Rotational Dynamics

Rotational dynamics

Peeter Joot January 29, 2008. Last Revision: Date : 2008/12/1204 : 41 : 44

1 GA introduction of angular velocity

By taking the ﬁrst derivative of a radially expressed vector we have the velocity

0 0 v = r rˆ + rˆ(rˆ ∧ r ) = rˆ(vr + rˆ ∧ v) Or, rvˆ = vr + rˆ ∧ v

rvˆ = vr + (1/r)r ∧ v Put this way, the earlier calculus exercise to derive this seems a bit silly, since it is probably clear that vr = rˆ · v. Anyways, let’s work with velocity expressed this way in a few ways.

1.1 Speed in terms of linear and rotational components

2 2 2 kvk = vr + (rˆ(rˆ ∧ v)) And,

(rˆ(rˆ ∧ v))2 = (v ∧ rˆ)rˆrˆ(rˆ ∧ v) = (v ∧ rˆ)(rˆ ∧ v) = −(rˆ ∧ v)2 = krˆ ∧ vk2

2 2 2 =⇒ kvk = vr + krˆ ∧ vk 2 2 = vr + krˆ ∧ vk

So, we can assign a physical signiﬁcance to the bivector.

1 krˆ ∧ vk = |v⊥| The bivector krˆ ∧ vk has the magnitude of the non-radial component of the velocity. This equals the magnitude of the component of the velocity perpendicular to its radial component (ie: the angular component of the velocity).

1.2 angular velocity. Prep.

Because krˆ ∧ vk is the non-radial velocity component, for small angles v⊥/r will equal the angle between the vector and its displacement. This allows for the calculation of the rate of change of that angle with time, what it called the scalar angular velocity (dimensions are 1/t not x/t). This can be done by taking the sin as the ratio of the length of the non-radial component of the delta to the length of the displaced vector.

krˆ(rˆ ∧ dr)k sin dθ = kr + drk

dr With dr = dt dt = vdt, the angular velocity is

1 sin dθ = krˆ(rˆ ∧ v)dtk kr + vdtk 1 = k(rˆ ∧ v)dtk kr + vdtk sin dθ 1 = krˆ ∧ vk |dt| kr + vdtk 1 = kr ∧ vk krk kr + vdtk

In the limit, taking dt > 0, this is

dθ 1 ω = = kr ∧ vk dt r2

1.3 angular velocity. Summarizing. Here is a summary of calculations so far involving the r ∧ v bivector

2 rˆ v = rˆv + (r ∧ v) r krk drˆ rˆ = (r ∧ v) dt r2 1 |v | = kr ∧ vk ⊥ krk dθ 1 ω = = kr ∧ vk dt r2

It makes sense to give the bivector a name. Given it’s magnitude the angular velocity bivector ω is designated r ∧ v ω = r2 So the linear and rotational components of the velocity can thus be expressed in terms of this, as can our unit vector derivative, scalar angular velocity, and perpendicular velocity magnitude:

dθ ω = = kωk dt v = rˆvr + rω = rˆ(vr + rω) drˆ = rˆω dt |v⊥| = r kωk

This is similar to the vector angular velocity (ω = (r × v)/r2), but instead of lying perpendicular to the plane of rotation, it deﬁnes the plane of rotation (for a vector a, a ∧ ω is zero if the vector is in the plane and non-zero if the vector has a component outside of the plane).

1.4 Explicit perpendicular unit vector. If one introduces a unit vector θˆ in the direction of rejection of r from dr, the total velocity takes the symetrical form

v = vrrˆ + rωθˆ dr dθ = rˆ + r θˆ dt dt

3 1.5 acceleration in terms of angular velocity bivector Taking derivatives of velocity, one can with a bit of work, express acceleration in terms of radial and non-radial components

0 a = (rˆvr + rω) 0 0 0 0 = rˆ vr + rˆvr + r ω + rω 0 0 0 = rˆωvr + rˆvr + r ω + rω 0 = rˆωvr + rˆar + vω + rω

But,

ω0 = ((1/r2)(r ∧ v))0 = (−2/r3)r0(r ∧ v) + (1/r2)(v ∧ v + r ∧ a) 2 = −(2/r)vrω + (1/r )(r ∧ a)

So,

a = rˆar − rˆωvr + vω + rˆ(rˆ ∧ a) = rˆar − (v − rω)ω + vω + rˆ(rˆ ∧ a)

2 = rˆar + rω + rˆ(rˆ ∧ a) 2 = rˆ(ar + rω ) + rˆ(rˆ ∧ a)

