On Fundamental Torque Fields in the Planck Scales

Discussion in 'Pseudoscience' started by DR. Nobody, Dec 9, 2014.

  1. DR. Nobody Registered Member

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    The best value for the square of the gravitational charge \(Gm^2\) came from manipulation of our best value for the CODATA charge to find the value for the charge was

    \(Gm^2 = \frac{\pm \alpha \hbar \hat{n}}{4 \pi \epsilon_0 \mu_o c}\)

    If we were in fact searching for the gravitational charge itself, we simply take the square root of the RHS. Interestingly in previous work, one can find a strong gravitational constant by making the gravitational interaction field dependent on a length scale. The square root of the fine structure could then be written in a different form knowing the following equational rule

    \(\omega_p \gamma(t - \frac{vx}{c^2}) = \sqrt{\alpha_G}\)

    \(\sqrt{G}m_p = \sqrt{\frac{\pm \hbar \hat{n}\omega_p \gamma(t - \frac{vx}{c^2})}{4 \pi \epsilon_0 \mu_o c}}\)

    Where, note, the use of ''\(\cdot\)'' is not a dot product. These have been written clearly under Planck parameters.

    Since an electric and a magnetic flux in MKS units are related to the magnetic and electric constants

    \(\Phi_e = \frac{e^2}{\epsilon_0}\) 1.

    \(\Phi_m = \frac{\phi_{0}^{2}}{2 \mu_0}\) 2.

    We can see that multiplication of \(\epsilon_0 \mu_0 c \mathbf{E}\) on equation 1. would give us

    \(\Phi_e \mathbf{B} = \phi^{2}_{0} \mu_0 c \mathbf{E}\)

    Because of the standard identity \(\mathbf{B} = \epsilon_0 \mu_0 c \mathbf{E}\)

    \(\mathbf{B} = \epsilon_0 \mu_0 c \mathbf{E}\)

    \(\frac{\mathbf{B}}{\mathbf{E}} = \epsilon_0 \mu_0 c\)

    then a quantization condition involving the magnetic and electric fields coupled to the gravitational charge is given as

    \(4 \pi \epsilon_0 \mu_0 c Gm^2 = 2 \alpha \pi \hbar\)

    This equation was obtained from our most accurate equation describing the CODATA charge. Combining the equations we obtain

    \(4 \pi \frac{\mathbf{e^2 B}}{4 \pi \epsilon \mathbf{E}} = 2 \alpha \pi \hbar\)

    In Gaussian units the expression \(4 \pi \epsilon\) would vanish. In the SI unit system the charge would be weighted down by \(e(4 \pi \epsilon)^{-1}\).

    We can take this as a quantization condition

    \(\frac{2}{\alpha} \frac{\mathbf{e^2B}}{\mathbf{E}} = \frac{1}{c} 4 \pi \epsilon Gm^2 = \hbar \hat{n}\)

    An action can be given in an integral as

    \(\int eA_{\mu}\frac{dX^{\mu}}{d\tau}\ d\tau = \int eA + eA_x V^x\)

    This reduces to

    \(\int e(A_0 - A \cdot v) dt\)

    where \(A_0\) is the time component of the electromagnetic potential.

    A difference \theta between two endpoints \(k = (i,j)\) can be given as

    \(\int eA_{\mu} \frac{dX^{\mu}}{d\tau} d\tau + \int \partial_{\mu} \theta dX^{\mu}\)

    with a transformation property of \(A_{\mu} \rightarrow A_{\mu} + e^{-1} \partial_{\mu} \theta\) on the potential.

    The gravimagnetic field given by motz in his paper [1]

    \(\frac{2 (\omega \times c)}{\sqrt{G}}\)

    which can be derived interestingly and simply by taking the coriolis force field and dividing it by the gravitational charge \(\sqrt{G}m\)

    \(F = 2m(\omega \times c) = 2(\omega \times p)\)

    \(\frac{F}{\sqrt{G}} = \frac{2(\omega \times c)}{\sqrt{G}}\)

    The inertial ficticious forces in two co-rotating frames are

    \(F_c + F_C = m \ddot{\mathbf{r}}\)

    Where \(F_c\) is the centriful force and \(F_C\) is the coriolis force.

