Angular motion done right

Henry Lin

November 19, 2013

My goal here is to talk about angular motion with just enough hand-waving.

1 Circular Motion

Here’s the key result: ~a is perpendicular to ~v is perpendicular to position.

Proof: x = (r cos θ)ex +(r sin θ)ey. Let θ = ωt. v = −(ωr sin θ)ex +(ωr cos θ)ey. Now v ·x ∝ − cos θ sin θ+sin θ cos θ = 0.

We called x position to be concrete, but it is really just a vector that is rotating. Think of it as a spinning arrow: it reallydoesn’t matter what the arrow represents.

Taking a time derivative of a rotating vector with angular speed ω just rotates it by 90 degrees and rescales it by ω.

2 Angular momentum

Consider a CCW rotation around the z axis:

(cos θ − sin θsin θ cos θ

)If we consider very small rotations θ = ε, we get

(1 −εε 1

)Acting on a column vector (x, y)T , we get x → x + εy and y → y − εx. In general, this small change in the coordinateswill change L by a little bit. But if it doesn’t we get a conserved charge.

∂L∂x

y +∂L∂y

x

If L = m2 (v · v) + V (x, y, z) = m

2 (x2 + y2 + z2) + V , we have

∂L∂x

y +∂L∂y

x = mxy +myx ≡ Jz

If we considered infinitesimal rotations around the x and y axis, we would get Jx and Jy. Now the magic is that thequantity (Jx, Jy, Jz) transforms like a vector! We will not prove this claim, just shift the burden of proof. Note that we

can write ~J = ~r × ~p. If you believe that ~r × ~p is a vector, we are done. Important note: ~r goes from the point ofsymmetry to the particle. For example, if we have a gravitating mass at some location ~x0, r = ~x− ~x0.

1

3 Angular velocity

Now that we see how easy it is to sweep subtleties under the words “cross product,” let’s try to do it again. Say that wehave a vector ~V rotating around with angular speed ω.

dV

dt= ~ω × ~V ,

where ~ω points along the axis of rotation and has magnitude ω.

4 A general, rotating rigid body

Let’s begin by studying something slightly more general than a single particle: a system of N particles. The Lagrangianhas the form L = T1 + T2 + · · ·+ TN − V , where Ti = mi

2 (xi · xi). Thus, the conserved quantity is J = J1 + J2 + · · ·+ JN .Thus, angular momentum adds.

Passing to the continuous case, we get J =∫~r × ~v dm.

5 Is ~ω parallel to ~J?

Generally, no. For example, consider a gravitating mass at x0. Now, at some distance away from the mass in the xdirection, we have a satellite going around in circles about a point x1 in the y direction. The angular velocity vector istotally constant; the angular momentum vector is certainly not.

In general, ~J = I(~ω), for some function I. Let’s see what we can say about I. I(~ω) =∫~r × ~v dm =

∫~r × (~ω × ~r) dm.

I(~ω1 + α~ω2) =

∫~r × (~ω1 + α~ω2)× ~r dm =

∫~r × (~ω1 × ~r) + α~r × (~ω2 × ~r) dm

Thus, we have I(~ω1 + α~ω2) = I(~ω1) + αI(~ω2), so I is a linear operator!

To answer the question “when is ~ω parallel to ~J?” we must find I~ω = λ~ω. In other words, J is parallel to ω when ω is aneigenvector of I.

5.1 Spectral theorem

We will not answer the question yet; instead, we will show that no matter what the object is, there are always threedirections we can rotate it such that ω is parallel to J .

We can say more about I. In particular, consider

ω′ · Iω = ω′ ·∫r × (ω × r) dm =

∫(ω′ × r) · (ω × r) dm =

∫(ω × r) · (ω′ × r) dm = ω ·

∫r × (ω′ × r) dm = ω · Iω′.

We have used the fact that a · (b×c) = (a×b) ·c to show that I is self-adjoint. It follows from the spectral theorem that wecan construct an orthonormal basis of R3 consisting of eigenvectors of I. Call the directions defined by these eigenvectorsprincipal axes.

5.2 Torque

We define a quantity τ = dJ/dt. Aside: For a general point particle, τ = r × F . Proof. τ = ddt (r × p) = v × p + r × F .

Since p is always parallel to v, we have τ = r × F . This is easily generalized to τ =∫r × adm.

We are now ready to answer our original question.

In general, τ = Iω + Iω. Suppose that I = τ = 0. We will show that this implies that ω is parallel to J . Obviously,Iω = 0. If I = 0, then ω is trivially parallel to J . Suppose I 6= 0. Unfortunately, we can’t say that I is invertible, becausewe imagine one-dimensional systems. But let us restrict ourselves to the real world and enforce the additional constraintthat I−1 must exist. Then ω = 0. On the other hand, ω =

(ddtI−1) J = −I−1ω × J = 0, so ω × J = 0.

In fact, we can be a little more general. All we need is that ω = 0; this will give us ω × J = 0.

2

6 Euler’s equations

Consider a rigid body instantaneously rotating around an axis ~ω. We want a fancy way of saying τ = dJ/dt. Morin’sderivation (page 393) is actually not bad. We will use his convention of writing L = J . Now, choose a basis of orthonormaleigenvectors {e1(t), e2(t), e3(t)}, which rotates with the body. This means that ei · I(ei) = Iii = constant.

We have L = Liei(t), so dLdt = dLi

dt ei + deidt L

i = dLi

dt ei + ω × eiLi = dLi

dt ei + ω × L. We have

dLi

dtei · ei =

d(Iω · ei)dt

=d

dt(Iωiei) · ei =

d

dt(ωiI(ei) · ei) = Iii

dωi

dt= Iω · ei,

so we get the Euler equations in component-free form

τ = Iω + ω × L.

At long last, we can answer our question. ω is parallel to L if and only if τ − Iω = 0. In the event that there is no torque,we need ω ∈ null I.

3