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Special Relativity and Black Holes

  1. Jan 26, 2009 #1
    Wouldn't an object traveling close the speed of light decrease in length and increase in mass, effectively creating a black hole?

    I attempted to do the calculations, but I don't know how to calculate the Schwartzchild radius of an accelerating body. Do you find out what the radius would be before and then at the top acceleration speed?

    I don't know if anyone has ever thought about this. What about the LHC?
     
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  3. Jan 26, 2009 #2

    JesseM

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    No, see this section of the Usenet Physics FAQ.
     
  4. Jan 27, 2009 #3
    In the text "ignores momentum and angular momentum as well as the dynamics of spacetime itself" Wouldn't "momentum and angular momentum" of a fast traveling object improve the conditions of a black hole forming?

    The explanation is vague and basically says "we don't know", except it's tremendously bias towards no.
     
  5. Jan 27, 2009 #4

    JesseM

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    It doesn't say "we don't know" what is predicted by the theory of general relativity (the theory of gravity that predicts black holes in the first place), it says that the answer in GR is definitely no. The only thing it says we don't know about is whether the conditions for some of the Penrose-Hawking singularity theorems apply in the real world.

    An obvious reason an object can't form a black hole just because it moves at high speed is because speed is relative to your choice of coordinate system (there is no absolute truth about whether any object is moving at close to light speed), whereas the formation of an event horizon is an objective physical truth that all coordinate systems agree on.

    If you want to see some previous discussions about this subject, go here or here or https://www.physicsforums.com/showthread.php?t=111857 [Broken] or here (where I quoted a longer version of the physics FAQ answer that used to be on that page) or here (where pervect gives a detailed analysis of the issue).
     
    Last edited by a moderator: May 3, 2017
  6. Jan 27, 2009 #5

    Dale

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    As JesseM and the FAQ mentioned, the short answer to your overall question is "no".
    There are two very common mistakes in this sentence.

    First, objects do not "increase in mass" according to typical physical useage. The unqualified word "mass" is always taken to refer to the invariant rest mass. The "relativistic mass" which increases without bound as v->c is the same as energy, so it is a redundant concept and is deprecated in modern physics.

    Second, regardless of if you use rest mass or relativistic mass, the source of gravity in GR is the stress-energy tensor, not mass. Energy (of which mass is normally the dominant source) is essentially only one component of the tensor, other components include momentum, pressure, and stress. But as v->c not only does energy increase without bound but also momentum increases without bound. That means that the other terms of the stress-energy tensor cannot be ignored. When you take them into account the result is that no black hole forms.
     
  7. Jan 29, 2009 #6
    I completely understand why you guys are wrong now. Wow how embarrassing for physics.

    You haven't provided a reservoir of energy that is needed to move an object at light speeds. You are only including the original mass of an accelerating object. More and more energy is needed to move an object close to light speed, in fact the amount of energy needed is infinite, so the mass would be infinite also, creating a black hole.

    It's impossible to do with jet fuel, or most likely impossible with any technology. But if we can get past this step i'll explain how to do it.

    Please don't ban me, just read, comprehend, and retort.
     
  8. Jan 29, 2009 #7

    JesseM

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    Do you understand that the speed of sublight objects is relative in both special and general relativity, so for any object it's possible to analyze its behavior from the point of view of a frame of reference where it's already moving at high speed, with no need to apply energy to accelerate it? For example, there is some frame where the Earth is right now moving at 0.9999999c, and this frame is no more or less correct than any other frame? (the laws of physics in this frame are no different than the laws of physics in a frame where the Earth is at rest, so if you believe a planet moving at 0.9999999c in the Earth's rest frame would be a black hole, then you should believe Earth is a black hole too)
     
  9. Jan 29, 2009 #8
    Also,

    From Wikipedia:

    "If an object is moving in any reference frame, then it has momentum in that frame. It is important to note that momentum is frame dependent. That is, the same object may have a certain momentum in one frame of reference, but a different amount in another frame."

    The momentum of an objecting accelerating toward C increases linearly, while the energy pushing the object increases exponentially, creating a black hole.
     
  10. Jan 29, 2009 #9

    JesseM

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    Did you read my previous post? There is no need to push an object in order to consider what happens when it's moving at a high fraction of light speed, since all speeds are relative to your choice of reference frame (and all frames are equally valid) you can just consider a frame where the object is already moving at a high fraction of light speed.
     
  11. Jan 29, 2009 #10

    Mentz114

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    Hilarious - keep going NewDesCartes, I really enjoy this sort of thing.

