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Z-bosons as Hawking radiation from BH

  1. Aug 6, 2015 #1
    Please check my logic.

    1. Hawking mechanism should give birth not only to photons, but also to their heavier analogs, Z for example.
    2. Contrary to photons, massive Z bosons are not gradually red shifted, low energy Z simply fall back to BH, so the "red" part of the black-body spectrum of Z is cut. It is also distorted because Z has non-zero invariant mass.
    3. Finally Z decay to photons, but the spectrum of these photons carry the "birth defect" - missing "red" part of the spectrum.
    4. As a result, total Hawking radiation from BH has NOT a black-body spectrum because of massive particles.

    I understand that effect is tiny, but is my logic flawed?
  2. jcsd
  3. Aug 6, 2015 #2


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    Staff: Mentor

    Distorted as in "non-existent" - unless the black hole is so small that its temperature gets comparable to 90 GeV. Such a small black hole will emit all particles lighter than the Z as well.
    That decay is very unlikely, it hasn't been observed so far. It would need at least three photons and a loop. A very old paper (before the top discovery) predicts a partial decay width of only 1.35 eV, or roughly 1 decay in a billion.
    It has a black-body spectrum by definition of that spectrum. At very high temperatures massive particles become part of that spectrum, if you have a possible decay you also have the possible recombination.
  4. Aug 6, 2015 #3
    1 I understand, this is like Scharnhorst effect, the value of effect is irrelevant.
    2 Hm...may be this is the answer...
    So Planck's law of black-body radiation is valid only on the low-energy limit?
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