The speed of sound is
340 meters per second--
it depends a little bit
on the temperature--
about 770 miles per hour.
When I speak to you, my sound
reaches you with that speed.
I produce
a certain frequency here,
a certain number
of oscillations per second.
They reach you,
your eardrum starts to oscillate
with the same frequency
and you hear that tone.
I have here a tuning fork
which oscillates
440 times per second.
(tuning fork produces
medium-pitched tone )
Your eardrum oscillates
440 times per second--
you hear this tone.
Here I have
256 oscillations per second.
(metal rod emits lower tone )
Your eardrum is now shaking,
going back and forth
256 times per second.
If you stay where you are
and you don't move
and I move these tuning forks,
you will hear
a different frequency
and that's what we call
Doppler effect.
If my sound source
approaches you,
you will hear a frequency
f prime which is larger
than the frequency
of the tuning fork.
If it moves away from you,
which I will call receding,
then f prime equals
lower... lower frequency.
For instance, I move to you
a sound source--
I call that a transmitter--
with a speed of
about one meters per second.
Transmitter is
the sound transmitter.
Then if it approaches you here,
you will hear f prime,
which is 1.003 times f.
This three here is
the one part out of 340
that you get an increase
in frequency.
If I move it away from you,
then f prime would be
0.997 times the frequency
of the source itself.
You stay where you are.
I have here a tuning fork
which generates 4,000 hertz,
a very high frequency.
If I move it to you with the
speed of one meter per second,
which I can do, then you get
an increase in pitch of 0.3%.
That makes it 4,012 hertz.
And when I move it away from you
there is a decrease of 0.3%.
And you can clearly hear
that difference.
I will first make you listen
to the 4,000 hertz
without my moving.
(tuning fork produces
very high pitched tone )
Can you hear it?
Very high frequency.
Is it painful, really?
High frequency.
Most of you are young enough
you should be able
to hear 4,000 hertz.
Okay, now I am going to move it
to you one meter per second
and away from you.
(high tone goes up and down
slightly in pitch )
Did you hear it? Once more.
(tone goes up and down
quickly again )
(class laughs )
When it comes to you, it's clear
that the frequency goes up,
and when it moves away from you,
the frequency is down.
Now imagine that I'm going
to rotate the sound source
around in a circle.
Now the sound that you receive,
the frequency that you receive
will change
in a sinusoidal fashion.
If this is that circle, and this
is the radius of that circle,
and if you are here
in the plane of the circle,
then when the source comes
straight to you
with the velocity v-- let's say
it's a uniform circular motion--
f prime will be larger than f
and it will, in this case,
reach a maximum.
When it is at 90 degrees
relative to you--
I don't have to give it
a vector notation--
f prime equals f.
When it moves away from you,
f prime is smaller than f,
you hear a minimum.
And when it is here again--
when the angle between the
velocity and your direction
is again 90 degrees--
then f prime equals f again.
And so this phenomenon
is called the Doppler effect.
So if I twirl it around,
you will hear a sinusoidal
fluctuation in f prime.
Suppose I plot,
as a function of time, f prime
the way you will receive it--
you sit still,
but I'm going to move the
sound source around like this.
Then you will have a curve
that looks something like this:
some sinusoidal-cosinusoidal
fluctuation of f prime.
This will be the value f
produced by
the sound source itself.
This will be f prime maximum and
this will be f prime minimum.
If you could record this, there
is an amazing number of things
that you can deduce
from this curve.
First of all, you can take...
You can measure
f prime max divided by F,
because you see this curve,
so you know what f is here,
you see what f prime max is,
and that should allow you
to retrieve immediately
v velocity
of the transmitter.
If that number were 1.003,
then you know
that the speed in the orbit
was one meter per second.
So this ratio immediately gives
you the transmitter velocity.
This time separation
gives you immediately
the period of rotation,
but since two pi R--
if R is the radius
divided by the velocity
of the transmitter--
since that is the... oh, I can
reverse it, it doesn't matter.
Two pi R divided
by the time to go around
is the velocity
of the transmitter.
Since you know the velocity of
the transmitter from this ratio
since you know the period,
which is this,
you now also find the radius R.
So from that curve--
and keep that with you,
because it's going to be
important in what follows--
we can derive three things:
the radius,
the period of rotation
and the speed of the object
as I twirl it around.
I have here what we call
a wind organ.
When I twirl this around,
it produces a particular tone.
We will talk later about 801
why it produces
a particular tone.
Sometimes you hear two tones.
I'll try to make you hear
only one.
And as I swirl it around,
the sound is coming...
the sound source, the
transmitter is coming to you.
This, when it goes like this,
it's 90-degree angle
so you should not hear
any Doppler shift.
When it is here,
it's moved away from you
and so you hear
a sinusoidal change in f prime.
Try to hear that.
(wind organ producing tone
that changes pitch )
Can you hear, when it's coming
to you, that it's higher-pitched
than when it's going
away from you?
Can you hear that?
Just say no
if you don't hear it.
Not very clear.
(wind organ again producing
varying tone )
For me, it's impossible to hear
because I'm standing
right under it, of course.
Well, I tried.
 
