Never twice the same

Posted on by arkajad

“Ever present, never twice the same;
Ever changing, never less than whole”

~Robert Irwin

If for an instant he dozed off, the image of Fantômas took shape in his mind, but never twice the same: sometimes he saw a colossal figure with bestial face and muscular shoulders; sometimes a wan, thin creature, with strange and piercing eyes; sometimes a vague form, a phantom—Fantômas!

Marcel Allain, Pierre Souvestre, Fantomas

In the last post, Taming the T-handle continued, we have learned how to model free rotation of an asymmetric rigid body in that very-very special case when the ratio of the angular momentum squared to the kinetic energy doubled is exactly I_2 – the middle moment of inertia. In that case only one flip of the axis happens. For an infinite time, in the past, and in the future, the T-handle rotates along the I_2 axis, or rather: very close to this axis. In its whole infinitely long life only once it flips, in our case near t=0. I say “near”, since what is, say 50, when compared with infinity?!

Of course in practice we will never be able to achieve this particular ratio of angular momentum and kinetic energy exactly! But who cares? In practice there are no perfect straight lines and there are no perfect spheres. So what? When the ratio is “slightly” different, when the parameter d that we used in all previous posts, d=2E_k/L^2 is “slightly” different from 1/I_2, then the motion is quasi-periodic, though I am not sure if it is indeed quasi-periodic according to the mathematical definition of quasiperiodicity. From Wikipedia:

In mathematics, a quasiperiodic function is a function that has a certain similarity to a periodic function. A function f is quasiperiodic with quasiperiod \omega if f(z+\omega )=g(z,f(z)), where g is a “simpler” function than f. What it means to be “simpler” is vague.

In our case there are flips occurring periodically, the pattern repeats itself is a very similar way (though not exactly the same) again and again, the period can be calculated using elliptic K-function. We have discussed it before, though we did not stress this phenomenon of “quasi” periodicity.

In Taming the T-handle continued we have derived several important formulas. We have

(1)   \begin{equation*}Q(t)=Q_1(t)Q_0(t),\end{equation*}

where

(2)   \begin{equation*}Q_0(t)=\begin{bmatrix}\frac{L_1(t)L_3(t)}{LL_p(t)}&\frac{L_2(t)L_3(t)}{LL_p(t)}&-\frac{L_p(t)}{L}\\-\frac{L_2(t)}{L_p(t)}&\frac{L_1(t)}{L_p(t)}&0\\\frac{L_1(t)}{L}&\frac{L_2(t)}{L}&\frac{L_3(t)}{L}\end{bmatrix},\end{equation*}

(3)   \begin{equation*}Q_1(t)=\begin{bmatrix}\cos\psi(t)&-\sin\psi(t)&0\\\sin\psi(t)&\cos\psi(t)&0\\0&0&1\end{bmatrix},\end{equation*}

and \psi(t) (for d<1/I_2) is given by:

    \[\psi(t)=c_1 t+\frac{c_2}{B}\,\Pi(-c_3;\am(t,m)|m).\]

Q_0(t) is periodic, with period 4K(m)/B. But Q_1(t) is not periodic at all. The functions \sin and \cos in Q_1(t) are periodic, but \psi(t) is a sum of two terms. The first term, linear in t, would make Q_1(t) periodic. with period 2\pi/c_1=2\pi I_3, but the second term spoils any periodicity whatsoever. That is why the flips, though governed by the periodicity 4K(m)/B from the elliptic functions \cn,\sn,\dn, never repeat exactly the same way.