Note that ω2 is a negative scalar, so as normal writing kωk2 = −ω2, we have acceleration in a fashion similar to the traditional cross product form:

2 a = rˆ(ar − rkωk ) + rˆ(rˆ ∧ a) 2 = rˆ(ar − rkωk + rˆ ∧ a)

In the traditional representation, this last term, the non-radial acceleration component, is often expressed as a derivative. In terms of the wedge product, this can be done by noting that

(r ∧ v)0 = v ∧ v + r ∧ a = r ∧ a

4 r a = rˆ(a − rkωk2) + (r ∧ v)0) r r2 1 d(r2ω) = rˆ(a − rkωk2) + r r dt

Expressed in terms of force (for constant mass) this is

F = ma 1 d(mr2ω) = rˆ(ma ) + (mr)ω2 + r r dt 1 d(mr2ω) = F + (mr)ω2 + r r dt

Alternately, the non-radial term can be expressed in terms of torque

rˆ(rˆ ∧ a) = rˆ(rˆ ∧ ma) r = (r ∧ F) r2 1 = (r ∧ F) r 1 = τ r

Thus the torque bivector, which in magnitude was the angular derivative dW dr of the work done by the force kτk = τ = dθ = F · dθ is also expressable as a time derivative

d(mr2ω) τ = dt d(mr ∧ v) = dt d(r ∧ mv) = dt d(r ∧ p) = dt

This bivector mr2ω = r ∧ p is called the angular momentum, designated J. It is related to the total momentum as follows

5 1 p = rˆ(rˆ · p) + J r So the total force is

1 dJ F = F + mrω2 + r r dt

dJ Observe that for a purely radial (ie: central) force, we must have dt = 0 so, the angular momentum must be constant.

1.6 Kepler’s laws example. This follows the [Salas et al.(1990)Salas, Hille, Etgen, and Etgen] treatment, mod- iﬁed for the GA notation. Consider the graviational force

mM ma = −G rˆ r2 rˆ rˆ a = −GM = −ρ r2 r2 Or, rˆ 1 dv = − r2 ρ dt The unit vector derivative is

drˆ rˆ = (rˆ ∧ v) dt r rˆ J = r2 m 1 dv = − J mρ dt d(− 1 vJ) = mρ dt

The last because J, m, and ρ are all constant. Before continuing, let’s examine this funny vector bivector product term. In general a vector bivector product will have vector and trivector parts, but the differential equation implies that this is a vector. Let’s conﬁrm this

6 vJ = v(r ∧ mv) = (mv2)vˆ (r ∧ vˆ ) = −(mv2)vˆ (vˆ ∧ r)

So, this is in fact a vector, it is the rejective component of r from the direction ofv ˆ scaled by −mv2. We can also calculate the product Jv from this:

vJ = −(mv2)vˆ (vˆ ∧ r) = −(mv2)(r ∧ vˆ )vˆ = −(r ∧ mv)v = −Jv

This antisymetrical result vJ = −Jv is actually the deﬁning property of the vector bivector “dot product” (unlike the vector dot product which is the symetrical parts). This vector bivector dot product selects the vector component, leaving the trivector part. Since v lies completely in the plane of the angular velocity bivector v ∧ J = 0 in this case. Anyways, back to the problem, integrating with respect to time, and intro- ducing a vector integration constant e we have 1 rˆ + vJ = e mρ Multiplying by r

1 r + rvJ = re mρ 1 r + (r · p + J)J = r · e + r ∧ e m2ρ

This results in three equations, one for each of the scalar, vector, and bivector parts

J2 r + = r · e m2ρ 1 (r · v)J = 0 mρ r ∧ e = 0

7 The ﬁrst of these equations is the result from Salas and Hille (integration constant differs in sign though).

J2 r − = r · e m2ρ

1.7 Circular motion

For circular motion vr = ar = 0, so:

v = rω v2 a = rˆ − + rˆ ∧ a r For constant circular motion:

a = vω + rω0 = vω + r(0) = r(ω)2 = −r kωk2

ie: the rˆ(rˆ ∧ a) term is zero... all accelartion is inwards. Can also expand this in terms of r and v:

a = r (ω)2 1 2 = r v r 1 1 = −r v v r r v2 = −r r2 1 = − v2 r

References

[Salas et al.(1990)Salas, Hille, Etgen, and Etgen] S.L. Salas, E. Hille, G.J. Etgen, and G.J. Etgen. Calculus: one and several variables. Wiley New York, 1990.