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Can kinetic energy lead to black hole?

  1. Sep 4, 2015 #1
    Dear PF Forum,
    I have a question in mind. But I'm not sure if this belong to SR forum, cosmology or classical physics.
    So I post it here.
    And perhaps as some of you might have known before or thought it over. It's about kinetic energy.
    Supposed this...
    A rocket, with a rest mass 1 ton.
    And the rocket can somehow grab matter and anti matter in interstellar medium. I know that anti matter just can't scattered freely in interstellar medium, but let's just dispense with the technical difficulties.
    So in its journey the rocket always accelerates 1 g. And in doing so it has consumed, say..., 10 solar mass of matter and anti matter annihilation as its energy source.
    Now after some times its speed is 99.9999% the speed of light (or should I add more nines?)
    Then this 1 ton rocket hits a neutron star, say 1.5 solar mass. Can this neutron star become a black hole if it is hit by this rocket, considering this 1 ton rocket relativistic mass must be about, say 3 solar mass (or more/less?), after spending much of its energy in the form of momentum to eject its propelant?
    Thanks for any clarification.
     
  2. jcsd
  3. Sep 4, 2015 #2

    Doc Al

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  4. Sep 4, 2015 #3
    Yes, thanks Doc Al for the link. I forgot I have read that link moments ago. And I read it again. But the situation is different here.
    The rocket stop now with all its potential energy. It is stopped by a neutron star. And I read that the maximum size of a neutron star is about 2 or 3 solar mass before it become a black hole. So even the mass of the rocket is 1 ton, it does spend a lot of energy to accelerate it near the speed of light wrt the neutron star right?
    I can visualize if a rocket travels near the speed of light wrt me, than I can't perceive it as a black hole. But, what if the rocket is stopped by other object with all of its potential kinetic energy? Is the situation the same as the link you give me?
     
  5. Sep 4, 2015 #4

    Doc Al

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    Yes, your situation is a bit different. I'll let others chime in.
     
  6. Sep 4, 2015 #5

    pervect

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    I believe the basic answer is yes. As discussed in http://link.springer.com/article/10.1134/S2070046612020069#page-1, if you have in theory, two hyper-relativistic particles colliding, they can form a black hole due to their kinetic energy even though neither one of the particles is itself a black hole.

    We can then imagine an appropriate transform to a frame where one of the particles is at rest to get something reasonably similar to the situation in your question, a collision between a moving object and a stationary one creating a black hole.

    The gravitational field solution of a hyper-relativistic particle in the extreme ultra-relativistic limit is called the Aichelburg-sexl ultraboost. https://en.wikipedia.org/w/index.php?title=Aichelburg–Sexl_ultraboost&oldid=551268593. So the process of colliding two hyper-relativistic particles can be "conveniently" modeled as the collision of two such "ultraboost" gravitational plane waves.

    I've never seen the detailed calculations of such a collision, but my recollection and the paper I cited above support the idea that such collisions are known to form singularities and black holes.
     
  7. Sep 4, 2015 #6

    PeterDonis

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    You've stated the situation vaguely, so I'm not sure if your intent was that the rocket, in its travels, had collected 3 solar masses' worth of rest mass, or just that its relativistic mass (i.e., total energy) was 3 solar masses, prior to colliding with the neutron star.

    In the first case (3 solar masses' worth of rest mass prior to collision), the rocket might well have become a black hole purely by absorbing enough matter, even before colliding with the neutron star. The only way it could have avoided doing that would have been to turn into something like a neutron star itself (or else igniting into an ordinary star).

    In the second case (3 solar masses' worth of total energy prior to collision), the rocket's rest mass could still be much smaller than that of a star (or even much smaller than a planet, if it's moving fast enough). But what happens in the collision doesn't depend on the rocket's rest mass; it depends on its total energy and momentum. It isn't what you're imagining; see below.

    No, it won't. We'll stick to the second case above, where we have a neutron star at rest with a rest mass 3 solar masses, being struck by a rocket (plus a bunch of accumulated debris) moving so fast that its total energy is 3 solar masses. We assume that after the collision, we end up with a single final object.

    The first thing we need to figure is the rocket's momentum. We don't know, from your statement of the problem, exactly how much rest mass the rocket has, but since it is moving at a highly relativistic velocity, we can assume that it's much, much less than its total energy. So to a first approximation, its momentum will be very close to its total energy, i.e., about 3 solar masses' worth of momentum.

    Since momentum is conserved, we must also have 3 solar masses' worth of momentum after the collision. Assuming, again, that the rocket's rest mass is negligible, the final object will have a rest mass approximately equal to that of the original neutron star, 3 solar masses. If it also has a total momentum of 3 solar masses, then its total energy is ##\sqrt{m^2 + p^2} = 3 \sqrt{2}## solar masses. In other words, the final object ends up with a relativistic ##\gamma## factor of ##\sqrt{2}##, which corresponds to a speed ##v## of ##1 / \sqrt{2} \approx 0.7## times the speed of light. This certainly can't be described as the rocket being "stopped" by the neutron star. :wink:

    As for whether the final object would be a black hole, it's hard to say. Some energy would probably be radiated away in the collision, so it could even be that the final object's rest mass would be smaller than that of the original neutron star. Even if it turned out to be larger, whether a black hole was formed or not would depend on how close the original neutron star was to the maximum neutron star mass. It would also depend on other details like how much the neutron star's structure was compressed by the collision process; it's possible that that could form a black hole even if the final object's mass was below the maximum mass limit for a neutron star.

