As Newton said once, all things attract all other things (I have it on audio if you don’t believe me), so if you are reckless enough to go for a warning shot, don’t aim too close to your target, or you might hit it by accident, because gravity. You have to make sure that the point of closest approach is not inside your target. Comets in the Kuiper belt or in the the Oort cloud aiming to zoom really close to the Sun often make this fatal mistake. Now, this blog’s author has an unhealthy fascination with black holes, so let’s try to near-miss one of those. Let’s shoot a projectile such that it passes real close to the event horizon before coming back out on the other side. How hard can it be, just keep aiming closer and closer…
Uh, no. Yes, you can aim closer and closer, up to a point. If you aim too close, you would notice that your tracer bullets, instead of circling the black hole decide to dive into it. If you shoot just close enough for the point of closest approach (peri-blackhole-ion?) to be somewhere around 2-3 black hole radii, depending on how speedy your bullet is, it (the imaginary bullet) circles and comes out, but if you shoot any closer, it does not. There is no trajectory such that the bullet comes close to grazing the event horizon before coming back out.
So, a regular bullet cannot come close to the horizon and live to tell the tale. How about a really fast bullet? If it flies faster, wouldn’t it be able to come closer compared to the slower one? After all, it spends less time in the danger zone. The answer is a qualified “yes”. The higher is the bullet’s velocity, the closer its peri-hole-ion is. But there is a limit of how fast things can fly. Even a near-light speed bullet can’t get all the way down and then back up. The best it can do is to come down to 1.5 black-hole radii. There it can linger a bit, make a turn or two or 10, then spiral back out. Any closer, and it’s a goner. This closest approach orbit is known as the photon sphere. Not that there are many extra photons there, they either fall in, or come out, or even both.
This is the black hole ballistics. But not being able to graze the event horizon with a bullet does not mean that you cannot come close to it, at least theoretically, with a powered star ship. You could, if your ship is powerful, if you can stand bone-crushing g-forces and if you are very careful.
For a solar mass black hole, if you want to stop and hover near the photon sphere, the g-forces you would feel are about 10 billion times the Earth’s gravity. Less if you want to have a quick flyby, but still probably more than a human can handle. Or a ship built out of any known material, for that matter. But what about inertial dampers? No, sorry, no messing with gravity while we are studying gravity, only non-gravitational physics allowed.
But suppose you and your ship are made out of spider silk or carbon fiber, and you have antimatter source for fuel. The matter and antimatter annihilate and the resulting radiation is emitted as propellant to provide reaction. That’s as efficient as one can get without warp drives and other gravity-based tech. Why would you need to be careful? Because… well, you might end up inside the horizon without realizing it, and then it’s too late. I’ll return to this question next time. Maybe.
Assuming Hawking radiation is detected some day, it is still unclear where it came from.
Let’s first recall what happens to a signal emitted by something falling into a black hole. The closer the source of emission gets to the event horizon, the weaker, slower and “redder” the signal detected by an outside observer becomes. Theoretically, it takes forever for an outside observer to see the infalling object reach the event horizon. More practically, the detected signal gets weakened and redshifted to undetectability exponentially quickly, so that after only a short time after the redshift becomes significant the signal all but disappears, its wavelength exceeding the size of the universe.
Now, the Hawking radiation would be seen coming straight out of a black hole, but, if you trace it back to its source, where would it be? Clearly it cannot have come from the black hole horizon itself, since nothing can leave the horizon. If the black hole in question is paired with a white hole, then Hawking radiation can conceivably come from the white hole. It would have to originate somewhere near the white hole singularity and then be hurled out through the white hole event horizon. But there is no clear reason why thermal radiation would be the only thing emitted. After all, the known laws of physics are likely to break down near the singularity and we have no idea what is going on there. Besides, the original calculation by Hawking and others assumes only that the black hole event horizon exists, and nothing more exotic, like white holes or singularities.
