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This question concerns the metric of a flat space:

$$ds^2=dr^2+cr^2\,\,d\theta^2$$ where $c$ is a constant. Why is it necessary to set $c=1$ to avoid singularities and to restrict $r\ge 0$?

Thanks.

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I don't know why one would have to restrict c = 1, since this is just a redefining of units. I can guess why r>0. Imagine a circle, where a point is given in polar coordinates (r,theta). Here r is the radius from the centre position. It makes no sense to define a point with negative radius, as the radius is always positive. –  Mew Feb 21 '13 at 14:34
    
@Chris: thanks! Yeah, those are exactly my thoughts. I am confused. –  Bert Feb 21 '13 at 14:35
    
At $r=0$ the metric $g_{ij}$ is no longer invertible. –  Qmechanic Feb 21 '13 at 14:37
    
@Qmechanic: Thanks! What does it mean physically when $g_{ij}$ is not invertible? –  Bert Feb 21 '13 at 14:42
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@Qmechanic: Also, we are setting $c=1$ as opposed to what? Why can't we have $c=2$, say? –  Bert Feb 21 '13 at 14:54

1 Answer 1

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The metric, in the form that you give, has no problems until you make the restriction $0 \leq \theta < 2\pi$, or allow $c < 0$. In either of these cases, you can run into problems when regularizing the metric. As has been stated, this metric is not invertible when $r=0$. Normally, this doesn't matter, because we're always free to make the substitution $x = r\cos \theta$ and $y=r \sin \theta$, and then you have transformed this metric into ordinary minkowski space.

If you have $0\leq \theta < 2\pi$ and $c \neq 1$, however, this transformation becomes $x = r \cos (\sqrt{c}\theta)$ and $y= r \sin (\sqrt{c}\theta)$, and then you're mapping more (or fewer $if 0<c<1$) points in your $r,\theta$ domain than there are in your $x,y$ domain, and the singularity is not resolved. In particular, if $0< c < 1$, you just get a sliver of the plane. You can remove badness at the boundary by identifying the point $(r,\theta_{max})$ with the point $(r,0)$, but when you do this, the space basically defines a cone, and the manifold is not smooth at $r=0$ Hence why these singularities are called conical singularities.

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And note that a cone is Gauss flat everywhere except at the singularity--I can turn a sheet of paper into a cone without folding it. –  Jerry Schirmer Feb 21 '13 at 15:00
    
Thank you, Jerry! This is very helpful. :) So setting $c=1$ doesn't actually remove the singularity right? it just has to do with the range of $\theta$? –  Bert Feb 21 '13 at 15:20
    
@Bert: that's one way of thinking of it. Basically, there will be no singularity if $\sqrt{c}\theta_{\rm max} = 2\pi$. If $\theta$ is not bounded, you get a space of infinite branch cuts. –  Jerry Schirmer Feb 21 '13 at 15:29

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