We have here, going back
to rotating objects...
I have an object here
that has a certain velocity v,
and it's going around
with angular velocity omega,
and a little later
the angle has increased
by an amount theta
and then the velocity is here.
We may now do something
we haven't done before.
We could give this object
in this circle an acceleration.
So we don't have to keep
the speed constant.
Now, v equals omega R, so
that equals theta dot times R.
And I can take now
the first derivative of this.
Then I get
a tangential acceleration,
which would be
omega dot times R,
which is
theta double dot times R,
and we call theta double dot...
we call this alpha, and alpha
is the angular acceleration
which is in radians
per second squared.
Do not confuse ever
the tangential acceleration,
which is along
the circumference,
with a centripetal acceleration.
The two are both there,
of course.
This is the one
that makes the speed change
along the circumference.
If we compare our knowledge
of the past of linear motion
and we want to transfer it
now to circular motion,
then you can use all your
equations from the past
if you convert x to theta,
v to omega and a to alpha.
And the well-known equations
that I'm sure you remember
can then all be used.
For instance, the equation
x equals x zero plus v zero t
plus one-half at squared
simply becomes
for circular motion
theta equals theta zero
plus omega zero t
plus one-half alpha t squared--
it's that simple.
Omega zero is then the angular
velocity at time t equals zero,
and theta zero is the angle
at time t equals zero
relative to
some reference point.
And the velocity was
v zero plus at.
That now becomes
that the velocity goes
to angular velocity omega
equals omega zero plus alpha t.
So there's really not much added
in terms of
remembering equations.
If I have a rotating disk,
I can ask myself
the question now
which we have never done before,
what kind of kinetic energy,
how much kinetic energy is there
in a rotating disk?
We only dealt
with linear motions,
with one-half mv squared,
but we never considered
rotating objects
and the energy
that they contain.
So let's work on that a little.
I have here a disk, and
the center of the disk is C,
and this disk is rotating
with angular velocity omega
that could change in time,
and the disk has a mass m,
and the disk has a radius R.
And I want to know
at this moment
how much kinetic energy of
rotation is stored in that disk.
I take a little
mass element here, m of i,
and this radius equals r of i
and the kinetic energy
of that element i alone
equals one-half m of i
times v of i squared,
and v of i is this velocity--
this angle is 90 degrees.
This is v of i.
Now, v equals omega R.
That always holds
for these rotating objects.
And so I prefer to write this
as one-half m of i
omega squared r of i squared.
The nice thing
about writing it this way
is that omega, the angular
velocity, is the same
for all points of the disk,
whereas the velocity is not
because the velocity of a point
very close to the center
is very low.
The velocity here is very high,
and so by going to omega, we
don't have that problem anymore.
So, what is now the kinetic
energy of rotation of the disk,
the entire disk?
So we have to make a summation,
and so that is omega squared
over two
times the sum
of m of i r i squared
over all these elements mi
which each have
their individual radii, r of i.
And this, now, is what we call
the moment of inertia, I.
Don't confuse that with impulse;
it has nothing to do
with impulse.
And this is moment of inertia...
So the moment of inertia is
the sum of mi ri squared.
In...
So this can also be written
as one-half I,
I put a C there--
you will see shortly why,
because the moment
of inertia depends
on which axis of rotation
I choose-- times omega squared.
And when you see that equation
you say, "Hey, that looks quite
similar to one-half mv squared."
And so I add to this list now.
If you go from linear motions
to rotational motions,
you should change the mass
in your linear motion
to the moment of inertia
in your rotational motion,
and then you get back
to your one-half mv squared.
You can see that.
So we now have
a way of calculating
the kinetic energy of rotation
provided that we know how to
calculate the moment of inertia.
Well, the moment of inertia
is a boring job.
It's no physics, it's pure math,
and I'm not going
to do that for you.
It's some integral, and if
the object is nicely symmetric,
in general you can do that.
In this case, for the disk
which is rotating about
an axis through the center
and the axis--
that's important--
is perpendicular to the disk--
that's essential-- in that case
the moment of inertia equals
one-half m times R squared.
And I don't even want you
to remember this.
There are tables in books,
and you look these things up.
I don't remember that.
I may remember it for one day,
but then, obviously, you forget
that very quickly again.
Needless to say, that
the moment of inertia depends
on what kind of object you have.
Whether you have a disk
or whether you have a sphere
or whether you have a rod
makes all the difference.
And what also makes
the difference--
about which axis
you rotate the object.
If we had a sphere,
a solid sphere,
then...
