Phi for All Posted September 21, 2013 Share Posted September 21, 2013 The opinion of a lay -- man. I eagerly hopped a break throw, I am sorry to tell that I am disappointed. Even I sensed that all this is nothing else but the old try to throw out of scene the concepts of mass - radius and to explain them with mean of pure energy (light), in your paper you continue to play with G*M. So like that you or not, the elephant stay in room. Some question from an ignorant: With who will interact your photon (h -- even with tire), to change direction in it’s movement? With it self ? And you think that this may create a particle? In fact you allude about a huge G. It comes from where? But maybe this is very subtle for a lay -- man. So I hush in this direction. When there are some hundred kinds of fields why not a Huge? I understand that my post will irritate you, I sincerely hopped that you may have an idea “to pacify” two extreme: h and Mplank (extrapolated toward fact electric charge). I hoped because your paper seems crafted by a specialist. And to give you a case to joke with an idiot like me , listen my suggestion: Try an equation of interactions between three photons the frequency of which are : fe = ( fplank * fg)^05 . here ---- fg = (G*mx / rx ) / ((2*pi/α) *rx) is a crazy speculation. Maybe the solving of this equation will give all kind of particles. ! Moderator Note Please, this is a section for discussing mainstream science. Your idea is speculative, and students come to these sections for mainstream explanations. Do NOT hijack other people's threads with your own speculation. We have a whole speculation section where concepts like this can be analyzed and discussed. Thank you. Link to comment Share on other sites More sharing options...

Kramer Posted September 22, 2013 Share Posted September 22, 2013 Moderator Note Please, this is a section for discussing mainstream science. Your idea is speculative, and students come to these sections for mainstream explanations. Do NOT hijack other people's threads with your own speculation. We have a whole speculation section where concepts like this can be analyzed and discussed. Thank you. Thanks for your Moderator Note, even I am somewhat puzzled: am I not in Speculation?My intention was to attire the OP in debate, about classic concepts in physic and modern concept of Mainstream. You say that students come in this section for discussing mainstream science. Is it taboo to compare both teaching --- for a stronger understanding?I am a lay-man obsessed with ’mass-particles’ hypothesis in the same way that O.P. has asserted an ‘energy – particles’ hypothesis. And the discuss will be useful for both: OP will learn a “training” how to hush opponent, (which will be very easy), and I will learn how to reconcile the ideas of “zero radius” and “surrogate of mass”.Where am I wrong?If OP has a complaint --- then I am sorry, very sorry. Link to comment Share on other sites More sharing options...

Phi for All Posted September 22, 2013 Share Posted September 22, 2013 Thanks for your Moderator Note, even I am somewhat puzzled: am I not in Speculation? ! Moderator Note It doesn't matter what section you're in, hijacking someone else's thread with a speculative idea is still hijacking. It's difficult enough dealing with one speculative idea at a time. Do NOT further derail the thread by responding to this modnote. If you feel it's in error, push the Report Post button and another staff member will discuss this privately with you. Link to comment Share on other sites More sharing options...

Mellinia Posted September 23, 2013 Share Posted September 23, 2013 Is there any place(link?) where I can read on Er=GM^2 ? Is M being the total mass of the system, and E being the total energy in the system? Link to comment Share on other sites More sharing options...

ajb Posted September 23, 2013 Share Posted September 23, 2013 Is there any place(link?) where I can read on Er=GM^2 ? Is M being the total mass of the system, and E being the total energy in the system? The units don't look right to me. Look up the force due to gravity and the gravitational potential in Newtonian gravity. Link to comment Share on other sites More sharing options...

Mellinia Posted September 23, 2013 Share Posted September 23, 2013 (edited) On the right side: Er = Joule meters = (kgms^-2 )(m^2) => ML^3 T^-2 On the left: GM^2 = (ms^-2 )(m^2)(kg) =>ML^3 T^-2by F= GMm / r^2, GM= Fr^2/ m = (kgms^-2 )(m^2)(kg^-1)=(ms^-2 )(m^2) LHS = RHS but I have no idea of the physics behind it. E_g = integral V_g over dr and V =- GM/ r but does GM^2 mean that the object is exerting a force on itself? Edited September 23, 2013 by Mellinia Link to comment Share on other sites More sharing options...