Answers to Reader’s questions

In a comment to the last past Bjab formulated the following hypothesis:

So now there is a hypothesis that:
velocity of the end of the leg of T-handle is always perpendicular to the plane of T-handle.
Always means: when m=1

Proving (or disproving) this hypothesis is a good exercise. My T-handle is defined by the following Mathematica code:

In fact I have two slightly shifted T-handles, of different colors, so that it looks like one but painted top red, bottom yellow. The shift is in z-direction, so that my T-handle is all in the plane (x,y). The leg is along the y axis, from -0.9 to 0.2. We can assume that the leg end is at p_0=(0,-1,0)=-\vec{e}_2. The point p_0 moves with time along the trajectory:

    \[p(t)=Q(t)p_0.$ The velocity vector $\vec{v}$ at time $t$ is given by \[\vec{v(t)}=\frac{d}{dt}p(t)=\dot{Q}(t)p_0=Q(t)W(t)p_0.\]

The plane of the T-handle is also rotated by the matrix Q(t), therefore the angle between the velocity vector and the plane is the same as the angle between W(t)p_0 and the (x,y) plane. But
W(t)p_0=-\vec{\Omega(t)}\times \vec{e}_2. Thus \vec{v}(t) has components (\Omega_3(t),0,-\Omega_1(t)). This vector is orthogonal to \vec{e}_2, but is at the angle \pi/3 to \vec{e}_1.
Therefore it seems that the hypothesis, though fruitful and interesting, should not be considered as proven.
Update: Why \pi/3? We know that L_1(t)=A_1/\cosh(Bt),L_3(t)=A_3/\cosh(Bt). Thus \Omega_1(t)=(A_1/I_1)/\cosh(Bt),\Omega_3(t)=(A_3/I_3)/\cosh(Bt). We want to find the angle, so we are interested in normalized vector. Therefore we can skip the common 1/\cosh(Bt) factor. Thus we see that the angle is constant in time – which is an interesting fact. In particular for I_1=1,I_2=2,I_3=3 we have

    \[A_1/I_1=1/2,\, A_3/I_3=\frac{1}{2\sqrt{3}}.\]

The square root of the sum of squares is \frac{1}{\sqrt{3}}. Therefore the normalized first component is \frac{1}{2}. We have \arccos \frac{1}{2}=\pi/3.

37 thoughts on “Never twice the same”

    1. Indeed. I have made a mistake with the components. Fixed. They are (-\Omega_3,0,\Omega_1) Coeeficients in \Omega_1 and Omega_3, in front of the same function, are A_1=1/2 and A_3=\sqrt{3}/2}. Their squares add to 1. Therefore \cos of the angle with \vec{e}_1 is 1/2. Therefore the angle is \pi/3.

      Impossible? Why?

  3. Anyway, the end of the leg is in (0,-1,0) when t = – infinity (I suppose)
    therefore
    \mathrm{cosh}(Bt) = infinty
    therefore
    then
    \Omega_1=0, \Omega_3=0
    therefore
    we can tell that \vec{v} is perpendicular to e_1

          2. I think that the problem is with using formulas where start of time (t=0) is set in different way.

          5. ” think that the problem is with using formulas where start of time (t=0) is set in different way.”

            The argument at the end of the post does not depend on the value of $Q(0).$ And I do not see where this argument can go wrong. But I will triple check. There is something fishy …
            I rechecked. Can’t find a fault.

    2. OK. I will make a separate animation showing just the normalized velocity and the moving e_1(t) vector, both on the sphere.

      Update. I did it. Numerically it comes out as constant pi/3. Need to check again the calculations in the note and fix them.

      Update. Done. Everything was OK, only in the last sentence I have exchanged the two components.

  8. Let’s talk about formula:
    q(t)=Q(t)q_0
    It is true for every point of the top. There is some corelation between t and q_0.
    So for some t Q(t) should be identity matrix.
    For which t Q(t) is identity matrix?
    We will know then how T-handle is position at that time and where its leg is at that time.

    1. Q(t) is a particular solution of a differential equation. On animation I am selecting it so that the conserved angular is along the x axis. There is no reason for one particular solution to be the identity matrix at some t. But we can always choose the unique solution \tilde{Q}(t) with the property that \tilde{Q}(0)=I.
      It is sufficient to define

          \[\tilde{Q}(t)=Q(0)^{-1}Q(t).\]

      Thanks for all corrections!

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