# I Can a cosmologist cross an event horizon?

1. May 31, 2017

### Tom Mcfarland

Suppose I am orbiting a black hole (BH) at some distance d outside its event horizon (EH), and with orbital velocity v.
I do not like where I am, so I try to increase d by using an amount E of energy to increase my orbital velocity to v', where v' is whatever is necessary to escape the BH and reach a velocity of zero at d = ∞ (relative to the BH) .

My understanding is that as the initial distance d shrinks to zero, the amount of energy E required to escape to distance d = ∞ increases without bound, that is, E approaches ∞ . If my "understanding" here is incorrect,
then the following question evaporates, but another question would then arise.

OK, now reverse this process......I start at (d = ∞) and fall toward the BH, converting my newly acquired potential energy into kinetic energy (or any other form of energy). I reacquire as kinetic energy the same E as that which I needed to escape, where E increases without bound as my d shrinks to zero. If I where close enough to the BH, I would acquire more kinetic energy than currently exists in the entire universe, and would continue upwards from there !

Since the universe is not flooded with energy, this scenario seems to imply that I can never reach the EH.
Yet BH forums frequently send cosmologists across an EH. Where is my misconception?

2. May 31, 2017

### kimbyd

There are three ways that I can see to tackle this issue:
1. Potential energy, in the way that you're using it, is from Newtonian physics. It doesn't apply to General Relativity, and thus has no bearing on black holes.
2. In order to escape the horizon, the particle doesn't just need really high speed. It needs to be traveling faster than light. It can't do this, at any energy. Even at infinite energy it would still only be moving at the speed of light.
3. In order to escape, it would need to be moving faster than the speed of light at the horizon. It won't have infinite energy at the horizon anyway: it'll only have a finite energy.

3. May 31, 2017

### nikkkom

Good example.
The solution to this is realization that in GR, there is no unambiguous way to define energy *globally*. (Which, in turn, makes it impossible to define its conservation, BTW). In GR, energy can be defined locally, and is locally conserved. That's all.

4. May 31, 2017

### Tom Mcfarland

For Kimbyd: In the question, we are outside the EH, not at or inside it

For Nikkom:: Is the difficulty of defining energy conservation due to time-dependence of "velocity"? Could I add up (or integrate) small local energy conversions?
Any definition of energy conservation should give equal total energies for any process and its time reversal, right? So the problem remains.

5. May 31, 2017

### PeroK

I would suggest that your fundamental misconception is that the direction of motion doesn't matter. If you are travelling away from a black hole at a certain speed you may escape, but if you are travelling towards it at the same speed you won't.

6. May 31, 2017

### Staff: Mentor

You should be aware that, if by "orbiting" you mean orbiting in free fall, this is only possible for $r > 3M$, i.e., $3/2$ of the Schwarzschild radius, and the free-fall orbits are unstable until you get to $r > 6M$, i.e., $3$ times the Schwarzschild radius. Note that these are radial coordinates, which are not the same as the distances $d$; those are finite, but computing them requires integrating from $r = 2M$ out to the $r$ coordinate of the orbit. (There are other issues involved as well.)

This $E$ is, of course, well-defined for any free-fall orbit, but in fact it is also well-defined for any state of motion whatsoever outside the horizon, i.e., at any $r > 2M$. Since the spacetime is static, you can use the concept of "potential energy" here--but you also have to be aware that this "potential energy" assumes you are at rest, i.e., $v = 0$ (and you will have to use rocket power to "hover"). So if you are in fact in a free-fall orbit, you already have $v > 0$, so the amount of energy that you need to gain to escape to infinity will be less than it would be for someone "hovering" with $v = 0$ at the same radius.

Also, the spacetime is only static outside the horizon, so all of this only works for motions that are restricted to the region outside the horizon. So you can't use the "potential energy" concept to analyze the motion of objects that fall through the horizon.

Yes, this is true, with the proviso that we are talking about the energy required for an observer with $v = 0$ (see above).

Yes (with the proviso that you started out with $v = 0$, as above). Which means that, once you have fallen to the same radius $r$ from which you originally started, you will have the same kinetic energy $E$, relative to an observer at rest there (i.e., with $v = 0$, the same state from which you started), as you originally had to gain to escape. And if you fall further, you will have still more kinetic energy, relative to observers at rest at the radial coordinates you pass. Which means you would have needed even more kinetic energy to escape from those radial coordinates.

You're missing a key point: kinetic energy is frame dependent. Notice that I carefully included a key qualifier just above: I said kinetic energy relative to observers at rest at the radial coordinates you pass. The reason your kinetic energy is so large relative to them is that you are moving very close to the speed of light relative to them. But, conversely, they are moving very close to the speed of light relative to you--so relative to you, they are the ones with all that kinetic energy. Kinetic energy is relative.

That means, for example, that there is no such thing as "more kinetic energy than currently exists in the entire universe". It's true that, for values of $r$ close enough to $2M$, the horizon, your kinetic energy relative to observers with $v = 0$ will be so large that it would be impossible to supply that amount of energy to one of those $v = 0$ observers at those radial coordinates by any known process--i.e., that those observers would basically be unable to escape. But there is nothing in the laws of physics that says escape must be possible in practice from anywhere, arbitrarily close to the horizon.

No, it just means you've failed to take into account that kinetic energy is frame dependent. See above.

7. May 31, 2017

### Staff: Mentor

This is true in the general case, but the particular spacetime we are considering here is static, and in static spacetimes there is an unambiguous way to define energy globally. However, using that unambiguous definition, the energy of an observer with rest mass $m$ who falls in from rest at $r = \infty$ is $m$. It doesn't change at all as the observer falls.

8. May 31, 2017

### kimbyd

In your situation, you are starting outside the event horizon. If the trajectory never intersects the black hole, then there's no problem. For the most part, it will follow a trajectory defined by Newtonian physics. If it passes very near the black hole, this trajectory will be deflected somewhat (how much and in which direction depends upon the rotation of the back hole). But otherwise it will behave as expected.

Things only get curious if the trajectory of the object actually intersects with the black hole's event horizon. In that case, it disappears into the horizon never to be seen again (though there is an after-image that gets redshifted to infinity over time).

No, energy isn't conserved in General Relativity. Here's a pretty in-depth description of this fact:
http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

Sean Carroll has another description:
http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

9. May 31, 2017

### Staff: Mentor

This is not correct. The correct trajectories when you get close to the hole are very different from what Newtonian physics would predict. For example, Newtonian physics does not predict that orbits inside $r = 6M$ are unstable, or that no free-fall orbits are possible at all inside $r = 3M$. Also, more generally, the correct formula for the free-fall orbital velocity differs significantly from the Newtonian formula unless you are very, very far from the hole.

The differences are much more than just this. See above. Most GR textbooks treat the correct trajectories for this problem in some detail. Carroll's online lecture notes are free and give a good treatment.

Not in general, but we are talking about a particular special case in which it is. See my post #7.

10. Jun 1, 2017

### kimbyd

I'm not sure what you think I said, but your response really doesn't apply. None of these situations you mention refer to a particle coming in from infinity. As long as the particle is far away from the black hole, which will be the case for the particle coming from infinity case, its trajectory will be described very precisely by Newtonian physics. Close in, things can get weird, but aren't too weird as long as the particle escapes again (i.e. doesn't intersect with the event horizon). For an outside observer who only sees the particle further away, most of the difference will just be that the particle will come out in an unexpected direction if it passed too close to the black hole (how unexpected depends upon how close).

There may be an additional effect due to any gravitational radiation emitted (causing the particle to have less momentum coming out than it had going in), but I doubt that will be significant for a single pass in all but the most extreme cases.

Sort of. It is true that if you're treating the black hole as static, as with the Schwarzschild or Kerr metrics, then yes, energy is conserved. But real black holes are not static. Sure, for the most part this is an irrelevant distinction, especially in the case of a particle whose mass is tiny compared to the black hole. But it's just not strictly true that energy is conserved in the case of a black hole, or black holes in general. The conservation is only ever approximate, because black holes aren't really static.

11. Jun 1, 2017

### Staff: Mentor

The time reverse of a black hole is called a white hole. It has an event horizon where things cross only from inside to out. So the time reverse of a cosmologist being dropped in to a black hole is a cosmologist being spat out of a white hole.

12. Jun 1, 2017

### Staff: Mentor

Only until it gets close to the hole. A particle falling in from infinity won't stay at infinity, or even at a large distance.

I don't think the OP would characterize the significant non-Newtonian effects I described as "not too weird", but in any case that's a subjective judgment, not physics. The physics is that the Newtonian equations are not accurate close to the hole.

Yes, this is true, but AFAICT the OP is talking about an idealized black hole that nothing else ever falls into.

13. Jun 1, 2017

### pervect

Staff Emeritus
Using some of the equations from "Orbits in Strongly Curved Spacetime" <<link>> (which are basically the same equations as in the textbook "Gravitation" , by Misner, Thorne, Wheeler), we can find an expression for the energy-at-inifnity for an object in a circular orbit. The energy-at-infinity is a conserved quantity that is a constant of motion of the orbiting particle, and can be thought of as the total energy of the orbiting particle, if we assume the orbiting particle has a unit rest mass.

The highlights of the results are as follows: The minimum energy-at-infinty occurs at $r =3 \,r_s$ , where r is the schwarzschild coordinate of the circular orbit, and $r_s$ is the Schwarzschild radius of the black hole. This is the closest stable circular orbit. Inside $r=3\,r_s$, the energy for a circular orbit starts to increases as r decreases further. The energy-at-infinity increases without bound as r approaches $\frac{3}{2}\,r_s$, the location of the photon sphere where light orbits the black hole.

The details, using the notation and geometric units of the webpage:

We set M=1/2 for convenience, making $r_s = 1$.

The condition for a circular orbit is $3\,L^2-2\,r\,L^2 + r^2 = 0$, which has a solution of $L = r/\sqrt{2r-3}$. Substituting this into the expression for energy, E, taking advantage of the fact that $dr/d\tau = 0$ for a circular orbit, we get

$$E=\frac{(r-1)^2}{r(r-\frac{3}{2})}$$

Plotting this the key features are that as r approaches infinty, E approaches 1. This is expected, it can be interpreted as meaning that the total energy of a mass m becomes equal to $1\,mc^2$ for a particle in a circular orbit far away from the black hole. E reaches a minimum of 8/9 at r=3. Below r=3, E starts to increase as r decreases, and asymptotically approaches infinity as r approaches 3/2.

[add]. It may also be interesting to plot the angular momentum, L vs r. L also has a minimum at $r=3\,r_s$. As the webpage mentions, for most value of L there are two values of r, one of which represents a stable orbit with $r > 3\,r_s$, the other an unstable orbit with $r< 3\,r_s$

This means that with $r < 3\,r_s$, the energy AND the angular momentum increase as r decreases, both becoming infinite at the photon sphere.

Last edited: Jun 1, 2017
14. Jun 3, 2017

### Tom Mcfarland

As a mathematician unaccustomed to doing physics, I find all above wonderful comments give me homework, especially from PeterDonis !

I am indeed guilty for failing to state my original question clearly, but in spite of this, most commenters seem to have re-formulated my question so as to be faithful to the original intent......thank you.

However, though the above comments seem to have given the original question a clearer formulation, and pointed out relativistic changes in Newtonian physics needed when the "orbiting" (or falling) particle is very close to the EH, I am not seeing a resolution to the original question....that is:

If the energy E needed to escape the BH rises without bound (approaches infinity) as the particle's starting position approaches the radius of the photon sphere (as in Pervect above), then that particle, starting far away from the BH, falling toward it, would need to re-acquire that energy, perhaps radiating some of it away as it fell. If the particle indeed reached the "photon sphere" or EH, it would have acquired and/or re-radiated away an infinite amount of energy. Since we don't see a universe flooded with energy, I am concluding that the particle (or cosmologist) cannot cross the EH, but may indeed throw a very large amount of energy into his/her vicinity while trying??

Can you describe my error in mercifully accessible language?

15. Jun 3, 2017

### Staff: Mentor

That's not what pervect said. He said that the "energy at infinity" of a particle in a free-fall circular orbit about the BH increases without bound as the orbital radius approaches the photon sphere. Equivalently, the kinetic energy of the particle, relative to a stationary observer (one not moving relative to the BH) at the same radius, increases without bound as the radius approaches the photon sphere.

Inside the photon sphere (which is not the same as the event horizon--see below), there are no free-fall orbits.

The photon sphere is not the event horizon. The event horizon is at $r = 2M = r_s$. The photon sphere is at $r = 3M = (3/2) r_s$.

16. Jun 3, 2017

### Staff: Mentor

Your basic error is the same as before: you are ignoring the fact that energy is relative. The "energy" that increases without bound as a particle falls radially into the black hole is the kinetic energy relative to the stationary observers that the infalling particle passes as it falls. That is not the same as the energy that the particle has "relative to the universe" (for example, relative to an observer at infinity). It is also not the same as the amount of energy (mass) that the particle adds to the mass of the hole when it falls in. Both of the latter energies are (assuming the particle does not radiate any energy away as it falls) equal to the rest mass of the particle.

17. Jun 3, 2017

### jartsa

Let's say we have one black hole, and one 100 kg oil barrel, far from the hole.

We lower the barrel near the hole. The released potential energy we convert to a 90 kg oil barrel.

Then we lower that barrel near the hole. The released potential energy we convert to a 81 kg oil barrel.

Then we lower that barrel near the hole. The released potential energy we convert to a 72.9 kg oil barrel.

.
.
.
And so on and so forth

Then all the oil hanging near the hole is used to fuel a machine that climbs up from the hole.

The point of this story is that it looks like one barrel of oil was used in the climbing process, and it also looks like many barrels of oil was used in the climbing process.

18. Jun 3, 2017

### Staff: Mentor

Let's consider a simpler scenario in which we can freely convert mass to energy and vice versa, without any losses.

We start with an object of rest mass $m$ at rest at infinity.

We let the object fall to some point close to the horizon, and bring it to rest there, extracting all of its kinetic energy and converting it into mass. So we now have, at this point close to the horizon, an object of rest mass $m$ and a much larger object of rest mass $M$, obtained from the kinetic energy of the first object.

Now we lift the much larger object out of the hole to rest at infinity. In order to do that, we must supply sufficient energy for it to escape. So we have to add more energy into the system from outside--from infinity. But your scenario assumed we didn't have to do that. So by this method, we can't do anything with this much larger object at all.

Alternately, we could keep the energy $M$ in the form of energy and try to transmit it out to infinity, say as radiation. But if we do that, the radiation will redshift, so by the time it makes it out to infinity, it will only have energy, relative to an observer at rest at infinity, of slightly less than $m$ (the exact formula will be $mM / (m + M)$). So we haven't gained any energy from this process; in fact we've lost a little bit (because we have to stop the infalling object and extract its kinetic energy at some finite altitude above the horizon). If we wanted to get back that last little bit, we could take the original object, still down close to the horizon, convert its rest mass into radiation, and send that back out to infinity--it would redshift so that at infinity the energy we obtained would be just enough to bring the total back up to $m$.

You could try to think of other ways to transmit the energy from close to the horizon back out to infinity, but they all have the same property as the radiation: the energy redshifts as it climbs back out. So there will never be a net gain of energy at infinity from such a process.

It is true that, close to the hole, you have much more energy, relative to observers at rest there; you have energy $m + M$ instead of just $m$. But as I said before, that's because energy is relative. Energy relative to observers at rest close to the hole is not the same as energy relative to observers at rest at infinity. That's all there is to it.

19. Jun 3, 2017

### pervect

Staff Emeritus
It doesn't. The point of the calculation is that a particle orbiting in a circular orbit near the photon sphere at r=3M, or equivalently at r = (3/2)$r_s$ , has more than enough energy to escape the black hole!

We can see this from the fact that E becomes greater than 1 sufficiently near the photon sphere, at least if we're familiar with what the numbers mean. I'll admit to skimping on explaining that part, but I'll try to explain a bit more. Doing so will entail a bit of a digression, however.

Rather than deal with your initial problem of a particle falling into a black hole, I'd like to switch to a different problem for a moment, to illustrate the details of the particular concept of energy that I'm talking about, the energy-at-inifinty, and how it works.

If you have a particle of unit rest mass at infinity, it's E value is equal to 1.

If you lower said particle by a winch, with some unphysical massless rope that doesn't take any energy to raise or lower the rope, so that all the energy goes to the particle, you can extract work from the particle as you lower it on the winch.

You measure the amount of energy you extract from said particle, at infinity (this turns out to be important) as you lower the particle into the black hole. And you find that ideally, by this rather unphysical means, that the amount of energy you can extract from the particle as you lower it close to the horizon is equal to it's rest mass (or rest mass * c^2 using non-geometric units).

This also means that you can raise the particle, from at rest near the event horizon, with the same amount of energy, by reversing the winch. But to do this, you basically need to supply almost enough energy via the winch to create the particle, you need to supply almost the total rest energy of the particle to get it back out.

Note that this is not compatible with your idea that it takes an "infinite amount of energy" to get out of the black hole. Using the notion of energy that I'm using, the energy-at-infinity - this is simply not true.

But it is still true that the particle can't escape the black hole. It may only take 1 unit of energy, as measured at infinity, for the particle to escape, but the total energy of the particle near the horizon approaches zero. So it can't lift itself out of the black hole, it doesn't have enough energy.

This suggests a rather informal notion of what's going wrong. There's a notion of energy 'at ininity" that's different from the notion of energy that the particle might have. I'll try to explain a little bit about this further, though I'm not sure how clear I can make this important point.

I mentioned that the particle had a "unit rest mass", and I at lest implied that we regard this notion of "rest mass" as not changing as we lower the particle into the black hole. But this notion of unchanging rest mass is different notion of energy than the "energy-at-infiinty". Unlike energy-at-infinty, the notion of "rest mass" is not a conserved notion of energy, because it doesn't account for gravitational binding energy, which must be accounted for if we are to have a conserved quantity. Perhaps the best way to explain this different noton of energy to say that it's measured in the frame of a particle, a locally Lorentz frame, as opposed to the conserved qunatity E, which is defined in a different manner.

Now we can go back to the original problem for a bit, the problem of the particle falling into the black hole. A particle won't naturally fall in a circular orbit - the webpage I mentioned presents some of the details of how the particle would fall. Basically, the calculation involves finding the differential equations of motion. To visualize these abstract differential equations, the authors use a method of the equivalent one-dimensional system that has the same differential equation for the radial motion. This technique, which is used for Newtonian mechanics as well, is called the "one dimensional effective potential" method, and can be regarded as a visual aid for interpreting the differential equations of motion.

While a particle won't - can't - natrually fall from infinity into a circular orbit near the photon sphere, we can analyze what's required to force the particle to move in such a manner. When we do this, we find that the particle has to move at nearly the speed of light, and that it takes an unboundedly large amount of energy to force the particle to do this. We also find that the resulting motion is unstable - this isn't too surprising, since there are lower energy states a particle with the same angular momentum can have than a circular orbit near the photon sphere.

Last edited: Jun 3, 2017
20. Jun 3, 2017

### jartsa

It sounds like you are assuming there must be something wrong with that oil barrel story.

With one barrel of oil the machine in my story can climb 1 meter. We managed to increase the number of barrels to 100 or something, which allows the machine to climb 60 meters or something. Is this not possible?

Let's use ideal fuel instead of oil:

10 grams of ideal fuel has 10 times more energy than 1 gram of ideal fuel. When the fuel is used to lifting an object, the increase of potential energy is the same as the energy of the fuel, so the change of altitude is 10 times larger when the amount of fuel is 10 times larger, assuming a constant gravity.

Oh, you are saying that we at infinity can not make use that large amount of oil we managed to create. I was not planning to do anything like that. I just wanted there to be surprisingly large amounts of energy available for an observer at a low position.

Last edited: Jun 3, 2017
21. Jun 3, 2017

### Staff: Mentor

Not necessarily. I'm just giving a simpler scenario that seems to illustrate the same phenomenon--the amount of energy down close to the horizon, relative to static observers there, is much larger than the amount of energy at infinity.

You had one barrel of oil at infinity, and you turned it into 100 barrels of oil very close to the horizon. (If you want to figure out how close, calculate what $r$ has to be in terms of $M$ for the quantity $\sqrt{1 - 2M / r}$ to be $1/100$.)

Sure, the machine can climb 60 meters from the point very close to the horizon, whereas at infinity it would only have been able to climb 1 meter.

That's correct, we can't. Trying to get it back out to infinity would use up all the extra energy that was present close to the horizon. So you would end up with just 1 barrel of oil at infinity again. (In any real process, of course, you would end up with less, because the conversions in between would not be 100 percent efficient. But we're ignoring that here.)

Yes, that's quite possible. But in practical terms it's not very helpful. The observer very close to the horizon has to be expending enormous amounts of energy to "hover" and avoid falling into the black hole. Closer than $r = 6M$ there are no stable free-fall orbits; closer than $r = 3M$ there are no free-fall orbits period; and closer than $r = (9/4) M$ there is no possibility even of building a static structure that can support itself against gravity without expending rocket power. The value of $r$ for which the "energy multiplier" is 100 is much closer to the horizon than any of those. So even multiplying the energy of barrels of oil thrown in by 100 won't help the observer very close to the horizon to "hover" for very long.

22. Jun 4, 2017

### pervect

Staff Emeritus
There's one other thing I'd like to mention, that will apply to the case of supporting some mass M via an idealized massless string, which I feel is easier to discuss and than the orbital case (though I have attempted to talk about the orbital case in my previous posts).

In Wald's text, "General Relativity", problem 4 in chapter six (pg 158), Wald discusses the mathematics of supporting a mass m via such an idealized massless string in a static space-time, such as the Schwarzschild metric of a black hole. The discussion is somewhat tehcnical, being phrased in terms of Killing vectors, which are a sort of symmetry that's present in any static space-time. But the result can be explained more simply, without the technical language. The interesting and non-intuitive result is that the force exerted by the string (the tension multiplied by the cross section) at infinity is different than the force exerted at the other end of the string where it is attached to the mass M. The ratio of the forces is the redshift factor V, which can be thought of as the reciprocal of the gravitational time dilation factor.

This redshift factor V approaches infinity as one approaches the event horizon. The force on the string to hold a mass M above the horizon as measured at the mass M approaches infinity as the mass approaches the horizon, but the force-at-infinity, does not, remaining finite. This is another way of understanding why one cannot pull an object out of a black hole. The string which supports the object will eventually break, being subject to infinite tension (i.e. tension that increases without bound) as one lowers the object towards the horizion. Even though the string must eventually break, one cannot extract infinite energy by lowering some mass M into a black hole. While the tension on the string near the black hole becomes infinite, the tension on the other end of the string, at infinity, remains finite, and cannot do infinite work.

23. Jun 4, 2017

### Staff: Mentor

I have always wondered how such idealized strings are best represented theoretically. They are almost guaranteed to violate one or more of the energy conditions.

24. Jun 4, 2017

### pervect

Staff Emeritus
As it was just an exercise, Wald only gave a few hints as to how to proceed. My interpretation of how to solve the problem is this. We represent the string by its stress energy tensor, $T^{ab}$. The condition that the string be "massless" is a coordinate dependent notion, it means that in static coordinates, $T^{00} = 0$.

We do expect the string to violate some of the various energy conditions, but we don't really care, we aren't trying to make the string act as a source of gravity.

I think it's reasonable to approximate the string as being under pure tension along it's length. Since $T^{00}$ is zero, this means that the dominant source of the stress-energy tensor is the tension along the length of the string, which can be represented by $T^{11}$. I feel it's reasonable to neglect non-axial stresses on the string (for instance, compression due to tidal forces) so that we can make the only non-zero component of the stress-energy tensor, $T^{11}$.

Then the stress distribution along the string can be solved by saying that the divergence of the stress-energy tensor of the string is zero.

It's possibly I'm not being abstract enough in my thinking, perhaps Wald intended us to take some different route to reach the conclusions he suggests in the exercise. But this is the way I think about the progblem.

25. Jun 4, 2017

### Staff: Mentor

Oh, that is an excellent point.