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

1. Sep 24, 2006

### 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.

2. Sep 24, 2006

### George Jones

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

3. Sep 24, 2006

### masudr

Heh, fair enough. So I did sit down and think about it for a bit, and is it this one?

$$\left( \begin{array}{clrr} r' \\ \theta' \end{array} \right) = \left( \begin{array}{clrr} 1 & 0 \\ 0 & \mbox{\frac{1}{r}} \end{array} \right) \left( \begin{array}{clrr} r \\ \theta \end{array} \right)$$

So that:

$$dx'^\mu = \left( \begin{array}{clrr} dr' \\ d\theta' \end{array} \right) = \left( \begin{array}{clrr} dr \\ \mbox{\frac{d\theta}{r}} \end{array} \right)$$

So that:

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

Last edited: Sep 24, 2006
4. Sep 24, 2006

### George Jones

Staff Emeritus
Careful, if, e.g.,

$$\theta' = \frac{\theta}{r},$$

then

$$d \theta' = \frac{\partial \theta'}{\partial r} d r + \frac{\partial \theta'}{\partial \theta} d \theta.$$

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.