I've heard the work a few times now, the most recent in the star trek film. Is a singularity a real thing? If so what is it?

  • $\begingroup$ Related: physics.stackexchange.com/q/60869/2451 $\endgroup$
    – Qmechanic
    Jun 28 '13 at 19:57
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    $\begingroup$ duplicate of physics.stackexchange.com/q/60869/2451 $\endgroup$
    – user4552
    Jun 28 '13 at 20:13
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    $\begingroup$ @BenCrowell I don't think it's a duplicate. This question asks whether singularities are "real things," this is distinct from "what is a singularity?" Granted, the OP does also ask the second question, and the title is rather damning. Perhaps the title should be edited. $\endgroup$ Jun 28 '13 at 20:20
  • $\begingroup$ I think that by a real thing he means something that scientists study whether it exists in reality or not, as opposed to a made up term for scifi stories. And the question is what a singularity is. I don't think that the question is a duplicate, but Ben's answer of the other question would be perfectly suitable here. $\endgroup$
    – MBN
    Jun 29 '13 at 11:16

The word "singularity" is generally used to denote that some quantity becomes infinite or in general becomes undefined (i.e. cannot be expressed as a finite real number).

One particularly common use is in general relativity. At the event horizon of a black hole (the surface of no return), often coordinates that were well behaved far away run into trouble. For example, the time as measured by a clock infinitely far away slows down and comes to a halt near that surface, so far-away observers see stuff falling into a black hole slower and slower, never quite making it.

However, the above is only what is known as a coordinate singularity - something went wrong with your coordinate system, but nothing physically "broke." For a more benign example, the North Pole of the Earth has a coordinate singularity if you use standard longitude and latitude. What is the longitude of the North Pole? Well, it's undefined. There is nothing physically important about this statement - it merely reflects the choice of a poor coordinate system for that region. (In fact it can be shown mathematically that no sphere can be covered entirely by a singularity-free coordinate system.)

There can also be "true" singularities. In general relativity you might compute some coordinate-independent quantity - the (scalar) curvature of spacetime for instance - and see that it goes to infinity. No change of coordinates will "fix" this, and it means physically meaningful quantities are behaving badly. In classical general relativity, it has been proposed (see the cosmic censorship hypothesis) that any such true singularities (as might exist inside the event horizons of black holes) are always shielded by event horizons, so information about them can never reach us. In models where this is true, it doesn't even make much sense to talk about the existence of such things, since their properties by definition can never be studied.

One can also cast singularities in terms of geodesics - the paths that objects, light, and anything else follow in spacetime when no other forces act on them. A singularity is a point where a geodesic just ends. You reach that point and then that's it - the laws of physics have nothing more to say about your position.

In these latter two cases, if the singularities are "real" and somehow not hidden from us, they signify either (1) the universe has stopped being well defined, and therefore all bets are off when it comes to predictability and science in general, or (2) our theory is incomplete and needs more work. Most people believe in choice (2).

  • $\begingroup$ How would you show the mathematical proof that no sphere can be covered entirely by a singularity-free coordinate system? $\endgroup$
    – Deep
    Jun 28 '13 at 20:15
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    $\begingroup$ @deep That would be a good question for math.stackexchange.com. A topological approach would note the different Euler characteristics between the sphere and some nice subset of $\mathbb{R}^2$. I'm not sure if there are simpler ways to do it. $\endgroup$
    – user10851
    Jun 28 '13 at 21:05
  • $\begingroup$ Isn't that basically a version of the Hairy Ball Theorem...? $\endgroup$
    – Thriveth
    Jun 28 '13 at 21:56

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