Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is the schwarzchild radius a radius of curvature?

  1. Jan 15, 2007 #1
    EDIT: Is the Schwarzschild coordinate a radius of curvature in the geodesic?

    And also, in physics, what do I make of a negative radius of curvature?
    Last edited: Jan 15, 2007
  2. jcsd
  3. Jan 15, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    No, the Schwarzschild coordinate is just a cordinate. You should probably read about "intrinsic" vs "extinsic" curvature. The 2-d surface of a sphere is an example of a surface with a positive (intrinsic) curvature, while the 2-d surface of a sadle-sheet is an example of a surface with a negative curvature.

    Intrinsic curvatures are of the sort of interest in GR, as they are indepenent of any particular embedding. See for instance


  4. Jan 15, 2007 #3
    Hmm, interesting definition, extrinsic and intrinsic curvature are defined here as mutually exclusive.
  5. Jan 15, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    They aren't really mutually exclusive, because a 3-d observer can calculate both sorts of curvature for an embedded 2-d manifold. The situation is that anyone can compute the intrinsic curvature, but only when one has an embedding diagram can the extrinsic curvature be defined.
  6. Jan 15, 2007 #5
    So intrinsic curvature is not recognized as an extrinsic curvature as well?
  7. Jan 15, 2007 #6
    When speaking about space-time it is desirable to define curvature intrinsically, to avoid having to assume that the universe is embedded in a higher dimensional space.
  8. Jan 15, 2007 #7
    I understand that. However the quote claimed that extrinsic curvature cannot be detected on the surface, so that implies that extrinsic curvature does not include intrinsic curvature.
  9. Jan 15, 2007 #8

    Chris Hillman

    User Avatar
    Science Advisor

    Curvature, curvature

    Hi all,

    As pervect already noted, the Schwarzschild radial coordinate r is a certain monotonic function on a spherically symmetric (four-dimensional) Lorentzian spacetime, i.e. one featuring a family of nested two-spheres inside members of an appropriate family of three-dimensional spatial hyperslices. This function is constant on the spheres and has the property that the surface area of the sphere labeled by [itex]r=r_0[/itex] is [itex]A = 4 \pi \, r_0^2[/itex].

    You mentioned "geodesic": note that in differential geometry geodesic curvature has a technical meaning closely related to path curvature of a curve, i.e. if [itex]\vec{X}[/itex] is the unit tangent vector to a curve, then [itex]\nabla_{\vec{X}} \vec{X}[/itex] is the acceleration vector, which is orthogonal to [itex]\vec{X}[/itex] and whose magnitude is the path curvature (the reciprocal of the radius of curvature).

    On the other hand, the Gaussian curvature of the sphere with surface area [itex]4 \pi \, r^2[/itex] is [itex]K=1/r^2[/itex]. The intrinsic and extrinsic curvature of submanifolds (of larger dimension than a curve) are "measured" in units of reciprocal area, i.e. sectional curvature. On the other hand, the path curvature of a curve is "measured" in units of reciprocal length. The path curvature of a great circle on a two-sphere embedded as an ordinary round sphere in [itex]E^3[/itex] would be [itex]1/r[/itex].

    Assuming you are asking about something like negative Gaussian curvature in a two-dimensional manifold, or negative sectional curvatures in a spatial hyperslice in a spacetime, this is associated, via the Jacobi geodesic deviation formula, with divergence of nearby and initially parallel geodesics. A positive Gaussian curvature is associated with convergence of nearby and initially parallel geodesics.

    For example, in on ordinary sphere (two dimensional Riemannian manifold with constant positive Gaussian curvature), longitude circles are geodesics, and two nearby longitudes are initially parallel at the equator. As you move North, they converge (positive curvature). In the hyperbolic plane (two dimensional Riemannian manifold with constant negative Gaussian curvature), two intially parallel geodesics diverge (look for pictures of the Poincare disk model in good textbooks). In the Schwarzschild vacuum solution, the sectional curvature associated with components [itex]R_{trtr}[/itex] is also negative, which means that initially parallel and radially outgoing null geodesics (in particular) will diverge; this leads to the phenomenon called "gravitational red shift".

