatever your
nuclear fuel is,
because you've got only a
limited amount of it in any one
star.
So, eventually,
the nuclear fuel runs out.
And then the star has many
adventures before it settles
down.
And for these,
I will have to refer you either
to a textbook or to some other
course,
because I'm not going to take
you through the whole exciting
life of a star once its nuclear
fuel is exhausted.
Suffice it to say that you know
in advance what the outcome has
to be, because there's no way it
can hold itself up,
in the long run,
because it doesn't have an
energy source down at the center
of the star.
So, the consequence of this has
to be that the star collapses.
Now, it doesn't necessarily
collapse all the way down to
being a black hole,
because there are other kinds
of pressure besides the pressure
exerted by an ideal gas.
So, at very high densities,
you get other kinds of
pressure.
In particular,
there's something called
"electron degeneracy pressure."
This is sometimes called Fermi
pressure, after the guy who
thought it up.
Degeneracy is another one of
those words that means something
different to physicists than
they do to ordinary,
normal people.
I remember the first time I
taught Astro 110,
right after I came here,
about fifteen years ago.
I started talking in the middle
of a class about degenerate
white dwarfs.
And you could feel this sort of
wave of something between
confusion and anxiety permeate
through the class,
and I had no idea what was
going on.
Some teaching assistant had to
pull me aside afterward and
remind me that these words mean
different things.
So, I apologize if this sounds
like a sort of adult version of
a Grimm's fairy tale,
you know, with degenerate white
dwarfs wandering around and
stuff, but such are the words we
have to work with.
Okay.
So you have this electron
degeneracy pressure--that's a
different kind of pressure –
and that can stabilize the star.
So it stabilizes a
star--stabilizes the star
at--around the radius of the
Earth.
So, that's very high density
material.
We can calculate the density.
You'll remember our equation
for density: mass over volume.
And so let's see,
how does this work?
This will be the mass of the
Sun, let's say,
divided by the volume of the
Earth.
That's 4/3πr^(3).
Radius of the Earth is (7 x
10^(6))^(3).
So, let's see,
(2 x 10^(30)) / 4.
7^(3)-- 7^(2) = 57,
cubed is 350.
Times 10^(18).
That's 6 x 3--yeah,
okay, we're doing all right.
And let's say that that's 2,000
x 10^(27) over,
I don't know,
1,400 x 10^(18).
Let's cancel those.
10^(9) kilograms per meters
squared.
That's about a million times
denser than water.
Water, you'll remember,
is 10^(3) kilograms per meters
squared.
And water is defined to have a
density of 1 gram for a cubic
centimeter.
So, if you were to pick up one
gram of this white dwarf--one
cubic centimeter of this white
dwarf,
it would be a million times
more massive than a gram.
That's about a ton.
So, a thimbleful of this stuff
weighs about a ton,
very dense.
And the Sun will end its life
as a white dwarf with this
electron pressure balancing the
gravity.
These--such stars are called
white dwarfs,
and there are many of them
known.
White dwarfs--this is the end
point of the Sun.
And then, in the 1930s,
what happened to make black
holes inevitable,
was that one of the great
theoretical astrophysicists of
the twentieth century,
a man named Subramanyan
Chandrasekhar,
discovered that electron
degeneracy pressure doesn't
always do the job.
So, in the 1930s,
Chandrasekhar discovers--proves
that if the mass of an object is
greater than 1.4 times the mass
of the Sun,
electron--this kind of electron
pressure is insufficient.
And the star continues to
collapse.
Now, Chandra was a graduate
student in England at the time
he figured this out,
and he presented this rather
dramatic result at a big meeting
of the Royal Astronomical
Society in London.
And then, every graduate
student's worst nightmare took
place.
Chandra's thesis advisor was a
very famous man named Arthur
Eddington, one of the great
physicists of the early
twentieth century.
And after Chandra had presented
his results to all these
assembled scientific
dignitaries,
Eddington got up and denounced
his own student,
and said, "This can't possibly
be true."
And Eddington made the famous
remark, "There ought to be a law
of nature to prevent stars from
behaving in this foolish
manner."
And the consequence of that is
that a lot of people didn't
follow up on Chandra's idea.
Chandra got miffed,
as you might understand.
He got on the boat to the
United States.
He spent the rest of his career
at the University of Chicago.
On the boat,
he wrote a great textbook still
in use on the structure of
stars,
in which he laid out in detail
all of the arguments that the
Chandrasekhar limit must
actually exist.
And fifty years later he got
the Nobel Prize for it.
But that--there was this kind
of time lag there,
and it is, I think,
important for those of us,
say, on the faculty to recall
that if Eddington had listened
to his student instead of to his
intuition,
the study of black holes would
probably be forty years more
advanced than it is.
So, another fable for our times.
Chandra was always very
gracious about this,
actually.
He would praise Eddington to
the skies as a wonderful
advisor, and then with this
little asterisk.
So, fable: Chandrasekhar's
limit is the title,
and the moral here is,
"believe your student,
not your intuition."
And actually,
the story of Eddington was
actually kind of interesting.
As Eddington got older,
he became more and more
convinced that,
you know, he could guess the
right answer.
And so, it wasn't just the
Chandrasekhar limit,
it was other things.
He got a little weird toward
the end of his life,
and he started believing his
intuition.
Einstein did this too, right?
Einstein famously discovers all
this great stuff,
and then the second half of his
life is completely useless,
scientifically,
because he becomes convinced
that his gut is telling him that
quantum mechanics is wrong.
This famous remark,
"God does not play dice with
the universe"--but it turns out,
that isn't true.
And so, there is this
probabilistic nature of reality
that quantum mechanics shows.
And so, Einstein spent the
second half of life trying to
prove that his intuition was
correct,
that quantum mechanics couldn't
really be true and thus,
did no physics worth doing for
about thirty or forty years.
So, you have to watch out for
this.
If you're too smart and start
believing yourself,
you can get into trouble.
All right.
Now, having said all that,
Eddington was partly right.
There is, sort of,
a law of nature that prevents
stars from behaving in this
foolish manner.
So, to understand that,
what happens when white
dwarfs--when the white dwarf
collapses?
It's got to get rid of its
electrons, because the problem
is, the way this electron
degeneracy pressure works--you
can't squeeze electrons any
closer together than they are in
a white dwarf.
So now, you have to get rid of
electrons.
And so what do you do?
You combine the electrons and
the protons and you turn them
into neutrons plus neutrinos.
These are neutrinos.
They stream out.
And so, you end up with
something that's made entirely
out of neutrons.
So, the whole star turns into
neutrons.
A chemist would think of this
as essentially turning the whole
star into one atom,
into one atomic nucleus.
An atomic nucleus with no--an
atom with no protons,
no electrons,
and 10^(57) neutrons.
And you could imagine putting
that somewhere on the Periodic
Table.
Astronomers call these things
neutron stars,
and they exist.
They were discovered in the
1960s.
And a typical neutron star,
a couple times the mass of the
Sun has--mass equals 2 times the
mass of the Sun.
Radius of about 10 kilometers.
And you can work out the
density for that.
Density is a billion times
greater--I'll leave this as an
exercise--greater than for white
dwarfs.
So, instead of a cubic
centimeter of the stuff weighing
a ton, it now weighs a billion
tons.
And you're having a tough time
moving it around.
But 10 kilometers – that's
getting close to the
Schwarzschild radius.
Remember, we calculated the
Schwarzschild radius of the Sun.
It was about 3 kilometers.
And, in fact,
if you calculate the
Schwarzschild radius of a star
in terms of the Schwarzschild
radius of the Sun--let's see,
you get 2GM /
c^(2),
where M is the mass of
the star,
divided by 2G mass of
the Sun, over c^(2).
So the Gs and the
cs all cancel,
here.
And you get
M_star /
M_sun.
So, if the Chandrasekhar mass
of the Sun is equal to 3
kilometers, as we calculated,
the Schwarzschild radius of a
star with--whose mass happens to
equal 3 times the mass of the
Sun,
is going to be 3 x three
kilometers, or 10 kilometers.
That's equal to the radius of a
neutron star.
So, a neutron star with mass
greater than 3 times the mass of
the Sun has a radius less than
its Schwarzschild radius.
And that's a black hole.
Remember?
And the key thing here is that
there are lots of stars with
mass more than 3 times the mass
of the Sun.
We don't see them as black
holes because they're still in
hydrostatic equilibrium.
But eventually,
they're going to run out of
nuclear fuel,
and they're going to collapse.
Now, in fact,
during the course of the star's
life, one of the things I
glossed over is stars tend to
lose mass as they live.
And so, they don't end up with
the same mass they started with.
But stars with initial masses,
at the beginning of their
lifetime, greater than--I don't
know,
something like thirty times the
mass of the Sun,
will end up with masses greater
than three times to the mass of
the Sun.
And then, there's nothing to
stop their collapse.
What happens is,
they turn into neutron stars,
but they turn into neutron
stars whose Schwarzschild
radius--whose radii are smaller
than their Schwarzschild radius,
and that is a black hole.
So, they collapse down into
black holes.
And so, you expect a large
number of black holes to
actually exist.
This is what happens to massive
stars at the end of their life.
And so, we expect there to
be--that there are many black
holes.
And so, the question we'll be
exploring in the rest of this
segment of the class is,
how can you find these things?
What are the properties of
these things from a theoretical
point of view?
What does Einstein's Theory of
Relativity suggest that these
things are going to behave like?
And then, the big question is,
once you've found some,
and you have a theory for what
they behave like--then you can
ask the question,
does the actual behavior of
these objects conform to the
theoretical expectations?
Another way of saying that is,
was Einstein right?
Is general relativity the
correct theory to describe these
very exotic objects?
And so, that's what we'll be
talking about in the rest of
this section of the course.
Now, let me turn back to the
previous section of the course,
which was--which culminated
last time in this little test.
And I think we're ready to hand
these back.
Is that true?
 
