I know that there's a lot of evidence for extremely compact bodies. But is there any observation from which we can infer the existence of an actual horizon?

Even if we are able to someday resolve directly the "shadow" of a possible black hole, isn't it still possible that it's just a highly compact body that's just gravitationally redshifting any radiation that it emits to beyond the detection capability of our telescopes? What would be indisputable evidence for a causal horizon?


3 Answers 3


At the galactic center, there is an object called Sagittarius A* which seems to be a black hole with 4 million solar masses. In 1998, a wise instructor at Rutgers made me make a presentation of this paper


by Narayan et al. that presented a successful 2-temperature plasma model for the region surrounding the object. The paper has over 300 citations today. The convincing agreement of the model with the X-ray observations is a strong piece of evidence that Sgr A* is a black hole with an event horizon.

In particular, even if you neglect the predictions for the X-rays, the object has an enormously low luminosity for its tremendously high accretion rate. The advecting energy is pretty "visibly" disappearing from sight. If the object had a surface, the surface would heat up and emit a thermal radiation - at a radiative efficiency of 10 percent or so which is pretty canonical.

Of course, you may be dissatisfied by their observation of the event horizon as a "deficit of something". You may prefer an "excess". However, the very point of the black hole is that it eats a lot but gives up very little, so it's sensible to expect that the observations of black holes will be via deficits. ;-)

  • $\begingroup$ In effect a solid body has a signature of a "splash," whereas an event horizon will not. $\endgroup$ Commented Jan 19, 2011 at 19:12
  • $\begingroup$ See my answer below. The "seems to be a black hole" above probably means that this determination depends on the validity of general relativity; i.e. the theory was used to make the determination. Then that wouldn't answer the question here. $\endgroup$
    – finbot
    Commented Mar 2, 2011 at 22:50
  • 2
    $\begingroup$ No, finbot, this argument about disappearing energy doesn't really "assume" general relativity: it pretty much proves one of its consequences, namely the event horizons. Moreover, GR is clearly valid. A major interpretation of the question is how to distinguish objects with horizons with those without horizons assuming the existing knowledge about physics - which surely includes the insights of GR. If you pretend that random pillars of physics are unknown, then you may prevent people from making arguments, indeed. But that wasn't really the case here. $\endgroup$ Commented Mar 14, 2011 at 6:36

This paper might be of interest:

The Rates of type I X-ray Bursts from Transients Observed with RXTE: Evidence for Black Hole Event Horizons

Ronald A. Remillard, Dacheng Lin, Randall L.Cooper, Ramesh Narayan


Abstract: We measure the rates of type I X-ray bursts, as a function of the bolometric luminosity, from a likely complete sample of 37 non-pulsing transients (1996-2004). Our goals are to test the burst model for neutron stars and to investigate whether black holes have event horizons. We find 135 type I bursts in 3.7 Ms of exposure for the neutron-star group, and the burst rate function is generally consistent with model predictions. However, for the black hole groups (18 sources), there are no confirmed type I bursts in 6.5 Ms of exposure, and the upper limits in the burst function are inconsistent with the model predictions for heavy compact objects with a solid surface. There are systematic spectral differences between the neutron-star and black-hole groups, supporting the presumption that physical differences underly the sample classifications. These results provide indirect evidence that black holes do have event horizons.

  • $\begingroup$ See my answer below. Aren't you assuming that the black hole is there in the first place, by using (depending on) general relativity in that determination? How else could you make that determination, when a black hole emits nothing to be detected? $\endgroup$
    – finbot
    Commented Mar 2, 2011 at 22:52

I think there is no observation from which we can infer the existence of an actual horizon. That's the nature of the beast, since the black hole emits nothing at all by definition.

It should be kept in mind that it's almost universal for texts to skip over the fact that their detection of the horizon depends on the validity of general relativity. That is, before the conclusion that a black hole (or its horizon) exists is reached, there's a step where the mass determined from some other observation (something outside the suspected black hole, like the orbit of some gas cloud--anything that's observed is necessarily outside of the horizon) is input into, say, a formula derived from the Schwarzschild metric, and the output tells whether sufficient mass is within a sufficiently small volume, such that the determination that a black hole (or its horizon) is made. The dependence on the metric is a dependence on the validity of general relativity. A different metric for a different theory of gravity could give a different result, such as to the existence of a black hole.

The Chandra X-ray Observatory FAQ notes that their discoveries of supposedly actual black holes depend on the validity of general relativity. So the black holes they found might not exist after all.

  • $\begingroup$ Shouldn't a black hole be emitting Hawking radiation at the event horizon? $\endgroup$
    – Gabe
    Commented Mar 2, 2011 at 8:43
  • $\begingroup$ @Gabe For astrophysical black holes, this is too small to be detected. $\endgroup$
    – dbrane
    Commented Mar 2, 2011 at 10:53
  • 1
    $\begingroup$ It is arguably not true that Black Holes follow from General Relativity per se. They follow from the Hawking-Penrose theorem and its assumptions (which are somewhat weaker than those of GR). For example HP theory does not assume 4 dimensions, nor does it assume much about the metric. One should also note that a Black-Hole-like object exists in Newtonian Gravity too. $\endgroup$ Commented Mar 15, 2011 at 14:38

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.