# Normal vector field of costant time Kerr slice

The the Kerr metric of the Kerr spacetime in Boyer-Lindquist coordinates is given by

$$ds^2=-\left(1-\frac{2mr}{\Sigma}\right)dt^2+\frac{\Sigma}{\Delta}dr^2+\Sigma d\theta^2+\left(r^2+a^2+\frac{2mra^2}{\Sigma}\mathrm{sin}^2\theta\right)\mathrm{sin}^2\theta\, d\phi^2-\frac{4rma\mathrm{sin}^2\theta}{\Sigma}dtd\phi.$$

I am considering a slice of constant time in the Kerr spacetime, i.e. a spatial hypersurface with induced Riemannian metric, where the $$dt$$-components of the metric from above vanish:

$$ds^2=\frac{\Sigma}{\Delta}dr^2+\Sigma d\theta^2+\left(r^2+a^2+\frac{2mra^2}{\Sigma}\mathrm{sin}^2\theta\right)\mathrm{sin}^2\theta\, d\phi^2.$$

My question is: what is the normal vector field in the spacetime to this hypersurface? Is it as simple as normalizing the vector field $$\partial_t$$?

My question is: what is the normal vector field in the spacetime to this hypersurface? Is it as simple as normalizing the vector field ∂t?

$$\partial_t$$ is not orthogonal to $$\partial_\phi$$. You need new vector field $$V=\partial_t+f\partial_\phi$$, such that $$g(V,\partial_\phi)$$ is zero ($$g(V,\partial_\theta)$$ and $$g(V,\partial_r)$$ are zero trivially), i.e. $$0=g_{t\phi}+fg_{\phi\phi}\Rightarrow f=\frac{-g_{t\phi}}{g_{\phi\phi}}$$

• Thanks a lot, only one thing: how does one know that the vector field $V$ has the factor $1$ for $\partial_t$ and not a different one, like $f$ for $\partial_\phi$? – aceituna Dec 2 '20 at 22:53
• @aceituna the vector field I have written is not normalized, you can rescale it as you wish – Umaxo Dec 3 '20 at 3:08
• of course, thanks! :) – aceituna Dec 3 '20 at 14:16

You're going to have to be careful with the whole 3+1 formalism in the case where you have nontrivial lapse and shift. (which is true here).

In that language, your 3-metric is given by:

$$\gamma_{ab} = n_{a}n_{b} + g_{ab}$$

where $$n_{a}$$ is your unit normal to the hypersurface found by choosing $$t =$$const. But, because your time variable and one of your angle variables are not orthogonal, time evolution will involve frame-dragging effects, and you have your time evolution vector equal to $$t^{a} = -\alpha n^{a} + \beta{a}$$, where $$\alpha$$ and $$\beta^{a}$$ are the lapse function and shift vector. Since $$t$$ has to be compatible with the condition $$t^{a}\nabla_{a}t = 1$$, this must mean that $$\alpha = \frac{1}{\sqrt{-t^{a}t_{a}}}$$

Now, that we've associated the time evolution with the $$t^{a}$$ vector, we can decompose

$$ds^{2} = g_{ab}dx^{a}dx^{b}$$

into

$$ds^{s} = g_{ab}\left(t^{a}dt + \gamma^{a}_{c}dx^{c}\right)\left(t^{b}dt + \gamma^{b}_{d}dx^{d}\right)$$

Which then becomes:

$$ds^{2} = -\left(\alpha^{2} - \beta^{i}\beta_{j}\right)dt^{2} + 2dt\,dx^{i}\beta_{i} + \gamma_{ij}dx^{i}dx^{2}$$

From which you can just read off the values of the Lapse and shift vectors compatible with your starting metric and choice of timelike foliation.