PROFESSOR: OK.
Here's an example that's
more or less for fun.
Because you'll see
me try to do it.
You can do it better.
I call the problem
the tumbling blocks.
Only in this example,
in my demonstration,
it's going to be
a tumbling book.
I'm going to take a
book, the sacred book,
and throw it in the air.
And I'll throw it
three different ways.
And the question is, is the
spinning book stable or not?
And let me tell
you the three ways
and then give you the three
equations that came from Euler.
So those are the
three equations.
You see that they're not linear.
And those are for
the angular momentum.
So there's a little physics
behind the equations.
But for us, those are
the three equations.
So the first throw
will spin around
the very short axis, just
the thickness of the book,
maybe an inch.
So when I toss that,
as I'll do now,
you will see if I can toss
it not too nervously I hope.
It came-- it was stable.
The book came back to
me without wobbling.
Of course, my nerves would
give it a little wobble,
and that wobble would continue.
It will be only
neutrally stable.
The wobble doesn't disappear.
But it doesn't
grow into a tumble.
OK.
So that's one axis,
the short axis.
Then I'll throw it also
around the long axis, flipped
like this.
I think that will be stable too.
And then, finally, on
the intermediate axis,
is middle length axis.
Notice the rubber band that's
holding the book together.
Holding so the pages don't open.
And this, we'll see, I
think, will be unstable.
And similarly,
throwing a football,
throwing other Frisbees,
whatever your throw.
Any 3D object has got these
three axes: a short one,
a medium one, and a long axis.
And the equations
will tell us short
and long axes should
give a stable turning.
And the in between
axis is unstable.
Well, how do we decide for
our differential equation
whether the fixed
point, a fixed point,
that's a critical
point, a steady state--
we have to find
this steady state,
and then for each steady
state we linearize.
We find the derivatives
at that steady state.
And that gives us a constant
matrix at that steady state.
And then the
eigenvalue is decided.
So first, find the
critical points.
Second, find the derivatives
at the critical points.
Third, for that
matrix of derivatives,
find the eigenvalues
and decide stability.
That's the sequence of steps.
OK.
The first time we've ever
done a three by three matrix.
Maybe the last time.
OK.
Let me, before I
start-- before I
find the critical points--
notice some nice properties.
If I multiply this
equation by x, this one
by y, this one by z, and
add, those will add to 0.
When there's an x there,
a y there, and a z there,
I get a 1 minus 2 and
a 1 they add to 0.
So x times dx dt.
y times dy dt. z
time dz dt adds to 0.
That's an important fact.
That's telling me that the
derivative of something is 0.
That something
will be a constant.
So I'm seeing here
the derivative
of that whole business
would be the derivative
of a half probably.
x squared, because the
derivative of x squared
will be with a half.
The derivative will be x dx dt.
And y squared and z squared
is the derivative is 0.
The derivative of that
line is just this line.
It's 0.
So this is a constant.
No doubt, that's
probably telling me
that the total energy, the
kinetic energy, is constant.
After I've tossed that
book up in the air,
I'm not touching it.
It's doing its thing.
And it's not going to change
energy because nothing
is happening to it.
It's just out there.
Now there are other-- so
that's a rather nice thing.
This is a constant.
Now there's another way.
If I multiply this one by 2x,
and I multiply this one by y,
and add just those
two, that cancels.
So 2x dx dt-- 2x times
the first one-- and y
times the second one gives 0.
Again, I'm seeing
something is constant.
The derivative of something,
and that something
is x squared plus 1/2 y
squared is a constant.
Another nice fact.
Another quantity
that's conserved.
And as I'm flying
around in space,
this quantity x squared plus
1/2 y squared does not change.
This sort of-- that
involved all of xyz.
And of course that's the
equation of a sphere.
So in energy space,
or in an xyz space,
our solution is wandering
around a sphere.
And this is the equation for,
I guess, it's an ellipse.
So there's an ellipse
on that's sphere
that it's actually
staying on that ellipse.
And in fact there's
another ellipse
because I could've multiplied
this one by 2z and this one
by y and added.
And then those
would have canceled.
Minus 2 xyz plus 2xyz.
So that also tells
me that it would
be probably z squared plus 1/2
y squared equals a constant.
That's another ellipse.
z squared plus 1/2 y squared.
You see this?
