General RelativityPhysics

Are Black Holes Eternally Collapsing Stars ?

Short answer : it depends who is asking the question !

During my many years on physics discussion forums, this is an old chestnut that keeps popping up again and again – the argument that “time stops” at the event horizon, and hence that nothing ever crosses that region, and ergo that the gravitational collapse itself is an infinitely long process. In essence, the argument is that black holes are just stars which are “frozen” in the process of gravitational collapse. But how physically valid is this ? Let us take a closer look at this.

Proponents of the “frozen star” argument usually start by assuming that the process of gravitational collapse is adequately described by the exterior Schwarzschild metric :

\displaystyle{d\tau ^2=\left ( 1-\frac{r_{s}}{r} \right )dt^2-\left ( 1-\frac{r_{s}}{r} \right )^{-1}dr^2-r^2\left ( d\theta ^2+\sin^2\theta d\varphi ^2 \right )}

One then proceeds by calculating the in-fall time of a test particle, i.e. the time it would take for a test particle to reach the horizon :

\displaystyle{t_0=-\int_{r_0}^{r_s}\frac{dr}{c\sqrt{2\left ( 1+GM/r \right )\left ( 1-\frac{r_s}{r} \right )}}=\infty }

Hence, the argument goes, nothing ever reaches the horizon, so a black hole does not form in finite time. Here’s the problem though : in curved space-times, the notion of “time” is purely local, and observer dependent. The above metric is a description of space-time from the point of view of a far-away, stationary observer, using his own local method of measuring lengths and intervals of time. It is not a valid description of space-time as seen and experience by a test particle in free fall towards the black hole ! To fix this, we need to adjust for the fact that “time” is always local; we do this either by performing a coordinate transformation of the Schwarzschild metric into a suitable set of coordinates ( e.g. Gullstrand-Painleve coordinates ), or by explicitly inserting the adjustment into the proper time integral :

\displaystyle{\tau _0=\int_{r_0}^{r_s}ds=\int_{r_0}^{r_s}dr\frac{d\tau }{dr}=-\int_{r_0}^{r_s}\frac{dr}{\sqrt{2\left ( 1+GM/r \right )}}}

which is finite, and well defined. This integral is obtained from the equations of motion, and evaluates to some fairly complicated expression involving hyperbolic functions. Nonetheless, the in-fall time as measured by the falling test particle itself is finite. The particle does indeed reach the horizon, and continues falling through it in finite time. Another, perhaps slightly more elegant, way of showing that space-time at the horizon is in fact smooth and regular, is to examine the curvature tensor, and the curvature scalars derived from it, for that region. The method of choice here is to calculate the Kretschmann scalar, being the complete contraction of the Riemann curvature tensor :

\displaystyle{R^{\alpha \beta \gamma \delta }R_{\alpha \beta \gamma \delta }=\frac{48G^2M^2}{c^4r^6}}

which, again, is finite and well defined everywhere, including at the horizon. Physically this means that space-time there is smooth and regular, and not at all geodesically incomplete; the singularity found in the original Schwarzschild metric at the horizon is purely a coordinate singularity, but not a physical curvature singularity. Therefore, there is no infinite curvature there, and time does not “stop” at the horizon. More formally, all light-like geodesics smoothly extend through the horizon surface, so black holes are not “frozen stars”; the misconception arises from erroneously attempting to describe local physics by using a coordinate chart which is not local to the region in question, and in fact does not even span that region at all. It’s a simple, but wide-spread and persistent error.

With all of this being said however, it is important to realise that the experience of in-fall into a black hole is observer dependent – as the Schwarzschild metric indicates, a far away stationary observer never measures anything to ever reach the horizon; he will see objects slow down and fade as they approach the horizon, but he will never see them reach it. In order to cross through the horizon and fall into the centre in finite time, one must place the measuring clock into the rest frame of the test particle itself.

Why is this so ? In General Relativity, many concepts we are used to from old Newtonian physics quite simply no longer apply – in particular, it is very important to realise that all notions of space and time are now purely local. Schwarzschild coordinates are used to describe an observer who is far away and stationary with respect to the central mass; this set of coordinates are his method of labelling events, based on far-away stationary clocks and far-away stationary rulers. As such, Schwarzschild coordinates cannot be used to model what physically happens in the local frame of a test particle freely falling towards a central mass or black hole; in order to do so, one must employ a coordinate basis which remains local to the test particle in free fall, everywhere along its trajectory. The most useful of these are Gullstrand-Painlevé coordinates, which reflect what clocks and rulers in the rest frame of the test particle would physically read. Schwarzschild and Gullstrand-Painlevé are related by a coordinate transformation, so these metrics describe the same space-time with the same geometry, just from the point of view of different physical observers. Performing the above calculation makes it evident that time-like geodesics extend across the horizon while remaining of finite length, so the “frozen star”concept is simply not physically nor mathematically tenable. For completeness, here is also a Penrose diagram of the geometry of this space-time :

 

https://i2.wp.com/i.stack.imgur.com/grB5f.png?resize=486%2C289

 

0 votes