ajb Posted September 23, 2013 Share Posted September 23, 2013 It looks to me that the units don't balance, check the units of length again. Link to comment Share on other sites More sharing options...

swansont Posted September 23, 2013 Share Posted September 23, 2013 Units look OK. E=GMm/r is the equation for potential energy, so Er = GM^2 is fine. Link to comment Share on other sites More sharing options...

ajb Posted September 23, 2013 Share Posted September 23, 2013 Units look OK. E=GMm/r is the equation for potential energy, so Er = GM^2 is fine. My bad, I was confusing potential energy with potential Link to comment Share on other sites More sharing options...

Mellinia Posted September 23, 2013 Share Posted September 23, 2013 Units look OK. E=GMm/r is the equation for potential energy, so Er = GM^2 is fine. so M^2 means that the object is experiencing a force by itself on itself (the author said that this was related to the energy of a system)? Link to comment Share on other sites More sharing options...

TrappedLight Posted September 23, 2013 Author Share Posted September 23, 2013 (edited) I'd like to point out, that GM^2/r is actually a gravitational self-energy. It can, as Swansont pointed out, is also a description of potential energy... depending on the text you read. In this work, it is a gravitational self-energy with a numerator describing the gravitational charge (the self-mass of the system). so M^2 means that the object is experiencing a force by itself on itself (the author said that this was related to the energy of a system)? The gravitational charge Gm^2 is related to the energy and the radius as Er. Actually, this is the gravitational charge squared. Perhaps importantly, knowing the energy is related to gravitational charge by a product with the radius [math]E r = GM^2[/math] The self-energy is calculated as [math]\frac{\hbar c}{(\frac{\hbar}{Mc})} = Mc^2[/math] So what we really have is [math]\frac{\hbar c}{(\frac{\hbar}{Mc})} \bar{r} = GM^2[/math] Or you can swap the definitions around [math]\frac{GM^2}{(\frac{\hbar}{Mc})} \bar{r} = \hbar c[/math] Means the same thing though Note, [math]\frac{\hbar}{Mc}[/math] is an important quantity known as Comptons reduced wavelength... considered to be the ''size'' of a particle. Edited September 23, 2013 by TrappedLight Link to comment Share on other sites More sharing options...

Mellinia Posted September 23, 2013 Share Posted September 23, 2013 I'd like to point out, that GM^2/r is actually a gravitational self-energy. It can, as Swansont pointed out, is also a description of potential energy... depending on the text you read. In this work, it is a gravitational self-energy with a numerator describing the gravitational charge (the self-mass of the system). The gravitational charge Gm^2 is related to the energy and the radius as Er. Actually, this is the gravitational charge squared. Taking a spherical cow...lol jking If I were to simplify this into a point mass, the point mass would exert an gravitational force on anything that enters it's field, but the force won't be exerted on itself, right? Drawing an analogy to EM, would a +ve charge be affected by it's own electric field? How does this cause "self-energy"? Link to comment Share on other sites More sharing options...

