I am trying to cross-check the Christoffel symbols and other very laborious geometric components in several metrics. In particular the Kerr metric is notoriously complex and results in expressions that are nontrivial to simplify and compare with tabulated results. I've been using two different references for checking Christoffel symbols, and I've found worrisome differences between the expressions. The first I started using was this one, however I was also checking this paper, as suggested by this answer.

One of the differences I spotted was with the off-diagonal $g_{03} = g_{t \phi}$ metric component: both sources seem to disagree by a factor of 2. In fact checking with CAS tools I've been able to confirm most of the Christoffel symbols of the Kerr metric with the first reference (not using the factor of 2 in the off-diagonal), and in fact the second reference agrees with the Christoffel symbols of the first reference, but not the metric component!

Also, on second reference, they report the inverse metric component $g^{00}$ as being equal to:

$$ - \frac{A}{\Sigma \Delta}$$

With the definitions $A = (r^2 + a^2)\Sigma + r_s a^2 r \sin^2(\theta)$ and the usual definitions of $\Sigma$ and $\Delta$ for which both references agree.

However I instead found the inverse metric (to the metric of 2nd reference with the factor of 2 on the off-diagonal) to be:

$$ g^{ 00 } = -\frac{ {( \Sigma a^{2}+ \sin(\theta)^{2} r_s r a^{2}+ \Sigma r^{2})} \Sigma}{ \Sigma^{2} r^{2}- \Sigma r_s r^{3}+3 \sin(\theta)^{2} r_s^{2} r^{2} a^{2}- \Sigma r_s r a^{2}+ \Sigma \sin(\theta)^{2} r_s r a^{2}+ \Sigma^{2} a^{2}} $$

I double-check that the result, multiplied by the metric does indeed equal to the 4x4 identity matrix. I add below the rest of the inverse metric components I obtained (sorry for the unwieldy expressions, I couldn't find ways to simplify any further)

$$ g^{ 03 } = -2 \frac{ \Sigma r_s r a}{ \Sigma^{2} r^{2}- \Sigma r_s r^{3}+3 \sin(\theta)^{2} r_s^{2} r^{2} a^{2}- \Sigma r_s r a^{2}+ \Sigma \sin(\theta)^{2} r_s r a^{2}+ \Sigma^{2} a^{2}} $$

$$ g^{ 11 } = \frac{\Delta}{\Sigma} $$

$$ g^{ 22 } = \frac{1}{\Sigma} $$

$$ g^{ 33 } = -\frac{ {( r_s r-\Sigma)} \Sigma}{ \Sigma^{2} \sin(\theta)^{2} a^{2}- \Sigma \sin(\theta)^{2} r_s r^{3}- \Sigma \sin(\theta)^{2} r_s r a^{2}+ \Sigma^{2} \sin(\theta)^{2} r^{2}+3 \sin(\theta)^{4} r_s^{2} r^{2} a^{2}+ \Sigma \sin(\theta)^{4} r_s r a^{2}} $$

All the other components are either $0$ or obtained by symmetry of the indices.

I also find other disagreements with the contravariant metric expressions but it is not helpful to write them here as it would only add clutter.

Unlike the Christoffel symbols, it is much easier to verify that the inverse metric is incorrect.

Can you help me verify where the error might be?

  • $\begingroup$ Are you aware that in line element form you have $ds^2 =\ldots + (g_{t\phi}+g_{\phi t}) dtd\phi + \ldots$ for the off-diagonal terms? $\endgroup$
    – TimRias
    Commented Jul 3 at 16:16
  • $\begingroup$ @TimRias, agreed, but that is already being accounted for when doing the contraction sums $\endgroup$ Commented Jul 3 at 16:20
  • 1
    $\begingroup$ Keeping that in mind both sources have exactly the same Kerr metric. $\endgroup$
    – TimRias
    Commented Jul 3 at 16:22
  • $\begingroup$ you are right @TimRias, in fact the 2nd source is writing it as a line element, so they both agree. However the contravariant metric still seems to disagree.. $\endgroup$ Commented Jul 3 at 16:23
  • $\begingroup$ @TimRias, check eq. 2.14.2 of 2nd reference $\endgroup$ Commented Jul 3 at 16:27

1 Answer 1


UnkemptPanda wrote: "both sources seem to disagree by a factor of 2."

The factor of 2 is wrong in some sources where they forgot that the crossterms are added twice in the ds² of the line element since gtφ and gφt are the same, but if you take them from here you have the input and the output so you can check for yourself that everything's correct.

  • $\begingroup$ yes, apparently I got burned by the factor of 2 in the contravariant line form $\endgroup$ Commented Jul 3 at 16:30

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