2

It is often said that the event horizon of a Schwarzschild black hole is lightlike. Is this correct and, if so, what exactly does this mean?

Intuitively, this may mean that any two points on the horizon surface are lightlike separated, but I am unable to see this from the metric. The temporal part of the Schwarzschild metric is zero at the horizon. The radial part approaches zero in the limit. However, the angular part remains well defined positive thus apparently making the interval spacelike.

What am I missing? I’ve searched the web, but couldn’t find anything relevant. I would appreciate if someone points me in the right direction.

safesphere
  • 12,640
  • The radial part approaches zero in the limit. How so, when $g_{rr}$ becomes infinite? – G. Smith Feb 02 '20 at 07:09
  • It’s very easy to find stuff like this stating that the horizon is lightlike (see the page numbered as 35), but they don’t use Schwarzschild coordinates to show it. – G. Smith Feb 02 '20 at 07:11
  • There are several equivalent definitions of a lightlike hypersurface. One is that the metric will be degenerate if you restrict it to the hypersurface. And, yes, the black hole horizon is lightlike. That is true for any black hole, not just Schwarzschild. In fact it is true for any causal boundary. – MBN Feb 02 '20 at 11:05
  • @G.Smith Try setting $r=r_s+dr$ for the radial part. You’ll see it goes to zero for $dr\to 0$. Thanks for the link. – safesphere Feb 02 '20 at 17:04
  • @safesphere I can’t speak for Ben. – G. Smith Feb 02 '20 at 17:39
  • @safesphere : I don't think it would be in Weinberg, but it should be in any good GR book. Try Hawking and Ellis. – MBN Feb 03 '20 at 07:47
  • https://en.wikipedia.org/wiki/Null_hypersurface – MBN Feb 03 '20 at 08:51
  • @safesphere No, that is not the terminology. The horizon is lightlike (or null) period. It isn't lghtlike in some direction and spacelike in others. – MBN Feb 03 '20 at 16:39
  • @ÁrpádSzendrei It does. Note that planetary disks don’t have to be flat. Even in our solar system, all planets rotate in a flat disk, except for one. Pluto rotates at an angle to the main planetary disk. – safesphere Jun 10 '22 at 03:25

2 Answers2

5

It is often said that the event horizon of a Schwarzschild black hole is lightlike. Is this correct and, if so, what exactly does this mean?

It means that only lightlike geodesics can stay stationary at the horizon, all other geodesics must fall in (or escape, assuming there are space-like geodesics as well) and the proper time intervall they can spend at the horizon is infinitesimal. If an observer on a timelike path emitts a radially outwards directed photon when he crosses the horizon, it will stay there forever. So the only stationary probes you can have at the horizon are light-like (and inside space-like, which means that only Tachyons, if they existed, could remain stationary inside the horizon). Outside of the horizon the stationary (with respect to the black hole) probes (called Fidos for fiducional observers) are on time-like paths.