TrappedLight Posted September 24, 2013 Author Share Posted September 24, 2013 You are right, it's not the same. I wrote in the original draft: ''Lloyd Motz, who was an American astronomer and cosmologist with a keen interest in gravity physics, began to describe elementary particles as bound pairs of photons, similar to a cooper pair only that the difference is that photons are in fact massless particles. But bound photons not only would cancel out the electric charge of a system but he speculated at the time that it could give rise to spin itself. He used this model consecutively to talk about the structure of neutrino's. In this work, we are not going to discuss the nature of other particles, instead we are going to focus on the electron. One particular part of his model involved describing mass of particles as a charge itself on the system, just like how an electron has an electric charge by moving in an electromagnetic field, the mass of an electron was also a charge on the system; the problem I suffered for quite a number of years was whether there was a direct analogy... ie. Is the mass of a particle then a charge gained by moving in a gravitational field? Gravity and the field itself is such a troublesome thing in physics at the moment. Not only does it appear to be significantly smaller by many magnitudes than the other three forces of nature but it also differs from the other forces that the force itself is not carried by a mediator particle. For instance, the electromagnetic force is mediated by photons, so what is the gravitational force mediated by? The speculation was that a field of ''gravitons'' was responsible for transmitting a gravitational signal from one place to another... but alas none have ever been found. In fact, the search for the fundamental mediator of gravity has been pretty much given up for not many scientists today believe there is even one. Instead, gravity was a pseudoforce, like the Coriolis force. Something experienced by a system in an inertial frame of reference. Of course, General Relativity which was formulated by Albert Einstein showed us that it was instrinsically related to the fabric of space and time and that objects accelerating in their frames of reference would not be able to tell whether they where falling or elevating. But what is gravity at the fundamental level? As I said, for many years I struggled to find an answer for myself why then there was not a complete analogy between the electric charge of a moving particle in an electromagnetic field and that with a mass charge moving in a gravitational field. The answer finally came to me by realizing that if the gravitational force wasn't actually a real force (ie. ficticious force) then it seems that mass had to come about another way. It can be as I found, still valuable to speak of the gravitational field and think of the particles charge of mass as something related to the curvature of spacetime.'' ... so to summarize and answer your question the gravitational self-energy is not caused by self-interaction with the gravitational field. The gravitational self-energy is characterized by the feature of the charge [math]\sqrt{G}M[/math] which may arise from fully electromagnetic features of the theory. The gravitational self-energy is about the effective mass of the system. The first time I wondered if [math]GM^2[/math] could be describes by electromagnetic features of the theory was when I evaluated a theoretical value for [math]GM^2[/math] by fitting it within the parameters of the trusted CODATA charge as I wrote ... Like all important fundamental dynamical processes in nature, physics often described them in terms of the constants of nature, such as the speed of light, permittivity, permeability, the gravitational constant ect. The CODATA elementary charge does exactly this. The point of the equation about to be shown, is to attempt to describe charge as a ratio of important fundamental constants [math]e = \sqrt{\frac{2 \alpha \pi \hbar}{\mu_0 c}}[/math] (1) We can be really theoretical about this and create a simple equation which will do the same thing for the gravitational charge. We first of all recall the relationship between the elementary charge and the gravitational charge [math]e^2 = 4 \pi \epsilon_0 GM^2[/math] Square equation (1) and set them equal, this gives [math]4 \pi \epsilon_0 \mu_0 c GM^2 = 2 \alpha \pi \hbar[/math] Solving for the gravitational charge and cancelling out some factors we have a fundamental relationship for the gravitational charge [math]\sqrt{G}M = \sqrt{\frac{\alpha \hbar}{2 \epsilon_0 \mu_0 c}}[/math] What is the interepretation of a dubious looking equation like this? I noticed a few things. First of all, in physics it is recognized that the angular momentum component [math]\hbar[/math] of a system is conserved through the fine structure constant [math]\frac{e^2}{4 \pi \epsilon_0 c} = \pm \alpha \hbar n[/math] Actually this is a special quantization condition. We should notice the positive and negative eigenvalues of increment [math]n[/math] and we may therefore imagine a same condition on our gravitational charge equation [math]\sqrt{G}M = \sqrt{\frac{ \pm \alpha \hbar n}{2 \epsilon_0 \mu_0 c}}[/math] This means that angular momentum is conserved in our definition of the gravitational charge itself. Notice also that our charge is proportionally dependant on the electric and magnetic resistance constants [math](\epsilon_0, \mu_0)[/math], something you might expect if mass itself was an electromagnetic phenomenon. Somehow, magnetic and electrical resistance, the mass and spin... and the charge are all fundamentally-related. Link to comment Share on other sites More sharing options...

