1. Prove or disprove that the Kerr metric can be expressed in a set of orthogonal coordinates over some coordinate chart.

Motivation for this question stems from my understanding that a metric can always be written in orthogonal coordinates if it exists on a flat spacetime. A metric written in an orthogonal set of coordinates has no off-diagonal terms.

As an example, the Kerr metric always seems to have at least one off-diagonal term. I understand physically why the off-diagonal terms are present in specific coordinates, but people seem to take it as a fact that it can never be transformed away or they say it is obvious that it can be.

So, since the Kerr spacetime has curvature, it makes sense that the Kerr metric cannot be written in a global set of coordinates. However, why can the metric not be written in a set of orthogonal coordinates?

Is it as simple as using a co-rotating timelike observer near infinity to be able to transform away the off-diagonal term(s) of the Kerr metric?

This question was similarly asked here, but the answers were unsatisfying because they are all stated as fact without citation or proof.

Lastly, would this be better for the Mathematics Stack Exchange? Is it purely a question of differential geometry?

  • $\begingroup$ Can you explain a little what it means to have "a set of globally orthogonal coordinates" to you? I would like to understand the context a little more. Generally speaking, for a curved space time, one can not find a (single) "global set of coordinates" at all - one can only find a set of overlapping coordinate charts which form an atlas over the manifold. Every coordinate chart will have singularities (or otherwise break down) somewhere. $\endgroup$ – enumaris Sep 25 '18 at 22:57
  • $\begingroup$ I have edited my question, after reading from mathpages.com/rr/s6-01/6-01.htm and which provides a heuristic for why there in principle is no global set of coordinates for non-flat spacetimes. This article provides a quote on the matter by Einstein on page 1 fizipedia.bme.hu/images/2/2e/05_GlobalLocalMetrics3.pdf $\endgroup$ – N. Steinle Sep 26 '18 at 2:41
  • $\begingroup$ So, I understand that the Kerr spacetime, since it has curvature, has no global set of coordinates, meaning the first question is bunk. Then my question regarding such a local set of coordinates still remains. What I think I mean by "a set of locally orthogonal coordinates" is that the metric can be expressed in these coordinates with only terms on its diagonal over some well defined coordinate chart (hence local). $\endgroup$ – N. Steinle Sep 26 '18 at 2:48
  • $\begingroup$ Please inform me if and why that's a poor definition. $\endgroup$ – N. Steinle Sep 26 '18 at 2:48

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