    Please explain something, if we have two travellers keeping pace and close ( so they are effectively not moving wrt to each other and can see each other), but both moving fast enough to become black holes relative to some distant observer, will they see each other turn into black holes ?
     
  12. Jan 29, 2009 #11
    I understand what you are trying to say, but you are trying to view earth from a black hole, which is impossible. Where are these frames in which you can view earth traveling at that speed? We use all frames of in the observable universe to decide whether something is in motion and what direction they are in, not just two frames of reference.

    Try to imagine viewing the earth from a planet orbiting a black hole moving .999999 C. An extreme redshift would take place, making the earth virtually invisible.
     
  13. Jan 29, 2009 #12
    How do you gain information from a black hole?
     
  14. Jan 29, 2009 #13
    I mean retrieve information, it is impossible. Light can not escape, so no information can be viewed.
     
  15. Jan 29, 2009 #14
    What I am stating is that the extreme Red Shift explains the frame reference phenomena. As for two objects moving fast enough to become black holes, they both are no longer viewable at any reference frame, so NO they don't see each other. All they see is black, because the velocity of space is C, heading right straight towards their face.
     
  16. Jan 29, 2009 #15

    JesseM

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    No, it's quite apparent that you don't.
    Frames aren't physical objects, they are just coordinate systems for assigning space and time coordinates to events. You can pick any coordinate system you like, including one where the Earth's has position x=0 light years at time t=0 years, and then later has position x=0.99 light years at time t=1 year (meaning it is traveling at 0.99c in this frame). And a basic idea in special relativity is that the laws of physics are exactly the same in all inertial frames, so there's no basis for seeing one inertial frame's perspective as more "correct" than any other's.
    This is a totally ignorant statement--we don't "decide whether something is in motion" in any absolute sense in relativity, one of the most basic ideas of relativity is that there is no absolute truth about which objects are in motion and which are at rest, all speed is RELATIVE to your choice of reference frame, and all inertial frames are on totally equal footing as far as the laws of physics go.
    It's a basic premise of relativity that any event can be assigned coordinates in any frame, practical difficulties aren't relevant to a description of a theoretical problem like a massive object moving at a large fraction of lightspeed, and in any case practical difficulties in observing certain events are never completely insurmountable.
     
  17. Jan 29, 2009 #16

    Mentz114

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    It's tempting to ridicule your ideas when you say things like 'the velocity of space is C'.
    You need to learn a bit more. Pay heed to what you're being told.
     
  18. Jan 29, 2009 #17

    JesseM

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    And what if an object passes right next to you at a high fraction of c? You could probably observe it as it passed an millimeter from your face, right? (especially since the moment after it passed you, it would be moving away from you rather than towards you and thus highly blueshifted rather than redshifted) Well, the theoretical method Einstein imagined for assigning coordinates to events involved only local observations by measuring devices spread out throughout space. The idea was to have a grid of rulers throughout space which were at rest in a given frame, with synchronized clocks attached to each ruler-marking, and assign coordinates based on local observations in this system--for instance, if a camera at the 5 light-second mark on the x-axis ruler noted an object passing right next to that mark when the clock there read 2 seconds, then that event on the object's path would be assigned coordinates x=5 light-seconds, t=2 seconds in the frame where this ruler/clock system was at rest. Assuming you agree this method would work in theory, then actually figuring out the coordinates of events in a given frame is purely a practical difficulty which is always in principle solvable, so it's of no consequence to a theoretical discussion.
     
  19. Jan 29, 2009 #18

    Objects are definitely relative to one another. If I am moving towards an object that is moving towards me, it looks like the object is getting closer. If I am accelerating toward an object that is moving away from me at V< A, it also appears that the object is moving closer to me.

    An object moving away from earth @ .999999% the speed of light creates a massive redshift in the light rays, effectively making the information from earth irretrievable on that object, a black hole. IF the velocity of that object decreases to .0000001% the speed of light, the earth would be viewable once again.

    Your comment "practical difficulties aren't relevant to a description of a theoretical problem" is a totally ignorant statement. Solving practical difficulties is the sole purpose we do physics.
     
  20. Jan 29, 2009 #19
    Let me Elucidate on MASSIVE REDSHIFT. When the object in motion approaches light speed, the earth becomes invisible.
     
  21. Jan 29, 2009 #20

    Mentz114

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    How would you know ? It is clear you can't do physics.

    The case of a body undergoing constant acceleration is well described by the Rindler space-time. No black hole is formed in any frame of reference.

