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Neutral particle falling into a black hole

  1. Apr 2, 2010 #1
    Probably there's a simple answer for the following question. What prevents a neutral particle falling into a black hole from beyond the event horizon - a point it would have to have an escape velocity greater than the speed of light to reach from the inside - from being accelerated past the speed of light on the way back in?
     
  2. jcsd
  3. Apr 2, 2010 #2

    bcrowell

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    Why do you say "on the way back in?" Why "back?"
     
  4. Apr 2, 2010 #3
    I shouldn't have used the word "back", because the particle's previous trajectory is not really relevant here. The point I was trying to make was that by starting beyond the event horizon a particle would be expected - under classical mechanics - to reach a terminal velocity greater than c when falling in. It would acquire mass relativistically with increasing velocity, but gravitation accelerates independently of this. What prevents the classical scenario?
     
  5. Apr 2, 2010 #4

    bcrowell

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    This argument is based on an attempt to understand general relativity in Newtonian terms. That doesn't work very well. You can't do all of GR just by adding corrections onto Newtonian physics. You can extend Newtonian physics with correction terms, but that is always going to fail when you get to extremely relativistic conditions.

    The over-all idea is that the "c" in general relativity isn't really the speed of light; it's a maximum speed of cause and effect. Therefore we don't expect the world-line of any particle to have v>c. If that did happen, it would violate causality, because there would be frames of reference in which the time-ordering of events along the world-line was reversed.

    If you want a specific proof that v>c can't happen in this particular situation, here's one. In general relativity, a free-falling particle moves along a geodesic, and one of the properties of a geodesic is that it preserves its own tangent vector. Parallel-transport won't change the norm of the tangent vector, and you can't change a velocity vector from <c to >c without changing its norm.
     
  6. Apr 2, 2010 #5

    Nabeshin

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  7. Apr 2, 2010 #6

    DaveC426913

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    It doesn't matter how the particle is accelerated - be it rocket propulsion or gravity (remember the principle of equivalence?) - its velocity is calculated as per the Lorentzian formula.

    And you'll see that no matter the acceleration, its final velocity will never reach c.
     
  8. Apr 2, 2010 #7
    I read your question as asking why can't the energy gained crossing the event horizon be used to allow a particle to then escape the black hole.

    If one naively examines the relativistic energy equation at speeds above the speed of light one notices that it becomes imaginary. One consequence of this is that inside a black hole the radial and time coordinates switch roles. Essentially the radius becomes the new time coordinate and the particle will proceed to the center like we proceed into the future.
     
  9. Apr 3, 2010 #8
    Thank you everyone for the very helpful responses.
     
  10. Apr 3, 2010 #9
    I think it's an interesting question. Here's an idea: gravitational mass, like passive electric charge, does not vary with speed. But inertial mass (or total energy) does. This being the case, the gravitational force experienced by the particle would be proportional to its gravitational (constant) mass, but its inertia would increase as its speed increases.
     
  11. Apr 3, 2010 #10

    bcrowell

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    Yes, I think that's the argument the OP had in mind. Does the explanation in #4 not work for you?
     
  12. Apr 3, 2010 #11
    If one examines an object in circular motion in the classical limit they can determine the required central force based only on the angular rotation, radius and mass of the object. If this system is viewed from a frame with relativistic speed along the axis of rotation the required central force can be determined using SR to be reduced by a factor of gamma.

    Now considering the initial case in the context of a weak gravitational field we get

    GMg1mg1/r2 = m0a

    The circular motion becomes motion along a cylinder with constant acceleration towards the axis with radius r1. Because the signal is time delayed the object will be felt at a distance increased by gamma. Only the radial component contributes to acceleration, causing the force to be reduced by gamma. The inertial mass increases by gamma, while the acceleration is decreased by gamma squared.

    The result is that

    1/gamma3 * GMg2mg2/r2 = 1/gamma * m0a.

    From this it is clear that gravitational mass is gamma * rest mass. This can also be expected by the fact that gravitation is caused by the stress-energy tensor, so should be related to the energy, not the rest mass.

    The gravitational force has an unobserved component in the axial direction that must be balanced by a fictitious force. This is the essence of frame dragging.

    If gravitation was proportional to rest mass the gravitational maximum during an eclipse would coincide with the optical eclipse. The gravitational maximum is directed towards the sun's present location.

    This lack of gravitational aberration can occur without violating the speed of light signaling by allowing space-time to carry a Taylor series of the Sun's position towards Earth. Satellite ranging data has verified the lack of aberration of the Earth towards the Sun up to third order derivatives, with measures for the fourth derivative consistent with no aberration, but near the limits of current technology.

    This is appears to be significant difficulty for quantum gravity that should produce aberration at some level.
     
    Last edited: Apr 3, 2010
  13. Apr 3, 2010 #12
    I knew I was on thin ice when I replied to this one. My knowledge of GRT is woefully deficient. Thanks for the pointers.
     
  14. Apr 3, 2010 #13
    Many thanks. I'll study your thread. Meantime, See #10. Best, GRD.
     
  15. Apr 3, 2010 #14

    bcrowell

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    Re utesfan100's #11: Ah, I see. I hadn't noticed that GRDixon's #9 seemed to be making an argument that gravitational and inertial mass are not always equal. But your arguments about the circular orbit and deflection of light seem shaky and overly complicated to me. Experimentally, we know that gravitational and inertial mass are always equal, because Eotvos experiments have been done with extremely high precision. Theoretically, this is baked into GR, since otherwise GR couldn't be formulated as a geometrical theory.


