# Black Hole Engine

1. Nov 2, 2005

### Longstreet

Someone posted an idea on another forum describing a scenario which basically extracts the self energy of a black hole. You add mass to the black hole but somehow you control the fall down to the Swartzchild radius giving a net energy gain through some type of dynamo.

My first attempt at how much energy you get out, just using non-relativistic energies, gave a pretty large number approaching the rest energy of the object you lowered into the black hole. However, I realized that you could somehow package the energy and then lower that down into the black hole, rinse and repeat. This was confusing because according to the energy equations I'm adding more energy to the black hole by doing this than if I just let the thing fall in on its own.

So then I altered it to include relativistic energies. However, now it says if I let the thing fall in on it's own it adds an inifinite amount of energy to the black hole when it reaches the Swartzchild radius!

Obveously the self energy of the black hole can't be infinite or else it would destroy the rest of the universe during it's formation. Also, the energy you could extract by lowering masses into the black hole should converge to something. I think the only way for this to converge is through the magic of GR, which I know very little about. Any help at understanding this is appreciated.

Thanks,

2. Nov 3, 2005

### Stingray

I think that what you're looking for is called the Penrose Process. It is a method of extracting energy from rotating black holes - not Schwarzchild.

3. Nov 3, 2005

### George Jones

Staff Emeritus
Let me elaborate a bit on what Stingray said.

For rotating black holes there is a conserved quantity that is: energy for observers far from the black hole; a component of spatial momentum for observers near the black hole.

Suppose particle A is far from a rotating black hole and has 10 for the value of this conserved quantity. Suppose further that particle A's trajectory takes it near a the black hole, and that explosives attached to A break into particles B and C. Part of 10 goes to B and part goes to C, say B picks up 13 and C picks up -3. Everything is OK because: 1) 13 + (-3) = 10, so the conserved quantity really is conserved; 2) near the black hole, the conserved quantity is a component of spatial momentum, so the fact that B's value is positive and C's value is negative just means that B and C move in roughly opposite (spatial) diretcions. Finally, suppose that C falls into the black hole, while B escapes to a place far from the black hole.

Note what happened:

1) initially A was far from the black hole with value 10 for the conserved quantity;

2) far from the black hole the conserved quantity is energy, so the initial energy far from the black hole is 10;

3) in the end, B was far from the black hole with value 13 for the conserved quantity;

4) far from the black hole the conserved quantity is energy, so the final energy far from the black hole is 13;

5) particle C fell into the black hole, thereby increasing the mass of the black hole.

How is it possible to increase energy from 10 to 13 and increase the mass of the black hole all at the same time? In effect, rotational energy has been extracted from the black hole.

Regards,
George

4. Nov 3, 2005

### Longstreet

No I'm not talking about anything like that. An analagy closer to home would be to drop a rock from orbit and somehow use all the heat it generates from falling through the atmosphere to drive a turbine. Of course, gravity doesn't result in much energy on earth, but near the swartzchild radius of a BH it seems to give quite a bit. I just haven't been able to accuratly determin how much.

Thanks,

5. Nov 3, 2005

### pervect

Staff Emeritus
Justifying the answer below would take a lot more work than just giving it.

If you imagine lowering an object of mass m into a non-rotating black hole with a string, eventually the string must break if it reaches the event horizon, and it will probably break long before that.

The net energy that you would get out "at infinity" via this process (after you put in the energy necessary to raise your string to it's original position) would be

(1-g_00) m c^2

where g_00 is the metric coefficient of space-time at the point at which you have to drop the mass "m" to prevent the string from breaking

g_00 = (1-2GM/c^2r)

in terms of the schwarzschild coordinate 'r' for a non-rotating black hole

You can see that the limit is the case where g_00 = 0, which happens at the event horizion. However, it's probably that any string you can actually construct, even carbon nano-fiber, will break before you can achieve even a fraction of that energy.

