# Rotation matrix with deficit angle

I need to find the rotation matrix for a space with a deficit angle. The question is as pictured

The following is my answer to the question:

If $\theta$ could vary between $0$ and $2 \pi$, $$R(\theta) = \begin{pmatrix} \cos(\theta) && \sin(\theta) \\ -\sin(\theta) && \cos(\theta) \end{pmatrix}$$ In this space, instead of rotating $2 \pi$ to get to the same point, we rotate $2 \pi - \phi$. So a rotation of $2 \pi$ (full circle) in this funny space is equivalent to a rotation of $2 \pi - \phi$ in ordinary space. So a rotation of $\theta$ in the ordinary space is equivalent to a rotation of $\frac{\theta}{1 - \frac{\phi}{2 \pi}}$ in the funny space. Thus, with the new metric, we let $\theta \rightarrow \frac{\theta}{1-\frac{\phi}{2 \pi}}$ and we have $$R(\theta) = \begin{pmatrix} \cos\Big(\frac{\theta}{1-\frac{\phi}{2 \pi}}\Big) && \sin\Big(\frac{\theta}{1-\frac{\phi}{2 \pi}}\Big) \\ -\sin\Big(\frac{\theta}{1-\frac{\phi}{2 \pi}}\Big) && \cos\Big(\frac{\theta}{1-\frac{\phi}{2 \pi}}\Big) \end{pmatrix}$$ $$\therefore R(0) = \begin{pmatrix} 1 && 0 \\ 0 && 1 \end{pmatrix}$$ and $$R(2 \pi - \phi ) = \begin{pmatrix} 1 && 0 \\ 0 && 1 \end{pmatrix}$$ This satisfies the requirement that $R(0) = R(2 \pi - \phi) = I_{2}$. Is this the correct rotation matrix and are my steps logical? Thank you.

• I think the result is correct, but you need to elaborate more on the "let $\theta \to \frac{\theta}{1-\frac{\phi}{2 \pi}}$" part. What is the definition of rotation? – Trebor Aug 20 '18 at 13:07
• @Trebor Thank you. I've added a few lines on why I let $\theta$ be $\frac{\theta}{1-\frac{\phi}{2 \pi}}$. – saad Aug 20 '18 at 14:31
• You need to show that the relationship is linear though... – Trebor Aug 20 '18 at 15:30

I think that your metric is not correct. why ?

your new polar coordinates are:

$x=r\cos \left( {\frac {2\pi \,\theta}{2\,\pi -\phi}} \right)$

$y=r\sin \left( {\frac {2 \pi \,\theta}{2\,\pi -\phi}} \right)$

The Jacobi matrix is:

$J=\left[ \begin {array}{cc} \cos \left( 2\,{\frac {\pi \,\theta}{2\, \pi -\phi}} \right) &-2\,r\sin \left( 2\,{\frac {\pi \,\theta}{2\,\pi -\phi}} \right) \pi \left( 2\,\pi -\phi \right) ^{-1} \\ \sin \left( 2\,{\frac {\pi \,\theta}{2\,\pi -\phi }} \right) &2\,r\cos \left( 2\,{\frac {\pi \,\theta}{2\,\pi -\phi}} \right) \pi \left( 2\,\pi -\phi \right) ^{-1}\end {array} \right]$

and the metric :

$g=J^T\,J=\left[ \begin {array}{cc} 1&0\\ 0&\,{\frac {4{\pi }^{2}}{ \left( 2\,\pi -\phi \right) ^{2}}r^2}\end {array} \right]$

If you know the equations for $x$ and $y$ you can calculate the transformation matrix $R$ with this equation:

$J=R\,H$ , with the matrix $H_{i,i}=\sqrt{g_{i,i}}\,,H_{i,j}=0$

$H= \left[ \begin {array}{cc} 1&0\\ 0&2\,{\frac {\pi \, r}{2\,\pi -\phi}}\end {array} \right]$

$R=J\,H^{-1}$

$R=\left[ \begin {array}{cc} \cos \left( 2\,{\frac {\pi \,\theta}{2\, \pi -\phi}} \right) &-\sin \left( 2\,{\frac {\pi \,\theta}{2\,\pi - \phi}} \right) \\ \sin \left( 2\,{\frac {\pi \, \theta}{2\,\pi -\phi}} \right) &\cos \left( 2\,{\frac {\pi \,\theta}{2 \,\pi -\phi}} \right) \end {array} \right]$

This is your transformation matrix.

Remark: I use symbolic Program MAPLE to do the calculation