Spinning Top – Space Odity

I start with quoting from Wikipedia:

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:

Major Tom: And there’s nothing I can do

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

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.

Geodesics of left invariant metrics on matrix Lie groups – Part 2 Conservation laws

In the last post, Geodesics of left invariant metrics on matrix Lie groups – Part 1,we have derived Arnold’s equation – that is a half of the problem of finding geodesics on a Lie group endowed with left-invariant metric.
[latexpage]
Suppose $G$ is a Lie group, and $g(\xi,\eta)$ is a scalar product (i.e. a nondegenerate bilinear form) on its Lie algebra $Lie(G)$. Then, using left translations $g$ defines a left invariant (Riemannian or pseudo-Riemannian) metric on the whole group $G.$ If $t\mapsto a(t)\in G$ is a path in $G,$ we use left translations to define the image $\omega(t)$ of the tangent vector $\dot{a}(t)$ in $Lie(G)$
\begin{equation}\omega(t)=a(t)^{-1}\dot{a}(t).\end{equation}
Usually it is written more carefully, using $L_{a^{-1}}$ instead of $a^{-1}$, but I am using a simplified notation, well adapted to dealing with problems for matrix groups.

If $a(t)$ is a geodesic for the metric $g$, then $\omega(t)$ satisfies the Arnold’s equation
\begin{equation}\dot{\omega}=B(\omega,\omega),\label{eq:ar1}\end{equation}
where
\[B:Lie(G)\times Lie(G)\rightarrow Lie(G)\]
is defined as
\begin{equation}B^k_{ij}=g^{kl}B_{ij,l}=g^{kl}C_{jl,i}=g^{kl}g_{im}C_{jl}^m,\end{equation}
and $C_{jl}^m$ are the structure constants of $G$, that is
\begin{equation}[\xi_i,\xi_j]=C_{ij}^k\, \xi_k\end{equation}
where $\xi_i$ form a basis for $Lie(G)$ and
\begin{equation}B_{ij,l}=g_{lk}B_{ij}^k,\,C_{ij,l}=g_{lk}C_{ij}^k.\end{equation}

Equation (\ref{eq:ar1}) is a system of nonlinear ordinary differential equations with constant coefficients. In this form it can be found in Arnold’s 1966 paper “Sur la g\’eom\’etrie differentielle des groupes de Lie de dimension infinie et ses applications \`a l’hydrodynamique des fluides parfaits”, Ann. Inst.
Fourier (Grenoble) 16 (1966), 319-361.

In his blog post “The Euler-Arnold equation” Terrence Tao mentions that some people call this equation the “Euler-Arnold” equation, while some other prefer to skip Arnold’s name completely and call it “Euler-Poincare equation”. Go-figure!

Solving equations (\ref{eq:ar1}) is just one half of the whole problem of finding a geodesic. Once $\omega^{i}(t)$ are known, we need to solve the linear differential equation with variable coefficients that results from the definition of $\omega$ (\ref{eq:do}):
\begin{equation} \dot{a}(t)=a(t)\omega(t),\end{equation}

For the case of the rotation group and free rigid body we did it in Taming the T-handle continued.

B. Kolev in his paper “Lie groups and mechanics. An introduction” shows that for the rotation group O(n) the equations of motion for the free rigid body are completely integrable. We will not need this result, but is useful to know that in general case there are always two quadratic constants of motion corresponding to “kinetic energy” and “square of the angular momentum”. We were discussing these constants of motion for the group O(3) in
“Asymmetric Spinning Top – The Hardest Concept To Grasp In Physics” – they are used in so-called Poinsot construction. We will discuss a version of it for the group O(2,1) in the future.

The first observation is that the “kinetic energy” $T(t)=g_{ij}\omega(t)^i\omega(t)^j$ is, in fact, a constant, independent of $t.$ To see that this is the case, we differentiate:
\begin{equation}\frac{dT(t)}{dt}=g_{kl}\dot{\omega}^{k}\omega^l+g_{kl}\omega^{k}\dot{\omega^l}=2g_{kl}\omega^k\dot{\omega}^l.\label{eq:dt}\end{equation}

In the previous post we wrote Eq. (\ref{eq:ar1}) as

\begin{equation}g_{kl}\,\dot{\omega}^l=C_{jk,i}\,\omega^{i}\omega^{j}.\label{eq:mk}\end{equation}

Substituting into Eq. (\ref{eq:dt}) we get

\begin{equation}\frac{dT(t)}{dt}=2C_{jk,i}\omega^{i}\omega^j \omega^{k}=0.\end{equation}

The result is zero because $C_{jk,i}$ is antisymmetric in $j,k$ while $\omega^j\omega^k$ is symmetric. That is an often used property: if $A^{ij}=-A^{ji}$ and $B_{ij}=B_{ji},$ then the contraction $A^{ij}B_{ij}=0.$ Indeed $A^{ij}B_{ij}=-A^{ji}B_{ij}=-A^{ji}B_{ji}=-A^{ij}B_{ij},$ where in the last equality we have exchanged dummy indices names $i\leftrightarrow j$. If a number is equal to its negative, it must be zero.