Yukterez
  • 11,553
  • "It means that only lightlike geodesics can stay stationary at the horizon" - This is true only in the radial direction, as described in my question. At constant angles ($d\theta=0$, $d\varphi=0$), the interval is zero ($ds=0$). However, this does not seem to be true in the angular direction where geodesics on the horizon appear to be spacelike, unless I am missing something. Do you know of any reference where this matter is discussed? – safesphere Feb 03 '20 at 05:35
  • That's why I talked about a "radially outwards directed photon", that does not imply all directions ("only likelike geodesics" does not mean "all lightlike geodesics". I don't have a reference at hand, but I would solve for $\ddot{r}=\dot{r}=0$ (which is at $r=2$ for radial and $r=3$ for transverse photons), using the equations at http://geodesics.yukterez.net. – Yukterez Feb 03 '20 at 07:58
  • @safesphere please take notice and possibly comment of my question https://physics.stackexchange.com/q/775158/281096 . – JanG Aug 07 '23 at 11:12
  • @Yukterez please take notice and possibly comment of my question https://physics.stackexchange.com/q/775158/281096 . – JanG Aug 07 '23 at 11:13
  • @JanG I’ve looked at the question, but I don’t have anything to contribute at this time, sorry. I’m glad that Simon has posted an answer. – safesphere Aug 10 '23 at 04:16
  • @safesphere Thanks anyway. The circular geodesics on event horizon is spacelike, with $k^2=0$ and $l^2=1$, and marginally unstable ($\ddot{V}_{eff}=0$. Thus, I think it means that the event horizon is space-like. – JanG Aug 10 '23 at 11:19
  • @JanG Yes, apparently qualifications, such as “spacelike” or “lightlike”, depend on direction. A beam of light also is spacelike across, but lightlike in the direction it points. – safesphere Aug 11 '23 at 02:47
  • @safesphere I agree. Therefore I speak about geodesics (geometry) and not about trajectories or orbits which describe movement of matter particles (tardyons, luxons, tachyons). – JanG Aug 11 '23 at 06:27
  • @safesphere I believe to have found that in the region $IV$ in my diagram there are no circular geodesics (space-, light- or time-like). In my understanding it proves that „behind“ the event horizon there is no spacetime. – JanG Aug 11 '23 at 06:32
  • @JanG Your argument may be a good confirmation, but the fact that there is no spacetime behind the horizon is pretty obvious. Nothing in the universe is older than the current cosmological time, by which the horizon hasn’t formed yet. To me all this is moot though. Things fall within the Planck length from the horizon in seconds. After that quantum effects take over and GR is no longer predictive. Whether or not the Hawking radiation exists, Hawking’s biggest accomplishment is showing that GR stops working at the horizon. – safesphere Aug 12 '23 at 09:39
  • @safesphere I understand your arguments. However, what I am trying to prove within the framework of GR is that the gravitational collapse story as told is not true. The Oppenheimer-Snyder model relies heavily on the nonphysical assumption of pressureless matter. The inevitable continuous contraction is the result. – JanG Aug 12 '23 at 14:29
  • Einstein field equations for static spherically symmetric perfect liquid spheres can be reduced to a second-order linear parametric differential equation. If you look at the parametric stability of the corresponding solution, you will get a different perspective on the black hole formation process. I try to prove that GR does not break down there – JanG Aug 12 '23 at 14:29
1

Lightlike simply means that your interval $ds^2=dx^2+dy^2+dz^2-c^2dt^2$ evaluates to zero.

It does not mean that real black hole would have light encircled forever in eternal loop over its event horizon. In fact, any non-zero energy object cannot do that. Real particles would lose energy because they would have accelerated orbit and gradually fall down. Even virtual particles are split and sucked under event horizon (BH is a heating engine).

sanaris
  • 875
  • 5
  • 8
  • Actually, light can orbit a black hole, see https://en.wikipedia.org/wiki/Photon_sphere But such orbits are unstable, and the photon sphere of a Schwarzschild black hole is at $3r_s/2$, not the event horizon. As Yuktetez states, a photon launched directly away from the event horizon exactly at $r_s$ just hovers there; that's also an unstable trajectory. – PM 2Ring Feb 02 '20 at 21:43
  • Schwarzschild coordinates have a coordinate singularity at the event horizon. You cannot use them to draw any conclusions from them about the horizon. Hence you'll have to "repair" the singularity first, which means switching to a coordinate system that works properly at the event horizon. – emacs drives me nuts Feb 03 '20 at 08:52
  • @PM2Ring what is the inner "circle" in the M37 BH image? The light horizon or the event horizon? Or none of the two but somehow near to one of the two? – Alchimista Feb 03 '20 at 09:50
  • 1
    @Alchimista There's good info about that in the answers here: https://astronomy.stackexchange.com/q/30317/16685 – PM 2Ring Feb 03 '20 at 10:14
  • @safesphere ah ok; the comment I referred to appears to have been deleted by now. – emacs drives me nuts Feb 03 '20 at 16:23