From Wikipedia's entry on Ricci tensor,

In differential geometry, the Ricci curvature tensor, named after Gregorio Ricci-Curbastro, represents the amount by which the volume of a geodesic ball in a curved Riemannian manifold deviates from that of the standard ball in Euclidean space.

Given we have a metric describing spacetime, does this mean that (a) there is an equation that tells you how your volume changes in curved space with respect to flat space and does this also imply that (b) the volume of the ball does not differ from that of a Euclidean ball when $R_{\mu\nu}=0$?

Just to clarify, I do not know what a geodesic ball is and I have tried to read it up but it is still a little fuzzy... I am just assuming that it is a curved space analogue of a Euclidean ball.

  • $\begingroup$ No it isn't enough, since you didn't say if you are talking about spacetime or not. The definition is different for a Lorentzian manifold and a Reimannian manifold. $\endgroup$
    – Timaeus
    Commented Oct 2, 2015 at 17:37
  • $\begingroup$ @Timaeus Is it not? Then I'll make the appropriate changes. Curiously why are they not the same... $\endgroup$
    – Horus
    Commented Oct 2, 2015 at 17:39
  • $\begingroup$ Maybe this will help pas.rochester.edu/~rajeev/phy413/Grav25.pdf ? $\endgroup$
    – bolbteppa
    Commented Oct 3, 2015 at 13:32

1 Answer 1


In a Riemannian manifold such as the surface of a sphere you could travel a distance $\epsilon$ in every direction and note that you get a different amount of surface area or volume than if you did so in a flat space. For instance if you walked a distance equal to the distance from the north pole and the south pole ($2\pi R$) you get an area of $4\pi R^2,$ whereas if you walked that same distance in a flat space you get $\pi (2\pi R)^2$ which is larger.

For a Lorentzian manifold such as a spacetime it is a trace of the Riemann curvature tensor. Both of these things and the fact that they are different are mentioned in the wikipedia page you cite.

edit to respond to edited question The full quote is

In differential geometry, the Ricci curvature tensor, named after Gregorio Ricci-Curbastro, represents the amount by which the volume of a geodesic ball in a curved Riemannian manifold deviates from that of the standard ball in Euclidean space. As such, it provides one way of measuring the degree to which the geometry determined by a given Riemannian metric might differ from that of ordinary Euclidean n-space. The Ricci tensor is defined on any pseudo-Riemannian manifold, as a trace of the Riemann curvature tensor. Like the metric itself, the Ricci tensor is a symmetric bilinear form on the tangent space of the manifold

Emphasis added.

Given we have a metric describing spacetime,

Not a Riemannian metric, just a nondegenerate symmetric bilinear form with a Lorentzian signature. What in GR you call a metric, but not what a mathematician would think you meant if you just said the word metric out of context, so not necessarily what they are talking about if they just say the word metric.

(b) the volume of the ball does not differ from that of a Euclidean ball when $R_{\mu\nu}=0$?

There is no obvious ball in spacetime, since intervals can be positive or negative. So what would be the ball about some event? Would it include every event that is lightlike separated, every timelike separated event, every spacelike one? One of those and then some more too?

If your goal is to understand the meaning of the Einstein Equation, there are better sources.

  • $\begingroup$ Hmm I see. This answer answers my question in the comments but it does not seem to answer my actual question. I may be a little slow since it is 2am here. $\endgroup$
    – Horus
    Commented Oct 2, 2015 at 17:56
  • 1
    $\begingroup$ @Horus Edited. But your question is still flawed because you read a bad definition (how can a rank two tensor be a ratio of volumes) that was not ever intended to apply to Lorentzian manifolds and then you say weird things about a non definition. Yes the volume you get when you go in all directions a certain distance is different when your Riemannian manifold is curved. And it doesn't even make sense in spacetime because some directions are spacelike and others are timelike. So you go back to the real definition for the Riemann tensor and have some traces of it that agree with other things. $\endgroup$
    – Timaeus
    Commented Oct 2, 2015 at 18:08
  • $\begingroup$ Hmm I see. I think I understand where my confusion came from. So if the wiki definition is bad, then what does the Ricci tensor describe then? $\endgroup$
    – Horus
    Commented Oct 2, 2015 at 18:13

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