### Spinning Top – Space Odity

Peggy Annette Whitson (born February 9, 1960) is an American biochemistry researcher, NASA astronaut, and former NASA Chief Astronaut. Her first space mission was in 2002, with an extended stay aboard the International Space Station as a member of Expedition 5. Her second mission launched October 10, 2007, as the first woman commander of the ISS with Expedition 16. She is currently in space on her third long-duration space flight and is the current commander of the International Space Station.

Two weeks ago we could see live how “Kimbrough and Flight Engineer Peggy Whitson of NASA reconnect cables and electrical connections on PMA-3 at its new home on top Harmony”

That is surely fascinating, but what we are particularly interested in the spinning top in zero gravity experiments on the board of ISS in 2013

At 0:42 in this video Peggy tells us that:

“Conservation of angular momentum keeps the top axis pointed in the same direction”

We look at it, and we compare with the other video featuring the Dzhanibekov effect

The same effect I have modeled with Mathematica:

is even better visible here:

There is also conservation of angular momentum there, but the axis of rotation evidently is not being kept all the time in the same direction. Once in while, quasi-periodically, it flips.

What’s going on?

The answer is: it is, as Chris Hadfield sings, a Space Odity

Can you hear me, Major Tom?
Can you “Here am I floating ’round my tin can
Far above the moon
Planet Earth is blue
And there’s nothing I can do

Indeed, there are laws of physics and sometimes they are odd. Major Tom can do nothing about it except of just watching:

But no, not exactly. We can do something about it. We can try to figure it out, why things happen the way they happen. Mathematics will help us when physical intuition does not suffice.

And that is our plan for the future. As P.A.M. Dirac wrote it on the blackboard during his lecture in Moscow in 1956:

Physical law should have mathematical beauty and we are watching this beauty while playing with geodesics of left-invariant metrics on Lie groups.

We will be looking into the spinning top – but relativistic one. These relativistic tops fly somewhere in space, but they are not yet mass-produced in factories in China. But soon….

It is all about “attitude”. Mathematically the attitude matrix satisfies nonlinear differential equations, and they have their odities. And, as Chris Hadfield explains it in his book “An Astronaut’s Guide to Live on Earth”:

In space flight, “attitude” refers to orientation: which direction your vehicle is pointing relative to the Sun, Earth and other spacecraft. If you lose control of your attitude, two things happen: the vehicle starts to tumble and spin, disorienting everyone on board, and it also strays from its course, which, if you’re short on time or fuel, could mean the difference between life and death. In the Soyuz, for example, we use every cue from every available source—periscope, multiple sensors, the horizon—to monitor our attitude constantly and adjust if necessary. We never want to lose attitude, since maintaining attitude is fundamental to success.
In my experience, something similar is true on Earth. Ultimately, I don’t determine whether I arrive at the desired professional destination. Too many variables are out of my control. There’s really just one thing I can control: my attitude during the journey, which is what keeps me feeling steady and stable, and what keeps me headed in the right direction. So I consciously monitor and correct, if necessary, because losing attitude would be far worse than not achieving my goal.

### SL2R as anti de Sitter space cont.

We continue Becoming anti de Sitter.
[latexpage]
Every matrix $\Xi$ in the Lie algebra o(2,2) generates one-parameter group $e^{\Xi t}$ of linear transformations of $\mathbf{R}^4.$ Vectors tangent to orbits of this group form a vector field. Let us find the formula for the vector field generated by $\Xi.$ The orbit through $y\in \mathbf{R}^4$ is
$$y(t)=e^{\Xi t}y.$$ Differentiating at $t=0$ we find the vector field $\Xi(y)$
$$\Xi(y)=\Xi y.$$
If $\Xi$ is a matrix with components $\Xi^{\mu}_{\phantom{\mu}\nu},$ then $\Xi(y)$ has components
$$\Xi^{\mu}(y)=\Xi^{\mu}_{\phantom{\mu}\nu}y^{\nu}.$$
Vectors tangent to coordinate lines are often denoted as $\partial_\mu$. Therefore we can write the last formula as:
$$\Xi(y)=\Xi^{\mu}_{\phantom{\mu}\nu}y^{\nu}\partial_\mu.$$
In the last post Becoming anti de Sitter we have constructed six generators $\Xi_{(\mu\nu))}.$ Their vector fields now become
$$\Xi_{(1,2)}=y^2\partial_1-y^1\partial_2,\Xi_{(1,3)}=y^3\partial_1+y^1\partial_3,\Xi_{(1,4)}=y^4\partial_1+y^1\partial_4,$$
$$\Xi_{(2,3)}=y^3\partial_2+y^2\partial_3,\Xi_{(2,4)}=y^4\partial_2+y^2\partial_4,\Xi_{(3,4)}=-y^4\partial_3+y^3\partial_4.$$
Bengtsson and Sandin in their paper “Anti de Sitter space, squashed and stretched” discussed in the previous note use coordinates $y^1=X,y^2=Y,y^3=U,y^4=V$. Our vector field $\Xi_{(1,2)}$ is the same as their $J_{XY}$, our $\Xi_{(1,3)}$ is the same as their $J_{XU}$ etc.

In SL(2,R) Killing vector fields in coordinates we introduced six Killing vector fields acting on the group manifold SL(2,R). How they relate to the above six generators of the group O(2,2)?

Vectors from the fields $\xi_{iL},\xi_{iR}$ are tangent to SL(2,R). We have expressed them in coordinates of the group SL(2,R) $x^1=\theta,x^2=r,x^3=u.$ The manifold of SL(2,R) is a hipersurface of dimension 3 in $\mathbf{R}^4$ endowed with coordinates $y^1,y^2,y^3,y^4$. What is the relation between components of the same vector in different coordinate systems? The formula is easy to derive and is very simple. If $\xi^{i}, (i=1,2,3)$ are coordinates of the vector in SL(2,R) and $\xi^{\mu},\, (\mu=1,2,3,4)$ are coordinates of the same vector in $\mathbf{R}^4,$ then

$$\xi^\mu=\frac{\partial y^\mu}{\partial x^{i}}\xi^{i}.$$

How $y^\mu$ depend on $x^{i}$? That is simple. In SL(2,R) vector fields in coordinates we have represented each matrix $A$ from SL(2,R) as

A=\begin{bmatrix}
r \cos (\theta )+\frac{u \sin (\theta )}{r} & \frac{\cos (\theta ) u}{r}-r \sin (\theta ) \\
\frac{\sin (\theta )}{r} & \frac{\cos (\theta )}{r}\end{bmatrix}.

On the other hand, Becoming Anti-de Sitter, we represented it as
$$A=\begin{bmatrix} V+X & Y+U \\ Y-U & V-X \end{bmatrix}.$$

Therefore coordinates $y^\mu$ are easily expressed in terms of $x^{i}$. It remains to do the calculations. I have used computer algebra software to make these calculations for me. My Mathematica notebook doing all calculations can be downloaded from here. The result of all these calculations is the expression of vector fields $\xi_{iL},\xi_{iR}$ in terms of the generators of O(2,2) used in the paper on anti de Sitter spaces. Here is what I have obtained:

\begin{eqnarray}
\xi_{1R}&=&-J_1=J_{XU}+J_{YV},\\
\xi_{2R}&=&J_2=J_{YU}-J_{XV},\\
\xi_{3R}&=&J_0=-J_{XY}-J_{UV},\\
\xi_{1L}&=&\tilde{J}_1=J_{YV}-J_{XU},\\
\xi_{2L}&=&\tilde{J}_2=-J_{XV}-J_{YU},\\
\xi_{3L}&=&\tilde{J}_0=J_{XY}-J_{UV}.
\end{eqnarray}

Bengtsson and Sandin introduce then their own parametrization of SL(2,R) and study the invariant metric on the group. We will find the connection between ours and their approaches in the next posts. We came to our problems starting from T-handles spinning freely in zero gravity. They are studying spinning black holes. It is interesting to see and to research similarities.

