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.

Circles of eternal return

Let me recall the particular trajectory that we are discussing. In fact the particular part of this particular trajectory where the flip occurs.


Click on the image to open gif animation. It will take time to load as it is 2.5 MB.

The flip is much like one of these flips that Russian cosmonaut Dzhanibekov observed in zero gravity, and was very much perplexed by the phenomenon.

We are investigating the jungle of math that we have to apply in order to simulate such a behavior of an ideal Platonic rigid body. At present we are analyzing the special conditions when only one flip happens. Apart of a rather short (as compared to eternity) period of time when the flip occurs, the body rotates uniformly about its middle axis. Its trajectory in the rotation group, represented by quaternions of unit norm, follows a circle. One circle before flip, another circle after the flip.

In this note I am taking a close look at these two circles. Can we find their exact coordinate representation in 3D after stereographic projection?
Yes, we can, and we will do it now.
We use the formulas from Meeting with remarkable circles.
For the body of our choice, with moments of inertia I_1=1, I_2=2,I_3=3, and on the trajectory where d=1/I_2, the constants that enter the solution have values

(1)   \begin{eqnarray*} A_1&=&\sqrt{\frac{I_1(I_3-I_2)}{I_2(I_3-I_1)}}=1/2,\\ A_2&=&1,\\ A_3&=&\sqrt{\frac{I_3(I_2-I_1)}{I_2(I_3-I_1)}}=\frac{\sqrt{3}}{2},\\ B&=&\frac{1}{I_2}\,\sqrt{\frac{(I_2-I_1)(I_3-I_2)}{I_1I_3}}=\frac{1}{2\sqrt{3}},\\ \delta&=&\frac{\sqrt{I_2(I_3-I_1)}-\sqrt{I_1(I_3-I_2)}}{\sqrt{I_3(I_2-I_1)}}=\frac{1}{\sqrt{3}}. \end{eqnarray*}

In the solution of the Euler’s equations we have functions \tanh (Bt) and \mathrm{sech} (Bt)

(2)   \begin{eqnarray*} l_1(t)&=& A_1\,\mathrm{sech }(B t),\\ l_2(t)&=&A_2\,\tanh( B t),\\ l_3(t)&=&A_3\,\mathrm{sech }(B t), \end{eqnarray*}

with \mathrm{sech }(x)=1/\cosh(x).
Here are the plots of these functions:


For |t|>100 we have l_1(t) smaller than 10^{-12}, and for |t|>200 smaller than 10^{-25}. We can consider it being zero for all practical purposes.
As for l_2(t), it becomes practically constant and equal 1 for t>100, and -1 for t<-100.
We then have the function \psi(t)

(3)   \begin{equation*} \psi(t)=\frac{t}{I_2}+2\arctan\left(\delta \tanh(B t/2)\right).  \end{equation*}

Again for all practical purposes for t>100 we have

(4)   \begin{equation*}\psi(t)\approx  t/2+\pi/3,\end{equation*}

and for t<-100,

(5)   \begin{equation*}\psi(t)\approx  t/2-\pi/3.\end{equation*}

Let us substitute these formulas into the solution for q(t)

(6)   \begin{eqnarray*} q_0(t)&=&\frac{\sqrt{1+l_1(t)}\cos\frac{\psi(t)}{2}}{\sqrt{2}},\\ q_1(t)&=&\frac{\sqrt{1+l_1(t)}\sin\frac{\psi(t)}{2}}{\sqrt{2}},\\ q_2(t)&=&\frac{l_3(t)\cos\frac{\psi(t)}{2}+l_2(t)\sin\frac{\psi(t)}{2}}{\sqrt{2(1+l_1(t))}},\\ q_3(t)&=&\frac{-l_2(t)\cos\frac{\psi(t)}{2}+l_3(t)\sin\frac{\psi(t)}{2}}{\sqrt{2(1+l_1(t))}}. \end{eqnarray*}

We obtain

(7)   \begin{eqnarray*} q_0(t)&=&\frac{\cos\frac{\psi(t)}{2}}{\sqrt{2}},\\ q_1(t)&=&\frac{\sin\frac{\psi(t)}{2}}{\sqrt{2}},\\ q_2(t)&=&\pm\frac{\sin\frac{\psi(t)}{2}}{\sqrt{2}},\\ q_3(t)&=&\frac{\mp\cos\frac{\psi(t)}{2}}{\sqrt{2}}. \end{eqnarray*}

The upper signs concern t>100 the lower signs concern t<-100.
In stereographic projection we are plotting (q_1,q_2,q_3) divided by 1-q_0. We see that the “future circle” is in the plane x=y, while the past circle in the orthogonal plane x=-y.

I did my geometric exercises and I have found that “the future circle” has origin at (0,0,-1), while “the past circle” has the origin at (0,0,1). Both circles have radius \sqrt{2}.

So, we have complete information about these “circles of eternal return”.
One more comment: in Another geodesic line I was showing what I thought was a subtle structure of the trajectory approaching the circle:

But now I know that it was a numerical artefact. As noticed in today’s post, for |t|>100 computer will see only one line, no fine structure. To be sure I asked my computer to take more points on the plot (100000 instead of 10000) and all the fine structure disappeared.

