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Photon energy and black holes

  1. Sep 19, 2012 #1
    First, a question. I assume that a photon can be captured into a stable orbit around a black hole. It seems it would just be a function of the radius of the event horizon and the velocity of the photon (always c I know but the vector is the important part). Where would this stable orbit be in regard to the event horizon? Would it be some small distance away from the event horizon, or at the very limit?

    If my first assumption is correct, that would indicate that there is a "shell" of photons in stable orbit around any black hole. Now, say there was some matter falling into the black hole. As it approached the event horizon, it would collide with and scatter these orbital photons. Would that account for some of the light emitted from a black hole as it consumes matter? It seems that for any black hole of significant age, there would be an unfathomable amount of energy locked in orbit around the black hole in the form of gamma and x-rays, not to mention visible light and radio waves. When that collided with infalling material, it seems like it would make a great show :)
  2. jcsd
  3. Sep 19, 2012 #2
  4. Sep 19, 2012 #3
    Hey, would you look at that :) I did try to answer my question first but didn't come across that article.

    So, that answers the first question but what about the second in regards to infalling matter? Could disruption of the photon shell manifest itself in something as energetic as a GRB?
  5. Sep 19, 2012 #4
    yes can be captured, no not stable. not much emitted as a result.

    I doubt very much since the photon sphere has so little light. Matter falling onto a black hole typically includes an accretion disk and heated by the friction of the infalling matter due to tidal forces, these form some of the brightest objects in the universe.
  6. Sep 19, 2012 #5
    Ok, that answers my question, I wasn't able to figure whether it would be stable or not. Thanks!
  7. Sep 20, 2012 #6
    Did you read the Wikipedia article??
    I had not when I posted above, but it's got some nice
  8. Sep 20, 2012 #7
    Yeah, very cool stuff. It seems odd to me that the location of the photon shell is 3/2 R. Intuitively, I would have guessed that it would orbit at the very limit of the event horizon. It doesn't mention this in the article, when it says "not stable", does that mean that photons in the photon shell can either continue to decay in past the event horizon or escape?

    Take an inactive black hole for instance, there would be relatively few massive particles for the photons to interact with so scattering could be discounted as a factor. It seems to me that if the photon's orbit was outside the event horizon, it wouldn't ever pass the event horizon. I say that just because of the layman's definition of event horizon as the point where not even light can escape the gravitational field.

    Don't laugh at my ignorance, I'm very much an amateur at all this :)
  9. Sep 23, 2012 #8
    It's a good question.

    I am unsure of the answer.
    The article makes it sound like the photons don't escape, but
    I have never read that Hawking radiation gets trapped there....
  10. Sep 23, 2012 #9
    Is Hawking radiation emitted in a range of vectors? What I mean is, does Hawking radiation always radiate out at 90° from the boundary of the event horizon? If that's the case, it seems like Hawking radiation would escape the black hole and even the photon shell by virtue of it's vector.

    The photons I would expect to be captured would be ones that are coming in at exactly the radius of the photon shell but from 180°. Ones coming in at < 180° compared to the tangent of the photon shell would eventually end up inside the event horizon. Ones coming in at 180° but at a distance further out than the photon shell radius would just be subject to gravitational lensing.

    "Tangent of the photon shell" might be an ignorant way to put it but it's the best way for me to explain the picture I have in my head.
  11. Sep 23, 2012 #10
    I have no clue what angle hawking radiation photons are emitted, but because I've never heard that being talking about I'd assume they are radiated at all angles. Meaning that there is a small chance any photon would be emitted into the photon shell.

    That being said, it seems to me the photon shell is an unstable equilibrium, so you would have to be emitted exactly in the right direction to fall into it, which would be near to improbable.
  12. Sep 23, 2012 #11
    I had figured that to have something even approximating a stable photon orbit would only be probable around a large black hole with a large event horizon. The slightest variance in "entry angle" would be magnified each orbit. So with a small event horizon say 10km in radius, the photon would blast around over a million times each minute and would quickly spiral in. But with an event horizon that is 1,000,000km in radius, it would only be something like 15 times a minute (if I did the calculations correctly). It would take more time to spiral in and might even appear to an outside observer to be in stable orbit.
  13. Sep 23, 2012 #12
    And interestingly enough, I'm getting that according to Hawking's research, black holes should radiate like a black body which would mean isotropically. I guess that says that Hawking radiation would be emitted at 90° in all directions? It would originate at a point at center mass of the black hole anyway so would necessarily be 90° right?

    *EDIT - Answered my own question, Hawking radiation comes from virtual pair production in close proximity to the event horizon. So it doesn't originate from center mass of the black hole so doesn't have to be emitted at right angles. Depending on the velocity of the virtual pair, it could be emitted in any vector but still isotropically from a distant observer's point of view. I think that's correct :)
    Last edited: Sep 23, 2012
  14. Oct 2, 2012 #13
    why is it that the photons cannot maintain a stable orbit?
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