### Geodesics on upper half-plane factory direct

This is a continuation of Einstein the Stubborn.
We have calculated the Christoffel symbols of the Levi-Civita connection of the SL(2,R) invariant metric on the upper half-plane.

We have used the standard formula, the same that physicists and astronomers are using in their calculations of Black Holes, White Holes, Big Bangs and Small Bangs:
[latexpage]
$$\Gamma^{i}_{kl}=\frac{1}{2}g^{im}\left(\frac{\partial g_{mk}}{\partial x^{l}}+\frac{\partial g_{ml}}{\partial x^{k}}-\frac{\partial g_{kl}}{\partial x^{m}}\right).$$

With the metric of upper half-plane hyperbolic geometry given by

$$g=\frac{1}{y^2}\begin{bmatrix}1&0\\0&1\end{bmatrix},$$

which is a very simple kind of metric, only four of the six Christoffel symbols are non-zero. They are:
\begin{eqnarray}
\Gamma^1_{21}&=&\Gamma^1_{12}=-\frac{1}{y},\\
\Gamma^2_{11}&=&\frac{1}{y},\\
\Gamma^2_{22}&=&-\frac{1}{y}.
\end{eqnarray}

In the previous post I have originally written “only three of the six Christoffel symbols are non-zero”, but I have forgotten about the symmetry, as in the first line above.

The Christoffel symbols, that is “the coefficients of the torsion free metric affine connection” serve as the tools for defining “parallel transport” of geometric objects along curves. The transport is, in general, path dependent, when there is a non-vanishing “curvature”.

Curvature is expressed in terms of Christoffel symbols and their derivatives. But “geodesics” are expressed directly in terms of the Christoffel symbols. Here are their equations:

$$\frac{d^2 x^i}{ds^2}= -\Gamma^{i}_{jk}\frac{dx^j}{ds} \frac{dx^k}{ds}.$$

One has to remember that the Einstein convention is being used, so that in the above formula summation over the dummy indices $j,k$ is implied.

Let us apply this general formula to our case, with $x^1=x,x^2=y.$ Let us calculate the right hand side for $i=1.$ With $i=1$ we have $\Gamma^1_{12}=\Gamma^1_{21}=-\frac{1}{y},$ therefore

\frac{d^2x}{ds^2}=\frac{2\frac{dx}{ds}\frac{dy}{ds}}{y}.\label{eq:d2x}
With $i=2$ we have $\Gamma^2_{11}=1/y$, $\Gamma^2_{22}=-1/y$. Therefore
$$\frac{d^2y}{ds^2}=-\frac{\left(\frac{dx}{ds}\right)^2-\left(\frac{dy}{ds}\right)^2}{y}.\label{eq:d2y}$$
These are the geodesic equations as they come directly from the factory, in their original shape.

In Geodesics on the upper half-plane – Part 2 circles we have derived the formula for geodesics from conservation laws, that is “second-hand”. We have obtained the following formulas:

\frac{\dot{x}}{y^2(t)}=\mathrm{const},\label{eq:gd1}

\frac{\dot{x}(t)x(t)+\dot{y}(t)y(t)}{y^2(t)}=\mathrm{const}.\label{eq:gd2}

They are simpler than Eqs. (\ref{eq:d2x},\ref{eq:d2y}). But they are consequences of (\ref{eq:d2x},\ref{eq:d2y}). Taking derivatives of the left hand sides of Eqs. (\ref{eq:gd1},\ref{eq:gd2}) we can easily check that they are automatically zero if Eqs. (\ref{eq:d2x},\ref{eq:d2y}) are satisfied! By using conservation laws we have simply taken a short way.

Once we got into the main objects of Riemannian differential geometry, in the next post we will calculate the curvature of our metric.