We're now entering
the part of 8.01 which is the most difficult
for students and faculty alike. We are going to enter the domain
of angular momentum and torques. It is extremely nonintuitive. The good news, however, is that
we will stay with this concept for at least four
or five lectures. Today I will introduce both
torque and angular momentum. What is angular momentum? If an object has a mass m
and it has a velocity v, then clearly it has
a momentum p. That's very well defined
in your reference frame, the product of m and v. Angular momentum I can take
relative to any point I choose. I choose this point Q
arbitrarily. This now is the position vector,
which I call r of Q. Let this angle be theta. And angular momentum relative to
that point Q-- it's a vector-- is the position vector relative
to that point Q cross p. So it is r of Q cross v,
and then times m. The magnitude of the angular
momentum, relative to point Q, is, of course, rmv, but then I have to take
the sine of the angle theta, so let's say it is
mv r sine theta and this I often call, shorthand
notation, r perpendicular. That r perpendicular is this
distance, relative to point C. What you just saw
may have confused you and for good reason, because I changed my index "Q"
to "C," and there is no C. The indexes should all be Q,
of course. So this r is the length
of this vector. It is the magnitude
of this vector. So this should have a "Q." And r of Q sine theta,
which I call r perpendicular, must have an index Q,
and that is this part here. This angle is 90 degrees and this here is r of Q
perpendicular. No Cs at all, only Qs--
I'm sorry for that. The direction of
the angular momentum is easy. You know how to do
a cross product. So in this case, r cross v would be perpendicular
to the blackboard and the magnitude is
also easy to calculate. Now comes the difficult problem
with angular momentum. If I chose any point
on this line, say point C, then the angular momentum
relative to point C is zero. Very obvious,
because the position vector, r, and the velocity vector, in this
case, are in the same direction. So theta is zero,
so the sine of theta is zero. So you immediately see
that angular momentum is not an intrinsic property
of a moving object, unlike momentum, whichis
an intrinsic property. If you sit there in 26.100, you see an object moving
with a certain velocity, it has a certain mass,
you know its momentum. What the angular momentum is depends on the point that you
choose, on your point of origin. If you had chosen this point D, then the angular momentum
would even be this way, because when you put here
the position vector in there you see r cross v is now coming
out of the blackboard. And this is why angular momentum
is such a difficult concept. But we will massage it in a way
that it will be very useful. Suppose I throw up an object
in 26.100 and at time t equals zero,
the object is here and at time t,
the object is there. So this, then, is
the position vector at time t. The object starts off
with a certain velocity v and a little later, here,
say, the velocity is like so. And there is, of course,
a force on it, mg, which makes this curve. What is the angular momentum relative to point C
at time zero? The angular momentum
is clearly zero, because the point itself,
the mass itself, is at point C. So the position vector
has no length, so it's clear that it's zero. What is the angular momentum at
time t when the object is here? Well, that angular momentum
is clearly not zero, because you see here position
vector and you see the velocity, so clearly the angular momentum
was changing. Now you will say, "Of course
it was changing-- big deal." Because angular momentum has
a velocity vector in it. And here the velocity vector
is changing all the time, so, obviously, you would say the
angular momentum is changing. Well, yes, that is
not a bad argument, but I will now show you a case where the velocity
is changing all the time, but where angular momentum
is not changing. I choose the Earth going
around the Sun. Here's the Earth, with mass m. At point C here is the Sun. This is the position vector
r of C and the Earth has
a certain tangential velocity and the speed never changes,
but the velocity does change. So this is the position vector
at a later point in time. What now is the angular
momentum of the Earth going around the Sun,
relative to point C? I pick C now. Well, that angular momentum... If I take the magnitude
of the angular momentum, because the direction is
immediately obvious... If the object is going
around like this-- this is the position vector-- then the direction will be
pointing out of the blackboard. That's easy. So I'm only worried now
about the magnitude. So the magnitude is
the mass of the Earth times the magnitude
of the cross product between these two vectors. And notice
the angle is 90 degrees. So I can forget about the cross,
the sine of theta is one, and so I simply get mrv,
v now being the speed. This is the case
