Effects of black holes on trajectories

[SciMan]

Okay, I meant "inflexible" in a purely classical sense.

In other words, it doesn't bend no matter how much force is applied to it, but because it is matter, it of course must follow space-time curvature.

More specifically, in a space-time of negligible curvature, holding one end of a rod of this material fixed and applying an infinite force to it on the other end would deflect that end by exactly 0. (or less than one planck length, if you want to get really technical)

Unless that contradicts itself, of course. If it is self-contradictory, please explain the contradiction to me, because I like to know when I'm wrong, but I like to know HOW I'm wrong even more.

[Z-Man]

Think about what would happen if you let two of your rods collide end to end at relativistic speeds. Think about how an observer stationary with respect to the center of mass sees the event and one observer sitting on the far end of each rod. From the perspective of the rod riders, the colliding ends will have to move and give way in reaction to the collision, no matter what.

[SciMan]

Because they can't deform in a classical sense, they would simply bounce off of one another, stop dead and emit a large amount of photons, or some combination of the two. Because there is no favored reference frame, all observers would see the same thing, after taking the appropriate reference frame transforms into account to correlate the results from one to the others.

I sense I'm missing something fundamental here.

Incidentally, over at Orions Arm one of the ways their sci-fi universe has to make black holes is to collide a large number of extremely thin rods of extremely dense matter at high relativistic speeds. That or they implode an asteroid to under it's Schwarzschild radius, using so much explosives that even Jeb might say it's too much.

[Z-Man]

I sense I'm missing something fundamental here.

The finite propagation speed of all information, yes. It works in any reference frame: before the tips of the rods collide, they can be perfectly rigid. However, it takes time for the information that a collision happened to propagate back to the rod ends. Until that happens, the rod ends need to keep moving forward, toward each other, at the same speed they had before. When they finally can react (and yes, for an as-rigid-as-possible material, they will directly bounce back), there simply is less space between them than the two rods would require, which means the rods must be compressed at this point.

I can make and upload crude time/space drawings later.

[Sky_walker]

His inflexible rod would basically shatter in a moment it touches another one.

Inflexible and indestructible rod is an impossible paradox. Good for cartoons, but that's pretty much where it ends.

[Z-Man]

That's another way to view it, yes.

The promised crudeness:

Time goes up, reference frame is the center of mass frame. Center of mass is the thick black line in the center. Before the collision, the rods are painted blue, L0 is the length of one of them while it moves freely (already including Lorentz contraction, its rest length would be larger still). At point A, the tips collide. The information about the collision can only spread along the black lines going at lightspeed from there. In the whole blue area, even above A, the rod material is blissfully unaware of the collision. At point B, the information arrives at the end of the rod, only then it can react. But by then, it has already contracted to length L1, clearly shorter than L0. In the yellow area, the rod is heavily compressed, probably oscillating and angry.

What happens above B in this picture is simlply the easiest way you can imagine an elastic collision: it's just the time reversal of what happened before.

I now can give an example of an "object" that never gets "destroyed", but is partially swallowed up by a black hole. It goes like this: Take a full gas tank. Make it hover above the black hole in a safe distance. Slowly release the gas. Do so that you never fully empty the tank, but always release some (say, let the amount left be m_0 + a/(t+t_0) at time t). The gas will never at any point in spacetime be torn apart and there will always be some left outside. Coming up with a hypothetical more solid material, object shape and movement pattern outside the horizon is left as an excercise for the reader An elastic, unbreakable rubber band you slowly lower into the black hole should do it.

Of course, it is debatable whether the infinite stretching at and inside the event horizon the material is inevitably going to suffer from is to be considered "breaking".

PS: In case it's not clear: in all of this we need to assume the material in quesion is not actually made of atoms. Atoms will always get torn apart. Instead, we need to assume it is truly continous.

[Everten P.]

I don't think we can be sure about these things. I mean do the gravity well of the stars cause time extensions( cause that would be news to me )? Anyways, I don't think that hypothesizing about something we have never seen or really learned about up close is a good idea by I will go with it. I am sure that the high speed ring of gas around the black hole is enough to tear up you ship but the gravity assist of the black hole should let you go through that within about 1/4 of a second. So you could use them to go faster indubitably.

[K^2]

I mean do the gravity well of the stars cause time extensions( cause that would be news to me )?

Yes. Planets too. In fact, Earth's gravitational time dilation is significant enough that it has to be accounted for by GPS satellites. In other words, if General Relativity was wrong, GPS would not work. There are a lot of other reasons that add up to us being more certain about General Relativity than tomorrow's sunrise. But these are mostly technical.