Aethelwulf Posted March 30, 2013 Report Posted March 30, 2013 (edited) In my previous work on the gravitational charge [1] definition given by numerical reference, I discovered that there was in fact two coupling field equations, one concentrated on the charge of the system and the other concerned with the definition of the gravimagnetic field aka. the Coriolis Field [math]-\frac{2 \omega \times v}{\sqrt{G}}[/math]. I found accidently at the time, this was related to a cross product of two correlated classical equations [math]\frac{F}{\sqrt{G}m} = 2m \epsilon_{ijk} \omega_j v_k[/math] [math]-\frac{2 \omega \times v}{\sqrt{G}} = \epsilon_{ijk} v_jB_k[/math] The two equations are unconventionally coupled by being complete results of each other [math]\frac{F}{\sqrt{G}m} = \frac{2 \omega \times v}{\sqrt{G}}[/math] Except for a sign change. You obtain this equation by understanding that the Coriolis Force was simply [math]F = 2m \times v[/math] Therefore you divide this by the square root of the gravitational charge [math]\sqrt{Gm^2}[/math] to obtain [math]\frac{F}{\sqrt{G}m} = \frac{2 \omega \times v}{\sqrt{G}}[/math] The sign change may appear in this theory concernning the two equations linking the gravitational charge and related coriolis cross product with the coriolis field definition and the Newtonian definition of gravity is becuase at this fundamental point, we are seeing (by no surprise at is well-known) changes in the magnetic field which orientate themselves in their line of path. There are of course, three components to the magnetic field [math](B_x, B_y, B_z)[/math] and they result in different conditions for each of the ''coupled equations'' (*). This will lead... and now I am using arithmatic of the brain hereb ;) ... there should be six solutions for the magnetic field for a free field particle, twelve for a relativistic case. * I use an asterisk to point outside of distracting my discussion, is that the term ''coupled equations'' is not to intend the highschool definition of the mathematical meaning. These equations are coupled through a so-called ''interaction term'' which will resemble something like [math]i e \phi[/math] but we will come to this soon enough in the position this was sited from. I will continue to use these asterisks to divert your attention to find them a few paragraphs away normally so I am not mixing too many subjects together. The gravitational charge is not only coupled to the magnetic field [constantly] but the magnetic field is a by-product of the gravitational charge itself. In an investigation I done recently, it seems very possble that mass is the same thing as talking about charge itself, in the sense that you cannot have a charge without mass. Both are dependantly relativistic properties of particles which have the property of inertia. In another investigation I made, I showed that inertia is caused by the inherent resisting of the changes of energy internally for these systems. If this is correct, (not including the inertial part this far ahead) but true for the coupling of the magnetic field to the gravitational through these cross products in the equations (1) [math]\frac{F}{\sqrt{G}m} = \frac{2 \omega \times v}{\sqrt{G}}[/math] (2) [math]-\frac{2 \omega \times v}{\sqrt{G}} = \epsilon_{ijk} v_jB_k[/math] ... then they must be doing so through the gravitational charge being proportional to the gravitational potential [math]\phi[/math]. Using intuition, not derivation, I found a component of such a coupling. It is slightly based on a little faith though.... we have known for a while there are similarities between the gravitational and electromagnetical fields. In the work I provided investigating Planck Particles I realized that equation (1) above which is the definition of the gravitational field (in an analogue sense) [math]\frac{F}{\sqrt{G}m}[/math] because this was the same as the definition of the electromagnetic file definition [math]\mathbf{E} = \frac{F}{q}[/math] Where [math]q[/math] is the electromagnetic charge and [math]\sqrt{G}m[/math] was the gravitational charge (which depends on you units because) [math]e = \sqrt{4 \pi \epsilon GM^2}[/math] Anyway, the fact it is analogue, we may assume that the coupling of the gravitational charge [math]\sqrt{G}m[/math] to the gravitational gradient [math]\phi[/math] is why a gravitational charge is experienced in the first place. Just like why an electron