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Light wavelenths longer than diameter of black hole?

  1. Aug 16, 2013 #1
    I was under the impression that one of the ways of representing a black body (for explaining the ultraviolet catastrophe) is as a 'resonant box'. Frequencies lower than the dimensions of the box cannot be contained within and the black body is thus 'transparent' to those frequencies of light. (Not unlike how you don't get ocean size swell in a coffee cup - the container simply isn't big enough to allow such low frequencies).

    I was thinking also how black holes have been compared to a black body for the purposes of hawking radiation. Also given the previous discussion on red/blueshift near black holes, it started me thinking about how very high frequencies of light interact with matter strongly, while very low frequencies of light interacts more weakly. In the case of light that is falling into a black hole, it is manifested by the more energetic light having a greater relativistic mass and is thus 'heavier' and is bent more by the gravity of the black hole.

    Given that there is no theoretical lower limit to light frequency, I assume it is possible to have a photon who's wavelength is the size of the observable universe, though I doubt that it would interact with anything much.

    So the question;

    Would a black hole be able to absorb a photon who's wavelength is longer than the diameter of the event horizon? If not, how about if we could generate light of such a long wavelength from inside the black hole, say if we threw a long-wavelength-light-generator into the black hole. Could the light therefore escape from inside the black hole if the wavelength of that light was greater than the diameter of the black hole it was in?

    Cheers
    Markus
     
  2. jcsd
  3. Aug 16, 2013 #2

    mfb

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    All light travels in the same way (as long as it is reasonable to define such a path) close to a black hole, independent of the frequency.
    Not necessarily. Visible light can be blocked by a thick sheet of paper, gamma rays cannot.

    With a small probability, but it is possible.

    No. In addition, your device does not have enough time to produce light with (only) such long wavelengths.
     
  4. Aug 16, 2013 #3

    Mentz114

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    Good question. I don't know the answer but I do know that geometrical optics formulated in GR defines 'rays' as null vectors that are perpendicular to surfaces of constant phase. The distance between these surfaces sounds like half a wavelength. This is an approximation and depends on the wavelength. But maybe geometrical optics is not the right tool in this case.

    [Edit:I think mfb has analysed this correctly]
     
  5. Aug 16, 2013 #4

    Bill_K

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    Note that for Hawking radation, the wavelength IS comparable to the size of the hole. This is why astrophysical black holes radiate very weakly. The radiation only becomes appreciable when the hole's mass and Schwarzschild radius are very small.

    Remember the Principle of Equivalence. The trajectory is a null geodesic and does not depend on the frequency of the light ray.
     
  6. Aug 16, 2013 #5

    PAllen

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    In ridiculously extreme cases, this is not strictly true. Consider that anti-parallel light beams attract, while parallel ones do not; and the attraction depends on the energy density of the beams. If you have a 'photon' so energetic in the BH frame that in affects the center of mass, then the closing speed will be greater than for an ordinary photon, and there will also be GW produced.
     
  7. Aug 16, 2013 #6

    Bill_K

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    Not to quibble, but I don't think this can happen, even as an extreme example. The minimum mass of a classical black hole is the Planck mass, whereas the maximum possible energy of a photon is the Planck energy. So unless you have a theory that goes beyond conventional GR, they cannot be comparable.
     
  8. Aug 16, 2013 #7

    PAllen

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    I was not aware of a maximum energy of a photon. What happens if, given a Planck energy photon, you approach it so it is blue shifted? Due to frame dependence of KE, I always assumed there cannot be any upper limit on photon energy. If this is incorrect, I would be very grateful for further explanation or sources on how this works.
     
  9. Aug 16, 2013 #8

    Bill_K

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    PAllen, If I had a theory of what happens at the Planck scale and beyond, I would publish it and become famous. As I'm sure you are aware, our current theories cease to be valid at that scale.
     
  10. Aug 16, 2013 #9
    Is this not generally true? I thought that the gamut of interactions at different energies ran from extremely high energy gamma rays that can break apart atoms, and even form matter from energy if two extremely energetic photons intereact, through longer wavelengths that cause the photoelectric effect, then even longer wavelengths that cant even dislodge electrons, then x-rays that travel some way through 'solid' materials. I figured eventually it must be possible to have light of long enough wavelengths that they would require an impracticably large detector to see because of incredibly weak interactions with matter. Of course, I'm presuming quite a lot here.

    Is this because of the uncertainty principle? Part of the probability wave lies inside the event horizon and part outside? Does this make the Event horizon 'fuzzy' and dependant on frequency of the light entering?

    You mean before it hits the singularity? Guess that makes sense. If there is a finite time before you hit the Singularity you would have to be travelling slower than C to have enough time to create a wavelength big enough. Even a long spiralling descent into the singularity would take the same amount of time as a direct path because time essentially stops at C. Is this more or less what you were getting at?

    Thanks
    Markus
     
  11. Aug 16, 2013 #10
    I thought the mechanism (when simplified) essentially came down to the capturing of virtual particles produced at the boundary? I know it's more complicated than that, but every explanation I have ever read on the subject seems to say that it's a highly mathematical concept. The closest real world approximation is the virtual particles one.