when the object is here, but when the object is here,
the situation has not changed. Again, r cross v, the magnitude,
is exactly the same, because the sine of the angle
hasn't changed. And so you see here a case whereby the velocity is
changing all the time but your angular momentum
relative to point C is not changing. Suppose I had chosen point Q. Is angular momentum changing
relative to point Q? You'd better believe it. There is a time that the object
will go through point Q. Well, then the angular momentum
is clearly zero because the position vector
is zero. If the object is here and you take the angular
momentum relative to point Q, for sure the angular momentum
is not zero. You have a position vector
and you have a velocity. So only relative to point C--
it's a very special case now-- is angular momentum
not changing. So angular momentum is conserved
in this special case, but only about point C. And I want to address that in
a little bit more general way. I take the angular momentum
and I choose a point Q, and I know that the definition is position vector relative
to point Q cross p. I take the derivative,
time derivative dL/dt relative to that point Q. It's always important that you state
which point you have chosen relative to which you take
the angular momentum. That is going to be dr/dt... excuse me-- cross p plus r of Q cross dp/dt. This is the way that you take the time
derivative of a cross product. We calculate the angular
momentum relative to point Q. So the index has to be Q
throughout the equation. The position vector,
relative to point Q. And in this equation, you see
the correct index Q here. You see the correct index Q
here, but I slipped up here and I put a "C" there. There is no "C" in this problem,
so this is also r of Q. Sorry for that. This, here, is the velocity of
the object, the velocity vector, which is always
in the same direction as p. So this is zero. dp/dt-- that is,
the force on the object-- we've seen that before in 8.01. And so now we have that dL/dt,
relative to a point Q, equals the position vector r
from that point cross F. And this, now, is
what we call torque. And we write for that the symbol
tau... it is a vector. And I put in that Q again. And this is one of
the most important equations that will stay with us
for at least five lectures. What this is telling you is that if there is
a torque on an object, the angular momentum must be
changing in time. If there is no torque
on the object, angular momentum will be
conserved. And now you get some insight into this situation
that we just discussed. The force, the attractive force, gravitational force exerted on
the Earth is in this direction. The position vector is in this
direction, so r cross F is zero. There is no torque
relative to this point C, because the angle between
the two vectors is 180 degrees and so the sine
of the angle is zero. Therefore, no matter
where you are on the circle, always r cross F will be zero. There is no torque
relative to point C. But if you take point Q
or you take here some point A, clearly, there is going to be a
torque, a changing torque even, and so there you will have
a change of angular momentum. So there's something
very special about that point C and I will come back to that,
of course. Now I want to expand the idea
of angular momentum from one point object
that moves in space to an object like a sphere
or like a disk which is rotating
about its center of mass. And I will start with a disk. Here we have a disk. The disk has mass M
and the disk has radius R, and at this point C is
the center of mass of this disk. It's rotating
with angular velocity omega and I want to know
what the angular momentum is of this rotating disk. The direction of the angular
momentum is going to be trivial. If it's rotating like this... If you take here a little
mass element, mass m of i, this is the position vector r
of i, relative to that point C, and here you have
the velocity, v of i. And you see immediately that r cross v
is coming out of the blackboard so that's easy. Angular momentum will be
in this direction, but what is the magnitude
of the disk as a whole? Well, let's first calculate
what the angular momentum is of this little mass element
about this point. So L of C
for mass element i equals... Oh, let's just
only worry about magnitude because already
we know the direction. So that is m of i and then the cross product
between r of i and v of i. But this angle is 90 degrees so I can forget
about the sine of theta. So I simply get r of i, v of i. r of i relative
to that point C times v of i-- this is the magnitude. Now, I hate to see v of i
in a rotating disk because the velocity will depend on how far you are
away from the center. The velocity here is zero. However, they all
have omega in common. Every single element that you
choose has the same omega. So I'm always going to replace--
in a case like this-- v by omega R. And so this then becomes
m of i, r of i of C. I get a square here
and I get omega. So I wrote down
