The cross product formula's given as a thought experiment and was provided with rigorous proofs of geometrical meaning using vector calculus. We found that the coupling of the electron in our thought experiment in a strong gravitational field in which the angular momentum of the particle was perpendicular to the radius of our black hole, which provided our frame of reference, using a radius which would have had to satisfy \(r_s < R\), where \(R\) was the radius of the black hole (and a distance) from the event horizon. This meant that in the thought experiment, if an electron was bound in a geodesic path around a black hole then it could contribute to the gyroradius of the electrons orbital trajectory. My thought experiment, changed some of the rules, for instance, on the LHS of equation: \(v_{B} \times (\nabla \times \mathbf{A}_{\mu}) = \frac{2(\omega \times v)}{\sqrt{G}}\) ... in order to derive the RHS of the equation, was obtained by taking the ''Coriolis Inertial Field'' as Prof. Lloyd Motz called it in his paper on the quantization of mass \(\frac{2(\omega \times v)}{\sqrt{G}}\). We never specified in detail, what the velocity on the right hand side can be interpreted as. We are in fact, taking it for the velocity of the electron as a particle example so I shouldn't have been as vague as that and rewrote it as \(v_e\), but you know now, so old habits may not die young in the name of simplicity. I actually arrived at a small mathematical discovery when I investigated the equations he asserted would hold true for primordial particles. Interestingly, he said the coupling to the classical gravi-magnetic fields are achieved through a cross product... take a look at the expression he presents \(\sqrt{G}M(\frac{V}{c}) \times (\frac{2(\omega c)}{\sqrt{G}})\) What was interesting, it that I came to realize through another derivation that this term \((\frac{2(\omega)}{\sqrt{G}})\) was equivalent to a magnetic field which was related to Coriolis field and also the Lorentz force. It normally contains a negative sign, but I later showed in previous work in my thought experiment that the sign isn't always necessarily negative because of a triple vector product. The signs of the equations flipped because of the cyclic nature of the cross products. Because I understood this, I could in fact equate the Coriolis inertial field, to the simple expression \(v \times \mathbf{B}\), which implies a motion in a three dimensional magnetic field. Keep in mind for those who actually dabble about with the electromagnetic Maxwell equations, know that \(B\) is itself a cross product \(\nabla \times \mathbf{A}\). The equating of the two expressions would imply that Coriolis fields are experienced when a spinning particle enters it. In the spirit of Motz' work is in fact by noticing that there is a gravitational feature in the equation \(\sqrt{G}\). Suppose now... we do it Motz' way. We take his suggested coupling term on the Planck quantization of mass, \(\sqrt{G}M \frac{\vec{V}}{c} \times \frac{2 \omega c}{\sqrt{G}}\) Then we take the equation I created (this time both velocities are referred to both sides as the electrons velocity)... \(v_e \times (\nabla \times \mathbf{A}_{\mu}) = \frac{2(\omega \times v_e)}{\sqrt{G}}\) ...we find that if we rearranged the equation into an expression that is similar to the one suggested by Motz, we have \(v_e \times (\nabla \times \mathbf{A}_{\mu}) \times \frac{2(\omega \times v_e)}{\sqrt{G}}\) Then it is also reasonable to assume that \(v_e \times (\nabla \times \mathbf{A}_{\mu})\) can be taken with the cross product of the same feature \(\sqrt{G}M \frac{\vec{V}}{c} \times v \times (\nabla \times \mathbf{A}_{\mu})\) This is itself, a triple cross product. or more simply \(\sqrt{G}M \frac{\vec{V}}{c} \times (v \times \mathbf{B})\) we have found a completely equivalent and simplified expression for Motz' coupling term of the fields. \(\sqrt{G}M\) is like a gravitational parameter on the particle which has a velocity \(v\) and moving in a magnetic field... which has three components. Because we know more equivalences from equations I have presented in work in the physics forum, we can further equate this with expression with \(\sqrt{G}M \frac{\vec{V}}{c} \times (v \times \mathbf{B}) = 2M(\omega \times v)\) In the spirit of seeing equations in their splender, let us cancel the velocity terms and we find a really neat equation concerning the mass and the angular velocity \(\sqrt{G}M \times \mathbf{B} = 2M\omega\) notes: 1\(^{\dagger}\) and one can also find that \(det[\sqrt{G}M \times \mathbf{B}] = \begin{vmatrix} \sqrt{\hbar c}_x & \sqrt{\hbar c}_y & \sqrt{\hbar c}_z \\ \hat{x} \partial_x & \hat{y} \partial_y & \hat{z} \partial_z \\ \mathbf{A}_x & \mathbf{A}_y & \mathbf{A}_z \end{vmatrix}\) using partial