Category: Stereographic projection

More recollections

Posted on by arkajad

The recollections in the last post were incomplete. So, here is the continuation.
We have derived the property

(1)   \begin{equation*}R\,W(\vec{v})\,R^t = W(R\vec{v}).\end{equation*}

On the other hand we stressed the fact that

given a unit vector \boldsymbol{\omega}:

(2)   \begin{equation*} Q(\boldsymbol{\omega},\theta)=\exp(\theta W(\boldsymbol{\omega}))\end{equation*}

is the rotation matrix that describes the rotation about the axis in the direction \boldsymbol{\omega} by an angle \theta.

From these two properties we can deduce the third one:

If R\in\mathrm{SO}(3) is a rotation, then rotation about the vector R\boldsymbol{\omega} by the angle \theta is given by the matrix

(3)   \begin{equation*} Q(R\boldsymbol{\omega},\theta)=R\, Q(\boldsymbol{\omega},\theta)\,R^t.\end{equation*}

To deduce this last property we use the fact that for any X and any invertible R we have:

    \[ e^{RXR^{-1}}=Re^XR^{-1}.\]

The proof of the above follows by expanding e^X as a power series, and noticing that (RXR^{-1})^n=RX^nR^{-1}.
I will use spin equivalent of Eq. (3) when explaining how exactly I got the two images from the last post that I am showing again below:


The advantage in using Eq. (3) is that it may be easy to calculate the exponential \exp(\theta W(\boldsymbol{\omega})), but not so easy the exponential \exp(\theta W(R\boldsymbol{\omega})).

Our aim is to draw trajectories in the rotation group \mathrm{SO}(3) made by the asymmetric top spinning about its largest moment of inertia axis, which in our setting is the third axis. As the rotation group \mathrm{SO}(3) is not very graphics friendly, we are using the double covering group \mathrm{SU}(2) or, equivalently, the group of unit quaternions. We then use stereographic projection to project trajectories from the 3-dimensional sphere S^3 to 3-dimensional Euclidean space \mathbf{R}^3.

But first things first. Before doing anything more we first have to translate the above properties into the language of \mathrm{SU}(2). Therefore another recollection is necessary at this point. In Pauli, rotations and quaternions we have defined three spin matrices:

(4)   \begin{equation*}s_1=\begin{bmatrix}0&1\\1&0\end{bmatrix},\quad s_2=\begin{bmatrix}0&i\\-i&0\end{bmatrix},\quad s_3=\begin{bmatrix}1&0\\0&-1\end{bmatrix}.\end{equation*}

In Putting a spin on mistakes we have explained that we have the map U\mapsto R(U) that associates rotation matrix R(U)\in\mathrm{SO}(3) to every matrix U\in\mathrm{SU}(2) in such a way that for every vector \vec{v} we have

(5)   \begin{equation*}U\,\vec{v}\cdot\vec{s}\,U^*=(R\vec{v})\cdot\vec{s}.\end{equation*}

It follows immediately from the above formula that U\mapsto R(U) is a group homomorphism, that is we have

(6)   \begin{equation*}R(UU')=R(U)R(U'),\quad R(U^*)=R(U)^t.\end{equation*}

At the end of Putting a spin on mistakes we have arrived at the formula:

(7)   \begin{equation*}\exp(\theta W(\vec{k}))=R\left(\exp(i\frac{\theta}{2}\vec{k}\cdot \vec{s})\right)\end{equation*}

which tells us that the matrix U(\vec{k},\theta) defined as

(8)   \begin{equation*}U(\vec{k},\theta)\stackrel{df}{=}\exp(i\frac{\theta}{2}\vec{k}\cdot \vec{s})\end{equation*}

describes the rotation about the direction of the unit vector \vec{k} by the angle \theta at the \mathrm{SU}(2) level.
The map U\mapsto R(U) is 2:1. We have R(U)=R(-U). Matrices U and -U implement the same rotation. When \theta changes from 0 to 2 \pi the matrix R in Eq. (7) describes the full 2\pi turn, but the matrix U becomes -I rather than I.. We need to make 4 \pi rotation at the \mathbf{R}^3 vector level to have 2\pi rotation at the spin level. That is described by \frac{\theta}{2} in the exponential in Eq. (7).
At the spin level we then obtain the following analogue of Eq. (\ref:rqr}):

(9)   \begin{equation*}V U(\vec{k},\theta)\,V^*=U(R(V)\vec{k},\theta),\quad V\in\mathrm{SU}(2).\end{equation*}

