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Visualizing Ricci curvature

  1. Feb 25, 2010 #1
    "Visualizing" Ricci curvature

    Can someone help me visualize the Ricci curvature?

    Since it is easier to visualize a surface bending in 3-D, let's try to view this as a sheet with one spatial dimension and one time dimension and embedding into euclidean 3-D.

    Since the metric can always be written locally as ds^2 = -dt^2 + dx^2, deviations to this must be quadratic or higher orders, and these quadratic terms are the "curvature" pieces, right?

    The examples I know are:
    Positive curvature would be like the top of a sphere.
    Negative curvature would be like a hyperbolic sheet.

    Now if we look at the Ricci tensor, what would be an example of something that has different "curvature" in different "directions"? The hyperbolic sheet already kind of looks like that to me, but that is constant curvature. So I'm having trouble visualizing it.

    Any help?
    Are there some good visual illustrations from websites or books that people can recommend?

    For a concrete example to help me visualize:
    Consider a spherical shell, with a large mass attached to the north and south poles. There is a lot of symmetry here. We should be able to use that to argue the form of the Ricci tensor at the origin. I want to be able to visualize curvature well enough that I can understand how to make such symmetry arguments myself.
  2. jcsd
  3. Feb 25, 2010 #2
    Re: "Visualizing" Ricci curvature

    In your polar mass scenario, I think I'd consider a couple of traffic cones, bulging as if inflated with their points away from the poles. Eliminating the base of the cone, and the perfect tapering of course.

    EDIT: So.. <> but rotated 90 degrees
  4. Feb 25, 2010 #3

    Ben Niehoff

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    Re: "Visualizing" Ricci curvature

    Forget Lorentzian signature and GR. Just look at this from a standard differential geometry perspective.

    In n-space, the Ricci tensor measures the local deformation of an n-sphere centered at a given point. This is a second-order effect (i.e., it goes as r^2, where r is a radius of the sphere).

    Take some point P and imagine all the geodesics emanating from P. Define a "sphere" of radius r to be the locus of points of distance r along a geodesic from P. As r grows, the sphere will begin to deviate in shape from the Euclidean notion of a sphere. The lowest-order deviation is quadratic in r, and describes the sphere distorting into an ellipsoid. The Ricci tensor is the quadratic form that describes the shape of this ellipsoid.

    If you diagonalize the Ricci tensor, you will find the principal axes of this ellipsoid. A bit of imagination should convince you that these MUST be the same principal axes as the metric. A 2-surface, e.g., has two principal directions, which correspond directly to the eigenvectors of Ricci and g.

    In general, the Ricci tensor is not trace-free, which means that there is a dilation involved in addition to a change in shape. Therefore, depending on the trace of the Ricci tensor, a small sphere might have more or less surface area than it is supposed to at a given r (as an r^2 correction as the sphere grows). So, the trace of the Ricci tensor (i.e., the Ricci scalar) measures the deficit angle. This is why integrating the Ricci scalar over a 2-manifold gives the Euler number (which is equal to the total deficit angle divided by [itex]2\pi[/itex]).

    Once you can see that, you can translate things back to spacetime. If you look out from point P along a geodesic, you can imagine a small area element perpendicular to this ray. In Euclidean 3-space, this area element would grow simply as r^2. In curved space, the Ricci tensor gives you corrections to this, making it grow faster or slower. In Lorentzian metrics, this same basic idea holds. If you have some future-pointing vector [itex]n^\alpha[/itex], then

    [tex]R_{\alpha \beta}n^{\alpha} n^{\beta}[/tex]

    will tell you the rate of growth of 3-spheres as time progresses along [itex]n^\alpha[/itex]. If this is nonzero at P, then it means that space is being compressed or dilated (as a function of time) at P. That is, an extended object will feel pressure as the actual volume of space is changing.

    Note that the Ricci tensor is zero everywhere except inside a gravitating body. Ordinary tidal effects are caused by the Weyl curvature, which distorts bodies but does not change their volume.
  5. Feb 25, 2010 #4
    Re: "Visualizing" Ricci curvature

    Oh god. It seems so obvious now, but I really did forget that. If the stress energy tensor is zero then so is the Ricci tensor.

    Spacetime is curved outside a blackhole, despite all components of the Ricci tensor being zero. So the Ricci curvature is missing way too much to tell us about things like initially parallel lines crossing and such?

    I wasn't expecting that.
    So what curvature is more appropriate to use than the Ricci tensor? You mention the Weyl curvature. How can one solve for that from the Einstein field equations, if the Rici and stress-energy tensor are zero? Do we need some other constraining equations?
    Last edited: Feb 25, 2010
  6. Feb 25, 2010 #5

    Ben Niehoff

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    Re: "Visualizing" Ricci curvature

    In empty space, the only curvature is due to the Weyl tensor. The Weyl tensor is what you get when you take the Riemann tensor and subtract ALL of its traces. Therefore, the Weyl tensor governs changes in shape, but NOT changes in size.

    In general, the Riemann tensor contains all information about curvature. In empty space, the Riemann tensor is equal to the Weyl tensor.
  7. Feb 26, 2010 #6
    Re: "Visualizing" Ricci curvature

    Since Einstein's field equations only refer to the Ricci tensor, how do we get the full Riemann tensor?
  8. Feb 26, 2010 #7


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    Last edited: Feb 26, 2010
  9. Feb 26, 2010 #8


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    Re: "Visualizing" Ricci curvature

    R^r_{mqs}=\Gamma ^{r}_{mq,s}-\Gamma ^{r}_{ms,q}+\Gamma ^{r}_{ns}\Gamma ^{n}_{mq}-\Gamma ^{r}_{nq}\Gamma ^{n}_{ms}
  10. Feb 28, 2010 #9
    Re: "Visualizing" Ricci curvature

    Can you explain some more? I'm not understanding how that answers the question.

    In vacuum, since Ricci curvature is zero, the Einstein field equations don't seem to provide enough constraint on the metric to solve for it. And without the metric, how can we solve for the Christoffel symbols to get the full Riemann curvature as you suggest above?

    I'm clearly not understanding something basic here.
    It seems to come down to:
    If Einstein's field equations only deal with the Ricci curvature, what determines the Weyl curvature freedom?
  11. Mar 1, 2010 #10


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    Re: "Visualizing" Ricci curvature

    One can solve for the metric for a vacuum solution by making an ansatz for the metric, then solving for a zero Ricci einstein tensor. This is how the Schwarzschild metric was found. The components of the Riemann tensor may be not all zero, but cancel on contraction to give a zero Ricci. For a vacuum solution, the Weyl tensor is just the Riemann tenesor.

    See the attached reprint.

    Attached Files:

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