Being a 4D butterfly – how does it feel?

It was yesterday that I started writing about the idea of more than one path. I imagine being a rigid body in zero gravity. I mean not like a dead human body floating in cosmic space, rather like the T-handle experiencing Dzhanibekov effect. I am floating in space.


[latexpage]
How does it feel being a T-handle? My space is the three-dimensional rotation group. Perhaps the group of unit quaternions, the sphere $S^3$. My perception, perhaps, projects it stereographically on 3-dimensional space, so it is much like for me, the human being thinking about being the T-handle. But, according to the analysis from yesterday’s post More than one path, my sky has two sets of privileged directions. If I start moving into such a privileged direction, very soon I will be close to performing eternal returns moving along essentially always the same circle.

But will it be the same circle for each of these special lines? Or they will be different circles? How do they look? These special trajectories? I am trying to figure it all out, and today I am reporting on my progress. It is all preliminary, there may be errors, perhaps my reasoning has mistakes? With time, and with help of my readers, I am sure we will get to the truth. So, here is what I think, and what I did because of my thinking.

Yesterday I was analyzing paths through the point -1. But, after some thinking and experimenting I decided that I do not like this point. We already have a very nice path, described by nice formulas, for instance in Meeting with remarkable circles

The path $q(t)$ in unit quaternions at $t=0$ passes through the point $q(0)=(\sqrt{3}/2,0,1/2,0).$ This point is in the middle between the two circles of eternal return. There is nothing wrong with this point. Therefore I want to consider all special paths that have the property that at $t=0$ they pass through this point. This is my task.
So, I went to work, and now I will show you the result. Which may be right, or may be wrong – I do not know yet. First pictures, then explanation how I got them:

The yellow point is the common starting point $q(0)=(\sqrt{3}/2,0,1/2,0).$ The red path is the same as in the picture before, the path that we already know. The other paths are also special, starting at the same point, but in different special directions. There are only 21 of them – to start with. What we are getting is something similar to a butterfly. It would not, probably, be able to fly in 3D, but it may be able to fly in 4D. If you click on the butterfly image animation will open showing it from different angles, so that you can get a better idea about its space structure.

Perhaps our souls, when flying, look like that?

How did I get it? I decided that other special paths I will get choosing other solutions of the same Euler’s equations. To choose a different solution I shifted time in the solution for $\mathbf{L}(t)$ (I did not play with signs yet). This way I get a different solution of the attitude equation. But this new solution has a different starting point. So I multiply by a constant quaternion from the left to move the starting point to $q(0)$ above.
I did not shift time too much, just $\pm 4$ with spacing $0.4,$ so I got 21 trajectories.
Now I have to check all that and, if necessary, fix what needs to be fixed.
When all is fixed, I will share the code.

11 thoughts on “Being a 4D butterfly – how does it feel?

  1. “Yesterday I was analyzing paths through the point -1”

    The special path you analysed in two previous posts did not go through q = -1, did it?

  2. “according to the analysis from yesterday’s post More than one path, my sky has two sets of privileged directions.”

    I don’t follow. What are the two sets of privileged directions?

    1. [latexpage]
      In “More than one path” we had
      $\Omega_3/\Omega_1=\pm 1/\sqrt{3}.$
      These are two planes. One for plus sign, one for minus.
      But what count is the direction, not the absolute value. It is the direction that determines the trajectory, the absolute value tells us only how fast we move along this trajectory. Thus what is important is the intersection of the plane with the sky sphere – the circle. So, we have two intersecting circles.
      If $q(0)$ is not -1, then these will be two different circles, but idea will be the same.

  3. “So I multiply by a constant quaternion from the left …”

    Does this operation make the trajectory not a true trajectory?

    Did another orientantion of laboratory frame require both – multiplication from left by q and multiplication from right by q* ?

    1. [latexpage]
      This is a natural question to ask.
      But the answer is: only from the left. Look at the equations of motion:
      \[\dot{q}(t)=\frac{1}{2}q(t)\widehat{\Omega(t)}\]
      There you see that multiplication of $q(t)$ from the left by a constant quaternion gives you another solution. But you should not multiply from the right, unless it commutes with all $\widehat{\Omega(t)}.$
      With matrices we were multiplying from the left. With quaternions the same, because the map from quaternions to matrices is the group homomorphism.
      Action of quaternions on vectors – that is another matter.

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