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Photon that "fits" into its schwarzschild radius

  1. Sep 21, 2014 #1
    Photons with smaller and smaller wave lengths have a higher and higher energy and these engeries have an increasing Schwarzschild radius [itex]r_s[/itex]. Consequently i can ask when half the wave length [itex]\lambda/2[/itex] is equal to [itex]r_s[/itex], such that one wave length fits into the sphere of the Schwarzschild radius.

    I did the calculation and came out with [tex]\lambda/2 = r_s = \sqrt{Gh/c^3} =\sqrt{2\pi}l_p[/tex] where [itex]l_p[/itex] is the Planck length. Incidently the mass of this photon is [itex]\sqrt{2\pi}\,m_p[/itex] with [itex]m_p[/itex] being the Planck mass.

    Now I wonder. Should I be at least a bit surprised about such extremely simple formulas or not. To put another way, is this as trivial as transforming [itex]ab=1[/itex] into [itex]a=1/b[/itex], or is there at least one physical statement needed between the Schwarzschild radius and this specific photon wave length? (Hmm, I hope someone can understand what I mean here. :confused:)
     
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  3. Sep 21, 2014 #2

    Orodruin

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    Photons are massless and their size is not a very well defined concept - and it is definitely not equal to the wavelength. Furthermore your statement is frame dependent as I can find an inertial frame of reference where your photon has a wavelength in the radio wave band.
     
  4. Sep 21, 2014 #3

    Vanadium 50

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    Before getting all wrapped up with photons and therefore quantum mechanics, why do you think the Swarzchild solution (boundary conditions being static and spherical) is an appropriate one for light?
     
    Last edited: Sep 21, 2014
  5. Sep 21, 2014 #4
    Since a photon has no inertial reference frame and is only a single object, it makes no sense to say that an extremely high energy photon is or is not a black hole. Making that classification is meaningless in the absence of some other object with which to interact. Now add an electron to the system.
    It is certainly possible that the photon + electron system will have enough energy in a small enough space to form a black hole. This scenario would have meaningful and observable consequences since neither particle would continue to exist after the intersection.
     
  6. Sep 23, 2014 #5
    Thanks for your answers, except I don't get it where you all are heading. I did not mention the term "black hole", I did not say that the photon has a certain size and I did not pronounce any appropriateness of the Schwarzschild solution for light. All i did was toy with some physical formulas out of curiosity and got an, at least for me, surprisingly simple result. My question is basically whether this is at least mildly surprising or completely trivial.
     
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