I now want to change
to electromagnetic waves.
Electromagnetic waves travel
with the speed of light,
which is 300,000 kilometers
per second.
And if you want to treat
that correctly,
you would have to use
special relativity.
In the case of sound,
I stressed repeatedly
that you in the audience
should not move
but that the sound source
is moving.
In the case
of electromagnetic radiation
when you deal
with the speed of light,
you don't have to ask
that question.
It is a meaningless question
in special relativity.
To ask whether you are moving
relative to me
or whether I am moving relative
to you, it doesn't matter.
All that matters
in special relativity
is the relative motion,
so you can always think
of yourself as standing still
and make the source
of electromagnetic radiation
move to you, or away from you,
relative to you.
Electromagnetic radiation
is optical light, infrared,
ultraviolet, radio,
x-rays, gamma rays.
All of that is
electromagnetic radiation.
If the velocity of the source
of electromagnetic radiation--
the transmitter--
if that is way, way smaller
than the speed of light,
then it is very easy to predict
the change in frequency
due to Doppler shift.
Let this be the transmitter
which produces frequency f,
and here is the receiver which
receives the frequency f prime.
And let the velocity
of the source of
electromagnetic radiation be v--
I could put transmitter here,
but we can drop that index--
and let this angle be theta.
Then this is the component
in your direction--
we call that
the radial component--
which is v cosine theta.
So I delete the tr.
This is just the velocity
of the source
relative to you at that angle.
 
If now we want to know
what f prime is,
then f prime equals f
times one plus v over c
times the cosine of theta.
What matters is only the radial
component of the velocity.
This is the radial component.
If theta is 90 degrees,
just like we had with sound,
then f prime equals f.
So 90 degrees, the cosine
of theta is zero,
f prime equals f.
If theta is smaller
than 90 degrees,
then it's coming towards you,
then f prime equals
larger than f.
If theta equals larger
than 90 degrees,
it's going away from you,
then f prime equals
smaller than f.
 
You would get
a similar equation for sound
by replacing this c
by the speed of sound.
But I want to stress
that this only holds for
electromagnetic radiation
if v over c is much,
much smaller than one.
 
Now, when we deal with sound,
there is something
mechanically oscillating.
Something is vibrating.
With electromagnetic radiation,
charges are vibrating.
Electrons are vibrating,
and they are vibrating
with a certain frequency,
and that means
there is a certain period
of one oscillation.
And that period of one
oscillation is, of course,
one over the frequency.
I can ask myself now
the question,
how far does
electromagnetic radiation,
how far does light travel in the
time of one period capital T?
Well, it goes
with the speed of light,
so in T seconds,
it moves a distance cT.
And that distance we call
the wavelength
of electromagnetic radiation,
lambda equals cT, for which you
can also write c divided by F.
So this is the wavelength of
the electromagnetic radiation--
the speed of light,
300,000 kilometers per second--
the period of one oscillation,
say, of the electrons,
and this is the frequency,
which you can give in hertz.
I could give you
a specific example.
I, for instance,
can take a period T
of two times
ten to the minus 15 seconds.
That would give me a wavelength
of about six times
ten to the minus seven meters--
six times ten
to the minus seven meters--
and that you would experience
as red light.
If I make the period shorter--
say, 1.3 times
ten to the minus 15 seconds--
I get a shorter wavelength.
I get four times
ten to the minus seven meters,
and you would experience that
as blue light.
 