Let us come back to how Hawking originally popularly explained the origin of the black hole radiation. He interpreted his calculations as follows. A wave packet (a short “ray of light”) going outward near the black hole horizon “splits into two”, one with “positive frequency” (and positive energy) going outward and one with “negative frequency” (and negative energy) going inward and getting trapped inside the black hole. Thus the black hole “sucks in” negative energy and loses mass. But where does the original wave packet come from? According to Hawking, the vacuum fluctuations are the source. A fluctuation with total zero energy gets split as described and the part with positive energy goes outward and becomes part of the thermal radiation. In essence, he treated black hole as an atom in a highly excited state, only with gravitational field instead of electromagnetic one.
So, according to this logic, Hawking radiation originates somewhere close to the black hole horizon, since the vacuum field fluctuations have to live long enough for the negative-energy part to be consumed by the black hole. But remember that the signal emitted from near the horizon gets extremely weakened when escaping the black hole. Which means that if it is still measurable, it must have been really really strong near the horizon. How strong? A typical thermal photon detected far away from a stellar black hole has been weakened over billion billion billion billion times. It must have been fantastically strong near its birthplace. It would have to be more energetic than anything we ever observed in the Universe. More energetic than anything LHC can ever hope to produce. Stronger than gamma-ray bursts coming out of supernova explosions or colliding neutron stars. One would be hard-pressed to attribute the birth of such energetic photons to mere vacuum fluctuations near horizon.
So, what on Earth (well, in space) is going on? There is no clear answer, as far as I know. The hints from the so called AdS/CFT Correspondence suggest that “black holes are time-reverse dual to themselves, and thus black hole and white hole are the same object”. How is this statement to be understood in a more macroscopic context? I hope to return to this question when I figure it out 🙂
Slightly more calculational details and mildly illuminating pictures are at http://www.scholarpedia.org/article/Hawking_radiation and http://physics.stackexchange.com/questions/22498/from-where-in-space-time-does-hawking-radiation-originate.
Continued from Part II.
Last time we saw that if someone decided to plunge into a white hole, there is nothing preventing them from reaching the horizon in the finite time. The question was, what happens next, given that there is no way to cross the white hole horizon inward, just like there is no way to cross the black hole horizon outward? To make progress in answering this question, let’s imagine what someone reaching the horizon would feel. He (to pick a gender pronoun at random) would notice the usual tidal gravity forces stretching him in the direction of the fall and squeezing him in the perpendicular directions.
And by the way, Einstein tells us that the squeezing force per unit distance in the two perpendicular directions together is equal to the stretching force per unit distance in the direction of the fall. (The “per unit distance” qualifier here is essential: tidal forces are always like that: the large the object, the stronger the tidal force.) And the closer you get to the horizon, the stronger these tidal forces get, whether it’s the black hole or the white hole horizon. Oh, and nothing special happens in your vicinity when you cross the horizon, you just keep falling. The horizon is a point of no return, to be sure, but you wouldn’t know it unless you actually tried to return.
So, you reach the white hole horizon, and you keep going, and the tidal forces keep kneading you into mush, eventually ripping you apart… Wait, what? I am describing the act of falling into a black hole, am I not. Well, right you are. So, what is this BS about white holes then? Ah. Remember, falling into the white hole, at least up until the horizon, is just like going away from a black hole, only with the time running backwards. And since there is nothing special about the horizon, you ought to go through it like there is no tomorrow… Which there won’t be, of course. But I digress.
Yes, falling into a white hole feels just like falling into a black hole, only you run a chance of being hit by the random junk coming out of the white hole and die somewhat prematurely, if not very much so. You see where it is going now, do you? Toward the singularity in the future of the infalling observer. Moreover, once you cross the horizon, you cannot get back. So, what I am really describing is that, from the point of view of the guy falling in, he is falling into a black hole. While from the point of view of a guy falling out, he is falling out of a white hole.
Which means that every white hole is also a black hole. Like Yin and Yang, they come in pairs. Let me qualify that. Any potential white hole would have to be paired up with a black hole. But it is not necessarily true that any black hole has to be paired with a white hole. In fact, it is most likely that they are not. Black holes are formed when massive stars collapse, and there is no obvious way to create a white hole in this process. In fact, we don’t know of any way to form a white hole. Maybe some could have formed during the Big Bang, and survived until the present day, who knows. This is a pure speculation. But if some did form and survived, they would most certainly be paired with black holes.
Now to summarize my answer to the title question: you cannot fall into a white hole because what you would really be falling into is a black hole!
But this is only scratching the surface, so to speak, of the tangled black hole/white hole relationship. For example, if something falls into a black part of the pair, and the black hole grows in size and mass, does the mass of the associated white hole change? Or, if you nudge the black hole with something really heavy, do you also move the white hole? if so, how can you, the white hole singularity is in the past, and you cannot change the past, can you? Also, what about the famous Hawking radiation? Does it come out of the white hole horizon? I hope to get to discussing these questions some day.
To learn more: eternal white holes and black holes are described in a semi-accessible way in the book Black Hole Physics: Basic Concepts and New Developments, by Valeri Frolov and Igor Novikov.
Continued from Part I.
So, the question is, what happens if you shine your flashlight into a white hole? Let us look at the black hole again, first. Shining light into a white hole is equivalent to shining light out of a black hole. Unfortunately, you cannot shine light from out of a black hole! Not from inside the horizon, anyway. But you can, from just outside. As far as an outside observer is concerned, it takes a very long time for the light to get out far from the gravity well. Just like it takes a long time for the light going in to get very close to the horizon. Light also blue-shifts a lot as it climbs out, just like it red-shifts when going in.
So it takes almost forever until you can see something coming out from near the horizon. But how long does it take for that something to get out by its own internal clock? After all, if you are falling into a black hole, you can cross the horizon pretty quickly by your own clock, even though it looks like it takes forever to anyone who is watching you fall in. And the answer is rather surprising: it takes exactly the same time to get out as it takes to fall in, assuming you have the same speed after you come out as you had when starting to fall in.
This is pretty easy to understand in the familiar Newtonian case: a ball bouncing against the floor doesn’t take any longer to bounce up than to fall down. The situation is the same in General Relativity: the bounce takes just as much time as the fall, if there is no energy loss. Of course, it is much harder to construct something to bounce off if you are dealing with an object so dense, it is on the verge of collapsing into a black hole. The only known objects with a hard surface which almost as dense as black holes are neutron stars. And you cannot bounce a ball off them, not really.
Now, back to the white hole case. What’s the counterpart of an object falling into a black hole? It’s an object coming out of the white hole. Nothing really special there. Just like something thrown into a black hole appears to merge with the horizon for an outside observer while crossing through the horizon and hitting the singularity shortly afterwards by its own clock, an object emerging from the white hole’s singularity and being hurled through the horizon and into the outside world, while appearing to slowly unglue from the horizon and zip outward.
Now consider the opposite case: an object leaving a black hole from close to the horizon, straight up, without engines. It is quite possible, if the object has enough speed to achieve the escape velocity. To escape, it would have to start traveling close to the speed of light relative to something hovering just above the horizon. Or it can be a light ray emitted by that something.
So far, so good. Now, again, let’s consider the time-reversed situation: a white hole and an object or a ray of light falling into it. To the outside observer, the object disappears near the horizon, just like it did in the case of falling into a black hole. But what happens from the point of view of the object itself? Remember, it takes as long to reach the white hole horizon as it takes to climb out of the horizon of a black hole, which is also as long as the time to fall to the horizon of a black hole. So, an person falling into a white hole will reach its horizon after a fairly short time… and then what? By definition of the white hole, which is a time-reversed black hole, nothing can fall through its horizon, no matter how hard it tries. But… what happens to that poor schmuck who reaches the white hole horizon at full speed and has nowhere else to go? What prevents him from going forward? Join us next time, for a surprising answer!