TrappedLight Posted September 24, 2013 Author Share Posted September 24, 2013 (edited) On the right side: Er = Joule meters = (kgms^-2 )(m^2) => ML^3 T^-2 On the left: GM^2 = (ms^-2 )(m^2)(kg) =>ML^3 T^-2 by F= GMm / r^2, GM= Fr^2/ m = (kgms^-2 )(m^2)(kg^-1)=(ms^-2 )(m^2) LHS = RHS but I have no idea of the physics behind it. E_g = integral V_g over dr and V =- GM/ r but does GM^2 mean that the object is exerting a force on itself? I just read this again and realized I never actually answered your question. There could be a self-force of particles, even in this model, it would be called a gravitational self-force. The Abraham-Dirac classical equation describing the self-force of a particle is a tad different perhaps, but it involves peculiar things like an increasing charge with a change in position! I wouldn't actually know how to calculate the self-force of a particle, especially in a relativistic manner as this would include retarded and advanced forms of the Green's function, the tool itself used to calculate self-energy. In a way, I suppose a gravitational self-force would exist as [math]F = \frac{GM^2}{r^2}[/math] Of course, inside a particle, in my model the force would take on an extremely large value to cancel out the forces due to the coloumb force. Therefore, the gravitational self-force has an order of [math]F_{self} = \frac{\Gamma M^2}{r^2}[/math] Where [math]\Gamma[/math] is the strong gravitational constant (not to be mistaken for a Christoffel symbol). This would be true if [math]E_{self} = \frac{GM^2}{r}[/math] is a gravitational self-energy. Edited September 24, 2013 by TrappedLight Link to comment Share on other sites More sharing options...

Mellinia Posted September 25, 2013 Share Posted September 25, 2013 I just read this again and realized I never actually answered your question. There could be a self-force of particles, even in this model, it would be called a gravitational self-force. The Abraham-Dirac classical equation describing the self-force of a particle is a tad different perhaps, but it involves peculiar things like an increasing charge with a change in position! I wouldn't actually know how to calculate the self-force of a particle, especially in a relativistic manner as this would include retarded and advanced forms of the Green's function, the tool itself used to calculate self-energy. Accelerated electrons emit photons as a consequence of the light speed limit, and mass particles may do the same, but this is a recoil force, describing the energy loss as a consequence of it's own EM field, and not the whole energy of the electron system, so while "self-force" is right (recoil, much?), self-energy...a bit suspicious? Anyhow, e^2 = 4 /pi /epsilon_0 GM^2, what is M? This could be an arbitrary mass? Though I did do some calcs and if e is the elementary charge, and other constants are what they are from the CODATA website, M should be around 1.857e-9 kg....? Link to comment Share on other sites More sharing options...

TrappedLight Posted September 26, 2013 Author Share Posted September 26, 2013 (edited) Accelerated electrons emit photons as a consequence of the light speed limit, and mass particles may do the same, but this is a recoil force, describing the energy loss as a consequence of it's own EM field, and not the whole energy of the electron system, so while "self-force" is right (recoil, much?), self-energy...a bit suspicious? Anyhow, e^2 = 4 /pi /epsilon_0 GM^2, what is M? This could be an arbitrary mass? Though I did do some calcs and if e is the elementary charge, and other constants are what they are from the CODATA website, M should be around 1.857e-9 kg....? I wasn't the one who mentioned the self-force, but there should be one in this theory for the interior of the particle. You also note that self-energy is a suspicious term? Not quite sure what you mean here, even pointlike particles have a self-energy, except that this energy is described as being infinite! Edited September 26, 2013 by TrappedLight Link to comment Share on other sites More sharing options...

Mellinia Posted September 26, 2013 Share Posted September 26, 2013 I wasn't the one who mentioned the self-force, but there should be one in this theory for the interior of the particle. You also note that self-energy is a suspicious term? Not quite sure what you mean here, even pointlike particles have a self-energy, except that this energy is described as being infinite! As in E=mc^2 infinite? M is included in your theory as the total mass in the system? Link to comment Share on other sites More sharing options...

