Attitude matrix and quaternion path for m>1

I noticed that somehow I did not finish with the case m>1. So today, without further ado, I am posting the algorithm.
We have the body with I_1<I_2<I_3, and we are considering the case with d>1/I_2, where d is, as always, the ratio of the doubled kinetic energy to angular momentum squared.
Then we define

(1)   \begin{eqnarray*} A_1&=&\sqrt{\frac{I_1(dI_3-1)}{I_3-I_1}},\\ A_2&=&\sqrt{\frac{I_2 (1-dI_1)}{I_2-I_1}},\\ A_3&=&\sqrt{\frac{I_3 (1-dI_1)}{I_3-I_1}},\\ B&=&\sqrt{\frac{(dI_3-1)(I_2-I_1)}{I_1I_2I_3}},\\ \mu&=&\frac{1}{m}=\frac{(1-dI_1)(I_3-I_2)}{(dI_3-1)(I_2-I_1)}. \end{eqnarray*}

(2)   \begin{eqnarray*} L_1(t)&=&A_1\,\dn(Bt,\mu),\\ L_2(t)&=&A_2\,\sn(Bt,\mu),\\ L_3(t)&=&A_3\,\cn(Bt,\mu). \end{eqnarray*}

(3)   \begin{equation*} \alpha=\frac{I_3-I_1}{\sqrt{\frac{I_1(dI_3-1)(I_2-I_1)I_3}{I_2}}}, \end{equation*}

(4)   \begin{equation*} \nu=\frac{I_3-dI_1I_3}{I_1-dI_1I_3}. \end{equation*}

(5)   \begin{equation*} \psi(t)=\frac{t}I_3-\arctan\left((A_2/A_1)\mathrm{sd}(Bt,\mu)  \right)+\alpha \Pi(\nu,\am(Bt,\mu),\mu),\ \end{equation*}

(6)   \begin{equation*} Q_1(t)=\begin{bmatrix}1-\frac{L_1(t)^2}{1+L_3(t)}&-\frac{L_1(t)L_2(t)}{1+L_3(t)}&-L_1(t)\\ -\frac{L_1(t)L_2(t)}{1+L_3(t)}&1-\frac{L_2(t)^2}{1+L_3(t)}&-L_2(t)\\L_1(t)&L_2(t)&L_3(t)\end{bmatrix}. \end{equation*}

(7)   \begin{equation*} Q_2(t)=\begin{bmatrix}\cos\psi(t)&-\sin\psi(t)&0\\\sin\psi(t)&\cos\psi(t)&0\\0&0&1 \end{bmatrix}. \end{equation*}

(8)   \begin{equation*} Q(t)=Q_2(t)Q_1(t). \end{equation*}

(9)   \begin{eqnarray*} q_0(t)&=&\sqrt{\frac{1+L_3(t)}{2}}\cos\frac{\psi(t)}{2},\\ q_1(t)&=&\frac{1}{\sqrt{2(1+L_3(t))}}\left(L_2(t)\cos\frac{\psi(t)}{2}+L_1(t)\sin\frac{\psi(t)}{2}\right),\\ q_2(t)&=&\frac{1}{\sqrt{2(1+L_3(t))}}\left(L_2(t)\sin\frac{\psi(t)}{2}-L_1(t)\cos\frac{\psi(t)}{2}\right),\\ q_3(t)&=&\sqrt{\frac{1+L_3(t)}{2}}\,\sin\frac{\psi(t)}{2},\\ q(t)&=&(q_0(t),q_1(t),q_2(t),q_3(t))=q_0(t)+\mathbf{i}\,q_1(t)+\mathbf{j}\,q_1(t)+\mathbf{k}\,q_3(t). \end{eqnarray*}

I use the above formulas to draw a stereographic projection of one particular path. So, I take I_1=1,I_2=2,I_3=3,d=0.5000001, and do the parametric plot of the curve \mathbf{r}(t) in \mathbf{R}^3,


I show below two plots. One with t\in(-1000,1000), and one with t\in(-10000,10000). For this selected value of d, the time between consecutive flips, given by the formula

(10)   \begin{equation*}\tau =4\mathrm{EllipticK}(\mu)/B= 116.472.\end{equation*}

So, for t\in(-10000,10000) we have somewhat less than 200 flips, and the lines are getting rather densely packed in certain regions.
Notice that is just one geodesic line, geometrically speaking the straightest possible line in the geometry determined by the inertial properties of the body.

It is this “geometry”that will become the main subject of the future notes.

Geodesic line for t between -1000 and 1000

Geodesic line for t between -10000 and 10000

How can such a line be “straight”? Well, it is….

