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It is my understanding that

  1. Hawking radiation is observed by external observers, and

  2. A necessary condition for having Hawking radiation is the formation of an event horizon during a gravitational collapse.

Since the emergence of an event horizon takes infinite time for an observer far away from the black hole, how is it possible that this observer sees thermal radiation coming from the black hole if a necessary condition for the existence of such thermal radiation is the presence of the event horizon?

Am I wrong in assuming that the formation and existence of the event horizon is necessary in order to have Hawking radiation?

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    $\begingroup$ "Since the emergence of an event horizion takes infinite time for an observer far away from the black hole," is false . The LIGO experiment saw gravitationally the merging of two black holes into one , the "saw" is the horizons merging $\endgroup$ – anna v Sep 20 at 10:00
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    $\begingroup$ @annav if that statement is false then general relativity is false cause it predicts that an external observer cannot see this. What we observe is an event where two black holes comes close togheter and after the event the metric tends to that of Kerr black hole, but you don't see the event horizon merging like you don't see a star coming into the black hole, that would take an infinite time for an external observer. $\endgroup$ – AnOrAn Sep 20 at 10:08
  • $\begingroup$ Please give a link for your statements. Experimentally it is wrong. we saw a black hole falling into a black hole, and their horizons merging. It followed the mathematics of black holes, which are GR $\endgroup$ – anna v Sep 20 at 10:27
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    $\begingroup$ @annav It's a standard computation in general relativity, done for example in the book right before the quote I'm writing here. In the excellent textbook Modelling black hole evaporation by Fabbri and Navarro-Salas when they describe the details of the gravitational collpase, in page 16 they say: In contrast, the external observer will never be able to notice the end-point of the collapse at r = 0. And even more surprising, the external observer will never see the star reaching its gravitational radius as this requires, according to Eq. (2.40), an infinitely large amount of time t. $\endgroup$ – AnOrAn Sep 20 at 10:41
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    $\begingroup$ Does this answer your question? From where (in space-time) does Hawking radiation originate? $\endgroup$ – Fattie Sep 21 at 19:09
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External observers and black hole formation

The event horizon is simply the delineation between the part of spacetime from which light can escape and the part of spacetime from which it cannot. In that sense, it is not directly observable, neither by external observers nor by infalling observers. Still, an external observer can observe the effects of the existence of a region from which nothing can escape.

An external observer can observe an object falling toward that region. The object's motion is increasingly slowed, and the light from that object is increasingly redshifted and increasingly reduced in intensity, until it is no longer observable for all practical purposes. The external observer never sees an object cross the event horizon, but the object quickly disappears from the external observer's senses because of the increasing redshift and decreasing intensity. This happens when the object is very near the event horizon.

That's true for any object falling toward the black hole, including the star itself — the star whose collapse forms the black hole. However, to say that the black hole never forms according to the external observer would be missing the point. The external observer sees the collapsing star quickly and smoothly disappear, again because of the rapidly increasing redshift as the "surface" of the star comes very close to the point of no return. In order for the distant external observer to continue detecting light from the star, larger and larger telescopes would need to be used in order to capture the ever-increasing wavelength and ever-decreasing intensity. When the redshifted wavelength exceeds the size of the universe, or when the intensity falls below one photon per age-of-the-universe, this clearly becomes hopeless. This occurs in a finite amount of time on the external observer's clock, so in this sense, the external observer does witness the formation of the black hole.

And remember that the event horizon delineates a region of spacetime. If we want to try to think of it as a region of space, then we need to remember that it can grow. The part of space where infalling objects become practically unobservable to the external observer at 2:00 can be larger than the part of space where infalling objects were becoming practically unobservable to the external observer at 1:00. If the external observer takes a video of objects falling toward a black hole, the video will show that the size of the crazy-region (around which the light from distant stars on the opposite side is bent in dizzying ways) is steadily growing as a result of the mass gained from the infalling objects — even though each infalling object becomes unobservable before reaching that (growing) region.

So yes, it's true that an external observer never sees an object cross the event horizon. And it's also true that an external observer does see the black hole form and grow, in the very real sense that the external observer could take a video and post it on the internet for the rest of us to watch (including seeing falling objects smoothly dwindle-and-disappear, as well as the dizzying effects on the background light from distant stars), all in a finite amount of time.

Hawking radiation

In contrast to the light emitted by the collapsing star, which is quickly redshifted to the point of unobservability, Hawking radiation persists. We can think of Hawking radiation as being emitted from just outside the event horizon (just outside the region from which nothing can escape), but unlike the light from the infalling star, Hawking radiation starts with arbitrarily short wavelengths, so that the wavelength received by the external observer is still finite despite the arbitrarily large redshift. Quantitatively, most of the Hawking-radiation wavelengths received by the external observer are comparable the size of the black hole. That's still a huge wavelength that would require incredibly sensitive instruments to detect (also because of the extremely low intensity), but it doesn't become increasingly difficult to detect (unless the black hole grows), in contrast to the light from the star which does become increasingly difficult to detect.