Now we can finally look back at the images of trajectories. In Seeing spin like an artist we noticed that rotation about the z-axis is described by the matrix

(10)   \begin{equation*}U(t)=\begin{bmatrix}\cos t/2+i\sin t/2&0\\0&\cos t/2 -i\sin t/2\end{bmatrix}=\begin{bmatrix}e^{i t/2}&0\\0&e^{-i t/2}\end{bmatrix},\end{equation*}

while rotation by the angle \phi about y-axis is described by the matrix

(11)   \begin{equation*}V(\phi)=\begin{bmatrix} \cos \left(\frac{\phi}{2}\right) & -\sin \left(\frac{\phi}{2}\right) \\ \sin \left(\frac{\phi}{2}\right) & \cos \left(\frac{\phi}{2}\right)\end{bmatrix}.\end{equation*}

U(t) describes uniform rotation of the top about its z-axis, when the z-axis of the top and of the laboratory frame are aligned. As observed in Towards the road less traveled with spin that is not an interested path for drawing. To make the path more interesting we tilt the laboratory frame. To this end we use, for instance, the matrix V(\phi) with \phi=\pi/6

(12)   \begin{equation*}V(\pi/6)=\frac{1}{2\sqrt{2}}\begin{bmatrix}1+\sqrt{3}&1-\sqrt{3}\\ \sqrt{3}-1&\sqrt{3}+1\end{bmatrix}.\end{equation*}

Now the axis of rotation of the top is tilted by 30 degrees with respect to the z-axis of the laboratory frame. We get new path of the evolution in the group \mathrm{SU}(2)

(13)   \begin{equation*} U'(t)=V(\phi/6)U(t)=\frac{1}{2\sqrt{2}}\begin{bmatrix}(\sqrt{3}+1)e^{ît/2}&(1-\sqrt{3})e^{-it/2}\\(\sqrt{3}-1)e^{it/2}&(\sqrt{3}+1)e^{-it/2}.\end{bmatrix}\end{equation*}

From Eq. (1) in Chiromancy in the rotation group we can now calculate real parameters W,X,Y,Z. For a general matrix U\in\mathrm{SU}(2) the formulas are:

(14)   \begin{align*} W&=\frac{1}{2}(U_{11}+U_{22}),\\ X&=\frac{1}{2i}(U_{12}+U_{21}),\\ Y&=\frac{1}{2}(U_{21}-U_{12}),\\ Z&=\frac{1}{2i}(U_{11}-U_{22}). \end{align*}

In our particular case we get:

(15)   \begin{align*} W&=\frac{1+\sqrt{3}}{2\sqrt{2}}\cos t/2,\\ X&=\frac{\sqrt{3}-1}{2\sqrt{2}}\sin t/2,\\ Y&=\frac{\sqrt{3}-1}{2\sqrt{2}}\cos t/2,\\ Z&=\frac{1+\sqrt{3}}{2\sqrt{2}}\sin t/2. \end{align*}

We then get rather awfully looking the parametric formulas for the stereographic projection:

(16)   \begin{align*} x(t)&=-\frac{\sqrt{2} \left(\sqrt{3}-1\right) \sin \left(\frac{t}{2}\right)}{\left(\sqrt{2}+\sqrt{6}\right) \cos \left(\frac{t}{2}\right)-4},\\ y(t)&=-\frac{\sqrt{2} \left(\sqrt{3}-1\right) \sin \left(\frac{t}{2}\right)}{\left(\sqrt{2}+\sqrt{6}\right) \cos \left(\frac{t}{2}\right)-4},\\ z(t)&=-\frac{\sqrt{2} \left(\sqrt{3}-1\right) \sin \left(\frac{t}{2}\right)}{\left(\sqrt{2}+\sqrt{6}\right) \cos \left(\frac{t}{2}\right)-4}. \end{align*}

The result (running t from 0 to r\pi) is not very interesting – just a circle:

It is better than the straight line with a non tilted top, but not a big deal. But the next thing we want to do is to rotate the tilt axis in the (x,y) plane, that is about the z-axis of the laboratory. If we want to rotate by an angle \psi, we should replace V(\phi) by U(\psi)V(\phi)U(\psi)^* – according to Eq. (9). That is we should look at the trajectory:

    \[U'(t,\psi)=U(\psi)V(\phi)U^*(\psi)U(t).\]

Now, U^*(\psi)U(t)=U(t-\psi). Since anyway we are are going to run with t through the whole (0,4\pi) interval, we can as well use U(t) instead of U(t-\psi). Therefore we can draw

    \[U'(t,\psi)=U(\psi)V(\phi)U(t).\]