differential notation. The gravitational feature of this equation, seems to be linked to the magnetic field, it's mass (obviously) and it's angular momentum. Motz continues to say, that you get the centrifugal force by multiplying \(\sqrt{G}M\) with the Coriolis inertial force field. He made a mistake in his paper, he actually meant the Coriolis force which is known as \(F = 2m(\omega \times v)\) What is probably more to the point, is that some of these equations may actually hold true in some semi classical models of the actual structure of particles themselves. We are of course assuming in this work, that the electron isn't really pointlike at all. An example might be the interaction energy of the electrons magnetic moment in a radial electric field was found by taking a triple scalar product, similar to how we performed it on the previous equation. The Scalar Triple product appears as \(\mu \cdot \frac{(r \times p)}{Mc^2}|\frac{E}{r}| = Mc^2\) That means we have three equations to deal with when solving them \(= \mu \cdot (r \times p)\) \(= r \cdot (p \times \mu)\) \(= p \cdot (\mu \times r)\) There is also a determinant which can be given \(det(\mu, r, p) = \begin{vmatrix} \mu_1 & \mu_2 & \mu_3 \\r_1 & r_2 & r_3 \\p_1 & p_2 & p_3 \end{vmatrix}\) A more general expression for these couplings is \(\epsilon_{ijk} \mu^{i}r^{j}p^{k}\) which describes the permutations of the properties of the triple scalar product, which Rpenner correctly pointed out in a previous thread. This determinant may treat the magnetic moment as intrinsic... \(r\) would represent the radius of curvature of the particle, which Motz in his paper describes using a three dimensional hypersphere using Gaussian curvature. 1\(^{\dagger}\) What about a relativistic momentum? We need to consider that \(\hbar c\) is related to the mometum. It so happens, I found a whole bunch of relationships using some matrix mechanics. The canonical relativistic momentum \(i\hbar \gamma^0\) as a matrix involving the spin \(\hbar\) is to be multiplied through by \(c\). \(i \hbar c \gamma^0 = \begin{pmatrix} iGM^2 & 0 & & 0 \\0 & iGM^2 & 0 & 0 \\0 & 0 & -iGM^2 & 0 \\0 & 0 & 0 & -iGM^2 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\) \((\gamma^0)^2\) gives us the unitary matrix. In squaring, we remove the imaginary parts \(-\hbar^2 c^2 \mathbb{I} = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2 \cdot \gamma^0\) This is beneficial because it has no imaginary coefficients. The matrix for \(\gamma^0\) can also be written as \(-\hbar^2 c^2 \begin{pmatrix} 0 & \sigma^1 \\ \sigma^2 & 0 \end{pmatrix} = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\) and since \(\gamma^0 = \beta\) (the standard beta matrix) then \(\beta a^k = \gamma^k\) so we can make \(-\hbar^2 c^2 \gamma^k = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2 \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}a^k\) \(\gamma^k\) can now be written in it's sub-matrix form \(-\hbar^2 c^2 \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix} = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2 \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}a^k\) working with \(\begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2 \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}a^k\) writing out the matrix case for \(a^k\) gives us \(a^k = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \end{pmatrix}\) using it we have \(\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \end{pmatrix}\) To solve the RHS we then have \(-\hbar^2 c^2 \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & -GM^2 & 0 \\0 & 0 & 0 & GM^2 \\0 & -GM^2 & 0 & 0 \\-GM^2 & 0 & 0 & 0 \end{pmatrix}^2\) We have used here, Pauli matrices \(\sigma_1\) and \(\sigma_3\). The equation above in a more simplified version: \(\begin{pmatrix} 0_2 & -\hbar^2c^2 \cdot \sigma^1 \\ \hbar^2 c^2 \cdot \sigma^3 & 0_2 \end{pmatrix} = \begin{pmatrix} 0 & 0 & -GM^2 & 0 \\0 & 0 & 0 & GM^2 \\0 & -GM^2 & 0 & 0 \\-GM^2 & 0 & 0 & 0 \end{pmatrix}^2\) where the full form of \(\begin{pmatrix} 0_2 & -\hbar^2c^2 \cdot \sigma^1 \\ \hbar^2 c^2 \cdot \sigma^3 & 0_2 \end{pmatrix}\) is \(\begin{pmatrix} 0 & 0 & 0 & -\hbar^2 c^2 \\0 & 0 & -\hbar^2 c^2 & 0 \\-\hbar^2 c^2 & 0 & 0 & 0 \\0 & \hbar^2c^2 & 0 & 0 \end{pmatrix}\) We can finally equate the two \(\begin{pmatrix} 0 & 0 & 0 & -\hbar^2 c^2 \\0 & 0 & -\hbar^2 c^2 & 0 \\-\hbar^2 c^2 & 0 & 0 & 0 \\0 & \hbar^2c^2 & 0 & 0 \end{pmatrix}=\begin{pmatrix} 0 & 0 & -GM^2 & 0 \\0 & 0 & 0 & GM^2 \\0 & -GM^2 & 0 & 0 \\-GM^2 & 0 & 0 & 0 \end{pmatrix}^2\) Both sides are conjugates of each other. If you multiply the RHS with the LHS, you get back real numbers, like this: \(\begin{pmatrix} 0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix}\) This is in fact a submatrix with entries \(\begin{pmatrix} \sigma^1 & 0_2 \\ 0_2 & \mathbf{1}_2 \end{pmatrix}\)
