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?

  • $\begingroup$ 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 $\phi$. $\endgroup$
    – Andrew
    Jul 18, 2022 at 11:17
  • $\begingroup$ 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. $\endgroup$
    – Andrew
    Jul 18, 2022 at 11:22
  • $\begingroup$ 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. $\endgroup$
    – Habouz
    Jul 18, 2022 at 11:22
  • $\begingroup$ " $x=\cos(π/4)ϕ$ 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. $\endgroup$
    – Habouz
    Jul 18, 2022 at 11:24
  • 1
    $\begingroup$ 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. $\endgroup$
    – Andrew
    Jul 18, 2022 at 13:33

1 Answer 1


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*}


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