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The temperature of acceleration (Unruh radiation)

  1. May 5, 2003 #1

    marcus

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    In 1976 Bill Unruh at British Columbia realized that
    acceleration has an intrinsic temperature---an accelerating observer will be bathed in radiation at that temperature.
    He published his finding in Physics Reviews Series D vol 14 page 870 and following.

    the Unruh temperature has been experimentally investigated at SLAC, and possibly elsewhere

    here is a SLAC press release about the work of P. Chen
    on Unruh radiation

    http://www.slac.stanford.edu/slac/media-info/20000605/chen.html

    The formula for the temperature associated with acceleration a is

    kT = hbar a/2πc

    Here is an example in planck units c=G=hbar=k=1

    Suppose the acceleration is 18E-30
    then to get the temperature we need to divide by 2π
    That gives about 3E-30---a bit hotter than the familiar boiling point of water.

    So if one accelerates a teakettle of water at 18 x 10^-30 planck, it will soon boil and one can brew tea with it.

    All that is necessary to find the temp is to divide by 2pi.

    this partily in response to dav2008, who in another thread mentioned the planck unit of acceleration

    Unruh's formula is similar to Stephen Hawkings formula for black hole temperature which depends in the same way on the surface gravity (acceleration at the event horizon).
     
  2. jcsd
  3. May 5, 2003 #2
    How do you accelerate a teakettle at such tremendous acceleration 10E-30?
     
  4. May 5, 2003 #3

    marcus

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    A thoughtful question suggesting that you like tea.
    It is a practical problem that would definitely have to be solved
    before one could brew tea by this method.
    18E-30 is almost exactly E22 gee----more precisely 1.02E22 times the earth sealevel gravity norm.

    Do you happen to know what size of black hole would
    have a Hawking radiation of the right temperature to
    boil water? I suppose it would be extremely small!

    Does anyone have a suggestion as to how to solve
    the practical problem posed by Alexander about the
    acceleration of teapots?
     
  5. May 5, 2003 #4

    marcus

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    the gravity feels like home

    Just thought of a related question. What size ordinary black hole has a familiar earth-like gravity at its event horizon?


    This is easiest to solve in planck c=G=hbar=1 terms. And then, to interpret the answer one should know that E38 planck lengths make a mile. And that one solar mass is about E38 planck mass.

    (more exactly the solar mass is 0.93E38 planck but E38 is close enough)

    Earth sealevel gravity is around 18E-52 planck----that's all you have to know to work the problem in planck.


    There is this useful universal constant force F and the surface gravity of a black hole is F/4m, where m is the mass of the hole.
    You know dividing a force by a mass gives an acceleration, so I'm really just telling you there is a 4 in the formula.

    So we just have to solve this for m

    F/4m = 18E-52

    F is the unit force so it is just equal to one

    1/m = 72E-52

    That means that m = 1.4E50 planck

    And its Schw radius is 2.8E50 planck.

    So we can say that in miles, radius is 2.8E12 miles
    or 2.8 trillion miles

    And in "solar" masses of E38 planck the hole is 1.4 trillion solar masses.

    It is not huge, just a thousand times larger than the solar system, but it is really massive----like a whole milkyway galaxy.

    The nice thing about this is the simplicity of the arithmetic, just had to multiply by 4 and find one over something. The hard thing (which is not all that hard) is needing to know stuff like E38 is a mile and roughly a solar mass.
     
  6. May 5, 2003 #5
    So, how to boil a teakettle (by acceleration)?
     
  7. May 5, 2003 #6

    marcus

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    tie a long rope to the handle of the teakettle
    and lower it down near the surface of a black hole
    which has the right acceleration of gravity at
    its event horizon


    it will probably have to be a very tiny teakettle
    to match the small size of the black hole

    or, if one only has a normal teakettle one must
    gather together many black holes to make
    an array of them the same size as the bottom
    of the teakettle, like a stove burner.

    I was thinking you might answer the question.
    Please you tell me. What mass of black hole
    has Hawking temperature to boil water?
    That is, around 3E-30 planck temperature (if
    you use planck units) or roughly 400 kelvin (if you use
    kelvin).
     
  8. May 5, 2003 #7
    Size? About a micron (2R~1.1x10-6m), or 0.4 Yg in mass (which is about mass of 0.5 Mm asteroid).
     
    Last edited by a moderator: May 6, 2003
  9. May 5, 2003 #8

    marcus

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    bravo

    I got (1/24π)E30 planck mass but did not
    work it out in microns, cant check this at the moment

    I must go to an appointment so must continue
    this very pleasant discussion later

    Now I'm back. That result for the mass means the
    Schw length (halfradius) is (1/24π)E30 planck length

    So in metric terms, Schw radius is 0.43 micron, and diameter (for comparison with what you calculated) is 0.86 micron.
    You got diameter about 1.1 micron which is in essential agreement. The difference is probably due to the fact that
    I used a temperature slightly hotter than conventional boiling because I wanted a round number input that would get the kettle boiling quickly.

    I used a temperature of 3E-30, while conventional boiling in planck terms is 2.63E-30. You probably used conventional boiling temperature 373 kelvin. So there would be a factor of 2.63/3 between the answers having to do with the hotter temperature that I assumed. Yeah, if I use boiling I get a diameter of right close to one micron.
     
    Last edited: May 5, 2003
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