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PROFESSOR: OK.

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Here's an example that's
more or less for fun.

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Because you'll see
me try to do it.

4
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You can do it better.

5
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I call the problem
the tumbling blocks.

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Only in this example,
in my demonstration,

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it's going to be
a tumbling book.

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I'm going to take a
book, the sacred book,

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and throw it in the air.

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And I'll throw it
three different ways.

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And the question is, is the
spinning book stable or not?

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And let me tell
you the three ways

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and then give you the three
equations that came from Euler.

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So those are the
three equations.

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You see that they're not linear.

16
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And those are for
the angular momentum.

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So there's a little physics
behind the equations.

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But for us, those are
the three equations.

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So the first throw
will spin around

20
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the very short axis, just
the thickness of the book,

21
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maybe an inch.

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So when I toss that,
as I'll do now,

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you will see if I can toss
it not too nervously I hope.

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It came-- it was stable.

25
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The book came back to
me without wobbling.

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Of course, my nerves would
give it a little wobble,

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and that wobble would continue.

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It will be only
neutrally stable.

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The wobble doesn't disappear.

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But it doesn't
grow into a tumble.

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OK.

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So that's one axis,
the short axis.

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Then I'll throw it also
around the long axis, flipped

34
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like this.

35
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I think that will be stable too.

36
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And then, finally, on
the intermediate axis,

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is middle length axis.

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Notice the rubber band that's
holding the book together.

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Holding so the pages don't open.

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And this, we'll see, I
think, will be unstable.

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And similarly,
throwing a football,

42
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throwing other Frisbees,
whatever your throw.

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Any 3D object has got these
three axes: a short one,

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a medium one, and a long axis.

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And the equations
will tell us short

46
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and long axes should
give a stable turning.

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And the in between
axis is unstable.

48
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Well, how do we decide for
our differential equation

49
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whether the fixed
point, a fixed point,

50
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that's a critical
point, a steady state--

51
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we have to find
this steady state,

52
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and then for each steady
state we linearize.

53
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We find the derivatives
at that steady state.

54
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And that gives us a constant
matrix at that steady state.

55
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And then the
eigenvalue is decided.

56
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So first, find the
critical points.

57
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Second, find the derivatives
at the critical points.

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Third, for that
matrix of derivatives,

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find the eigenvalues
and decide stability.

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That's the sequence of steps.

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OK.

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The first time we've ever
done a three by three matrix.

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Maybe the last time.

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OK.

65
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Let me, before I
start-- before I

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find the critical points--
notice some nice properties.

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If I multiply this
equation by x, this one

68
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by y, this one by z, and
add, those will add to 0.

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When there's an x there,
a y there, and a z there,

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I get a 1 minus 2 and
a 1 they add to 0.

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So x times dx dt.

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y times dy dt. z
time dz dt adds to 0.

73
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That's an important fact.

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That's telling me that the
derivative of something is 0.

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That something
will be a constant.

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So I'm seeing here
the derivative

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of that whole business
would be the derivative

78
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of a half probably.

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x squared, because the
derivative of x squared

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will be with a half.

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The derivative will be x dx dt.

82
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And y squared and z squared
is the derivative is 0.

83
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The derivative of that
line is just this line.

84
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It's 0.

85
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So this is a constant.

86
00:05:13,780 --> 00:05:16,910
No doubt, that's
probably telling me

87
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that the total energy, the
kinetic energy, is constant.

88
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After I've tossed that
book up in the air,

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I'm not touching it.

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It's doing its thing.

91
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And it's not going to change
energy because nothing

92
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is happening to it.

93
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It's just out there.

94
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Now there are other-- so
that's a rather nice thing.

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This is a constant.

96
00:05:43,060 --> 00:05:47,560
Now there's another way.

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If I multiply this one by 2x,
and I multiply this one by y,

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and add just those
two, that cancels.

99
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So 2x dx dt-- 2x times
the first one-- and y

100
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times the second one gives 0.

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Again, I'm seeing
something is constant.

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The derivative of something,
and that something

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is x squared plus 1/2 y
squared is a constant.

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Another nice fact.