If I take the
derivative of that,
I have 2z times dz dt
plus y times dy dt.
Adding give 0.
The derivative is 0.
The thing is a constant.
But!
But, but, but!
If I subtract this
one from this one,
take the difference
of these two.
Suppose I take this
one minus this one.
The 1/2 y squared will go.
So that will tell me
that x squared minus z
squared is a constant.
Oh, boy!
I haven't solved
my three equations.
But I found out a whole
lot about the solution.
The solution stays on the
sphere, wanders around somehow.
It also at the same time
stays on that ellipse.
And it stays on that ellipse.
But this is not an
ellipse, not an ellipse.
That's the equation
of a hyperbola.
And that's why-- which, of
course, goes off to infinity.
And that's why the-- well,
it goes off to infinity,
but it has to stay
on the sphere.
It wanders.
This will be responsible
for the unstable motion.
Professor [INAUDIBLE], who would
do this far better than me,
his great lecture in 1803,
Differential Equations,
was exactly this.
The full hour to tell you
everything about the tumbling
box.
So I'm going to do
the demonstration
and write down the main facts
and understand the stability,
the discussion of stability.
I'm ready to move on to the
discussion of stability.
Again, here are my
three equations.
We're up to three
equation, so we're
going have a three
by three matrix.
And first I have to find
out the critical points,
the steady states
of this motion.
How could I toss it so
that if I toss it perfectly
it stays exactly as tossed?
And the answer is,
around the axis.
If I toss this perfectly,
with no nerves,
it'll just spin exactly
as I'm throwing it.
The x, y, and z will
all be constant.
Now, when I toss
it on that axis.
I'm looking for-- here
are my right hand side.
YZ, minus 2XZ, and XY.
And I wrote those
in capital letters
because those are going
to be my steady states.
Now I'm looking for are points
where nothing's happened.
If those three right hand
sides of the equation are 0,
I'm not going to move.
xyz will stay where they are.
So can you see solutions
of those three equations?
Well, they're pretty
special equations.
I get a solution when, for
example, solutions could be 1,
0, 0/
If two of the three--
if y and z are 0.
y is 0, z is 0, y
and z are 0, I get 0.
So that is a certainly
steady state.
x equal 1, y and
z equal 0 and 0.
And that steady state is
spinning around one axis.
And, actually, I could have
also a minus 1 would also be.
So I've found, actually, two
steady states with y and z 0.
Then there'll be two
more with x and z 0.
And this could be--
that'll be spinning
around the middle axis.
And then 0, 0, 1
or minus 1, that
would be spinning around the
third axis, the long axis.
So those are my steady states.
And I guess, come
to think of it, 0,
0, 0 would also
be a steady state.
I think I found them all.
These are the xy's.
These are the x,
y, z steady states.
OK.
So now once you know the steady
states, that's usually fun,
as it was here.
Now the slightly less fun step
is find all the derivatives,
find that Jacobian
matrix of derivative.
So I've got three equations.
Three unknowns, xyz.
Three right hand sides.
And I have to find-- I'm
going to have a three by three
matrix of derivatives.
This Jacobian matrix.
So J for the Jacobian, the
matrix of first derivatives.
So what goes into the
matrix of first derivative?
Let me write Jacobian.
It is named after Jacoby.
It's the matrix of
first derivatives.
On the top row are
the derivatives
of the first function
with respect to x.
Well, the derivative
with respect to x is 0.
The derivative with
respect to y is z.
The derivative with
respect to z is y.
Those were partial derivatives.
They tell me how much the
first unknown x moves.
They tell me what's happening
with the first unknown
x around the critical
point whichever it is.
OK.
What about the
partial derivatives
from the second equation?
it's partial derivatives
will go into this row.
So x has a minus 2z.
y derivative is 0.
z derivative is minus 2x.
And the third one, the
z derivative is 0 here.
The y derivative in x.
And the x derivative is y.
I've found the 3 by 3
matrix with the nine
partial first derivatives.
OK.
It's the eigenvalues of
that matrix at these points
that decide stability.
So I write that down.
Eigenvalues of J at
the critical points x,
y, z that's what I need.
That's what decides stability.
Let me just take the
first critical point.
What is my matrix?
I have to figure out what
is the matrix at that point?
And I'll just take 1, 0, 0.
1, 0, 0.