In astronomy,
in optical astronomy
we cannot measure the period or
the frequency of optical light.
All we can measure
is the wavelength.
And so if I want
to use this equation,
then I have to replace f prime
by c divided by lambda prime
and f I have to replace
by c divided by lambda.
And when I do that,
I get the following result.
I get lambda prime equals lambda
times one minus
v over c cosine theta.
If this is a plus,
this is a minus.
Check that for yourself.
You have to use
the small number approximation,
the Taylor expansion,
namely that v over c is
much, much smaller than one.
So you can see now
if the object comes to you,
in other words, if
f prime is larger than f,
if the frequency is higher, then
the wavelength will be smaller.
And so let me write that down.
When the cosine of theta...
so the object is coming to you--
when the cosine of theta
is larger than zero,
the object is approaching you,
then the wavelength lambda prime
will be less than lambda.
And that has a name--
we call that blue shift.
And the reason why we call
that blue shift is
that if the wavelengths
become shorter,
it moves towards
the blue end of the spectrum,
because blue has
a lower wavelength than red.
If cosine theta is negative,
then the object
is receding from you,
then lambda prime is larger
than lambda,
and we call that red shift.
These are the terms that
astronomers use all the time.
When you make a spectrum
of a star--
you can do that using prisms
or by other means--
and you look
at the light intensity
as a function of wavelength--
so here is the light intensity
as a function of wavelength--
then you may expect to see
some kind of a continuum.
But, in fact,
what you do see is...
Superimposed on a continuum
you see sometimes
very sharp absorption lines--
black, missing light,
called absorption lines
And these absorption lines
correspond
to elements in the atmosphere
of the star.
In fact, if you see the
absorption lines, you can tell
what kind of elements
are present in the star.
Some are very characteristic
absorption lines:
some for hydrogen, some
for calcium, some for silicon,
some for magnesium and so on.
It's actually interesting
that when you look
at the spectrum of the sun--
when people did that first,
when they had the means
of doing that,
they found absorption lines
in the spectrum of the sun
which could not be identified.
They had never been seen here
on Earth, these lines,
and so they called these lines
after the sun.
The sun is Helios,
and so they called it helium.
So helium was
first discovered on the sun
before it was
later discovered on Earth
by looking at the absorption
lines of the solar spectrum.
If a star moves to you,
then all the lines--
every single line--
will be blue-shifted.
And if the star moves
away from you,
all the lines will be
red-shifted.
If you take an example:
With lambda prime
divided by lambda
and you pick
any one of those lines--
it doesn't matter
which you pick
because they will all do
exactly the same.
If this were,
for instance, 1.00333--
I just pick a very nice number;
that means lambda prime, as
you see, is larger than lambda;
the wavelengths get longer,
so we have a red shift--
and you substitute that
in that equation,
then you'll find
that the velocity
at which that star is moving
relative to you--
that gives you
immediately the answer there--
equals minus 0.00333
times the speed of light, c,
and that is
minus 100 kilometers per second.
And the minus, then, reminds you
that the object
is receding from you.
So that gives you a red shift.
 
I just wrote down
that the velocity v is
minus 100 kilometers per second.
It's, of course, v cosine theta
that is minus 100
kilometers per second.
It's theradial velocity--
that's all you can measure.
You have
no information on theta.
So it is this component,
v cosine theta--
which we call
the radial velocity--
that is minus 100 kilometers
per second.
 
Half of all the stars
in the sky are binaries,
and so when you look
at the spectra,
you will see them go
around each other.
And so you, in principle,
can measure the red shifts
and the blue shifts
as they go around each other.
You see Doppler effect.
If they come to you,
you see blue shift.
If they go away from you,
you see red shift.
So, in principle,
you can determine
for each one of those stars
the velocity in orbit,
the radius of their orbit
and, of course, the period
of the binary system.
So it's an extremely
powerful tool in astronomy
if you have a binary system,
when the stars exactly do this,
to determine
all these quantities
that you would like to know.
I first would like to show you
now some slides.
 
The first slide... oh, I have
to lower the screen, by the way.
That would help, wouldn't it?
(chuckles )
The first slide is a spectrum
made in the laboratory
of hydrogen, helium
and calcium and sodium.
It shows you emission lines,
no absorption lines.
These lines are produced
by lamps,
and the frequencies
are very well known.
Here you see
the famous sodium yellow lines.
So here is the red part
of the spectrum
and there is the blue part
of the spectrum.
 