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Another quantity
that's conserved.

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And as I'm flying
around in space,

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this quantity x squared plus
1/2 y squared does not change.

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This sort of-- that
involved all of xyz.

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And of course that's the
equation of a sphere.

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So in energy space,
or in an xyz space,

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our solution is wandering
around a sphere.

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And this is the equation for,
I guess, it's an ellipse.

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So there's an ellipse
on that's sphere

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that it's actually
staying on that ellipse.

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And in fact there's
another ellipse

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because I could've multiplied
this one by 2z and this one

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by y and added.

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And then those
would have canceled.

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Minus 2 xyz plus 2xyz.

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So that also tells
me that it would

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be probably z squared plus 1/2
y squared equals a constant.

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That's another ellipse.

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z squared plus 1/2 y squared.

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You see this?

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If I take the
derivative of that,

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I have 2z times dz dt
plus y times dy dt.

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Adding give 0.

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The derivative is 0.

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The thing is a constant.

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But!

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But, but, but!

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If I subtract this
one from this one,

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take the difference
of these two.

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Suppose I take this
one minus this one.

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The 1/2 y squared will go.

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So that will tell me
that x squared minus z

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squared is a constant.

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Oh, boy!

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I haven't solved
my three equations.

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But I found out a whole
lot about the solution.

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The solution stays on the
sphere, wanders around somehow.

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It also at the same time
stays on that ellipse.

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And it stays on that ellipse.

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But this is not an
ellipse, not an ellipse.

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That's the equation
of a hyperbola.

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And that's why-- which, of
course, goes off to infinity.

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And that's why the-- well,
it goes off to infinity,

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00:08:59,640 --> 00:09:01,800
but it has to stay
on the sphere.

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It wanders.

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This will be responsible
for the unstable motion.

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00:09:08,080 --> 00:09:15,830
Professor [INAUDIBLE], who would
do this far better than me,

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his great lecture in 1803,
Differential Equations,

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was exactly this.

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The full hour to tell you
everything about the tumbling

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box.

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So I'm going to do
the demonstration

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and write down the main facts
and understand the stability,

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the discussion of stability.

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I'm ready to move on to the
discussion of stability.

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Again, here are my
three equations.

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We're up to three
equation, so we're

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going have a three
by three matrix.

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And first I have to find
out the critical points,

164
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the steady states
of this motion.

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How could I toss it so
that if I toss it perfectly

166
00:10:07,550 --> 00:10:10,640
it stays exactly as tossed?

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And the answer is,
around the axis.

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If I toss this perfectly,
with no nerves,

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it'll just spin exactly
as I'm throwing it.

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The x, y, and z will
all be constant.

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00:10:27,530 --> 00:10:30,145
Now, when I toss
it on that axis.

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I'm looking for-- here
are my right hand side.

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YZ, minus 2XZ, and XY.

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00:10:47,500 --> 00:10:50,320
And I wrote those
in capital letters

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because those are going
to be my steady states.

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00:10:54,560 --> 00:11:00,610
Now I'm looking for are points
where nothing's happened.

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If those three right hand
sides of the equation are 0,

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I'm not going to move.

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xyz will stay where they are.

180
00:11:09,350 --> 00:11:13,340
So can you see solutions
of those three equations?

181
00:11:13,340 --> 00:11:16,040
Well, they're pretty
special equations.

182
00:11:16,040 --> 00:11:23,450
I get a solution when, for
example, solutions could be 1,

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0, 0/

184
00:11:26,230 --> 00:11:30,690
If two of the three--
if y and z are 0.

185
00:11:30,690 --> 00:11:34,400
y is 0, z is 0, y
and z are 0, I get 0.

186
00:11:34,400 --> 00:11:38,830
So that is a certainly
steady state.

187
00:11:38,830 --> 00:11:42,740
x equal 1, y and
z equal 0 and 0.

188
00:11:42,740 --> 00:11:49,310
And that steady state is
spinning around one axis.

189
00:11:49,310 --> 00:11:52,790
And, actually, I could have
also a minus 1 would also be.

