When it reached the center, I'd expect that its gravity well would pull the rest of Earth into the black hole, unless it somehow found itself in a vacuum large enough to not create a gravity well. That would presumably be a sphere the radius of the amount of core material that is equivalent to the mass of Mt Everest, plus whatever it accreted on its descent.
So really, the question is whether or not a black hole that we could create would be able to make it to the floor of the lab... the article sorta touches on this:
The gravitational pull of a few micrograms of matter, regardless of how it is arranged, is never dangerous; you wouldn’t get pulled inside out if you ate it. However, you wouldn’t get the chance, since any black hole that we could reasonably create would already be mid-explosion.
Since the proposed method of creating the hole is to collide particles in an accelerator, we would expect the hole, when formed, to be moving the way particles created in an accelerator move, i.e., at close to the speed of light. So the hole would fly off into space, since it would be moving much, much faster than escape velocity. (Also, as the article notes, the hole would explode due to Hawking radiation soon after it was created.)
> unless it somehow found itself in a vacuum large enough to not create a gravity well
Not sure what you mean by this. The "gravity well" of the hole is the same whether it's surrounded by vacuum or matter. The only difference is what can possibly fall in.
And....wouldn't it interact with everything along it's path?
I mean..it wouldn't be quantum tunneling correct? (I've only taken rudimentary physics related to standard mechanics and electricity/magnetism).
But it seems to me, since the LHC is underground, it would have a decent probability of hitting something along it's path to space, or am I way off?
I guess it depends on the length of the "drill hole", the mass of the original black hole before it left the vacuum in the lab, the size of the event horizon and the original velocity?
[1] Technically I suppose it has some momentum due to the rotation of the Earth but I assume that's not significant. If it were significant, it would just mean that there's are "good" and "bad" directions for the escape so a worse case scenario could only get slightly worse, not better.
The mass would increase (unless the hawking radiation is fast enough to counteract the mass it is absorbing, which I don't know), so unless something had other forces on it, it would slow down somewhat... but unless some stuff outside of it has momentum significantly changed, I would think that the momentum would stay the same?
(assume a frame of reference where the matter it is passing through has roughly no momentum to start with)
Momentum is mass times speed, the momentum would stay the same, but since the mass is going up the speed has to go down.
It would stop pretty quickly actually.
"Friction" type interactions with matter it did not eat would also bleed energy.
Only in the non-relativistic limit. We're talking about a hole that starts out with ultrarelativistic velocity, so its momentum is not mv, it's gamma * mv, where gamma = 1 / sqrt(1 - v^2). For v close to 1 (meaning the speed of light), gamma can easily be a thousand or a million, meaning the actual momentum of the object is a thousand to a million times what the non-relativistic formula would make you think.
> It would stop pretty quickly actually.
I think you need to show some calculations. If the hole starts out moving at the typical velocities of particles in an accelerator experiment, it would need to absorb thousands or millions of times its starting mass to be slowed down to less than Earth escape velocity, let alone to be stopped. And at its starting speed it would take less than a twentieth of a second to pass through the entire Earth.
That's only if use non-relativistic mass which you should not be doing anyway. In any case it's still proportional to mass and velocity - the gamma factor does not change that.
> meaning the actual momentum of the object is a thousand to a million times what the non-relativistic formula would make you think.
You are missing parts of the interaction. Say it eats an atom - it has to accelerate that atom to the same relativistic velocity - meaning that new atom now has EXACTLY the same gamma as the black hole, and steals just as much momentum as if gamma was 1.
Or in other words the gamma factor is completely irrelevant for these calculations, it cancels out.
> it would need to absorb thousands or millions of times its starting mass to be slowed down to less than Earth escape velocity
As I've shown you, this is incorrect. It only needs to absorb mass similar to its own rest mass.
> And at its starting speed it would take less than a twentieth of a second to pass through the entire Earth.
That is also irrelevant. The only thing that matters is its "width", i.e. how many of the particles it interacts with does it eat.
Because otherwise you could say anything is proportional to velocity, but with another term which counteracts that factor?
Also, I thought the view was shifting to using rest mass instead of relativistic mass but I might have that backwards.
Yes. At ultrarelativistic velocities, momentum can change a lot for a very, very small change in velocity, so the behavior is highly nonlinear.
Not quite the same, but almost, yes. The final velocity of the hole after it eats the atom will be equal to the velocity of the center of mass of the hole + atom before the atom is eaten. But if the hole has the rest mass of Mount Everest, the motion of the center of mass of the system will be almost the same as the motion of the hole, since the rest mass of Mount Everest is probably about 40 orders of magnitude larger than the rest mass of an atom.
