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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.

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  • $\begingroup$ The radial part approaches zero in the limit. How so, when $g_{rr}$ becomes infinite? $\endgroup$
    – G. Smith
    Commented Feb 2, 2020 at 7:09
  • $\begingroup$ 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. $\endgroup$
    – G. Smith
    Commented Feb 2, 2020 at 7:11
  • $\begingroup$ 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. $\endgroup$
    – MBN
    Commented Feb 2, 2020 at 11:05
  • $\begingroup$ @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. $\endgroup$
    – safesphere
    Commented Feb 2, 2020 at 17:04
  • $\begingroup$ @safesphere I can’t speak for Ben. $\endgroup$
    – G. Smith
    Commented Feb 2, 2020 at 17:39

2 Answers 2

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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.

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  • $\begingroup$ "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? $\endgroup$
    – safesphere
    Commented Feb 3, 2020 at 5:35
  • $\begingroup$ 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 geodesics.yukterez.net. $\endgroup$
    – Yukterez
    Commented Feb 3, 2020 at 7:58
  • $\begingroup$ @safesphere please take notice and possibly comment of my question physics.stackexchange.com/q/775158/281096 . $\endgroup$
    – JanG
    Commented Aug 7, 2023 at 11:12
  • $\begingroup$ @Yukterez please take notice and possibly comment of my question physics.stackexchange.com/q/775158/281096 . $\endgroup$
    – JanG
    Commented Aug 7, 2023 at 11:13
  • $\begingroup$ @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. $\endgroup$
    – safesphere
    Commented Aug 10, 2023 at 4:16
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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).

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  • $\begingroup$ Actually, light can orbit a black hole, see 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. $\endgroup$
    – PM 2Ring
    Commented Feb 2, 2020 at 21:43
  • $\begingroup$ 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. $\endgroup$ Commented Feb 3, 2020 at 8:52
  • $\begingroup$ @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? $\endgroup$
    – Alchimista
    Commented Feb 3, 2020 at 9:50
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    $\begingroup$ @Alchimista There's good info about that in the answers here: astronomy.stackexchange.com/q/30317/16685 $\endgroup$
    – PM 2Ring
    Commented Feb 3, 2020 at 10:14
  • $\begingroup$ @safesphere ah ok; the comment I referred to appears to have been deleted by now. $\endgroup$ Commented Feb 3, 2020 at 16:23

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