You might be wondering why I picked 10g as the star drive’s acceleration and whether it is gives you much vs a leisurely 1g or less. Let us deal with this question.
First, assuming perfect mass-to-energy conversion, how long can we keep accelerating before all the fuel is used up? (We will not be considering ramjet-style refueling by scooping interstellar medium.) An honest calculation would require some calculus, but we can skip that by noting that when rate of change as a fraction of magnitude is a constant, the magnitude decreases exponentially. A well-known example is radioactive decay.
So, suppose we start with a rocket ship with mass M. If every T second we eject m kg of fuel with velocity v, then the thrust pushing the rocket forward is and its acceleration is . After T seconds the rocket’s mass is , so to keep the acceleration constant we need to reduce thrust correspondingly, and continue to do so as the rocket gets lighter and lighter. As a result, the rocket’s mass decreases slower and slower with time, getting to about 37% of its original weight after seconds (and to 37% of that after another t seconds). For a light-speed exhaust this works out to be only 35 days at 10g acceleration.
This should give you pause. It sure made me check this simple calculation, just in case. With our star drive we would have to basically annihilate 60% of the star in just one month. For scale, the Sun converts not even 10% of its mass into energy by burning Hydrogen then Helium for over 10 billion years! And we plan to burn that much in one week! Or at least one week by the ship’s clock, as we would be close to the speed of light after only a few weeks, when time dilation firmly sets in. Still, this is very much comparable with supernova explosions, only going on for weeks non-stop.
Now would be a good time for environmental considerations. If the exhaust consists mostly of electromagnetic radiation, then whatever is in the path of the exhaust would not fare well, not within a few light years, anyway. Says Wikipedia:
Gamma rays induce a chemical reaction in the upper atmosphere, converting molecular nitrogen into nitrogen oxides, depleting the ozone layer enough to expose the surface to harmful solar and cosmic radiation
And that’s a few dozen light years away. Using near-light speed massive particles, like Hydrogen plasma jets is just as bad. CEPA, the Cosmic Environmental Protection Agency, might be a bit put out with us for destroying all life on a planet or ten.
Fortunately, there is a way out. Kind of. If the exhaust mostly consisted of neutrinos, the impact on the surrounding Cosmos would be minimal. Neutrinos go through everything almost completely unaffected, so even at a few million km the radiation exposure from them is quite small.
Interestingly enough, most of the energy released in a core-collapse supernova, the kind where a large star runs out of gas and collapses on itself until everything in it turns into neutrons, is in the form on neutrinos. However, this is just way too weak by our standards, as most of the energy is not released, but remains as its neutron core or the resulting black hole. Yes, that’s right, one of the most dreadful star explosions we observe, visible a few galaxies over, just doesn’t produce enough energy to power our star drive.
There are good reasons that it is very hard to turn normal matter into neutrinos completely, most of them are related to conservation laws. Specifically, we currently do not know of any way to turn “quarks”, the stuff of which atomic nuclei are made into “leptons”, which is what neutrinos are. It is entirely possible that as-yet-unknown laws of physics come into play at very high energies, but, if so, this energy would have to be higher than anything a mere supernova can produce, since we don’t see supernovae disappear in a puff of neutrinos.
One bit of good news is that this is not an issue for black/white holes, as quark number is not conserved (or at least not visible) after the black hole is formed. Only mass, angular momentum and electric charge are preserved by the collapse. So there is a hope that if we give up on “real” stars, like white dwarfs and neutron stars, and build our star drive out of the black/white hole combo instead, then there is no restriction on the type of material emitted.
Anyway, back to our calculations. Suppose we have run at full power for some time, and burned the fraction X of our star drive. How fast are we going now? Well, almost at the speed of light if X is large enough. What is more interesting is not the exact speed, but the time dilation/space contraction factor. Because this factor is what tells you how fast you get to where you are going. Like, if the factor is 10, then you get across 10 light years distance in only in one year local time.
So the space contraction factor for a 100%-efficient rocket with light-speed exhaust is . While the expression itself is simple, I have not found an easy way to derive it, without cranking through the relativistic rocket calculations, so please forgive me for just dropping it here without a proper motivation. And if you know of a way, please comment.
Let’s plug in some numbers! As we have seen earlier, after one month of travel at 10g (or after almost a full year at just 1g) we are left with 40% of the fuel we had at the start. Plugging in X=0.4 in the time-dilation formula above, we get , not a very impressive number. It only reduces our subjective travel time by barely 30%. If we give it another month, we are down to X=0.16 and . Only when we burn away 95% of the star we get 10 times for our time dilation effect. How long will it take at 10g? About 3 months ship time.
So, we accelerate until our time dilation factor is 10, then what? Turn off the star drive and cruise? But then we are left weightless for most of the flight, until we start decelerating, which kind of defeats the purpose of our ingenious setup to provide a steady 1g for the crew. Also, how long would we have to cruise weightless? With the time dilation factor of 10 it would take full 3 years to get to, say, Vega, if you count acceleration and deceleration. Including 2.5 years with the drive off. And after decelerating we are left with 5% of 5% or just one quarter of one percent of the original star. Somewhat wasteful, is it not? We used up a whole star in a most efficient way imaginable to get not across the Universe, not across the Galaxy, but across less than one one-thousandth of the Milky Way. And it took us three years to do it.
To summarize the disappointing answer to our original question, 10g acceleration gets us the time dilation/space contraction factor of 10 if we accelerate for 3 months and burn away 95% of the fuel. But wait! Not all is lost quite yet. This result is for a conventional rocket, not a black-hole joy ride, where the effects of General Relativity are also important. We will see if it makes any difference next time.
Last time we designed our literal-star-drive-based space ship:
and figured out that to provide 10g of acceleration in the frame of the star while surfing its gravitational wake in free-fall, we would have to trail the star drive by about a million kilometers (or miles, the calculations were not very precise). We also noticed that the tidal gravity from the star limits the size of the crew module to tens of kilometers, or the size of a really small moon, like Phobos. Anything larger is going to be stretched beyond breaking point and ripped into pieces. This size is certainly nothing to sneeze at, but it has certain limitations. For example, if you want to travel comfortably without leaving your home planet and instead use it at the crew module, it is probably too large to remain in one piece. So even the largest possible crew module does not provide enough natural gravity to feel comfortable.
Of course, gravity does not have to be natural. One can always produce enough centrifugal forces by spinning the module fast enough. Many sci-fi stories rely on that. A ring of 1 km in radius only needs to make a full turn once a minute to give the illusion of 1g gravity on the inside of its surface. “Only”? Seems a bit dizzying, to see the world around you complete a full rotation in barely one minute. Just over 3 min for a 10 km ring. Still not a lot of fun. Imagine all those distant stars zipping around you every minute. So, just for fun, let’s consider other alternatives. After all we’ve been playing with gravity for some time in this series, why stop now?
But first… Let us return to the issue of stability. If you ever surfed, skied or even walked (and I hope you have done at least that last activity, otherwise you are probably just a brain in a jar), you realize that it requires some effort to remain on the sweet spot of the wave. If you goof, you are either too far down or too far up and probably off your board. And that’s assuming you don’t fall over sideways. This is because the equilibrium you achieve is unstable. It’s more like balancing a long stick vertically on your finger than holding it hanging down. You can do it, but you have to constantly compensate for the damn thing trying to fall. It is the same with our star-drive setup. If the crew module lags a bit behind the sweet spot, it would tend to keep lagging farther and farther, until it is left behind in the darkness, forever lost. If the crew module slides a bit forward, the star’s gravity will pull it harder and it will plunge toward it and disintegrate in the short order, which is also a suboptimal outcome. So we have to maintain constant vigilance and compensate for random deviations from the optimal surfing location. How would we do that? One obvious solution is to have small rocket engines — maneuvering thrusters — to provide course corrections. This means having to carry extra fuel, maintain the engines, etc. While inescapable in other circumstances, using on-board engines for course correction seems a bit silly when we are literally blowing up a star for fuel just ahead of us.
Note that in the diagram above I carefully drew the exhaust skirting the crew module. Mainly because I did not want the crew members to be incinerated by a stream of radiation strong enough to accelerate a star-massed object at freaking 10 gravities! We will get to the question of how terrifyingly awesome this exhaust must be later in this post series. For now let us note that if the star drive goes out of alignment only a little bit and the propellant hits the crew module, the crew does not stand a chance in hell, because hell is what it would feel like for the scant few seconds until they get evaporated. However, in homeopathic doses this same exhaust might be helpful, rather than harmful. (Who said homeopathy doesn’t work?)
If only a small amount of exhaust hits our crew module, it nudges us backward. This can be good if we stray too close to the star. A useful side effect is that it provides some acceleration, which to the people inside would feel like gravity. Isn’t it nice to kill two birds with one stone? Umm, this metaphor might be a bit cruel and dated. Throwing stones at birds might have been considered good clean fun a couple of centuries ago, but not by the enlightened standards of the present day. Anyway, I digress. We want to use a small amount of starlight to stabilize the crew module and to simulate gravity. Again, this idea is not new, using solar sail for interplanetary travel has been discussed for some 100 years. A German-Russian-Soviet scientist and engineer Friedrich Zander was apparently one of the first to do a serious calculation of the solar sail propulsion, during his downtime between designing one of the first liquid-fuel rockets (well, second, after Goddard in the US, but before von Braun in Germany) and figuring out the details of a gravitational slingshot, a now-standard technique for interplanetary travel.
There is at least a couple of ways we can design our solar sail: we can put it in front of the crew module, like a rocket:
or behind it, like a parachute:
There are advantages and disadvantages in both cases, as evident by the fact that both approaches are in use. A parachute-style sail provides stability for free, while a crew module equipped with a reflector would have to be actively stabilized, the way planes and rockets are. On the other hand, a heavy module precariously suspended on something like ropes… What if one of them breaks? And how will they fair while being bombarded by the exhaust rays? In either case, as you can see, the direction toward the star becomes “down”. The preferred location of the crew module would be somewhat ahead of the free-fall point. How far ahead? Not very much. If the star drive provides 10g acceleration, we only need to get to the 11g zone to get 1g apparent gravity. And since the gravitational force is proportional to the inverse square of the distance, we only need to be 5% closer to the star to get that extra 1g, or around 50,000 km closer.
Interestingly, the solar sail provides us something like a neutral equilibrium: if we get closer to the star, its gravity gets stronger, but so does the exhaust density, which also falls off as inverse square of the distance, or close to it. So we still need to do some course correction, but not nearly as much. There are also some other small effects which affect the crew module, like a slight blueshift of the exhaust during the time it took for it to reach the sails, the gravitational effects of the exhaust left behind, and a few others. However, we cannot properly address them without doing a fully general-relativistic analysis, and they are too small to matter, anyway.
So, as promised, we have improved our star ride to make it more stable, easier to control and more comfortable by diverting a tiny fraction of the ejecta toward the solar sail. We have not yet calculated how big or strong the sail would have to be. Hopefully next time. We will also look into how far 10g acceleration gets us and how fast. In the meantime, please feel free to comment or ask questions!
Last time we started by trying to reduce the effects of g-forces on the crew and ended up considering a star corpse as our star ship. Here is a schematic drawing of how it looks:
Yes, this is a 5-min drawing in Google docs. If you are inspired to do better, let me know and I will gratefully replace this “drawing” with yours, and give you credit, of course.
Here is what is happening on this picture: A star is used as a propulsion source controlled by as-yet-unspecified technology and emits its content at or near light speed in one direction. As a result, it is pushed in the opposite direction, like a rocket. A crew module is positioned at a respectable distance (which we will shortly calculate) and is in a free fall toward the star. While the star in question keeps getting away. If you position the crew ship just right, it will be stationary relative to the star, even if the acceleration of the star itself is many times Earth’s gravity.
Let’s do a few simple calculations to see how far from the star we should place our ship. We assume that the star used as a star-drive is roughly one solar mass. This is not an unreasonable assumption, as all white dwarfs and neutron stars and even newly formed stellar-remnant black holes are in that range. Instead of dealing with the large and inconvenient numbers, like the gravitational constant, or the Sun’s mass, I will use only a few basic small ones I remember:
- Free-fall acceleration at earth’s surface (1g).
- Earth’s speed in its orbit around the Sun (v = 30 km/s).
- Distance from Earth to Sun (r = 1 astronomical unit = 150 million kilometers).
- Newton’s law of gravitation: the attraction force falls as the square of the distance between two bodies.
For now, we will do only the basic Newtonian physics, no relativistic corrections whatsoever. Because none are needed just yet: we are dealing with relatively slow speed and weak gravity. First, let’s calculate the centripetal acceleration due to the Earth-Sun attraction force: it is So, to get a 1g acceleration we need to get 40 times closer to the Sun,down to only 4 million kilometers from its center, or barely 3 million km from its surface, 10 times closer than Mercury. Clearly this is not a healthy place to be in. That’s one reason that a white dwarf would work better. Hopefully. If we want to have our star drive to work at something like 10g, we’d need to be 3 times closer still, down to just over 1 million km from the center. This close even relatively dim objects like white dwarfs and neutron stars would probably make your life uncomfortable, no matter how good your radiation shielding is. So, we are likely down to our last star drive candidate, the black hole. Unless we use a huge crew module… like an asteroid-sized one? Made from an actual asteroid, maybe? But will a relatively large object like that withstand the tidal forces exerted on it by the star’s gravity? Let’s do another simple calculation!
Given gravitational acceleration and distance from the massive object, what is the tidal force per unit length? Well, dimensional analysis to the rescue! There is only one way to get acceleration per unit length from these two numbers, and it is to divide them: ! (An honest calculation which involves a small amount of calculus gives us an extra factor of 2.) For an asteroid 100 km in diameter the resulting tidal gravity between its “bow” and “stern” is . This is a rather inconvenient number. On one hand, it is too small to provide a comfortable level of gravity to the crew, on the other hand, it is large enough to tear our gravitationally-bound asteroid apart, or at least make it shed most of its mass, until it is small enough for the cohesion forces to dominate self-gravity, a few dozen kilometers at best. How did I calculate that last number? I didn’t. I cheated. I looked up shapes of some moons and asteroids, like Phobos, and picked the size where they stop being more-or-less round.
Let’s review what we have figured out so far. If we get a white dwarf-powered star drive provide a 10 g acceleration, we can make a Phobos-sized crew module surf its gravitational wake some 1 million km behind while nearly weightless. This answers our first question posed in the previous post: “How far behind the star-ship is the sweet spot for the crew?”. We shall discuss the next question, how stable the ride is, in the next post. We will also talk about ways to provide Earth-like gravity for the crew without expending a lot of on-board fuel.
We have been talking about white holes and black holes and how one cannot exist without the other (or at least white without black). But let us return to the slow and familiar world of Sir Isaac Newton for the moment.
Suppose you have a massive star ship. Maybe one carved from an asteroid or even a planetoid. To protect the crew from the dangers of the outer space. And it has plenty of material for propulsion, assuming some day the rocks and metals can be used as fuel and/or propellant.
Now, once in flight at a 1 g acceleration the crew inside would feel the normal Earth’s gravity. However, someone on the surface of the ship would feel either heavier or lighter, depending on where exactly they are standing. On the bow of the ship the asteroid’s gravity would add to the ship’s acceleration, while on the stern the apparent gravity would be less than the ship’s acceleration by the same amount. Just take care to not get too close to the engines and exhaust. Imagine how powerful and probably hot they must be to accelerate something like a Death Star at 1g.
Now, the surface gravity of an asteroid is pretty small. Even on the largest known one in the solar system, Ceres, it is not even 1/30 of that on Earth, so it would not significantly affect the apparent gravity on the ship. But if the asteroid was made of a denser material (and be correspondingly smaller for the same mass), its gravity would be stronger. Same mass in half as much radius means four times as much gravity, etc.
This would present an opportunity. Suppose we want to accelerate to the cruising velocity faster than just at 1g. Instead of subjecting the crew to strong g-forces for an extended period of time, we could relocate them toward the stern of the ship, where the ship’s own gravity would counteract the g-forces. Unfortunately, no known material is dense enough to provide 1g worth of surface gravity in a relatively small object. So, unless we want to just take a whole Earth-sized planet, our hope of counteracting the effects of acceleration are in vain. Plus, any self-respecting planet keeps itself in shape thanks to its own gravity. It would quickly rearrange its shape or even fall apart if we were to do something as violent to it as accelerating beyond the gentlest nudge.
But wait, not all is lost! Why think small? We know of a few of celestial objects which are dense enough and durable enough to withstand a bit of rough handling. Alas, none lend themselves easily to carving a star ship out of. But let’s see where this leads us. One such dense object is a white dwarf, a remnant of a sun-like star. Its surface gravity is some millions of g, so an extra g required to accelerate it would hardly make a dent. Literally. Of course, there is the small matter of making the white dwarf into a rocket engine, but let’s suppose we solved this minor problem and the cooling star corpse is made to emit its guts in the right direction at something close to the light speed to provide acceleration.
Now, our crew cannot, of course, make their quarters inside or on the surface of the dying star, it is still way too hot, Sun’s surface hot. It is also as massive as the Sun. So our initial plan to ride an asteroid has been inflated somewhat. Still, size-wise a white dwarf it is relatively small compared to a “real” star, it is only maybe Earth-sized. So we need to keep our distance, or radiation and gravity will do us in.
Before we do some calculations regarding the practicality of star-riding, let’s look at other options. One object even denser than a white dwarf is a neutron star. They are only slightly heavier than white dwarfs, but much denser and smaller (a city-size, rather than a planet-size) and so emit a lot less radiation in total, even though they are hotter. Also, turning one into a rocket could be somewhat problematic, given how strongly it is gravitationally bound. On the other hand, pulsars manage to fling a lot of energetic matter and radiation into space, so maybe the task of rocketizing a neutron star is not as impossible as it looks at the first glance.
The last option for high-acceleration star riding is, of course, the object dear to my heart, the amazing black and white hole combo. Black holes are roughly as massive as the other two super-dense objects, only smaller, maybe a dozen city blocks in size. And black holes usually do not emit anything, so that is both convenient and annoying. On one hand, you don’t need to block the potentially harmful radiation from the star, on the other hand you cannot harness this radiation as a source of energy. On the third hand, if our project involves milking stars for fuel, the scale of the energy sources required to do that is probably rather larger than what solar batteries can provide.
So, we are back to black holes, bye-bye Newton, hello Einstein. So, how the heck would we extract energy from a black hole, let alone shape it as radiation emitted in a specific direction as exhaust? Well, the actual mechanism is a bit fuzzy, just like it is for white dwarfs and neutron stars, but the god news is that, just like Tsiolkovsky was the first one (well, not really first, but hey, it’s Stigler’s world out there) who figured out the rocket equation for the non-relativistic propulsion, one William Kinnersley did that for relativistic one, at least when the propellant is massless. This is known as the Kinnersley photon rocket. And it just so happens that it describes the last case we discussed, at least to some extent: light, or something like light is emitted from a black/white hole preferentially in one direction, accelerating the hole in the opposite direction.
Now it is almost time for some basic calculations. But first let’s review what we have figured out so far. We wanted to use starship’s natural gravity to reduce the effects of g-forces on the crew and quickly realized that the ship would have to be very dense for this to work. And, as far as we know, dense means heavy, Sun-mass heavy. So we’d have to turn a star into a star-ship, literally. And this means we have to keep our crew a ways out, behind the star-ship, surfing its wake. Maybe it should be called star-surfing? Things we still need to figure out are manifold, but here are some of them, in no particular order:
- How far behind the star-ship is the sweet spot for the crew?
- How large and stable is that sweet spot?
- How fast can we accelerate and how far can we travel until the star-ship is all used up?
- The twin paradox and all, how much will the crew age during a round trip, compared to those left behind?
- Is it even ethical to kill stars for fuel, even if they are already [almost] dead?
To be continued…
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!
We all know that black holes are dangerous, if it swallows you, you are a goner. You will be stretched, ripped in half, folded and squeezed, again and again, until not a single particle of your body remains intact, before meeting an unfortunate demise of being absorbed by the singularity that eats all that falls in. You have no chance to survive… and, unlike in the old meme, you cannot make your time. The only safe option is to study the beast from afar. But, as usual, this coin, too, has a flipside. The mighty theory of relativity predicts that the time-reversed version of a black hole is just as valid an object. Enter the white hole.
If you thought black holes are weird, white holes are even weirder, trust me. Everything you learn about a black hole applies to the white hole, only with time running backwards. And this is where the weirdness starts. You cannot fall into a white hole, but something (or someone) might fall out of it. Yet, white holes have mass, just like black holes, so they attract things. Wait, what? They attract you but you cannot fall into them? How does that work? Can you get close to one? I mean, it attracts you, so you better be able to. From afar, you cannot tell without looking if an object that attracts you is a black hole, a white hole, or a garden-variety star, all you know is that something massive pulls you in. So, what magic kicks in when you get too close? Do you suddenly feel an overwhelming repulsion (this is a physical term, not a emotional one) and bounce right back? Do you try to fall in and somehow miss it? Does something even stranger happen? Let’s have a look.
Suppose you point your flashlight toward a black hole. What happens to the light beam? Here I have traced the rays and got the following picture (click to enlarge):
On this picture the light rays enter stage right, woosh around and disappear. The ones which do not get close to the black hole are barely deflected and exit stage left. As the light rays come closer to a black hole, they get deflected more and more. Actually, this happens with every massive object. Our own Sun deflects light, if only a little bit, and that was how the Einstein’s celebrated General Relativity Theory was first tested. But black holes bend light like there is no tomorrow. Which there won’t be for any daring ray of light which comes too close. You can see how the few really close ones hit the black hole and never come back out. If you were to shine light directly at the black hole, it would, of course, disappear without any deflection. There is one circle on that picture which shows that, if done just right, you can almost make a light ray spin around the black hole forever. “Almost” because you can’t, this circular orbit is unstable and before you can say “unstable equilibrium” the circling light ray will either fall in or spiral away and out to infinity.
This picture might remind you of the comets coming from the edges of the Solar system, circling the Sun and disappearing back to the edge, to not be seen again for a long long time, often forever. Black holes have strong enough gravity to do to light what the Sun does to comets.
Now, remember that white hole is black hole backwards in time? Well, this means that on that picture, you can run every light ray in reverse like a video run backwards, and you get the light deflection by white holes out of it for free, no need to do more calculations. So, for every light ray which comes from far away, spins around a bit and go back to the infinite void, there is another light ray which comes on the exact opposite trajectory.
So far, so good. Just like with black holes, light rays come in, light rays come out. There are also some light rays which come out of the white hole and disappear into infinity. That’s the “white” part, of course, you can never tell what will come out next, if anything. And now riddle me this: what happens if I shoot a ray of light straight into a white hole? Will it fall into it? Well, it cannot… nothing can fall into a white hole, just like nothing can escape a black hole. So… what happens?
Let’s think about it for a moment. Every possible light ray trajectory for the white hole has its counterpart trajectory for the black hole. What would be the black hole counterpart of “shooting a ray of light straight into a white hole”? Well, let’s time reverse it: it would be a ray of light coming “straight out of a black hole toward you”. And we know this is impossible. So, what goes wrong? Why can’t you shine light at a white hole? Does some magic cosmic force subvert your aim? After all, you can shoot what you can see, right? So, if you were to see a white hole somewhere in the sky, you just aim with your flashlight and voila! Only, apparently, not. Yes, relativity is weird, and general relativity is generally more so. Stay tuned.