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Homework Help: Spacetime Curvature

  1. Nov 20, 2009 #1
    in this article
    it says in the first paragraph that the curvature of spacetime is related to the stress energy tensor i.e. the matter in the universe. i.e. the curvature at a particular point depends on the matter present at that point.

    however in this article
    it says in the section "mathematics of spacetimes" that the metric dictates the geometry of the spacetime. i assume this means that since the metric is dictating the geometry, it is also dictating the curvature.

    anyway, which is correct?
    from the Einstein field equations, it would appear to me that both influence the curvature.

    my interpretation is that since the metric is defined over the entire spacetime it defines the way in which matter will influence the curvature. and then since the mater is obviously different at each point, the matter at that particular point will dictate the curvature(i.e. geometry) at that particular point (with respect to the spacetime metric). is this correct?
  2. jcsd
  3. Nov 20, 2009 #2


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    Both articles are correct. The Einstein equation reads:
    [tex]G_{\mu \nu} = 8 \pi T_{\mu \nu}[/tex]

    What this says is that the stress-energy tensor gives us the form of the Einstein tensor. If you're familiar with the Einstein tensor, it's a jumble of christoffel symbols, but it boils down to just some horrible combination of multiplications and differentiations of the metric. So, the stress-energy tensor determines the metric. And, as you understand, the metric is really what tells us the geometry of the spacetime. I would say it's more correct to say it describes the geometry of the spacetime, whereas the matter distribution dictates it.
  4. Nov 20, 2009 #3
    thanks. i have 2 quick questions:

    (i)is this a suitable defn of a singularity:

    "A singularity in general relativity is a point in spacetime at which the scalar (invariant) curvature becomes infinite. We note that singularities are often associated with geodesic incompleteness, that is where geodesics cannpt be smoothly extended due to unbounded curvature on their incomplete ends"

    (ii) In the above when it talks about the scalar curvature is it referring to the contracted Ricci tensor i.e. [itex]R=R_{ab}R^{ab}[/itex] or the contracted Riemann tensor [itex]R_{abcd}R^{abcd}[/itex]?

  5. Nov 20, 2009 #4


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    First, a note on singularities... There are two types of singularities in a theory such as GR. The first type are coordinate singularities, which arise because of a particular set of coordinates (Think the schwarzschild metric at R=2M, the event horizon). In general, there's nothing special about such locations (or the curvature associated with them). And then, there's the real singularities.

    i) Yeah, that definition sounds suitable.

    ii) This is a good question, that I don't know the answer to. http://en.wikipedia.org/wiki/Gravitational_singularity suggests it is the square of the riemann tensor, but I think the ricci scalar is invariant as well, so it should suffice as well.
  6. Nov 20, 2009 #5
    ok. if we look more closely at the two types of singularity that can occur in GR just so I can straighten this out in my head.

    (i) For said coordinate singularity at r=2M in the Schwarzchild metric, here all the components of the curvature tensor remain finite and continuous and so it isn't the same as the curvature singularity that occurs at r=0. Is there any physical interpretation for a coordinate singularity or does it just meant he choice of coordinates is unsuitable at this point?
    *disclaimer: i'm kind of self-teaching myself this stuff and haven't looked at the Schwarschild solution yet as that's going to be covered in my course enxt semester.

    (ii) when you talk about a real singularity, i assume you mean a curvature singularity.

    (iii) when i was discussing this with my supervisor, he also told me there were two types of singularity:
    curvature singularities
    and also singularities that occur due to a breakdown in causal structure. I think the example of this that he gave (assuming I remember it correctly) is if we identify two points at either "edge" of our spacetime such that if a particle travels to one fo these points it will reappear at the other point on the other side of the spacetime. If we take a geodesic between the two points that takes time 1 to get back to the starting point on the first lap, time 1/2 to get back to the starting point on hte second lap etc (ie.e its' speed doubles each lap) then it will perform infinite number of cycles in under time 2. But then you have either a complete geodesic that's not continuous or an incomplete geodesic that's also continuous (I can't remember which it was...I reckon probably the latter). Hopefully I've remembered that example ok...
    Anyway my question is, would the "breakdown in causal structure" singularities represent a third type of singularity?

    thanks again
    also apologies for the slightly dodgy example I used.
  7. Nov 20, 2009 #6


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    i) Not in general, no, there is no physical interpretation for coordinate singularities. Obviously in the schwarzschild solution there is "something" special about R=2M, but spacetime is still smooth at that point (as opposed to r=0). If you want to know a little bit more about this particular example, you can look up an alternative coordinate description of the metric around a schwarzschild black hole (http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates). If you think about it, I can take flat space and do a coordinate transofrmation to u=1/x or something, which will be singular for x=0. However, as you probably know, change of coordinates does not change the geometry of the spacetime.

    ii) Yeah. I mean a place where the metric actually goes to infinity, regardless of choice of coordinates.

    iii) I haven't heard of that before, perhaps someone better versed in GR can say something about a situation like that.
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