# 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 $$\partial_t$$?

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 "spatial evolution in the coordinates", 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.

Once you have the lapse and shift vectors, then it's just a matter of inverting your equation for $$t^{a}$$ to compute the normal vector.

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 $\partial_\phi$? Dec 2, 2020 at 22:53
• @aceituna the vector field I have written is not normalized, you can rescale it as you wish Dec 3, 2020 at 3:08
• of course, thanks! :) Dec 3, 2020 at 14:16