In which frame does the unit 2-sphere look locally flat?

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Discussion Overview

The discussion revolves around the coordinate transformation needed to make a given metric locally flat. Participants explore the nature of the metric provided and its relation to the geometry of a 2-sphere versus a 2-dimensional plane in polar coordinates.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant requests clarification on the coordinate transformation that would make the metric ds^2=dr^2 + r^2d\theta^2 locally flat.
  • Another participant suggests that the transformation is familiar, implying a common understanding of the topic.
  • A subsequent reply proposes a specific transformation involving a change of variables for r and θ, leading to a simplified expression for the metric.
  • Another participant cautions about the implications of the transformation, noting that the metric initially presented resembles that of a 2-dimensional plane in polar coordinates rather than a 2-sphere, which introduces confusion regarding the original question.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the interpretation of the metric in question, with some viewing it as applicable to a 2-sphere and others arguing it represents a 2-dimensional plane. The discussion remains unresolved as to the correct characterization of the metric.

Contextual Notes

There is a lack of consensus on the nature of the metric and its implications for the transformation, with participants expressing uncertainty about the correct interpretation and application.

masudr
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Can someone please tell me, the coordinate transformation that will make the following metric:

ds^2=dr^2 + r^2d\theta^2

look locally flat? Many thanks.
 
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I think that the transformation is already very familiar to you. :smile:
 
Heh, fair enough. So I did sit down and think about it for a bit, and is it this one?

\left( \begin{array}{clrr} r&#039; \\ \theta&#039; \end{array} \right) = <br /> \left( \begin{array}{clrr} 1 &amp; 0 \\ 0 &amp; \mbox{$\frac{1}{r}$} \end{array} \right)<br /> \left( \begin{array}{clrr} r \\ \theta \end{array} \right)

So that:

dx&#039;^\mu =<br /> \left( \begin{array}{clrr} dr&#039; \\ d\theta&#039; \end{array} \right) =<br /> \left( \begin{array}{clrr} dr \\ \mbox{$\frac{d\theta}{r}$} \end{array} \right)

So that:

ds^2 = dr^2 + d\theta ^2
 
Last edited:
Careful, if, e.g.,

\theta&#039; = \frac{\theta}{r},

then

d \theta&#039; = \frac{\partial \theta&#039;}{\partial r} d r + \frac{\partial \theta&#039;}{\partial \theta} d \theta.

Also, I'm now a bit confused. When, I first responded, I didn't really read the title of the thread. The metric you give in the original post seems not to be the metric for a 2-sphere. It looks more like the metric 2-dimensional plane written in terms of polar coordinates, and hence my first reply.
 

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