    Newtons second law in a co-rotating frame can be reduced to the Euler Langrange equations. For example, the radial equation is:

    \(m\ddot{\mathbf{r}} = m r \dot\theta^2 - \frac{\mathrm{d}U}{\mathrm{d}r}\)

    In three dimensions the rotation matrix is

    \(\begin{alignat}{1} R_x(\theta) &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\[3pt] 0 & \sin \theta & \cos \theta \\[3pt] \end{bmatrix} \\[6pt] R_y(\theta) &= \begin{bmatrix} \cos \theta & 0 & \sin \theta \\[3pt] 0 & 1 & 0 \\[3pt] -\sin \theta & 0 & \cos \theta \\ \end{bmatrix} \\[6pt] R_z(\theta) &= \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\[3pt] \sin \theta & \cos \theta & 0\\[3pt] 0 & 0 & 1\\ \end{bmatrix} \end{alignat}\)

    Newtons force law in spherical coordinates is

    \(F = m((\ddot r - r \dot \theta^2) \hat{r} + (r \ddot\theta + 2 \dot r \dot\theta) \hat{\theta})\)

    where \(F\) is a central force, as \(F(r) \hat{r}\). Also \(\pm F(r)\) can be true and tells us whether the central force is inward (negative) or outward (positive) - gravity acts as an inward force and so \(-F(r)\) is negative. However inside the classical particle of dimensions \(R^3\) the gravitational force becomes an outward positive force according to Motz in his paper on gravitational charge and so, it would cancel out the Poincare Stress thought to haunt the classical physics of stable spherical particles so that \(G\) becomes positive and massive to the scale of \(G \cdot 10^{40}\).

    Newtons force law of motion in the radial direction reduces to

    \(F(r) = m(\ddot r - r \dot\theta^2)\)

    Previously I explained that the inertial ficticious forces in two co-rotating frames are

    \(F_c + F_C = m \ddot{\mathbf{r}}\)

    Where \(F_c\) is the Centrifugal force and \(F_C\) is the coriolis force.

    Writing the LHS side fully we really have in all the components of the combined fields (in the style of Machian theory)

    \(m\Omega \times (\Omega \times \mathbf{r}) + 2m \Omega \times \mathbf{v}_{\mathrm{r}} = m \ddot{\mathbf{r}}\)

    Lorentz force equal to the coriolis force is interesting, earlier we said that by dividing the gravitational charge \(\sqrt{G}m\) yielded the gravimagnetic field \(\frac{2(\omega \times c)}{\sqrt{G}}\). Interestingly if we set the Lorenz force equal to the Coriolis force field (in which the Coriolis force is a gravitational link to the frame dragging of magnetic phenomena and the Lorentz force is an electromagnetic phenomenon but we have reduced it totally to the magnetic part) it yields

    \(e( v \times \mathbf{B}) = 2m(\omega \times c)\)

    Do the same thing as we did before to find Motz' gravimagnetic field by dividing it by the Weyl gravitational charge and simplify by cancelling out the velocity terms

    \(\frac{e\mathbf{B}}{\sqrt{G}m} = \frac{2\omega }{\sqrt{G}}\)

    Now \(e\) does equal \(\sqrt{G}m\) if we work in Gaussian units, which is fine, so this gives a further simplification and a direct equivalence between the magnetic field and a gravitational interpretation of it as well it seems.

    \(\mathbf{B} = \frac{2\omega}{\sqrt{G}}\)

    replace this magnetic field in our equation

    \(e( v \times \frac{2\omega}{\sqrt{G}}) = 2m(\omega \times c) \rightarrow e( v \times \frac{2\omega}{\sqrt{G}}) = 2m(\omega \times c)\)

    Notice that we can rearrange this as

    \(\rightarrow e(\frac{2\omega \times c}{\sqrt{G}}) = 2m(\omega \times c)\)

    for the velocity set to the celeritas \(c\) and the term \(\frac{2\omega \times c}{\sqrt{G}}\) crops up again, the gravimagnetic field of Motz.

    Which shows there are two different ways to have obtained the gravimagnetic field and is the first of its kind it appears from not finding it in literature.

