I was reading up on the Kerr metric (from Sean Carroll's book) and something that he said confused me.

To start with, the Kerr metric is pretty messy, but importantly, it contains two constants - $M$ and $a$. $M$ is identified as some mass, and $a$ is identified as angular momentum per unit mass. He says that this metric reduces to flat space in the limit $M \rightarrow 0$, and is given by $$ds^2 = -dt^2 + \frac{r^2 + a^2 \cos^2\theta}{r^2 + a^2}dr^2 + \left(r^2 + a^2 \cos^2\theta \right)d\theta^2 + \left(r^2 + a^2 \right)\sin^2\theta d\phi^2 $$

and $r$, $\theta$ and $\phi$ are regular spherical polar co-ordinates.

But I don't understand why this space is obviously flat. The Schwarzschild metric also contains terms involving $dt^2$, $dr^2$, $d\theta^2$ and $d\phi^2$ but is curved. I always thought that a metric with off diagonal elements implied a curved space, but clearly I was very wrong.

Question: How do you tell if a metric is curved or not, from it's components?

  • $\begingroup$ Are you sure you wrote the metric down correctly? It doesn't seem to match Carroll 6.75. $\endgroup$ – joshphysics Apr 8 '13 at 16:23
  • $\begingroup$ Yup... I'd inverted the co-efficient of $dr^2$. Thanks. :) $\endgroup$ – Kitchi Apr 8 '13 at 17:07
  • $\begingroup$ Hopefully someone can correct me if I'm wrong (cough @joshphysics cough), but it's also important that $a \rightarrow 0$ --- which is required by $M \rightarrow 0$. $\endgroup$ – DilithiumMatrix Apr 8 '13 at 17:14
  • $\begingroup$ @zhermes - I believe Carroll defines $a = J/M$, $J$ is total angular momentum. So when $M \rightarrow 0$, we keep $a$ fixed. $\endgroup$ – Kitchi Apr 8 '13 at 17:16
  • $\begingroup$ @zhermes I agree with Kitchi that $a$ is being kept fixed in the limit $M\to 0$ being taken by Carroll. $\endgroup$ – joshphysics Apr 8 '13 at 20:13

You tell if a space (or spacetime) is curved or not by calculating its curvature tensor. Or more unambiguously one of the curvature scalars (e.g. Ricci, or Kretschmann) since these don't depend on the coordinate system, but all of the information in the scalars is also contained in the Riemann tensor.

It is not necessarily obvious whether a given metric is curved or flat. You can take a perfectly flat spacetime and express it in some bizarre coordinate system, in which the metric has nonconstant off-diagonal terms. It's a simple exercise to take flat space and use the tensor transformation laws for the metric, with some arbitrary weird coordinate transformation that you just made up. You'll see what I mean.

  • $\begingroup$ Is it curvature scalars because calculating the tensors are harder? I thought the condition that $R=0$ corresponded to a sourceless metric? Or is that wrong as well? $\endgroup$ – Kitchi Apr 8 '13 at 15:45
  • 1
    $\begingroup$ It's just that the values of the scalars are independent of coordinates. It is possible for the curvature scalars to vanish and yet the full curvature tensor doesn't vanish. An example of the latter situation is the Schwarschild solution in GR. You're right, $R=0$ corresponds to source-free solutions in GR, which need not be flat. It is the full curvature tensor that tells you whether a space is flat or not, and indeed the curvature tensor for Schwarschild spacetime is nonzero. But $R\neq 0$ implies curvature unambiguously. $\endgroup$ – Michael Brown Apr 8 '13 at 15:55
  • $\begingroup$ Actually, just remembered, $R=0$ in GR corresponds to any source with $\rho - 3 p = 0$ (from the trace of $T_{\mu\nu}$). The general source free situation is $R_{\mu\nu}=0$, which of course implies $R=0$ as well. $\endgroup$ – Michael Brown Apr 8 '13 at 16:00

The flat space time refers here to the spacetime of Minkowski written with the spherical coordinates (I think one of your sign is wrong in your equation) $$ ds^2 = -dt^2 + dr^2 + r^2 d\theta^2 + r^2 sin^2 \theta d\phi^2. $$ where the metric is diagonal and has constant coefficients $g_{\mu \nu} = ( -1,1,1,1)$. I would say that the conditions for a flat space-time, regarding only its metric, refer to its diagonal shape and constant (at least constant)

  • 3
    $\begingroup$ that's not right. you can find transformations of the coordinates $t,x,y,z$ where the metric becomes non-diagonal - but it still describes the same Minkowskian space. As for entries being constant - the example you provide already has coefficients that vary depending on the coordinates! $\endgroup$ – nervxxx Apr 8 '13 at 14:37

In the limit where $M \to 0$, the Kerr metric reduces to the spherical coordinates form of the Minkowskian metric. In that sense, we recognize it and say it is 'obvious' that is is flat. (The Schwarzschild metric is also flat in the limit $M \to 0$.)

But to show that any given metric is curved or not we have to calculate a curvature invariant. For example usually we calculate the Ricci curvature $R= R^i{}_i = R^{ki}{}_{ki}$ where the first $R$ is the Ricci curvature, the second $R$ the Ricci curvature tensor and the third $R$ the Riemann tensor. If it's $0$ the space is curved, otherwise it's not. Carroll has it in his book.

  • $\begingroup$ This isn't quite right - if the Riemann curvature tensor is identically zero than the metric is flat. But it can be curved and still have contractions $R_{\mu \nu} = 0$ or $R = 0$. $R_{\mu \nu} = 0$ only tells you that there isn't any stress-energy at that point in spacetime, but there could still be strong curvature (for example, the Schwarzchild metric has $R_{\mu \nu} = 0$ everywhere except for the singularity), and $R = 0$ only tells you that the stress-energy tensor is traceless at that point (for example, this is true in any spacetime filled with a radiation-dominated fluid). $\endgroup$ – tparker Jun 22 '16 at 5:22

Carroll points out that, after the (a=fixed,M->0) limit "we recognize the spatial part of this as flat space in ellipsoidal coordinates", so in order to had realized that it was a flat space by inspection you should have been aware of the aspect of a flat metric in ellipsoidal coordinates, touche.


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