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The wave equation in a Black Hole.

  1. Aug 24, 2009 #1
    The Shroedinger equation defines the time evolution of the wave function. If we observe a region of large gravitational fields where observed time has slowed, the wave function will be observed to evolve slowly. In the limit of a Black Hole it will stop evolving altogether.

    Still quantum mechanical probabilities exist. They just do not change. So in the primordial past, before time, quantum mechanical fluctuations finally lined up to create the Big Bang.

    I naively invite comments.
     
  2. jcsd
  3. Sep 7, 2009 #2
    That's an interesting question, I don't know if time even has a definition within the horizon of a black hole where space-time is infiinitely curved. If it were - time no longer exists, but I don't think anyone knows. I've been trying to find papers on the qm properties of neutron stars (basically the wave equation for a large number of bosons) where the potential is primarily gravitational and not coulomb.
     
  4. Sep 7, 2009 #3
    sounds interesting. Would you like to educate me a lttle?
     
  5. Sep 8, 2009 #4
    Well they have to change, and again the definition of time within a black hole is not clearly understood but surely we know that once matter passes through the horizon it surely moves from the horizon to the center of mass, and does it in a certain amount of time. Our idea of reference frames and length and width may not mean anything anymore but surely these things happen over some distance and time starting from a point where presumable time stops (the horizon)... Some parts of quantum mechanics may have an answer for these questions, but I haven't ever read or seen anything on it yet. What I'm talking about is the existence of true singularities - zero volumn, infinite density systems. Though most physicits these days adhere to the singularity nature of BH's, I fundamentally do not. Both relativity and qm theory are well tested and have been proven, and two things have never been shown false - mass velocities cannnot exeed c, and the uncertainty in position and velocity (or momentum) cannot fall below planks constant over two. What does this mean for BH's? Well it means that if a singularity were possible (i.e. the volume = 0, thus the uncertainty of any one particle in any direction would be "exactly" known with infinite precision) the uncertainty in velocity would have to be infinite. But thats not possible - Einstien put an uppeer limit on relativistic speeds for everything - c. This would force the uncertainty of a black holes volumn to reach a lower limit, that which cannot be exeeded. On another note, no astorphysicist really knows why jets fly out of BH's (quarasrs, blasars, pulsars, etc.) and I think QM may have the answer for that too, related to the minimum volumn I just descibed. Hiesenebergs uncertainty principle allows for KE = Mc^2 as the black hole reaches it's minimum volumn, where M is the black hole mass, this energy could be responsible for the relativistic speeds seen in the jets.
     
  6. Sep 8, 2009 #5
    So ... as a star collapses it heats up quantum mechanically because the positions of its particles are confined to smaller and smaller volumes. At some point particles will have velocities close to the speed of light. The speed of light limit on particle velocities will then prevent further collapse.

    On the other hand, the star may be so large that speed of light velocities are too slow to balance the collapse and the star either continues to collapse and either Relativity or QM fails.

    Did I understand you right?
     
  7. Sep 8, 2009 #6
    Ya you understood me right, but now I'm unsure if my assumption of momentum is correct. I'm wondering if a infinite uncertainty in momentum is allowed because the particle direction has rotated 90 degrees - in that case the uncertainty is infinite... For example shoot a laser beam through a single slit. If you try to localize the photons to smaller and smaller regions (decrease the size of the slit), the interference patter on a sheet of paper past the slit will widen - the particles don't necessarily speed up, they just begin to increase in velocity in the perpendicular direction to the laser beam, and slow down in the direction of the laser beam, to preserve that the photon speed is still c. Now let's let the slit width go to 0, assuming the laser beam lies along the x-axis, (the interference pattern width in the y) the uncertainty in momentum will increase, the interference patter on the sheet will approach inf in length (i.e. the x component of the photon velocity is 0 and the y component is c and those photons wouldn't even hit the paper at all). In this case the BH collapses and QM does nothing to stop it. I will say this. The jets of a BH are OBVIOUSLY breaking a rule that nothing can escape a BH, things OBVIOUSLY do escape and escape at relativistic speeds. I still believe QM holds the answer but this probably isn't it. You should think about that - do some research on general relativity and very quickly rotating objects. Nuetron stars of about 2 solar masses rotate about 700 times per second and have a diameter of ~20km. Put that mass in the volume 4/3*pi*(h_bar/(M*c)^3) and see what happens to gravity. Along the axis of rotation it may be very possible that antigravity affect is responsible.
     
  8. Sep 8, 2009 #7
    Ok ... you are a little over my head. So let me ask a question.

    Why does infinite uncertainty in momentum imply velocities greater that the speed of light? Wouldn't this mean for a particle with mass that the center of the wave function would be moving faster than the speed of light and not necessarily preclude a large uncertainty in momentum? I thought that a particle's velocity was the velocity of the mean ( for a symmetrical wave function).

    Second, what about the Pauli exclusion principle? Wouldn't it be violated in a BH? Does the Pauli exclusion principle imply unbounded particle velocities in a collapsing star?
     
  9. Sep 9, 2009 #8

    Demystifier

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  10. Sep 9, 2009 #9

    alxm

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    Well no. Particles can exist at the same location, as long as they're not in the same state. If they move to a higher energy level, they can still be in the same place, spatially. (E.g. An atom can only have two electrons in an orbital, after which they have to be in a higher orbital. Hence we have a periodic table, and not just a bunch of electrons hanging about in 1s. The different atomic orbitals do overlap spatially)

    So what the Pauli principle leads to is a big increase in energy as things get compressed. So it works as a pressure (degeneracy pressure). This is what keeps a neutron star from collapsing in on itself. If the mass is big enough so that gravity can overcome this pressure, (the Tolman-Oppenheimer-Volkoff limit) that's when you get a black.

    I'm not a cosmologist at all, but I don't think anyone really knows exactly how this breakdown happens, though. Or even what's going on inside a neutron star.
     
  11. Sep 9, 2009 #10
    that's interesting. So in a Black hole the particle's energies theoretically become unbounded?
     
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