v equals omega R, which, of course,
holds in general. It would have been better,
perhaps, if I had written down v of i equals
omega times r of i, because each element
little "i", which has
a position vector r of i, has a velocity which is given
by v of i equals omega r of i. But I condensed that, sort of, in one equation--
v equals omega R. But this is the connection
that will make it, perhaps, easier for you
to understand what follows. So that is the angular momentum
for this little mass element. But now I want to know what the entire angular momentum
is about that point C as an axis going
through the center of the mass, through the center of the disk
perpendicular to the blackboard. And now, of course,
I have to do the summation of all these elements i. I can bring the omega outside,
and I would have, then, the summation of m of i
r of i relative to that
point C squared. And you see immediately-- I
hope that you see immediately-- that this is
the moment of inertia for a spin around the center
of mass for that point C. And so I can write for this,
I of C times omega. Now comes the question--
so this is the magnitude-- now comes the question,
is this angular momentum different, for instance,
for this point A? And your first reaction
will be, "Yeah, of course, because it depends
on the point you choose." Well, the remarkable thing is
that if you have a rotation about the center of mass
which I have chosen here, then even if you calculate the angular momentum
relative to this point-- or any other point,
even this point in space-- you will always find
the same answer. But only in case that there is a rotation
about the center of mass, and we call that
the spin angular momentum. The spin angular momentum is an
intrinsic property of an object regardless of
which point you choose relative to which you calculate
the angular momentum. So in the case that an object is spinning
about its center of mass, you no longer have to specify the point that you have chosen,
your point of origin. You can really talk now
aboutthe angular momentum. The Earth is spinning
about its center of mass, so the Earth has an intrinsic
spin angular momentum. In addition, it has
an orbital angular momentum. If you want to talk about the orbital angular
momentum of the Earth, however, you'd better do it
relative to that point, otherwise it would be
changing in time. It's only uniquely defined
if you take this special point, because only about that point, which is the location
of the Sun, is the angular momentum-- the orbital angular momentum
of the Earth-- not changing. I'm going to do a daredevil
experiment with you and that is called
ice-skater's delight. You will see that
it is not a delight at all. But in any case, it is
definitely a fun experiment. I have here a turntable--
very little friction-- and I'm going to rotate
the turntable about the center and I'm going to stand
on that turntable and I will hold in my hand
two weights, these two. They're each
about 1.8 kilograms. So these weights m,
1.8 kilograms... My entire mass-- including the turntable
and my body, let's say-- is about 75 kilograms. And I'm going to ask someone
to give me a little twist to rotate me
about this axis of symmetry. Rotate me, say,
if you look from below, let's say
I'm being rotated clockwise. So we have here a situation
of a rotation about the center of mass so we can talk about
the intrinsic angular momentum of this rotating system. And the angular momentum vector will obviously
be pointing upwards. That's clear. If you rotate clockwise
from below... Remember, here you were
rotating counterclockwise; it was coming
out of the blackboard. Here you rotate clockwise,
it will be going up. So far, so good. There is a force on me
due to gravity-- mg, no concern. There is an equally
strong force, normal force up, and the two cancel
each other out. Once I have been given a certain rotation,
a certain angular velocity I'm going to pull my arms in
and pull my arms out and pull my arms in,
and when I do that, that does not cause
a net torque on the system. I can keep doing that
all the time and there is no net torque. And so angular momentum as we have specified
for a spinning object must be conserved,
cannot change. L equals I omega. But as I pull my arms in, my
moment of inertia will go down. And if my moment of inertia
goes down, then if this product
has to remain constant, my angular velocity must go up. And vice versa, so when I pull
my arms in, I will go faster and when I do this,
I will go slower. And I want to be a little bit
quantitative with you. I simplify my own body
by a geometric object for which I can calculate
the moment of inertia which is a cylinder. I may not look like a cylinder, but close enough
for all practical purposes. And this cylinder has a radius
of about 20 centimeters-- not too bad,
it sort of fits me-- and I'm going to rotate
this cylinder around this axis and I can calculate now
what the moment of inertia is. The cylinder has
a mass of 75 kilograms, has a radius of 20 centimeters,
