# Finding Locally flat coordinates on a unit sphere

I know this is more of a math question, but no one in the Mathematics community was able to give me an answer, and since physicists are familiar with General Relativity, I thought I might get an answer.

Imagine a unit sphere and the metric is:

$$ds^2 = d\theta ^2 + \cos^2(\theta) d\phi^2$$

I want to find Locally Flat Coordinates (I think they're called Riemann Normal Coordinates) on the point $$(\frac{\pi}{4}, 0)$$, so what I need are coordinates such that the metric would reduce to the Kronecker Delta and the Christoffel Symbols should vanish. I start by the following translation:

$$\theta' = \theta - \frac{\pi}{4}$$

then do the following substitution by guessing:

$$\frac{f(\theta')}{\cos(\theta)} d\phi' = d\phi$$

And the condition is $$f(0)$$ should be 1, so the metric becomes:

$$ds^2 = d\theta' + f^2(\theta')d\phi'$$

And it is a matter of finding $$f(\theta')$$. I calculate the Christoffel Symbols:

$$\Gamma^{\lambda}_{\mu\nu} = \frac{1}{2} g^{\lambda \alpha}(\partial_{\mu}g_{\alpha \nu} + \partial_{\nu}g_{\mu \alpha} - \partial_{\alpha}g_{\mu \nu})$$

And make them vanish.

So what I get is:

$$\frac{f'(0)f(0)}{f^2(0)} = 0$$

Obviously, $$f(\theta')=\cos(\theta')$$ is a solution which is the thing I know is correct. However, there are infinite functions that satisfy the above conditions. Are all of these functions eligible to make the new coordinates Riemann normal coordinates?

• You expand about the point $(\theta,\phi)=(\pi/4, 0)$, so when you say there should be an infinite number of locally flat coordinate systems you can find, I suppose you mean you are looking for coordinate systems related by a rotation through an axis going through that point? To find the most general coordinate system you need to do a more general coordinate transformation of the form $\theta'=f(\theta, \phi)$, $\phi'=g(\theta, \phi)$. At the moment it seems you are effectively assuming $\phi'=f(\theta)\phi$, but a rotation will mix $\theta$ and $\phi$. Jul 18, 2022 at 11:17
• To be honest I just wouldn't use the method you are proposing. I would construct a Cartesian coordinate system with $x=\cos(\pi/4) \phi$ and $y=\theta$ so the metric is $dx^2+dy^2$ (please double check this coordinate transformation works as advertised, I am writing this very quickly and didn't check). Then rotations in the $x,y$ coordinate system are obvious, and you can always return back to $\theta, \phi$ by substitution. Jul 18, 2022 at 11:22
• I didn't quite get what rotations have to do with the question. "I suppose you mean you are looking for coordinate systems related by a rotation through an axis going through that point?" No, I just want a coordinate system on the point $(\pi/4, 0)$ such that it looks flat at that point. What I did is guess a coordinate transformation that makes the metric look like the euclidean metric at the respective point, and make the christoffel symbols vanish as well at that point. Jul 18, 2022 at 11:22
• " $x=\cos(π/4)ϕ$ and $y=θ$ so the metric is $dx^2+dy^2$" Yes these work but I don't think those are Reimann normal coordinates. The make the metric reduce to the euclidean metric, but the christoffel symbols do not vanish. Jul 18, 2022 at 11:24
• Fair enough, you might need to add some term to the $x$ and $y$ coordinates to make the Christoffel symbols vanish, but this is just a detail. My point is that the metric will locally look like $dx^2+dy^2$ in locally flat coordinates. Once in that form, rotations about the point $(\pi/4, 0)$ will move between different sets of locally flat coordinates. Jul 18, 2022 at 13:33

starting with components of the unit sphere : \begin{align*} &\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}=\left[ \begin {array}{c} \cos \left( \phi \right) \sin \left( \theta \right) \\ \sin \left( \phi \right) \sin \left( \theta \right) \\ \cos \left( \theta \right) \end {array} \right] \end{align*} from here
\begin{align*} &\begin{bmatrix} dx \\ dy \\ dz \\ \end{bmatrix}=\underbrace{\left[ \begin {array}{cc} \cos \left( \phi \right) \cos \left( \theta \right) &-\sin \left( \phi \right) \sin \left( \theta \right) \\ \sin \left( \phi \right) \cos \left( \theta \right) &\cos \left( \phi \right) \sin \left( \theta \right) \\ -\sin \left( \theta \right) &0\end {array} \right]}_{\mathbf J}\, \left[ \begin {array}{c} d\theta \\ d\phi \end {array} \right] \end{align*} and the metric \begin{align*} &\mathbf{G}=\mathbf J^T\,\mathbf J=\left[ \begin {array}{cc} 1&0\\ 0& \left( \sin \left( \theta \right) \right) ^{2}\end {array} \right] \end{align*}
now we are looking for the transformation matrix $$~\mathbf{T}~$$ that transformed the metric to unit matrix
\begin{align*} &\mathbf{T}^T\,\mathbf{G}\,\mathbf T=\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}\quad\Rightarrow\quad \mathbf{T}=\begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{\sin(\theta)} \\ \end{bmatrix} \end{align*} hence \begin{align*} &\begin{bmatrix} dx \\ dy \\ dz \\ \end{bmatrix}\mapsto \underbrace{\mathbf{J}\,\mathbf T}_{\mathbf{T}_n}\, \left[ \begin {array}{c} d\theta \\ d\varphi \end {array} \right] \end{align*}
and the neue metric is: \begin{align*} &dx^2+dy^2+dz^2\mapsto d\theta^2+d\phi^2 \end{align*}
where $$~\mathbf{T}_n~$$ is a function of $$~\theta~,\phi~$$
\begin{align*} &\mathbf{T}_n(\theta=\pi/4~,\phi=0)= \left[ \begin {array}{cc} \frac 12\,\sqrt {2}&0\\ 0&1 \\ -\frac{1}{2}\,\sqrt {2}&0\end {array} \right] \end{align*}