    Now using first and second quantization on a discrete space of charge using the Weyl quantization of \(Gm^2 = \hbar c\)

    In a space with fields \((\psi,(q) \hat{\psi}(q))\) where \(q\) is a generalized coordinate second quantization leads to a description of the creation and annihilation operators \((a^{-},a^{+})\)

    \(\psi(q) = \sum_k a^{+}(k) e^{-ikq}\)

    \(\hat{\psi}(q) = \sum_k a^{-}(k) e^{ikq}\)

    This is the quantization of the length. Written in the discrete form, for \(k = \frac{2\pi n}{\ell}\) and a periodic space interval of

    \(\int^{\frac{1}{2}}_{-\frac{1}{2}} e^{ikq} dq = 2\pi \sum_{k} a^{+}(k) \delta_{nm}\)

    with a value of zero at the centre (simple topological space example). Hitting them with a momentum operator can yield a complexified version of the momentum space in increment of unit length \(n\) - for the case of \(\delta_{nm}\) the operators work on both

    \(-i \hbar \frac{\partial}{\partial t} \cdot 2 \pi \sum_k <n|a^{+}|m> \delta_{nm}\)

    Where \(\delta_{nm} = <\phi_n|\phi_m>\) and the increments of \(n\) are acting on the operator \(a^{+}\) which yields the raising operation \(<n|m + 1> \sqrt{n+1}\) ... the same can be done for lowering.

    In the Schwinger quantization of charge context, \(\hbar\) is measured in terms of increment \(<n>\) for the square of the charge with using the discrete basis picture, we have the relation \(<\phi_n|\phi_m>= \delta_{nm}\). This discrete version of the delta function \(\delta(x−y)\) is best seen as the identity matrix \(\mathcal{I}\), where \(\delta_{nm} =\mathcal{1}\) if \(m=n\) , \(\delta_{nm}\) goes to zero if and only if \(m \ne n\).
     
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  3. DR. Nobody Registered Member

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    If \(n\) acts on the creation operator, we can rewrite it in forms where \(m=0\).

    In this way, we can our quantization and second quantizated discrete fundamental space and rewrite an equivalence to the raising operator to a smooth periodic function with discplacement always at zero when they converge.

    To write this we can give it in a new set of forms, first of all with \(m = 0\) and \(k\) taking on the forms \(\frac{2 \pi n}{\ell}\) would give us the simpler form of this picture we have adopted one might notice that we are dealing with

    \(i \hbar \frac{\partial}{\partial t} \cdot 2 \pi(n \cdot n)\sum_k a^{+}(k) \delta_{nm}\)

    An unit vector dot product with another unit vector gives the value of the identity matrix and they effectively vanish.

    \(-i \hbar \frac{\partial}{\partial t} \cdot 2 \pi \sum_k a^{+}(k)|n> \delta_{nm}\)

    Because the the ket acts on the creation operation space \(a^{+}(k)|n>\) this can also be written as \(<n|m + 1> \sqrt{n+1}\) with \(m\) not acting on \(m\) not acting on the topological space in this case for the creation operator gives us

    \(-i \hbar \frac{\partial}{\partial t} \cdot 2 \pi \sum_k a^{+}(k) \delta(q) = \alpha \hbar n\)

    Notice the righthandside is the same form of the Schwinger quantization expression for his quantization of the charge and then used magnetic charge - we know today there are no magnetic monopoles or at least if any exist, inflation made them aloof as they were diluted during the verocious inflational stages of the universe, thanks to the work of Alan Guth. The equation presented before is possible because


    \(\sum_k a^{+}(k) \int^{\frac{1}{2}}_{-\frac{1}{2}} e^{-ikq} \delta(q)\)

    and \(\sum_k a^{-}(k) \int^{\frac{1}{2}}_{-\frac{1}{2}} e^{ikq} \delta(q)\)

    which are equal to the simple expression \(2 \pi \delta(q) = \displaystyle \binom{-\sin \theta}{-\cos \theta}\).

    Concerning the quantization of space using the second and first quantization methods in this equation

    featured on page 1.

    If \(n = -\sin \theta\) and \(r = -\cos \theta\)

    then

    \(-i \hbar \frac{\partial}{\partial t} \cdot \begin{pmatrix} n \\ r \end{pmatrix} \sum_k a^{+}(k) \delta(q) = \alpha \hbar n\)

    Well \(2 \pi\) is in fact a column matrix giving the charge a second geometric meaning

    \(-i \dot{\hbar} \cdot \begin{pmatrix} n \\ r \end{pmatrix} \sum_k a^{+}(k) \delta(q) = \alpha \hbar n\)

    Where \(\dot{\hbar}\) signifies the torque.