and so the moment of inertia-- in the situation that,
for instance I have these two objects
next to my body here or I have them like here, so
this is my shorthand notation-- equals simple
one-half M R squared. Remember, that was
the moment of inertia-- we discussed that last time--
of a rotating disk which rotates
about the line of symmetry. And so, if I put in the
numbers here, the 75 kilograms, and I take a radius
of 20 centimeters, then I found that this is about 1.5
in our mks units. But now I'm going to put
my arms like this and now the moment of inertia
will go up. And I'll make a very crude
calculation how much it goes up. My arm length is
about 90 centimeters. The weights here
are 1.8 kilogram. So I just assume
that my arms have no weight for simplicity, that all the
weight is in these two objects. It's a simplification, but you
will see it's a dramatic change and that's
all I want you to see. So now the moment of inertia,
when my arms are like this. Of course, there's my body,
which is the 1.5. That is still there but now there is
an additional component: one from this mass,
which is M R squared, and one from this mass,
which is M R squared, where this is now that radius r. And so I get twice that mass and then I have to take R
squared, which is 0.9 squared, and I have to take the 1.8, because that's the moment
of inertia of this object about this point. It is M R squared, I assume
that my arm has no mass. And when you add this up,
you'll find 4.5 in mks units, kilograms meters squared. And now you see, if I go
from this situation to this, my moment of inertia goes down
by a factor of three and if my moment of inertia goes
down by a factor of three, my angular velocity must go up
by a factor of three. And vice versa. I want to do this experiment but this experiment
is not without danger. The problem with this experiment is that the moment
that you pull your arms in you get immediately
extremely dizzy and you can lose your balance and you can fall
flat out on the floor. And I have just talked
this morning with some student here
who did that in high school and he told me that, indeed, one of
the teachers went flat down and I'll try not
to do that today. So I need really assistance
from someone whom I can trust. Do you think I can trust you? Not you. (class laughs ) LEWIN:
That's an honest answer. You're a strong man--
can I trust you? The first thing I want you to do
is to help me get on here, because even getting on here
is not easy. If I just step on here,
I will probably fall. Okay, so stand there,
put your arm around my neck. Support me strongly, yeah, okay. All right, there we go. Now, stay with me for a while,
okay, just stay there. All right, now you give me
a reasonable angular velocity, whatever you think
is reasonable. I'll tell you if it's
completely unreasonable. (class laughs ) LEWIN:
Give me a push, that's fine. Wow... is it fine? Now, you walk a little bit away. If I fall, try to catch me. (class laughs ) LEWIN:
Okay, my arms go in now. My arms go out. My arms go in. My arms go out. Okay, now
I'm completely dizzy now. This is no joke,
so stop me, yeah? Just hold it. (class laughs ) LEWIN:
No, just hold me, hold me. Okay, get my hand. Okay... okay,
you passed the course. (class laughs ) (class applauds ) Whew! Sacrifice
for the sake of science. (groans theatrically ) All right, I've done worse. If we have a collection
of many points-- like we earlier discussed
with momentum-- points that interact
with each other... They could be stars
who gravitationally interact. They could be objects which
are connected with springs. They have internal interactions
which go on all the time. They bounce off each other,
they collide, they break up in pieces,
internal friction, anything. Then if I take two
of these objects, if this one, for instance,
is attracted towards this one, then action equals
minus reaction and these two forces are
identical in magnitude. So if I take any point Q here,
no matter where you choose it, that willnever put a torque
on that system because the two forces cancel
each other out. And so now we get the final
conservation of angular momentum in all its glory if only we add, here,
one little word-- "external." The angular momentum
of asystem... This was angular momentum
of just one object; this is the angular momentum
of a system of many particles. They could be connected
with springs. There could be chemical
explosions going on. They could plow into each other. They could break each other up. The angular momentum
will not change if there is no net
external torque on that system, because all
the internal torques cancel out because action equals
minus reaction. So if we now compare
conservation of angular momentum with conservation of momentum, then in the case of
the conservation of momentum, remember, when we have
a system of objects, in the absence of an external
force on the system as a whole, the net external force,
momentum was conserved. Now we have...
