4
$\begingroup$

Initial considerations

Consider a metric of the form

$$ ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}=-\alpha^2(r,t)\,dt^2+a^2(r,t)\,dr^2+r^2\,b^2(r,t)\,d\theta^2+r^2\,b^2(r,t)\,\sin^2\theta\,d\varphi^2\ .\quad (1) $$

This metric is written in terms of the coordinates $x^\mu=(\,t\,,\,r\,,\,\theta,\,\varphi\,)$. Then, by looking at the metric, we know that $g_{\mu\nu}=g_{\mu\nu}(t,r,\theta)$. Thus, there is a Killing vector

$$ \eta^{\mu}=(\,0\,,\,0\,,\,0\,,\,1\,)\Rightarrow\eta_{\mu}=g_{\mu\nu}\eta^\nu=(\,0\,,\,0\,,\,0\,,r^2\,b^2\,\sin^2\theta\,)\ ,\quad (2) $$

which is the Killing vector associated with rotations around the $\varphi$ axis (from what I understand). The norm of this Killing vector, $X$, is defined as

$$ X\equiv\eta_\mu\eta^\mu=g_{\mu\nu}\eta^\mu\eta^\nu=r^2\,b^2\,\sin^2\theta\ .\quad (3) $$

Then, consider the condition

$$ \left.\frac{\partial_{\mu}(X)\partial^\mu(X)}{4X}\right|_{r\to r_0}=1\ .\quad (4) $$

This condition, as stated in equation (19.3) of Exact Solutions of Einstein's Field Equations, by Stephani et al, is said to indicate that the spacetime is regular at the considered point, $r_0$. For the case of the metric (1) above and the Killing vector (2), condition (4) holds at $r_0=0$ as long as (I will include the calculation below but will finish the question first as to avoid interrupting the narrative)

$$ a(0,t)=b(0,t)\ .\quad (5) $$

Question

What is the physical meaning of the object on the LHS of (4) and why does that condition imply regularity at the point $r_0$?

Appendix: Going from (4) to (5)

We have

\begin{align*} \partial_{\mu}X&=\left(\,2r^2b\dot{b}\,,\,2\sin^2\theta\,(rb^2+r^2bb^\prime)\,,\,2r^2b^2\sin\theta\cos\theta\,,\,0\,\right)\ ,\\ \partial^{\mu}X&=\left(\,-\frac{2r^2b\dot{b}}{\alpha^2}\,,\,\frac{2\sin^2\theta\,(rb^2+r^2bb^\prime)}{a^2}\,,\,2\sin\theta\cos\theta\,,\,0\,\right)\ , \end{align*}

then

\begin{align*} \frac{\partial_\mu(X)\partial^\mu(X)}{4X}&=\frac{1}{4r^2b^2\sin^2\theta}\left\{-\frac{4r^4b^2\dot{b}^2}{\alpha^2}+\frac{4\sin^4\theta(rb^2+r^2bb^\prime)^2}{a^2}+4r^2b^2\sin^2\theta\cos^2\theta\right\}\\ &=\sin^2\theta\left[\frac{b(r,t)}{a(r,t)}\right]^2+\cos^2\theta+\mathcal{O}(r)\ . \end{align*}

Then in the limit $r\to0$

\begin{align*} \left.\frac{\partial_{\mu}(X)\partial^\mu(X)}{4X}\right|_{r\to 0}=\sin^2\theta\left[\frac{b(0,t)}{a(0,t)}\right]^2+\cos^2\theta\ .\quad (\star) \end{align*}

Demanding that $(\star)=1$ then implies

$$ b(0,t)=a(0,t)\ . $$

$\endgroup$
  • $\begingroup$ It might be simpler to ask this question in two dimensions, getting rid of the $\theta$ and $t$ coordinates entirely. The regularity condition you get at the end would reduce to the statement that the metric, near the origin, is a rescaling of the usual flat metric in polar coordinates, $\sim a(r, \phi)^2 (dr^2 + r^2 d \phi^2)$. I've always found physical meanings a bit elusive in GR, but you may be able to rephrase (4) in freshman physics language and centripetal acceleration. Using the definition of a Killing vector, $\partial_\mu X = -2 \eta^\nu \nabla_\nu \eta_\mu$. $\endgroup$ – user2309840 Oct 25 '17 at 3:07
  • $\begingroup$ While I understand the consequences of condition (4), I have no idea what motivates it. If you tell me the motivation is because we want to impose flatness at the origin, then my question is rephrased as "why is (4) the way to do that? Where does it come from?". $\endgroup$ – L. Werneck Oct 26 '17 at 12:29

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.