Without goals everything is a propos of nothing

We are still traveling in the land of 4D butterflies made of remarkable circles such as those introduced in the note Meeting with remarkable circles three days ago. We did not travel too far during these three days, but we managed to familiarize ourselves with one particular family of these peculiar creatures in the last post Being a 4D butterfly – how does it feel? It was supposed to be an illustration of More than one path, which I illustrated with the wisdom of Miyamoto Musashi, expert Japanese swordsman and rōnin:

But, in fact, my many geodesics 4D butterfly picture of yesterday did not fit the Buddhist philosophy of many paths to the top of the mountain:

4D butterfly

What we see are many paths, all of the same type, but each leading to similar, but a different mountain. It is, perhaps, a proper illustration for the saying that:
Happiness is a journey, not a destination.

It sounds amusing, but, as you can read in Don’t think of a black cat” by Peter Wright,

Every scenario or paradigm and situation should have a goal or a collection of goals, outcomes that are worked towards. Without goals everything is a propos of nothing (as with Dirk and Deke … or maybe All I Wanna Do by Sheryl Crow).

All I Wanna DoSheryl Crow
This ain’t no country club
And it ain’t no disco
This is New York City
1, 2, 1, 2
“All I wanna do is have a little fun before I die, ”
Says the man next to me out of nowhere
It’s apropos of nothin’
He says, “His name is William”
But I’m sure he’s Bill or Billy or Mac or Buddy
And he’s plain ugly to me
And I wonder if he’s ever had a day of fun in his whole life

I promised to explain in detail how exactly the 4D butterfly is created. If only to have some fun. Yesterday I was not completely sure that my idea was good. Today I think it is OK, therefore worth sharing.

The key is the algorithm described in Meeting with remarkable circles. It produces a path q(t) in the 3D sphere in 4D space, the 3-sphere of quaternions of unit length. The function q(t) is a very particular solution of the evolution equation for a free rigid body rotating in zero-gravity space about its center of mass – like in Dzhanibekov effect. It is a very special solution, with just one flip, and also with time origin t=0 arranged so, that it is exactly in the middle of the flip, between one story of almost uniform rotation one way in the past, and another story of almost uniform rotations, another way, in the future. For our particular path we have

(1)   \begin{equation*}q(0)=(\frac{\sqrt{3}}{2},0,\frac{1}{2},0),\quad\mbox{i.e. } q(0)=\frac{\sqrt{3}}{2}+\frac{1}{2}\,\mathbf{j}.\end{equation*}

The function q(t) is a solution of the quaternion evolution equation (cf. Quaternion evolution)

(2)   \begin{equation*}\dot{q}(t)=\frac{1}{2}q(t)\widehat{\boldsymbol{\Omega} (t)},\end{equation*}

where \Omega_i(t)=L_i(t)/I_i, and \mathbf{L} is a solution of Euler’s equations:

(3)   \begin{eqnarray*} \frac{dL_1}{dt}+(\frac{1}{I_2}-\frac{1}{I_3})L_2L_3&=&0,\\ \frac{dL_2}{dt}+(\frac{1}{I_3}-\frac{1}{I_1})L_3L_1&=&0,\\ \frac{dL_3}{dt}+(\frac{1}{I_1}-\frac{1}{I_2})L_1L_2&=&0. \end{eqnarray*}

There are two simple ways in which we get, from our particular solution q(t), other solutions. The first way is to multiply q(t) by a constant unit quaternion u. If q(t) is a solution, then uq(t) is another solution. This follows by multiplying both sides of Eq. (2) by u form the left, and by noticing that u, being constant, enters under the symbol of differentiation d/dt, indicated by the dot over q.

The second method is by shifting the origin of time. For any fixed s the function t\mapsto q(t+s) is also a solution of Eq. (2) if q(t) is. Formally it follows from the rule of differentiation of a composite function (the Chain rule) and from the fact that \frac{d}{dt}(t+s)=1.

For producing the family of solutions used in the 4D butterfly image we use both methods as follows. Let
q(t,s) be defined as follows:

(4)   \begin{equation*}q(t,s)=q(0)q(s)^*q(s+t).\end{equation*}

Then, for any fixed s, the function t\mapsto q(t,s) is a solution of Eq. (2) with the property that q(0,s)=q(0). that is all trajectories have the same origin, namely q(0). In verifying this last property we use the fact, that q(s) are unit quaternions, therefore q(s)^*q(s)=1.

To produce the butterfly on the picture above I used s from s=-4 to s=4, with step 0.4. Today I went further on, with s between -10 and 10, step 0,5, to produce this image:

An interesting symmetry shows up when looking from far away, between circles that are “bridges”and circles that are “attractors”.

