Light Speed?

[Diche Bach]

This thread needs a theme song

One of the best geeky space rock songs ever . . .

[qeveren]

Actually, from a distant observer's point of view it should be entirely possible for a spacecraft to exceed c; you've forgotten about the ergosphere. In the vicinity of a rotating black hole, spacetime itself is dragged around it; the ergosphere begins at the point where spacetime is rotating at such a rate that in order to remain 'stationary' with respect to the outside universe, you must travel at the speed of light. An object entering the ergosphere should appear superluminal to a distant observer.

[NGTOne]

Actually, from a distant observer's point of view it should be entirely possible for a spacecraft to exceed c; you've forgotten about the ergosphere. In the vicinity of a rotating black hole, spacetime itself is dragged around it; the ergosphere begins at the point where spacetime is rotating at such a rate that in order to remain 'stationary' with respect to the outside universe, you must travel at the speed of light. An object entering the ergosphere should appear superluminal to a distant observer.

I don't think that changes the overall mechanic, in the same way that a warp drive doesn't. As far as I understand, when you enter the ergosphere, yes, you are travelling faster than light to a sufficiently distant observer (i.e. someone outside the black hole's gravity well). However, within your own frame of reference, you are still travelling at less than c - it's spacetime around you that is moving faster, and dragging you with it. For example, if I teleported a stationary spaceship to within the ergosphere of a rotating black hole, it would appear to travel faster than light if I was standing (floating?) far enough away. However, relative to local spacetime, the spacecraft is still stationary. It's exactly the same principle as generating a warp field (see my last post for more on that one) - the only difference is that you're moving rotationally as opposed to forwards.

I think, before we proceed much further into this discussion, we need an adequate explanation of spacetime (and spacetime warping).

Basically, imagine spacetime as a flat rubber sheet with a 2D Cartesian grid printed on it. Now, let's say I place a bowling ball on the sheet (analogous to a planet or other massive body). The sheet (and the grid) distort, thus producing an indent in the sheet that any object leaving the surface of the bowling ball (such as a rocket) must climb out of in order to escape (this is the idea of escape velocity - the minimum dV required to reach the undistorted area of the sheet). An orbit is simply not having enough energy to reach the undistorted area, but rolling down the indent and back around, repeatedly (assuming, under our analogy, no rolling resistance or other such forces). A black hole creates an infinitely deep indentation in the sheet, as it has infinite mass (basically, imagine that the sheet curves nicely, albeit steeply, down to a circular hole cut deep below the undistorted plane of the sheet - this the event horizon).

So, now, we get to the crux of it. Let's say I take our black hole, and make it rotate (and distort the grid to match - i.e. if a point on the grid matches a point on the perimeter of the event horizon, it will always match that point). I then take a pen, and make a mark on the grid near the rotating black hole (representing our rather unfortunate spacecraft). Because the grid is distorting, the co-ordinates of the space that the spacecraft occupies do not change (i.e. the spacecraft is not moving relative to absolute spacetime). However, because the grid is distorting, the spacecraft appears to move, because to someone standing on the flat part of the grid, the area of spacetime containing our spacecraft is moving, and carrying the spacecraft with it.

Edited August 17, 2013 by NGTOne

[qeveren]

I'm not entirely sure what the objection here is. No object will ever measure its own velocity to be locally greater than c, because no object is actually moving in its own frame of reference in the first place.

[Person012345]

I'm not entirely sure what the objection here is. No object will ever measure its own velocity to be locally greater than c, because no object is actually moving in its own frame of reference in the first place.

I don't know what he's talking about but we already think that faster than light speeds are theoretically possible with spacetime warping anyway, but that's "cheating". Distorting the fabric of spacetime so that something arrives somewhere faster than a beam of light outside that spacetime isn't really "faster than light" in the sense that any beam of light passing through the same area will be observed moving faster than the "faster than light" object.

Right? Or am I wrong?

[Spyritdragon]

Tell me where i'm wrong here (i know i am), but:

Regardless of going lightspeed, can anyone explain why exactly the escape speed being the speed of light forms the border it forms? In my memory, one can escape from a body without ever achieving escape velocity? Why is this specific border related to velocity, not energy, and why does exactly that and nothing before it create a time-dilation effect?

