Aethelwulf Posted September 2, 2012 Report Posted September 2, 2012 (edited) Hi Guys, I know I haven't been here in a while, but I have finished some work and have decided to post it here. I speculated a lot of this work here before but I have created a more solid foundation for it now which I wish to share. Abstract The aim of this paper is to investigate the quantum possibility of Planck-like particles. To do this, I have attempted to take a different approach by studying the fundamental equations which describe Planck particles and by studying them I will attempt to bring some light on different but related topics, topics like the fine structure constant and speculations on what caused the background temperatures of spacetime. To do this, we will need to study the work of Lloyd Motz, who has wrote a number of papers dedicated to the investigation of Planck parameters for objects and for the study of what is called the gravitational charge [math]\sqrt{G}M[/math]. This quantity will be studied with great depth in this paper. The Concept of Planck Particles Planck particles are very interesting subatomic systems which is generally regarded as a type of miniature black hole. This kind of black hole only exist for very small amounts of time, but the physics behind such an exotic object are interesting to say the least. A Planck particles wavelength is usually set equal or approximated to its Schwarzschild radius. It would be a very small particle indeed, with a very large mass on the scale of the Planck Mass. We can obtain the relationship directly between the wavelength and the Scwarzschild radius directly by inferring first on a very special quantization condition which is given as [math]\hbar c = GM^2[/math] This is a quantization of charge and one may see this because of the Heaviside relationship [math]e = \sqrt{4 \pi \hbar c}[/math] Because [math]\hbar c[/math] is set equal to [math]GM^2[/math] I determined that this relationship is also true [math]e = \sqrt{4 \pi GM^2}[/math] where [math]\sqrt{G}M[/math] is the gravitational charge. We often see the mass as the Planck mass so the reader must keep this in mind. Now, one can derive very easily the relationship for the Planck particles wavelength and it's Schwarzschild radius by dividing the inertial energy by the quantization condition as [math]\frac{\hbar c}{Mc^2} = \frac{GM^2}{Mc^2} = (\frac{\hbar}{Mc} = \frac{2GM}{c^2} = r_s)[/math] where [math]r_s[/math] is the Scwarzschild radius and the factor of 2 comes from the usual convention for it. The Work of Lloyd Motz In Motz' work (1), he set the Guassian curvature equal to the radius by equation [math]8 \pi \rho (\frac{G}{c^2}) = 6(\frac{\hbar}{Mc})^{-2}[/math] My approach is to exchange the right hand side for the gravitational charge, which can be easily done. 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. A Gravitational Energy When considering a gravitational charge [math]\sqrt{G}M[/math] one may consider also a graviational energy which can be set equal to this with an extra term of the Schwarzschild radius [math]\sqrt{E_gr_s} = \sqrt{G}M[/math] I call it a gravitational energy (over an inertial energy) because we are talking about a gravitational charge on our system. However, in all simplistic sense of the words, inertial quantities and gravitational quantities are the same thing. These are ''intrinsic properties'' in this work. A photon may exert gravitational influences by curving spacetime around it, but it contains no instrinsic gravitational charges, nor does it intrinsically relate to any gravitational energies. This new distinction helps solve the problem highlighted in this paper: http://www.tardyon.de/mirror/hooft/hooft.htm In this paper, the authors argue there is a problem concerning what we consider as mass and a system which exerts a gravitational influence on the surrounding vicinity of the spacetime vacuum. I show, that by discerning a difference when talking about intrinsic properties, this problem can be easily avoided. Interestingly, if you make the radius go to zero in this equation [math]\lim_{r_s \rightarrow 0} \sqrt{E_gr_s} = \sqrt{G}M = 0[/math] then the gravitational charge (and thus), the gravitational energy (in the intrinsic sense is zero). Obviously then, we don't mean that an object without a gravitational energy cannot exert gravitational forces which stays strictly within the predictions of relativity. Because of the relationship then: [math]e^2 = 4\pi GM^2 = 4 \pi \hbar c[/math] we can also see that the elementary charge itself is equal to [math]e = \sqrt{4 \pi E_g r_s}[/math] which is an entirely new relationship derived for this paper. It is interpretated as saying, that the gravitational energy covers the length of the radius, which defines the charge of your system. 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 and what we might call the inertial mass ? The above equation, given by the asterisk *, is to be taken to mean that the elementary charge [math]e[/math] is something associated to the gravitational energy [math]E_g[/math] in terms of the gravitational charge [math]\sqrt{G}M[/math] - 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 . 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. 