TrappedLight Posted September 26, 2013 Author Share Posted September 26, 2013 (edited) As in E=mc^2 infinite? M is included in your theory as the total mass in the system? Yes as in infinite. This is one of the problems dealing with pointlike particles. The equation goes like this [math]U = \int_{|r| < R} \frac{\epsilon_0}{2} E^2 dr = \frac{e^2}{8 \pi \epsilon_0 R}[/math] When the radius [math]R \rightarrow 0[/math] the energy [math]U[/math] is infinite, note that [math]E[/math] is the electric field! This means that classical physics predicts an infinite energy for the electron (which would result in an infinite bare mass). It's ludicrous to think that is even a physical property of the world, so something appears to be either: A) something wrong with the mathematical approach or B) particles are not actually pointlike. And yes [math]M[/math] is the mass and in the theory, it characterizes the possible mass of all particles. There was even more I wished to write about but I kept it from the OP. One of them being an approach by Hestenes involving zitter motion, and how the original paper http://www.cybsoc.org/electremdense2008v3.pdf is closely related also to zitter motion. I wrote before, and now I quote: '' The ''zitter clock'' was varified experimentally. This was an idea created by deBroglie and was confirmed in a channelling experiment. The period of the frequency of the electron clock is given as [math]\psi(\tau) = e^{i \omega \tau}[/math] It has been noted by ref 1. in my main paper, that the electron clock can be attributed to the toroidal motion of our charged massless system. In fact, the electron in it's most elegantly explained form, is that it is a lightlike particle which follows in which the spin is determined by the internal helical dynamics where it even has a curvature and a frequency... and as I showed in my own work, may even be attributed to a phase related to the intrinsic gravitational charge. If the electron clock is determined by the circular motion at the speed of light, we may attribute this to a invariant proper time operator [math]\frac{\bar{r}}{c}[/math] which would be related to the ''rest'' energy of the system as [math](\frac{\bar{r}}{c}) Mc^2 = \hbar[/math] Now this invariant time operator coefficient term on the energy giving us the quantization condition [math]\hbar[/math] Can actually be related to as Motz explains; the proper time must be treated as an operator which is canonically conjugate to the rest mass of particles, which was discussed in ref 1 below. It actually brings us back to the electron clock, where we are dealing with the proper time of the electrons history. The zitter motion and the invariant proper time operator therefore might be doing the same thing in respects of giving rise to angular momentum component [math]\hbar[/math]. ref 1. L Motz, phys. Rev., 93, 901 (1954) There is a problem which exists though, concerning the modelling of the zitter motion as a proper time invariant operator, that is that the electron is modelled by Hestenes as a lightlike curve [math]c^2 \Delta t^2 = \Delta r^2[/math] thus [math]s^2 = 0[/math] and proper time cannot be defined in such ways. Hestene does however mention this and that a physical definition of the time parameter [math]\tau[/math] must be determined by other features of the model FQXi - Foundational Questions Institute However, as I pointed out, ''The zitter motion and the invariant proper time operator therefore might be doing the same thing in respects of giving rise to angular momentum component'' and this is meant to be taken seriously that the time parameter derives from the assumption that the electron has an intrinsic angular momentum (which is as we have covered), the toroidal topology of our massless charge The spin [math]S[/math] in this sense, becomes a function of the time parameter [math]S(\tau)[/math] which as the mathematical jargon of Hestene points out adequately, ''is a null bivector.'' This is how he reaches the potential zitter term [math]\Phi[/math], by introducing the relevant equations of motion for the velocity [math]\dot{u}[/math], momentum [math]\dot{p}[/math] and the spin [math]\dot{S}[/math]. Very ingenuous, so that the potential zitter interaction term becomes a function [math]\Phi(\tau, z)[/math]. In this respect from this model, we can talk about the proper time experienced between two events - the incremental events which defines the electron clock. And as we have seen, the electron clock may be written as a function or it could actually be represented as a dynamical time invariant operator on the energy term which in return tells us it has an angular momentum. Or we can write it as an integral [math]Mc^2 \int_{P_t} d\tau = \hbar[/math] Where the integral defines a proper time interval. '' Another part I wish to make known '' There is then a relationship to the uncertainty in the energy and the uncertainty in the length of the system. [math]\Delta E \leq \frac{GM^2}{ \lambda}[/math] The physical meaning of this is that the internal energy flux is bound by the maximal uncertainty in the mean transport of radius which may take the role of the wavelength but we have a correction order. [math]\lambda = \alpha \lambda_C[/math] As the paper http://www.cybsoc.org/electron.pdf recites, the limitation of the speed of light means that only paths within the radius can provide a contribution of inertial energy (mass) to the system. The uncertainty relationship then to satisfy the toroidal system is [math]\Delta E \leq \frac{GM^2}{ \alpha \lambda_C}[/math] Shows us that the energy depends on the varying uncertainty of the Compton wavelength up to a maximal radius of transport. '' This relationship might be more important than one realizes as even Motz deduced that there was an uncertainty relationship between the mass of a particle and it's radius. The more you attempted to measure the radius of the particle, the more uncertain it's mass became and vice versa. He spoke about this relationship in (ref. 2) but I will quickly cover it. He showed that you can equate the Compton wavelength to the Gaussian curvature [math]8 \pi \rho_0(\frac{G}{c})[/math] where [math]\rho_0[/math] was the proper density and [math]\frac{G}{c}[/math] is the Schwarzschild constant. It can obtain the equation [math]\frac{K}{6} = (\frac{\hbar}{Mc})^{-2}[/math] Introducing the radius of curvature, which would equate to our use of the curvature of transport this would be, keeping in mind that the radius of a three dimensional hypersphere is [math]\frac{1}{6}K = (\frac{\hbar}{Mc})^{-2}[/math] you obtain [math]R^2 = (\frac{\hbar}{Mc})^{-2}[/math] and from this he states that one gets [math]RMc = \hbar[/math] He says that one can look upon this as an uncertainty relationship where the mass term and the radius anticommute. In a similar fashion, I have shown above there is an uncertainty relationship with the internal radius of transport and it's gravitational charge. '' And for the case where [math]t_0 \rightarrow t_p[/math] where the Planck time becomes the proper time for an interval of distance traversed by a light particle, the phase of both the orbital rotation and the internal photon merge to form a gravitational fine structure [math]\phi = \omega_C t_p[/math] where [math]t_p[/math] is the proper time where [math]t_p = \gamma(t - \frac{vx}{c^2})[/math] The phase of the internal photon can be written as [math]\omega_C t_p = \omega_C \gamma(t - \frac{vx}{c^2}) = \sqrt{\alpha_G}[/math] Where simply [math](\omega_C t_p)^2 = \alpha_G[/math] Of course, [math]\frac{GM^2}{\hbar c} = \alpha_G[/math] Where [math]M[/math] is not considered the Planck mass per se, but an electron mass with a twist. In fact, the definition of [math]\alpha_G[/math] is surprisingly simple: it is the square of the electron mass measured in units of Planck mass. This fine structure relation to the internal phase of the photon is very important to the gravitational physics behind the theory. Edited September 26, 2013 by TrappedLight Link to comment Share on other sites More sharing options...

