Dzhanibekov effect – Part 2

Posted on by arkajad

“Everything is repeated, in a circle. History is a master because it teaches us that it doesn’t exist. It’s the permutations that matter.” So wrote Umberto Eco in his Focault’s Pendulum. Somehow we, humans, are helping these repetitions to happen. I can see it quite clearly looking at my own adventures. In the previous post we have started our adventure with nuts flipping in space. But very soon I moved to quaternions. Frankly speaking quaternions are not absolutely necessary. We could avoid them and stay with real 3×3 real rotation matrices, known to every engineer. So, why did I call quaternions into teh game? Sure, they are somewhat exotic and cool. That is a good reason. But the truth is that with quaternions I am relating my today’s adventure with Dzhanibekov’s effect to my old adventure with Quantum Fractals. In my monograph on Quantum Fractals
Arkadiusz Jadczyk, Quantum Fractals: From Heisenberg's Uncertainty to Barnsley's Fractality you can find the following images from quantum theory of pure spin 1/2. These images represent what mathematicians often call ‘Villarceau circles’ and ‘Hopf fibration’.

Tori of constant probability

Villarceau circlkes
Hopf fibration

That was in my quantum past. I really liked the Villarceau circles. They seem to have some magic in them:
Villarceau circles, Strassbourg Notre Dame
I also like Penrose, Hopf fibration and Clifford parallels

Clifford parallels

And today, at the end of this post you will see the same structures reappearing – when quaternions are used for rotating the winged nut in space.
Therefore, let us continue from the last post.

In order to visualize the trajectories we do stereographic projection from S^3 with removed one point, q=1, onto \mathbf{R}^3. Denoting (x,y,z) the coordinates in \mathbf{R}^3 we take

(1)   \begin{eqnarray*} x=\frac{X}{1-W},\\ y=\frac{Y}{1-W},\\ z=\frac{X}{1-W}. \end{eqnarray*}

The inverse transformation is given by

(2)   \begin{eqnarray*} W&=&\frac{r^2-1}{r^2+1},\\ X&=&\frac{2x}{r^2+1},\\ Y&=&\frac{2y}{r^2+1},\\ Z&=&\frac{2z}{r^2+1}, \end{eqnarray*}

where r^2=x^2+y^2+z^2.

We now take a point (x,y,z) in \mathbf{R}^3, transform it into a point (W,X,Y,Z) on \mathbf{R}^3, apply the right shift to obtain W(t),X(t),Y(t),Z(t) and project to get x(t),y(t),z(t). The result is

(3)   \begin{eqnarray*} x(t)&=& \frac{2(x \cos(t)+y\sin(t))}{1+r^2+(1-r^2)\cos(t)+2z\sin(t)},\\ y(t)&=& \frac{2(y \cos(t)-x\sin(t))}{1+r^2+(1-r^2)\cos(t)+2z\sin(t)},\\ z(t)&=& \frac{2z \cos(t)-(1-r^2)\sin(t)}{1+r^2+(1-r^2)\cos(t)+2z\sin(t)}. \end{eqnarray*}

First consider the special trajectory through the origin r=0, therefore x=y=z=0. It is given by x(t)=y(t)=0,\,z(t)=\frac{-\sin(t)}{1+\cos(t)}.
Using trigonometric identities we get z(t)=-\tan (t/2). The trajectory is therefore the z-axis. It is the stereographic projection of the curve q(t)=-u_3(t), connecting q=-1 with q=1.

The second special case is when r=1 and z=0. We get the unit circle in the plane z=0

(4)   \begin{eqnarray*} x(t)&=& x cos(t) +y\sin(t),\\ y(t)&=& y \cos(t)-x\sin(t),\\ z(t)&=&0. \end{eqnarray*}

Apart of these two special cases every trajectory intersects the plane z=0 twice. Therefore it is enough to restrict to the trajectories originating at points with z=0. We introduce polar coordinates on the plane z=0. With \rho=\sqrt{x^2+y^2}, we set x=\rho \cos (\alpha),\, y=\rho \sin(\alpha). With z=0 the equations for trajectories become:

(5)   \begin{eqnarray*} x(t)&=& \frac{2\rho\cos(\alpha-t)}{1+\rho^2+(1-\rho^2)\cos(t)},\nonumber\\ y(t)&=& \frac{2\rho\sin(\alpha-t)}{1+\rho^2+(1-\rho^2)\cos(t)},\\ z(t)&=& \frac{-(1-\rho^2)\sin(t)}{1+\rho^2+(1-\rho^2)\cos(t)}\nonumber. \end{eqnarray*}

Right quaternion trajecties
Family of trajectories for \rho=0.2,04,0.6,0.8,1.0 front view.

Right quaternion trajectories
Family of trajectories for \rho=0.2,04,0.6,0.8,1.0 top view.

8 thoughts on “Dzhanibekov effect – Part 2”

  3. Yes I am plotting Eq. (5)
    {\frac {-2\,\rho\,\cos \left( \alpha \right) \cos \left( t \right) -2 \,\rho\,\sin \left( \alpha \right) \sin \left( t \right) }{\cos \left( t \right) {\rho}^{2}-1-{\rho}^{2}-\cos \left( t \right) }}

    {\frac {-2\,\rho\,\sin \left( \alpha \right) \cos \left( t \right) +2 \,\rho\,\cos \left( \alpha \right) \sin \left( t \right) }{\cos \left( t \right) {\rho}^{2}-1-{\rho}^{2}-\cos \left( t \right) }}

    {\frac {-\sin \left( t \right) {\rho}^{2}+\sin \left( t \right) }{\cos \left( t \right) {\rho}^{2}-1-{\rho}^{2}-\cos \left( t \right) }}

    plt2 := plot3d([x(t), y(t), z(t)], t = 0 .. 2*Pi, alpha = 0 .. (1/8)*Pi, style = wireframe, colour = purple, scaling = constrained)

    1. What I did:
      x := 2*rho*cos(alpha-t)/(1+rho^2+(-rho^2+1)*cos(t)); y := 2*rho*sin(alpha-t)/(1+rho^2+(-rho^2+1)*cos(t)); z := -(-rho^2+1)*sin(t)/(1+rho^2+(-rho^2+1)*cos(t));

      x02 := subs(rho = .2, x); y02 := subs(rho = .2, y); z02 := subs(rho = .2, z); x04 := subs(rho = .4, x); y04 := subs(rho = .4, y); z04 := subs(rho = .4, z); x06 := subs(rho = .6, x); y06 := subs(rho = .6, y); z06 := subs(rho = .6, z);

      plot3d({[x02, y02, z02], [x04, y04, z04], [x06, y06, z06]}, t = 0 .. 2*Pi, alpha = 0 .. Pi, grid = [100, 100])

  4. Thank you,
    Nicer, grid option makes the resolution much better I was playing around with numpoints.
    So what does this represent? I see there is a circle that is tilted over depending on the value of rho. t & alpha determine how much of the whole structure is plotted.
    I see the circle is ofset wrt to the orign. This is buried in the details to the equations. I probably need to understand the detail. Could you shed some light on that aspect?
    How did you place the picture directly in the reply?
    Ah… So many levels of details to learn.
    Edit:- New editor is better

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