Category: Spin and rotations

Mathematics and sex

Posted on by arkajad

Nowadays mathematics is almost everywhere. To know some cool mathematics (or mathematician) is nowadays sexy.

Therefore I am helping you here, on this blog, to have better relations in your lives. Just looking at the equations can already make you feel better. When you look for a little at Euler’s equations, you will sleep better. That is why today I recall some of these pretty formulas from the dynamics of spinning bodies.

Repetitio est mater studiorum – repetition is the mother of study/learning, that is what they say. So let us repeat, recollect, before traveling another road. If we climb high enough, then going down will be easy.

It is not that climbing up is difficult. It is not. We can take stairs. But it takeS time. And that is what we are going to do now: take time, do it easy way. Enjoying happy sliding down afterwards.

The two essential ingredients af the asymmetric top spinning in zero gravity are:

  1. Euler’s equations
  2. Attitude matrix equations

Euler equations, written in terms of the angular momentum vector \mathbf{L}, with components (L_1,L_2,L_3) are

(1)   \begin{eqnarray*} \frac{dL_1(t)}{dt}&=&(\frac{1}{I_3}-\frac{1}{I_2})L_2(t)L_3(t),\\ \frac{dL_2(t)}{dt}&=&(\frac{1}{I_1}-\frac{1}{I_3})L_3(t)L_1(t),\\ \frac{dL_3(t)}{dt}&=&(\frac{1}{I_2}-\frac{1}{I_1})L_1(t)L_2(t). \end{eqnarray*}

I_1,I_2,I_3 are principal moments of inertia. We assume that our spinning body is asymmetric, therefore I_1,I_2,I_3 are all different from each other, and we will order them as I_1<I_2<I_3. Perhaps it is worthwhile to mention that physics of ordinary materials requires that all three numbers are strictly positive, and that I_1+I_2\geq I_3.  I have my pet rigid body that is essentially flat, with I_1=1,I_2=2,I_3=3. It looks as on the picture below

It is kind of simple but it shows all the nontrivial behavior that a rigid body can have.

Remark: As noted by Bjab in the discussion of this post, the sentence above is incorrect. At the other end of the spectrum of rigid body shapes there are one-dimensional objects. These are flying rods with I_1=0,I_2=I_3\neq 0. They deserve a separate study.

But that is not important now.  What is important is that the components (L_1,L_2,L_3) are the components of the angular momentum vector with respect to the noninertial frame that rotates with the body. Its components with respect to the inertial laboratory frame are constant in time. This is “conservation of angular momentum”- one of the most fundamental “laws” of classical mechanics. Why such a law holds, or at least approximately holds, in our Universe is not known. Physicists and philosophers are debating about “the origin of inertia”.  Some say that “because of space”, some say “because of distant parallelism”, some other say “because of distant matter”. We will not worry about these problems now. We have our laboratory inertial frame, we have frame rotating with the body, and we have rotation matrix Q(t) that maps coordinates in the rotating frame to coordinates in the laboratory frame.

Recall from Angular momentum

Let \mathbf{E}_1(t), \mathbf{E}_2(t), \mathbf{E}_3(t) be an orthonormal frame corotating with the body, and aligned with its principal axes, and let \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 be an inertial laboratory frame, both centered at the center of mass of the body. The two frames are related by time-dependent orthogonal matrix A(t)

(2)   \begin{equation*} \mathbf{E}_i=A_{ji}\mathbf{e}_j,\quad X_i= A_{ij}x_j.\end{equation*}

The inverse of A is denoted by Q

(3)   \begin{equation*} Q= A^{-1}=A^t,\end{equation*}

and it is often called the attitude matrix. For a rotating body, if X_i are coordinates of a fixed point in the body, then its coordinates in the laboratory system change in time:

(4)   \begin{equation*} {\bf x}(t)=Q(t){\bf X}.\end{equation*}

Matrix Q(t) satisfies differential equation that is essentially nothing more than the definition of the angular velocity vector:

(5)   \begin{equation*} Q(t)^{-1}\frac{dQ(t)}{dt}=W(\boldsymbol{\Omega}(t)),\end{equation*}

thus

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

where for any vector \vec{v}

(7)   \begin{equation*}W(\vec{v})=\begin{bmatrix}0& -v_3&v_2\\v_3&0&-v_1\\-v_2&v_1&0.\end{bmatrix}.\end{equation*}

We have

(8)   \begin{equation*}L_1=I_1\,\Omega_1,\quad L_2=I_2\,\Omega_2,\quad L_3=I_3\,\Omega_3.\end{equation*}

A couple of comments are due at this point. First of all a careful Reader will notice that in Angular momentum we had

