0
$\begingroup$

Let's restrict to the radial direction, so the metric can be expressed as

$ds^2=-(1-r_S/r)dt^2+(1-r_S/r)^{-1}dr^2$

with $r_S$ the Schwarzchild radius. Expressed in Kruskal coordinates, the metric is

$ds^2=4(r_S^3/r)e^{-r/r_S}(-dT^2+dR^2)$

Further coordinate transformation brings the infinity into finite coordinate values:

$\tan\frac{\tau+\rho}{2}=T+R=\sqrt{r/r_S-1}e^{(r+t)/2r_S}$

$\tan\frac{\tau-\rho}{2}=T+R=\sqrt{r/r_S-1}e^{(r-t)/2r_S}$

You can verify that the ranges of $\tau$ and $\rho$ are

$\tau\in(-\pi/2,\pi/2), \rho\in(-\pi,\pi)$

As we know that, the spatial infinity is defined at two points: $\tau=0, \rho=\pm\pi$. Now,I want to verify that all spacelike curves ($t(r),r$) parametrized by $r$ terminate at the spatial infinity and finally, I end up with counterexamples.

Because the curve $(t,r)$ is spacelike, its length is positive,

$-(1-r_S/r)(\frac{dt}{dr})^2+(1-r_S/r)^{-1}>0\Rightarrow \frac{dt}{dr}<(1-r_S/r)^{-1}$

only considering a outgoing spacelike curve outside of the horizon. So

$dt<\frac{dr}{1-r_S/r}=dr_*$

Here, $r_*$ is the tortoise coordinate. We can assume $dt=dr_*+f(r)dr$. $f(r)$ should be positive definite and not a constant function. Now, in terms of $f(r)$, we have

$\tan\frac{\tau+\rho}{2}=(r/r_S-1)e^{r/r_S}\exp[r_S^{-1}\int^r_{r_0} f(r')dr']$

$\tan\frac{\tau-\rho}{2}=-\exp[-r_S^{-1}\int^r_{r_0} f(r')dr']$

We can thus consider the behavior of the above express at $r\rightarrow\infty$. Defintely, the first expression is $+\infty$, but since $\int^r_{r_0} f(r')dr'$ is not necessarily infinite (for example, $f(r)=1/r^2$), the second expression is not necessarily $-\infty$.

Therefore, we cannot conclude that all spacelike curves terminate at the spatial infinity $i_0$ in the case of a Schwarzchild spacetime.

But all the books state the opposite. What is wrong with my calculation?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Do you really mean all spacelike curves, not all spacelike geodesics, or all spacelike inextensible curves, or all complete spacelike geodesics? If the question is really about all spacelike curves, then there are easy counterexamples, including closed spacelike curves, the spacelike curve from Chicago to LA, and spacelike curves that terminate on the singularity. But all the books state the opposite. What books do you have in mind? Please quote exactly what they claim is true. $\endgroup$
    – user4552
    Commented Nov 18, 2014 at 0:14
  • $\begingroup$ I would guess that the correct claim is that all spacelike geodesics terminate at $i^0$, in which case your counterexample might fail because it probably isn't a geodesic. $\endgroup$
    – user4552
    Commented Nov 18, 2014 at 0:27
  • 1
    $\begingroup$ I was such a careless reader. Spacelike geodesics terminate at $i_0$. I just worked out the condition for a spacelike geodesics with the help of Killing vector $\partial_t$, that is, $g_{\mu\nu}(\partial_t)^\mu s^\nu=Const.$. Here, $s^\mu=(\frac{dt}{dr},1)$. So it turns out $t\propto r_*$, but because of being spacelike, $t=kr_*$ with $k<1$. So $\tan\frac{\tau-\rho}{2}=-(r/r_S-1)^{(1-k)/2}e^{(1-k)r/2r_S}\rightarrow -\infty$. Yes, indeed, the spacelike geodesics terminate at $i_0$. $\endgroup$
    – Drake Marquis
    Commented Nov 18, 2014 at 2:30

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.