To get the formula for the second quadratic invariant we need to return to the Ad-invariant scalar product that we have denoted $\mathring{g}$ in Killing vectors, geodesics, and “Noether’s theorem”:
\begin{equation}\mathring{g}(\eta_1,\eta_2)=\mbox{const}\, \frac{1}{2}Re(\mbox{Tr}(\eta_1\eta_2))).\end{equation}
The fact that $\mathring{g}$ is Ad-invariant implies an important relation between the matrix $\mathring{g}_{ij}$ and the structure constants $C_{ij}^k$
that we are going to use. Ad-invariance means that:
\begin{equation}\mathring{g}(e^{t\eta}\xi_1 e^{-t\eta},e^{t\eta}\xi_2 e^{-t\eta})=\mathring{g}(\xi_1,\xi_2)\end{equation}
for all $t\in\mathbf{R}$ and $\xi_1,\xi_2,\eta\in Lie(G).$
Differentiating at $t=0$ we get
\begin{equation}\mathring{g}([\eta,\xi_1],\xi_2)+\mathring{g}(\xi_1,[\eta,\xi_2])=0.\end{equation}
Setting $\xi_1=\xi_i,\xi_2=\xi_j,\eta=\xi_k$ we get
\begin{equation}C_{ki}^l\mathring{g}_{lj}+C_{kj}^l\mathring{g}_{li}=0.\end{equation}
Multiplying both sides by $g^{im}g^{jn}$ we obtain
\begin{equation} C_{ki}^n\,g^{im}+C_{kj}^m\,g^{jn}=0\label{eq:adj}\end{equation}
We can now derive the second conservation law. The angular momentum $m_i$ is defined as
\begin{equation}m_i=g_{ij}\omega^{j}.\label{eq:m}\end{equation}
Notice that $m$ is a covector, a one-form on $Lie(G)$, it is in the dual $Lie(G)^*$ of $Lie(G).$ It is the metric that connects the space to its dual. While vectors in $Lie(G)$ play an active role, they generate transformations, elements in the dual, one-forms from $Lie(G)^*$, are “passive”, they evaluate vectors to numbers. It is the metric that is the third element here, that allows the active principle to connect to the passive principle. The metric depends on the mass distribution. In application to rigid bodies the inertia tensor is encoded in the metric on the rotation group.

The second conservation law states that the square of the angular momentum evaluated with the Ad-invariant metric $\mathring{g}$ is constant:
\begin{equation}
m_0^2(t)=\mathring{g}^{ij}m_i(t)m_j(t)=\mbox{const}.\end{equation}
To verify we differentiate and use Eq. (\ref{eq:mk}) rewritten as
\begin{equation}\dot{m}_k=C_{jk}^{i}\,m_{i}\omega^{j}.\label{eq:mk1}\end{equation}
\begin{equation}\frac{dm_0^2(t)}{dt}=2\mathring{g}^{kl}\dot{m}_k m_l=2\mathring{g}^{kl}C_{jk}^{i}m_im_l\omega^j.\end{equation}
Now, according to Eq. (\ref{eq:adj}) $\mathring{g}^{kl}C_{jk}^{i}$ is antisymmetric in $(i,l)$, while the product $m_im_l$ is symmetric, therefore we get zero:
\begin{equation}\frac{dm_0^2(t)}{dt}=0.\end{equation}

In the following posts we will first return to the case of the rotation group in three dimensions and the rigid body, and then try to apply a similar reasoning to the case of the Lorentz group O(2,1) in 2+1 dimensions.

Geodesics of left invariant metrics on matrix Lie groups – Part 1

An elegant derivation of geodesic equations for left invariant metrics has been given by B. Kolev in his paper “Lie groups and mechanics. An introduction”.
[latexpage]
Here we will derive these equations using simple tools of matrix algebra and differential geometry, so that at the end we will have formulas ready for applications. We will use
the conservation laws derived in the last post Killing vectors, geodesics, and Noether’s theorem. We will also use the same notation. We consider matrix Lie group $G$ with the Lie algebra $Lie(G).$ The tangent space at $a\in G$ is denoted $T_aG$.
Thus $Lie(G)=T_eG.$ On $Lie(G)$ we assume nondegenerate scalar product
denoted as $g(\xi,\eta),\, \xi,\eta\in Lie(G).$ We propagate it to the whole group using left translations as in Eqs. (8,9) of Killing vectors, geodesics, and Noether’s theorem
\begin{equation}g_a(\xi,\eta)=g_e(a^{-1}\xi,a^{-1}\eta),\end{equation}
which implies for $\xi,\eta\in T_bG$
\begin{equation}g_{ab}(a\xi,a\eta)=g_b(\xi,\eta),\,a,b\in G.\end{equation}
The metric so constructed is automatically left-invariant, therefore for each $\xi\in Lie(G)$ the vector field $\xi(a)=\xi a$ is a Killing field.

