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Black hole, Penrose Process, SL, TL-Vector quick question

  1. Dec 8, 2014 #1
    ##E= -T^uP_u ##,

    where ##T^u## is the time-like killing vector associated with the Kerr Metric and ##P_u## is the 4-momentum of the particle. ##E## is the energy.

    Outside the ergosphere ##T^u## is time-like and inside the ergosphere it is space-like. Therefore it can be arranged within the ergosphere that ##E= -T^uP_u <0. ## *

    The condition for which this can be arranged is described so far as : particle's motion must be against the angular momentum of the black hole..

    My Question:

    I looked at a quick proof justifying that assuming negative energy of a particular particle within the ergosphere, ##E'##, its angular momentum ##L'## is negative (against the black hole's rotation) using the fact that the particle crosses the event horizon and by defining a new killing vector - linear combination of the time-like and angular killing vector, with the angular velocity of the event horizon -##\omega## - being a multiplicative constant:

    ##L'< \frac {E'}{\omega} ##

    Question:

    But, if ##L'## is negative, we can make no conclusion on the sign of ##E'##. So my interpretation so far is that the particles motion to be opposing the angular momentum of the black hole is a necessary but not sufficient condition for the particle to have negative energy.

    My question then is, what further conditions must we impose so that * is space-like - and what do they physically correspond to ? I.e- I see that the scalar product of a TL vector with a SL can be null, SL or TL- how do you formulate the relationship / relationships between the components of the vectors/vectors forming the scalar product so that we know what the resulting nature will be and , if this is formulated for * does this give the corresponding physical conditions that must additionally be imposed so that the particle's energy is negative?

    (I've had a google around but can't find anything).

    If anyone could help me ut or lead me to some helpful resources that would be extremely appreciated ! Thank you !!!


     
  2. jcsd
  3. Dec 8, 2014 #2

    PeterDonis

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    First of all, ##L'## being negative does not mean the particle must be moving in a retrograde (against the hole's rotation) orbit. It only means that the particle's angular velocity must be less than that of a zero angular momentum observer, or ZAMO. The ZAMO's angular velocity is zero at infinity, and gradually rises until it approaches ##\omega##, the angular velocity of the horizon, as the horizon is approached. Any observer with an angular velocity less than that of a ZAMO at the same position, even if the angular velocity is still positive, will have a negative angular momentum ##L'##.

    This is important because, within the ergosphere, retrograde orbits (negative angular velocity) are impossible! In fact, even zero angular velocity (staying at the same angular coordinates forever) is impossible. All timelike worldlines have positive angular velocity, because of the "frame dragging" due to the hole's rotation.

    (Some sources do use the term "counter-rotating" to describe orbits within the ergosphere; what they actually mean is orbits with angular velocity less than the ZAMO angular velocity. As noted above, this angular velocity must still be positive within the ergosphere.)

    That said, if you could give a reference to your "quick proof", it would be very helpful. It's hard to know what you need help with if we don't know what you've already seen.

    I'm not sure what you mean by "spacelike" here. The Killing vector field ##T^u## is spacelike everywhere inside the ergosphere. The 4-momentum ##P_u## is never spacelike; it's timelike (at least for any object with nonzero invariant mass). The energy ##E## is a scalar; it's just a number, not a vector, so it isn't spacelike or timelike. So what exactly are you asking?

    If you're asking what the conditions are for ##E## to be negative, I think it will be easier to answer if you give a reference to the quick proof you mention.

    A note: this exhibits the same confusion as above. Scalars are not spacelike or timelike or null; they're just numbers. I think what you're really trying to say here is that the scalar product of a timelike vector and a spacelike vector can be zero, positive, or negative.
     
  4. Dec 10, 2014 #3
    A note: this exhibits the same confusion as above. Scalars are not spacelike or timelike or null; they're just numbers. I think what you're really trying to say here is that the scalar product of a timelike vector and a spacelike vector can be zero, positive, or negative.[/QUOTE]

    Sorry my bad here ! Yhep I meant just pos/neg/zero.
    Source: (page 214) : http://arxiv.org/pdf/gr-qc/9712019.pdf
     
  5. Dec 10, 2014 #4

    PeterDonis

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    Ah, ok. This source (unfortunately) uses the confusing language I referred to above; when it says "moving against the hole's rotation", it means negative angular momentum, but not negative angular velocity (which, as I said before, and as this source agrees, is impossible within the ergosphere).

    Equation 7.144 here shows that, if the energy at infinity ##E## is negative, the angular momentum ##L## must also be negative; but it does not show what is required for ##E## itself to be negative. I don't see anywhere that it gets more specific than equation 7.138, which is just the equation for ##E## in terms of the appropriate Killing vector field. To get any more information about what it takes for ##E## to be negative, you would have to expand out equation 7.138 in terms of known quantities like the hole's mass and angular momentum. I don't have time to go into that now, but you might want to try it to see what you come up with.
     
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