Altogether, a distant observer can detect the Hawking radiation even though that observer never sees any part of the star cross the (growing) event horizon. In fact, the spacetime of a collapsing star that is used to derive Hawking radiation predicts the experience of the distant observer that was described above.

Most importantly, the derivation of Hawking radiation does not rely on the perspective of any particular observer. The derivation takes all of spacetime into account, not just the part that a distant observer can see. Infalling objects cross the horizon in a finite amount of time on their own clocks, and the derivation of Hawking radiation "knows" this — just like it "knows" that distant observers never see those same infalling objects reach the horizon.

By the way, Hawking radiation can be — and originally was — derived using quantum field theory in classical curved spacetime, and that's the model assumed in this answer. This answer didn't use quantum gravity, which isn't necessary for deriving Hawking radiation and isn't necessary for this question.

Technical note about time and black hole formation

A more technical note for those who are comfortable with the concept of a spacelike hypersurface:

It is sometimes said that the emergence of an event horizon takes infinite time for a distant observer, but we need to be careful when talking about "time" in relativity. The distant observer never sees anything cross the horizon, because light cannot escape. However, there are spacelike hypersurfaces that include stuff behind the horizon and that also intersect the distant observer's worldline. In that sense, the horizon forms in finite time on the observer's clock, even though the observer can never see it. We can construct a continuous sequence of spacelike hypersurfaces (called a foliation), each one intersecting the distant observer's worldline at a particular time on that observer's clock, and each one intersecting the inside of the black hole. The black hole grows along this sequence of spacelike hypersurfaces, and this formation happens in finite time on the distant observer's clock.$^\dagger$

$^\dagger$ The details of the timeline are ambiguous, of course, because we can also construct (infinitely many!) other sequences of spacelike hypersurfaces instead. This is one of relativity's most basic lessons: "simultaneous" is generally ill-defined. We can't use a clock in one place to unambiguously assign times to events that occurred in a different place.

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    $\begingroup$ I completely agree with everything that happens in a finite time for an external observer. Those objects we can see in a finite time are well described by GR and we can approximate one of them for example with the Schwarzschild metric from our POV. My point is that, even if the object we observe, looks, in a finite time, practically undistinguishable from a General relativistic theoretical black hole, it is not such an object, it's an object that tends to the General relativistic black hole. Now, in the derivation of the hawking radiation we use the theoretical black holes, not the things $\endgroup$ – AnOrAn Sep 20 at 18:16
  • $\begingroup$ that in the real world look like the theoretical black holes. As you said in a finite time we observe the star reaching its gravitational radius and not sending light to us anymore, we can say a black has formed. My point is, since the hawking radiation needs the event horizon (does it?) and the event horizon never forms then the external observer never see the Hawking radiation. $\endgroup$ – AnOrAn Sep 20 at 18:19
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    $\begingroup$ Maybe I should rephrase it, whatever we call a black hole, I can ask, can I detect Hawking radiation from an object that collapsed and never quite reached its gravitational radius from my point of view? $\endgroup$ – AnOrAn Sep 20 at 18:24
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    $\begingroup$ @AnOrAn I don't think Hawking-like radiation requires an event horizon. The effect is famous because of its profound implications when it's associated with an event horizon, but more generically, particles can potentially be produced in dynamic curved spacetimes. The paper Essential and inessential features of Hawking radiation comments on this, and the paper Hawking-like radiation from evolving black holes and compact horizonless objects gives more detail. $\endgroup$ – Chiral Anomaly Sep 25 at 0:42
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    $\begingroup$ @AnOrAn ... Maybe more importantly, as far as the detector is concerned, Hawking radiation is just thermal radiation with a very low temperature. I don't know if a distant observer could reliably distinguish between radiation produced by the Hawking effect and radiation that a horizonless compact object might emit due to other effects. $\endgroup$ – Chiral Anomaly Sep 25 at 1:05
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Hawking radiation comes from the space outside of the event horizon. And the event horizon forms as the actual black hole forms. So what is necessary first, is the formation of the black hole, which also forms an event horizon, then phenomena such as Hawking radiation can be considered. And for a distant observer anything that is happening at the event horizon appears to take an infinite amount of time because the wavelength of photons emitted near the event horizon are stretched to almost infinity. This does not mean that there is nothing happening local to the black hole event horizon.

But all is not lost since for huge cosmic events (like black holes merging) we can “see” these things due to the detection of gravitational waves at LIGO.

