Effects of black holes on trajectories

[theend3r]

From my basic understanding of orbital mechanics I concluded that:

a) there are four basic trajectories in an inertial system of two bodies: circular, elliptical, parabolic, hyperbolic

if you're on an escape trajectory, you need to do a breaking maneuver or aerobreak (or lithobreak ) in order to not escape the celestial body

That would mean that going beyond the event horizon of a black hole should be no different, accelerating you immensely before the periapsis and decelerating you accordingly thereafter, leading to your escape. By the definition of event horizon, that should be impossible though. So either the black hole can change your trajectory without applying any additional force to your craft or a) is wrong.

If a) is wrong then just someone say so and we can close this discussion.

[GoSlash27]

Interesting conundrum!

Technically a) is false. the Newtonian model works well enough to get by on, but it's not actually true according to relativity. The presence of something outside the kinematic realm (such as a black hole) distorts time and space so badly that the Newtonian model no longer holds. In fact, according to the eggheads the errors of the simple Newtonian model are already quantifiable.

So a black hole doesn't add any energy, but simply warps the space and time around it.

Best,

-Slashy

Edited July 5, 2014 by GoSlash27

[Tex_NL]

GoSlash is correct. When dealing with black holes strange things happen. Orbital velocity at the event horizon will be light speed. To escape from below the horizon you would have to travel faster which is according to modern science impossible.

[theend3r]

Interesting conundrum!

Technically a) is false. the Newtonian model works well enough to get by on, but it's not actually true according to relativity. The presence of something outside the kinematic realm (such as a black hole) distorts time and space so badly that the Newtonian model no longer holds. In fact, according to the eggheads the errors of the simple Newtonian model are already quantifiable.

So a black hole doesn't add any energy, but simply warps the space and time around it.

Best,

-Slashy

Thanks, that's what I wanted to hear

GoSlash is correct. When dealing with black holes strange things happen. Orbital velocity at the event horizon will be light speed. To escape from below the horizon you would have to travel faster which is according to modern science impossible.

I know, I only wanted to confirm if a) was wrong.

[GoSlash27]

GoSlash is correct. When dealing with black holes strange things happen. Orbital velocity at the event horizon will be light speed. To escape from below the horizon you would have to travel faster which is according to modern science impossible.

Now having said that, I should point out that I don't personally subscribe to it.

I don't believe that exceeding the speed of light is actually impossible and I don't believe that relativity is a logically sound construct. It happens to work, but so do a lot of other irrational arguments.

/going to geek hell

-Slashy

[K^2]

Take a look at this article. Don't get hung up on math. The main point is that when you get close to the black hole, you end up with weird trajectories. Things can spiral in, or do a pretzel sort of twist before escaping. And, of course, any trajectory that takes you to the event horizon takes you down to the center of a black hole, never to escape again. So as people said, assumption a) is wrong for trajectories near a black hole.

[Aethon]

Ok K^2, I always enjoy picking your brain- riddle me this : When we speak of an 'event horizon' (http://arxiv.org/pdf/1401.5761v1.pdf), or the point of no escape, we're talking about a ballistic trajectory. If I understand correctly, a spaceship (with suitable propulsion) around a massive enough black hole, could dip below the 'event horizon' since it has the ability to change it's velocity (or delta it's v). Could you elaborate on this?

[Tex_NL]

Ok K^2, I always enjoy picking your brain- riddle me this : When we speak of an 'event horizon' (http://arxiv.org/pdf/1401.5761v1.pdf), or the point of no escape, we're talking about a ballistic trajectory. If I understand correctly, a spaceship (with suitable propulsion) around a massive enough black hole, could dip below the 'event horizon' since it has the ability to change it's velocity (or delta it's v). Could you elaborate on this?

No you can not. At the event horizon the orbital velocity will be equal to the speed of light. Any orbit that dips below the event horizon would require a speed higher than that.

[GoSlash27]

No you can not. At the event horizon the orbital velocity will be equal to the speed of light. Any orbit that dips below the event horizon would require a speed higher than that.

Technically, not so. Else, it would be impossible for matter to ever enter a black hole. More properly, time and space are warped by the gravity in such a way that every possible direction from the event horizon is inward. It would be more proper to say that escape velocity from an event horizon exceeds the speed of light, although that's not necessarily the important part. The important part is that escape velocity exceeds infinity since there is no possible direction to escape.

Best,

-Slashy

[K^2]

Ok K^2, I always enjoy picking your brain- riddle me this : When we speak of an 'event horizon' (http://arxiv.org/pdf/1401.5761v1.pdf), or the point of no escape, we're talking about a ballistic trajectory. If I understand correctly, a spaceship (with suitable propulsion) around a massive enough black hole, could dip below the 'event horizon' since it has the ability to change it's velocity (or delta it's v). Could you elaborate on this?

It can "delta its v", but it won't be exiting the event horizon. Just slightly alter the in-falling trajectory.