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-particle interactions. This is called the parapositronium decay. Such a phase transition is given as [math]\gamma \gamma \rightarrow e^{-}e^{+}[/math] and [math]e^{-}e^{+} \rightarrow \gamma \gamma[/math] This relationship is best seen in light of my equation [math]\lim_{r_s \rightarrow 0} \sqrt{E_gr_s} = \sqrt{G}M = 0[/math] The radius is concerned with bodies of mass. If the radius shrinks to zero, then what we have is a similar transition phase between systems with mass and massless systems. Therefore, gravitational energy is obtained through the presence of the Scwarzschild radius. The fundamental reasons behind mass, of course, will be much more complicated. This is only an insight into mass in terms of the radius itself and gravitational energy when in relation to the gravitational charge (or inertial mass) of the system. The Origin of the Quantization of Charges Here is a nice excerpt about the charge quantization method adopted by Motz http://encyclopedia2.thefreedictiona...tion+condition What we have is [math]e \mu = \frac{1}{2} n \hbar c[/math] where your constants in the paper have been set to natural units and the angular momentum component comes in multiples of [math]n[/math]. It seems to say that [math]\mu[/math] plays the role of a magnetic charge - this basically squares the charge on the left handside, by a quick analysis of the dimensions previously analysed. Here, referenced by Motz, you can see the magnetic charge is given as [math]g[/math], the electric charge of course, still given by [math]e[/math]. http://wdxy.hubu.edu.cn/ddlx/UpLoadF...0535154942.pdf what we have essentially is [math]\frac{e\mu - e \mu}{4\pi} = n \hbar c[/math] In light of this, one may also see this can be derived from the Heaviside relationship since it has the familiar [math]4 \pi[/math] in it. The Radius We may start with the quantum relationship [math]\hbar = RMc[/math] Knowing the quantized condition [math]\hbar c = GM^2[/math] we may replace [math]\hbar[/math] for [math]\frac{GM^2}{c} = RMc[/math] Rearranging and the solving for the mass gives [math]\frac{Rc^2}{G} = M[/math] we may replace the mass with the definition of the Planck mass in this equation, this gives [math]\frac{Rc}{G} = \sqrt{\frac{\pi \hbar c}{G}}[/math] Actually, this is not the Planck mass exactly, it is about a factor of [math]\pi[/math] greater, however, the Planck mass is usually an approximation so what we have here is possibly the correct value. After rearraging and a little further solving for the radius we end up with [math]R = \sqrt{\frac{\pi G \hbar}{c^2}}[/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. Motz' Uncertainty And so, I feel the need to explain an uncertainty relationship Motz has detailed in paper. But whilst doing so, I also feel the need to explain that such a black hole particle is expected to give up its mass in a form of Unruh-Hawking radiation. The hotter a black hole is, the faster it gives up it's radiation. This is described by the temperature equation for a black hole [math]T \propto \frac{hc^3}{4\pi kGM}[/math] The time in which such a particle would give up the energy proportional to the temperature is given as [math]t \propto \frac{\hbar G}{c^5}[/math] That is very quick indeed, no experimental possibilities today could measure such an action. Uncertainty in both energy and time is given as [math]\Delta E \Delta t \propto \hbar[/math] Motz explains that the smallest uncertainty in [math]RMc \leq \hbar[/math] is [math]\frac{\hbar G}{c^3}M^2c^2 \leq \hbar^2[/math] This leads back to the quantization condition, Motz explains [math]\hbar c = GM^2[/math] Planck Particles and the Background Temperatures 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. To understand intuitively how this happens, these primordial particles would have been borne from the intense gravitationally-dense region we associate to the ''very first instant of time'' at the creation of the big bang. At this point, we are often told in Cosmology that the universe would have been flooded with primordial black holes. In this paper, we believe that the black holes could have been very small and gave up their radiation about a planck time after the big bang appeared. Taking enough of these particles, we can assume the universe was floaded in radiation which was then smoothed out by the inflationary phase. The CODATA charge In the CODATA method of understanding charge, the idea is simply this: treating charge not as a independent quantity, but rather but a relationship of fundamental constants. This is a wise move, since we are often taught that the relationships in nature are not by accident, that there may be some fundamental written set of rules which determine charges for systems. Because of this, we may understand that perhaps the gravitational charge is also strictly governed by similar principles. The charge due to fundamental relationships is given as [math]e = \sqrt{\frac{2\alpha \pi \hbar}{\mu_0 c}}[/math] Therefore, because of the relationship we have already covered: [math]e^2 = 4\pi GM^2 = 4 \pi \hbar c[/math] Gives rise to an understanding of the gravitational charge determined by the same fundamental constants of nature [math]\sqrt{G}M = \sqrt{\frac{\alpha \hbar}{2 \pi \mu_0 c}}[/math] But this is not where this speculation ends. The quantity [math]\sqrt{\alpha \hbar}[/math] is actually quite important. A curious interpretation of [math]e^2 = \alpha \hbar c[/math] is that the angular momentum component is in fact conserved by the fine structure constant by stating [math]\frac{e^2}{c} = \alpha \pm \hbar[/math] this means in my equation we actually have [math]\sqrt{G}M = \sqrt{\frac{\pm \alpha \hbar}{2 \pi \mu_0 c}}[/math] this means the gravitational charge itself can be understood as depending on the conservation of angular momentum as well. This was quite an important discovery, I felt. Two Famous Relationships Barrow and Tipler calculated a type of gravitational charge in their equations, when considering the ratio of an electric charge with the gravitational charge: [math]\frac{\alpha}{\alpha_G} = \frac{e^2}{GM_pM_e} = \frac{e^2}{GM^2}[/math] The