Actually, some months ago I wrote down some equations when I first started looking at the magnetic and gravitational fields as geometric artifacts of fundamental properties of particle dynamics. I found a Hamiltonian, which described the dynamics of the spin of the particle as a dot product with the magnetic field, \(\mathbf{H} = \frac{e}{M c}\frac{\hbar}{2} \cdot \mathbf{B}\) I found that taking the cross product of the radius of curvature on the magnetic term on the right hand side of the equation, produced \(\frac{e}{M c} \frac{\hbar}{2} \cdot \mathbf{B} \times R = \mu_B \cdot \mathbf{A}_{\mu}\) Note we are using the gauge \(\mathbf{A} = \mathbf{B} \times R\) where \(\mu_B\) is the Bohr Magneton. It was interesting, because \(\mu_B \cdot \mathbf{A}_{\mu}\) had the same dimensions as \(\hbar c\). This can only be correct, since the energy term \(\frac{e}{M c}\frac{\hbar}{2} \cdot \mathbf{B}\) is actually connected to the definition \(GM^2 \equiv Er\) which as most know, is just the rearranged form of the gravitational energy \(U = \frac{GM^2}{r}\). So it would hold true then that \(\mu_B \cdot \mathbf{A}_{\mu}\) and \(\hbar c\) is related at least in some cases.
I will start in the next week to describe how my model might fit into general relativity, and this post will start that investigation through a series of calculations. There is a strong classical relationship to General Relativity I am about to derive for you, concerning a primordial particle. In the classical sphere model of the particle, the density of a three dimensional hypersphere should be \(\rho_0 = (\frac{M}{\frac{4}{3} \pi R^3})\) The general relativistic relationship to the principle curvatures \(K = k_1 k_2\) is found as \(8 \pi \rho_0(\frac{G}{c^2})\) sidenote *the term \(8 \pi\) arises often because it simplified several of the field equations describing gravity The expression above was noted by Motz on his paper of quantization, and by a reference I will have below this post. Let us plug in and solve the equation. Be careful to remember your algebra when solving, when dividing a whole number by a fraction. I calculate this as \(\frac{m}{6R^3}\) A three dimensional hypersphere is normally written as \(\frac{K}{6}\) when equated with the Compton wavelength. Equating my expression with this term we have \(\frac{m}{6R^3}(\frac{G}{c^2}) = \frac{K}{6}\) Cancelling out the 6's, we end up with a general relativistic interpretation of how the geometry of a particle relates to the proper density and the Gaussian Surface Curvature \(K = \rho_0 (\frac{G}{c^2}) = \rho_0 (\frac{L_P}{M_P})\) Where the last equality comes about from a knowing that \(\frac{G}{c^2}\) has dimensions of length over mass... the equations are usually written in the Planck form. http://books.google.co.uk/books?id=...planck length divided by planck mass?&f=false http://tpb.physik.rwth-aachen.de/Dohm/FP_on_ISS_GRG.pdf
In my original work, these equations simplified to perhaps, an understanding of gravity and particle dynamics. Let me give you a brief understanding why I am doing all this. This is a pet theory, based on... much forgotten work by Sciama on the Origins of Inertia and work by Motz who was truly underrated at this moment in history as not even many people have heard of him. What they have shown though, is that there is a possibility questions about how gravity couples with magnetism at the fundamental level may have serious implications about the very origins of our universe itself. Today, there is the traditional Higgs Boson model (perhaps it isn't in the standard sense), nonetheless, the question of inertia itself is still a persistent question and how gravity acts at the fundamental scale isn't fully understood. Part of the plan of physicists, even to this day is create a unification between the classical, quantum and relativistic theories. They noticed, this involved three important constants of nature \((G, c, \hbar)\). In quantum theory, we have \(\hbar\) the angular momentum and the speed of light \(c\). In the classical theory, we have angular momentum but is a classical rotation, we do however have the gravitational feature \(G\) in the Newtonian theory. In General Relativity, we have the gravitational constant \(G\) and \(c\) as being important within it's framework. The trouble is how to find a full \((G,\hbar,c)\)-theory describing all three of these in a consistent way. The assumption is that to describe it, we require a full understanding of how these three constants play their parts in other respective fields of physics. In my work, I have attempted to do some of this using the equations provided by Motz and Sciama. While the Gaussian curvature above doesn't describe the full fundamental theory of