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There's the following exercise in Hubbard's book:

In the Singapore public garden, there is a statue consisting of a spherical stone ball, with the diameter perhaps 1.3m, weighing at least a ton. This ball is placed in a semispherical stone cup, which it fits almost exactly; moreover, there is a water jet at the bottom of the cup, so the stone is suspended on a film of water, making the friction of the ball with the cup almost 0; it is easy to set it in motion, and it keeps rotating in whatever way you start for a long time.

Suppose now you are given access to this ball only near the top, so that you can push it to make it rotate around any horizontal axis, but you don't have enough of a grip to make it turn around the vertical axis. Can you make it rotate around the vertical axis anyway?

I thought of answering it this way:

Let $\mathbf B$ be a ball in $\Bbb R^3$, we want to prove that a rotation of $\alpha$ around the $x$-$axis$ composed with a rotation of $\beta$ around the $y$-$axis$, or the other way round, is equivalent to a rotation of $\theta$ around the $z$-$axis$. Let $\mathbf v$ be a vector from the origin to a point $P$ on the surface of the ball. The matrices $$[R_{\theta_{z}}]=\begin{bmatrix}\cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 &1\end{bmatrix} [R_{\alpha_{x}}]=\begin{bmatrix} 1 & 0 & 0\\ 0 & \cos\alpha & -\sin\alpha \\ 0 & \sin\alpha & \cos\alpha \\\end{bmatrix} [R_{\beta_{y}}]=\begin{bmatrix}\cos\beta & 0 & -\sin\beta \\ 0 & 1 &0 \\ \sin\beta & 0 & \cos\beta \end{bmatrix}$$ represent the rotations around z, x and y respectfully. Now we want to prove that there exists $\alpha$ and $\beta$ such that

$$ [R_{\alpha_{x}}][R_{\beta_{y}}]=\begin{bmatrix} 1 & 0 & 0\\ 0 & \cos\alpha & -\sin\alpha \\ 0 & \sin\alpha & \cos\alpha \\\end{bmatrix} \begin{bmatrix}cos\beta & 0 & -\sin\beta \\ 0 & 1 &0 \\ \sin\beta & 0 & \cos\beta \end{bmatrix}= \begin{bmatrix} \cos\beta & 0 & -\sin\beta \\ \sin\alpha \sin\beta & \cos\alpha & -\sin\alpha \cos\beta \\ \cos\alpha \sin\beta & \sin\alpha & \cos\alpha \cos\beta \\\end{bmatrix}= [R_{\theta_{z}}]=\begin{bmatrix}\cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 &1\end{bmatrix} $$ or

$$ [R_{\beta_{y}}] [R_{\alpha_{x}}]=\begin{bmatrix}\cos\beta & 0 & -\sin\beta \\ 0 & 1 &0 \\ \sin\beta & 0 & \cos\beta \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & \cos\alpha & -\sin\alpha \\ 0 & \sin\alpha & \cos\alpha \\\end{bmatrix} = \begin{bmatrix} \cos\beta & -\sin\beta \sin\alpha & \sin\beta \cos\alpha \\ 0 & \cos\alpha & \sin\alpha \\ \sin\beta & -\cos\beta \sin\alpha & \cos\beta \cos\alpha \\\end{bmatrix}= [R_{\theta_{z}}]=\begin{bmatrix}\cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 &1\end{bmatrix} $$ Analising the first case we get $ \sin\alpha =0 \\ \ \sin\beta =0 \\ \ \sin\theta= 0$ which implies $ \alpha = n_1 2 \pi + {\pi \over 2} \\ \beta = n_2 2 \pi + {\pi \over 2} \\ \theta = n_3 2 \pi + {\pi \over 2} $, for some integers $n_1,n_2,n_3$. Thus the composition of these maps does not get to all points of the circumference traced by the rotation around the $z$-$axis$. It's similar for the other composition. There's no way to rotate the in the vertical axis for all angles.