So here you have a solid sphere,
and I rotate it about an axis
through its center.
Then the moment of inertia,
I happen to remember,
equals two-fifths mR squared
if R is the radius
and m is the mass of the sphere.
My research is in astrophysics.
I deal with stars, and stars
have rotational kinetic energy.
We'll get back to that
in a minute--
not in a minute but today--
and this is the one moment
of inertia that I do remember.
If you have a rod,
and you let this rod rotate
about an axis
through the center,
and this axis is
perpendicular to the rod--
the latter is important,
perpendicular to the rod--
and it is length l
and it has mass m,
then the moment of inertia--
which I looked up this morning;
I would never remember that--
equals 1/12 ml squared.
And all these moments
of inertia
you can find in tables
in your book on page 309.
So the moment of inertia
for rotation about this axis
of a solid disk
is one-half mR squared.
But it's completely different,
the moment of inertia,
if you rotated it
about this axis.
So you take the plane
of the disk.
Instead of rotating it this way,
you rotate it now this way.
You get a totally different
moment of inertia.
And most of those you can find
in tables, but not all of them.
Tables only go so far,
and that is why I want
to discuss with you
two theorems which will help you
to find moments of inertia
in most cases.
Suppose we have a rotating disk,
and I will make you see
the disk now with depth.
So this is a disk,
and we just discussed
the rotation
about the center of mass.
And I call this axis l.
And so it was rotating like this
and was perpendicular
to the disk.
This is the moment of inertia.
But now I'm going
to drill a hole here,
and I have here an axis l prime
which is parallel to that one.
And I'm going to force
this object
to rotate about that axis.
I can always do that--
I can drill a hole
have an axle,
nicely frictionless bearing
and I can force it
to rotate about that.
What now is
the moment of inertia?
If I know the moment of inertia,
then I know how much rotational
kinetic energy there is.
That's one-half I omega squared.
And now there is a theorem
which I will not prove,
but it's very easy to prove,
and that is called
the parallel axis theorem.
And that says
that the moment of inertia
of rotation about l prime--
provided that l prime is
parallel to l--
is the moment of inertia
when the object rotates
about an axis l
through the center of mass
plus the mass of the disk
times the distance d squared.
So this is the mass.
And that's
a very easy thing to apply,
and that allows you
now in many cases,
to find the moment of inertia
in situations
which are not very symmetric.
Imagine that you had
to do this mathematically,
that you actually had to do
an integration of all these
elements mi from this point on.
That would be
a complete headache.
In fact, I wouldn't even know
how to do that.
So it's great.
Once you have demonstrated,
once you have proven
that this parallel
axis theorem works,
then, of course, you can always
use it to your advantage.
Notice that
the moment of inertia
for rotation about this axis--
which is not through a center
of mass-- is always larger
than the one through the center.
You see, you have this md
squared; it's always larger.
There is a second theorem
which sometimes comes in handy,
and that only works when you
deal with very thin objects,
and that is called
the perpendicular axis theorem.
If you have some kind
of a crazy object--
which of course
we will never give you;
we'll always give you a square
or we'll give you a disk...
But it has to be a thin plate.
Otherwise the perpendicular
axis theorem doesn't work.
And suppose I'm rotating it
about an axis perpendicular
to the blackboard
through that point.
I call that the z axis.
It's sticking out to you.
That's the positive z axis.
I can draw now any xy axis where
I please, at 90-degree angles,
anywhere in the plane
of the blackboard.
So I pick one here,
I call this x,
and I pick one here
and I call that y.
So z is pointing towards you.
Remember, I always choose
a positive right-handed
coordinate system.
My x cross y is always
in the direction of z.
I always do that.
And so you see that here,
x cross y equals z.
Now, you can rotate
this thin plate about this axis.
You can also rotate it
about that axis.
And you can also rotate it
about the z axis.
And then the
perpendicular axis theorem,
which your book proves
in just a few lines,
tells you
that the moment of inertia
for rotation
about this axis here
is the same as the moment
of inertia for rotation about x
plus the moment of inertia
for rotation about the axis y.
And this allows you
to sometimes...
in combination with
the parallel axis theorem
to find moments of inertia in
case that you have thin plates
which rotate about axes
perpendicular to the plate
or sometimes
not even perpendicular.
Sometimes you can use... if
you know this and you know this,
then you can find that.
So both are useful,
and in assignment 7
I'll give you a simple problem
so that you can apply
the perpendicular axis theorem.
There are applications
where energy is temporarily
stored in a rotating disk,
and we call those disks
flywheels.