Just dumping matter into the accretion disk at letting it radiate electromagnetic radiation back out to infinity will give something like a 30% matter/energy conversion efficiency according to some recent estimates

http://www.universetoday.com/am/publish/early_black_holes_grew_quickly.html?1762005

6. Nov 3, 2005

### Longstreet

I can beleive that. However, my main problem is that if g_00 = 0, then you basically have two objects (say, obj_0 and obj_1) giving a total energy of 2*(mc^2). obj_0 is in the black hole and obj_1 is at infinity where we initially had obj_0. However, now say I drop obj_1 into the black hole in the same way resulting in obj_2 being made at infinity, giving a total energy of 3*(mc^2) with two of them inside the black hole. Obveosly we aren't gaining any energy at infinity, we still only have mc^2 there. My problem is seeing how energy is being conserved here.

Logically the total amount of energy after dropping obj_0, and creating obj_1, should equal the total energy of obj_0 had we INSTEAD simply let it fall from infinity into the black hole. In the system we have the energy of obj_i, obj_(i+1), and the potential energy of obj_(i+1). But the potential energy of obj_(i+1) seems to be undefined.

Ok, so basically my question is this; what is the total energy of an object if you allow it to fall from infinity to the swartzchild radius with no string attached? My guess would be somewhere in between 0 and infinity; it's not infinity, right?

The reason I ask is when I tried to see this I got this as the total energy assuming newtonian type gravity, and I don't think it's right:

$$E=\frac{mc^2}{\sqrt{1-\frac{2GM}{rc^2}}}$$

Last edited: Nov 3, 2005
7. Nov 4, 2005

### pervect

Staff Emeritus
Heh - I just dodged a similar question, but it looks like there's a lot of interest in it.

Anyway...

Basically, a Newtonian analysis isn't going to give you the right answer.
I can tell you what the GR results are, but you're going to have to take quite a bit on faith, or do quite a bit of reading to understand this problem.

There is a conserved quantity E_0 for any object falling into a Schwarzschild (non-rotating) black hole (which I am analyzing because it's the simplest possible case). This is commonly called "the energy at infinity". For an object with no angular momentum, it is equal to

m ((dr/dtau)^2+1-2GM/r)

m is the invariant mass of the object that is falling. dr/dtau is a rapidity (a measure of velocity) of the infall. M is the mass of the black hole. c is the speed of light, r is the radial Schwarzschild coordinate, and tau is the "proper time" of the infalling particle (not to be confused with the Schwarzschild coordinate t").

a web source:

http://www.fourmilab.ch/gravitation/orbits/, conversion made to standard units from geometric units.

a better source: MTW, "Gravitation" pg 655-658 (a textbook).

This conserved quantity E_0 turns out to be related to the Komarr mass, which is a quantity that's conserved in any static space-time -the space-time of a Schwarzschild black hole is static, so the Komarr mass applies. (MTW doesn't discuss the Komarr mass, but other books do, though they may not directly relate it to this problem).

Basically, if an object is falling into a black hole, if it follows a geodesic its "energy at infinty" stays constant during its fall (it's conserved) and the mass that is added to the black hole is E/c^2 where E is the energy at infinity of the infalling object.

see MTW pg 906 for a confirmation that the black hole mass is increased by delta-E , where E is the "energy-at-infinity" of the infalling particle. Note that the factor of c^2 is omitted in MTW, which uses "geometric units".

If the infalling does not follow a geodesic, the "energy-at-infinity" changes during the fall, because the object is "doing work" on something else and not falling freely. The mass that gets added to the black hole is the "energy-at-infinty" that the particle has at the event horizon of the black hole.

Note that the quantity E_0, the "energy-at-infinty" is not the same as the energy E in the energy-momentum 4-vector. The relation is

E_4vector = sqrt(|E_0 E^0|)

Last edited: Nov 4, 2005
8. Nov 4, 2005

### Longstreet

See if this translation is correct. Even though the object is accelerating in space, space itself is becoming more stretched. So is it like stretchy space kind of cancels out increased velocity? So even though an object might logically (ie newtonianly) approach the speed of light (ie infinite energy), the space there is so stretched that the relative energy actually converges to this E_0 value?

Thanks,

9. Nov 4, 2005

### pervect

Staff Emeritus
That's a rather lose translation of what happens, but if you identify "stretchy space" with the metric coefficients, it might actually work. At this point, I'm not too sure, I dont really think in terms of "stretchy space". I do know there's at least one person on here who hates the whole idea of "stretchy space", but unfortunately I've never been able to take the concept seriously enough to see the points either for or against.

Taking (r,t) seriously as a coordinate system is going to give one headaches with infinites at the horizon, this is the well known "coordinate singularity with Schwarzschild coordinates.

It turns out that (r,tau) where tau is proper time of an infalling particle, doesn't have this problem. Note that the quantity E_0 above is defined in terms of the rapidity (dr/dtau) for the "velocity" component of energy, so it doesn't suffer from the problems with infities due to a bad choice of coordinates.

Note that dr/dt turns out to be zero for the infalling object in Schwarzschild coordinates.

Using the coordinate system of "hoovering observers" who have an infinite acceleration at the event horizon also leads to problems with infinities - these are the only class of observers who see an object fall into a black hole at the speed of light. The "observer at infinity" sees the object falling at zero velocity, the "hovering observer" with infinite acceleration sees the object falling at 'c'. The infinite discrepancy between the two obsevers is due to the coordinate singularity which is IMO the fundamental problem you are running into.

We know that particles can "fall in" to a black hole with differeing values of E-at-infinty, they could fall into the black hole from "at rest", or they could fall in already having a large velocity at infinity, or they could fall in from some distance shorter than infinity, being bound and unable even to reach infinity, winding up with an energy-at-infinty that is less than the rest energy of the particle.

Thus, we would expect that "kinetic energy" of an infalling particle would be variable, using a naieve Newtonian viewpoint to divide up energy into "kinetic" and "potential". (This naive viewpoint still works in this specific instance, I suppose I should add a warning that it won't always work). The fact that dr/dt always equals zero in Schwarazschild coordinates, or that dr'/dt' always equals c in the frame of a "hovering" observer, does not seem compatible with such a "variable kinetic energy". The resolution of this "paradox" is that these coordinates are singular and hence no good, but a good coordinate system like (r,tau) does give us a "variable kinetic energy" expressed as a function of the rapidity, dr/dtau.

And now for something completely different....

I don't suppose I could interest you (Longstreet) n Noether's theoerem, which shows how time-translation symmetry in any physical system based on an "action principle" gives rise to a conserved energy? :-)

http://en.wikipedia.org/wiki/Noether's_theorem

The static black hole has such a time-translation symmetry (at least in the exterior region of the black hole), this is what gives rise to the conserved energy.

It is the principle that objects fall in such a way as to maximize their proper time that satisfies the "action principle" part of the argument.

Last edited: Nov 4, 2005
10. Nov 6, 2005

### Longstreet

We actually just started talking about the relationship between symmetries and conservation in my physics class. For example, linear translation symmetry gives rise to conservation of momentum, time translation symmetry gives rise to energy conservation, etc. So I wouldn't be surprised at all that such a symmetry in time translations, even in relativity, would result in some kind of conserved energy. I'm guessing that our infinite energy for the "hovering observers" is somewhat similar to what happens at the origin of spherical coordinates. We just need a different way of describing that spot to make sense out of it.

11. Nov 6, 2005

### George Jones

Staff Emeritus
Note: c = G = 1 in what follows.

Your answer is correct for a hovering observer. If a particle is dropped from rest at infinity, an observer hovering at a fixed value of r will measure the speed of the praticle, as the praticle rushes by, to be

$$v= \sqrt{\frac{2M}{r}}$$.

Both Pervect and I derived this result in another thread. My post is
https://www.physicsforums.com/showpost.php?p=621802&postcount=32". Take E = 1 in this post. You might also want to look at Pervect's posts in the same thread.

The hovering observer will thus measure the total relativistic energy of the particle to be

$$m \left( 1 - v^2 \right)^{-\frac{1}{2}}$$

Regards,
George

Last edited by a moderator: Apr 21, 2017