9
$\begingroup$

I've seen the Kerr metric usually presented in the Boyer-Lindquist coordinates where there is a cross term in the $d\phi$ and $dt$ term. I've done a good bit of searching and cannot find any coordinates which express the Kerr metric in an orthogonal fashion. Is there ANY choice of coordinates that eliminates all cross terms/off-diagonal terms for the Kerr metric?

If not, is it just a fact that you can never find such a coordinate transformation because of the inherent geometry of the Kerr space-time?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ No. No such coordinate system exists $globally$. The cross-term in the Kerr metric describes a physically important property of the Kerr metric - that it is not static. In particular, the solution is not symmetric under $t \to - t$ (which it would be if you could find a diagonal coordinate system). $\endgroup$
    – Prahar
    Commented May 23, 2015 at 12:29

2 Answers 2

Reset to default
9
$\begingroup$

A static spacetime can be defined as one that is stationary and such that coordinates exist in which the metric is diagonal. It's interpreted as a spacetime that (1) is stationary, and (2) has no rotation. The Kerr metric is stationary and clearly has rotation, so it can't be static, and you won't be able to put it in diagonal form.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Could you please give a reference that motivates this definition? I'm rehabilitating my GR after 20 years and it sounds as though understanding this definition would be a good foundational thing to know. I'll ask it as a "real" question if it isn't simple a matter of a reference and if it's going to take you a little bit of work to explain. $\endgroup$
    – Selene Routley
    Commented Aug 8, 2013 at 1:54
  • $\begingroup$ @WetSavannaAnimalakaRodVance: You could try my own book: lightandmatter.com/genrel . IIRC this definition is the primary one used in Rindler, Essential Relativity: Special, General, and Cosmological, 1979. $\endgroup$
    – user4552
    Commented Aug 8, 2013 at 2:53
  • $\begingroup$ This doesn't seem right. Diagonal metric is not enough for a static spacetime. For example FRWL metric. $\endgroup$
    – MBN
    Commented Aug 8, 2013 at 12:15
  • $\begingroup$ @MBN: Oops, thanks for pointing out my mistake. It's also necessary for it to be stationary. I've tried to fix my answer. $\endgroup$
    – user4552
    Commented Aug 8, 2013 at 20:04
  • $\begingroup$ Georges Lemaître says from on high, "Now say three hail marys" :) Thoroughly wonderful book by the way - I can tell you enjoyed writing it a great deal and I was browsing it until the small hours of this morning. $\endgroup$
    – Selene Routley
    Commented Aug 9, 2013 at 1:07
1
$\begingroup$

Any metric is locally equivalent to a symmetric matrix, which is always diagonalizable. Just taking a reference system co-rotating with the black-hole and you will get a diagonal form to the Kerr metric.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.