Choptuik pioneered numerical studies of the gravitational collapse of massless scalar fields in spherically symmetric, asymptotically flat spacetime. In his papers (e.g. https://arxiv.org/pdf/gr-qc/9803075.pdf ), he chooses the metric

$$ds^2=-\alpha^2(r,t)\,dt^2+a^2(r,t)\,dr^2+r^2\,d\Omega^2\, ,\quad (1)$$

in 4 (3+1) spacetime dimensions. The metric above is a particular choice of coordinates of the more general metric

$$ds^2=\left[-\alpha^2(r,t)+\beta^i\beta_i\right]\,dt^2+2\beta_i\,dt\,dx^i +a^2(r,t)\,dr^2+r^2b^2(r,t)\,d\Omega^2\, ,\quad (2)$$

where $\alpha$ is the lapse function and $\beta$ is the shift vector. Now, Choptuik (and other authors) states that there is an elementary flatness condition which reads $$g^{\mu\nu}\partial_\mu(rb)\partial_\nu(rb)|_{r=0}=0\Rightarrow a(0,t) = b(0,t)\,.$$ I've searched for references on this but I do not understand where this condition comes from nor how $a=b$ follows from it. Here is my attempt:

I consider a coordinate system where $\beta_i=(\beta(r,t),0,0)$, i.e. the shift vector only has a radial component. It then follows that,

$$g^{tt}=-\alpha^{-2}\ ,\quad g^{rt}=g^{tr}=\frac{\beta}{\alpha^2}\quad \text{and}\quad g^{rr}=a^{-2}\ ,$$

so that

$$g^{\mu\nu}\partial_\mu(rb)\partial_\nu(rb)=\left[\frac{(rb)^\prime}{a}\right]^2+2\frac{\beta}{\alpha^2}(r\dot{b})(rb)^\prime-\left[\frac{r\dot{b}}{\alpha}\right]^2\ .$$

To take the limit $r\to0$, I then take into account the regularity conditions

$$\alpha^\prime(0,t)=b^\prime(0,t)=a^\prime(0,t)=\beta(0,t)=0\ ,$$

but these seem to imply that

$$\left.\left(\frac{b}{a}\right)^2\right|_{r=0}=0\ ,$$

which doesn't make any sense. Any help is greatly appreciated!

  • $\begingroup$ You seem to be using $\beta_i\beta^i=0$; why is that? $\endgroup$ – Demosthene Oct 17 '17 at 12:59
  • $\begingroup$ From what I understand, we can always assume $\beta_i(r,t)=(\beta(r,t),0,0)$, i.e. $\beta_i$ only has a radial component. Then, choosing $\beta(r,t)=0$ is a particular coordinate condition (not really sure what it means) that ensures maximal slicing of the spacetime. To go from (2) to (1), we must further impose the condition $b(r,t)=1$ which ensures that surfaces of constant $r$ and constant $t$ have proper area $4\pi r^2$. $\endgroup$ – L. Werneck Oct 17 '17 at 15:45
  • $\begingroup$ There was a typo in the references I've found and the correct condition is $g^{\mu\nu}\partial_{\mu}(rb)\partial^{\mu}(rb)|_{r=0}=1$, which then implies $a(0,t)=b(0,t)$ as desired and computed above. $\endgroup$ – L. Werneck Oct 23 '17 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.