with a system of particles in the absence
of a net external torque, angular momentum is conserved. In the case
of the ice-skater's delight, when you pull your arms in, the moment of inertia goes down
and so your frequency goes up. When a star shrinks,
its radius goes down, its moment of inertia goes down, and therefore its angular
velocity must go up. Moment of inertia goes
with R squared. What determines the size
of a star? If this is a star, then inside this star is
a furnace, nuclear furnace. Nuclear fusion is going on. That produces heat and pressure,
which wants to expand the star. On the other hand,
there is gravity, which says,
"Sorry, you can't do that. I want to hold you together." In fact, gravity would like
to collapse the star. And nature finds a balance
between the gravity and this pressure due
to the nuclear furnace. Now, there comes a time that the nuclear furnace
has been completely consumed. For our Sun, that takes an
additional five billion years. The Sun has already been
burning nuclear fuel for five billion years. It has another
five billion to go. And once the nuclear fuel
has been consumed, there are three end-products of
the dead star that is left over. And these three end-products
are the following: Number one is called
a white dwarf. It has a radius approximately
the same as the Earth, some 10,000 kilometers, and the mass of a white dwarf... There's a whole range of them,
but a typical number, say, is half the mass of the Sun. So that's
one possible end-product. This will be the fate
of our Sun, by the way. The density of
such an object is quite high-- some ten to the rho-- will be roughly ten to the 6th
grams per cubic centimeter. Another possibility is that
you end up with a neutron star. A neutron star has a radius
of about ten kilometers, and it has a mass of roughly
1.5 times the mass of the Sun, and its density is about ten to the 14 grams
per cubic centimeter, which is even higher
than the density of nuclei. And then there is a possibility,
which is even more bizarre, that you end up
with a black hole. I will not talk
about black holes today but I will get back
to that later in 8.01. And a black hole,
for all practical purposes, has no size at all. The mass of the black hole
must be larger than we think-- three solar masses-- and so
the density is infinitely high. Whether you end up
to be a white dwarf, a neutron star or a black hole depends on the mass
of the progenitor-- of the star that collapsed when the fuel, when
the nuclear fuel was gone. And in order to form
a neutron star, you would have
to start off with a star of probably at least ten solar
masses, maybe even more. So our Sun will not become
a neutron star, but our Sun will ultimately
become a white dwarf. Now, it would be
a reasonable question to ask, Why do you end up only with
these three possibilities? Why is there nothing in between? Look, there is a huge difference from 10,000 kilometers
to ten kilometers. Is there nothing in between? And the answer to that
lies in quantum mechanics, which is not part of this course
but you will see that in 8.05. Why are there only these two? And then if you get
into general relativity, then you will understand why there is, then, this third,
very bizarre possibility. When a star collapses,
two things happen. First of all, there is a huge amount of
gravitational potential energy that is released
in the form of kinetic energy. The stuff falls in-- we call
it gravitational collapse. And that gravitational
potential energy converts to kinetic energy and that ultimately converts
to heat and to radiation. If I take an object here,
a piece of chalk, and I drop that, that you can
call gravitational collapse. Gravitational potential energy
is converted to kinetic energy and, ultimately,
it goes to heat. Here we're talking about a star
which is imploding, collapsing, and the amounts of
gravitational potential energy that become available
are enormous. In addition to this
huge amount of energy release, the star must spin up, because
its moment of inertia goes down and therefore
the angular velocity must go up. I want to do a little bit
of quantitative work on this. And I want to take
an object like our Sun and I would like
to collapse that object from its present radius-- of the Sun, which is
about 700,000 kilometers-- I want to collapse that
to a neutron star with a radius of ten kilometers, even though I know
and I told you that the Sun will
not become a neutron star. It's just to get
some feeling for the numbers. So we take an object
like the Sun, which has a radius