    In short, it is certainly possible for a black hole to be formed in this type of process, but I don't think we can say very much about the detailed conditions that would determine whether one does in a specific case.
     
  8. Sep 4, 2015 #7

    PAllen

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    If you make the problem extreme enough, you don't need to worry about the dynamics Peter pointed to, if one is willing to assume the (unproven but plausible) hoop conjecture. Then, if in the COM frame, the total energy of neutron star + rocket would correspond to a Schwarzschild radius larger than the neutron star + rocket, then a black hole will form. Note, this is a substantially higher threshold than the Chandrasekhar limit.
     
  9. Sep 4, 2015 #8

    Vanadium 50

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    Stephanus, as usual you have way, way, way overcomplicated the situation with all this talk of antimatter and the interstellar medium.

    Your question is whether an inelastic collision between an object of mass m1 at rest and one of mass m2 with momentum p can produce an object of invariant mass m3. That's a straightforward question in relativistic mechanics, and depending on m1, m2 and p, it could end up either above or below m3.
     
  10. Sep 4, 2015 #9
    Thanks PeterDonis for your answer.
    What I mean is this. Supposed there are anti matters and matters enough in interstellar medium. And every time the rocket "grabs" say 1 mili gram (or gram?) matter/anti matter and annihilate them to propel the rocket. After some times, the rocket has "grabbed" perhaps tens of solar mass matter and anti matter.
    Now, I was just thinking. To accelerate the rocket to that velocity, the rocket has consumed tremendous amount of energies.
    Its kinetic energy ##E_k = \frac{1}{2}mv^2## is bigger than ##E = mc^2##?? So, the energy of this 1 ton rocket is bigger than ##1 ton * c^2## I guess. That's what I mean. That's my answer to your above paragraph. I have read your next paragraphs, there are good answers there.

    Come on...
    Can kinetic energy be bigger than E=mc2?
    What is "p" here? "mv"?
    Yeah, there's no irresistible force and there's no immovable object either. I think if we fire a bullet right in the heart of a rigid body like neutron star, the neutron star will move at least 1 micrometer or 1 nanometer, or less?

    Thanks for the answer. This post is just my reply to your post PeterDonis. But I have to (and am) read(ing) your post all over again.
     
  11. Sep 4, 2015 #10
    Particle?? Ahh, I read that CERN is creating (tiny) black hole on daily basis, by colliding particles. Thanks for your reply pervect. I'll reread your post again. And thanks for the links.
     
  12. Sep 4, 2015 #11
    Chandrasekhar? It's 1.4 solar mass, right.
     
  13. Sep 4, 2015 #12
    Could you elaborate more? What is m3? Is it m1+m2, or below it?
     
  14. Sep 4, 2015 #13

    pervect

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    Let's clear up a few things that other posters have mentioned

    You didn't specify what happens to the extra mass that the rocket "picks up". I assumed that the rocket "burned" this extra mass, in the form of matter-antimatter fuel, expelling it out the rear of the rocket, and that we are to assume the rocket does all this with sufficient efficiency that the thrust generated by burning this fuel more than compensates for any momentum penalty involved in the "pick up" process. Furthermore I assumed that the rocket burned and expelled all the fuel, so that the rocket's final rest mass was the same as its initial rest mass, 1.e. 1 ton.

    To specify the end result (rather than the process of obtaining it), my assumption was that we are to consider the rest mass of the rocket as still 1 ton, the same as it started out as, and the kinetic energy of the rocket (relative to its starting frame) is about 3 solar masses * c^2.

    Can you confirm if this was your intent?

    This particular form of the problem is unnecessarily difficult. Would you be satisfied with considering a simpler problem, two such rockets, colliding head on?

    With this modification, the center-of-mass of the collision is stationary, and the center-of-mass energy density is what is immportant.

    Since 1 solar mass is about 1.5 km, a rocket smaller than 1km would easily produce the energy densities required to create a black hole.

    While you didn't ask this question, I don't believe it would even be necessary for the two rockets to actually collide, passing within, say, the previoiusly mentioned kilometer of each other should be easily good enough to create a black hole.

    The p-p wave approach I mentioned previously is probably a bit too advanced, I hope this simpler scenario is easier to grasp.
     
  15. Sep 4, 2015 #14

    PeterDonis

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    But none of this increases the rest mass of the rocket; the matter and antimatter is annihilated and ejected out the back of the rocket. So the total mass that it has "grabbed" is not relevant except that it contributes to the rocket's total energy in the frame of the neutron star.

    Assuming ##m## is rest mass, certainly it can. There is no limit in principle to how much kinetic energy an object can have. In this case, if the rest mass ##m## of the rocket is 1 ton all through the scenario, and its total energy is 3 solar masses, that equates to kinetic energy being about ##10^{27}## times ##m##.

    ##p## is momentum. The correct relativistic formula is not ##p = mv##. It's ##p = \gamma m v##, where ##\gamma = 1 / \sqrt{1 - v^2}## (in units where ##c = 1##).

    Yes, in principle the neutron star will always move by some amount--more precisely, it will gain some nonzero velocity in the frame in which it was originally at rest. But for an ordinary object like a bullet hitting it, that velocity will be way too small to detect--many orders of magnitude too small. The reason I pointed it out in the case you described was that the neutron star's speed after the collision is certainly not too small to detect--it's going to be a substantial fraction of the speed of light.
     
  16. Sep 4, 2015 #15

    DrGreg

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    By the way. that's the wrong formula. In relativity it's ##E_k = (\gamma - 1) mc^2##
     
  17. Sep 5, 2015 #16
    Thanks pervect for your post. I use matter/anti matter in this example not chemical because I want to know if ##E_k = \frac{1}{c}mv^2## can be bigger than ##E = mc^2##
    If I use chemical reaction, then we'are talking with joulse not mass/energy equivalent. So, let's say that the rocket grabs 1 gram matter and 1 gram anti matter and annihilate it. The mass of the rocket? Not affected, still 1 ton. Why I don't use joule? Because it will be difficult to convert it to mass, with matter/anti matter, we just can say that the rocket consumes energy equivalent to say... 3 solar mass :wideeyed:
    Yes.
    Yes
    Yes
    Can you confirm if this was your intent? I did

    Yeah, should have thought it over. Remember particle-particle collision in LHC, or accelerator?

    Yes, I didn't ask that. But I think it over lunch just now. I was just thinking, still with neutron star. If the neutron star is 3 solar mass, and the rocket with its 3 solar mass potential energy pass by the rocket less then 10 km should be enough. Then I go back to my computer and see your quote. Doesn't actually collide, for 3 solar mass it will be... 3 * 3 * 3 = 81 km?
     
  18. Sep 5, 2015 #17
    Yes, it is ejected. The mass of the rocket still 1 ton when it hit the neutron star. I only want to know, in the small compact radius, smaller than Schwarzschild Radius, with only 3 solar mass, what if we put additional energy in that radius, not mass, mass is neglible 1 ton compared to 3 solar mass, right.. Can it become a black hole? Because the potential kinetic energy of the rocket is 3 solar mass.
    Oh
    Hmm..., okay. DrGreg post below states it. I'll read it.

    Yes, in principle the neutron star will always move by some amount--more precisely, it will gain some nonzero velocity in the frame in which it was originally at rest.
    Nonzero, yes.
    But for an ordinary object like a bullet hitting it, that velocity will be way too small to detect--many orders of magnitude too small.
    Too small, I agree. How small? Have to calculate it for solar mass vs 10 grams :wink:
    The reason I pointed it out in the case you described was that the neutron star's speed after the collision is certainly not too small to detect--it's going to be a substantial fraction of the speed of light.
    Yes, that's right. I already suspect that. The neutron star will move a substantial fraction the speed of light. But over lunch, I thought. The rocket doesn't necessarily HIT the neutron star. Just when the rocket arrives at the vicinity of the neutron star and their combined mass is below the schwarzschild radius, it can become a black hole. Then I read pervect post.
    Thanks for your confirmation PeterDonis
     
  19. Sep 5, 2015 #18
    So, what is ##\frac{1}{2}mv^2## This is what you get when you accelerate an object with newtonian law right?
    Suppose I push an object 1 kg for 1 n for 50 meters, its velocity will be 10 m/s, etc, etc. I don't have to tell you this DrGreg :smile:. You must have known this inside out!
    Now, if the object suddenly stops. Then it's kinetic energy is ##\frac{1}{2}mv^2##, right. Because it's the amount of energy spent to move that object.
    Hmmm, I'll study your formula later.
     
  20. Sep 5, 2015 #19

    Drakkith

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    No, that equation is only useful when speeds are very low when compared to the speed of light. The work done to accelerate an object actually follows the equation that DrGreg gave. If you were to accelerate an object to 0.9c, you would find that 1/2 mv2 is incorrect and gives a kinetic energy of less than 1/3 the actual value.
     
  21. Sep 5, 2015 #20

    DrGreg

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    $$
    E_k = (\gamma - 1)mc^2 = \left( \frac{1}{\sqrt{1 - v^2 / c^2}} - 1 \right) mc^2
    $$
    When ##v## is very small compared with ##c## and you do the maths, it turns out that ##E_k## is approximately ##\tfrac{1}{2}mv^2##.

    If you want to know why the relativistic formula is exactly correct and the Newtonian formula is only an approximation, you'll need to study a textbook as there's no simple one-paragraph explanation. But as a plausibility argument you should see that ##E_k## increases towards ##\infty## as ##v## approaches ##c##, so no matter how much energy you supply to a mass, it never reaches the speed of light. If the Newtonian formula were correct, you'd only need to supply ##\tfrac{1}{2}mc^2##.to reach the speed of light.
     
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