190
00:11:52,790 --> 00:11:58,080
So I've found, actually, two
steady states with y and z 0.

191
00:11:58,080 --> 00:12:02,980
Then there'll be two
more with x and z 0.

192
00:12:02,980 --> 00:12:05,890
And this could be--
that'll be spinning

193
00:12:05,890 --> 00:12:07,870
around the middle axis.

194
00:12:07,870 --> 00:12:13,090
And then 0, 0, 1
or minus 1, that

195
00:12:13,090 --> 00:12:16,500
would be spinning around the
third axis, the long axis.

196
00:12:16,500 --> 00:12:18,840
So those are my steady states.

197
00:12:18,840 --> 00:12:21,130
And I guess, come
to think of it, 0,

198
00:12:21,130 --> 00:12:26,240
0, 0 would also
be a steady state.

199
00:12:26,240 --> 00:12:28,810
I think I found them all.

200
00:12:28,810 --> 00:12:30,180
These are the xy's.

201
00:12:30,180 --> 00:12:37,880
These are the x,
y, z steady states.

202
00:12:37,880 --> 00:12:39,360
OK.

203
00:12:39,360 --> 00:12:42,310
So now once you know the steady
states, that's usually fun,

204
00:12:42,310 --> 00:12:44,750
as it was here.

205
00:12:44,750 --> 00:12:50,800
Now the slightly less fun step
is find all the derivatives,

206
00:12:50,800 --> 00:12:54,970
find that Jacobian
matrix of derivative.

207
00:12:54,970 --> 00:12:58,190
So I've got three equations.

208
00:12:58,190 --> 00:13:01,350
Three unknowns, xyz.

209
00:13:01,350 --> 00:13:03,180
Three right hand sides.

210
00:13:03,180 --> 00:13:09,440
And I have to find-- I'm
going to have a three by three

211
00:13:09,440 --> 00:13:12,010
matrix of derivatives.

212
00:13:12,010 --> 00:13:14,050
This Jacobian matrix.

213
00:13:14,050 --> 00:13:19,680
So J for the Jacobian, the
matrix of first derivatives.

214
00:13:19,680 --> 00:13:24,300
So what goes into the
matrix of first derivative?

215
00:13:24,300 --> 00:13:26,710
Let me write Jacobian.

216
00:13:26,710 --> 00:13:30,380
It is named after Jacoby.

217
00:13:30,380 --> 00:13:32,270
It's the matrix of
first derivatives.

218
00:13:32,270 --> 00:13:35,650
On the top row are
the derivatives

219
00:13:35,650 --> 00:13:38,950
of the first function
with respect to x.

220
00:13:38,950 --> 00:13:42,400
Well, the derivative
with respect to x is 0.

221
00:13:42,400 --> 00:13:45,226
The derivative with
respect to y is z.

222
00:13:45,226 --> 00:13:49,910
The derivative with
respect to z is y.

223
00:13:49,910 --> 00:13:51,350
Those were partial derivatives.

224
00:13:54,150 --> 00:14:00,160
They tell me how much the
first unknown x moves.

225
00:14:00,160 --> 00:14:02,400
They tell me what's happening
with the first unknown

226
00:14:02,400 --> 00:14:07,940
x around the critical
point whichever it is.

227
00:14:07,940 --> 00:14:08,440
OK.

228
00:14:08,440 --> 00:14:13,860
What about the
partial derivatives

229
00:14:13,860 --> 00:14:15,390
from the second equation?

230
00:14:15,390 --> 00:14:19,040
it's partial derivatives
will go into this row.

231
00:14:19,040 --> 00:14:22,256
So x has a minus 2z.

232
00:14:22,256 --> 00:14:26,280
y derivative is 0.

233
00:14:26,280 --> 00:14:29,470
z derivative is minus 2x.

234
00:14:29,470 --> 00:14:33,600
And the third one, the
z derivative is 0 here.

235
00:14:33,600 --> 00:14:35,910
The y derivative in x.

236
00:14:35,910 --> 00:14:37,790
And the x derivative is y.

237
00:14:41,570 --> 00:14:45,950
I've found the 3 by 3
matrix with the nine

238
00:14:45,950 --> 00:14:48,800
partial first derivatives.

239
00:14:48,800 --> 00:14:49,540
OK.

240
00:14:49,540 --> 00:14:54,410
It's the eigenvalues of
that matrix at these points

241
00:14:54,410 --> 00:14:56,460
that decide stability.

242
00:14:56,460 --> 00:14:58,280
So I write that down.

243
00:14:58,280 --> 00:15:05,782
Eigenvalues of J at
the critical points x,

244
00:15:05,782 --> 00:15:09,230
y, z that's what I need.

245
00:15:09,230 --> 00:15:11,270
That's what decides stability.

246
00:15:11,270 --> 00:15:19,030
Let me just take the
first critical point.

247
00:15:19,030 --> 00:15:20,440
What is my matrix?

248
00:15:20,440 --> 00:15:24,490
I have to figure out what
is the matrix at that point?

249
00:15:24,490 --> 00:15:27,160
And I'll just take 1, 0, 0.

250
00:15:27,160 --> 00:15:28,600
1, 0, 0.

251
00:15:28,600 --> 00:15:37,050
If x is 1-- so I'm getting,
this is at the point x equal 1.

252
00:15:37,050 --> 00:15:41,090
y and z are 0.

253
00:15:41,090 --> 00:15:45,900
So if x is 1, then that
that's a minus 2 and a 1.

254
00:15:45,900 --> 00:15:47,830
And I think
everything else is 0.

255
00:15:52,070 --> 00:15:56,030
So it'll be the eigenvalues
of that matrix that

256
00:15:56,030 --> 00:16:03,800
decide the stability 1,
0, 0 of that fixed point.

257
00:16:03,800 --> 00:16:08,930
And remember, that's the
toss around the narrow axis.

258
00:16:08,930 --> 00:16:14,810
That's the toss
around the short axis.

259
00:16:14,810 --> 00:16:16,010
OK.

260
00:16:16,010 --> 00:16:19,110
What about the eigenvalues
of that matrix?

261
00:16:19,110 --> 00:16:24,420
Well, I can see here that
really it's three by three.

262
00:16:24,420 --> 00:16:26,930
But really, with
all those 0s, that

263
00:16:26,930 --> 00:16:29,700
gives me an eigenvalues of 0.

264
00:16:29,700 --> 00:16:33,290
So I'm going to have an
eigenvalue of 0 here.

265
00:16:33,290 --> 00:16:35,270
And then I'm going
to have eigenvalues

266
00:16:35,270 --> 00:16:40,310
from the part of that
matrix, which is two by two.

267
00:16:40,310 --> 00:16:43,560
So I'll have a
lambda equals 0 here.

268
00:16:43,560 --> 00:16:46,230
And two eigenvalues from here.

269
00:16:46,230 --> 00:16:51,290
And I look at that,
and what do I see?

270
00:16:51,290 --> 00:16:54,740
Now this is a two
by two problem.

271
00:16:54,740 --> 00:16:59,550
I see the trace is 0.

272
00:16:59,550 --> 00:17:00,420
0 plus 0.

273
00:17:00,420 --> 00:17:03,650
My eigenvalues are a
plus and minus pair

274
00:17:03,650 --> 00:17:05,859
because they add to 0.

275
00:17:05,859 --> 00:17:08,339
They multiply to
give the determinant.

276
00:17:08,339 --> 00:17:12,530
The determinant of
that matrix is 2.

277
00:17:12,530 --> 00:17:15,319
The determinant of
that matrix is 2.

278
00:17:15,319 --> 00:17:16,069
OK.

279
00:17:16,069 --> 00:17:18,470
So it has a positive
determinant.

280
00:17:18,470 --> 00:17:20,359
That's good for stability.

281
00:17:20,359 --> 00:17:22,680
But the trace is only 0.

282
00:17:22,680 --> 00:17:24,079
It's not quite negative.

283
00:17:24,079 --> 00:17:25,099
It's not positive.

284
00:17:25,099 --> 00:17:26,579
It's just at 0.