    Where there are six trigonometric functions at play. The simpler set of versions is

    \(-(i \hbar \frac{\partial}{\partial t} \cdot f( \theta) \sum_k a^{+}(k) \delta(q) = \alpha \hbar n\)

    \(-(i \hbar \frac{\partial}{\partial t}) \cdot \cos \theta \sum_k a^{+}(k) \delta(q) = \alpha \hbar n\)

    because \(f(\theta) = \cos \theta = 2 \pi\)

    Only for these cases it is forbidden to have a period less than \(\pi\)

    \(cos(\theta) > 0 \rightarrow \theta \in (\frac{\pi}{2}, \frac{\pi}{2}) cos(\theta) < 0 \rightarrow \theta \in (\frac{\pi}{2}, \frac{3\pi}{2})\)

    The space topological charge which appears to indicate in previous Weyl quantization equation that there is also a presence of a topological charge-space through a relationship to working with the Weyl quantization equivalence to the generation of mass (aka. gravitational charge) via \(\hbar c = Gm^2\) as mention before from a Motz paper on quantization of the mass \(\hbar = \frac{Gm^2}{c}\) where the numerator plays a synonymous role of the quantization method of Schwinger \(\frac{e^{2}_{1} - e^{2}_{2}}{c} = \pm \alpha \hbar n\) where this time it seen in the more accurate light of taking positive or negative values on the quantization of the angular momentum component. When dealing with the wave functions dictating the operations on the first and second quantized space, it must be kept in mind that they strictly work in the Banach space.
     
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  5. DR. Nobody Registered Member

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    On The Matrices
    Which ''Pilot'' a Rebuttal
    To Bells Inequality Allowing
    Hidden Variables


    There are pilot matrices which can determine the gamma matrices to allow a local hidden variable interpretation. Doing so leads to new quantization conditions of the spin.

    Reciting a matrix to start us off:

    \(\begin{pmatrix} 0_2 & -\hbar^2c^2 \cdot \sigma^1 \\ \hbar^2 c^2 \cdot \sigma^3 & 0_2 \end{pmatrix}\)

    If the components where all valued \(1\), we can hit one of either sides of this with the matrix contained in this paper, on page (11.), for photon polarization; http://arxiv.org/pdf/1306.6292.pdf yields for the charge \(\hbar c\) case

    \(\begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix}\)

    Is in fact \(\gamma_3\), or you can change sign \((-,+)\) by flipping the index for \(\gamma^3\) whilst when we calculate the RHS we get

    \(\begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\-1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix}\) [2]

    These are close to Chirality, except for some sign changes, the chirality \(\gamma^5\) is in fact a product of the unit psuedoscalar \(i\) with the four [common] gamma matrices \(\gamma_0\gamma_1\gamma_2\gamma_3\). Some more work needs to be looked at, but when you multiply [1] with [2] you get back an ''almost'' negative diagonal components, which means there can't be far off an important symmetry.

    This appears after all to yield some symmetries after working on it for a while and came to the conclusion that \(\gamma^{3}\gamma_{3}\) produces a Hermitian matrix

    \(\begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\-1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix}\) [1]

    Is in fact \(\gamma_3\), or you can change sign by flipping the index for \(\gamma^3\), so I came to call it the spin connections, that are seen to oscillate in the real and complex domain.

    It might seem strange using notation like this to ''switch'' signs in the components of the matrix, but they could very well indicate a relationship to Chirality, as they show intelligible resemblance to the Dirac Basis

    \(i \gamma^0 \gamma^1 \gamma^2 \gamma^3 = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix}\)

    In fact, the two are symmetric with respect to entry positions.

    To obtain say the matrix \(\gamma^3\) we would introduce another matrix

    \(i \gamma^0 \gamma^1 \gamma^2 \gamma^3 = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix}\)

    \(\gamma^3 = i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i} = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1\\-1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix}\)

    To find \(\gamma_3\) you simply calculate

    \(\gamma_3 = i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}_i = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\)

    The Operating Matrices are given by:

    \(\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix}\)

    and

    \(\begin{pmatrix} -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\)

    are interesting for another reason, when they compute they give a Diagonally Dominant Matrix, if all its diagonal elements are negative, then the real parts of its eigenvalues are negative. These results can be shown from the Gershgorin circle theorem. This further indicates that there are two solutions in which the possible eigenvalues \(\pm\) can take this as a property of Chirality, which should be thought of as a ''handedness'' to particle systems.

