# A Question on Singularity

I am not aware of GR, but due to curiosity i have a question in my mind. Please let me know if it is inappropriate to ask here.

My question is about singularity. I am under the assumption that singularity means in mathematical terms equivalent to a discontinuity in a function. My question is what type of discontinuity are the ones corresponding to Penrose-Hawking singularity theorems and also to the the naked singularity ? by type i mean

1. removable (left and right limits exist and are equal but not equal to the function value at that point).

2. a jump type discontinuity (left and right limits exists and are not equal to each other).

3. Oscillating discontinuity (as described as described in the above given link)

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Note: Depending on one's definition of the discontinuity (whether one allows talking about the continuity of the function at a point where the function is undefined), pole might also be considered an infinite discontinuity.

Singularity is certainly not equivalent to a discontinuity (not even mathematically). It can also mean pole and this is the usual meaning it has in GR. I.e. the curvature of the space-time blows up.

For example, Schwarzschild metric's curvature diverges at the origin as may be checked by looking at Kretschmann invariant. There might also exist coordinate singularities but these are unphysical and can be removed by a better choice of coordinates.

Note: What does one mean by curvature?

The basic object associated with the metric tensor is a connection and for that one can define a curvature form. Now this is quite complicated object and it is not quite clear how to detect singularities in it. One easy way is to define invariants (such as the above-mentioned Kretschmann invariant) associated with this form.

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@Marek : Is curvature a derivative of some function ? – Rajesh Dachiraju Dec 4 '10 at 15:13
@Rajesh D: it's a metric en.wikipedia.org/wiki/Metric_(mathematics) – Sklivvz Dec 4 '10 at 15:19
@Rajesh: I added a little note about curvature. But I don't think it's fruitful to explain more than this if you don't know GR. It's a vast topic that requires whole books to learn. At the very least, one should understand basics of differential geometry and special relativity. – Marek Dec 4 '10 at 15:23
@Sklivvz @Marek : but the curvature is equal to the derivative of an integral over it.Is there any meaning for the integral over the curvature ? – Rajesh Dachiraju Dec 4 '10 at 15:29
@Rajesh: yes, you can integrate curvature and this can give you topological information about the manifold. See e.g. Gauss-Bonnet theorem. But I don't think this is related to your question anymore, so if you want to know more about curvature ask a separate question. – Marek Dec 4 '10 at 15:37

The formal definition is 'a point through which a geodesic cannot be extended'. It can be a point where the curvature is infinite, as said above, but the concept is a little more general--it is a point at which there is no smooth way to extend a path into the future. Formally, what the singularity theorems show is that, when the curvature obeys certain conditions, and you have positive energy, then certain geodesics must end in a finite proper time.

Note the contrast with Minkowski space, where a particle moving in a straight line is free to continue moving in a straight line forever, always tracing out more and more proper time.

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Hm, if you want to include this definition of singularity then you should not say path but geodesic and also say what a geodesic is, right? That's why I was reluctant to talk about this. In Minkowski example you talk (implicitly) about geodesics. – Marek Dec 6 '10 at 10:08

I suspect mathematically singularity means $\frac{1}{0}\rightarrow\infty$. The size goes to zero and the math blows up. In fact, the size does not have to be zero for singularity to exist. If you think of trying to force a wave to exist in a sub-harmonic wavelength (size), it also makes the math blow up.

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