So we know these frequencies,
we know these wavelengths
very well.
And here you see
the spectrum of the sun
with all these absorption lines
that I mentioned to you.
It's plastered
with absorption lines,
and each of them
can be identified.
These are due to calcium,
iron, hydrogen and so on.
Here is the blue part
of the spectrum,
here is the green part,
the green part,
and here is the red part
of the spectrum.
 
And here you see the basic idea
behind a binary system.
Suppose you have a binary system
that only one star is visible
and the other one is invisible
and the one star shows you
three clear absorption lines.
Then as the star moves
around the center of mass,
you see that all these lines
drift in unison.
And out of this information
you get the radius,
the velocity and the period,
assuming that you are on Earth
in the plane of the orbit
of the stars.
If you have a binary system
whereby both stars are visible
so you get
the spectrum of both stars,
then you see the Doppler shift
of both stars in the spectrum.
Here we have a simple case
that we only have two absorption
lines, not to confuse the issue,
and so in one...
in the case of one star,
the shift will be towards
the left of the two lines,
but the other star, the shift
will be to the right,
because if you have
a binary system
when one star comes to you,
the other star goes away
from you, and vice versa.
So now you are very lucky,
now you have an ideal situation
that you can find for both stars
the radius of the orbit,
the velocity in orbit
and the period for each star,
which, of course,
is the same for both.
 
And here you see real data.
You see here, first of all,
the emission lines which are
measured in the laboratory
that I just showed you.
They are always done
simultaneously
with the measurements.
You always must be sure
that you have a good calibration
of your wavelength.
And this spectrum a,
the top spectrum, is of a star,
a binary system, that has
a period of 20.5 days.
And you see here single lines,
if you have good eyes.
That means at this very moment
both stars move relative to you
at angles of 90 degrees,
so you don't see
any Doppler shift.
But now look here.
Later in time, you see that
this line has split in two lines
and this one has also split
in two lines.
Clearly, one component
is coming to you
and the other component
is moving away from you.
And so you get all this
useful information in astronomy
by making the Doppler shift
measurements of binary systems.
 
I want to pursue the idea
of binary stars.
They give us
not only the information that
we want regarding the orbits,
but there is even more
that we can get out of it
which is even more exciting.
So I will remind you what
a binary system looks like.
 
Remember the second exam.
I'm sure you will never forget
that second exam
and maybe never forgive me
for that.
 
Binary system: star one,
radius r1, mass m1, velocity v1,
and star two-- going about
their common center of mass--
mass m2, radius r2
and velocity v2.
m1 r1 equals m2 r2.
That's the way the center
of mass is defined.
Imagine that you as an observer
are somewhere in the plane of
this orbit, and you are here.
And you are observing
the system going around.
Kepler's Third Law,
which you derived on your exam
as well as on an assignment:
the period squared equals
four pi squared times r1 plus r2
to the power three.
divided by G times m1 plus m2.
Let me check that to make sure
I have that right.
Yes, that is correct.
Imagine now you can make
the Doppler shift measurements
of both stars.
You make the Doppler shift
measurement of star number one,
so you measure lambda one prime
as a function of time.
Out of that pops immediately
the period of rotation.
Out of that pops the velocity,
as we discussed.
Out of that pops the radius, r1.
And now you measure
the Doppler shift of star two
as a function of time.
Out of that pops the period
which, of course,
better be the same.
Out of that pops
its velocity in orbit
and out of it pops its radius.
All these things come out of
the Doppler shift measurements,
but if you know r1
and you know r2,
then you also know r1 plus r2,
so you know this part
in Kepler's Third Law.
Since you also know the periods,
you can find what m1 plus m2 is.
So now you get an extra bonus.
You know now what the sum
of the mass of the two stars is
in the binary system, but you
also know that m1 r1 is m2 r2.
So now you have two equations.
You know what m1 plus m2 is
and you know this equation,
and you can solve for m1 and m2,
which is an amazing thing
when you come to think of it.
So we finally end up
with the mass of star one
and the mass of star two.
All this comes out
of Doppler shift measurements:
the velocities,
the radii, the periods
and even the masses
of these objects.
Now, if you
as an observer on Earth
are not exactly
in the plane of the orbit,
then the situation is
a little bit more complicated,
and I will not discuss
that here today
because, in principle,
it doesn't affect the idea
behind Doppler shift.
But for astronomers,
it is very important.
It's really a nuisance,
but I will not discuss that
in any detail.
I want to discuss
a fascinating application
that we have in x-ray astronomy.
Namely, we have x-ray binaries.
 