New wine in new bottles

In my recent post I was presenting new formulas for the attitude matrix, different from those of a month ago. And I did not explain in detail how I got them, and why they look like being in disagreement with the previous ones. This caused a legitimate concern from Bjab, when he noticed the apparent disagreement. And indeed putting new wine into old bottles would lead to a trouble.

Mar 2:21 No man also seweth a piece of new cloth on an old garment: else the new piece that filled it up taketh away from the old, and the rent is made worse.
Mar 2:22 And no man putteth new wine into old bottles: else the new wine doth burst the bottles, and the wine is spilled, and the bottles will be marred: but new wine must be put into new bottles.


What we did is this: we poured new wine into new bottles. And here is how it is done.

We consider free rigid body with normalized (unit length) angular momentum vector, and the doubled kinetic energy equal to d=1/I_2. Euler’s equations can be then solved explicitly and we choose the following solution:

(1)   \begin{eqnarray*} A_1&=&\sqrt{\frac{I_1(I_3-I_2)}{I_2(I_3-I_1)}}=1/2,\\ A_2&=&1,\\ A_3&=&\sqrt{\frac{I_3(I_2-I_1)}{I_2(I_3-I_1)}}=\frac{\sqrt{3}}{2},\\ B&=&\frac{1}{I_2}\,\sqrt{\frac{(I_2-I_1)(I_3-I_2)}{I_1I_3}}=\frac{1}{2\sqrt{3}},\\ \end{eqnarray*}

(2)   \begin{eqnarray*} L_1(t)&=& A_1\,\mathrm{sech }(B t),\\ L_2(t)&=&A_2\,\tanh( B t),\\ L_3(t)&=&A_3\,\mathrm{sech }(B t), \end{eqnarray*}

where \mathrm{sech }(x)=1/\cosh(x).
We will need the angular velocity vector with components

(3)   \begin{equation*}\Omega_i(t)=\frac{L_i(t)}{I_i}.\end{equation*}

We want to solve now the differential equation for the attitude matrix Q(t):

(4)   \begin{equation*}\frac{dQ(t)}{dt}=Q(t)W(t),\end{equation*}

where

(5)   \begin{equation*}W(t)=\begin{bmatrix}0&-\Omega_3(t)&\Omega_2(t)\\\Omega_3(t)&0&-\Omega_1(t)\\-\Omega_2(t)&\Omega_1(t)&0\end{bmatrix}.\end{equation*}

The strategy is this: We seek the solution of the form Q(t)=Q_2(t)Q_1(t), where Q_1(t) is a certain matrix cleverly constructed out of L_i(t), and Q_2(t) is a simple rotation matrix by some angle \psi(t).
There are two methods: old one, and new one.

The old method.

In the old method, as used for instance in Taming the T-handle continued (we use now Q_1,Q_2 instead of Q_0,Q_1 there), we set:

(6)   \begin{equation*} L_p(t)=\sqrt{L_1(t)^2+L_2(t)^2},\end{equation*}

(7)   \begin{equation*}Q_1(t)=\begin{bmatrix}\frac{L_1(t)L_3(t)}{L_p(t)}&\frac{L_2(t)L_3(t)}{L_p(t)}&-L_p(t)\\-\frac{L_2(t)}{L_p(t)}&\frac{L_1(t)}{L_p(t)}&0\\L_1(t)&L_2(t)&L_3(t)\end{bmatrix},\end{equation*}

(8)   \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*}

Then we solve for \psi(t):

(9)   \begin{equation*}\psi(t)=\frac{t}{I_2}+\arctan\left( \sqrt{\frac{(I_2-I_1)I_3}{I_1(I_3-I_2)}}\tanh\sqrt{\frac{(I_2-I_1)(I_3-I_2)}{I_1I_2^2I_3}}t  \right).\end{equation*}

The new method.

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

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

(12)   \begin{equation*} \psi(t)=\frac{t}{I_2}+2\arctan\left(\delta \tanh(B t/2)\right), \end{equation*}

where

(13)   \begin{equation*}\delta&=&\frac{\sqrt{I_2(I_3-I_1)}-\sqrt{I_1(I_3-I_2)}}{\sqrt{I_3(I_2-I_1)}}.\end{equation*}

Comparing old and new wines

In both cases we get a solution of the attitude differential equation (4). But these are different solutions. If Q(t) is a solution, and if R is a fixed rotation matrix of determinant one, then RQ(t) is another solution. This is the same body rotating, but observed from two different laboratory frames.
Let Q_{old}(t) and Q_{new}(t) be the two solutions, one obtained with the old, the other one with the new method.
With some little algebra it can be verified that Q_{new}(0)=RQ_{old}(0), where

(14)   \begin{equation*}R=\begin{bmatrix}0&0&1\\0&1&0\\-1&0&0\end{bmatrix}.\end{equation*}

.
Therefore for all t

(15)   \begin{equation*}Q_{new}(t)=R Q_{old}(t).\end{equation*}

I verified it numerically. The above comparison was for rotation matrices. For quaternions it is similar, though the quaternion case deserves a comment. I will give it in another note.

Update: Mathematica notebook verifying Eq. (15) numerically is available here.