# Is the event horizon locally detectable?

There are two conflicting ideas.

According to the traditional view, determining the location of the event horizon of a black hole requires the knowledge of the whole future behaviour of the black hole solution. Namely, given a partial Cauchy surface, one cannot find where the event horizon is without solving the Cauchy problem for the whole future development of the surface. (Cf. Hawking - Large Scale Structure of Space-Time, p. 328). Therefore, locally one cannot tell whether one is passing through the horizon.

However, according to Karlhede and arXiv:1404.1845, there is a certain scalar, now called the Karlhede's Invariant, which changes sign as one crosses the event horizon. A local measurement of this scalar can tell whether one has encountered an event horizon as of yet or not.

So, who is right? If both are right, how do we reconcile the apparent conflict?

Thanks.

• +1 I'd like to see an answer as well. One huge common misconception is "the frame of the infalling observer". The event horizon is lightlike. When an observer "crosses" the horizon, he doesn't have a frame, because there are no lightlike frames. Thus the world line of the observer ends at the horizon with a zero mass-energy in any valid reference frame. All mass-energy of infalling objects is converted to the energy of the gravitational field making the empty horizon larger. – safesphere Aug 2 '18 at 17:10
• @safesphere: One huge common misconception is "the frame of the infalling observer". The event horizon is lightlike. When an observer "crosses" the horizon, he doesn't have a frame, because there are no lightlike frames. No, this is wrong. The observer's velocity vector is timelike, not lightlike, before, during, and after passing through the horizon. One way to see this is that a geodesic is a curve that parallel-transports its own tangent vector, and parallel transport conserves norms. – user4552 Aug 2 '18 at 23:19
• Thus the world line of the observer ends at the horizon with a zero mass-energy in any valid reference frame. No, the Schwarzschild spacetime has the same Killing vectors everywhere, so the conserved quantities for motion of test particles remain valid all along the geodesic. – user4552 Aug 2 '18 at 23:21
• @BenCrowell Thanks Ben, I will give it a thought. – safesphere Aug 2 '18 at 23:31

Toth is a competent person who does good work, but IMO this is one of those cases where sometimes a good scientist writes a bad paper.

If you give me global information about my spacetime, plus the ability to measure local information about my own environment, I can find out all kinds of things about my location, and I don't need to resort to the Karlhede invariant. For example, if you tell me that I'm in the Schwarzschild spacetime with mass $m$, I can measure the Carminati-McLenaghan invariant $W_1$, and because I know that $W_1=6m^2r^{-6}$ for the Schwarzschild spacetime, I can immediately determine my $r$, even if I'm locked inside a closet. If $r=2m$, I know I'm at the horizon.

So the information processing looks like this:

Tell me I'm in the Schwarzschild spacetime with mass $m$, and let me do local measurements on the gravitational field --> I can find out if I'm at the horizon (by measuring $W_1$).

Tell me I'm in the Schwarzschild spacetime, and let me do local measurements on the gravitational field --> I can find out if I'm at the horizon (by measuring the Karlhede invariant).

So the only difference between using the Karlhede invariant and using some other invariant is that I need less global information -- but I still need global information (that I'm in the Schwarzschild spacetime). And keep in mind that the Schwarzschild spacetime does not actually exist. It's not the metric of an astrophysical black hole.

Curvature scalars give you only very limited information about what's going on in a spacetime. For example, all curvature scalars vanish for a gravitational plane wave, so although LIGO can tell you there's a wave passing through your location, you'll never get that information from curvature scalars.

Curvature scalars are also difficult to measure and do not affect laboratory physics except with extremely sensitive hypothetical measurements that we can't actually do. (We currently do not have any technology capable of doing a practical measurement of any curvature scalar in any gravitational environment that we have access to.) So it's absurd IMO to attribute violent physical effects like a firewall to the behavior of a particular curvature scalar.

The short answer is that the event horizon is detectable in principle locally if a freely falling observer measures the redshift of light emitted far away (more precisely at infinity) with $z = 1$.