What is an x-ray binary?
Well, it is a binary system.
This is a star
not unlike our sun.
It has as certain mass,
has a certain radius,
and it is in orbit, let's say,
with a neutron star,
even though it could be
a black hole.
But for now, let's just assume
it is a neutron star.
And if these two masses
are the same,
which I only use
for the sake of simplicity--
in practice,
they could be very different--
then there is a point between
these two, right in the middle,
whereby the gravitational pull
in one direction is the same
as the gravitational pull
in the other direction.
And we call that
the inner Lagrangian point.
In other words,
if you were there,
the neutron star would pull
at you
with exactly the same force
as the other star.
So you wouldn't know
where to go.
If this inner
Legrangian point lies
below the surface of this star,
that means
if the stars are a little closer
than I have drawn them here,
then the matter of this star
will fall
towards the neutron star,
because the pull in this
direction is, then, larger
than the pull in this direction.
Now, of course,
this system is a binary system.
They go around in the plane
of the blackboard, say,
and so this matter
cannot fall radially in
but it will fall in
and spiral in
and forms what we call
an accretion disk
around the neutron star.
This is called the accretor
and this is called the donor.
There is mass transfer from
the donor to the neutron star.
Oops, I just noticed I
misspelled the word "accretion."
There is an "r" in "accretion."
And as that occurs,
there is a tremendous amount
of energy that is released.
I want to blow up
the neutron star.
Very simple
801 considerations, now.
What comes is
extremely pedestrian.
This is the mass
of the neutron star
and this is the radius
of the neutron star.
And I take a little bit
of matter m,
and I drop it from a large
distance onto the neutron star.
At what speed
will that little piece of matter
reach the neutron star?
You should almost be able
to close your eyes
and give me that answer
right now.
The kinetic energy when
it reaches the neutron star
equals one-half m v squared.
That is the speed
at which it will crash
onto the neutron star,
and that must be
mM neutron star G divided by
the radius of the neutron star.
You always lose your m,
and so you find
that the speed at which
it reaches the neutron star
is the square root of two
M neutron star
times G divided
by R neutron star.
You should remember
this equation.
This was the equation
that we had for escape velocity.
If you were here,
and you go back to infinity,
you reach exactly that speed,
so if you fall in from infinity
that is exactly the speed
at which you reach
the neutron star.
It should obviously
be the same number.
And you don't really have to be
infinitely far away;
you just have to be
much further away
than the radius
of the neutron star.
When this matter crashes
onto the neutron star,
the kinetic energy that is
released is one-half mv squared.
It is converted to heat,
and to give you some feeling
for the incredible power
of a neutron star,
if you make this little m
as little as ten grams--
think of it as a pretty
full-sized marshmallow--
and you throw a marshmallow
from a large distance
onto a neutron star,
the energy that is released
is comparable
to the atomic bomb
that was used on Hiroshima.
A ten-gram object thrown
onto a neutron star--
the reason being
that this velocity
becomes enormously high.
If you put in
for the neutron star
a mass of about three times
ten to the 30th kilograms,
and you take for the radius
of the neutron star
about ten kilometers,
you will find
that that velocity becomes
about two times ten
to the eighth meters per second,
which is about 70%
of the speed of light.
And because
of thisenormous speed--
one-half mv squared
is horrendously high--
it is a conversion
of gravitational potential
energy to kinetic energy
and then ultimately to heat.
Now, nature is transferring mass
at an extraordinarily high rate
in many of these binary systems.
There are at least
some hundred or so that we know
in our own galaxy.
The mass transfer rate,
which I call dm/dt--
so that is the transfer rate
from the donor
onto the neutron star--
that transfer rate
is roughly ten to the 14th
kilograms per second.
It is ahorrendous
mass transfer rate.
You can calculate--
by multiplying it
with one-half v squared--
how many joules per second
are released
in the form of kinetic energy.
That means in the form of heat.
And I call that
the power of that neutron star,
and that, then,
for this mass transfer rate,
that's about two times ten
to the 30th joules per second,
which is watts.
And that is
about 5,000 times larger
than the power of our own sun.
But the temperature
of this neutron star--
because of this enormous
amount of energy released,
the temperature
would reach values
of about
ten million degrees Kelvin,
and at that high temperature,
the neutron star would emit
almost exclusively x-rays.
You and I are very cold bodies,
only 300 degrees Kelvin.
We radiate
electromagnetic radiation
in the infrared part
of the spectrum.
We have warm bodies.
When you hold someone
in your arms, you can feel that.
If I would heat you up
to 3,000 degrees Kelvin,
you would become red-hot.
And you actually...
we could turn the light off
and I would see you.
You're just emitting red light.
If I would heat you up
to three million degrees,
you would start to begin
to radiate in x-rays.
You may not like it,
but that's a detail, of course.
So I want you to appreciate
the fact that the...
the kind of radiation that
you get depends strongly
on the temperature,
and at ten million degrees
you're dealing almost
exclusively with x-rays.
 
