If you are practising spiritual discipline under the guidance of a Master, it is always advisable to give up your connection with other paths. If you are satisfied with one Master but are still looking for another Master, then you are making a serious mistake.
Be All You Can Be: Don’t Choose One Path, Choose Multiple Paths
We will follow the second option, we will look for other paths, other than the one we already know. Other but of the same quality
This project has two parts. The first part is pure algebra. No pictures. Pictures will come in the second part, when we will already know what to picture.
Consider this scenario: we are looking at all possible trajectories of the dynamics of our free rigid body with in the quaternion group Consider all trajectories with the property that Or, better, with the property that Why is better? Because under stereographic projection is mapped into infinity, while is mapped into the origin of the coordinate system. Easier to draw.
From our point the trajectory can go in any direction. The direction of the trajectory is described by the derivative We know that every trajectory is a solution of the equation
Let us assume that the angular momentum vector is normalized, it has unit length. We are interested in those very-very special trajectories for which They are special, they encode the essence of the flip in Dzhanibekov effect. In what follows, to simplify the notation I will write instead of
The doubled kinetic energy is
The length of the angular momentum vector, assumed to be 1, is
We want our path to be special, that is we want
Eq. (3) gives then
Comparing this with Eq. (4) the term cancels out and we get
Thus in order for the trajectory to be special the ratio must be special:
For our particular rigid body that we often use, with we should have
For our special trajectory that we were plotting in the last post Circles of eternal return
So it is just one particular way of satisfying Eq. (9). There are, however other ways. These other ways need to be researched.
We will do it in the next posts.
Let me recall the particular trajectory that we are discussing. In fact the particular part of this particular trajectory where the flip occurs.
Click on the image to open gif animation. It will take time to load as it is 2.5 MB.
The flip is much like one of these flips that Russian cosmonaut Dzhanibekov observed in zero gravity, and was very much perplexed by the phenomenon.
We are investigating the jungle of math that we have to apply in order to simulate such a behavior of an ideal Platonic rigid body. At present we are analyzing the special conditions when only one flip happens. Apart of a rather short (as compared to eternity) period of time when the flip occurs, the body rotates uniformly about its middle axis. Its trajectory in the rotation group, represented by quaternions of unit norm, follows a circle. One circle before flip, another circle after the flip.
In this note I am taking a close look at these two circles. Can we find their exact coordinate representation in 3D after stereographic projection?
Yes, we can, and we will do it now.
We use the formulas from Meeting with remarkable circles.
For the body of our choice, with moments of inertia and on the trajectory where the constants that enter the solution have values
In the solution of the Euler’s equations we have functions and
Here are the plots of these functions:
For we have smaller than and for smaller than We can consider it being zero for all practical purposes.
As for it becomes practically constant and equal 1 for and for
We then have the function
Again for all practical purposes for we have
Let us substitute these formulas into the solution for
The upper signs concern the lower signs concern
In stereographic projection we are plotting divided by We see that the “future circle” is in the plane while the past circle in the orthogonal plane
I did my geometric exercises and I have found that “the future circle” has origin at while “the past circle” has the origin at Both circles have radius
So, we have complete information about these “circles of eternal return”.
One more comment: in Another geodesic line I was showing what I thought was a subtle structure of the trajectory approaching the circle:
But now I know that it was a numerical artefact. As noticed in today’s post, for computer will see only one line, no fine structure. To be sure I asked my computer to take more points on the plot (100000 instead of 10000) and all the fine structure disappeared.
and so Eq. (1) follows.
While description of rotations in terms of orthogonal matrices in principle suffices in classical mechanics, sometimes it is convenient to use the group of quaternions of unit norm, the group isomorphic to the group used in quantum mechanical description of half-integer spin particles. Quaternions have some advantages in numerical procedures (as for instance in 3D computer games), but they are also convenient for graphical representation of the intrinsic geometry of the rotation group. And this is what interests us, when we plot trajectories representing history of a spinning rigid body.
Quaternions of unit norm, representing rotations, form a 3-dimensional sphere in 4-dimensional Euclidean space, and we are projecting stereographically this sphere onto our familiar three-dimensional space, where we orient ourselves in a usual way known from everyday experience.
The question therefore arises: how the equation describing the time evolution looks like when represented in the quaternion setting?
To derive it we have to return to the fundamental relation between quaternions and rotations of vectors in space.
For every vector with components denote by the pure imaginary quaternion defined as:
Then to unit quaternion that is such that there corresponds rotation matrix such that for all the following identity holds
The second interesting property of the hat map is that for all we have
where is the commutator. The property follows directly form the definitions and from the quaternion multiplication rules. Every pure imaginary quaternion is of the form for some
As shown on plaque by the Royal Canal at Broome Bridge in Dublin,
the essence of the quaternions is contained in the four equations
From these, multiplying from the left and/or from the right by or or we derive
From the definition of the hat map we have
When we expand the parenthesis and use the multiplication rules in Eqs. (9),(10), we find that the terms with squares cancel out, while the terms with products like etc. can be organized as follows
The coefficients in the parentheses on the right are now exactly the components of the cross product
Suppose now that is a trajectory in the space of unit quaternions, while is the corresponding trajectory in the rotation group. Thus we have
for all and all We now differentiate both sides with respect to On the left we use the standard product rule for differentiation, but we pay attention so as to preserve the order, because multiplication of quaternions is non commutative. On the right we enter with the differentiation under the “hat”, because it is a linear operation. We obtain:
Assume now that is a solution of Eq. (1), so that We have
Note: It may be that the final formula can be derived in a shorter way, but I do not know how. It is this last formula that I was using when verifying that the algorithm provided in Meeting with remarkable circles gives indeed a solution of the evolution equation.