# Magnitude of basis vectors in Kerr metric

I am following some work in David McMahon's 'Relativity Demystified'.

In it, he gives an example of some line element in flat spacetime in polar coordinates

$$\mathrm ds^2 = \mathrm dr^2 + r^2 \mathrm d\theta^2 +r^2 \sin^2 \theta\, \mathrm d \phi^2 .$$

Since in a coordinate basis the basis vectors satisfy

$$e_a \cdot e_b = g_{ab}$$

for metric $$g_{ab}$$, it is then simple to get the magnitudes of the basis vectors (e.g. $$|e_{\phi}| = r \sin \theta$$)

That is all fine and makes sense to me.

Now, what happens if we instead consider the Kerr metric in Boyer Lindquist coordinates?

Specifically, what happens to the cross $$t,\phi$$ terms? Is it still true that $$e_{\phi} \cdot e_{\phi} = g_{\phi \phi}$$? The vierbein one-forms seem to suggest that this is no longer OK, but it is not clear to me why.

Given a manifold $$M$$ and a coordinate system $$x^\mu$$, the metric tensor $$g_{\mu \nu}$$ is given as the scalar product of the coordinate basis vectors $$e_\mu = \partial_\mu$$, that is $$g_{\mu \nu} = e_\mu \cdot e_\nu$$.
• So does it then follow that for the coordinate basis, $|e_{\phi}| = \sqrt{g_{\phi \phi}}$? – user1887919 Aug 3 '19 at 6:43
Typically, when picking a tetrad basis, you can't choose a tetrad that is ALSO a coordinate basis. In the case of a non-diagonal metric like the Kerr solution in Boyer-Linquist coordinates, you'll end up with tetrad vectors that have more than one component. Obviously, there are many choices you can make, but ordinary linear algebra tells you how you can go about choosing a set of (labeled by indices I=0 to 3) $$e^{\mu}{}_{I}$$ such that
$$e^{\mu}{}_{I}e^{\nu}{}_{J}g_{\mu\nu} = \eta_{IJ}$$
One common choice is to choose $$e^{\mu}{}_{0}$$ to be the timelike killing vector of the kerr spacetime and $$e^{\mu}{}_{3}$$ to be the normal vector to that one that also spans $$t, \phi$$ (which will be the angular killing vector), and then pick the two remaining normal vectors that have alternately zero $$\theta$$ and $$r$$ components.