    That would be an horizon, not a black hole.
     
  22. Jan 29, 2009 #21
    There are no observers in a black hole, if there is, we can never know, unless the black hole lowers its gravity, or slows down.

    I think it is ironic you are trying to explain relativity to me, when you don't understand relativity.

    Earlier you quoted "For example, there is some frame where the Earth is right now moving at 0.9999999c, and this frame is no more or less correct than any other frame? (the laws of physics in this frame are no different than the laws of physics in a frame where the Earth is at rest, so if you believe a planet moving at 0.9999999c in the Earth's rest frame would be a black hole, then you should believe Earth is a black hole too)"

    You think like you do about relativity because you don't understand that BOTH frames are correct. The earth is a black hole, and is not a black hole. From the reference frame of the object moving away the speed of light, it is for general speaking purposes a black hole. Using additional reference frames as a guide, an observer moving away at very close to the speed of light, could infer that they are moving away. This is due to the observer seeing objects in direction A moving towards them at < C, and the objects in opposite direction B are moving away from the observers.
     
  23. Jan 29, 2009 #22

    JesseM

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    When did I say otherwise? My point was that there is no absolute truth about which of two objects is moving at a greater speed, whereas there is an absolute truth about whether a given object satisfies the definition of a black hole, it's not a frame-dependent thing (see below).
    V < A doesn't even make sense as an equation--are you unfamiliar with dimensional analysis? You can't have an inequality where one side involves a velocity and the other involves an acceleration, because which side is bigger will depend on what choice of units you happen to make. For example, would you say V < A if V=60 meters/minute and A=360 meters/minute^2? But wait, 60 meters/minute translates to 1 meter/second, whereas 360 meters/minute^2 translates to 0.1 meters/second^2. So are you going to say V < A if we use units of meters and minutes, but V > A if we use units of meters and seconds? That doesn't sound very physical.

    And even if we settle on a particular system of units like meters and seconds your statement isn't true. Suppose in some frame at time t=0 I am at rest (v=0) and another object is moving to the right at 5 m/s. If I am accelerating to the right at 10 m/s^2, then my velocity as a function of time will be v(t) = 10*t, so my velocity will match that of the other object at t=0.5 s when my velocity will be 5 m/s to the right. Between t=0 s and t=0.5 s, my velocity will be less than 5 m/s, so since the other object is moving to the right at a constant speed of 5 m/s, prior to t=0.5 s the distance between me and the other object is increasing, not decreasing. Only after t=0.5 s am I getting closer to the object as time passes.
    No, part of the definition of the term "black hole" is that it's an object with an "event horizon" such that it is impossible for light emitted from events inside the event horizon to escape to the outside, whereas you're talking about a situation where the light from an object can reach us but it's just so redshifted that it'd be very hard to detect in practice. And you didn't reply to my more recent post where I pointed out you can always use a network of local observers who measure the object as it passes right next to them.
    Theoretical physics involves taking results of practical experiments and abstracting them into a general set of mathematical rules. As long as a given experiment would be possible in principle it can be a topic for a theoretical discussion about what these mathematical rules would predict for the results, regardless of the practical difficulty of actually carrying out the experiment (we can always hope that better technology would make previously-impractical experiments practical). You should be aware that the very concept of a "black hole" is something that originally arose purely from the theory of general relativity rather than any actual observations at the time, and that even today we can't verify with any great certainty that any real astrophysical objects have all the properties predicted by the theory (including the event horizons which are part of the definition of black holes).
     
    Last edited: Jan 29, 2009
  24. Jan 29, 2009 #23

    JesseM

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    For a genuine black hole, local observers just outside the event horizon could still assign coordinates to the event horizon at any given moment, even if no one can assign coordinates to events inside the horizon and communicate with the outside world. But in any case, an object moving at high speed does not have an actual event horizon, you're just talking about objects whose light can escape but is highly redshifted.
    No, you just don't understand the definition of "black hole". There is no frame in which the Earth has an event horizon from which it is absolutely impossible for light to escape even in theory--for an observer moving towards the Earth at high speed the light is highly redshifted and may be difficult to see, but the light is still reaching them even if they don't have sufficiently good equipment to detect it.
     
  25. Jan 29, 2009 #24
    It is clear you can't do philosophy.
     
  26. Jan 29, 2009 #25
    It is highly red shifted @ .999999% C. It flat lines @ C. We never catch an object moving @ C, nor can our probing light rays.
     
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