The Nature Of Gravity And Mass Through Investigation Of Planck-Like Particles

[Aethelwulf]

The Nature of Gravity and Mass through Investigation of Planck-like Particles

 

One begins with the quantization condition

 

[math]\hbar c = GM^2[/math] 1]

 

which is a famous relationship. Divide through by [math]Mc^2[/math] gives

 

[math]\frac{\hbar}{Mc} = \frac{GM^2}{Mc^2} = (\frac{GM}{c^2} = r_s)[/math] 2]

 

One might want to put a factor of 2 in the numerator where the Scwarzschild radius is it's compton wavelength

 

[math]\frac{\hbar}{Mc} = \frac{2GM^2}{Mc^2}[/math]

 

The left is recognized as the Compton wavelength and the right is recognized as the Schwarzschild radius. I derive a new kind of equation from a differentt equation provided by Motz (1)

 

[math]8\pi \rho (\frac{G}{c^2}) = \frac{2GM^2}{Mc^2}[/math] 3]

 

A Planck Particle is of course, a particle whose Schwartzschild radius is equal to it's Compton Wavelength. Keeping this is mind, we will follow a new derivation.

 

We begin with the equation from quantum theory

 

[math]\hbar = RMc[/math] 4]

 

The quantization condition from 1] then can be rearranged and the angular momentum component can be re-written as

 

[math]\frac{GM^2}{c} = RMc[/math] 5]

 

Multiply by c on both sides, divide by M on both sides then divide off G on both sides gives

 

[math]\frac{Rc^2}{G} = M[/math] 6]

 

M is usually taken to be the Planck Mass from the quantization condition. Because of this, we can set this relationship above directly to the Planck Mass

 

[math]\frac{Rc^2}{G} = \sqrt{\frac{\pi \hbar c}{G}}[/math] 7]

 

Actually, this is not quite the Planck Mass as it has a value of [math]\sqrt{\pi}[/math] larger. However, this may be the true value of the Planck Mass since the Planck Mass is not exactly an equation, it is usually a proportionality.

 

Square everything in equation 7 and remove the square root and rearrange to get

 

[math]\pi G\hbar = R^2c^3[/math]

 

Solve for R gives

 

[math]R = \sqrt{\frac{\pi G \hbar}{c^3}}[/math]

 

which makes the radius the Compton wavelength. In a sense, one can think of the Compton wavelength then as the ''size of a particle,'' but only loosely speaking.

 

 

(1) http://www.gravityresearchfoundation.org/pdf/awarded/1971/motz.pdf

Edited July 31, 2012 by Aethelwulf

[Aethelwulf]

A Planck Particle is considered not only a particle in it's own right, but also a miniature black hole of types. It's mass is equal to the Planck Mass and it's length is about the Planck Length. To date, we cannot probe such lengths accurately enough to know what is happening at this extremely small level. However, there is a minimal uncertainty which can be extracted from the Planck Length. It is doubtful they exist for any long periods of time - they are likely to give up their energy in the form of Unruh-Hawking Radiation.

 

It is generally considered important that these particle do not exist below the threshold of the Planck Lengths - their energy would of course become exceedingly large under the Planck Scales, not only this, but they may also stable under Extremal Black Hole theories, which is not very well-accepted.

 

The hotter a Planck Particle is, the quicker it will give up its energy in the form of radiation. This is predicted by it's temperature:

 

[math]T \propto \frac{hc^3}{4\pi kGM}[/math]

 

The time in which it would give this energy up would be proportional to the Planck Time

 

[math]t \propto \frac{\hbar G}{c^5}[/math]

 

So experimentally-speaking, these objects exist for the shortest time which is possible - that's a tremendous discharge of energy in such a small period of time since it's mass is very large (Planck Mass). There is of course, an uncertainty principle between energy and time given by

 

[math]\Delta E \Delta t \propto \hbar[/math]

 

As Motz has shown, the uncertainty principle leads to the quantization condition

 

[math]RMc \leq \hbar[/math]

 

The smallest uncertainty in the length is given by

 

[math]\frac{\hbar G}{c^3}M^2c^2 \leq \hbar^2[/math]

 

which of course leads to

 

[math]GM^2 = \hbar c[/math]

 

Our quantization condition.

 

Whilst a Planck Particle has never been observed, it could be a source of the radiation present in the vacuum when the universe was very young. These would be a kind of primordial black hole bath of particles.

Edited August 3, 2012 by Aethelwulf

[Aethelwulf]

Why did my computer stop a virus coming through when I posted this? Secondly, why have most post doubled up...? Thirdly, why can't I edit the page?

[Aethelwulf]

Also, I need to make corrections. What I have written in the post is not the Planck Time. I need to write it for the Planck Time, not what I have written there.

[Aethelwulf]

However, just in case anyone is wondering, that time is correct for the evaporation time, just not for something with a Planck Mass.

[Aethelwulf]

Finally. Fixed.

[Aethelwulf]

Now what is wrong with it... my equations are missing!

 

and now... they are back. ??

Edited August 3, 2012 by Aethelwulf

[Aethelwulf]

Right...now, I might be a few factors out of my equation

 

[math]8\pi \rho (\frac{G}{c^2}) = \frac{2GM^2}{Mc^2}[/math]

 

The reason why is because Motz calculated this for the Gaussian Curvature which would have a value of [math]6(\frac{\hbar}{mc})^{-2}[/math]. This means I need to treat the right hand side of my equation in the same manner since

 

[math]\frac{\hbar}{mc} = \frac{GM}{c^2}[/math]

 

So that perhaps the correct form of the equation is

 

[math]8\pi \rho (\frac{G}{c^2}) = 6(\frac{2GM^2}{Mc^2})^{-2} = (\frac{12GM}{c^2})^{-2}[/math]

Edited August 3, 2012 by Aethelwulf

[Aethelwulf]

The Gaussian curvature of a 2D surface is the product of two principle curvatures. Each principle curvature is equal to [math]1/R[/math], the Gaussian curvature has dimensions [math]1/(\ell^2)[/math] and the three dimensional case of a hypersphere is [math]K/6 = R^{-2}[/math] which is how these extra factors come up where [math]R[/math] is the radius of curvature.