a uniton, it does give what we should expect to be the correct General Relativistic description of the curvature of radius of our system. Because of this, we simply go back to the same cross products we encountered before, this time explicitely noting that every time a length is involved we must be asserting a system with a compton wavelength equal to the curvature of radius described by the product of the principles bundles \(K\). Thus we find a whole serious of rather... obvious equations. With \(K = \rho_0 (\frac{G}{c^2})\) We may have \(K \times \mathbf{B} = \mathbf{A}_{\mu}\) and \(\mu_B \cdot(K \times \mathbf{B}) = \hbar c\) Notice this last equation is interesting, it appears that the charge is related to a similar interaction to an energy term we derived earlier using a triple scalar product, except we explicitly have an equation describing the magnetic field rather than a radial electric field (albeit one is present). Also, we knew this was an obvious equation because we already identified previously that \(\mu_B \cdot A_{\mu} = \hbar c\) which can be elegantly proved if anyone asks for it but it requires a gauge and it also requires an assumption there is an internal momentum inside of the particle. To employ the \((G, \hbar, c)\) dynamics in the equation, we must return back to the equations that where derived from the very beginning of my written work in the physics forum. \(\mu_B \cdot(K \times \frac{2\omega}{\sqrt{G}}) = \hbar c\) Here we have regained the gravitational feature in the magnetic term describing the gravimagnetic field; an important term in itself because it came from the rational equating of the Coriolis Force to the Lorentz force and then removing the linear velocities. I was also careful to mention that the gravimagnetic interpretation \(\frac{2\omega}{\sqrt{G}}\) isn't always true for the exception of a simple sign change, properties we realized due to cyclic order of cross product permutations. It may in fact imply deeper sign changes in spin dynamics, specially-speaking for spin \(\frac{1}{2}\) fermion particles. There is an important thing however to consider when we look a bit closer at the dynamics going on here. The linear velocity would appear from the term \(2(\omega \times K)\). Even though we are talking about an internal curvature relationship with the angular velocity, there is nothing which suggests there is no reason that such fundamental properties are repsonsible for their own motions. In fact, we already know that their ''intrinsic spin'' is related to their orbital trajectories when the spin couples in magnetic fields. The angular velocity is simply \(dP = K d\theta\) The linear velocity is obtained by \(\vec{v} = \omega \times K\) Remember that the cross product of a variable with itself is zero which is how we obtain this term. Interestingly, any calculation in the classical sense that computes the internal electric field finds \(\nabla \cdot \mathbf{E} = 0\) We have real life macroscopic experiments which offer support for the validity of this fact for a classical spherical shell of Planck charge. However, we have also shown that in an equally classical sense, that it doesn't make a coherent argument to talk about an angular momentum \(dP\) without the implication of an internal geometry \(\theta\) related also to the area of the Gaussian surface \(4 \pi r^2\). Intuitively, we cannot just presume this will be the case, since quantum theory often surprises us how different particles behave to macroscopic systems in general. It is an open question yet to be resolved. Again... intuitively one might have an idea how this field would arise in the equations. If the charge \(\hbar c\) must feature in our equation, the the electric field might be \(\mathbf{E} = \sqrt{\frac{\hbar c}{\epsilon_0 K^4}}\) This is just a thematic guess however. It lacks the gravitational feature \(G\), but this obviously has room to work with, aside from the relationship we have already discussed, \(\hbar c = GM^2\).
I have no idea what this thread is about. It's like jumping into the middle of somebody's mathematics blog.
I haven't really got around to where it all fits in together... I'm trying to do this slowly because I have lots of work to talk about.
Why don't you work in private then post your coherent results once you have something to show? Nobody wants to hear you thinking out loud.
I'm sure they don't, which is why that isn't my intention. Each post has been designed to give you an overview of a much more complete theory I have been working towards. I am being totally honest here... this isn't something I can express 'coherently' in a single post, there are lots of derivations and tying the knots is an arduous process. Writing skills might not be my best skills... but I am trying.