    The operating matrices play a crucial role in the works. The operating matrices together help produce a diagonally negative result, like the antithesis of the Hermitain. I remember work Motz claimed to have worked on with Schwinger concerning a very suitable model mathematically and gave the right results, par one conclusion their professor said to them, that their Hermitian wasn't real. Whether Motz or Schwinger knew what was being proposed and they may have even protested, but do not quote me, on the opinions of the one who has to credit the work for publication.

    In this model, it's interesting because the matrices \(\gamma^3\gamma_3\) that act as the basis for a Chirality coupling, with our operating matrices. These operating matrices, might be factors about the geometry of space-time we yet don't understand. Are he matrices like a hidden variable which determines information about the system? (see Notes at the end)

    Thus, the back-lash of the operating matrices is that they yield relationships which can be described by projecting our new representation of the Chirality components

    \(\psi_{R,L} = \frac{1 \pm \gamma^3\gamma_3}{2}\psi_{R,L}\)

    This holds true naturally because it's eigenvalues depend on whether the diagonal entries are negative, again to do this we must use our operating matrices to dictate it.

    \(\gamma^3\gamma_3 = i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i} \cdot i \gamma_0 \gamma_1 \gamma_2 \gamma_3 \mathbf{A}_{i} = \mathbf{I}_4\)

    When you separate the left handedness from the right handedness in the equations above you find the operating matrices part of the famous anticommutator with the four gamma matrices

    \(<\i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i}, \gamma^{\mu}> = (i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i}) \gamma^{\mu} + \gamma^{\mu} (i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i})\)

    Which as most who study physics knows, computates the so-called ''intrinsic spin'' of the system. I have had some keen interest on Ricci Rotations and curvature terms during these studies and hope to write something up soon about it.
     
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  7. DR. Nobody Registered Member

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    Thus, the back-lash of the operating matrices is that they yield relationships which can be described by projecting our new representation of the Chirality components

    \(\psi_{R,L} = \frac{1 \pm \gamma^{\hat{3}}\gamma_{\hat{3}}}{2}\psi_{R,L}\)

    (see Notes) - here I have changed the notation to \(\gamma^{\hat{3}}\gamma_{\hat{3}}\) to show they are actually different when faced with the ''pilot matrices'' which becomes a convenient name for the expression \(\mathbf{A}^i\mathbf{A_i}\) which act on the Dirac Basis in such a way, that they dictate the spin before the measurable has been collapsed.

    This holds true because its eigenvalues depend on whether the diagonal entries are negative, again to do this we must use our operating matrices to dictate it.

    \(\gamma^3\gamma_3 = i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i} \cdot i \gamma_0 \gamma_1 \gamma_2 \gamma_3 \mathbf{A}_{i} = \mathbf{I}_4\)

    When you separate the left handedness from the right handedness in the equations above you find it's Eignenvalues satisfying \(\pm 1\) because of \((i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i})^2 = \mathbf{I}_4\) This is the same as saying \((\gamma^{5+})^2 = \mathbf{I}_4\), The term \(i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i}\) should also anticommute with the four gamma matrices.

    (see Notes)

    \(<\i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i}, \gamma^{\mu}> = (i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i}) \gamma^{\mu} + \gamma^{\mu} (i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i})\)

    This explains an electrons spin due to internal changes based on polorization on the operating matrices using the pilot gamma matrices, which are based on only symmetry within the matrices themselves, but they operate accordingly for \(\gamma^{5+}\) for the classical sphere. If the operating matrices are some hidden symmetry, they may have a role in ''spooky action at a distance.''

    Using said principles, we can find relationships between the Dirac matrices.

    \(\gamma^{\hat{3}} \gamma^1 \gamma^0 = \begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & 1\\1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix} = \gamma_{\hat{3}}\)

    Let's define the operator matrices:

    \(\mathcal{O}^a = \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix}\)

    and

    \(\mathcal{O}_b = \begin{pmatrix} -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\)

    \(\gamma_3 \mathcal{O}_a = i\gamma^0 \gamma^1 \gamma^2 \gamma^3 = \gamma^{5}\)

    \(\gamma^3 \mathcal{O}^b = i\gamma^0 \gamma^1 \gamma^2 \gamma^3 = \gamma^{5}\)

    and the anticommutator relationship is

    \(<\ \gamma^3 \mathcal{O}^b, \gamma^{5}\ > = (\gamma^3 \mathcal{O}^b) \gamma^{5} + \gamma^{5}(\gamma^3\mathcal{O}^b)\)

    Likewise you should be able to do the same for \(\mathcal{O}_a\).

