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Is the curvature of space-time a smooth function everywhere (except at black holes) in view of General relativity. By 'smooth' it is meant that it possesses derivatives of all order at a given point.

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This question is so wrong... first, it doesn't ask under what assumptions are you working (what distributions of energy-stress tensor). Second, as I already told you, there are lots of different notions of curvature so you have to specify which one of them are you using. –  Marek Dec 4 '10 at 20:33
But except for that, the answer is a simple no. As in basically every other mathematical and physical theory (i.e. there is no problem in using math on the class of non-smooth functions or indeed even non-continuous functions). Why are you still asking these questions? They have nothing to do with physics. –  Marek Dec 4 '10 at 20:36
@Marek : I think that for a theory to be called successful it cannot choose different versions/notions of the basic concept like curvature to explain different things. The theory should be general enough. SWo I did doesn't make sense to argue as "which notion are you working?" –  Rajesh D Dec 5 '10 at 13:19
@Rajesh: because it is not meant for that. Every physical theory has a domain of applicability. For example, you can't take heat equation and expect that it will predict how gravity works. In the very same way, GR doesn't tell you about anything else than large-scale space-time. –  Marek Dec 5 '10 at 13:43
@Rajesh: it doesn't matter what you think. For a physical theory to be successful it suffices that it agrees with an experiment. The theory itself doesn't even have to be described by mathematics (but all our major theories do, fortunately). As to the curvature: there are various notions of curvature in GR. But I think you can't really appreciate this if you don't even know GR (as you stated before yourself) ;-) –  Marek Dec 5 '10 at 13:46

3 Answers 3

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No, not at the boundary of a solid object like a planet. There's a step function in the stres-energy tensor, and so you'll have a step function in the Riemann tensor.

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@DarenW : Is there any situation in GR where the curvature at one point is the derivative of curvature at some other point ? –  Rajesh D Dec 4 '10 at 19:39
@Rajesh D: except for the dimensional problems comparing an entity with its derivative, I guess you can find a whole class of points where this is true. It might be useful if you could describe why you pose these questions. Instead of trying to understand a new theory it might be much easier to explain the things you know and ask other people to help for transferring their knowledge to your area of research - the main idea of collaboration, I suppose. –  Robert Filter Dec 5 '10 at 15:15
@Robert: big +1. My point exactly. It makes no sense to ask about smoothness of some object if one doesn't even know what the object is. And that's just mathematics. It makes even less sense to ask these unmotivated questions at a physics site. –  Marek Dec 5 '10 at 16:21
@Robert Filter : I am asking questions just out of curiosity...I do intend to learns GR but want to get a feel for it(in mathematical nature) to begin with. As you have mentioned that "except for the dimensional problems comparing an entity with its derivative, I guess you can find a whole class of points where this is true" which means i think your statement supports that curvature is always smooth !...please correct me if i am mistaken. –  Rajesh D Dec 6 '10 at 13:36
@Robert Filter, @DarenW, @Marek : do you mean to say that there are geodesics which take a $90^0$ (right angle) turn at a particular point (even if it is not a black hole or any such unexplainable thing)....I think its unfair. –  Rajesh D Dec 6 '10 at 13:40

Well other than black hole singularities, there are other cosmological topological defects like cosmic strings, domains, and textures (and magnetic monopoles) that would not be considered as smooth. These are regions where there is or has been a phase transition. Of course, these things may not exist, but string theory predicts monopoles, so they had better ;)

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I actually believe the answer to this question is yes, and the reason is that a fundamental assumption of GR is that the gravitational field is constant over small regions of space time. Another way of phrasing this is to say the GR is only valid in the limit of geometrical optics, which is the limit for which the wavelength of light is small compared with the length scale of the problem under consideration.

It's very hard to do any calculations in GR, and the 2 well known cases (black holes and cosmology) both involve smooth functions for all intents and purposes. Actually I think it's generally accepted that almost all functions that arise in physics are analytic (i.e. can be represented by power series), which except for some pathological examples means the functions are smooth.

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Wrong, wrong and wrong. First, GR can deal with things like planets (as pointed out by DarenW). Second, characteristic surfaces of gravitational waves are discontinuous (by very definition of characteristic surface). Third, even the BH you mention are obviously not smooth by the very definition of singularity. Fourth, GR can be used purely as a mathematical theory and you can work with whatever functions you want. Shall I continue? –  Marek Dec 6 '10 at 1:13
@Marek: Supposing we have a stress energy tensor with a step function in it like for a planet, then the space time manifold also has some sort of step function in it. Do you think this metric with a step function satisfies the principle of equivalence at the point of discontinuity? –  Matt Calhoun Dec 6 '10 at 2:49
@Matt: Thank you for your effort but you might have mixed things up. First, I do not know your fundamental assumption. If I remember correctly, you can transform the metric at one point to Minkowski one with higher order terms. But this does not necessarily mean the field is constant; only for one point. Also, there are more (also physical) solutions known to GR than you mentioned. Further, it might be much more likely that nature is discrete at some scales, not at all analytic. –  Robert Filter Dec 6 '10 at 6:55
@Matt: what does this have to do with principle of equivalence? We know it's true and it's build into GR. So anytime you are working with GR it has to hold (because geodesics never depend on mass). –  Marek Dec 6 '10 at 10:38
@Marek: My entire point is that the equivalence principle, (constant scalar curvature at small distance scales) is a condition that makes the space time metric a smooth function. The whole problem for your step function example, is that GR is not valid at the length scale for which the discontinuity in the step function kicks in. GR is not just mathematics, it is a physical theory which is not valid in all situations; in fact GR is only valid in the limit of geometrical optics. In this limit, any non-smooth properties of light paths are "smoothed" out. –  Matt Calhoun Dec 6 '10 at 16:52

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