285
00:17:26,579 --> 00:17:30,500
So this is going to be a
case of neutral stability.

286
00:17:30,500 --> 00:17:36,440
The eigenvalues will be-- I'll
have a 0 eigenvalue from there.

287
00:17:36,440 --> 00:17:40,130
The eigenvalues from this
two by two will be-- there'll

288
00:17:40,130 --> 00:17:45,220
be a square root of
2 times i and a minus

289
00:17:45,220 --> 00:17:47,610
the square root of 2 times i.

290
00:17:47,610 --> 00:17:50,220
I think those are
the eigenvalues.

291
00:17:50,220 --> 00:17:54,760
And what I see there is
they're all imaginary.

292
00:17:54,760 --> 00:17:57,040
This is a pure oscillation.

293
00:17:57,040 --> 00:17:59,320
The wobbling keeps wobbling.

294
00:17:59,320 --> 00:18:00,570
Doesn't get worse.

295
00:18:00,570 --> 00:18:02,360
Doesn't go away.

296
00:18:02,360 --> 00:18:06,360
It's neutral stability
at this point.

297
00:18:06,360 --> 00:18:12,240
So neutral stability is what
we hopefully will see again.

298
00:18:12,240 --> 00:18:13,270
Yes.

299
00:18:13,270 --> 00:18:18,820
And I think, also, if I
flip on the long axis.

300
00:18:18,820 --> 00:18:19,320
Good.

301
00:18:19,320 --> 00:18:22,600
Did you see that
brilliant throw?

302
00:18:22,600 --> 00:18:24,300
It's neutral stability.

303
00:18:24,300 --> 00:18:30,310
It came back without
doing anything too bad.

304
00:18:30,310 --> 00:18:38,240
And I finally have to do the
axis that we're all intensely

305
00:18:38,240 --> 00:18:42,170
waiting for, the middle axis.

306
00:18:42,170 --> 00:18:45,960
And the middle axis is when
the book starts tumbling,

307
00:18:45,960 --> 00:18:48,590
and it's going to be a question
of whether I can catch it

308
00:18:48,590 --> 00:18:49,460
or not.

309
00:18:49,460 --> 00:18:50,610
May I try?

310
00:18:50,610 --> 00:18:55,440
And then may I find-- what am I
expecting on the neutral axis?

311
00:18:55,440 --> 00:18:57,920
I'm expecting instability.

312
00:18:57,920 --> 00:18:59,980
I think actually it
will be a saddle point.

313
00:18:59,980 --> 00:19:03,820
But there'll be a
positive eigenvalues.

314
00:19:03,820 --> 00:19:05,660
There will be a
positive eigenvalue.

315
00:19:05,660 --> 00:19:10,770
And it is responsible for the
tumbling, the wild tumbling

316
00:19:10,770 --> 00:19:12,280
that you will see.

317
00:19:12,280 --> 00:19:16,420
And it's connected
with the point staying

318
00:19:16,420 --> 00:19:20,670
on this hyperbola that
wonders away from-- so it's

319
00:19:20,670 --> 00:19:22,920
this one now that I'm doing.

320
00:19:22,920 --> 00:19:27,440
This guy is the-- I'll put
a box around-- a double box

321
00:19:27,440 --> 00:19:28,490
around it.

322
00:19:28,490 --> 00:19:34,590
That's the unstable one, which
I'm about to demonstrate.

323
00:19:34,590 --> 00:19:36,020
Ready?

324
00:19:36,020 --> 00:19:38,771
OK.

325
00:19:38,771 --> 00:19:39,270
Whoops.

326
00:19:39,270 --> 00:19:39,769
OK.

327
00:19:39,769 --> 00:19:42,130
It took two hands to catch it.

328
00:19:42,130 --> 00:19:44,820
Let me try it again.

329
00:19:44,820 --> 00:19:52,420
The point is it starts tumbling,
and it goes in all directions.

330
00:19:52,420 --> 00:19:56,860
It's like a football, a
really badly thrown football.

331
00:19:56,860 --> 00:20:05,010
It's like a football being
thrown that goes end to end.