Let $a(t)$ be a geodesic for this metric. We denote by $\omega(t)\in Lie(G)$ the tangent vector left translated to the identity:
\begin{equation}\omega(t)=a(t)^{-1}\dot{a}(t).\end{equation}

Then, from the conservation laws derived in the last post, we know that the scalar product of $\dot{a}(t)$ with $\xi a(t)$ is constant. That is
\begin{equation}g_{a(t)}(\xi a(t),\dot{a}(t))=\mbox{const}.\end{equation}
The metric is left-invariant, therefore $g_e(a(t)^{-1}\xi a(t),a(t)^{-1}\dot{a}(t))=\mbox{const}$, or
\begin{equation}g_e(a(t)^{-1}\xi a(t),\omega(t))=\mbox{const}.\label{eq:tod}\end{equation}
We will differentiate the last equation with respect to $t,$ but first let us notice that by differentiating the identity $a(t) a(t)^{-1}=e$ we obtain
\begin{equation}\frac{d}{dt}a(t)^{-1}=-a(t)^{-1}\dot{a}(t)a(t)^{-1}=-\omega(t)a(t)^{-1}.\end{equation}
Now, differentiating Eq. (\ref{eq:tod}), and using also $\frac{da(t)}{dt}=a(t)\omega(t)$ we obtain
\begin{equation}g_e([a^{-1}\xi a,\omega],\omega)+g_e(a^{-1}\xi a,\dot{\omega})=0.\label{eq:a1}\end{equation}
We now need a certain bilinear operator on Lie(G) that is defined using the commutator and the scalar product. The commutator $[\xi_1,\xi_2]$ itself is such an operator
from $Lie(G)\times Lie(G)\rightarrow Lie(G).$ But using the scalar product we can define another operator $B(\xi_1,\xi_2)$ by the formula:
\begin{equation}g_e(B(\xi_1,\xi_2),\eta)=g_e([\xi_2,\eta],\xi_1),\quad \xi_1,\xi_2,\eta\in Lie(G). \label{eq:b1} \end{equation}
The right hand side is linear in $\eta,$ and owing to the nondegeneracy of the scalar product every linear functional is represented by a scalar product with a unique vector. Therefore $B(\xi_1,\xi_2)$ is well defined, and evidently is linear in both arguments.

Let $\xi_i$ be a basis in $Lie(G),$ so that the structure constants are $C_{ij}^k$
\begin{equation}[\xi_i,\xi_j]=C_{ij}^k\,\xi_k.\end{equation}
We can also write $B$ as
\begin{equation}B(\xi_i,\xi_j)=B_{ij}^k\,\xi_k.\end{equation}
Then Eq. (\ref{eq:b1}) gives
\begin{equation}
g_e(B(\xi_i,\xi_j),\xi_k)=g_e([\xi_j,\xi_k],\xi_i)\label{eq:b2}\end{equation}
or
\[ B_{ij}^l g_{lk}=C_{jk}^lg_{li},\]
which can be solved for $B$ using the inverse metric:
\begin{equation}B_{ij}^m=g^{mk}C_{jk}^lg_{li}.\end{equation}
On the other hand, if we agree to lower the upper index of $B$ and $C$ with the metric, we can write Eq. (\ref{eq:b2}) as
\begin{equation}B_{ij,k}=C_{jk,i},\label{eq:b3}\end{equation}
which is easy to remember.

We can now return to Eq. (\ref{eq:a1}) and rewrite it as
\[g_e(a^{-1}\xi a,\dot{\omega})=g_e([\omega,a^{-1}\xi a],\omega)=g_e(a^{-1}\xi a,B(\omega,\omega)).\]
Since $\xi$, and therefore also $a^{-1}\xi a$ is arbitrary, we obtain
\begin{equation}\dot{\omega}=B(\omega,\omega),\end{equation}
or, using a basis and Eq. (13)
\begin{equation}g_{kl}\,\dot{\omega}^l=C_{jk,i}\,\omega^{i}\omega^{j}.\end{equation}