    So are you saying the radiation of hawking radiation can also be seen as analogous to radiating wavelengths that are longer than the black hole can contain?

    Not sure I know enough to understand you. Still struggling to get an intuitive understanding of the diffferent geodesics on Lorenzian manifolds. I get that matter follows timelike geodesics, whereas massless particles follow null geodesics; is this essentially another way of saying matter must travel through time, whereas for a massless particle time is irrelevant (from the point of view of the particle) due to the fact it is travelling at C? And by the equivalence principle which equates inertial and gravitational mass, are you meaning the energy of the photon is the same as inertial mass?

    So in a nutshell, an observer near a black hole would not expect light from stars to 'smear out' like in a prism, but rather for the light to remain coherent, but but lengthen or shorten as appropriate. So the physics for gravitational lensing are different than for a physical lens, where differing wavelengths are bent to greater or lesser degree based on frequency?

    Thanks
    Markus
     
  12. Aug 17, 2013 #11

    mfb

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    ??

    In order of decreasing energy:
    Gamma rays - can penetrate even thick sheets of metal
    X-rays (note their position) - can penetrate thin sheets of metal (and the whole human body)
    UV - gets absorbed quickly by most materials, including window glass, clothes and so on.
    Visible light - can be absorbed by thin sheets of many materials. Metal sheets reflect them if they have a size at least of the order of the wavelength (this is true for all following entries as well).
    Infrared - can be absorbed by thin sheets of many materials, including window glass. With longer wavelengths, it can go through the human skin.
    Terahertz radiation - very material dependent
    Microwave radiation - very frequency and material dependent. Microwave ovens are an example, they heat water but other materials can stay cold.
    Radio waves - they are reflected by some materials and can pass through other materials, and their wavelength is so long that they can go around most obstacles

    There is no clear trend of "more absorption" in one direction.

    No, and you can understand this with a fully classical picture - a part of the wave "hits" the black hole, other parts miss it.

    It does not matter how the object moves inside, it has a limited time until it hits the singularity. As far as I remember, acceleration (independent of the direction) just reduces the time it has left.

    How is that in conflict with Bill_K's post?
    (It is not)

    Right. A black hole has no dispersion.
     
  13. Aug 17, 2013 #12
    Thanks, MFB

    Right. He's saying that hawking radiation is proportional to the size of the black hole. Im asking if the radiations of wavelengths longer than the size of the black hole is another way of visualising what is a highly abstract mathematical concept.

    But wait... if true, doesn't that mean that some light can escape the Event Horizon if of a sufficiently low frequency? I thought we established that the event horizon was not frequency dependant? Now I'm confused....

    Markus
     
  14. Aug 17, 2013 #13

    Bill_K

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    Hawking radiation doesn't come from inside the hole. Not from the surface of the hole either. Must remember that it's a quantum effect, so we shouldn't try to pinpoint exactly where it does come from. It's a global phenomenon, and all we can say is that it comes primarily from the vicinity of the hole.

    Should also point out that Hawking radiation is not all the same frequency. It's black body radiation with a continuous frequency spectrum. The maximum is where the wavelength is about the same size as the hole.
     
  15. Aug 17, 2013 #14

    ZapperZ

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    The problem here is that you also can't say that what you stated is also "generally true". There Just because a photo has a higher energy/shorter wavelength, it doesn't mean it will interact more strongly than something that has a lower energy/shorter wavelength. The fact that x-ray and gamma rays can travel through matter more easily than lower-energy photons clearly shows experimentally that the shorter wavelength interacts LESS, in these cases, than the lower wavelengths.

    What this forces you to do is not to make one, general blanket statement. Rather it forces you to pay attention to the situation, and evaluate the physics that are relevant for that situation. This means that you have to pay attention to not only the EM radiation, but also to the nature of the material involved. There are two actors in this play, and they both have to be centerstage at the same time, not just one. You not only have to know about the EM radiation, but you also have to know the properties of the material it is interacting with! This is not trivial, and this is why we study the properties of materials in condensed matter.

    Zz.
     
  16. Aug 18, 2013 #15
    If you compare a wavelength to the BH diameter, don't you need to measure that diameter along the "line of sight" of the photons' path, not as subtended arc measured from off at a known distance?

    Is there a wavelength so long it exceeds the distance from the EH on one side of a BH to the EH on the other side?
    If you use rods and clocks to measure that distance, what do you get?

    Seems like the distance between two points outside the EH on opposite sides of a BH, both lying in a line that is a principle axis of the BH, must be infinite. Maybe not so far for points off that line?
     
  17. Aug 18, 2013 #16

    Bill_K

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    You can't think of the Hawking radiation as particles emitted from the surface of the hole, traveling along geodesics, getting red-shifted, etc. Details of the Schwarzschild geometry at partcular points do not matter.

    Think instead in terms of a wavefunction. Hawking radiation is observed at infinity, and that's where the wavelength is measured.
     
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