Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parapositronium question and ''does this seem reasonable?''

  1. Sep 18, 2012 #1
    In work I have been studying, I came to the conclusion that the gravitational charge [tex]\sqrt{G}M[/tex] is in fact due to the inertial/gravitational energy within some radius of a particle treated as a sphere (doesn't need to be a classical sphere, just a sphere of smaller magnitude.) My conclusions where


    I contended that the relationship

    [tex]E_g r_s = GM^2[/tex]

    was a type of definition of what gravitational mass was. Whilst [tex]E_g[/tex] still had units of energy, I defined it specially as a ''gravitational energy.'' It was a new name I derived for a new type of concept which attempted to bridge the idea of what inertial mass is and what Motz had in mind when he considered inertial mass as a gravitational charge.

    Since in my work, I defined the gravitational charge in a spin relationship given as

    [tex]\sqrt{G} \frac{e}{\mu} \frac{\hbar}{2} \vec{\sigma} = \sqrt{G}M[/tex]

    we will now concentrate on the left hand side in a completely new way, but by assuming that the quantity [tex]E_g r_s[/tex] is not entirely complete, instead, should be seen as an integral over the energy in respect of the Schwarzschild radius [tex]r_s[/tex] and by setting it equal to the left hand side expression we just mentioned. This then takes the form of

    [tex]\int E_g\ dr_s = \sqrt{G} \frac{e}{\mu} \frac{\hbar}{2} \vec{\sigma}[/tex]

    Before, I explained

    [tex]\lim_{r_s \rightarrow 0} E_g r_s = GM^2 = 0[/tex]

    How if the radius goes to zero, we cannot speak about a gravitational intrinsic mass. Now what is interesting from the approach made in the equation above is that if

    [tex]\int_{r_s}^{0} E_g\ dr_s = \sqrt{G} \frac{e}{\mu} \frac{\hbar}{2} \vec{\sigma}[/tex]

    then this directly effects the right hand side by saying that

    [tex]r_s \rightarrow 0[/tex]

    (which we established) then so does the elementary charge

    [tex]e \rightarrow 0[/tex]

    via this equation

    [tex]e = \sqrt{4\pi \epsilon GM^2}[/tex]

    but if charge goes to zero, the consequence of this equation is that it cannot either describe [tex]\frac{\hbar}{2} \vec{\sigma}[/tex] since the coefficient of the angular momentum is the charge and if the charge is zero, then so is the spin which strictly describes spin 1/2 particles. All these quantities which are related to the gravitational charge implies there can be no inertial gravitational charge, or inertial mass, whatever one wishes to call mass. The implications can be seen in a more productive sense, because if the radius goes to zero, then we cannot talk about Planck Particles, or any type of particles described by spheres because the bare mass of such a particle would become infinite. There for, only massless radiation can be described as being pointlike systems, because they do not contain a bare mass. Renormalization procedures are often adopted to solve this incongruity, by turning the bare mass of a particle negative. However, if we treat particles with mass as a special condition synonymous with the Schwarzschild radius, then we can say it does not have an infinite bare mass without resorting to altering the bare mass as a negative quantity. Photons do not have such a radius and therefore should not be subject to infinite energies.

    The justification for such a move, is that of course, massless systems have no charge and they do not have Schwarzschild radius in which any mass can be contained within. Therefore if there was no radius, there cannot be a gravitational charge - also it directly effects the spin, which would need to be compensation for, let's say a spin 1 boson.

    source http://www.physicsgre.com/viewtopic.php?f=11&t=4765#p42184
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted