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Photons that orbit at the event horizon.

  1. Jul 24, 2008 #1
    My question is: Photons that orbit at the event horizon, require relativistic time to circumvent the circumference of the black hole relative to an observer external to the black hole, keeping Mr. Einsteins second postulate in mind, that being the velocity of light is constant through vacuum ( or free space), and keeping in mind, dilated space-time is considered free space or in vacuum. What is the relativistic time required for the photon to complete one revolution at the event horizon, relative to an external observer of the black hole, would it be the same as the time required light to travel an equivalent distance as observed in free space external to the black hole, or would it take a much longer relativistic time?
     
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  3. Jul 24, 2008 #2

    George Jones

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    Photons don't orbit at the event horizon, they orbit on the photon sphere at a (coordinate) distance of 1.5 that of the event horizon.
     
  4. Jul 24, 2008 #3
    George Jones: Would the existance of the photon sphere, being outside the event horizon, prevent protons between the photon sphere and the event horizon, escaping the black hole?
     
  5. Jul 25, 2008 #4

    George Jones

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    No. If a hovering laser located between the event horizon and the photon is fired within a certain range of directions, its light will escape the black hole. For lasers located closer and closer to the event horizon, the range of directions becomes more and more restricted.
     
  6. Jul 25, 2008 #5
    Thank you for your answers, they fit very well.
     
  7. Jul 25, 2008 #6

    DrGreg

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    As the laser's direction varies, some photons escape, some do not. So what happens at the borderline between those two possibilities? Naively I might expect photons to orbit at that borderline angle, but I have heard this is not possible.
     
  8. Jul 25, 2008 #7

    George Jones

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    The border is orbit on the photon sphere!

    When [itex]c=G=1[/itex], the event horizon is at [itex]r=2M[/itex] and the photon sphere is at [itex]r=3M[/itex]. Suppose the laser hovers at any [itex]r>2M[/itex], and let [itex]\phi[/itex] be the firing angle with respect to vertically up, in the laser's frame. When

    [tex]
    \sin\phi = \sqrt{3} \frac{3M}{r} \sqrt{1 - \frac{2M}{r}},
    [/tex]

    the light will asymptotically approach orbit on the photon sphere as [itex]t \rightarrow \infty[/itex].

    Note that as [itex]r \rightarrow 2M[/itex], [itex]\phi \rightarrow 0[/itex], and that as [itex]r \rightarrow 3M[/itex], [itex]\phi \rightarrow 90[/itex] degrees.

    If you want, you can try

    https://www.physicsforums.com/showthread.php?p=1091901#post1091901

    For example, I was just playing with [itex]r=38[/itex], [itex]v=1.0[/itex], which results in a border angle (with respect to horizontal) of

    [tex]
    \cos^{-1} \left( \sqrt{3} \frac{45}{38} \sqrt{1 - \frac{30}{38}} \right) = 19.760326661 \mathrm{ degrees}.
    [/tex]
     
  9. Jul 27, 2008 #8
    George Jones: I appreciate your help and suggestions, however my original thought was not very well composed, so I will try again, I have noticed that the velocity of propagation of electromagnetic energy has been postulated to remain constant in vacuum or free space, per Mr. Einstein, to all observers, but only the relativistic energy of the electromagnetic wave depends on the relativistic dynamics of the frame of reference of the observer, and can in theory range from zero to infinity (without any limiting laws ) and not the velocity of the electromagnetic wave. I use this concept to explain to myself the reason the photon can not escape the black hole and allows the gravitational boson to have the ability to mediate its interaction with space-time outside the black hole, from within the black hole. I would like to know if there is, a better explaination in the world of physics? Hope you can help.
     
  10. Jul 27, 2008 #9
    George Jones : One of the most basic questions about the Schwarzschild radius on my mind is, was the premise of Mr. Schwarzschild equation based on the escape velocity of the interaction of photon boson, or on the premise of the relativistic energy content of the photon boson reaching null geodesics.
     
  11. Jul 27, 2008 #10

    George Jones

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    Black holes can have electrical charge as well as mass. Consequently, a black hole can interact electromagnetically (attraction or repulsion) with a charged particle outside the black hole's event horizon. Read D.08 and D.08 of the FAQ

    http://sciastro.astronomy.net/sci.astro.4.FAQ
     
    Last edited: Jul 27, 2008
  12. Jul 27, 2008 #11
    George Jones : Thank you for your reference, this problem first came to me while I was in college in Texas, I managed to speak to several Professors, each of which indicated another that was more knowledgeable than the previous, until I was introduced to a Dr. Remler from the University of Texas dept. of Relativity, whom at the time was presenting a theory of his pertaining to the visible edge of the Universe, he refered to his concept as Horizons. He told me, I had a great imaginative mind. He liked some of my concept.
     
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