The simple answer: Its because Gravity of Black Hole there doesn't allow it. See also this and this Phys.SE posts.

Isn't it a classical answer? When we're unable to connect Gravity with Quantum Theory, how can we apply classical force concept on a denizen of Quantum realm?

My main concern is Quantum Tunneling. Why can't a photon just break the barrier based on Uncertainty principle?

I know about Hawking Radiation, but it also doesn't tell anything about it. Cosmic fluctuations of vacuum at event horizon boundary creates pair of particles; one goes inside and one is released with Quantum Tunneling breaking gravitational barrier at that point.

What's the real explanation of light not being escaped from inside Black Hole?

  • $\begingroup$ possible duplicate of Why is a black hole black? $\endgroup$ Commented Dec 11, 2012 at 17:35
  • 1
    $\begingroup$ You might want to watch this video about the "Black Hole War" from Prof. Susskind. $\endgroup$
    – Dilaton
    Commented Dec 11, 2012 at 17:35
  • $\begingroup$ Hawking radiation (including photons) can be thought of as a tunneling phenomenon. The tunneling transmission coefficient, of course, is incredibly tiny. So I don't understand what you find dissatisfying about the Hawking radiation description for escaping photons. $\endgroup$
    – twistor59
    Commented Dec 11, 2012 at 19:53
  • $\begingroup$ @twistor59 Nothing is dissatisfying me about Hawking radiation. Actually, I am asking about normal light waves produced by outside source. Hawking Radiation talks about suddenly appeared particles in vacuum. $\endgroup$ Commented Dec 11, 2012 at 20:25
  • $\begingroup$ I see - basically the question is "why can't real photons tunnel out?". I suspect that they can - effectively - the mechanism can be: virtual particle/anti particle (photon here) pair created just outside horizon, negative energy partner tunnels in, where it can annihilate a real photon. So a real photon has disappeared from the interior and the BH mass has decreased. But I'm a little unsure so won't post it as an answer. $\endgroup$
    – twistor59
    Commented Dec 11, 2012 at 21:08

4 Answers 4


Short answer: Force is generally a bad way to think about things in relativity. It can occasionally be useful, but this is most commonly the case when you are closely approximating Newtonian cases or you're already well within the relativistic framework.

In general (and special) relativity, all curves are of one of three types: timelike, spacelike or null. For timelike curves, the magnitude of the $\dot t$ component of the 4-velocity exceeds the spatial velocities. For spacelike curves, the opposite is true. Null curves are the boundary between the two. In special relativity, material particles follow timelike curves, and light follows null curves. Spacelike curves represent spatial seperations as measured in a particular coordinate frame.

It turns out that, in the case of a black hole, for all points inside the horizon, there are no timelike or null curves that point into the future and that leave the horizon. All future-pointing timelike and null curves are therefore trapped. Leaving the inside of the horizon would require that they cease being timelike/null, which would be contrary to their character in the theory.

  • $\begingroup$ For my question, you'll take time-like/light-like interval between two spacetime points.. one lying outside event horizon too. I don't see why leaving horizon would be impossible by Quantum Tunneling. $\endgroup$ Commented Dec 11, 2012 at 19:27
  • 2
    $\begingroup$ @SachinShekhar: read my answer. it's not meaningful to talk about quantum tunneling in this case. Quantum tunneling happens when there are two classically allowed regions seperated by a classically forbidden zone. Here, you have two incompatible notions of time that makes travel in one direction meaningless. They're different scenarios. Abandon your classical intuition that a black hole is gravitationally binding the things in its interior, and that is why you have an EH. That's not the best way to look at black holes. $\endgroup$ Commented Dec 11, 2012 at 19:41
  • 1
    $\begingroup$ @SachinShekhar Even in tunneling across a barrier, particles move forward in time: it starts on this side and ends on that side later in time. From the horizon in forward in time is towards the singularity. $\endgroup$ Commented Dec 11, 2012 at 19:44
  • $\begingroup$ @ JerrySchirmer Gravity is also there for binding in modern concepts. Only, its affecting Spacetime on which we are doing calculations. Anyway, back to the Quantum Tunneling: Quantum tunneling happens when there are two classically allowed regions separated by a classically forbidden zone. ~> If I minimize the time interval (at a cost of energy of photon being unpredictable based on Uncertainty principle), wouldn't then everything be your classical? Why can't I really talk Quantum Tunneling? $\endgroup$ Commented Dec 11, 2012 at 20:08
  • 1
    $\begingroup$ Not a potential in the conventional sense, nevertheless people do treat escaping radiation as a tunneling phenomenon (action gets an imaginary part) e.g. here $\endgroup$
    – twistor59
    Commented Dec 11, 2012 at 21:12

Here's one simple way to look at it classically: Photons have energy, and by $m = E/c^2$, we know that energy means mass. So, photons are attracted by gravity just like ordinary matter. A black hole is simply a gravitational field that is so strong that photons can't reach escape velocity.

Now, from a quantum viewpoint, that means this: Any photon that makes it up to the event horizon will have zero energy left, and so has nothing left to tunnel through the horizon with. Or equivalently: While you are correct that photons can in principle tunnel through an event horizon from the inside out, if the only way they can reach that event horizon is by losing all their energy, the option of tunneling means nothing. So, an object within the black hole cannot send a photon out via tunneling.