has an electric charge, it is because it is moving through and electromagnetic field. So since the definition of the gravitational charge (is mass itself) could it also be arising because of a coupling to the gravitational field when in motion? Keep in mind, no particle is ever truly at rest but photons are an exception to this coupling because they have not been spontaneously broken within the symmetries of the mathematics describing their gauge fields - but this is of course abstract but this is what physicists mean when they talk about the differences to why a photon does not have a mass and therefore an inertia. The coupling equation which makes most sense (since it is similar to the coupling of the electric charge to the electromagnetic four potential [math]A_{mu}[/math] which describes its own field) would be replaced by a gravitational four-field, also a gradient potential. If we adopt units [math]e = \sqrt{\epsilon \mu Gm^2}[/math] then we have the gravimass coupling as [math]-e\phi \partial_{\mu} \psi = 2i\sqrt{\pi \epsilon Gm^2} \phi_{\mu} \frac{\partial \psi}{\partial X^{\mu}}\psi[/math] The idea then, is that the gravitational field provides the particles with a charge of mass. The gravitostatic equation of contribution of energy to mass would be [math]E = \frac{1}{2} \frac{GM^2}{r_s}[/math] This keeps as the gravitational analogue of [math]E_{EM} = \frac{1}{2} \frac{e^2}{4 \pi \epsilon R_{classical}}[/math] The contribution of mass to a system which has undergone a symmetry change mathematically in my equation is gained through the as [math]M = \frac{1}{2} \frac{GM^2}{r_s c^2}[/math] where [math]GM^2[/math] is the squared gravitational charge. Now going back to a similar process to Wein (1900), the attraction of the gravitational field can be understood as [math]G \frac{\frac{1}{2} \frac{GM^2}{r_s c^2} M}{R}[/math] It is therefore a condition under this theory that the mass of a system [math]M[/math] depends on the gravitational strength of the gradient [math]\phi[/math] [math]M\phi = \frac{G M^2(\phi)}{r_s}[/math] Where [math]GM^2[/math] is the gravitational charge (mass) squared and [math]r_s[/math] is the Schwartzchild radius. Also from Motz' paper, 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] I correct this equation for a more accurate representation. I know physicists like to simplify equations as best as possible. Taking the general gist of his argument, [math]R[/math] the curvature can be more understood in terms of the second order Dirac operator, namely [math]D \psi(x)= \hbar^2 R^{-2}\psi(x)[/math] If one factors this equation in terms of the gamma matrices four tuple [math]\gamma^{\mu}[/math] we obtain [math]i\gamma^{\mu}\nabla_{\mu}R = 1[/math] from which Motz deduces that [math]R[/math] must be the squared spacetime interval. Now as I explained, for a Planck Particle, one must set the Compton wavelength equal to the Schwarzschild radius. This means that the Gaussian curvature equation can be rewritten in this form [math]8 \pi \rho (\frac{G}{c^2}) = (\frac{12GM}{c^2})^{-2}[/math] (note this condition does not hold for Motz' outdated Uniton particle) The factor of [math]12[/math] arises because there is a coefficient of [math]2[/math] on the gravitational parameter which makes the Schwarzschild radius. Since this simply has dimensions of [math]R^{-2}[/math] then this means we can rewrite the Dirac Operator as [math]D \psi(x)= \hbar^2 (\frac{12GM}{c^2})^{-2}\psi(x)[/math] To finish up this little essay, the canonical relativistic momentum is given as [math]i\hbar \gamma^0[/math] (using the famous gamma matrices). Multiplying invariantly through by the speed of light, you obtain the gravitational charge [math]i \hbar c \gamma^0 = \begin{pmatrix} iGM^2 & 0 & & 0 \\0 & iGM^2 & 0 & 0 \\0 & 0 & -iGM^2 & 0 \\0 & 0 & 0 & -iGM^2 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}[/math] [math](\gamma^0)^2[/math] just punches out the unitary matrix, basically ignorable then, and squaring we remove the imaginary parts [math]-\hbar^2 c^2 \mathbb{I} = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2 \cdot \gamma^0[/math] This finds a relation through [math]\hbar c[/math] and [math]GM^2[/math] than simply setting the equal. We find they are related through a kind of symmetry which can be seen within a matrix analysis: [math]a^k = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \end{pmatrix}[/math] so calculating it we get [math]\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \end{pmatrix}[/math] To solve the right hand side completely, the final matrix form will look like [math]-\hbar^2 c^2 \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & -GM^2 & 0 \\0 & 0 & 0 & GM^2 \\0 & -GM^2 & 0 & 0 \\-GM^2 & 0 & 0 & 0 \end{pmatrix}^2[/math] Now this is just one result, by involving pauli matrices [math]\sigma_1[/math] and [math]\sigma_3[/math]. We should solve the rest of matrix possibilities - this will take up a lot of space, so I will keep the calculations at a minimum. Before I even undertake that task, I want to show the equation above in a more simplified version: [math]\begin{pmatrix} 0_2 & -\hbar^2c^2 \cdot \sigma^1 \\ \hbar^2 c^2 \cdot \sigma^3 & 0_2 \end{pmatrix} = \begin{pmatrix} 0 & 0 & -GM^2 & 0 \\0 & 0 & 0 & GM^2 \\0 & -GM^2 & 0 & 0 \\-GM^2 & 0 & 0 & 0 \end{pmatrix}^2[/math] where the full form of [math]\begin{pmatrix} 0_2 & -\hbar^2c^2 \cdot \sigma^1 \\ \hbar^2 c^2 \cdot \sigma^3 & 0_2 \end{pmatrix}[/math] is [math]\begin{pmatrix} 0 & 0 & 0 & -\hbar^2 c^2 \\0 & 0 & -\hbar^2 c^2 & 0 \\-\hbar^2 c^2 & 0 & 0 & 0 \\0 & \hbar^2c^2 & 0 & 0 \end{pmatrix}[/math] An interesting symmetry revealed itself in these matrix equations. [math]\begin{pmatrix} 0 & 0 & 0 & -\hbar^2 c^2 \\0 & 0 & -\hbar^2 c^2 & 0 \\-\hbar^2 c^2 & 0 & 0 & 0 \\0 & \hbar^2c^2 & 0 & 0 \end{pmatrix}=\begin{pmatrix} 0 & 0 & -GM^2 & 0 \\0 & 0 & 0 & GM^2 \\0 & -GM^2 & 0 & 0 \\-GM^2 & 0 & 0 & 0 \end{pmatrix}^2[/math] The quantity [math]\hbar c[/math] which is purely quantum mechanical is symmetrically related to [math]GM^2[/math] the quantity we have well-established as the gravitational charga squared. If I didn't know better, I would say this is an accident, but being a bit more scientifically-minded than my other self ;) I don't believe this symmetry appears by accident but instead the math is telling me it could be a symmetry of nature. [math]\hbar c[/math] is called the ''Planck Charge'' and therefore [math]GM^2[/math] must be it's mass equivalent. the gravitational constant however [math]G[/math] can take on either high or low values. If [math]\hbar c = GM^2[/math] then [math]G = \frac{\hbar c}{M^2}[/math] (MUST be) taking on large values inside of the particle (***). In this work we would normally consider a classical radius. Since gravity is always positive in nature [math]|\Gamma|[/math], this would essentially mean that gravity takes on a huge value inside of it's own sphere of radius/curvature [math]\frac{e^2}{Mc^2}[/math], which on a warm summers night last year I realized could be actually the real form of the Poncare Stress which neutralizes the electromagnetic stress inside of the particle, apparently stopping it from ripping apart. (**) The justification to think we might be dealing with two solutions is that it is common for physicists to take the WDW-equation for the famous ''scale factor'' of the Friedman Universe Model and obtain two phase matching equations [math]e^{i\omega t}[/math] and [math]e^{ikx}[/math]. In the situation of gravity, we might be looking at a similar situation even under all the asbtract pretense. Another thing I wish to note is the existence of a mathematical procedure which many mathematicians are unaware of, called super-complex numbers. They basically and on cannot miss, effectively keep a system positive. This was a line I will probably investigate in the future. (***) http://www.gravityresearchfoundation.org/pdf/awarded/1971/motz Apart from my matrix mathematics discovery, it means that Planck Charge and the Gravitatioinal Charge have components which are anticommutative. A perfect anti-commuting property as a demonstration [math][\hat{P},\hat{x}] = i\hbar[/math] The [....] will be given as Poisson brackets since I don't know how to write in latex This uses a mathematical notation called Poisson Brackets describing commutators. I won't go too much into how this quantization method is accomplised, only that the operators of momentum [math]\hat{p}[/math] and position [math]\hat{x}[/math] do not commute. My matric analysis kind of proves at least mathematically, that the charge definitions [math]\hbar c[/math] and [/math]GM^2[/math] are not equal without a consideration of them being non-commutative.... To be commutative can mean several concepts or variations of commutative operations. The one concentrated on here, the anticommutativity can be relatively easy to understand that the gravitational charge and the Planck Charge are belonging in different groups using set theory. Not my specialist point by far, but I know that if [math]\mathcal{G}[/math] is a group pertaining [math]\sqrt{G}m[/math] the inertia of any of the systems contained therewithin equal also to its inverse allows a double index [math](i.j)[/math]. Both indexes are real under a superpositioning [math](i + j)[/math] but are subject to the Kronecker delta as my coupling expression had tried to explain. Certainly I am no expert on mathematics, but I do see a clear suspicion within the context of things. For there to be cross products in the equations.... (1) [math]\frac{F}{\sqrt{G}m} = \frac{2 \omega \times v}{\sqrt{G}}[/math] (2) [math]-\frac{2 \omega \times v}{\sqrt{G}} = \epsilon_{ijk} v_jB_k[/math] It seems then, that quantum action (spin) is not only related to the magnetic field but is intrinsically related to gravitational field. The more mass you have, the more of a coupling you have to the charge itself [math]M = \sqrt{\frac{\hbar c}{M}}[/math] influencing the hierarchy of the different mass scales, or at least in this primitive theory. But perhaps the most and important thing to note is that (1*) [math]\gamma^5 = i\bar{\gamma}^{1_{2}} \bar{\gamma^{1_{2}}} \gamma^2 \gamma^3[/math] The matrices not only dictate that [math]\hbar c[/math] and [math]GM^2[/math] anticommutate in an uncertainty relationship which we will get to like momentum and position, but the matrix equations from (1*) shows that it is inherently entangled into the beauty of gamma matrices spitting out perhaps why Chirality even appears in fundamental systems. If you want the derivation to this breakthrough, you must ask and I will provide. Right now, I want to change tracks to a different subject still related to equation denoted (1*). To end, perhaps the most amazing feature of investigating Pauli Matrices on the gravitational charge definition [math]\hbar c = GM^2[/math] found [math]M = \begin{pmatrix} 0 & 0 & 0 & 1 \\0 & 0 & 1 & 0 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix}[/math] My gamma matrix was given the name [math]\gamma^{1_{2}} = \begin{pmatrix} \sigma^1 & 0_2 \\ 0_2 & \mathbf{1}_2 \end{pmatrix}[/math] to represent the entry of the submatrix [math]\sigma^1[/math] and it's other entry, [math]\mathbf{1}_2[/math]. To someone even as mathematical as I am this doesn't make much sense, but it comes from a beautiful symmetry in theory where you can find this by calling it [math]M[/math] for our matrix - the matrix [math]M[/math] is then just [math]\begin{pmatrix} \mathbf{1}_2 & 0_2 \\ 0_2 & \sigma^1 \end{pmatrix}[/math] So it's diagonal entries are reversed to the the other new gamma matrix [math]\begin{pmatrix} \sigma^1 & 0_2 \\ 0_2 & \mathbf{1}_2 \end{pmatrix}[/math] Edited March 31, 2013 by Aethelwulf Quote
Aethelwulf Posted March 30, 2013 Author Report Posted March 30, 2013 One way [math]\hbar c[/math] could be anticommutative with [math]Gm^2[/math] through a dynamical operator which does not commute with the mass [math]m[/math]. http://www.gravityresearchfoundation.org/pdf/awarded/1971/motz.pdf In this work, we find from Motz that from a three dimensional hypersphere for a particle of mass, there is a relationship with the radius of curvature given as [math]Rmc = \hbar[/math] He says we can think about expressing this as an uncertainty relationship between the rest momentum and the radius. [math]\frac{R}{c}Mc^2 = \hbar[/math] [math]\frac{R}{c}[/math] is the time invariant operator. Therefore since [math]\hbar[/math] is proportional to the mass in [math]m = \sqrt{\frac{\hbar c}{G}}[/math] which can be rearranged to find the relationship [math]\hbar c = Gm^2[/math]. So if this is right there must be anticommutative properties between them. Quote
Aethelwulf Posted March 30, 2013 Author Report Posted March 30, 2013 (edited) The actual uncertainty principle proposed by Motz is given as [math]\Delta R Mc \geq \hbar[/math] When taking the smallest uncertainty in the length (Planckian) we find [math]\frac{\hbar G}{c^3}M^2c^2 \geq \hbar[/math] He then further states this leads back to the definition of the gravitational charge [math]\hbar c = Gm^2[/math] So there are indeed antisymmetric properties involved here, or simply, they do not commute. Edited March 30, 2013 by Aethelwulf Quote
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