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Given the Kahn-Penrose metric: $$ ds^2=2dudv-(1-u)^2dx^2-(1+u)^2dy^2 $$ I calculated the Riemann Tensor and found that all elements equal 0.

Is there some symmetry principle by which I could have easily deduced zero curvature or other properties directly from the metric without actually doing the calculation?

EDIT: I would be grateful for an explanation about the connection between the curvature and symmetries of any metric, not just the Kahn-Penrose example.

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  • $\begingroup$ It seems wrong. The output for the Christoffel symbols is 0 for all, but even $\Gamma ^a_{aa}$ is non-zero. $\endgroup$ – MBN Jan 12 at 12:08
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One way to see that the curvature tensor is zero is to start with the fact that there is no dependence of metric components on null coordinate $v$. So performing Kaluza–Klein reduction along the Killing vector $\partial_v$ we would obtain a Newton–Cartan spacetime with two spatial coordinates ($x$ and $y$), while $u$ now becomes Galilean time coordinate.

The time evolution of spatial metric corresponds to the following law of motion for comoving observers: $$ X(u)=(1-u)\, X_0,\qquad Y(u)=(1+u)\,Y_0. $$ But these are equations for linear motion along a straight line. So there is no Newtonian gravity here and $(u,x,y)$-spacetime is just a reparametrization of an empty Galilean spacetime with zero NC curvature and correspondingly the original Kahn–Penrose metric (which would be the Bargmann lift of NC solution) would also have zero Riemann tensor.

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  • $\begingroup$ Thnx. (i tried to upvote but dont have enough reputation=( ). So are you saying that i would have to derive the laws of motion and then see that they are linear? $\endgroup$ – Daniel Jan 12 at 14:33
  • $\begingroup$ The laws of motion here give us transformation for Cartesian coordinates of Minkowski spacetime. But the fact that they correspond to linear motion of Galilean observers give us intuition about the physical meaning of such coordinates. $\endgroup$ – A.V.S. Jan 12 at 15:59

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