The force acting between two bodies of mass m1 and m2 is calculated as F = G*m1*m2/r^2, right? This force, put in an equation with distance and integrated, leads to work needed to escape. Energy, not velocity. Going lightspeed means one needs infinite energy, so how can this be the escape velocity needed for any body? The body of the black hole must have a finite mass, therefor the amount of energy needed for me to counter its gravity must be finite, seeing as we're working with a linear equation. If i managed to place a huge, very very charged magnetic ball outside of the event horizon, so my mass is a extremely small fraction of its mass, then put on a suit of oppositely charged material, so the force of the black holes gravity and the magnetic force would counter eachother. If i were now to push a small piece of mass towards the black hole, i'll gain a small momentum, say 1 m/s, which i will keep because of the magnetic ball cancelling the black holes gravity. I'll keep moving away form the black hole at 1 m/s, why don't i eventually reach the event horizon (ignoring the fact i might have died in the meantime. In any case my suit remains there and it remains charged).

[SargeRho]

Spacetime falls into a black hole at the speed of light, if not faster. You would need to move at c or above through spacetime to escape the black hole.

[Spyritdragon]

But why? If i go at a consistent 1 m/s, i'll still be moving away from the black hole, right?

[Bacterius]

I think the point is that you can't maintain a consistent 1 m/s away from the black hole, because if you were capable of doing so, then you would be able to move faster than the speed of light without the influence of a black hole.

Edited August 17, 2013 by Bacterius

[SargeRho]

But why? If i go at a consistent 1 m/s, i'll still be moving away from the black hole, right?

No, because you have to move faster than light to escape a black hole, and exactly the speed of light to maintain your position once past the event horizon. Spacetime falls into a blackhole at the speed of light due to gravity. You have to move faster than that if you want to leave, and you cannot do that. It takes infinite energy to accelerate to the speed of light. There are no paths that lead out of the event horizon.

[Person012345]

I think the point is that you can't maintain a consistent 1 m/s away from the black hole, because if you were capable of doing so, then you would be able to move faster than the speed of light without the influence of a black hole.

This. Also I'm not aware that spacetime falls into a black hole, seems like an odd way of phrasing it.

[Person012345]

If you enter the event horizon at the speed at light (in an orbital path as per the OP), then you already have infinite energy. If that infinite energy can't carry you through, then how much energy will you burn to pull you out? Infinity+1?

Edited August 17, 2013 by Person012345

[NGTOne]

I'm not entirely sure what the objection here is. No object will ever measure its own velocity to be locally greater than c, because no object is actually moving in its own frame of reference in the first place.

No objection - the way you phrased it just sounded like you were implying that the object would be travelling faster than c under its own power/the power of gravity, as opposed to the power of spacetime warping.

I don't know what he's talking about but we already think that faster than light speeds are theoretically possible with spacetime warping anyway, but that's "cheating". Distorting the fabric of spacetime so that something arrives somewhere faster than a beam of light outside that spacetime isn't really "faster than light" in the sense that any beam of light passing through the same area will be observed moving faster than the "faster than light" object.

Right? Or am I wrong?

No, you're right - the light inside the spacetime warp will still be travelling at c, from the PoV of someone inside the warp bubble. To someone on the outside, it would be travelling at c + the speed of the warp bubble. I'm not sure of the visual implications of this (i.e. how it would look).

This. Also I'm not aware that spacetime falls into a black hole, seems like an odd way of phrasing it.

It doesn't "fall in" - it distorts, just as it would with any other gravity well. A rotating black hole drags spacetime with it in a rotary fashion, but does NOT pull it in - basically, so far as I understand, there is a region of spacetime that becomes infinitely stretched surrounding the rotating black hole. You fall into a rotating black hole by gravity, just as you would with a non-rotating one (or any other massive body, really). Unless you've moving fast enough relative to local sideways (which is just another way of saying you're in orbit). And, yes, I believe it would be possible to orbit a rotating black hole inside the spacetime distortion zone - it might look funny to an outside observer, but to the spacecraft, the mechanics haven't changed in the slightest.

If you enter the event horizon at the speed at light (in an orbital path as per the OP), then you already have infinite energy. If that infinite energy can't carry you through, then how much energy will you burn to pull you out? Infinity+1?

The problem is REACHING the speed of light in the first place. If you're using classical physics, you CAN'T, end of story. You can get infinitely close, but the amount of energy you need to get closer increases asymptotically towards infinity as you approach it.