The how fast the ship is moving is actually not really a question. When you are falling into a black hole, velocity is a subjective thing. It depends on your choice of coordinate systems. In standard, Schwarzschild coordinates, yes, you'd be moving at the light speed as you pass the event horizon, and faster as you keep falling. But then there are coordinate systems according to which you have a nice, sub-light journey all the way to the true singularity at the center.

What's important is that any time-like trajectory bellow event horizon leads to the center of the black hole. (Well, rotating black holes might be a bit more interesting, but that's a separate story.) What this means in layman's terms is that the only kind of ship that can escape a black hole after "dipping in", is the kind that is capable of FTL speeds in normal space. Because both of these feats require ability to follow a space-like trajectory.

[trekkie_]

It's simple to come to terms with, without getting super technical.....light is basically 'mass-less', and traveling at well, the speed of light.....so if light can't escape it, neither can a ship with tons of mass even if it is going at the speed of light.

[Duxwing]

What if one flew an Alcubierre-driven ship under the event horizon?

-Duxwing

[GoSlash27]

It can "delta its v", but it won't be exiting the event horizon. Just slightly alter the in-falling trajectory.

The how fast the ship is moving is actually not really a question. When you are falling into a black hole, velocity is a subjective thing. It depends on your choice of coordinate systems. In standard, Schwarzschild coordinates, yes, you'd be moving at the light speed as you pass the event horizon, and faster as you keep falling. But then there are coordinate systems according to which you have a nice, sub-light journey all the way to the true singularity at the center.

What's important is that any time-like trajectory bellow event horizon leads to the center of the black hole. (Well, rotating black holes might be a bit more interesting, but that's a separate story.) What this means in layman's terms is that the only kind of ship that can escape a black hole after "dipping in", is the kind that is capable of FTL speeds in normal space. Because both of these feats require ability to follow a space-like trajectory.

Heck, not even FTL travel will get you out of an event horizon. Even a ship capable of instantaneous travel between any 2 points in the universe can't escape a black hole. "out" is a direction that doesn't exist AFA an event horizon is concerned. "Out" is just another word for "in" in that construct.

Going back to the OP, I have a model that illustrates the problem.

Imagine you have a horizontal stretched piece of fabric. This would represent "space" in the 3 dimensional sense. Now if you place a mass on the fabric so that it distorts the fabric, it represents matter distorting space. If you place another object on the fabric and move it, it will behave exactly in accordance with Newtonian kinematics (so long as you neglect friction), leading you to conclude that the newtonian model is correct. So far so good...

An event horizon would be represented by a depression in the fabric so deep that it's sides are parallel. A weighted soup can hanging from a hole in the fabric. Once you cross the lip of the can, how fast do you have to go to get back out? Of course, you can't. There is no such thing as a speed fast enough to get a marble out of that can.

The poor marble that stumbled across the lip of that can would continue to follow the laws of Newtonian physics... right into the bottom of the can.

The can doesn't have to add any energy to the marble to trap it and there's nothing wrong (fundamentally) with physics where the model breaks down. It's just that the *space* has been distorted.

Edited July 7, 2014 by GoSlash27

[K^2]

Heck, not even FTL travel will get you out of an event horizon. Even a ship capable of instantaneous travel between any 2 points in the universe can't escape a black hole. "out" is a direction that doesn't exist AFA an event horizon is concerned. "Out" is just another word for "in" in that construct.

That's wrong. When people say that "out doesn't exist", they are talking about time-like trajectories only. There have to be trajectories leading out unless there is a true singularity on the path. The only true singularity in a black hole is at the center. The event horizon is merely a coordinate singularity. So there are trajectories going out. They just all happen to be space-like trajectories. So ability to go "instantly" along, say, radial direction would be just the thing to get you out of a black hole. Well, "instantly" is poorly defined in relativity. But any motion that's instant in your coordinate system of choice is a space-like trajectory in any other. So it will do the trick just fine.

[GoSlash27]

That's wrong. When people say that "out doesn't exist", they are talking about time-like trajectories only. There have to be trajectories leading out unless there is a true singularity on the path. The only true singularity in a black hole is at the center. The event horizon is merely a coordinate singularity. So there are trajectories going out. They just all happen to be space-like trajectories. So ability to go "instantly" along, say, radial direction would be just the thing to get you out of a black hole. Well, "instantly" is poorly defined in relativity. But any motion that's instant in your coordinate system of choice is a space-like trajectory in any other. So it will do the trick just fine.

I disagree. trajectories don't have to lead "out" in order to not lead "in". An infinite acceleration inside an event horizon would theoretically approach the event horizon, but would never cross it.

Of course, I've never taken a vacation inside a black hole....

-Slashy

[K^2]

It doesn't really matter if you agree or disagree. The geometry of Schwarzschild black hole gives you trivial escape trajectories. Such as r(Ä) = r₀+ vÄ; t, θ, Æ constant. It's a geodesic of Schwarzschild metric, it is space-like, and it can start with r < r_s and end with r > r_s. There is nothing more to it.