Mellinia Posted September 26, 2013 Share Posted September 26, 2013 Yes as in infinite. This is one of the problems dealing with pointlike particles. The equation goes like this [math]U = \int_{|r| < R} \frac{\epsilon_0}{2} E^2 dr = \frac{e^2}{8 \pi \epsilon_0 R}[/math] When the radius [math]R \rightarrow 0[/math] the energy [math]U[/math] is infinite, note that [math]E[/math] is the electric field! This means that classical physics predicts an infinite energy for the electron (which would result in an infinite bare mass). It's ludicrous to think that is even a physical property of the world, so something appears to be either: A) something wrong with the mathematical approach or B) particles are not actually pointlike. And yes [math]M[/math] is the mass and in the theory, it characterizes the possible mass of all particles. Thus are you saying that the intrisic charge is related to it's mass? FOR e^2 =4 /PI /EPSILON G M^2, should I take M as eletron mass? Link to comment Share on other sites More sharing options...

TrappedLight Posted September 26, 2013 Author Share Posted September 26, 2013 (edited) Thus are you saying that the intrisic charge is related to it's mass? FOR e^2 =4 /PI /EPSILON G M^2, should I take M as eletron mass? I'm saying that mass is a charge. Usually when people speak about mass, they simply use the term [math]M[/math]. There is a new parameter in this work, that mass isn't an independent quantity, that the mass of a system is better calculated as a charge [math]\sqrt{G}M[/math] with the gravitational feature of [math]G[/math]. Interestingly, there are no known cases outside of the neutrino where there is a particle with neutral charge and has a mass. My early investigation into this theory, suggested to me that perhaps mass and charge where in fact the same thing. This was made apparent when I noticed (in Gaussian units) [math]e = \sqrt{\hbar c}[/math] can be rewritten as [math]e = \sqrt{GM^2}[/math] which implied that the mass was like a charge on the system. Therefore I took it to mean that charge cannot be without a mass present and that the two quantities [math]e[/math] and [math]\sqrt{G}M[/math] where somehow interdependent. The mass in the equation of course, now going back to SI units [math]e = \sqrt{4 \pi \epsilon GM^2}[/math] the mass of course, can be taken as an electron mass. In fact, we frequently do so when we want to obtain the fine structure contsant [math]\frac{e^2}{4 \pi \epsilon \hbar c}[/math] since [math]e[/math] is describing the charge of an electron. The equations where meant to be generic. You can apply them to different conditions assuming you are willing to bring into consideration all the factors. ......................... Ajb You know how you wondered why I called it the Heaviside relationship [math]e = \sqrt{4 \pi \epsilon \hbar c}[/math] ? I have now found a reference to it being called that here http://en.wikipedia.org/wiki/Natural_units except it is written as [math]e = \sqrt{\alpha \hbar c}[/math] * This relationship is the Lorentz-Heaviside units, which are different from the units I use clearly, so the equation I use is not called the Heaviside relationship. Albeit, it is very similar. ............................................... Another part I wished to make known '' It's been part of the task to describe mass with purely electromagnetic features of the theory and we have looked at that by discussing about how the gravitational charge can be related to magnetic and electric properties. The gravitational charge [math]GM^2[/math] can also arise from the magnetic and electric flux! In MKS units, the electric and magnetic flux are [math]\Phi_e = \frac{e^2}{\epsilon_0}[/math] and [math]\Phi_m = \frac{\phi_{0}^{2}}{2 \mu_0}[/math] one can obtain [math]GM^2 = \frac{\sqrt{2}}{2 \pi} \sqrt{\Phi_e \Phi_m}[/math] again, [math]GM^2[/math] can be obtained with a RHS describing electromagnetic properties. http://arxiv.org/vc/hep-ph/papers/0306/0306230v2.pdf The author of the main equation states that it says that the magnetic and electric flux have the same dimensions. It is actually very similar to the Gaussian-cgs units in which the electric and magnetic fields have the same dimension - in fact if my history is correct, MKS units are the alternative Gaussian-cgs units. Edited September 26, 2013 by TrappedLight Link to comment Share on other sites More sharing options...

ajb Posted September 26, 2013 Share Posted September 26, 2013 [math]e = \sqrt{4 \pi \epsilon \hbar c}[/math] Assuming that the RHS all have their standard meaning ([math]\epsilon = \epsilon_{0}[/math] = permittivity of free space, right?) then the LHS is the Planck charge. You can relate this to the electron charge via the fine structure constant [math]\frac{e}{\sqrt{\alpha}} = \sqrt{4 \pi \epsilon \hbar c}[/math]. You then use the Planck mass to get [math]\hbar c = G M_{p}^{2}[/math] and we get [math]\frac{e}{\sqrt{\alpha}} = \sqrt{4 \pi \epsilon G M_{p}^{2}}[/math], which is fine. It seems to relate gravity to electric charge, but all I have really done is mess about fundamental constants in such a way that the units make sense. I don't see there is any real physics in this. You claim to have done something different as you have a variable G theory. So, what does G depend on and how? Link to comment Share on other sites More sharing options...

TrappedLight Posted September 26, 2013 Author Share Posted September 26, 2013 Assuming that the RHS all have their standard meaning ([math]\epsilon = \epsilon_{0}[/math] = permittivity of free space, right?) then the LHS is the Planck charge. You can relate this to the electron charge via the fine structure constant [math]\frac{e}{\sqrt{\alpha}} = \sqrt{4 \pi \epsilon \hbar c}[/math]. You then use the Planck mass to get [math]\hbar c = G M_{p}^{2}[/math] and we get [math]\frac{e}{\sqrt{\alpha}} = \sqrt{4 \pi \epsilon G M_{p}^{2}}[/math], which is fine. It seems to relate gravity to electric charge, but all I have really done is mess about fundamental constants in such a way that the units make sense. I don't see there is any real physics in this. You claim to have done something different as you have a variable G theory. So, what does G depend on and how? Right, you can do that. Of course, the equations are generic and are meant to have broad ranging applications. Yes, it relates the source of gravity [math]\sqrt{G}M[/math] to the charge, the physics might be simple as you cannot have mass without a charge on the system, here, the mass term is not singular, mass arises as a charge by definition. And [math]G[/math] plays the exact same role as the strong gravitational constant, especially when inside the interior of the particle which has a non-zero radius. It's different because, we don't consider gravity inside the particle, however, if it truly is a shell, then we can assume gravity takes on a large value of [math]\frac{\hbar c}{M^2}[/math]. The gravitational constant has a number of different values for it's strong case and that is noted by a number of different authors. The constant however, depends on the distance. The idea is that gravity becomes strong at shorter and shorter distances. Believe it or not, but the strength of gravity on very small distances (involving interactions which would involve specifically [math]\alpha_G[/math]) are extremely unknown, with many uncertainties. Link to comment Share on other sites More sharing options...

