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

[masudr]

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.

 

[George Jones]

I think that the transformation is already very familiar to you.

 

[masudr]

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

 

[George Jones]

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.