Edited August 3, 2012 by Aethelwulf

[Aethelwulf]

One can also come to realize why [math]\sqrt{G}M[/math] should be the intrinsic gravitational charge of a system by looking at the Gaussian charge relationship, that being

 

[math]e = \sqrt{\alpha \hbar c}[/math]

 

Naturally, one might recognize the right handside's [math]\sqrt{\hbar c}[/math], this is normally set equal to [math]\sqrt{G}M[/math] for the quantization condition. In a separate work, I evaluated that

 

[math]GM^2 = E_gr_s[/math]

 

was the squared gravitational charge, by equating it with the gravitational energy times the Schwarzschild radius of a system. The gravitational energy is the contribution of mass due to energy, or it's gravitational energy is it's rest mass. This is an interesting thing to consider for a particle, because if the Schwarzschild radius is made to go to zero, then the inertial gravitational energy is zero also

 

[math]\sqrt{G}M = \lim_{r_s \rightarrow 0} \sqrt{E_gr_s} = 0[/math]

 

This is not meant to imply however that the entire energy of our system is zero, only that the presence of inertial energy is absent.

 

Taking this relationship into consideration, one may even see a charge relationship of the form

 

[math]e = \sqrt{\alpha E_g r_s}[/math] (*)

 

Why this might be interesting is because there is no such thing as a charged massless system. The Yang Mills Equations once predicted massless charge bosons but the approach has been generally considered today as a false one. Is this failure important? Is there an intrinsic relationship between the elementary charge [math]e[/math] and what we might call the inertial mass [math]M_i[/math]?

 

The above equation, given by the asterisk *, is to be taken to mean that the elementary charge is something associated to the gravitational energy - if we are dealing with classical sphere's, the idea is that the gravitational energy would take on this value inside the sphere of radius [math]\frac{e^2}{Mc^2}[/math]. Of course, the charge itself is also distributed over such a radius.

 

Indeed, if the Scwartzschild radius goes to zero again, then so does the gravitational energy and thus the elementary charge must also vanish [math]e = 0[/math]. There are such cases in nature where this might be an important phase transition. Gamma-gamma interactions can either give up mass in the form of special types of decay processes or, vice versa, the energy can come from antiparticle interactions. Such a phase transition is given as

 

[math]\gamma \gamma \Leftrightarrow e^{-}e^{+}[/math]

 

If we take the notion seriously that charge is a property of systems with mass, and that massless charged bosons don't exist in nature then we can assume that in this specific phase transition charge appears and disappears. If one does not know the process of how matter can be somehow a type of trapped form of light energy, then one would almost assume that it would be by magic. There must be a dynamical and fundamental explanation to how this can happen, indeed, Glaswegian authors J. Williamson and M. Van der Mark have already questioned whether the electron is really a photon caught up in a type of toroidal path http://www.cybsoc.org/electron.pdf.

 

(By the way, clearly one of the authors are foreign to Glasgow, but a Glaswegian is a ''resident'' of Glasgow)

Edited August 16, 2012 by Aethelwulf

[maddog]

Van der Mark have already questioned whether the electron is really a photon caught up in a type of toroidal path http://www.cybsoc.org/electron.pdf.

I was not able to navigate to this link. Would you update it please. Thank You.

 

maddog

[Aethelwulf]

Sure, but I checked the address of the page and it seemed fine.. you can google it easy:

 

 

[PDF]

Is the electron a photon with toroidal topology?

www.cybsoc.org/electron.pdf

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Published in: Annales de la Fondation Louis de Broglie, Volume 22, no.2, 133 (1997). Is the electron a photon with toroidal topology? J.G. Williamson(a) and ...

[Pincho Paxton]

The PDF doesn't work, it has a bad font code.

[Aethelwulf]

It works for me?

[Aethelwulf]

As I said however, you can search for the paper yourself, it is easy to find :)

[Aethelwulf]

Anyway, back to the equation

 

[math]GM^2 = E_g r_s[/math] 1.

 

... You can get the gravitational energy (inertial mass) by stating

 

[math]\frac{GM^2}{r_s} = E_g[/math] 2.

 

Now, there is something to understand by this relationship. There is a hidden relationship on the left, the [math]\frac{GM}{r_s}[/math] part can in fact be understood as [math]\phi[/math] which can be understood as the gravitational gradient. This means, we have a coupling in equation 2. in the form of [math]M\phi[/math]. The mass, is ''coupled'' to the potential [math]\phi[/math]. This means equation 2. is really just

 

[math]\frac{GM^2}{r_s} = M\phi[/math]

 

I wonder sometimes, if this coupling is a hint that mass comes about from some kind of potential. In fact, the prototypical case of color charges in quark confinement get's a mass from a potential energy. Is the same for mass in general? Does mass come about from some kind of potential?

Edited August 16, 2012 by Aethelwulf

[Aethelwulf]

Just a quick example, that the gravitational energy can be the inertial mass using natural units stating very simply that

 

[math]E_g = M_ic^2[/math]

 

setting [math]c=1[/math] then

 

[math]E_g = M_i[/math]

 

meaning the gravitational energy, measured in Planck units of energy, equals the inertial mass of a particle, measured in Planck units of mass