of about 700,000 kilometers, and we're going to collapse
that to a neutron star which has a radius
of about ten kilometers. The mass of the Sun is two
times ten to the 30 kilograms, and for those of you
who are good at math, they can calculate-- when you collapse this object
without losing any mass, you keep all the mass, but you
shrink it to ten kilometers-- how much gravitational
potential energy is released? And that is a staggering number,
and I call that delta u. It is a loss of gravitational
potential energy which is about ten
to the 46 joules. And this number is
truly mind-boggling. This is converted
to kinetic energy and then it is converted to heat
and all forms of radiation. To give you a feeling for how
absurdly large this number is, if you take the Sun and you take all the energy
that the Sun produces in its ten billion years
that it will live, the total energy output of
the Sun is a hundred times less than this number, and this comes
out in a matter of seconds. So it is a mind-boggling idea
that the Sun is producing in ten billion years,
the lifetime of the Sun... It is producing less energy
than what happens during a stellar collapse
to a neutron star. Hundred times less. So, when this in-fall occurs and this huge amount of energy
is released, the outer layer bounces off
the inner core and is expelled and that's what we call
a supernova explosion. The outer layers are thrown off with speeds typically some
10,000 kilometers per second. Our Sun will not become
a neutron star, but it will become
a white dwarf. We talked about
the Crab Nebula last time, and the Crab Nebula is a remnant of a supernova explosion which
occurred-- believe it or not-- on the Fourth of July
in the year 1054. Talking about fireworks. The supernova explosion was
noticed by Chinese astronomers. They called this a guest star. Chinese astronomers were
very prestigious. The reason for that was that these Chinese astronomers
advised the emperor. They looked at the sky
and they derived from the sky information that was key
for the emperor. They knew how to interpret
the occurrence of comets or, for instance,
shooting stars, or a particular
line-up of planets and certainly the appearance
of a guest star. And they would know
that, for instance, a comet in a certain part
of the sky might mean that there would be hunger
or there would be diseases, there would be famine, or it would be a good time
for a battle or it would be a bad time
for a battle. And that's
what these people were doing. They were advising the emperor and therefore they were keeping
a very close eye on the sky. No pun implied. This star was visible for weeks
during the day when it exploded, and it was the brightest star
in the sky for years to come. It is a complete puzzle
why there isn't a single report by any European astronomer on the occurrence
of the supernova of 1054. It is very puzzling. Now, you can argue that in
the Netherlands and in England there are always clouds,
it's always raining, so you can't see the sky--
okay, I grant you that. But then we have Italy, and we
have Spain and we have France. And it is very strange. It must have been
a cultural thing. Somehow in the 11th century, somehow, Europe was
not interested in looking at the sky. This is something
they could not have missed, but they didn't write it down. I now want to pursue
the spin-up of this star when it collapses from 700
kilometers to ten kilometers. If we round the numbers
off a little bit, then the reduction in radius
is about 100,000. It's really only 70 --> but
let's just make that 100 -->
0:37:56 --> 00:38:02
ten to the 5th. That means R square goes down
by a factor of ten billion. And if R square goes down
by a factor of ten billion, then the moment of inertia goes
down by a factor of ten billion, and so omega must go up
by a factor of ten billion. If you started off with a star that rotated about its own axis
in a hundred days, it ends up rotating around
in one millisecond when it has become
a neutron star. A neutron star, ten kilometers. It has about the same mass
as the Sun, a little more. And it spins around
in one millisecond. At the equator
of the neutron star, you reach about 20%
of the speed of light. We know of hundreds
of neutron stars in the sky. Two of them have, in fact, rotational periods
of 1.5 milliseconds-- many of them much slower-- and we discussed last time
why that is. Because remember, in the case
of the Crab Pulsar, the pulsars slow down. Nature is tapping on the
rotational kinetic energy of these pulsars and is converting it
into other forms of energy-- in the case of the Crab Pulsar, radio emission,
optical emission, gamma rays, X rays and even jets. The Crab Pulsar was slowing down
every day 36.4 nanoseconds, which led to
a staggering power output. I still remember the number. I think
six times ten to the 31 watts. In 75 years, the pulsar
slows down by one millisecond, so the 33 milliseconds would become 34 milliseconds
in 75 years. If the star has an original... There's no star,
that was a disk, right? If the star has
an original magnetic field which most stars do-- oh, I lost my star,
but that's okay-- then in the collapse the magnetic fields
will become stronger. And this is something
you will learn about in 8.02, why it becomes stronger. So most of these neutron stars
have strong magnetic fields and most rotate very fast. And for reasons
that we don't quite understand, many of them blink at us. They blink at us
in radio emission. We believe that there are
two beams of radio emission like a lighthouse going out from the two magnetic
poles of the neutron star, and as the neutron star rotates
and you are on Earth, if it sweeps over you you see radio emission,
radio emission, you see nothing. You see radio emission,
you see radio emission. So many pulsars whose beam
doesn't sweep over the Earth we would never be able to see,
of course. And in the case
of the pulsar in the Crab it is even more special, because that pulsar
also blinks at us in the optical, in the X rays
and in the gamma rays. And so now I would like
to show you some slides and discuss in a little bit more
detail the supernova explosions and the fabulous light output
and the spin-up, so I have to make it quite
dark... make it completely dark. There we go. And so the first slide is
simply an artist's conception-- don't take this too seriously-- of these beams
of radio emission. This is, then,
the rotating neutron star, and if the axis of rotation
doesn't coincide with the magnetic dipole axis, and if you have
these radio beams-- which we do not understand
how they are formed-- there you can see,
when they rotate how they can sweep over you. Now you may say, "Well,
that's a little bit artificial, "because why would
the axis of rotation be different from
the magnetic dipole axis?" Well, that is not an exception
at all in astronomy. The Earth itself has
a magnetic dipole axis which doesnot coincide
with the axis of rotation. In fact, almost all the planets
in our planetary system
have a magnetic dipole axis which makes a large angle
with the axis of rotation, so that's the rule in astronomy,
rather than the exception, even though it may not be easy
to understand that. And these blue lines, then,
represent magnetic field lines. You will see more of them than
you like when you take 8.02. And here is Jocelyn Bell. Jocelyn Bell was
a graduate student under Anthony Hewish
in Cambridge, England, and she discovered pulsars. She found in the radio data-- which were obtained
using a new telescope that Anthony Hewish had built-- she found in there
periodic signals-- pulses,
if you want to call them-- you see some of them
here at the bottom, and they were 1.3 seconds apart. And she reported that
to Anthony, and Anthony said, "Well, they've got to be
nonsense, of course. "I mean, there's
not an object in the sky "that is going to give us pulses with a separation
of 1.3 seconds." So they just assumed that
it was caused by an elevator, by maybe milking cow machines
or things of that nature, motorcycles... and so they
did every conceivable thing to check whether, indeed,
this was a man-made phenomenon. But they could not
find anything, and it was Jocelyn, through
her incredible brilliance, who was able to convince Anthony that indeed this is
an object that is in the sky and that the radiation
doesn't come from the Earth. And when they realized that,
they realized that this would be the discovery of not
only the century but of all of mankind,
because they said, "Well, who could possibly
send radio beams at us "and modulate them
with a period of 1.3 seconds? Only intelligent life
can do that." And so they called this first
object "Little Green Man." But just before they published-- they discovered it
in 1967, by the way-- they found a second pulsar which had a slightly
different frequency than the 1.3 second period,
and so then they realized ~that it was probably
not intelligent life but that it was
an astronomical object. So they gave that second object
the name "Little Green Man II," but they abandoned
that idea very quickly. Now comes the sad part
of the story.