And the rotational kinetic
energy can be consumed, then,
at a later time,
so it's very economical.
And this rotational
kinetic energy
can then be, perhaps,
converted into electricity
or in other forms of energy.
And there are really remarkably
inventive and intriguing ideas
on how this can be done.
Of course,
whether it is practical
depends always
on dollars and cents
and to what extent it is
economically feasible.
But I have always,
even when I was a small boy...
I remember when I was seven
years, it already occurred to me
that all this heat
that is produced
when cars slam their brakes--
all you're doing is
you produce heat;
you lose all that kinetic energy
of your linear motion--
whether somehow that couldn't
be used in a more effective way.
And this is what I want
to discuss with you now
and see where we stand.
This is actually
being taken seriously
by the Department of Energy.
So I want to work out
with you an example of a car
which is in the mountains and
which is going to go downhill.
And the mountains are
very dangerous-- zigzag roads--
and so he or she can
only go very slowly.
And the maximum speed
that the person could use is
at most ten miles per hour--
without killing him or herself--
which is about
four meters per second.
And so here is your car,
and let's assume
you start out with zero speed.
And let's assume
that the mass of the car--
we'll give it nice numbers--
is just 1,000 kilograms.
And so you zigzag
down this road.
Let us assume
that the height difference h--
let's give it a number,
500 meters...
And you arrive here at point p.
And you later have to go
back up again.
What is your kinetic energy
when you reach point p?
Well, you have a speed
of four meters per second,
and as you went down, you've
been braking all the time.
One way or another,
you got rid of your speed
and that's all burned up--
heat, you heat up the universe.
So when you reach point p, your
kinetic energy at that point p
is simply one-half mv squared.
m is the mass of the car,
so that is 500 times 16--
v squared--
so that is 8,000 joules.
Now compare this
with the work that gravity did
in bringing this car down.
That work is mgh,
and mgh is a staggering number.
1,000 times ten times 500--
that is five million joules!
And all of that was converted
to heat using the brakes.
It actually even gives you also
wear and tear on the brakes.
So who needs it?
Is there perhaps
a way that you can salvage it
or maybe not all of it,
maybe part of it?
And the answer is yes,
there are ways.
At least in principle
there are ways.
You can install
a disk in your car,
which I would call,
then, a flywheel,
And you can convert the
gravitational potential energy.
You can convert that
to kinetic energy of rotation
in your flywheel.
And to show you that
it is not completely absurd,
I will put, actually,
in some numbers.
Suppose you had a disk
in your car
which had a radius
of half a meter.
That's not completely absurd.
That's not
beyond my imagination.
That's a sizable disk.
And I give it a modest mass--
so that the mass of the car
is not going to be too high--
200 kilograms.
That's reasonable.
That would be a steel plate
only five centimeters thick,
so that's quite reasonable.
And the moment of inertia
of this disk if I rotate it
about an axis through the center
perpendicular to the disk--
that moment of inertia,
we know now, is one-half m...
oh, we have a capital M--
R squared, and that equals 25.
The units are kilograms,
if you're interested,
kilograms/meter squared.
So we know
the moment of inertia.
Now, what we would like to do
is we would like to convert
all this gravitational
potential energy
into kinetic energy
of that disk.
If you think of a clever way
that you can couple that--
people have succeeded in that--
then you really would like
one-half I omega squared...
You would really like that
to be five times
ten to the six joules.
And so that
immediately tells you
what omega should be for
that disk, and you find, then,
if you put in your numbers,
which is trivial...
you find that omega is
about 632 radians per second,
so the frequency of the disk
is 100 hertz,
100 revolutions per second.
I don't think that that is
particularly extravagant.
So as you would come
down the hill,
you would not be braking
by pushing on your brake,
you would not be heating up
your brakes,
but you would somehow
convert this energy
into the rotating disk
and that would slow you down.
So the slowdown,
the "braking" is now done
because of a conversion
from your linear speed--
which comes from gravitational
potential energy--
to the rotation of the disk.
And when you need that energy,
you tap it.
So you should also be able
to get the rotational
kinetic energy out
and convert that again
into forward motion.
And if you could really do this,
then you could go back uphill
and you wouldn't have to use
any fuel.
All your five million joules
can be consumed, then,
in an ideal case, and you
would not have to use any fuel.
Now, you can ask yourself
the question,
is this system only useful
in the mountains
or could you also use this
in a city?
Well, of course
you can use it in a city.