332
00:20:05,010 --> 00:20:09,260
The whole flight breaks
up, and the ball is a mess.

333
00:20:09,260 --> 00:20:11,660
Catching it is ridiculous.

334
00:20:11,660 --> 00:20:14,590
And I'm doing it with a book.

335
00:20:14,590 --> 00:20:15,230
Yes.

336
00:20:15,230 --> 00:20:19,520
You saw that by
watching really closely.

337
00:20:19,520 --> 00:20:21,480
OK.

338
00:20:21,480 --> 00:20:23,140
Better if you do it.

339
00:20:23,140 --> 00:20:27,360
I'll end with the
eigenvalues at this point.

340
00:20:27,360 --> 00:20:29,400
So the eigenvalues
at that point--

341
00:20:29,400 --> 00:20:32,330
can I just erase my matrix?

342
00:20:32,330 --> 00:20:36,570
So this was a neutrally
stable one, a center

343
00:20:36,570 --> 00:20:38,810
in the language of stability.

344
00:20:38,810 --> 00:20:42,430
That's a center which you just
go around and round and round.

345
00:20:42,430 --> 00:20:47,640
But now I'm going to just take
x and z to be 0 and y to be 1.

346
00:20:47,640 --> 00:20:52,240
So can I erase that
matrix and take--

347
00:20:52,240 --> 00:20:57,890
If x and z are 0, and y is
1-- so I get a 1 down here.

348
00:20:57,890 --> 00:21:00,040
And I get a 1 up there.

349
00:21:00,040 --> 00:21:01,280
And nothing else.

350
00:21:01,280 --> 00:21:02,540
Everything else is 0.

351
00:21:05,810 --> 00:21:08,470
OK.

352
00:21:08,470 --> 00:21:10,240
That's my three by three matrix.

353
00:21:10,240 --> 00:21:12,300
What are its eigenvalues?

354
00:21:12,300 --> 00:21:16,240
What are the eigenvalues
of that three by three very

355
00:21:16,240 --> 00:21:18,010
special matrix?

356
00:21:18,010 --> 00:21:25,860
This is now the-- this was
the first derivative matrix,

357
00:21:25,860 --> 00:21:29,950
the Jacobian matrix, at
this point, corresponding

358
00:21:29,950 --> 00:21:31,670
to the middle axis.

359
00:21:31,670 --> 00:21:32,420
OK.

360
00:21:32,420 --> 00:21:40,810
Again, I'm seeing some 0s.

361
00:21:40,810 --> 00:21:47,150
I'll reduce this to that two
by two matrix and this matrix.

362
00:21:47,150 --> 00:21:52,190
Really, I have this two
by two matrix in the xz,

363
00:21:52,190 --> 00:21:54,590
and this one in the y.

364
00:21:54,590 --> 00:21:56,400
How about that guy?

365
00:21:56,400 --> 00:22:01,710
You recognize what we're
looking at with this matrix.

366
00:22:01,710 --> 00:22:08,830
So with that matrix, I can
tell you the eigenvalues.

367
00:22:08,830 --> 00:22:10,820
We can see the trace is 0.

368
00:22:10,820 --> 00:22:13,480
The eigenvalues add to 0.

369
00:22:13,480 --> 00:22:15,560
They multiply to
the determinant.

370
00:22:15,560 --> 00:22:18,870
And the determinant is minus 1.

371
00:22:18,870 --> 00:22:22,640
So the eigenvalues
here are 1 and minus 1.

372
00:22:22,640 --> 00:22:25,140
And then this guy gives 0.

373
00:22:25,140 --> 00:22:30,250
And it's that eigenvalue
of 1 that's unstable.

374
00:22:30,250 --> 00:22:32,980
That eigenvalue
of 1 is unstable.

375
00:22:32,980 --> 00:22:33,660
OK.

376
00:22:33,660 --> 00:22:38,270
So mathematics shows
what the experiment

377
00:22:38,270 --> 00:22:44,430
shows: an unstable
rotation tumbling

378
00:22:44,430 --> 00:22:47,150
around that middle axis.

379
00:22:47,150 --> 00:22:48,950
Thank you.