> meaning that new atom now has EXACTLY the same gamma as the black hole
Yes. But this gamma will be slightly less than the gamma of the hole before it ate the atom.
> and steals just as much momentum as if gamma was 1.
I see what you're thinking, but it's still wrong; you haven't calculated the effect correctly. See below.
> It only needs to absorb mass similar to its own rest mass.
No. Say the hole starts out with rest mass M, and is moving at velocity v, where v is almost 1 (the speed of light), so the gamma factor is, say, 1000. Then the total momentum is about 1000M.
Now the hole absorbs enough atoms to have rest mass 2M, and each atom is at rest before it's absorbed, so each atom contributes zero momentum to the total, so the total momentum is still 1000M after all the atoms are absorbed. That means the gamma factor is now (to a very good approximation) 500 (rest mass 2M times gamma of 500 times v approximately 1 gives momentum 1000M), which still corresponds to an ultrarelativistic velocity.
For the velocity to be less than Earth escape velocity, the total momentum of 1000M would have to be carried by a hole with a rest mass of about 3 million M, since Earth escape velocity is about 1/3000 of the speed of light (11.2 km/s vs. 300,000 km/s)--i.e., we would have rest mass 3 million M times gamma of approximately 1 times v of 1/3000, to give total momentum 1000M.
> That is also irrelevant. The only thing that matters is its "width", i.e. how many of the particles it interacts with does it eat.
The cross section (what you are calling "width"--what fraction of particles encountered does the hole eat) will depend on the speed of the hole; the faster the hole is traveling, the lower the cross section will be.
OK, I see your point. At those speeds a tremendous reduction in speed-energy (AKA relativistic mass) is only a small reduction in speed.
> the faster the hole is traveling, the lower the cross section will be.
At these masses gravitational attraction (that could pull in matter if given some time) is basically nil. I am assuming direct front-end collisions between the black hole and matter, so the speed doesn't make much difference.
Although thinking about it, I suspect the cross section (event horizon radius) of a proton massed black hole would be smaller than a neutrino.
Ah, ok; yes, in this approximation (which I agree is a pretty good one--there is another subthread on this that I've been posting in), the number of particles eaten is basically the number of atoms that can be lined up end to end across the diameter of the Earth (in the worst case scenario where the hole flies straight down through the Earth). That number is about 10^17, which is many, many orders of magnitude less than the number of atoms in Mount Everest (the very rough back-of-the envelope number I come up with for that is 10^40).
> I suspect the cross section (event horizon radius) of a proton massed black hole would be smaller than a neutrino.
I thought the hole's mass was assumed to be about that of Mount Everest, at least for this discussion. The horizon radius of such a hole would be about 1.5 x 10^-14 meters, or about the size of a large atomic nucleus. I'm not sure we have a good number for the "size" of a neutrino.
There were two discussions at once - Mount Everest, and a particle from the LHC.
That depends on the initial size of the hole; if it starts out small enough, it would radiate away mass fast enough that it would evaporate before it had a chance to travel far.
> unless some stuff outside of it has momentum significantly changed
Which it would.
Highly unlikely. If the hole starts out moving at nearly the speed of light, it will spend less than a twentieth of a second passing through the Earth. That's not enough time for it to absorb much matter, and it would have to absorb many times its starting mass (quite possibly thousands or even millions of times) to be slowed down from nearly the speed of light to less than escape velocity.
That's exactly it -- if the gravity well is a vacuum, then by definition there is nothing to fall into it, and thus nothing to increase the mass, and so the black hole would not grow.
Not quite. The proposed method involves rotating two [sets of] particles in opposite directions, and then ramming them together. Yes, both sets of particles have super-high velocities- but if they were rammed together, the resulting net velocity should be zero-ish (plus or minus experimental error).
Not necessarily. This is how current accelerator experiments work, and the particles coming out of those do not have zero velocity.
However, it is true that there is a key difference between current experiments and a hypothetical experiment that forms a black hole. The rest mass of the particles formed in current experiments is fixed, so any energy pumped into the particles over and above that becomes kinetic energy. But the rest mass of a black hole is not fixed, so it would be possible for all of the energy pumped in to become rest mass of the hole, instead of kinetic energy. (Possible, but still not guaranteed, as far as I can see; but since nobody really has a detailed theory of how all this would work, it's all speculation anyway.)
As a layperson, I don't understand this. Are you suggesting it would somehow pull in nearby mass and increase the density of that mass somehow? How would it keep adding mass while sustaining the necessary high density?
Any additional mass that enter the event horizon falls to the center and the density is still infinity.
(The event horizon get bigger when you add mass. You can image the black hole as a big sphere that is almost empty, except for a few things that are quickly falling to the center. And you have at the center all the remainder mass gathered at one point.)