You wouldn't be braking
like this, then,
but again, you would slow down
by taking out kinetic energy
of linear forward motion,
dump that into kinetic energy
of rotation of your flywheel
and that would slow you down.
And when the traffic light turns
green, you convert it back--
rotational kinetic energy
into linear kinetic energy--
and you keep going again.
Now, of course, this is
all easier said than done,
but it is not complete fantasy.
People have actually made
some interesting studies,
and I would like to show you
at least one case
that I am aware of,
that I found on the Web,
that shows you that United
States Energy Department
is taking this quite seriously.
This view graph is also
on the 801 home page.
And so you see here
the idea of mounting such
a flywheel under the car here.
And it has the location
of the "flywheel energy
management power plant."
Wonderful word, isn't it?
And here you see
a close-up of this flywheel.
I didn't get any numbers on it.
I don't know
which fraction of the energy
can be stored in your flywheel,
but it's an attempt.
People are seriously thinking
about it.
And it may happen
in the next decade
that cars may come on the market
whereby some of your energy,
at least, can be salvaged.
Instead of heating up
the universe, use it yourself,
which could be very economical.
I have here a toy car--
I'll show it on TV first.
And this toy car has a flywheel.
Do you see it?
That the flywheel itself is
the wheel of the car,
but the idea is there.
In this case, I cannot convert
linear motion into the flywheel.
I could do that, but I'm going
to do it in a reverse way.
I'm going to give this flywheel
a lot of kinetic energy
of rotation,
and you will see shortly
how I do that.
And then I will show you
that that can be converted back
into forward motion--
in this case, it's very easy
because the flywheel itself
is the wheel.
So let me try
to... power this car.
I do that
with this plastic... okay.
So I'm going to put
some energy into this wheel,
into this flywheel,
and then we'll see
whether the car can use that
to start moving.
(sweeping sound)
Great that
my lecture notes were there.
So, you see, it works.
And, of course, if
you could reverse that idea,
that when the car...
before it stops,
get it back into the flywheel,
then you have the idea
that I was trying to get across.
Very economical,
and definitely that will happen
sometime in the future.
Flywheels are used more often
than you may think.
MIT, at the Magnet Lab, has
two flywheels which are amazing.
They have a radius,
I think, of 2.4 meters--
that is correct--
and each one of those flywheels
has a stunning mass
of 85 tons, 85,000 kilograms...
and they rotate
at about six hertz.
You can calculate
the moment of inertia.
They rotate about their center
axis perpendicular to the plane.
You know now what
one-half I omega square is,
and so you can calculate
the kinetic energy of rotation.
And that kinetic energy
of rotation is, then,
a whopping 200 million joules
in each of
those rotating flywheels.
Now, they use this
rotational kinetic energy
to create very strong
magnetic fields
on a time scale
as short as five seconds.
So they convert
mechanical energy of rotation
to magnetic energy,
which is not part of 801
so I will not go
into how they do that.
This is part of 802,
and I'm sure all of you
are looking forward to 802,
and that's when you will see
how you can convert mechanical
energy into magnetic energy.
We have already seen
a demonstration in class
whereby we converted
mechanical energy
when someone was rotating,
into electric energy.
I think that was you, wasn't it?
And we got these light bulbs on.
Well, you can also convert it
into magnetic energy.
And then when they have created
these strong magnetic fields
that they do their research with
and when they want
to get rid of them,
they go the other way around
and they dump that energy,
that magnetic energy,
back into the flywheels,
who then start spinning again
at six hertz.
Needless to say
that huge amount
of rotational kinetic energy
must be stored in planets
and in stars,
and I would like to spend
quite some time on that.
It's a very interesting subject.
I will first discuss with you
the sun and the earth
and see how much rotational
kinetic energy is stored
in the earth and in the sun.
This is also on the 801
home page, so don't copy this.
Let's first look at the sun.
We have the mass of the sun,
we have the radius of the sun
so you can calculate the moment
of inertia of the sun.
I have used
my two-fifths mR squared,
which is really
a crude approximation,
because the two-fifths m
R squared for a solid sphere
only holds if the mass
is uniformly distributed
throughout that sphere.
With a star, that's not the
case; not with the earth either,
because the density is higher
at the center.
But this sort of gives
you a crude idea.
So we have there
the moment of inertia,
which is easy to calculate
with that
two-fifths mR squared,
and I get the same
for the earth.
This is the radius of the earth,
and you see the moment
of inertia of the earth.
Now I want to know
how much kinetic energy
of rotation these objects have.
