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What would be the mass of matter that it captures as it passes through the Earth? That matter has a momentum of zero relative to the Earth[1] and the total mome

by rlpb 11y ago

What would be the mass of matter that it captures as it passes through the Earth? That matter has a momentum of zero relative to the Earth[1] and the total momentum of the black hole would therefore be reduced as it drills through. Would this loss of momentum be enough to reduce its velocity below Earth's escape velocity? 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.

drdeca 11y ago
Are you sure that the total momentum would be reduced? 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)
- ars 11y ago
  > I would think that the momentum would stay the same? 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.
  - pdonis 11y ago
    > Momentum is mass times speed 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.
    - ars 11y ago
      > Only in the non-relativistic limit. 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.
      - drdeca 11y ago
        If the gamma is expressed in terms of velocity, wouldn't that suggest that the momentum isn't exactly proportional to velocity? 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.
        pdonis 11y ago
        > If the gamma is expressed in terms of velocity, wouldn't that suggest that the momentum isn't exactly proportional to velocity? Yes. At ultrarelativistic velocities, momentum can change a lot for a very, very small change in velocity, so the behavior is highly nonlinear.
      - pdonis 11y ago
        > Say it eats an atom - it has to accelerate that atom to the same relativistic velocity 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.
- pdonis 11y ago
  > unless the hawking radiation is fast enough to counteract the mass it is absorbing 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.
pdonis 11y ago
> Would this loss of momentum be enough to reduce its velocity below Earth's escape velocity? 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.

by rlpb 11y ago

drdeca 11y ago

Are you sure that the total momentum would be reduced? 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)

ars 11y ago
> I would think that the momentum would stay the same? 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.
- pdonis 11y ago
  > Momentum is mass times speed 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.
  - ars 11y ago
    > Only in the non-relativistic limit. 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.
    - drdeca 11y ago
      If the gamma is expressed in terms of velocity, wouldn't that suggest that the momentum isn't exactly proportional to velocity? 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.
      - pdonis 11y ago
        > If the gamma is expressed in terms of velocity, wouldn't that suggest that the momentum isn't exactly proportional to velocity? Yes. At ultrarelativistic velocities, momentum can change a lot for a very, very small change in velocity, so the behavior is highly nonlinear.
    - pdonis 11y ago
      > Say it eats an atom - it has to accelerate that atom to the same relativistic velocity 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.
pdonis 11y ago
> unless the hawking radiation is fast enough to counteract the mass it is absorbing 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.

pdonis 11y ago

> Would this loss of momentum be enough to reduce its velocity below Earth's escape velocity? 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.