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Quantum gravity with constants

  1. May 2, 2004 #1
    I would like to hear comments from anyone concerning the below. Honest and critical feedback from qualified individuals is most welcome. Thanks for your reviews!

    This simple calculation bases the quantum effects of gravity on the electromagnetic frequency across ½ the Compton wavelength of the electron.

    2.998 x 10^8 ms-1
    ½ (2.4263 x 10^-12 m) = 2.471 x 10^20 s-1

    This is multiplied by Planck’s constant to arrive at energy in joules.

    (2.471 x 10^20 s-1) (6.626 x 10^-34 kgm2s-1) = 16.373 x 10^-14 kgm2s-2

    This value is multiplied by the square of the radius of a planet like Earth and the product divided into Planck mass.

    2.1767 x 10^-8 kg
    (16.373 x 10^-14 kgm2s-2) (6.378 x 10^6 m)2 = .003268 x 10^-6 m-4s2

    In the meantime, the number of Planck lengths along a geometric line across the Schwartzchild radius of the planet is determined. First, the Schwartzchild radius:

    2(6.673 x 10^-11 m3kg-1s-2) (5.974 x 10^24 kg)
    SRearth = (2.998 x 10^8 ms-1)2 = 8.87 x 10^-3 m

    The Schwartzchild radius is divided by Planck length to get the total number of lengths along the planet radius.

    8.87 x 10^-3 m
    1.616 x 10^-35 m = 5.489 x 10^32

    This value is energized in joules by multiplying by the energy equivalent of the electron mass.

    (5.489 x 10^32) (8.187 x 10^-14 kgm2s-2) = 44.938 x 10^18 kgm2s-2

    This value is multiplied by that arrived at above in the first part.

    (.003268 x 10^-6 m-4s2) (44.938 x 10^18 kgm2s-2) = 1.469 x 10^11 kgm-2

    This value multiplied by G predicts the standard acceleration of gravity at 6,378,000 radial meters and a mass of 5.974 x 10^24 kg.

    (6.673 x 10^-11 m3kg-1s-2) (1.469 x 10^11 kgm-2) = 9.8 ms-2

    As an additional for instance, it accurately predicts g on Venus to be
    8.8 ms-2 at a radius of 6,052,000 meters and mass of 4.869 x 10^24 kg. Actually, the data of any planet plugged into the formula above yields answers at 100% the accuracy rate of the classical g = Gm/r2.

    (this is copyrighted material) © March 2004
  2. jcsd
  3. May 3, 2004 #2
    wow, someone might steal it!
  4. May 3, 2004 #3


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    Perhaps if you wrote out the calculation, using different symbols for things like the speed of light, the mass of the electron, etc and then did some 'cancellation of terms', you might get some insight into why the numbers come out as they do.
  5. May 4, 2004 #4
    Thank you for the response! Yes, this has been done and the results are lackluster, at least on the surface. Such an exercise produces classical acceleration times 1 or g = Gm/r2 x 1.
    Does this negate any significance to you?
    Is a multiplication by 1 to be expected?
    In abstract terms, could/should quantum gravity equal or hold a value of 1 or very nearly 1?
    If Newtonian gravity “works” in the universe, will quantum gravity, when it is discovered, turn out to be classical Newtonian times the mathematical identifier “1” ?
    (Your answers are very important to me and will be carefully considered. Thanks for any additional input.)
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