Category: Spin and rotations

The mysterious paths on the three-sphere

Posted on by arkajad

Wherever you are, whatever you do, there is a certain special direction that takes you out of the infinite labyrinth, and leads to the reincarnation cycle getting closer and closer to the ideal path. Ordinary people do not know about it. Warriors do know.

Beyond a certain point there is no return. This point has to be reached.
Franz Kafka

I am discussing geodesic lines on the three-sphere, geodesics of the left-invariant metric determined by asymmetric rigid body. There is a special, very special class of geodesics there. At every point there is a special direction. If you start the geodesic in this special direction – it has, at both ends, in the future and in the past – a limit cycle/circle. Enough esoteric talk. Let’s go to the math. The following math describes one such geodesic line. Other are obtained by left translations.

I_1,I_2,I_3 are moments of inertia of our rigid body, ordered as I_1<I_2<I_3.
We define

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

Define:

(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).
Define

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

Define:

(4)   \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*}

The story is this. The body can rotate uniformly about its middle axis forever. Either left or right. These are two circles in the rotation group that, using its double cover, topologically is S^3. Stereographic projection maps circles into circles. So they become two circles in \mathbf{R}^3. These circles:

But there is another possibility, when this uniform rotation along the middle axis is only asymptotic. It happens in the infinite past and in the infinite future. But only approximately with the real life, and excluding the short metamorphosis period. This is the trajectory described by the formulas above. Here is the trajectory for time t from t=-20000 to t=20000:

And here is this trajectory together with asymptotic circles. Here the red circle is for the bad past, the blue circle, for the good future. Under microscope it reveals rich structure – infinite mystery.

When China Rules the World

Posted on by arkajad

The title of the post can be found on Wikipedia with additional information. But this post is more personal. Indeed China rulez! Few weeks ago I received the following email:

Quantum World-2017

Time: 16th-18th October, 2017

Place: Changsha, Hunan Province, China

Dear Dr. Arkadiusz Jadczyk,

On behalf of the organizing committee of CQW-2017, we sent you a letter a few days ago invited you to join us and give a speech at first Annual Conference of Quantum World (CQW-2017), which will be held on 16th-18th October, in Changsha, Hunan Province, China. It seems you have not received that letter yet, so I am writing again to extend our sincere invitation. As we have learnt your valuable contribution to Asymptotic Formula for Quantum Harmonic Oscillator Tunneling Probabilities…, we believe your inspirational speech and participation will highlight this congress a lot!

Under the theme “From E=MC2 to Quantum Industry”, the first Annual Conference of Quantum World (CQW-2017) aims at 200+ oral presentations in Quantum Physics and Mechanics, Quantum Information Science, Quantum Chemistry, Quantum Optics, Quantum Materials, and Quantum System, Quantum Engineering and Application, which cover hot topics with both theoretical and experimental contributions.

The conference venue Changsha, as the capital of Hu’nan province, is a beautiful, creative, historical and cultural city with comfortable climate, unique scenery and convenient transportation. It will give you a special experience on the colligation and integration of the Huxiang Culture with the modern civilization.

Worth mentioning that partial scientific program with speakers’ profile and excellent speech titles has been updated on website, kindly click here to view and give us your valuable advice.

Look forward to your kind reply with positive response.

Sincerely yours,

CQW-2017 Organizing Committee

Of course I was surprised, because I do not expect anybody but few experts in the whole world care about my paper. Looks to me like a huge conference industry.

China opens world’s ‘highest and longest glass bottomed bridge’

In my blog post Lorentz transformation from an elementary point of view – from blogging to science publishing I wrote about a paper that came as the result of blogging. This is the second paper that I wrote together with prof. Jerzy Szulga, a continuation of our previous paper, A Comment on “On the Rotation Matrix in Minkowski Space-time” by Ozdemir and Erdogdu, http://arxiv.org/abs/1412.5581, Reports on Mathematical Physics, 74(1), 2014, 39-44,
DOI: 10.1016/S0034-4877(14)60056-2. Today I received from my coauthor a message with a copy of another invitation from China:

Dear Dr. ….,

I’m writing to follow-up my last invitation as below, would you please give me a tentative reply? Thank you very much. I apologize for the inconvenience if the letter disturbed you more than once.

It is our great pleasure and privilege to welcome you to join the 8th World Gene Convention-2017, which will take place in Macao, China during November 13-15, 2017. We would like to welcome you to be the chair/speaker in Theme 902: Agriculture, Food and Plant Biotechnology while presenting about A Comment on “On the Rotation Matrix in Minkowski Space-Time” by Ozdemir and Erdogdu…….

If the suggested thematic session is not your current focused core, you may look through the whole sessions and transfer another one that fit your interest (more info about the program is available athttp://www.bitcongress.com/wgc2017/ProgramLayout.asp

Under our SAB members’ contributions and endeavor, BIT’s 7th World Gene Convention-2016 (WGC-2016), successfully held in Shanghai, China during November 13-15, 2016. Totally, there were nearly 300 world-renowned experts, professors, laboratory principals, project leaders and representatives of well-known enterprises attended the WGC-2016. Participants from the international enterprises, academic and research institutions enjoyed the three days scientific program. Depending on the warmly support and good suggestions from all of the participants, we are confident in organizing WGC-2017 which would be better and more successful than WGC-2016.

WGC-2017 features a very strong technical program, mainly focused on: breakthroughs in gene, advances genomics & genetics, new research of DNA and RNA, focus on basic research, the frontier research of life sciences, new biotherapy discovery, emerging areas for medicine applications, robust technology development, and cutting-edge Biotechnology. It aims to provide a platform for all experts from academia, industry and national labs to discuss latest hot researches and achievements. Attendees will hear world-class speakers discussing the challenges and opportunities facing the gene, biotechnology and life sciences field. The business & academic experts who are from home and abroad will give excellent speeches.

In addition to the dynamic scientific program, you will benefit from the wonderful experience in Macao, China. Macao is an international free port. It’s famous for light industry, tourism, and hotel. Macao is also one of the most developed and richest regions in the world, this is a city of amazing and fascinating cultural wealth. The unique blend of European and Oriental cultures existing here creates a pleasurable and laid back atmosphere in a truly beautiful city. We hope you will enjoy your stay in this beautiful city with all its feature, beauty, architecture and hospitality!

We expect your precious comments or suggestions; also your reference to other speakers will be highly appreciated. We look forward to receiving your replies on the following questions:

1. What is the title of your speech?

2. Do you have any suggestions about our program?

For more information, please visit the conference website http://www.bitcongress.com/wgc2017/default.asp

We look forward to see you in Macao in 2017 for this influential event.

Sincerely yours,

Ms. Teresa Xiao

Organizing Committee of WGC-2017

It is only Chinese people can have such a broad and brave imagination and vision of the future. To connect Lorentz and spin groups of matrices with genes and agriculture – it is a real feat.

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.