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In spherical coordinates the flat space-time metric takes:


where $r^2d\Omega^2$ come from when the signature of metric $g_{\mu\nu}$ is (-,+,+,+)?

what is signature of spherical metric?

this is signature of spherical coordinates: $(1,r^2,r^2\sin^2\theta$)

let me ask in this way, Isn't it metric of spherical coordinates?

$$g_{\mu\nu}= (-1,+1,+r^2,+r^2\sin^2\theta)$$

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OK, let's take a step back. What is the signature of polar coordinates $ds^2 = dr^2 + r^2 d \theta^2$ or spherical coordinates $ds^2 = dr^2 + r^2 d\Omega^2$ ? – Vibert Apr 22 '13 at 17:38
The signature is defined in such a way that it's a sequence of positive and negative 1's, so coordinates should not be written in the signature. – joshphysics Apr 22 '13 at 18:51

The trickiness is what you mean by a spherical metric. What you've written down is the metric of flat space in spherical coordinates, which can be thought of as a warped product of the flat minkowskian two space $(t,r)$ with the unit sphere. This space is equivalent to the normal $(t,x,y,z)$ coordinates of standard special relativity under a coordinate transformation.

There are other choices, however. In particular, you can have a space where the constant-time hypersurfaces are 3-spheres, rather than 2-spheres. Here, the metric will be:

$$ds^{2} = -dt^{2} + d\psi^{2} + \sin^{2}\psi d\theta^{2} + sin^{2}\psi\sin^{2}\theta d\phi^{2}$$

You will find that this space is NOT equivalent to flat space. There is other trickery you can do with spheres to get yet more inequivalent spaces.

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Metric signature is a coordinate-invariant notion. Given a metric, one computes the number of positive and negative eigenvalues that it has, and this gives its signature. For a diagonal metric, like the metric $$ ds^2 = dr^2 + r^2 d\theta^2 $$ both diagonal components are positive, so the metric has precisely two positive eigenvalues, and its signature is therefore $(+,+)$.

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