Well, the sun rotates
about its axis in 26 days,
the earth in one day,
and so I finally convert
everything to MKS units
and I find these numbers for
the rotational kinetic energy.
Now, look at the number
of the sun--
1½ times ten to the 36th joules.
Our great-grandfathers
must have been puzzled
about where
the solar energy came from--
the heat and the light,
where it came from.
And conceivably
it came from rotation.
Maybe the sun is spinning down,
is slowing down
and maybe the energy that we get
is nothing but
rotational kinetic energy.
If that were the case, however,
since the sun produces four
times ten to the 26th watts--
four times ten to the 26th
joules-- per second,
it would only last 125 years.
So you can completely
forget the idea
that the energy from the sun
that we now know,
of course, is nuclear,
but our great-grandparents
didn't know that--
that the energy would be tapped
from kinetic energy of rotation.
Let's look at the earth.
2½ times
ten to the 29th joules.
Well, let me try something...
some fantasy on you,
some crazy, some ridiculous idea
and I'm telling you first,
it is ridiculous.
Remember that
the world consumption...
Six billion people
on earth
consume about four times ten
to the 20th joules every year.
So if somehow...
I thought if you could tap
the rotational energy
of the earth
by slowing the earth down,
maybe we could use it to satisfy
the world energy consumption.
Um, I wouldn't know
how to do it,
and it is, of course,
complete fantasy.
All you would have to do
is slow the earth down
by about... 2.4 seconds.
After one year...
So you slow it down.
After one year,
the day wouldn't last...
Day and night wouldn't last
24 hours
but only 2.4 seconds longer.
But, of course,
after a billion years, then,
you would have consumed up all
the rotational kinetic energy
and then the earth would
no longer be rotating.
It is, of course, a crazy idea
but sometimes it's cute
to speculate about crazy ideas.
There is an object
which we call the Crab Pulsar.
It is a neutron star and it is
located in the Crab Nebula.
The Crab Nebula is the result
of a supernova explosion
that went off in the year 1054,
and during my next lecture I
will talk a lot more about that.
For now, I just want
to concentrate
on this neutron star alone.
And so here you have
the data on the Crab Pulsar.
The mass of the Crab Pulsar
is not too different
from that of the sun.
It's about 1½ times more.
The radius is
ridiculously small--
it's only ten kilometers.
All that mass is compact in
a ten-kilometer-radius sphere.
It has a horrendous density
of ten to the 14th grams
per cubic centimeter.
So, of course, the moment
of inertia is extremely modest
compared to the sun,
because the radius is so small
and the moment of inertia goes
with the radius squared.
However, if you look
at rotational kinetic energy,
the situation is very different,
because this
neutron star rotates
in 33 milliseconds
about its axis.
So it has a phenomenal
angular velocity.
And so if now you calculate
one-half I omega squared,
you get a fantastic amount
of rotational kinetic energy.
You get an amount which is
more than a million times more
than you have in the sun.
And this object,
this pulsar in the Crab Nebula
is radiating copious amounts
of x-rays, of gamma rays.
There are jets coming out
of ionized gas,
and we are certain
that all that energy
that this object is producing
comes from
rotational kinetic energy.
And I will give you
convincing arguments
why there is no doubt
about that.
If you take the Crab Pulsar
and you calculate
how much energy comes out
in x-rays and gamma rays
and everything that
you can observe in astronomy,
then you find
that it has a power
roughly of about six times
ten to the 31st watts.
It's a phenomenal amount
if you compare that
with the sun, by the way.
The sun is only four times
ten to the 26th watts.
So the Crab Pulsar
alone generates
about 150,000 times
more power than the sun.
We know the period of the pulsar
to a very high degree
of accuracy.
The period of rotation
of the neutron star
is 0.0335028583 seconds.
That's what it is today.
I called my radio astronomy
friends yesterday
and I asked them,
"What is the rotation period
of the neutron star
in the Crab Nebula?"
and this was the answer.
Tomorrow, however, it is
longer by 36.4 nanoseconds.
So tomorrow,
you have to add this.
That means it's slowing down.
The Crab Pulsar is slowing down.
That means omega is going down.
That means one-half I omega
square is going down.
And when you do your homework,
which you should be able to do--
to compare the rotational
kinetic energy today
with the rotational
kinetic energy tomorrow--
you will see
that the loss of energy
is six times ten to the 31st
joules per second,
which is exactly the power
that we record
in terms of x-rays, gamma rays
and other forms of energy.
So there's no question
that in the case
of this rotating neutron star,
all the energy that it radiates
is at the expense
of rotational kinetic energy.