If x is 1-- so I'm getting,
this is at the point x equal 1.
y and z are 0.
So if x is 1, then that
that's a minus 2 and a 1.
And I think
everything else is 0.
So it'll be the eigenvalues
of that matrix that
decide the stability 1,
0, 0 of that fixed point.
And remember, that's the
toss around the narrow axis.
That's the toss
around the short axis.
OK.
What about the eigenvalues
of that matrix?
Well, I can see here that
really it's three by three.
But really, with
all those 0s, that
gives me an eigenvalues of 0.
So I'm going to have an
eigenvalue of 0 here.
And then I'm going
to have eigenvalues
from the part of that
matrix, which is two by two.
So I'll have a
lambda equals 0 here.
And two eigenvalues from here.
And I look at that,
and what do I see?
Now this is a two
by two problem.
I see the trace is 0.
0 plus 0.
My eigenvalues are a
plus and minus pair
because they add to 0.
They multiply to
give the determinant.
The determinant of
that matrix is 2.
The determinant of
that matrix is 2.
OK.
So it has a positive
determinant.
That's good for stability.
But the trace is only 0.
It's not quite negative.
It's not positive.
It's just at 0.
So this is going to be a
case of neutral stability.
The eigenvalues will be-- I'll
have a 0 eigenvalue from there.
The eigenvalues from this
two by two will be-- there'll
be a square root of
2 times i and a minus
the square root of 2 times i.
I think those are
the eigenvalues.
And what I see there is
they're all imaginary.
This is a pure oscillation.
The wobbling keeps wobbling.
Doesn't get worse.
Doesn't go away.
It's neutral stability
at this point.
So neutral stability is what
we hopefully will see again.
Yes.
And I think, also, if I
flip on the long axis.
Good.
Did you see that
brilliant throw?
It's neutral stability.
It came back without
doing anything too bad.
And I finally have to do the
axis that we're all intensely
waiting for, the middle axis.
And the middle axis is when
the book starts tumbling,
and it's going to be a question
of whether I can catch it
or not.
May I try?
And then may I find-- what am I
expecting on the neutral axis?
I'm expecting instability.
I think actually it
will be a saddle point.
But there'll be a
positive eigenvalues.
There will be a
positive eigenvalue.
And it is responsible for the
tumbling, the wild tumbling
that you will see.
And it's connected
with the point staying
on this hyperbola that
wonders away from-- so it's
this one now that I'm doing.
This guy is the-- I'll put
a box around-- a double box
around it.
That's the unstable one, which
I'm about to demonstrate.
Ready?
OK.
Whoops.
OK.
It took two hands to catch it.
Let me try it again.
The point is it starts tumbling,
and it goes in all directions.
It's like a football, a
really badly thrown football.
It's like a football being
thrown that goes end to end.
The whole flight breaks
up, and the ball is a mess.
Catching it is ridiculous.
And I'm doing it with a book.
Yes.
You saw that by
watching really closely.
OK.
Better if you do it.
I'll end with the
eigenvalues at this point.
So the eigenvalues
at that point--
can I just erase my matrix?
So this was a neutrally
stable one, a center
in the language of stability.
That's a center which you just
go around and round and round.
But now I'm going to just take
x and z to be 0 and y to be 1.
So can I erase that
matrix and take--
If x and z are 0, and y is
1-- so I get a 1 down here.
And I get a 1 up there.
And nothing else.
Everything else is 0.
OK.
That's my three by three matrix.
What are its eigenvalues?
What are the eigenvalues
of that three by three very
special matrix?
This is now the-- this was
the first derivative matrix,
the Jacobian matrix, at
this point, corresponding
to the middle axis.
OK.
Again, I'm seeing some 0s.
I'll reduce this to that two
by two matrix and this matrix.
Really, I have this two
by two matrix in the xz,
and this one in the y.
How about that guy?
You recognize what we're
looking at with this matrix.
So with that matrix, I can
tell you the eigenvalues.
We can see the trace is 0.
The eigenvalues add to 0.
They multiply to
the determinant.
And the determinant is minus 1.
So the eigenvalues
here are 1 and minus 1.
And then this guy gives 0.
And it's that eigenvalue
of 1 that's unstable.
That eigenvalue
of 1 is unstable.
OK.
So mathematics shows
what the experiment
shows: an unstable
rotation tumbling
around that middle axis.
Thank you.