Some mystery is hidden there, waiting for being discovered in the future.

But, all of the above was in the spirit of “Happiness is a journey, not a destination” philosophy. What about many paths to the top of the mountain?

That would be keeping the attractor circles fixed (these are the two tops of mountain, one in the past, one in the future), and changing only the bridges connecting these limit circles.

This is, in fact, much easier. We just shift the time origin, and skip the part of multiplication from the left. Here are images produced this way.

s between -1 and 1

s between -4 and 4

s between -8 and 8

The two eternal return circles are the same, only the bridges connecting them are shifting.

This last image resembles images from more realistic Dzhanibekov’s many-flips histories. It is time now to study them… if only to have some fun.

More than one path

On one hand we easily find this kind of an advice:

Is it possible to follow more than one path?

If you are practising spiritual discipline under the guidance of a Master, it is always advisable to give up your connection with other paths. If you are satisfied with one Master but are still looking for another Master, then you are making a serious mistake.

On the other hand we find this

Be All You Can Be: Don’t Choose One Path, Choose Multiple Paths

We will follow the second option, we will look for other paths, other than the one we already know. Other but of the same quality
This project has two parts. The first part is pure algebra. No pictures. Pictures will come in the second part, when we will already know what to picture.

Consider this scenario: we are looking at all possible trajectories of the dynamics of our free rigid body with I_1<I_2<I_3, in the quaternion group S^3. Consider all trajectories q(t) with the property that q(0)=1. Or, better, with the property that q(0)=-1. Why q(0)=-1 is better? Because under stereographic projection q(0)=1 is mapped into infinity, while q(0)=-1 is mapped into the origin of the coordinate system. Easier to draw.
From our point q(0)=-1 the trajectory can go in any direction. The direction of the trajectory is described by the derivative \dot{q}(0). We know that every trajectory is a solution of the equation

(1)   \begin{equation*} \dot{q}(t)=\frac{1}{2}q(t)\widehat{\boldsymbol{\Omega} (t)}.\end{equation*}

Therefore, since q(0)=-1,

(2)   \begin{equation*}\dot{q}(0)=-\frac{1}{2}\widehat{\boldsymbol{\Omega}(0)}.\end{equation*}

Let us assume that the angular momentum vector is normalized, it has unit length. We are interested in those very-very special trajectories for which d=1/I_2. They are special, they encode the essence of the flip in Dzhanibekov effect. In what follows, to simplify the notation I will write \boldsymbol{\Omega} instead of \boldsymbol{\Omega}(0).
The doubled kinetic energy is

(3)   \begin{equation*} 2 E_k=I_1\Omega_1^2+I_2\Omega_2^2+I_3\Omega_3^2.\end{equation*}

The length of the angular momentum vector, assumed to be 1, is

(4)   \begin{equation*} 1=I_1^2\Omega_1^2+I_2^2\Omega_2^2+I_3^2\Omega_3^2.\end{equation*}

We want our path to be special, that is we want 2 E_k=1/I_2.
Eq. (3) gives then

(5)   \begin{equation*}1=I_1I_2\Omega_1^2+I_2^2\Omega_2^2+I_2I_3\Omega_3^2.\end{equation*}

Comparing this with Eq. (4) the term I_2^2\Omega_2^2 cancels out and we get

(6)   \begin{equation*} I_1^2\Omega_1^2+I_3^2\Omega_3^2=I_1I_2\Omega_1^2+I_2I_3\Omega_3^2, \end{equation*}


(7)   \begin{equation*}I_3(I_3-I_2)\Omega_3^2=I_1(I_2-I_1)\Omega_1^2.\end{equation*}

Thus in order for the trajectory to be special the ratio \Omega_3/\Omega_1 must be special:

(8)   \begin{equation*} \frac{\Omega_3}{\Omega_1}=\pm\sqrt{\frac{I_1(I_2-I_1)}{I_3(I_3-I_2)}}.\end{equation*}

For our particular rigid body that we often use, with I_1=1,I_2=2,I_3=3, we should have

(9)   \begin{equation*} \frac{\Omega_3}{\Omega_1}=\pm \frac{1}{\sqrt{3}}.\end{equation*}

For our special trajectory that we were plotting in the last post Circles of eternal return
we had

(10)   \begin{eqnarray*} \Omega_1(0)&=&l_1(0)/I_1=A_1=\frac{1}{2},\\ \Omega_3(0)&=&l_3(0)/I_3=A_3/3=\frac{1}{2\sqrt{3}}. \end{eqnarray*}

So it is just one particular way of satisfying Eq. (9). There are, however other ways. These other ways need to be researched.
We will do it in the next posts.