An observer who is hovering at constant r-coordinate will see light emitted far away blueshifted. In free fall relativistic doppler redshift has to be taken into account additionally. Both frequency shifts combined yields $\lambda'/\lambda = 1 + sqrt(r_S/r)$, with $\lambda'$ the wavelenght the freefaller measures and $\lambda$ the wavelenght of the light emitted far away. At the event horizon $r = r_S$ and hence the redshift $z = 1$.

• Could you provide either the details of the calculation or a link to where this calculation is published? – safesphere Aug 2 '18 at 21:02
• This solution relies on the existence of a light source in the infinite future of the universe. In addition the source must obey conflicting requirements of being located at infinity, yet with no redshift due to the expansion of space. – safesphere Aug 2 '18 at 21:30
• Measurement of light emitted by distant objects is not a local measurement. – user4552 Aug 2 '18 at 23:22
• @safesphere with the gravitational blue-shift at shell $r$ $f'/f = 1/sqrt(1-r_S/r)$ combined (means multiplied) with the relativistic doppler shift $f'/f =sqrt{(1+v/c)/(1-+v/c)}$ - where $v = -sqrt(r_S/r)$ is is the free fall velocity relative to the shell at $r$ - you obtain the desired formula, after simplifying using binomial formulas. – timm Aug 3 '18 at 7:54
• " yet with no redshift due to the expansion of space." Remember, the Schwarzschild metric doesn't expand. With the light source at a finite r-coordinate the expression for the redshift at the horizon gets a bit more complicated. – timm Aug 3 '18 at 8:18

There is an observer called the Swartzschild observer.

Let's agree that nothing can (no information) pass through the event horizon from inside the black hole.

Let the observer orbit around the black hole outside the event horizon at distance d.

The person falling into the black hole keeps counting upwards, from 0, and sends EM wave signals to the orbiting observer. The observer sends these signals right back after receiving them.

They keep on doing this all the way, counting from 0.

The person falling in starts by sending 0 signal to the orbiting observer, and while he (the person falling in) is still outside the event horizon, receives back the signal 0 from the orbiting observer.

Now this goes on until the person falling in reaches the event horizon. At this point, the orbiting observer sees the person falling in frozen at the event horizon, and no signals coming from the person falling in any more, so the orbiting observer will have nothing to send back.

From the view of the person falling in, to answer the question, at this point everything seems like normal, he still tries to send signals to the orbiting observer. But there is no signal coming back from the orbiting observer. So the person falling in knows he reached the event horizon.

• @NanashiNoGombe "Therefore, locally one cannot tell whether one is passing through the horizon." This is your statement. I think I wrote down how one is able to tell they passed through the event horizon. The second statement of yours must be right, since when the Karlhede's Invariant changes sign is when my description gives you the same. – Árpád Szendrei Aug 2 '18 at 17:19
• I don't think querying an outside observer counts as "local". – Acccumulation Aug 2 '18 at 18:36
• @ÁrpádSzendrei The Schwarzschild observer is not orbiting around the black hole. It's sitting at infinity, outside the black hole's gravitational influence. – probably_someone Aug 2 '18 at 19:48
• An external observer will never see the falling observer frozen and will receive his continuous signal forever (disregarding quantization for a moment), just more and more redshifted. If the signals are sent at finite intervals, then there would be the last signal (e.g. the last photon emitted from outside the horizon) arriving to the external observer at an arbitrary moment of the distant future (depending on the exact location from where it was sent), but this would not be an indication that the event horizon is reached. It is never reached in the view of the external observer. – safesphere Aug 2 '18 at 21:23
• @safesphere I agree that it is never reached to an external observer. The contradiction is where you say " and will receive his continuous signal forever (disregarding quantization for a moment)". That is where I see the contradiction between GR and QM. The signal cannot be seen for an external observer forever, because that would require an infinite amount of photons. – Árpád Szendrei Aug 2 '18 at 23:12

user4552 answer is too hasty. let us give that the spacetime is schwarzchild, but we dont know M. if the metric is spherically symmetric it has to be schwarzchild, even astrophysically.certainly in numerical simulations i can assume that and raise this question. karlhede invariant should be useable to locate the EH. if hawking and ellis say it cant be done without knowing whole future behaviour, then there seems to be a conflict. all other considerations like whether it is easy to measure this invariant are irrelevant for this in principle discussion.