    Notes


    [1] I would like the reader to keep in mind, I am selecting a new geometry by using the matrices rotating their real eigenvalues into complex ones by seeing the operating matrices as the ''hidden variable'' if you'd like to call it, which may have dictated the pilot wave function and ''knew'' which state it would always be in. This would solve to an argument against superluminal connections, but not superdeteriminism.

    [2] In this work, \(\gamma^{\hat{3}}\gamma_{\hat{3}}\) doesn't act like it normally does, in ordinary theory, they normally do commute to a Hermitain matrix, but by conjecture of new matrices to explain why their components can fluctuate eigenvalues in the real and complex plane, and so really have a new notation for them, an example above is given with \(\hat{3}\) symbolizing rotations in the real and complex plane and perhaps even the imaginary.

    [3] I use notation \(\gamma^{5+}\) to show if we would like to work in a spatio-region without time included.



    Using The Euler Characteristic
    With Genus \(g\)
    For the General Relativistic Gaussian Curvature \(K\)


    In the modified manifold space by simplifying terms by using the definition of the Gaussian curvature bundle described in General Relativistic terms \(8 \pi \rho_0 (\frac{G}{c^2}) = K^{-6}\) allowed me to extract a simplified measure of that theory using the Euler Characteristic \(\chi\) for a spherical particle with genus \(g\). It was shown that we are dealing in a closed surface when dealing with a particle is actually still topologically a sphere meaning the simplified version is (quickly rehashing some work):

    A three dimensional hypersphere is normally written as \(\frac{K}{6}\) so that we have

    \(\frac{m}{6 \pi r^3}(\frac{G}{c^2}) = \frac{K}{6}\)

    canceling out like terms we have a simplified version of our first equation

    \(\int_M \rho_0 (\frac{G}{c^2})\ dA + \int_{\partial M}k_g\ ds = 2 \cdot \chi(M)\)

    as an exact term on the manifold, where our \(\pi\) terms also cancel out, and the Euler characteristic \(\chi\) is set equal to \(2-2g\) which specifies we are dealing with a closed surface, or sphere with genus \(g\) (for the case of no boundary), is still topologically a sphere with handles attached which means the term \(\int_{\partial M}k_g\ ds\) can be omitted.

    However, line elements are required to understand topological charge.

    The geodesic curvature is associated to the line elements on our manifold, while the Gaussian curvature relies on the element area \(A\).

    A line is only a special case of a curve in which an object similar to a line but is not straight when \(k_g = 0\). This is the most general idea of a curve in one dimensions \(E^2\), this means we are left with an even deeper meaning of the Gauss Bonnet Theorem of how the proper density and the Schwarzchild constant \((\frac{G}{c^2})\) are intrinsically related to the curvature of the metric quadratically

    Hidden inside this object, we have a null curvature form, which basically says even a straight line is not required to be straight at all. The deformation of the curve comes from changes in an angle \(\theta\). The vector field of the photon is \(c(t) = x\) which travels along a unit speed of curve with a geodesic curvature \(k_g(t)\) is parallel along the curve satisfies

    \(\dot{\theta}(t) = -k_g(t)\)

    which means we are dealing with a piecewise geometry, the angle remains constant because \(k_g = 0\) which means specifically the system is traveling a geodesic, with each segment constant to some central singularity (speculated at zero radius conditions).

    \(\int_M \rho_0 (\frac{G}{c^2})\ dA = 2 \cdot \chi(M)\)

    \(\chi(M)^n = 1 + (-1)^n = \begin{cases} 2 & n\text{ even}\\ 0 & n\text{ odd}. \end{cases}\)

    And therefore, there are no non-vanishing sections of the tangent bundle of even spherical particles mathematically.