We can calculate W,X,Y,Z and then x,y,z. Here are the results from my Mathematica code:

(17)   \begin{align*} W&=\cos \left(\frac{\phi }{2}\right) \cos \left(\frac{t+\psi }{2}\right),\\ X&=-\sin \left(\frac{\phi }{2}\right) \sin \left(\frac{\psi -t}{2}\right),\\ Y&=\sin \left(\frac{\phi }{2}\right) \cos \left(\frac{\psi -t}{2}\right),\\ Z&=\sin \left(\frac{\phi }{2}\right) \cos \left(\frac{\psi -t}{2}\right). \end{align*}

(18)   \begin{align*} x(\psi,\phi,t)&=\frac{\sin \left(\frac{\phi }{2}\right) \sin \left(\frac{\psi -t}{2}\right)}{\cos \left(\frac{\phi }{2}\right) \cos \left(\frac{t+\psi }{2}\right)-1},\\ y(\psi,\phi,t)&=\frac{\sin \left(\frac{\phi }{2}\right) \cos \left(\frac{\psi -t}{2}\right)}{1-\cos \left(\frac{\phi }{2}\right) \cos \left(\frac{t+\psi }{2}\right)},\\ z(\psi,\phi,t)&=\frac{\cos \left(\frac{\phi }{2}\right) \sin \left(\frac{t+\psi }{2}\right)}{1-\cos \left(\frac{\phi }{2}\right) \cos \left(\frac{t+\psi }{2}\right)}. \end{align*}

And here is the resulting of plotting 30 such circles, each for \phi=\pi/6, with \psi increasing from 0 t0 180 degrees, every 6 degrees:

It looks like half of a torus. With \phi=\pi/3, we get another, larger torus …
In the next posts we will disturb the motion of the top to see what kind of trajectories we will be getting then.

Seeing spin like an artist

Posted on by arkajad

In the last note, Towards the road less traveled with spin, we were looking at the asymmetric top spinning about its z-axis, and we have analyzed the path taken by the top, using the stereographic projection. The result turned out to be really boring. A straight half-line, not even worse of looking at. But who or what is the guilty party? The object of the painting, the spinning top, or the painter?

On the net you can find 3 Handy Tips For Learning to See Like an Artist. The second of these tips concerns us:

2. Turn Your Subject Upside Down

We tend to have the most difficulty drawing what we see when we are drawing things we are very familiar with, like facial features, body parts etc.

Because we know we are drawing an eye, for example, instead of relying on what we are seeing, our brain takes over and starts saying “this is what an eye should look like”.

But if we turn the reference (and our drawing) upside down, the features appear much less familiar, and we essentially trick our brain into staying quiet so that we can get on with drawing what we see.

Obviously if you’re drawing from life, you can’t turn your model upside down, but even just turning your drawing around in 90 degree steps can help you see areas where your drawing isn’t quite accurate. You can tilt your head 90 degrees too, to get a fresh look at the model.

And that is what we are going to. But why 90 degrees? We will first try 30, the 60 degrees and see what comes out? And we will also turn around, for even better experience.

You will see that the road becomes more interesting, we will experience something similar to the passengers of the Serra Verde Express

Here are the details.

Suppose we tilt our head 30 degrees left. The effect is the same as if the whole universe, including our spinning top, would tilt right. It is rather difficult to tilt the whole universe without disturbing it. But we can do it in our mind. We do not want to touch the top while it spins. If we do, it will change its rotation, probably it would start to nutate. Like on this movie at 1:18

If we want to have our top rotating about its z-axis, but tilted, we should first tilt it, and only then set in motion respecting its orientation. It is safer to tilt our head or, more scientifically, to tilt the laboratory frame.

Suppose we want to tilt the laboratory frame by 30 degrees about its t-axis. In Towards the road less traveled with spin) we have found that rotation about the z-axis is described by the matrix

(1)   \begin{equation*}U(t)=\begin{bmatrix}\cos t/2+i\sin t/2&0\\0&\cos t/2 -i\sin t/2\end{bmatrix}.\end{equation*}

Using the same method we can find that rotation by the angle \phi about y-axis is described by the matrix

(2)   \begin{equation*}V(\phi)=\begin{bmatrix} \cos \left(\frac{\phi}{2}\right) & -\sin \left(\frac{\phi}{2}\right) \\ \sin \left(\frac{\phi}{2}\right) & \cos \left(\frac{\phi}{2}\right)\end{bmatrix}.\end{equation*}