    I don't think this is right, although I may be misunderstanding you. This recent thread may be helpful: https://www.physicsforums.com/showthread.php?t=294328
     
  16. Apr 3, 2010 #15
    You can't mix Newtonian mechanics with SR in the hope of answering GR questions. Besides, your SR formulas are incorrect as well.


    Multiple experiments show the above to be false. Starting with Eotvos and continuing with modern reenactments. This is not even wrong.
     
  17. Apr 3, 2010 #16
    So it is wrong to examine GR in the Newtonian limit for consistency and a sanity check?

    I would enjoy corrections to my understanding of SR. I am not sure how you can state my equations are wrong when I did not list them.

    The original reference frame is at non-relativistic speed so the classical relations should hold. m is the invariant mass throughout.

    a=Rw2
    P=mRw
    F=ma

    Since the velocity is normal to the rotation the R is unchanged. w acts like a clock and is dilated by gamma.

    R'=R
    w'=gamma * w
    a'=R'w'2 = gamma2 * Rw2 = gamma2 a

    Momentum in the direction of the plane is unchanged, with an additional constant momentum in the direction of motion.

    Pxy'=Pxy=mRw
    F' = dP'/dt' = Pxy'w' = gamma * mRw2 = gamma * F

    In terms of the observer's frame:
    F' = Pxy'w' = mR'ww' = mR'w'2 / gamma = ma' / gamma

    If my SR above is out of order, I would like to learn how.

    If the acceleration is caused by a force that propagates at c then trigonometry shows that the distance between the emission of this force and its absorption is extended by gamma. This force has a component towards the axis with a coefficient of 1 / gamma and opposing the motion of v/c.

    Mass in GR is tricky. Gravitational and inertial mass are equivalent. But what in GR is inertial mass? (In addition to rest mass and the mass energy one can define longitudinal and transverse mass for a force.)

    If this force is caused by a weak gravitational field GR approaches Newtonian approximations. (Let M be the mass of the mass at the barycenter of a Kepler problem required to cause the motion and a subscript g indicate gravitational mass.)

    F ~ GMgmg / R2
    F' = ma' / gamma ~ GMgmg / R2 / gamma3

    ma' ~ GMgmg / R2 / gamma2

    If one were to use charge rather than mass one would find the gamma2 term in the denominator results in the electric field plus the magnetic field. If gravitational mass was the invariant mass their would be no difference between the optical and gravitational maximum during an eclipse.

    This would also go against the claim that in GR gravitation is caused by the stress energy tensor. The energy of a moving mass is gamma * m. If this is not the gravitational mass I don't know what would be.
     
  18. Apr 3, 2010 #17
    F=G*M1*M2/r^2 does not exist in GR.
    Neither does F=ma


    There is no rotation in the original problem.


    w can never "act like a clock" and will not be "dilated by gamma". Besides, since there is no rotation, there is no "w" to talk about. The infalling particle moves in a straight line.
    If you insist in talking about how w transforms, its transformation is dictated by the "Transverse Doppler Effect" and the correct formula is :

    w'=w/gamma

    totally contrary to the wrong formula you are trying to make up.



    No, see above : w'=w/gamma (not that it matters since your whole derivation is nonsense)

    Also no, acceleration does not transform this way. Transformation in rotating frames, since you insist, is much more complicated.

    "Force propagates at c"? "Distance between emission of the force and its absorbtion"? Are you making up your own brand of physics?

    Force does not transform this way in SR. Besides, it isn't even clear what you are trying to accomplish. The solution to the problem has nothing to do with any coordinate transformations. You need to solve the Euler-Lagrange equations. In order to do that, you need to form the Lagrangian by "reading" it off the Schwarzschild (or Kerr) metric. This is the correct way of solving this problem.


    There is no much sense continuing since there are so many errors right off the bat.
     
    Last edited: Apr 3, 2010
  19. Apr 6, 2010 #18
    I was under the impression they existed as the limit as the field goes to 0, which I stated as a condition for my side diversion.

    The original problem discussed gravitational mass. It was stated that gravitational mass is the rest mass. I am arguing that gravitational mass is energy/c^2, or gamma*mo for massive particles.

    This argument is presented by considering circular motion in the Newtonian limit and then applying SR to an axial component to show that in SR in the weak field limit gravity is not the rest mass.

    This implies that in the OP gravitational mass is also gamma*mo.

    I will defend the positions challenged above, but I agree that since the scenario is different from the OP the circular motion scenario should be its own topic.

    w is the angular velocity. This is proportional to the period of rotation of the stars. If this rotation rate of the stars is not a clock, I don't know what is.

    w=d theta / dt
    w'=d theta' / dt' = dtheta / dt * dt /dt' = w / gamma

    Thank you for the correction.

    This is why I used uniform circular motion, so the non-radial components of acceleration are 0.

    I am applying the law of causality being limited by the speed of light. This should be true for all forces.

     
  20. Apr 6, 2010 #19
    The above error renders your whole "proof" incorrect.
     
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