    To avoid confusion, a saddle surface or hyperboloid of one sheet is a two-dimensional Riemannian manifold embedded in [itex]E^3[/itex] which has nonconstant negative Gaussian curvature.

    No, no, not at all. For example, a three-dimensional spatial hyperslice in a Lorentzian manifold has both extrinsic curvature, which is measured by a symmetric rank two tensor field (this tensor field is essentially the negative of the expansion tensor of a timelike unit vector field which agrees with the normals to the slice on the slice itself), and intrinsic curvature, which is measured by a rank four tensor field, the three-dimensional Riemann curvature tensor of the hyperslice.

    The extrinsic curvature measures how the normal vectors are changing as we move around the slice, i.e. how the slice is "bending" in the bigger manifold. The intrinsic curvature doesn't depend upon the embedding, only on the metric tensor induced in the hyperslice by restricting the metric tensor of the bigger manifold.

    MeJennifer, you should probably look at a good textbook on surface theory, since it would be difficult to imagine a more fundamental distinction in "metric geometry" than that between extrinsic and intrinsic curvature.

    Consider a two dimensional Riemannian manifold embedded in [itex]E^3[/itex]. There are many ways of embedding a two-manifold so that it has a given intrinsic geometry (determined from the induced metric tensor) but dramatically different extrinsic geometry. For example, the Dover reprint of the classic text by Struik shows some nice pictures of two embedded surfaces which both have vanishing Gaussian curvature, but which have very different extrinsic curvature and which look very different.

    They are completely distinct concepts, as should apparent from the fact that one is measured by a fourth rank tensor and the other by a second rank tensor!

    It depends upon what you mean by "detected". For example, the normal vectors for a spatial hyperslice are not part of the intrinsic geometry of that hyperslice. But if this hyperslice arises as one of the slices orthogonal to an irrotational timelike congruence, then we consider this congruence as the world lines of a certain family of observers, and then the expansion tensor of this congruence can in principle be measured by these observers.
    Last edited: Jan 15, 2007
  10. Jan 15, 2007 #9
    Well let me refrase it, can we conclude that if the symmetric rank two tensor field, the field that measures external curvature, shows a presence of curvature that that particular curvature can under no circumstances be detected by the Riemann tensor?
    I (think) I understand te difference between internal and external but I am not sure if we could say that there is zero overlap.
  11. Jan 15, 2007 #10

    Chris Hillman

    User Avatar
    Science Advisor


    Hi again, MeJennifer,

    I don't think I understand your question. Are you trying to ask what implications a nonzero extrinsic curvature tensor has for the three-dimensional Riemannian curvature of the hyperslice? (None: in particular, there are many ways of embedding a locally flat Riemannian three-manifold inside any four dimensional Lorentzian manifold, such that the extrinsic curvature is nonvanishing. Similar statements hold for Riemannian geometry, or in other dimensions; see for example horospheres in [itex]H^n[/itex].) Or what implications a nonzero intrinsic curvature tensor (three dimensional Riemannian curvature tensor, computed from the metric tensor induced on the hyperslice) has for the extrinsic curvature? (None, in the sense that one can find Lorentzian manifolds admitting hyperslices which are locally isometric say to [itex]S^3[/itex] but which have vanishing extrinsic curvature tensor, for example in the so-called "Einstein static universe".)

    I don't know what you mean by "zero overlap", but I think you should look at say Spivak's five volume textbook on differential geometry, which should help you to understand intrinsic versus extrinsic geometry. If you just want to see a bit about three dimensional Riemannian curvature (of a spatial hyperslice in a Lorentzian manifold) versus four dimensional Riemann curvature (of the big Lorentzian manifold), a standard citation for physics students is Hawking and Ellis, Large Scale Structure of Space-Time. A good key phrase here is Gauss-Codazzi equations.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook