Aethelwulf Posted June 8, 2012 Share Posted June 8, 2012 (edited) Abstract Mass in our most up-to-date theory was beleived to be a by-product of a Boson and its respective field. Finding this particle, called the Higgs Boson has proven difficult. It seems likely that scientists will have to find a new mechanism for mass - and in this work, we will advocate one of those theories. What is Mass? So, what do physicists mean when they speak of a mass, or rather, a mass term? For a while, physicists tried to answer this question a number of ways. There was such a thing as an electromagnetic mass [1] at one point in physics. The was the idea that the electromagnetic field interacted with a quantum object in such a way that it gave a contribution of mass to a system. This was called the self-energy of a system -- in fact, so mainstream the idea was at one point, it was in fact considered as an explanation for inertial mass itself. It is quite curious to mention that there was even an electromagnetic inertia which was experienced by particles which where accelerated, this was explained through what is called the Larmor Equation, but given as an expression here: [math]\frac{2}{3}\frac{e^2}{c^3}a^2[/math] This would in fact describe how much radiation a charged particle would give off as it was accelerated to higher and higher speeds. The faster you made your particle move at, the more power was required to bring it to that speed. This was in fact analogous to inertia which was experienced by particles with a charge: today, our new definition is that inertia is strictly a property of mass. As much as this idea has faided, some scientists still use the equations describing the EM-mass. In a sense, mass has been shown to be an electrical phenomenon derived from first principles. There maybe some important clues however that it may still have importance, especially since the current model, the Higgs Boson is now finally faiding into the past as scientists have narrowed the possible energy levels for the Higgs Boson to a very small corner. Some of these clues, may exist in the fact that charged massless bosons don't exist. The Yang Mills Equations once predicted massless charge bosons but the approach has been generally considered today as a false one. But why should this prediction and how it failed important? Charge is considered by physicists as an ''intrinsic property'', something which is inherent with particles of mass. When we think about charge, we usually talk about electromagnetic charges [math]e[/math], which will sometimes be found as a coefficient to the electromagnetic potential in equations [math]ieA_{\mu}[/math]. What is important here, is that the electric charge is always found with mass, so maybe electromagnetic mass contributions will arise from the shadows again? Who knows, but what does seem clear that the Higgs Boson is now failing our theory so we will need to revise the equations again. So what is a Higgs Boson? How does it enter our theories? Our theory of Gauge Fields [2] where [math]\phi[/math] is a Gauge Boson, has a structure in the equations which invariance. This has been called, Gauge Invariance and this invariance allows physicists to make certain transformations to fields. The laws of physics always remain the same under gauge invariance, indeed, that is what it's all about! It means there is no such thing as an absolute position in physics - the only thing which does count is relative positions. Another feature of the gauge fields is that they retain a symmetry of the theory. A symmetry can be understood from the most simplest langrangian term [math]\partial\phi^{*} \partial \phi[/math] Suppose we had a transformation [math]\partial \phi' \rightarrow e^{i\theta}\phi(x)[/math] In this transformation, the derivatives of [math]\phi'[/math] are concerned only with [math]\phi[/math]. This transformation will look like [math]\partial \phi'(x) = e^{i \theta} \partial \phi(x)[/math] [math]\partial \phi'^{*}(x) = e^{-i \theta} \partial \phi^{*}(x)[/math] Therefore, when you multiply [math]\partial \phi '[/math] with [math]\partial \phi ^{*}[/math] the [math]e^{i\theta}[/math] and [math]e^{-i\theta}[/math] will cancel out because that is how you would compute them with their conjugates. Thus In other words, our field [math]\theta[/math] is constant, and does not require the same derivatives as our boson. In return, we would just get [math]\partial \phi^{*}\partial \phi[/math] and viola! This was the most simplest demonstration of a symmetry conserving field. An even simpler demonstration would be: [math]\frac{d(x +C)}{dt}[/math] is in fact simply the same as [math]\frac{dx}{dt}[/math] this means that the equations where symmetry existed and what we have in these symmetries are extra constants always remain the same, just as our Gauge example above. But what if [math]\theta[/math] also was a function of position [math]\theta (x)[/math]? Well let's see shall we. [math]\partial \phi' = (\partial \phi + i \phi \frac{\partial \theta}{\partial x})e^{i \theta}[/math] and our conjugate would be [math]\partial \phi'^{*} = (\partial \phi^{*} - i \phi^{*} \frac{\partial \theta}{\partial x})e^{-i \theta}[/math] Multiplying the two, we need to factorize it [math](\partial \phi + i\phi\frac{\partial \theta}{\partial x})(\partial \phi^{*} - i\phi^{*}\frac{\partial \theta}{\partial x})[/math] which gives [math]=\partial \phi^{*} \partial \phi + i(\phi \partial \phi^{*} - \phi^{*} \partial \phi)\frac{\partial \theta}{\partial x} + \phi^{*} \phi (\frac{\partial \theta}{\partial x})^2[/math] The reason, again why this equation turned out to be such a mess was because our [math]\theta[/math]-field now depended on position, so the derives of our boson field also included it. This motivated scientists to find a symmetry again in the equations, and to do so required the use of te Covariant Derivative which originally came form the work on fibre bundles. To restore symmetry, we need to define the Covariant Derivative as [math]A_{\mu}' \rightarrow A_{\mu} - \partial_{\mu} \theta[/math] Here, we can see our four-vector potential again [math]A_{\mu}[/math] - you can basically build the electromagnetic fields form this. It's time component in [math]A_0[/math] but that is really not relevant right now. Our Covariant Derivative and the respective conjugate fields are usually denoted as [math]D_{\mu} \phi = \partial_{\mu} \phi + iA_{\mu} \phi[/math] and [math]D_{\mu} \phi^{*} = \partial_{\mu} \phi^{*} - iA_{\mu} \phi^{*}[/math] So calculating it all together, we just define the whole thing again as: [math]D\phi' = (\partial \phi + i\phi \frac{\partial \theta}{\partial x})e^{i\theta} + i(A_{\mu} - \partial_{\mu} \theta) \phi e^{i \theta}[/math] Well, with this, we can see straight away that some terms cancel out. The [math]i\phi[/math] terms cancel, and [math]\frac{\partial \theta}{\partial x}[/math] is in fact the same as [math]\partial_{\mu} \theta[/math]. So what we are really left with is [math]D\phi' = D \phi e^{i\theta}[/math] and so its conjugate is [math]D\phi^{*} = D\phi^{*} e^{-i\theta}[/math] Again, the latter terms cancel out when you multiply these two together and so what you end up with is [math]D \phi^{*}' D \phi ' = D \phi^{*}D \phi[/math] and so by using the Covariant Derivative, we have been able to restore the lost [math]U(1)[/math] symmetry where [math]U(1)[/math] symmetries deal with rotations. But when physicists talk about a mass, we don't want to retain symmetry in these fields. In fact, the very presence of a mass term will imply an explicit symmetry breaking. The process of course, is a little more complicated however. It involves another boson, called a Goldstone Boson, which can be thought of as a ground-state photon which lives in the minimum of a Mexican Hat potential. Something which exists in the minimum is the same as saying our system does not contain a mass term and so [math]\phi=0[/math]. In such a potential, we may describe our field as [math]\phi = \rho e^{i\alpha}[/math] Here, [math]\rho[/math] is a deviation from the ground state. [math]\rho[/math] is in fact our Higgs Boson and [math]\alpha[/math] is our Goldstone Boson. If [math]\alpha[/math] is a frozen (constant) field, then there are no changes in the equations. But if there is a deviation of the Goldstone Boson from the minimum of our potential, then we are saying that it costs energy to do so, and this energy is what we mean by particles like a photon obtaining a mass. In fact, the Goldstone Boson is gobbled up by the Higgs Boson which gives the system we speak about a mass. Our flucuation from the minimum has the identity [math]f \ne 0[/math] where [math]f[/math] plays the role of mass. Let's discuss this mass term in terms itself of the electromagnetic field tensor. Such a tensor looks like: [math]F_{\mu \nu}F^{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}[/math] here, [math]F_{\mu \nu}F^{\mu \nu}[/math] is in fact just [math]F^2[/math] and it makes up the langrangian. [math]F_{\mu \nu}[/math] is an antisymmetric object with respect to swapping its indices. It is like a four dimensional curl. It is Gauge Invariant, but checking that, remember our transformation [math]A' \rightarrow A - \partial \theta[/math] plugging that into the tensor gives: [math]F_{\mu \nu}' = \partial_{\mu}A'_{\nu} - \partial_{\nu}A'_{\mu}[/math] That is [math]\partial_{\nu} \partial_{\mu} A_{\nu} - \partial_{\mu} \partial_{\nu} \theta - \partial_{\mu} \partial_{\nu}A_{\mu} + \partial_{\nu} \partial_{\mu} \theta[/math] Since the order of partial differentiation doesn't matter, the [math]-\partial{\mu} \partial_{\nu}\theta[/math] and [math]\partial_{\nu}\partial_{\mu} \theta[/math] cancel and what you are left with is invariance. There are some numerical factors left out of this, such as a quarter, but that really isn't all that important in this demonstration. And so, we may have a completely Gauge Invariant term [math]D_{\mu}\phi D_{\mu} \phi^{*} - V(\phi^{*}\phi) + F_{\mu \nu} F^{\mu \nu}[/math] One should keep in mind that the potential term here [math]V(\phi \phi^{*})[/math] is also manifestly invariant. What becomes interesting however, is the question of what cannot be added to this but still can remain Gauge Invariant. What we cannot add to the electromagntic part ([math]F^2[/math])-part, is this: [math]\frac{M^2}{2}A_{\mu}A^{\mu}[/math] This is just the same as [math]\frac{M^2 A^{2}_{\mu}}{2}[/math]. The reason why, is because a mass-term for a photon would typically look like this. This actually breaks local invariance. In nature there is only one massless spin-1 boson - the photon - then it seems that if we where to use a local gauge group that is less trivial than [math]U(1)[/math], then the symmetry would be broken. It's interesting, that when we speak about symmetry breaking for a photon, we are in fact talking about an electromagnetic term which is squared in the Langrangian. It seems, by no far shot, that physicists will not abandon the use of symmetry breaking to describe mass for particles, but it must be noted that our usual, and perhaps, most insightful way out of the problem, the Higgs Boson, may need some drastic changes. As I presented at the beginning of this thread, we have dealt with theories involving electromagnetic theories of mass. If this idea is not recollected, what will we turn to? There are some alternatives. These actually go by the specific name of ''Higgless models''. We have actually a reasonable list of different theories which can be read about here:http://en.wikipedia....Higgsless_model . But what theory could entertain a gravitational charge best? I have decided that there is a theory which might be able to satisfy the speculations of treating mass as a charge on a system. Long before I learned about the Self-Creation Cosmology, I realized that this was an idea I had developed in my own head independantly, which is why it has me very excited. The development of Self-Creation Cosmology (SCC) should historically stem from the work of Nordstrom, who developed a relationship for matter and the gravitational field. In his relation, matter actually depended on the gravitational field: [math]\Box \phi = 4 \pi \rho[/math] In this equation, we can replace the density for the stress-energy tensor [math]\Box \phi = 4\pi \lambda T_{\mu \nu}[/math] In the (SCC) approach, we may have a trace free solution [math]\nabla_{\mu}T_{M \nu}^{\mu} = 4\pi f(\phi) = f(\phi) \Box \phi[/math] The is almost identical to an approach made by Ni [math]\eta_{\mu \nu} \partial_{\mu}\partial_{\nu} \phi = 4 \pi \rho k(\phi)[/math] Where the function [math]k(\phi)[/math] is written slightly different, but which is still related to the metric. The density of a particle, I once speculated, might just be seen as [math]\frac{1}{4\pi\lambda}\Box \phi = T_{\mu \nu}[/math] Essentially, the gravitational field gives rise to the creation of mass. Interestingly, my own argument for a similar model was that a particle coupled to the gravitational field will experience a gravitational charge, similar to how a particle experiences an electromagnetic charge when moving in an Electromagnetic field. The Gravitational Charge The gravitational charge has already been given with the expression of [math]\sqrt{G} M[/math] by Motz in his paper ''On the Quantization of Mass'' [3] In my own studies, I derived the gravitational charge as the following relationship: [math]\sqrt{G}M = \sqrt{E_gr_s}[/math] where [math]r_s[/math] is the Schwarzschild radius and [math]E_g[/math] is the gravitational energy. The gravitational energy is interpreted as the intrinsic gravitational energy due to charge, which for a photon would be zero. The gravitational charge then, can be thought of as the quantity of inertia inherent within a particle and you may even derive the relationship of the proper density of a particle which will be shown soon. The energy therefore can be given as [math]E_g = \frac{GM^2}{r_s}[/math] This may even be seen as a potential energy equation [math]\frac{GM^2}{r_s} = M\phi[/math] The question then is whether mass comes from a potential energy? The prototypical case of color charges in quark condensate also get masses from a potential energy. Is it possible that when we speak of a system getting a mass, is the mass a by-product of potential gravitational energy? One thing to keep in mind, is that the Schwartzschild metric describes bodies with mass. Therefore, it might be slightly interesting to realize that when you make [math]r_s \rightarrow 0[/math] in a limit, it may describe the non-presence of a gravitational charge of something, [math]\sqrt{G}M = \lim_{r_s \rightarrow 0} \sqrt{E_g r_s} = 0[/math] Would be a true statement for massless radiation. Photons do not decay spontaneously in space. However, using a catalyst like an electron or maybe even a photon collision, particles with matter can be created [math]\gamma \gamma \rightarrow e^{-}e^{+}[/math] So it seems it is not entirely unreasonable to believe that such photon collisions create a matter under what would be a special type of energy transition. Given enough energy in KeV, the energy transition will create particles with mass. The Planck relationship [math]\hbar c = GM^2[/math] Can yeild the Compton wave length and the Schwartzschild radius simultaneously by dividing both sides of this equation by [math]Mc^2[/math] [math]\frac{\hbar c}{Mc^2} = \frac{GM^2}{Mc^2} = (\frac{\hbar}{Mc} = \frac{GM}{c^2})[/math] The proper density of a particle can be given in a relationship as [math]8\pi \rho_0 (\frac{G}{c^2})[/math] and this can be set equal with the radius or the Compton wave length. The gravitational charge from the energy radius relationship can derive the following equation [math]\frac{GM^2}{Mc^2} = r_s[/math] which is the gravitational case analogue of the classical electron radius [math]\frac{e^2}{Mc^2} = R[/math] Increasing the gravitational charge on a system would be effectively increasing the inertial mass quantity of a system. In the Heirarchy problem, we do realize that there are many different masses in the Standard Model. During the energy transition, we may take a Taylor series on the gravitational energy and increase it according to the amount of energy which is contributing to the mass of the system. In this work, I have not derived an understanding mathematical how one might be able to approach the idea that mass is a charge as a charged particle would be as it moved through its respective field - in this case, the gravitational field. That is a difficult task but I do have a proposition to head in that direction. A mass operator takes into account of the interaction of a particle with it's own field but also including other fields if necessery. It has been shown already that in many senses, mass is an electromagnetic phenomenon from first principles and that charged particles do possess what is called an electromagnetic inertia when being accelerated through spacetime. In fact, we may believe that a portion of the mass of a system might exist as a contribution of electromagnetic phenomena. Then some of the rest of the inertial mass is purely the charge of system moving in it's respective field. So it would be a good approach to use a mass operator to describe both the gravitational field contribution of mass and the respective electromagnetic field. I will most likely write up more later, but there is so much in this work it might be best to chew over the first installment first. [1] - http://en.wikipedia....romagnetic_mass http://en.wikipedia....romagnetic_mass [2] http://www.wiley-vch...7408355_c01.pdf [3] http://www.gravityre...d/1971/motz.pdf (The work in which these investigations stemmed from) I see this place uses [math] tag latex. I will change it now. Sorry for the delay. Has this post been moved, I can't find it now? Edited June 8, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

studiot Posted June 8, 2012 Share Posted June 8, 2012 How does this theory, based on Larmors equation account for the mass of the neutron? Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 9, 2012 Author Share Posted June 9, 2012 (edited) How does this theory, based on Larmors equation account for the mass of the neutron? I'll be honest. I don't know because I have not studied these equations for it. If I come across one, which I am looking into as we speak, I will be sure to tell you of one. How does this theory, based on Larmors equation account for the mass of the neutron? Are you aware of any technique in which the Larmor equation accounts for the mass of any particle? As far as my memory recollects, the Larmor equation is [math]\omega = \gamma B_0[/math] Or do you speak of Larmors formula even? That would make more sense because it has been spoken about in the work, very early on? (I called it Larmor's equation, but it can be a bit confusing as different authors call it by different names.) Edited June 9, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

juanrga Posted June 9, 2012 Share Posted June 9, 2012 Mass in our most up-to-date theory was beleived to be a by-product of a Boson and its respective field. Finding this particle, called the Higgs Boson has proven difficult. It seems likely that scientists will have to find a new mechanism for mass - and in this work, we will advocate one of those theories. There are rumours that the announcement of the experimental finding of the Higgs boson will be made this year. It seems that the LHC has found a Higgs particle about the 125 GeV region. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 9, 2012 Author Share Posted June 9, 2012 There are rumours that the announcement of the experimental finding of the Higgs boson will be made this year. It seems that the LHC has found a Higgs particle about the 125 GeV region. It will be surprising if they have. Rumours are just that though... rumours. I'll be very interested to see if they hold any truth. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 10, 2012 Author Share Posted June 10, 2012 Where is this thread in the forum, I can't find it anywhere? Link to comment Share on other sites More sharing options...

Spyman Posted June 11, 2012 Share Posted June 11, 2012 Where is this thread in the forum, I can't find it anywhere? Science Forums > Sciences > Physics > A forgotten theory of mass Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 13, 2012 Author Share Posted June 13, 2012 (edited) This next derivation, as I continue this work, will be concerned with the quantization condition on the mass of a particle (the gravitational charge). This was discussed by Motz in his paper, which is referenced in post 1. We begin with the equation [math]G = \frac{rc^2}{2Gt^2}[/math] which can be found in [1]. Taking the square root of both side and rearranging yields and then multiplying through by [math]\sqrt{M}[/math], [math]\sqrt{G}M = \sqrt{\frac{Mr^3}{2t^2}}[/math] Which is the gravitational charge. As shown in the beginning of this work, I equated the gravitational charge to a gravitational energy multiplied by the schwartschild radius, [math]\sqrt{G}M = \sqrt{E_gr_s}[/math] we simply substitute now [math]\sqrt{\frac{Mr_{s}^{3}}{2t^2}} = \sqrt{E_g r_s}[/math] and let [math]r = r_s[/math] on the left hand side of the equation. Now multiply through [math]2t^2[/math] [math]\sqrt{Mr_{s}^{3}} = \sqrt{E_g r_s 2t^2}[/math] Seeing the gravitational energy for it's relativistic equality [math]Mc^2[/math] and then cancelling out some of our terms on both sides of the equation yields [math]\sqrt{r_{s}^{2}} = \sqrt{c^2 2t^2}[/math] We can fudge this result for [math]r_{s} = \sqrt{c^2 2t^2}[/math] since there is no such thing as a negative radius. Notice this derivation is the mathematical equivalent of the Minkowski relation, where here we are swapping the radius for the affine length [math]x=ct[/math] Using the relationship [math]c = \omega r_s[/math] (which I could derive on request) we have [math]r_{s} = \sqrt{2t^2 \omega^2 r_{s}^{2}}[/math] We may take the cross product for the frequency and the radius [math]r_{s} = \sqrt{2t^2 (\omega^2 \times r_{s}^{2})}[/math] Multipling energy on both sides gives the square of the gravitational charge again [math]E_g r_{s} = Mc^2\sqrt{2t^2 (\omega^2 \times r_{s}^{2})}[/math] Square everything we have [math]G^2M^4 = M^2c^4 2t^2 (\omega^2 \times r_{s}^{2})[/math] Rearrange [math]\frac{G^2M^4}{2M^2c^4 t^2} = (\omega^2 \times r_{s}^{2})[/math] This is just [math]\frac{G^2M^4}{2\hbar^2} = (\omega^2 \times r^2)[/math] Halving this all out we have finally the quantization condition on the mass with the exception that the right handside is expressed in terms of a cross product between the angular frequency and the radius [math]\frac{GM^2}{\hbar} = (\omega \times r_s)[/math] Here is a link which explains the cross product between the angular frequency and the radius. http://www.physics.s.../360notes12.pdf [1] http://arxiv.org/ftp/arxiv/papers/0810/0810.1629.pdf Edited June 13, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 18, 2012 Author Share Posted June 18, 2012 Here's a good paper on electromagnetic mass - a topic spoke about in the OP. http://ivanik3.narod.ru/EMagnitizm/JornalPape/ParadocsCullwick/hnizdo_ajp_65_55_97.pdf Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 18, 2012 Author Share Posted June 18, 2012 (edited) [math]\frac{GM^2}{\hbar} = (\omega \times r_s)[/math] Since [math]\omega[/math] will lye along the axis of rotation, and if [math]r_s[/math] defines the radius of our system which could or may not be set equal to the Compton Wavelength, and if this is related to the quantized condition [math]\hbar = \frac{GM^2}{c}[/math] Then I ask, is there some kind of spin relationship between the square of the gravitational charge of a system (perhaps along the same lines as a Gravi-electromagnetic force) with the angular spin of a particle? We'd have to start thinking about inertia as well, even in a classical sense. One derivation I came to followed the lines: [math]M=\frac{r^3}{2Gt^2}[/math] multiply M on both sides and then rearrange [math]2t GM^2 =Mr^3[/math] What we established from my derivation in the OP was [math]GM^2 = E_gr_s[/math] Set [math]r=r_s[/math] again, and substitute meanings and then divide by radius we get [math]2Et^2 = Mr^2[/math] It turns out after a little investigation that [math]Et^2[/math] is defined as an inertia of bodies from two separate sources: http://books.google....squared&f=false http://www.worldscin...5902000638.html (except) in my derivation, we have an interesting factor of [math]2[/math] showing up. The reason why this could have important relevance, (in this derivation), is because [math]2Et^2[/math] has been set equal with [math]Mr^2[/math]. Anyone even with a slight knowledge of ''inertia physics'' will know that [math]Mr^2[/math] is in fact the rotational inertia of a body. (and just to add) Here is a paper by Schwinger who was noted by lloyd motz in his own derivation of [math]\sqrt{G}M[/math] (the square root of the gravitational charge) - an interesting read on similar approaches by Lloyd and a direct importance with my own work http://wdxy.hubu.edu...10535154942.pdf I also derive a generic integral form of [math]\frac{GM^2}{\hbar} = \frac{\int \sqrt{Fds}}{\sqrt{\frac{1}{v}} \int \sqrt{Fdt - m_0}}[/math] To define the inertia of a moving body where [math]M_0[/math] is a rest mass term. This would mean then that the gravitational charge can be computed in terms of a rest mass given as (or understanding it with terms of rest mass) as: [math]GM^2 = \hbar(\frac{\int \sqrt{Fds}}{\sqrt{\frac{1}{v}} \int \sqrt{Fdt - m_0}})[/math] Albeit, it looks like a complicated mess. http://mathpages.com/rr/s2-03/2-03.htm Edited June 18, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 19, 2012 Author Share Posted June 19, 2012 (edited) Writing the above in a simpler form, [math]GM^2 = \hbar( \frac{E}{m-m_0})[/math] The extra energy of any moving particle is directly related to an increase of the energy of a system - Indeed, this is what led Einstein to the idea that perhaps all inertia is is the energy of some system. I believe this is true as well. I defined it as a special type of energy, the gravitational energy of the system [math]E_g[/math], to differentiate from non-rest energies. This must mean that the gravitational charge must vary as well proportionally to moving particles. Edited June 19, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 19, 2012 Author Share Posted June 19, 2012 (edited) Since the gravitational charge is the mass which can change due to velocity, I feel the need to quote Taylor and Wheeler here: ''The concept of 'relativistic mass' is subject to misunderstanding. That's why we don't use it. First, it applies the name mass - belonging to the magnitude of a 4-vector - to a very different concept, the time component of a 4-vector. Second, it makes increase of energy of an object with velocity or momentum appear to be connected with some change in internal structure of the object. In reality, the increase of energy with velocity originates not in the object but in the geometric properties of space-time itself.'' If they are correct, then the gravitational charge cannot change due to velocity because the gravitational charge is treated in this work as an intrinsic internal property. Instead, the gravitational charge only appears to change because of a change in energy which is resultant from the spacetime geometry. Which is a fascinating thought. Edited June 19, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 24, 2012 Author Share Posted June 24, 2012 Now, just as Motz had made clear in his paper, the Schwinger Quantization method for charged particles in fields takes on a remarkable similarity to the gravitational charge [math]Gm^2 = \hbar c[/math]. The link here will explain the charge condition: http://encyclopedia2.thefreedictionary.com/Dirac-Zwanziger-Schwinger+quantization+condition What we essentially have is [math]\frac{e\mu}{c} = \frac{1}{2}n\hbar[/math] The gravitational charge quantization is [math]\frac{Gm^2}{c}= \hbar[/math] So motz was absolutely correct in stating the importance between the two equations. Link to comment Share on other sites More sharing options...

ydoaPs Posted June 24, 2012 Share Posted June 24, 2012 There are rumours that the announcement of the experimental finding of the Higgs boson will be made this year. It seems that the LHC has found a Higgs particle about the 125 GeV region. The results from this year are actually just confirmation that they found it last year. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 24, 2012 Author Share Posted June 24, 2012 The results from this year are actually just confirmation that they found it last year. ''Think'' they found it and none of the results are conclusive. It will need more tests. http://www.guardian.co.uk/science/2011/dec/13/higgs-boson-glimpsed-cern-scientists Link to comment Share on other sites More sharing options...

ydoaPs Posted June 24, 2012 Share Posted June 24, 2012 (edited) ''Think'' they found it and none of the results are conclusive. It will need more tests. http://www.guardian.co.uk/science/2011/dec/13/higgs-boson-glimpsed-cern-scientists Two 5 sigma results confirming the 3 sigma results from last year is a FAR cry from inconclusive. Edited June 24, 2012 by ydoaPs Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 24, 2012 Author Share Posted June 24, 2012 Two 5 sigma results confirming the 3 sigma results from last year is a FAR cry from inconclusive. If the results where conclusive,we would not be sitting here debating this. It would have been on the screens of news reports and I wouldn't have created this thread. No doubt, the information gathered will have to pass a great many minds before anything conclusive is brought forward. Link to comment Share on other sites More sharing options...

ydoaPs Posted June 25, 2012 Share Posted June 25, 2012 If the results where conclusive,we would not be sitting here debating this. I'm gonna call BS on that one. Have you never heard of Climate Change Deniers or Young Earth Creationists? Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 25, 2012 Author Share Posted June 25, 2012 I'm gonna call BS on that one. Have you never heard of Climate Change Deniers or Young Earth Creationists? If the Higgs Boson has been found conclusively, will you show me some conclusive evidence for this statement, otherwise, I call BS on you. Link to comment Share on other sites More sharing options...

ydoaPs Posted June 25, 2012 Share Posted June 25, 2012 If the Higgs Boson has been found conclusively, will you show me some conclusive evidence for this statement, otherwise, I call BS on you. How on earth do you think that two 5 sigma results confirming two 3 sigma results is inconclusive? Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 25, 2012 Author Share Posted June 25, 2012 (edited) How on earth do you think that two 5 sigma results confirming two 3 sigma results is inconclusive? Did you find me a link yet saying they have definitely found a Higgs Boson...No? I thought so. Start understanding what the meaning of the word ''conclusive'' means. When scientists are absolutely sure, they will not hesitate publishing such a find... outside of rumors. I will not believe rumors until I see something which collaborates your own assertions here. I don't deny that Higgs may have been found, but as I said, nothing conclusive can be said about it because no one has announced it. Edited June 25, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

studiot Posted June 28, 2012 Share Posted June 28, 2012 In a sense, mass has been shown to be an electrical phenomenon derived from first principles. Please explain the relationship of this statement to particles without electric charge. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 28, 2012 Author Share Posted June 28, 2012 Please explain the relationship of this statement to particles without electric charge. Particles without an electric charge still have a magnetic charge. The neutron is electrically neutral but still possesses a magnetic charge as a magnetic moment. I don't really know what you are getting at. So while the Neutron for instance, may not have an electric charge, it is not fundamental. The stuff it is made of does. Link to comment Share on other sites More sharing options...

studiot Posted June 28, 2012 Share Posted June 28, 2012 So I ask you once again. You have stated in so many words exactly and precisely that mass is an electrical phenomenon. I disagree and ask you to prove your claim. Please address the question asked not something else. Link to comment Share on other sites More sharing options...

pmb Posted June 28, 2012 Share Posted June 28, 2012 So I ask you once again. You have stated in so many words exactly and precisely that mass is an electrical phenomenon. I disagree and ask you to prove your claim. Please address the question asked not something else. A counter example is the contribution to the mass of a body due to thermal energy. That mass increase is due to kinetic energy, not electromagnetic energy. Link to comment Share on other sites More sharing options...

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