2
$\begingroup$

I do still not really understand the solution for a radially falling body in the Schwarzschild metric:

One the one hand, in terms of proper time $\tau$ the equation of motion can be solved and gives a solution $r(\tau)$ for $r\ge 0$. The body crosses Schwarzschild radius $r_s$ without any problems and the solution is valid also for $r\le r_s$.

On the other hand, the same calculation, carried out in coordinate time t shows, that the particle is never reaching $r_s$ in finite time.

The body definitely crosses the Schwarzschild boundary (although is cannot be proven, because nobody can come back), but why is there no solution for $r\le r_s$? What is the deeper reason, that this region is not mapped by using t? I would expect, that a switch to a new coordinate system maps a point into another unique point. For points inside $r_s$ this seems to be not possible anymore, but, on the other hand, I can easily mark an inside point in the t-r-Diagram and the Schwarzschild Metric is, from its definition, not restricted to $r>r_s$.

I know, that its an artifact due to geometry, but how can it be, that the mapping of inside points between the two coordinate systems fails? Naively speaking, there should be a solution also for the region within Schwarzschild radius because moving astronauts COULD get there and make a meeting. This Meeting, taken as a real event is not visible in the t-r- coordinates. Thinking about that makes me crazy...

$\endgroup$
3
  • 1
    $\begingroup$ youtube.com/watch?v=vNaEBbFbvcY $\endgroup$ Commented Aug 9, 2021 at 10:59
  • 2
    $\begingroup$ In many coordinate systems, falling bodies reach the Schwarzschild radius in finite coordinate time. Conversely even in flat spacetime, you can choose a coordinate system (which, like traditional Schwarzschild coordinates, don't cover the whole spacetime) so that an object that does pass through an event $p$ doesn't reach that event in any finite coordinate time. Coordinate systems are arbitrary. Physically meaningful statements are independent of which coordinate system we use. Are you asking why a hovering observer never sees a falling body reach the Schwarzschild radius? $\endgroup$ Commented Aug 9, 2021 at 13:11
  • 1
    $\begingroup$ Related physics.stackexchange.com/questions/556746/… $\endgroup$
    – Mauricio
    Commented Sep 8, 2021 at 11:16

1 Answer 1

3
$\begingroup$

The causal structure, defined by light cones can be shown in t-r plane. The slope of the cones given by

\begin{equation}\tag{1} \frac{dt}{dr} = \pm \frac{1}{(1-\frac{r_S}{r})} \end{equation}

increases to infinity for $r\rightarrow r_S$. (first picture below) Hence light rays asymptotically 'reaches' Schwarzschild radius in this coordinate system. The idea of tortoise coordinate is to make $\frac{dt}{dr}$ smaller. Just by integrating (1), we get $r^* = r+r_S \text{ln}(\frac{r}{r_S}-1)+\text{const}$. We can now map $r < r_S$ using tortoise coordinates ($t,r^*$) in which Schwarzschild metric beomes,

\begin{equation} ds^2 = -(1-\frac{r_S}{r})(dt^2 + dr^{*2}) + r^2d\Omega^2) \end{equation}

(because $dr^* = dr/(1-\frac{r_S}{r}))$

Now $dt/dr^*$ is a constant hence we have light cones which are not asymptotic in $t-r^*$ plane (second picture).

The proper time and coordinate time can be related (from geodesic equation) as,

\begin{equation}\tag{2} \frac{d\tau}{dt} = (1-\frac{r_S}{r})^{1/2} \end{equation} Hence, distant observer (named B) will observe light coming from infalling observer (name him A), red shifted by (one over) this factor (and also we cannot define once $r<r_S$ - physically A appears not only to be still but gets reddened and hence eventually dimmer to B).

A reaches $r_S$ in finite proper time but for B at rest, this would take infinite time. In other words, $r_S$ forms the Cauchy Horizon beyond which we have unique geodesics but within it, is a singularity (which does not belong to the Lorentzian manifold) with no future null-like geodesics.

So from (2) it is clear that, even though, A crosses $r_S$, B cannot see this event. Even though A can hold meeting inside the Horizon, B will never know of it. Is there a transformation $t\rightarrow\tau$? All such transformations are affine, in other words, proper time (atleast for null-like geodesics) are affine parameters which appear in the geodesic equations. But from the very definition of singularity, future null geodesics do not exists.

t-r for Schwarzschild metric

t-r for tortoise coordinates

$\endgroup$
0

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.