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

while now we have

(10)   \begin{equation*}\frac{dQ(t)}{dt}=Q(t)W(\boldsymbol{\Omega}(t)).\end{equation*}

What is going on?
Several things at once. First we have the map \vec{v}\mapsto W(\vec{v}) that associates a matrix to every vector. In fact it associates an antisymmetric matrix. The association has the property that is easy to verify: for every vector \vec{w} we have

    \[W(\vec{v})\,\vec{w}=\vec{v}\times\vec{w}.\]

Acting with the matrix W(\vec{v}) is the same as taking cross product with vector \vec{v}. That such an association should exist should be not a surprise. Taking cross-product with a fixed vector is a linear operation, and every linear operation on vectors is implemented by a matrix. However many students who learn about cross products,

do not learn about their relation to antisymmetric matrices and bivectors. And many students that are learning about matrices, operations with them, eigenvalues and eigenvectors, do not learn about cross-product of vectors. Perhaps because the cross-product is particular to 3D?
Anyway, we have this association, and this association has a very nice property of “covariance”: for every orthogonal matrix R of determinant one we have:

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

That is a very important and very useful property. It can be equivalently expressed as the “covariance of the cross-product”

(12)   \begin{equation*}(R\vec{w})\times (R\vec{v})=R(\vec{w}\times\vec{v}).\end{equation*}

But equivalent expression does not replace the proof, and to prove it needs a bunch of simple but lengthy algebraic calculations and taking into account the definitions of cross-product and determinant. It can be easily done with any computer algebra software. I skip it now.
The second thing is the difference between \boldsymbol{\omega} that represents the angular momentum vector in the laboratory frame, and \boldsymbol{\Omega} that represents the same vector in the body frame. Our matrix Q, by definition, connects the two representations (see Eq. (4) above, but now we skip the time-dependence):

    \[\boldsymbol{\omega}=Q\,\boldsymbol{\Omega}.\]

Now, going back to Eq. (9), we have (again we skip the time dependence)

    \[W(\boldsymbol{\omega})Q=W(Q\boldsymbol{\Omega})Q=Q\,W(\boldsymbol{\Omega})Q^t\,Q=Q\,W(\boldsymbol{\Omega}),\]

which solves our puzzle.
Remark: Above we have used \boldsymbol{\Omega} and \boldsymbol{\omega} to distinguish between numerical representation of the same vector with respect to two different frames. But when there is no possibility of a confusion, we can denote a vector by any convenient letter whatsoever.
Let us see how the above recollection of facts can be applied. In Towards the road less traveled with spin there was the following statement:

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 equations 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})).

I promised to answer the question why is it so.

In general, when we have matrix differential equation of the type:

    \[\frac{dM(t)}{dt}=M(t)X,\]

where X is a constant (i.e. time independent) matrix, we can verify that M(t)=\exp(tX) is a solution. In fact, it is the unique solution with the initial data M(0)=I. To prove it, one would have to expand e^{tX} into power series, take care about the convergence of the infinite series (no care needed, it is absolutely convergent), and then differentiate term by term. Let us take it for granted that is the case. In the quote from the other post above I used the letters \vec{\omega}=(0,0,\omega_3), instead of “more adequate”, \vec{\Omega}=(0,0,\Omega_3), but, as in the remark above, we are supposed to adjust the meaning to the context. For the matrix X in the attitude equation we have W(\vec{\omega}), so the solution with the property W(0)=I is Q(t)=\exp(tW(\vec{\omega})), which, because \vec{v}\mapsto W(\vec{v}) is a linear map, is the same as Q(t)=\exp(W(t\vec{\omega})).
In fact here we have the opportunity to make another application of today’s recollections. Suppose that our spinning top is completely symmetric – a perfect ball spinning one of its axes. Perfect ball means I_1=I_2=I_3. If so, the right side of the Euler’s equations is automatically zero. Therefore any constant vector \boldsymbol{\Omega} is a solution. Let us choose \boldsymbol{\Omega} a unit vector, so that the absolute value of the angular velocity is 1. That means our top makes a complete rotation about the axis along the vector \boldsymbol{\Omega} every 2\pi units of time. As noticed above the solution of the attitude equation is \exp(tW(\omega})). Therefore \exp(tW(\boldsymbol{\omega})) is the rotation matrix that describes the rotation about the axis in the direction \boldsymbol{\omega} by an angle t. We have already used this fact before, but now it was a good place for recollecting.
After all these recollections we are now ready to look again at the roads that are less traveled.

Of course it is not Albert Einstein who is the author, but, in the age of fake news, who cares?

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.