## Monday, 30 August 2021

### Question

The two-dimensional metric for a flat sheet of paper in polar coordinates is $\left(r,\theta\right)$ is$${ds}^2={dr}^2+r^2{d\phi}^2$$or in modern notation$$\mathbf{g}=\mathbf{d}r\otimes\mathbf{d}r+r^2\mathbf{d}\phi\otimes\mathbf{d}\phi$$Presumably the coordinates are $\left(r,\phi\right)$ not $\left(r,\theta\right)$.
(a) Calculate the connection coefficients using 8.24.
(b) Write down the geodesic equation in $\left(r,\phi\right)$ coordinates.
(c) Solve these equations for $r\left(\lambda\right)$ and $\phi\left(\lambda\right)$ and show that the solution is a uniformly parameterized straight line. ($x\equiv r\cos{\phi}=a\lambda+b$ for some $a$ and $b$, $y\equiv r\sin{\phi}=j\lambda+k$ for some $j$ and $k$).
(d) Verify that the noncoordinate basis $\mathbf{e}_{\hat{r}}\equiv\mathbf{e}_r=\frac{\partial\mathcal{P}}{\partial r},\ \mathbf{e}_{\hat{\phi}}\equiv r^{-1}\mathbf{e}_\phi=r^{-1}\frac{\partial\mathcal{P}}{\partial\phi},\ \ \mathbf{\omega}^r=\mathbf{d}r,\ \mathbf{\omega}^{\hat{\phi}}=r\mathbf{d}\phi$ is orthonormal, and that $\left<\mathbf{\omega}^\alpha,\mathbf{e}_{\hat{\beta}}\right>=\delta_{\ \ \hat{\beta}}^{\hat{\alpha}}$. Then calculate the connection coefficients of this basis from a knowledge [part (a)] of the connection of the coordinate basis.

I think 1) there are hats missing from omega indices and 2) 'modern notation' might not be very modern. I find it surprising that such an old, respected book has so many misprints.

a,b,c were straightforward. (d) contained the surprises. The $\left(\hat{r},\hat{\phi}\right)$ system (as we might call it) was orthonormal. The $\left(r,\phi\right)$ system was not, it was only orthogonal. The connection coefficients of the $\left(\hat{r},\hat{\phi}\right)$ system are not all symmetric in the lower two indices: $\Gamma_{\hat{\phi}\hat{r}}^{\hat{\phi}}=0\neq\Gamma_{\hat{r}\hat{\phi}}^{\hat{\phi}}=\frac{1}{r}$ which we prove. The method of calculating the coefficients is a great exercise in the piercing counter $\left<,\right>$.

${\hat{e}}_r,{\hat{e}}_\phi$ form a noncoordinate basis because, if we use them, the same point can have different coordinates. We show that is true. $e_r,e_\phi$ do not suffer from this problem.

Additionally we calculate the commutators (or Lie derivatives) $\left[e_r,e_\phi\right]=\left[\partial_r,\partial_\phi\right]$ and $\left[{\hat{e}}_r,{\hat{e}}_\phi\right]$. The first vanishes the second does not. This is a proof that the first is a coordinate (holonomic) basis and the second a noncoordinate (anholonomic) basis where you can't use coordinates. So you can't say $\left[{\hat{e}}_r,{\hat{e}}_\phi\right]=\left[\partial_{\hat{r}},\partial_{\hat{\phi}}\right]$. even though you can use the indices as in $\Gamma_{\hat{\phi}\hat{r}}^{\hat{\phi}}=0$.

Tiptoe through noncoordinate minefield at 8.5 Exercise Plane polar coordinates.pdf  (14 pages)