Rotation of the laboratory frame is described by acting from the left. Therefore rotation of the top observed from the tilted reference frame is described by the product V(\phi)U(t). That way we will obtain a trajectory. The axes of the tilted top will never coincide with the axes of the laboratory frame. Therefore the stereographically projected trajectory will never escape to infinity. We can plot it all. But one trajectory would be boring. It is better to tilt the laboratory frame (all the time by 30 degrees) using 50 different tilting axes (all in the xy plane. The following picture is then produced:

If we add trajectories from 60 degree tilts and look from the top, we get something that is not much worse than Serra Verde Express:

We can also try this tip: By simply squinting your eyes as you look at the subject, you can eliminate a lot of the detail, making it much easier to see simple shapes and values.

In the following posts we will be squinting our top.

Towards the road less traveled with spin

Posted on by arkajad

We will start with a simple road. Like that in Nebraska in the last post,

except that we take a nice blue sky with some puffy clouds:

Instead of T-handle (as in Taming the T-handle ) we take the nice, asymmetric, but as much symmetric as possible, spinning top, with principal moments of inertia I_1=1,I_2=2,I_3=3.
The x-axis, with the smallest moment of inertia, is along the two bronze spheres. The y-axis, with the middle moment of inertia, is along blue-red line. The z-axis, whose moment of inertia is the sum of the other two, is vertical.

The vertical z-axis is the natural axis to try to spin the thing. Imagine our top is floating in space, in zero gravity. We take the z-axis between our fingers, and spin the device. If our hand is not shaking too much, our top will nicely spin about the z-axis. This is the most stable axis for spinning.

The corresponding solution of Euler’s equation is \vec{\omega}=(0,0,\omega_3), where \omega_3 is a constant. The solution of the attitude matrix equation is Q(t)=\exp(W(t\vec{\omega})).
The square of angular momentum vector is I_3^2\omega_3^2, the doubled kinetic energy is I_3\omega_3^2, the parameter d that we were using in the previous notes is 1/I_3. The parameter m is just zero, m=0. In what follows for simplicity we will take \omega_3=1.

We will start with describing the spin history in \mathbf{R}^3 using stereographic projection described in Chiromancy in the rotation group.

In fact, we already did it in Dzhanibekov effect – Part 1, and Dzhanibekov effect – Part 2, but that was at the very beginning of this series, and we did not know yet what we were doing! So, now we do it again, in a more “legal” way.

Rotation about an axis \mathbf{k} by an angle \theta can be described by (see Eq. (4) in Putting a spin on mistakes) an \mathrm{SU}(2) matrix:

(1)   \begin{equation*}U(t)=\exp(i\frac{t}{2}\vec{k}\cdot \vec{s})=\cos\frac{t}{2}I+i\sin\frac{t}{2}(\vec{k}\cdot\vec{s}).\end{equation*}

In fact in Putting a spin on mistakes I have made a mistake that I corrected only at this moment, when checking it again: I have forgotten the imaginary i in the formula!

We will be rotating about the z-axis, so we take \mathbf{k}=(0,0,1). Then \vec{k}\cdot\vec{s}=s_3. Therefore

(2)   \begin{equation*}U(t)=\begin{bmatrix}\cos t/2+i\sin t/2&0\\0&\cos t/2 -i\sin t/2\end{bmatrix}.\end{equation*}

Comparing with Eq. (1) in Chiromancy in the rotation group, we get

(3)   \begin{equation*} \begin{align} W(t)&=\cos t/2,\\ X(t)&=0,\\ Y(t)&=0,\\ Z(t)&=\sin t/2. \end{align} \end{equation*}

The stereographic projection (Eq. (4) in Chiromancy in the rotation group ) is

(4)   \begin{equation*} \begin{align} x(t)&=0,\\ y(t)&=0,\\ z(t)&=\frac{\sin t/2}{1-\cos t/2}. \end{align} \end{equation*}

We can use trigonometric formulas to simplify:

(5)   \begin{equation*}z(t)=\cot t/4.\end{equation*}

Here is the plot of \cot t/4 from 0 to 2\pi:

At t=0 we have z=\infty. That is OK, because at t=0 the frame of the body coincides with the laboratory frame. The rotation Q(t) is the identity, its stereographic image is at infinity.
At t=2\pi we get x=y=z=0. That is the point representing the matrix U=-I. It also describes the identity rotation in \mathbf{R}^3. We get a straight path, the positive part of the z-axis. It is like the road in Nebraska at the top.


In the next posts we will learn how to travel more dangerous, less traveled roads.