But there is still one more way to interpret your question: Can pairs of virtual photons form at the event horizon, with one falling in (with negative energy) and the other escaping outwards (with positive energy)? The answer to that one is: Sure! In fact, I'm pretty sure that is the dominant mechanism by which Hawking radiation is produced in a quite smallish (e.g. dust-particle down to molecule sized) but not-quite-ready-to-go-boom black hole. The reason of course is that photons can be produced in pairs at any level of energy, whereas cases such as electron-positron pairs require a lot higher minimum level of energy to participate significantly in Hawking radiation events.

So, just as a hot piece of iron starts radiation first with photons, and has to get a lot hotter (understatement!) to start giving off matter-antimatter pair creation radiation, a black hole that is getting small enough to start radiating non-trivially will begin by giving of simple thermal black-body radiation (wow, that's almost like a pun, isn't it?).

Finally, notice the important distinction in my two answers: Photons traveling from within the black hole cannot tunnel through, and so cannot convey information about the interior. However, photons created with quantum randomness at the event surface, and thus not containing information about the interior (though there's been a lot more theorizing about this since Hawking first proposed his ideas) can radiate outward, but without providing specific information about the interior of the black hole.

Or so it is all theorized. Direct experimentation is, ah, difficult for this domain.

  • 2
    $\begingroup$ By this concept, alpha particles would never be able to escape Nucleus of radioactive element because its escape velocity can't beat Strong Force. Don't you see, I mentioned Quantum Tunneling in the question? $\endgroup$ Commented Dec 11, 2012 at 17:34
  • $\begingroup$ Well, after a couple more careful readings of your question, I do think I finally see what you are really trying to ask. See my addition above to my earlier answer. $\endgroup$ Commented Dec 16, 2012 at 4:31
  • $\begingroup$ I am unable to see any update... You are still saying the same thing. $\endgroup$ Commented Dec 16, 2012 at 4:56
  • $\begingroup$ If you have seen what I just wrote and still think I am saying the same thing... well, then I've nothing more to say. $\endgroup$ Commented Dec 16, 2012 at 5:08
  • $\begingroup$ Now, I can see texts after first paragraph. At the time of my previous comment, only 1st paragraph was there. $\endgroup$ Commented Dec 16, 2012 at 5:09

Your question highlights quantum tunneling and seems to ask for a hand-waving explanation.

First of all, strict quantum tunneling is not relevant here. Rather the hand-waving argument should rely on quantum uncertainty. Secondly, the question to ask is not "why can't a photon escape from behind the horizon", but rather "why can't a photon disappear behind a horizon". Key here is to chose a frame of reference. Our frame of reference is that of a stationary distant observer.

Suppose a distant observer throws an object into a supermassive black hole. As the object approaches the black hole horizon, he will observe the photons from the object to become red-shifted. The object falls slower and slower and never really reaches the horizon. The photons that reach the observer become more and more redshifted, and at a certain stage it is ever long radio waves that reach him. This process continues until the photons reach wavelengths comparable to the black hole horizon. At that point the quantum uncertainty of the photons gets too large to keep them localized around the horizon. At his point the object has merged with the (stretched) horizon and is just radiating ultra long wavelength photons and evaporating.

Two disclaimers apply:

1) the above is very hand-waving in character, and an attempt to describe phenomena deducted from QFT in curved spacetime in a simple semi-classical way. Keep this in mind and don't interpret all of this too literally.

2) One can't deduct from this the events the object undergoes. Black hole complementarity enters the picture here. A comment above suggests to have a look at Susskind's video (or to read his book). I echo that remark.


The book General Relativity from A to B by Geroch has a pretty clear explanation. I think i remember the key parts.

Things are basically stuck inside the event horizon because ALL paths curve back towards the center. Suppose you are at the North Pole. Start traveling any which way so long as you are approaching due south. You can even slow down your southward velocity, just so ong as its always towards south. Eventually you hit the South pole regardless of whatever side to side direction you were going. Once you hit the south pole, continue at the same velocity northward that you had when you arrived at the south pole. Again continue with the same north south direction if not speed. Now you are northward bound. Stay northward and again however you move, you eventually end up back at the north pole.

Now it doesn't actually matter where you start so long as you move in this way, you always end up at the north pole eventually.

In a sense, everything is always moving this way. Everything is always moving forward in time. The rate might change according to different observers, but it's always positive. In your rest frame, your 4 velocity is 100% in the time direction. In this sense nothing is stationary in space-time.

In the sense that you are always moving in time in your rest frame, you are moving through space and time in some other frame. There being a component of your four velocity not towards time, you have less than 100% 4 velocity in the "time direction", so your clocks tick more slowly to some other observer. Gravity "mixes" space and time. So the local trajectory of inertial motion has a non-zero change in velocity over time. Traveling South West on the surface of the earth you have a vector pointing in a constant direction from your point of view. From a cartesian coordinate perspective your direction is ever changing, always being a mix of the standard basis vectors. Change in a unit vector implies an angular change in direction. If that has a first derivative with time, that implies change in curvature from place to place in space time. This curved motion implies a pseudo force.

So things can't escape because there are no exits from inside the event horizon. All paths go back to the singularity in the center.


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.