So these binary systems are
very potent sources of x-rays.
The neutron stars rotate around,
we discussed that earlier--
conservation
of angular momentum--
and they have
strong magnetic fields.
The matter that falls
onto the neutron star
already heats up
during the infall
because there is gravitational
potential energy released,
and so the matter is so hot
that, in general,
it's highly ionized.
And highly ionized material
cannot reach
a magnetic neutron star
in all locations
that it prefers to do so.
In 802, you will learn
why that's the case.
However, the matter can reach
the neutron star
at the magnetic poles.
And so what you're going
to see now is
you're going to have a neutron
star with magnetic poles,
so the matter streams in
onto the magnetic poles,
which gives you two hot spots.
And if the axis of rotation
doesn't coincide
with the line
through the two hot spots,
if the neutron star rotates,
you're going to see
x-ray pulsations.
When the hot spot is here,
you will see x-rays,
and when the hot spot is here,
you will not see x-rays.
And so we observe from these
systems x-ray pulsations.
Now think of the following.
The x-ray pulsations
are a clock.
It is the clock
of the rotating neutron star.
If the neutron star
in a binary system--
because all of these are
in a binary system,
these x-ray binaries--
if it's coming to you,
you see Doppler shift.
The ticks of the clock come
a little closer together.
If the neutron star moves
away from you,
the ticks of the clocks are
a little bit further apart.
That is exactly what
Doppler shift is all about.
So by timing the pulses
of the x-rays,
you can get a handle
on the Doppler shift
of the neutron star.
That means you can get
the speed of the neutron star,
you can get the radius
of the orbit,
and you can get the period,
just like we discussed before.
But now you take an optical...
The x-ray observations,
by the way, have to be made
from outside
the Earth's atmosphere,
because x-rays are absorbed
by the Earth's atmosphere.
Now you take
an optical telescope
and you look from the ground,
and now you see the optical
spectrum of the donor.
And what do you see
in the donor?
You see these absorption lines.
And as the donor moves
around the center of mass,
these absorption lines move
back and forth.
The Doppler shift of the donor.
So you know the velocity
of the donor,
you know the radius
of the donor...
not the radius of the donor--
you know the radius of the
orbit, and you know the period.
So now we have a situation
that I just described earlier
that you have the Doppler shift
of both objects,
and remember, I told you
that you also get the masses.
You get the mass of the donor
and the mass of the accretor.
 
Before I go ahead,
let me show you some slides.
So we have to lower this again,
if that's possible-- yep.
 
I want to show you
an artist's conception
of such a binary system.
 