ajb Posted September 26, 2013 Share Posted September 26, 2013 Right, you can do that. Of course, the equations are generic and are meant to have broad ranging applications. Yes, it relates the source of gravity [math]\sqrt{G}M[/math] to the charge, the physics might be simple as you cannot have mass without a charge on the system, here, the mass term is not singular, mass arises as a charge by definition. What about neutrinos, they do not carry electric charge, but have a small mass? (They carry weak hypercharges, but this is not what we are describing here) Astrophysical bodies are near enough neutral, but have gravitational fields that are not proportinal to their net charge? And [math]G[/math] plays the exact same role as the strong gravitational constant... I am not familiar enough with strong gravity to comment here. Just for anyone else reading this, strong gravity is an alternative approach to gravity at both cosmological and particle scales. It was shown to produce confinement and asymptotic freedom even though it has an inverse-square law. So it reproduces some of the features of QCD, which of course has grown into one of the corner stones of the standard model of particle physics. This should not be confused with strong gravitational fields within general relativity. Believe it or not, but the strength of gravity on very small distances are extremely unknown, with many uncertainties. Yes I am aware of this. Link to comment Share on other sites More sharing options...

TrappedLight Posted September 26, 2013 Author Share Posted September 26, 2013 What about neutrinos, they do not carry electric charge, but have a small mass? (They carry weak hypercharges, but this is not what we are describing here) Astrophysical bodies are near enough neutral, but have gravitational fields that are not proportinal to their net charge? The neutrino needs to be a pair of bound photons, so that one topological charge exactly cancels the other out. It's a special case of this theory for the neutrino. It is allowed a mass, but it's charges are internally cancelled out. In fact, Motz got to this theory before me! I was surprised how similar our theories actually meshed. http://www.gravityresearchfoundation.org/pdf/awarded/1966/motz.pdf The neutrino presents itself in this theory, as a gravitationally-bound pair of photons. However, his theory has the neutrino with a zero-rest mass. I believe that can be overcome by realizing that there is still a small contribution of mass contributed by a non-zero magnetic moment. The energy associated to the moment, is analogous to the idea that the electric or magnetic features (just like a charge) can contribute mass to a system, the so-called electromagnetic mass. If there isn't a small correction to the mass through this, then I am not entirely sure how to remedy that problem. And yes, galactic bodies are electrically neutral. Galactic bodies however, are not fundamental and they are comprised of many individual systems which do have charges and mass. Link to comment Share on other sites More sharing options...

ajb Posted September 26, 2013 Share Posted September 26, 2013 The neutrino needs to be a pair of bound photons, so that one topological charge exactly cancels the other out. It's a special case of this theory for the neutrino. It is allowed a mass, but it's charges are internally cancelled out. In fact, Motz got to this theory before me! I was surprised how similar our theories actually meshed. http://www.gravityresearchfoundation.org/pdf/awarded/1966/motz.pdf The neutrino presents itself in this theory, as a gravitationally-bound pair of photons. However, his theory has the neutrino with a zero-rest mass. I believe that can be overcome by realizing that there is still a small contribution of mass contributed by a non-zero magnetic moment. The energy associated to the moment, is analogous to the idea that the electric or magnetic features (just like a charge) can contribute mass to a system, the so-called electromagnetic mass. If there isn't a small correction to the mass through this, then I am not entirely sure how to remedy that problem. This is vastly different to how QFT sees the neutrino and don't forget the amazing experimental accuracy that we have with the syandard model. Motz's paper from 1966 is before we had such a good handle on QFT. Link to comment Share on other sites More sharing options...

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