Also, you CAN'T escape the event horizon - this is the "point of no return", where not even emitted light escapes (hence the "black" part of black hole). Unless you've got a warp drive handy, you're not getting away (and we don't really know how spacetime behaves on the other side, so that might not work either).

Also, the reason the OP is wrong in his theory is that orbital paths just... well, in short, cease to work inside the event horizon. See this post:

Won't work. Basically, when you get really close to the black hole, Kepler's Laws break down. In particular, you no longer follow an elliptic trajectory. Once you cross event horizon, you'll just keep falling in towards the singularity at the center of the BH.

There are coordinate systems in which the object bellow the event horizon is traveling faster than the speed of light, but in GR this is a very loose statement. Yes, you are traveling faster than light relative to remote observer if you draw the coordinates a certain way. But it doesn't really have any physical meaning. What's important about the speed of light limit is that it's a true limit locally. That means you can't be moving faster than light past something. And that's still going to be true even after you deep bellow the event horizon.

You can't "orbit out of" the event horizon, as classical mechanics starts to, for lack of a better term, break down below it.

Edited August 17, 2013 by NGTOne

[qeveren]

No objection - the way you phrased it just sounded like you were implying that the object would be travelling faster than c under its own power/the power of gravity, as opposed to the power of spacetime warping.

Oh, I gotcha. Yeah, you're not going to be able to beat a local beam of light to a finish line under any known circumstances. It cheats.

[WestAir]

Can the space time warping effect of an alcubbierre drive let you warp your way out of the black holes event horizon?

[NGTOne]

Can the space time warping effect of an alcubbierre drive let you warp your way out of the black holes event horizon?

Not sure. While I don't see why not (assuming that spacetime and spacetime warping work the same way inside the event horizon as they do outside), I really have no idea. If it were possible, it would likely require a radically different warp field geometry than for ordinary Alcubierre-drive travel.

[SargeRho]

Possibly. The problem is, you will be torn to shredds before ever reaching the event horizon.

[NGTOne]

Possibly. The problem is, you will be torn to shredds before ever reaching the event horizon.

I think we're basically operating under the assumption that our spacecraft (and its luckless pilot) are indestructible, for the purposes of a physics demonstration.

[Spyritdragon]

No, because you have to move faster than light to escape a black hole, and exactly the speed of light to maintain your position once past the event horizon. Spacetime falls into a blackhole at the speed of light due to gravity. You have to move faster than that if you want to leave, and you cannot do that. It takes infinite energy to accelerate to the speed of light. There are no paths that lead out of the event horizon.

So.... to stay exactly where i am, i need to be moving fast? That sounds... very weird.

[SargeRho]

Think of a boat on a rapid river. The boat has to go exactly at the speed the water moves, relative to the water, upstream to stay stationary. More or less the same thing happens here.

[WestAir]

The tidal forces at the event horizon of a supermasaive rotating black hole are not lethal. An astronaut will not be torn apart by entering the EH of the black hole at the galaxy core, for instance, because the EH is so far away from the singularity.

[Diche Bach]

Uhhh, black holes are not something for which we have 100% empirical confirmation, right?

My understanding is that they are an inferred phenomenon that fits with existing theory and could account for a lot of observations. But there are evidently alternative explanations, which are more speculative?

[NGTOne]

Uhhh, black holes are not something for which we have 100% empirical confirmation, right?

My understanding is that they are an inferred phenomenon that fits with existing theory and could account for a lot of observations. But there are evidently alternative explanations, which are more speculative?

Right on all counts.

However, there has to be SOMETHING that's large enough out there to cause the gravitational lensing that has been noted by some telescopes, and to create the enormous gravity well that generates galaxies (without it, there would be no discrete galaxies - stars and other large bodies would be far more evenly distributed across the universe than they are). After all, our sun is in orbit of SOMETHING rather large (putting it mildly) somewhere at the heart of the Milky Way.

[Person012345]

Uhhh, black holes are not something for which we have 100% empirical confirmation, right?

My understanding is that they are an inferred phenomenon that fits with existing theory and could account for a lot of observations. But there are evidently alternative explanations, which are more speculative?

They're also predicted by relativity.

[NGTOne]

They're also predicted by relativity.

So far as I'm aware, they're not so much PREDICTED by GR as one possible solution for a part of it. There are other possible mathematical solutions, which are also equally valid (mathematically - which ones are correct physically remain to be seen).