What is an example for a spacetime that is non-stationary that is considered as a description of something in nature?

So far all the spacetimes I encounted have always been stationary (Schwartzschild, FRW, Kerr, etc.).

  • $\begingroup$ The most general, non-static axially symmetric metric can be reduced to the form $ds^2=-e^\alpha dr^2-e^\beta d\theta^2-e^\gamma d\phi^2 + e^\delta dt^2$ where $\alpha,\beta,\gamma$ and $\delta$ are functions of $r,\theta,t$ alone. $\endgroup$ – JamalS Jun 12 '14 at 9:02
  • $\begingroup$ @JamalS Kerr is non-static, axisymmetric and does have off-diagonal terms. $\endgroup$ – auxsvr Jun 12 '14 at 14:59
  • $\begingroup$ @auxsvr: link.springer.com/article/10.1007%2FBF02450444 $\endgroup$ – JamalS Jun 12 '14 at 15:00
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    $\begingroup$ @JamalS This is wrong, quoting from Wald, p. 119: "A stationary but nonstatic metric unavoidably must have $dt dx^\mu$ cross terms in any coordinate system which uses the Killing parameter as one of the coordinates" and $t$ is a Killing parameter for a stationary metric. We infer that the inexistence of hypersurfaces orthogonal to the orbits of isometry generated by the Killing vector $\partial_\phi$ implies that you must have cross terms of the form $d \phi dx^\mu$ in the metric. The cross terms have physical significance, you cannot remove them in general. $\endgroup$ – auxsvr Jun 12 '14 at 17:06
  • $\begingroup$ Also, see this for details. $\endgroup$ – auxsvr Jun 12 '14 at 17:07

The simplest physical example is just given by FRW (differently from what you wrote!) models describing the large scale of the universe: $$ds^2 = -d\tau^2 + a(\tau)^2 d\Sigma^2\:.$$ If $a$ is not constant (as it happens in our universe) $\frac{\partial}{\partial \tau}$ is not a Killing vector. In general, these metrics do not admit local or global timelike Killing vectors (even different from $\frac{\partial}{\partial \tau}$). Another case is the spacetime of a collapsing star in the absence of spherical symmetry (otherwise Birkhoff's theorem would imply that the metric is stationary in a region).

  • $\begingroup$ Birkhoff not Birkov $\endgroup$ – MBN Jun 12 '14 at 12:53

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