It's a mind-boggling concept
when you think of it.
And if the neutron star
in the Crab Nebula
were to continue to lose
rotational kinetic energy
at exactly this rate,
then it would come to a halt
in about 1,000 years.
Now I would like to show you
a few slides,
and I might as well
cover this up
so that we get it
very dark in this room.
I want to show you
the Crab Nebula,
and I think I will also show you
the beautiful flywheels
in the Magnet Lab.
Now I need a flashlight.
I need my laser pointer.
I need a lot of stuff.
Okay, there we go,
so I'm going to make it dark.
You ready for that?
Okay, if I can get
the first slide.
What you see here are these
flywheels at the Magnet Lab.
These are the wheels
that have a mass of 85 tons
and that have a radius
of 2.5 meters--
an incredible, ingenious device,
and you can store in there
200 million joules,
and you can dump it
into magnetic energy
and in five seconds dump it back
into kinetic energy of rotation.
It is an amazing accomplishment,
by the way.
And here you see
the Crab Nebula.
The Crab Nebula is
at a distance from us
of about 5,000 light-years.
It is the remnant of a supernova
explosion in the year 1054--
much more about that
during my next lecture--
and what you see here is
not stuff that is generated
at this moment in time
by the pulsar.
This, by the way, is the pulsar,
and the red filaments
that you see here is material
that was thrown off
when the explosion occurred.
The explosion,
the supernova explosion throws
the outer layers of the star
with a huge speed--
some 10,000 kilometers
per second-- into space,
and that is what you are seeing.
From here to here is
about seven light-years
to give you an idea
of the size of this object.
This pulsar alone, however,
generates the...
six times 31... watts.
And we do know that it is
this star that is the pulsar
and we know
that it is not that star.
And the way that
that was observed,
that that was measured,
is as follows.
A stroboscopic picture,
a stroboscopic exposure was made
of the center portion
of the Crab Nebula.
And a stroboscopic picture means
that you are using a shutter
which opens and closes.
In this case,
you have to open and close it
with exactly the same frequency
as the rotation
of the neutron star.
This neutron star-- for reasons
that is not well understood--
is blinking at us.
It blinks at us
at exactly the frequency of
its rotation, 33 milliseconds.
That means 30 hertz.
Roughly 30 times per second
you see the star become bright
and then go dim again.
If now you set your frequency
of your shutter of your...
in front of your photographic
plate at exactly that frequency
and you expose
the photographic plate
only when the star is bright,
then you will see
a very bright star
when you develop your picture.
If now you take another picture,
expose it
the same amount of time,
but the shutter is open
when the star is dim
and you develop that picture,
the star is dim.
But the beauty is
that all other stars
in the vicinity, of course,
will show up on both
photographic plates
with exactly the same strength
because they are
not blinking at you,
since they don't blink at us
with a period
of 33 milliseconds.
That is what you will see
on the next slide,
which is
a stroboscopic exposure.
This star is clearly
not the pulsar
as it is about equally bright
on both exposures.
This is not the pulsar,
but this one is.
You see, this one
is missing here.
And so this is
beyond any question
that we know exactly
which the pulsar is.
A very new observatory
was launched only recently,
and that is called
the Chandra X-ray Observatory.
And Chandra made a picture
very recently
of the Crab Nebula,
of the pulsar,
and that's what
I want to show you now.
It's on the Web,
and I show you a picture
that many of you probably
haven't seen yet,
which is the center part
of the Crab Nebula,
and the pulsar is located here.
And all this is x-rays, nothing
to do with optical light.
This is all x-rays,
and you see there is a huge
nebula here around this pulsar
which is about
two light-years across,
and all that energy in x-rays
is all at the expense
of rotational kinetic
energy of the pulsar,
which is quite amazing.
And when this picture was made
with Chandra X-ray Observatory,
they discovered immediately that
the pulsar also produces a jet.
Maybe you can see that
from where you are sitting.
There is a jet coming out here,
and with a little bit
of imagination
you can see
this jet going out there.
And all that energy is
at the expense
of rotational kinetic energy.
MIT has a big stake, by the way,
in the Chandra Observatory,
and not only MIT
but Cambridge as a whole.
The Center for Astrophysics
and MIT are running
the Chandra Science Center,
from which
all radio commands are given,
which is here just across
the street, a few blocks away.
And many MIT scientists
have dedicated
the major part of their careers
in this endeavor.
And these are one
of the wonderful results
that have come out.
So, see you Friday.
