Is the Surface of a Sphere Locally Flat?

[Jufa]

TL;DR Summary

Something our teacher explained me regarding the metric and local flatness doesn't match a simple example of a sphere

Given a certain manifold in $R^3$ I've been told that at every location $p$ it is possible to encounter a reference frame from which the metric is the euclidean at zero order from that point and its first correction is of second order. This, nevertheless does not match with the following example:
Consider the surface of a sphere $S^2$ at a certain polar angle $\theta_0$ and azimutal angle $\phi_0$. The metric of that surface is
$g_{\phi \phi} = sin^2(\theta)$ and $g_{\theta \theta} = 1$. Let as define a new local reference frame such that $\theta' = \theta$ and $\phi' = \frac{\phi}{sin(\theta_0)}$. It is easy to see that the new metric in this reference frame will be euclidean at $p$. But what happens in its neighbourhood?

$g_{\phi \phi} = \frac{sin^2(\theta_0 + \delta \theta)}{sin^2(\theta_0} = 1 + 2sin(\theta_0)cos(\theta_0)\delta \theta + cos^2(\theta_0)(\delta \theta)^2$

which clearly has a non-vanishing correction at first order.

Where is the problem? Is my teacher wrong?
Thanks in advance.

 

[Orodruin]

Summary:: Something our teacher explained me regarding the metric and local flatness doesn't match a simple example of a sphere

Where is the problem?

You picked the wrong coordinates. The statement is just that those coordinates exists where the metric is Euclidean to zeroth order around p and the first correction is at order two, not that this will be the case whatever coordinates you pick.

 

[Jufa]

You picked the wrong coordinates. The statement is just that those coordinates exists where the metric is Euclidean to zeroth order around p and the first correction is at order two, not that this will be the case whatever coordinates you pick.

I can't imagine a set of coordinates that fulfills the non-zero first order correction condition. Indeed, once you impose that the metric is euclidean at p, the new coordinates become fixed and, thus, my choice was the only one possible to be made.

 

[Ibix]

I can't imagine a set of coordinates that fulfills the non-zero order condition.

Polar coordinates with the equator passing through your location should do the trick.

 

[Dale]

But what happens in its neighbourhood?

$g_{\phi \phi} = \frac{sin^2(\theta_0 + \delta \theta)}{sin^2(\theta_0} = 1 + 2sin(\theta_0)\delta \theta + (\delta \theta)^2$

which clearly has a non-vanishing correction at first order.

Where is the problem? Is my teacher wrong?

As @Orodruin mentioned you just picked bad coordinates, and as @Ibix mentioned (people are fast on this forum!) you can just use standard coordinates with the equator going through the point of interest. However, in general you can always construct such coordinates even if they have nothing to do with some original starting coordinates.

 

[Jufa]

Polar coordinates with the equator passing through your location should do the trick.

That definitely solves it. Many thanks!

 

[PeterDonis]

once you impose that the metric is euclidean at p, the new coordinates become fixed

No, they don't. There is more than one possible set of Cartesian coordinates; you can rotate the axes any way you like.

 

[Jufa]

As @Orodruin mentioned you just picked bad coordinates, and as @Ibix mentioned (people are fast on this forum!) you can just use standard coordinates with the equator going through the point of interest. However, in general you can always construct such coordinates even if they have nothing to do with some original starting coordinates.

Sure, I only mention the first set of coordinates in order to define the metric tensor.

 

[Orodruin]

I can't imagine a set of coordinates that fulfills the non-zero first order correction condition. Indeed, once you impose that the metric is euclidean at p, the new coordinates become fixed and, thus, my choice was the only one possible to be made.

That you cannot imagine a different set of coordinates does not mean it doesn't exist. It is fairly simple to see that having orthonormal basis vectors at one point does not fix the coordinate system, not only because you can always rotate your coordinates, but also because any coordinate change for which $\partial x^a/\partial y^b = \delta^a_b$ at the point of interest will not change the components of the metric tensor. Furthermore, among those coordinate changes you can find one that makes the Christoffel symbols vanish at that point because the coordinate change of the Christoffel symbols also depends on the second derivatives of the coordinate change functions, which are not fixed by fixing the metric components.

 

[Jufa]

I know it. I did not mean that the fact that I cannot imagine it means that it does not exist. I was just expressing my ignorance of such a coordinates system.
Thanks.

 

[Ibix]

I know it. I did not mean that the fact that I cannot imagine it means that it does not exist. I was just expressing my ignorance of such a coordinates system.

I'm guessing English isn't your first language? Unfortunately, "I can't imagine that..." in English is usually taken to mean "It is impossible that...", which may have caused some miscommunication here. I think "I can't think of a set of coordinates..." would probably be closer to what you were aiming for. You could probably do with an "I think" somewhere in the second sentence of post #3 too, I'm afraid.