fine structure constant can also be given as [math]\alpha_G = \frac{\hbar c}{GM^2} = (t_p \omega)^2[/math] A New Relationship Between the Gravitational Charge and the Bohr Model One can actually derive a graviational charge relationship from the derivation of a Bohr Model. It wasn't completely obvious, but was to me considering the equations we have covered thus far. Using normal convention, you usually set the centrifugal force equal to the Coulomb force by stating [math]M_0 \frac{v^2}{r}= \frac{e^2}{4\pi\epsilon r^2}[/math] 1) Now we may decide to view the LHS purely in terms of the gravitational energy as found in this equation [math]E_g r_s = GM^2[/math] by dividing through by [math]r^2[/math] which gives [math]\frac{E_g}{r} = \frac{GM^2}{r^2}[/math] This means we can set it equal to equation 1) when [math]v=c[/math] [math]\frac{GM^2}{r^2}= \frac{e^2}{4\pi\epsilon r^2}[/math] Now, the orbital length is equal to 1 x the de Broglie wavelength [math]2\pi r = \frac{h}{Mc} \times 1[/math] Now from the charge relationship [math]e = \sqrt{4 \pi \hbar c}[/math] one can get [math]c = \frac{e^2}{4\pi \frac{GM^2}{c}}[/math] where [math]\frac{GM^2}{c} = \hbar[/math] is the quantized angular component. The ratio of this velocity is equal to the fine structure constant with the twist that it is related to the gravitational charge of the system [math]\beta = \frac{e^2}{4\pi GM^2} = \alpha[/math] where [math]\beta = \frac{v}{c}[/math]. The ratio of velocity to obtain the fine structure constant is not new, however, no one as far as I know have ever written it in terms of a relationship for the gravitational charge. This is certainly a new way to view the relationship between the constant and the charge. An Ambiguity of the Black Hole Charge There does exist, a certain ambiguity for those who are familiar with the Black Hole charge I wish to discuss - that being, why is the gravitational charge this, and not the integral of taken over a two dimensional surface defining the gravitational charge as [math]M_g = \frac{1}{4\pi} \int g ds[/math] Well, the reason has to do with the quantization condition, also bearing in mind that we often simply deal with pointlike systems rather than objects which truly have 2-dimensional surfaces. The equation above is true for your usual standard black hole, but to describe Planck Particles, you would need to quantize the gravitational charge. This can be understood if we (just for now) defined the gravitational charge as [math]\mu_g[/math] and saw that [math]\frac{\mu_g}{Mc^2} = \ell_P[/math] is in fact the gravitational analogue of [math]\frac{e^2}{Mc^2} = R_{classical}[/math] which defines the classical electron radius, so the squared gravitational charge certainly has it's right place. Not to mention, the other relationships we have covered so far. Conclusions The gravitational charge [math]\sqrt{G}M[/math] may be very important when considering inertia and the concept of mass itself. Clearly, we have seen, that the nature of mass is questioned into the light of understanding gravitational energies in two forms: Those which have gravitational influences on the surrounding fields and those which have instrinsic gravitational energies (those attributed to the systems with inertial mass descriptions). The fine structure constant is also a mystery in physics, but we have also seen that the idea of gravitational masses may be determined somehow by the fine structure constant in a number of ways. We saw how the gravitational charge of a system was in fact conserved by the fine structure constant. We also showed how if the elementary charge of a system truly depended on different fundamental values, then what we have is a gravitational elementary charge also depending on similar constant values in nature. The gravitational charge may be so fundamental, that it can itself be an explanation perhaps, of the Heirarchy problem - that being, if the fine structure constant plays a role in different masses, then different masses arise because they are just finely tuned to the different fundamental constants. References (1) http://www.gravityresearchfoundation.org/pdf/awarded/1971/motz.pdf Edited September 3, 2012 by Aethelwulf Quote
Aethelwulf Posted September 2, 2012 Author Report Posted September 2, 2012 (edited) Oh we use math-tex code here eh?Ok... here comes the dubious process of me correcting it all. Edited September 2, 2012 by Aethelwulf Quote
Aethelwulf Posted September 2, 2012 Author Report Posted September 2, 2012 right, That is me. I have converted it properly now to readable text. Quote
LaurieAG Posted September 2, 2012 Report Posted September 2, 2012 Hi Aethelwulf, Actually, this is not the Planck mass exactly, it is about a factor of [math]\pi[/math] greater, however, the Planck mass is usually an approximation so what we have here is possibly the correct value. After rearraging and a little further solving for the radius we end up with [math]R = \sqrt{\frac{\pi G \hbar}{c^2}}[/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.The difference between the Compton wavelength and the reduced Compton wavelength is a factor of 2 * Pi and that is the same difference between the Planck length and its reduced form. Quote
Aethelwulf Posted September 2, 2012 Author Report Posted September 2, 2012 Hi Aethelwulf, The difference between the Compton wavelength and the reduced Compton wavelength is a factor of 2 * Pi and that is the same difference between the Planck length and its reduced form. That's correct. This is why we might see a pi stuck in front of an hbar to remove the pi in its denominator. Quote
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