So this is
what it may look like.
You see the donor there
and you see the neutron star
right here,
so small, of course,
that it's invisible.
And this is the accretion disk.
Swirls in, the matter
ends up on the neutron star.
And this is another view that
gives you an idea of the donor
and then the swirl of matter,
and then it swirls in and ends
up here on the neutron star.
And here you see data
that were obtained in 1971.
It's clear evidence
for the existence
of these rotating neutron stars
with these x-ray hot spots.
You see here the observed x-ray
intensity as a function of time.
And the actual data are
these very thin lines.
And this bold line was drawn
over by the authors
to convince you that you see a
signal which is highly periodic.
The time from here to here
is 1.24 seconds.
This object was called
Hercules X-1.
So this is one
of the magnetic poles
and this is
the other magnetic pole.
One magnetic pole
and the other magnetic pole.
So you see here unmistakably
the rotation of the neutron star
and the x-ray pulsations.
Here you see data
from the same object,
but now the time scale is
very different.
From here to here is one day.
This is two days.
And when you look at this data
alone, forget this for now,
notice that you see
the source is active in x-rays--
the 1.42-second oscillations
you cannot see,
of course, anymore
because the time scale
is different,
but notice here
there are no x-rays at all:
1.7 days later,
no x-rays at all.
1.7 days later, no
x-rays at all.
And so what
you're looking at here
are what we call x-ray eclipses.
When the neutron star moves
behind the donor star,
all the x-rays are absorbed
by the donor star
and you get x-ray eclipses.
In other words, you get...
Independently
from the Doppler shift
you also get the period of
the orbit by the x-ray eclipses.
And this really changed
our whole concept of astronomy,
the existence of these
neutron star binaries.
And now comes a part, what are
the masses of these objects?
I already alluded you
to the idea
of the possibility
that there may be black holes.
All the mass measurements
that have been done to date
of these neutron stars
where you see the pulsations...
all of them are very close
to 1.4 solar mass.
And there's a good reason for
that-- that's not an accident.
In 1930, the physicist
Chandrasekhar predicted
that white dwarfs
could not exist
if their mass is larger
than 1.4 solar mass.
It was a quantum
mechanical calculation
for which he received
in 1983 the Nobel Prize.
Remember we discussed
white dwarfs earlier.
A white dwarf has about
a radius of 10,000 kilometers,
about the same as the Earth.
And imagine
that you have a white dwarf,
and you add matter
to the white dwarf
and you pass
the 1.4-solar-mass mark.
Then the white dwarf
will collapse
and becomes a neutron star.
And so when we measure
the masses of neutron stars,
it turns out,
maybe somewhat by surprise,
that they're all
very close to 1.4.
If you could add more matter
to the neutron star
by accreting
more and more matter
and you reach the point
that the neutron star becomes
as massive as three times
the mass of the sun,
we believe that the neutron star
can no longer support itself
and becomes a black hole.
And so now comes the question,
what is a black hole?
A black hole is the most bizarre
object that you can imagine,
and it is something
that you want
to stay away from, too.
A black hole has no size,
unlike a neutron star.
It has no size, but it does have
mass, and it has a lot of mass--
three times the mass of the sun,
ten times the mass of the sun,
a hundred times the mass
of the sun.
So it has mass,
but it has no size.
We identify around the black
hole a sphere with radius R
which we call the event horizon.
 
Imagine you are
at the event horizon
and you want to get away
from the black hole.
What kind of speed do you need?
You should be able to give me
that answer immediately.
The escape velocity must be
2 MG divided by the radius
of that event horizon.
In other words, the radius
of the event horizon itself
equals 2 MG
divided by c squared.
If you tell me what m is,
I will tell you what the radius
of the event horizon is.
 
I went a little fast here.
I skipped an important step.
v is the escape velocity
from the event horizon,
which is at a distance capital R
from the mass M.
So we see that here.
Now, this escape velocity
can never be larger
than the speed of light, so
the maximum value possible is c.
And if now you look
at this part of this equation
and you take the radius
on one side,
you'll get that the radius
of the event horizon
equals 2 MG divided
by c squared,
and that's how I found
that equation.
Sorry that I went
a little too fast.
 
If M is the mass of the Earth,
the radius of the event horizon
is one centimeter.
If M is the mass of the sun,
the radius of the event horizon
is three kilometers.
If M is three times
the mass of the sun,
the radius of the event horizon
would become ten kilometers.
It scales linearly
with the mass.
If you were inside
the event horizon,
you could never escape
the black hole
because you would need
a speed which is larger
than the speed of light.
Therefore, you can never escape
from inside the event horizon.
Nothing can get out of it,
not x-rays, no radio emission,
no light, nothing.
Once you're inside the event
horizon, you've had it.
You cannot escape it.
And so the question now
that comes up:
Can we see x-rays
from a black hole?
Because if nothing can come
out of a black hole,
how can we see x-rays?
And the answer is yes, we can,
because as long as the matter
that swirls in
is outside the event horizon,
it would still be very hot.
Because gravitational
potential energy
would already have been
released, it would be very hot
and it would emit x-rays.
So we can see x-rays
outside a black hole.
However, you will
never see pulsations,
because a black hole has no
surface, like a neutron star.
So there's no such thing
as two hot spots
which rotate around.
And so now comes the problem
for astronomers:
How can you determine
the mass of the accretor
if the accretor is
not a pulsating neutron star
but if the accretor
is a black hole?
Well, you can only now measure
the Doppler shift of the donor,
because the donor, in general,
is quite well visible.
It's an optical star.
But you will not be able
to measure
the Doppler shift of the
black hole-- no pulsations.
If, however, an astronomer
can make an estimate
of the mass of that donor,
then you will find
the mass of the accretor.
In other words,
instead of having
the Doppler shift measurements
of both stars--
the neutron star
and the donor star,
which gives you the mass
of two stars--
now you have to settle for the
Doppler shift of only the donor
and the mass
of the donor itself.
And if you have a reasonable
idea of what that mass will be,
then you can find the mass
of the accretor.
And there is a very famous case
that was the first one
discovered in the early '70s,
which is called Cygnus X-1.
Cygnus X-1 is an x-ray binary
which has an orbital period
of 5.6 days.
The Doppler shift measurements
of the donor were made,
and astronomers simply looking
at the spectrum--
at the absorption lines
and the structure
of the absorption lines
and the kind
of absorption lines--
were able to say, "Yeah,
the mass of the donor
is probably
approximately 30 solar masses."
And with that information
and with the Doppler shift,
you can now arrive
at the mass of the accretor,
and that is, in this case...
oh, by the way
there is an r missing
in the word "accretion" there--
that mass turns out to be
about 15 solar masses.
 
Now, when this was found
in the early '70s,
most people concluded
this has to be a black hole.
It is a very compact object.
Otherwise it wouldn't emit
x-rays in the first place.
And clearly, if the mass
of that compact object
is way larger
than three solar masses,
then there is no doubt in our
minds that this is a black hole.
Since that time,
many black hole x-ray binaries
have been discovered.
So, if I summarize,
the amazing thing is
from studying the Doppler shift
of binary systems
like x-ray binaries,
you can derive the orbital
parameters, orbital radius,
orbital periods, the speed
of the stars in orbit,
but you can also find
the masses.
And whenever you make
a measurement of the mass
when it is a neutron star
when you see
the x-ray pulsations,
you almost always find
that it is very close to 1.4
times the mass of the sun.
But in a few cases,
you will find
that the mass is
substantially larger.
Admittedly you have to do
without the Doppler shift, then,
of the accretor, but you have
to use some other information,
and then you can conclude
in most cases
with pretty good confidence
that you're dealing
with something like...
bizarre as a black hole,
which you can only define
the event horizon...
And you can never escape
a black hole
when you're inside
the event horizon,
because that is
when the escape velocity
would be larger
than the speed of light.
So this is the escape velocity.
If you set that equal to c,
then you can solve for
the radius of the event horizon,
and out of it pops
this equation.
I would like to show you now
a slide of Cygnus X-1,
which is the oldest known
black hole x-ray binary.
 
I have to lower the screen.
 
And there it comes.
 
This was really a bombshell
when this was discovered.
I still remember reading
that first publication.
Two people discovered this
independently, by the way.
They came independently
to the same conclusion.
Tom Bolton
and it was Paul Merlin--
two independent groups.
 
All right, here is an optical
picture-- it is a negative,
so you see the stars dark
and you see the sky bright--
and right here is the star
that is Cygnus X-1.
It is the donor.
It is a very large star,
a supergiant, huge radius,
and it is believed to have
a mass of 30 times
that of the sun.
You see here the close-up.
This is not the companion,
believe me.
This is just an image
of that star.
The position was... it was hard
to get an accurate position.
Various groups made
a major contribution
to finding the position.
One of the rocket flights of MIT
found a position
that is quite precise
and there was no doubt later...
When the orbital period
was found of 5.6 days,
there was no doubt that
this was the x-ray source.
And so this is a system
whereby you can only see
the donor in the optical light.
You can measure the
Doppler shift of the donor,
and by looking at the spectrum
of this star alone,
you come to the conclusion
that the mass must be
about 30 solar masses,
and then you can argue
that the invisible x-ray source
must be a black hole.
 
 