In 1974, Anthony Hewish
was awarded the Nobel Prize for this discovery. And Jocelyn,
who more than deserved it, whoreally was the discoverer, who was the person who proved
that this was astronomical, did not share
in the Nobel Prize. It is upsetting, it is sad. I have discussed it
with Jocelyn several times-- I know her quite well-- and she takes it actually very
lightly, too lightly, I think. Still, people
feel unhappy about it and still, after so many years-- the Nobel Prize
was awarded in 1974-- every time that I think about
this magnificent discovery and I think about Jocelyn, I think about
this gross injustice. Here we see the Crab Nebula
again, we've seen it before. The red filaments that you see
here are the result of matter that was blown off
when the implosion occurred and when the outer layers
bounced off the inner core. And originally they had a speed of about 10,000 kilometers
per second and by now-- it is
about 1,000 years later-- these speeds have
been reduced somewhat. But here at the center
you see the pulsar, and last time I showed
you convincing evidence that this is the pulsar,
because it's blinking at us. It has a diameter
of about seven light-years. It is a distance from us
of about 5,000 light-years. In terms of angular size
in the sky, it's about five arc minutes
in size, which is about one-sixth of the
angular diameter of the Moon. This is a drawing, a cave
drawing, made by Navajo Indians, and some people have
speculated to what extent the Navajo Indians may have seen
the supernova in 1054. It's unclear,
but it is a possibility. The Moon certainly gets
very close to the supernova, but it also gets very close to
Venus, and so this is something that is not well established,
but it is a possibility. Here you see a galaxy
in which a supernova occurred, and when the supernova occurs
at its brightest, it can be brighter than
all hundred billion stars in the galaxy. That's how much energy
is released in optical light. You see that it is at least
as bright as the entire galaxy. From this picture to this
picture is about one year. This occurred in 1972
and over the period of one year, you can still see this star
quite clearly, but it has diminished
in strength quite a bit. And then a great thing happened
in February 1987, on the 23rd of February. The supernova went off
in the Large Magellanic Cloud, which is a satellite galaxy
to our own galaxy. It's a distance of
about 150,000 light-years, and there was an astronomer who
was observing in South America. His name is Ian Shelton, and he left the dome
to look at the stars and he decided
to take a pee outside. And as he was taking a pee--
these were his own words-- he looked at the Large
Magellanic Clouds and he said, "Hey, that is funny! That star
is not supposed to be there." And he was the discoverer
of what is now known as 1987A, an enormous supernova going off
so close to where we live. And the next slide shows you the same portion
of the Large Magellanic Clouds and you can clearly see that
there is a very bright star. He could see this
with his naked eye. This is a picture made
by the Hubble Space Telescope of supernova 1987A. The inner ring that you see is the result of matter
that was thrown off by the star before it went
into supernova explosion some 25,000 years earlier. It expelled gas in its equator. This is really a circle,
although it looks like an ellipse because
of the projection effect. And this ring of matter
moved out with a speed of about
eight kilometers per second and it has a radius
of about eight light-months. And so eight months after
the supernova explosion, the ultraviolet light and
the X rays from the supernova caught up with this ring
of matter and they excited it and it became visible. Before the supernova explosion,
this ring was not visible. We are expecting in a few years
that the matter itself that was thrown off
with a much more modest speed of about 10,000 kilometers
per second... that that matter will also plow
into this ring and then we expect
some real fireworks again. There is no explanation
that people agree upon for these two
called "hourglass" rings. They are quite mysterious and
there are papers written on it and people disagree
on their origin. Supernovae explosions
in our galaxy and in the Large Magellanic
Clouds are quite rare. We expect no more than
about one in a hundred years. The previous one that could be
seen with the naked eye was in the year 1604. It's called Kepler supernova. And 1987A was really the first that could be studied
with modern equipment-- radio observatories,
X-ray observatories in orbit around the Earth. With some luck, you may see
a supernova explosion naked eye in your life. The chance is no better than ten
percent, so maybe it will help if occasionally
you take a pee outside and you become as famous
as Ian Shelton did, who is now a very famous man. See you Monday.
Start from the beginning. (IF the video starts at a certain point)
The second 1250 midterm was the worst. The average for our class was under a 60.
Just curious, does physics 1250 cover all the material analogous to 8.01?