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Curvature (Ricci things)

  1. May 14, 2010 #1
    If the Ricci scalar R happens to be zero (everywhere according to our metric), is that the definition of a 'flat' space time?

    And how are flat space times related to Minkowski space precisely? ARE they the SR space exactly?

    Thanks. Just trying to understand why 'flat space time' is a definition worth having around.

    (It's nice that R is actually a scalar (like, invariant) I suppose. Since like, it doesn't matter if we switch to a 'bent' metric to represent the universe - R isn't affected)
  2. jcsd
  3. May 14, 2010 #2


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    That is not the definition of a 'flat' space time. For example, the Scwarzchild spacetime has R=0 everywhere, but it is certainly not flat. A better definition would be a Riemann tensor that is everywhere 0.
  4. May 14, 2010 #3
    Ah, thanks.

    So you need each R_abcd=0.
    Is that the simplest test for 'flatness'?

    (Oopsie, I was following an example set in 2D space-time, where R_abcd has only one indep. component anyway...so the vanishing of R would imply all of R_abcd=0, which isn't usually the case)
  5. May 15, 2010 #4


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    A vanishing Ricci tensor only implies a vanishing Riemann tensor in 2 and 3 dimensions :)
  6. May 15, 2010 #5
    haushofer - I believe the above posters were referring to the curvature scalar, not the Ricci tensor itself.

    As for OP's question, you could also use the metric tensor in the region to determine the "flatness" of space; in the linearized regime, we typically say that the metric tensor is

    g_{uv} = \eta_{uv} + \kappa h_{uv}

    where the latter term is a small perturbation. Even if it is not small, one can still use this deviation from the Minkowski metric to get a decent idea of how warped the spacetime is.
    Last edited: May 15, 2010
  7. May 16, 2010 #6
    1- A spacetime is called "flat" if its Riemann tensor vanishes. This also means that both the Ricci tensor and Ricci scalar vanish equivalently. A very famous example of this is the Minkowski spacetime or the Rindler metric.

    2- If the curvature scalar vanishes so does the Ricci tensor and therefore the spacetime is said to be Ricci-flat. Examples of this are all spacetimes that satisfy the vacuum solutions of the Einstein's field equations. Here the Riemann tensor may be non-zero, so the spacetime is not flat of the above type.

    3- If the Ricci tensor vanishes only so does the curvature scalar and thus we are back to the second status. Here it is not necessary to have a vanishing Riemann tensor.

    Not really. Not all of the flat spacetimes are Minkowski-like spacetimes. If there is an explicit coordinate transformation (or are coordinate transformations) that can turn the metric into Minkowski, then both metrics are the same provided that the region wherein the metric is defined is given and actually there both spacetimes coincide. Remember that it is possible to see the dynamics of the two metrics differring from each other, but they definitely are two sprits in one body. Example of this is the Rindler metric. (See Wald's treatment in his book.) Nevertheless, all spacetimes satisfying 1 above are related to the Minkowski spacetime somehow.

  8. May 16, 2010 #7
    If you have a geometry that doesn't map to a Euclidean geometry and has a non-vanishing Ricci tensor, you have a non-euclidean geometry.

    Or, I believe you can say that if the curvature tensor vanishes, then covariant derivatives will commute, which indicates a coordinate system that can be transformed into a Minkowskian system.
    I think that's correct anyway....
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