[Comrade Jenkens]

What if one flew an Alcubierre-driven ship under the event horizon?

-Duxwing

If one can be built I'm pretty sure it wouldn't work within the event horizon. It has to bend space in a certain way to work and a black hole would distort that so severely that the bubble just wouldn't form.

(All the is pure waffle I have no idea what would happen! )

[YNM]

What if one flew an Alcubierre-driven ship under the event horizon?

Weirdness... Inside the event horizon, black holes bend the spacetime to a perfect "vertical" (people say, space become timelike, time become spacelike),and alcubierre drive just lifts and bent the already "vertical" space around it, preserving the already-bended sheet of space inside the drive. Really weird. If that thing ever comes out from the blackhole you can move some properties of spacetime around a blackhole to another place... and what ? What the heck is that ?

But it's more likely that the bubble never forms, due to all the weirdness...

Edited July 7, 2014 by YNM

[K^2]

If one can be built I'm pretty sure it wouldn't work within the event horizon.

It would, provided that the space-time isn't too curved in the immediate vicinity of the warp ship. For example, Alcubierre Drive should allow you to dip slightly bellow event horizon of a supermassive black hole. Smaller black holes would have too much curvature at the horizon, and any black hole causes too much curvature close to its center. But if the black hole is large enough and you don't go too deep, there is no reason why AD or related warp drive shouldn't work there.

[theend3r]

Another question: If there was an enormous object made from an indestructible material (so it wouldn't just be torn apart) and only an infinitesimal part of it dipped below the event horizon, would the whole object be inevitably swallowed?

[Z-Man]

That is a good question. The answer appears to be yes. Rationale:

1. We know that locally, the horizon is nothing special. No infinite tidal forces appear to a local observer. So the object will not be torn apart there. For large holes, that is true even for regular materials.

2. We know that anything going inside (or past) the horizon can't come back.

Since the part "dipping in" can't come back and the object will not be torn apart, it follows that the whole object is swallowed.

And it doesn't matter how large the object is. What pulls the bits far outside the horizon in is not the (supposedly) infinite force from the bits inside the horizon, that force does not exist (the bits inside cannot have a causal effect on the bits outside), it's the infinite force from the bits very close to the horizon on the outside.

What you need to understand here is that the horizon is not a surface like a wall in your house; it is 'lightlike'. Locally, as you cross it, the horizon swoops over you at the speed of light. Therefore, to get away from it once you gotten too close, you essentially have to run away from something moving at the speed of light. Which is possible with finite acceleration. Even in flat space, you can run away from a wall of light chasing you by applying finite acceleration for the rest of your life, and in the black hole case, the curvature of space allows you to get away completely after a bit. But: The closer you are to the horizon, the more total delta-v you need, and it diverges to infinity as you approach.

[K^2]

That's correct. The only thing I can add is that if you try and pull on the part of the object that's outside, the black hole will always pull harder. So it will either pull the object in or, if you apply enough force trying to "rescue" that object, rip it apart. In any case, no matter how strong the object is, you can't pull out the part that crossed the event horizon already.

[SciMan]

Okay, so what would happen if you take a rod of this indestructible and inflexible material that is twice as long as the diameter of the event horizon of the black hole, and connect the ends with a ring of the same material so that the rod goes thru the center point of the ring. Then cut out one half event horizon radius from the center of the rod on both sides. Finally, maneuver the center of the rod to be exactly the same as the center of the event horizon.

What happens? The force pulling the rod into the black hole is the exact same on both sides, and because the ring of material connecting the ends of the rod exists purely outside the event horizon, any forces acting on one part of this construction act on all of it.

Would it be possible to extract the portion of the rod that is inside the black hole's event horizon in this situation, or have I discovered a way to anchor down a black hole? Aside from the fact that it would need to be made from unobtainium, of course.

Edited July 8, 2014 by SciMan

[Sky_walker]

There is no indestructible and inflexible material. All material is made of matter, even if your material would be made of quarks or anything else - it'd still be sucked in a black hole - in your example: it'd simply pull inside the ends that are sticking out of the event horizon.

[Z-Man]

There is no indestructible and inflexible material.

Correct. "Inflexible" is incompatible with special relativity already, there can be no rigid bodies if all information about impacts and forces can only travel at the speed of light. One can theorize about indestructible materials, though, they just need appropriate reactions to deformations.

Also, mind that for an outside observer, nothing ever passes the horizon. So while from the perspective of the bit of the object that has passed the horizon, the rest is inevitably pulled in after it*, from the perspective of the bits outside, that may very well happen only after infinite time has passed. Whether that's the case may depend on the specific properties of the hypothetical material, though.

*: Edit, thinking a bit more, it also can't be ruled out in general that from the perspective of every bit of the object that crosses the horizon, it hits the singularity (and not even fantasy material survives that) before all of the rest is pulled in. Again, this may depend on the material properties.

Edited July 8, 2014 by Z-Man