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    $\begingroup$ But for an external observer the event horizon never forms $\endgroup$ – AnOrAn Sep 20 at 10:03
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    $\begingroup$ @AnOrAn Hawking assumed a static Schwarzschild geometry, but that's not really necessary. It suffices that the infalling member of a virtual particle pair is sufficiently delayed that it cannot interfere with the outgoing member. Of course, there isn't the slightest experimental evidence for Hawking radiation or for black holes of infinite depth, so this is math, not physics. $\endgroup$ – John Doty Sep 20 at 20:21
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Micro Black Holes

If micro black holes are possible, they should generate massive amounts of Hawking radiation which should be easily observable at any "reasonable" distance from the MBH, immediately, no matter whatever else you wish to say about event horizons.

Imaging Event Horizons

When you say that:

the emergence of an event horizion [sic] takes infinite time for an observer far away from the black hole

there is only one sense in which this is true: if a glowing object falls into the BH from the direction of the observer, it takes a theoretically infinite time for the observer to "see" the object cross the event horizon. But from this fact, you derive the incorrect conclusion that it therefore takes an infinite amount of time for the event horizon to form.

Your conclusion is false, because the unbounded redshift which delays your "observation" of the event horizon can only be present if the event horizon already exists. That is, there is a distinct difference between the existence of the event horizon and your ability to detect it. Just because you cannot "see" the event horizon doesn't mean it hasn't already formed. It just means your abilities as an observer are quite limited. Don't feel bad. This is true of most processes in the universe. For instance, you can't observe gamma rays being emitted in the core of Proxima Centauri, even though it is a mere 4 ly away. That doesn't mean the gamma rays don't exist, or that the core of Proxima Centauri doesn't exist. It just means that process is unobservable for you, little one.

As others have noted, there are many other signals that you can observer to infer the existence of the event horizon long before you see your glowing astro-buddy cross it forever. If you witness the BH forming from a star, you will see the luminous radius of the star contract in quite finite time. If there are other bright stars nearly behind the BH from your perspective, you can change your position until the stars are occluded by the BH.

Conclusion

There are more photons in the world than the ones being emitted from an object falling into a BH. If you choose to only look at those photons, you will have a very misleading understanding of black holes, and you may recklessly decide it is safe to fly your starship through one, since obviously the black hole has not finished forming yet. If you look at other photons, such as the ones emitted by the star as it collapses into the BH or photons emitted from behind the BH, you will get a very timely picture of the shape and location of the BH...no infinite wait necessary.

Note that photons which pass close to the event horizon will obviously have their trajectories altered, and thus, gravitational lensing will prevent you from forming a crisp image of the event horizon (unless perhaps you can form a Dyson shell around the BH to use as your imaging sensor). Even so, you should be able to infer the extent of the event horizon with "prompt photons", if you will, to a precision that is more than adequate for starship navigation.

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  • $\begingroup$ So, if I understand well, if we look at the black hole with a telescope, and say it's eating a star, we won't see that it has an event horizon because this takes an infinite amount of time to "form" according to us. However we can know that it already formed because of its Hawking radiation (assuming that such a thing is measurable of course). Pretty cool, if I understood well. $\endgroup$ – AccidentalBismuthTransform Sep 21 at 10:05
  • $\begingroup$ You don't need to see the Hawking radiation to infer the event horizon. In fact, we have not unambiguously detected Hawking radiation as yet, but we have "imaged" a black hole: nasa.gov/mission_pages/chandra/news/…. Only the photons emitted very close to the event horizon take a long time to escape. If there are no intervening stars/gas/etc., a bright light source illuminating a BH from behind will cause a conspicuous absence of photons in the region of the event horizon. $\endgroup$ – Lawnmower Man Sep 21 at 17:13
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The answer by Chiral Anomaly is correct and tells you most of what you wish to know. I will add a few details.

I think your question is asking for a calculation that does not assume a horizon is already there, but which rather considers a dynamic spacetime in which a horizon can appear and grow, but obeying all the usual properties. For example, matter takes an infinite time to reach and cross a horizon, as recorded by the Schwarzschild time coordinate. It is my understanding that Hawking's original calculation concerned just such a dynamic situation.

I think the best way to see this is to divide the problem into two parts. First one calculates the Hawking radiation by a method which does not involve a coordinate singularity at the horizon. One thus obtains a stress-energy tensor for the electromagnetic field outside the horizon, including at infinite distance away. Then the second step is to interpret the result.

In the second step one interprets not by asking "has any matter crossed the horizon at the time recorded on a distant clock?" but by asking "is there any radiation coming to the distant observer at finite times?" The answer to the second question is yes. If you like you can interpret by saying that virtual photons started outside the horizon and then followed spacelike trajectories to inside it, thus accessing a region of spacetime which non-virtual matter could not access so quickly. The horizon itself should be seen as an aspect of spacetime not just space.

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