Which seems to imply that there's no composition that does it. (The next exercise made me think this) But think of a telescope which can only rotate around it's base and up&down, I can imagine a way to rotate around a 3rd axis.This is a composition of two other rotations. How can it be? What am I missing in the proof I made?

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    $\begingroup$ So... many.. operatornames... $\endgroup$ – ÍgjøgnumMeg Oct 6 '16 at 23:28
  • $\begingroup$ Sorry, I'm not sure how else would I do it. $\endgroup$ – GoetheGrimm Oct 6 '16 at 23:38
  • $\begingroup$ not to worry. Special functions like sine and cosine can be written \sin and \cos to give $\cos, \sin$ $\endgroup$ – ÍgjøgnumMeg Oct 6 '16 at 23:41
  • $\begingroup$ I've edited it. $\endgroup$ – GoetheGrimm Oct 6 '16 at 23:48
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You’re constraining yourself by only considering two orthogonal axes of rotation, but it appears that this is impossible by composing only two rotations, anyway.

The calculations are a bit less tedious if you use quaternions to represent the rotations. Recall that a quaternion can be thought of as a scalar-vector pair $q=(w,\mathbf v)$ and that in this representation the product of two quaternions is $$q_1q_2=(w_1w_2-\mathbf v_1\cdot\mathbf v_2,w_1\mathbf v_2+w_2\mathbf v_1+\mathbf v_1\times\mathbf v_2).$$ If $\mathbf u$ is a unit vector, then the quaternion $q(\mathbf u,\theta)=\left(\cos\frac\theta2,\mathbf u\sin\frac\theta2\right)$ represents rotation through an angle of $\theta$ about the axis given by $\mathbf u$. Composition of rotations corresponds to multiplication of the corresponding quaternions.

We wish to know whether or not there is a pair of unit vectors in the $x$-$y$ plane such that a composition of rotations about them is equal to a rotation about the $z$-axis. Let these two vectors be $\mathbf u=(u_x,u_y,0)$ and $\mathbf v=(v_x,v_y,0)$. Then the vector part of $q(\mathbf u,\alpha)q(\mathbf v,\beta)$ is $$\left(u_x\sin{\frac\alpha2}\cos{\frac\beta2}+v_x\cos{\frac\alpha2}\sin{\frac\beta2},u_y\sin{\frac\alpha2}\cos{\frac\beta2}+v_y\cos{\frac\alpha2}\sin{\frac\beta2},\pm\sin{\frac\alpha2}\sin{\frac\beta2}\right).$$ For this to always represent a rotation about the $z$-axis, the first two components must vanish for all values of $\alpha$ and $\beta$, which can only happen if $u_x=u_y=v_x=v_y=0$.

As regards your example of a telescope, it’s true that two axes of rotation are sufficient to move a single point on a sphere to any other location on that sphere (think spherical coordinates), but that’s not the same as simultaneously rotating all points on the sphere about some other fixed axis.

On the other hand, if you compose more than two rotations about horizontal axes, then it is possible to do this, as demonstrated in Rahul’s answer. Essentially, you convert a rotation about the $y$-axis into one about the $z$-axis by finding a rotation about some horizontal axis that brings the $z$-axis onto the $y$-axis, perform the desired rotation, and then rotate back up. You might recognize this as a change-of-basis operation.

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  • $\begingroup$ I should look into quarternions, rotations do look a lot simpler this way. In fact, I'm doing it right now. Also, @Rahul 's answer and your clarification answers the question. It's change of basis, a rotation, then back to the original frame. $\endgroup$ – GoetheGrimm Oct 7 '16 at 1:44
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You are unnecessarily restricting yourself to two consecutive rotations. Consider:

  1. Rotate about the $x$-axis by $-\pi/2$.

  2. Rotate about the $y$-axis by $\theta$.

  3. Rotate about the $x$-axis by $\pi/2$.

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  • $\begingroup$ Thank you, this made me think. I thought, wrongly, that it would be analogous to $\Bbb R^2$ and only two rotations would suffice. $\endgroup$ – GoetheGrimm Oct 7 '16 at 1:48

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