    The parameterized curve of a smooth interval is given as

    \(s = \int_{\lambda_1}^{\lambda_2} d\lambda \sqrt{ g_{ij}\frac{dq^i}{d\lambda}\frac{dq^j}{d\lambda}}\)

    It's not too dissimilar a method when calculating the Jacobian for a timeless action principle by making the trial curves \(\lambda\) through reparameterization (See Julian Barbour's work at Platonia.com)

    For these curvilinear coordinates the definition of the square of the line element \(ds\) is

    \(ds^2 = d\bold{q}\cdot d\bold{q} = g(d\bold{q},d\bold{q})\)
     
  8. DR. Nobody Registered Member

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    We can write this in terms of the mass tensor yielding the Kinetic energy as well, using Maupertuis' principle - this principle is generally called the principle of stationary action,

    The kinetic energy is denoted as \(T\) in normal convention and is quadratic in the generalized velocities \(\dot{\mathbf{q}}\)

    \(T = \frac{1}{2} \frac{d\mathbf{q}}{dt} \cdot \mathbf{M} \cdot \frac{d\mathbf{q}}{dt}\)

    although the mass tensor \(\mathbf{M}\) may be a complicated function of the generalized coordinates \(\mathbf{q}\). For such systems, a simple relation relates the kinetic energy, the generalized momenta and the generalized velocities

    \(2T = \mathbf{p} \cdot \dot{\mathbf{q}}\)

    So long as the potential energy \(U(\mathbf{q})\) does not involve the generalized velocities. We can write the metric as

    \(ds^{2} = d\mathbf{q} \cdot \mathbf{M} \cdot d\mathbf{q}\)

    Which was really shown to show there is a different take on the equation.
    The metric \(ds\) has the line elements;

    \(ds = \frac{\partial \bold{r}}{\partial q^i}\cdot\frac{\partial \bold{r}}{\partial q^j}\)

    so the square of the line element is given as

    \(ds^2 = \frac{\partial \bold{r}}{\partial q^i}\cdot\frac{\partial \bold{r}}{\partial q^j}dq^idq^j\)


    The final part

    Schwinger Quantization and Weyl Procedure Method in a Nutshell


    (with added work extending Motz' paper, see references)

    In my mind, Motz was definitely onto something by equating the Cwavelength to the Gaussian curvature \(K\).

    Taking in mind everything I have said so far about this curvature, it can yield very interesting things

    \(\Delta = \gamma_{\mu} \partial^{\mu}\) [1]

    \(i \gamma^{\mu} \Delta \gamma_{\mu} R = 1\) [2]

    Repeating indices cancel out

    \(i \gamma^{\mu} \gamma_{\mu} \partial^{\mu} \gamma_{\mu} R = 1\) [3]

    so what we really have is

    \(i \gamma^{\mu}\partial^{\mu} R = 1\)

    Which is the square root of the space time interval.

    In an antisymmetric product, the spin matrices is given as

    \(2 \sigma_{\mu \nu} = \gamma_{\mu} \gamma_{\nu} - \gamma_{\nu} \gamma_{\mu}\)

    This is redundant view in a geometric algebraic-sense, and... Hestenes of course mentions this in his own work.

    The unit two form is

    \(dx \wedge dy\)

    This term is contracted under the sigma, so it already implies the term \(\sigma_{\mu \nu}\). The reason lies within making sure for instance, \(\sigma_{\mu} \sigma_{\mu}\) is \(0\) and not \(1\). This means it has a property which allows it to square to a negative \(-1\), indeed, applying this rule we find a rotation in the imaginary axis

    \(i^2 \gamma^{\mu} \partial^{\mu} R = -1\)

    It's almost like a special case of a wick rotation. A contraction tensor should be able to reveal this identity as well for equation [3].

    Let's return now to the radius \(r^{-2} = K\) (remember, previous coefficients of \(6\) could be removed by applying the curvature to a three dimensional sphere)

    The Weyl theory of gauge invariance applied to the Ricci tensor \(R_{ik}\) and by using the Feynman sum over paths procedure yields

    \([\frac{e\hbar}{2c} \sigma^{\mu \nu} F_{\mu \nu} + (\frac{\hbar}{i} \frac{\partial}{\partial q_{\nu}} - \frac{e}{c}A^{\nu})(\frac{\partial}{\partial q^{\nu}})]\psi(q) = \hbar^2 K \psi(q)\)

    The curvature \(r^{-2}\) is in fact treated like an operator than usually an eigenvalue thus we have

    \(-\nabla_{\mu} \nabla^{\mu} \psi = r^{-2} \psi\)

    Factoring using the gamma matrices one can obtain the equation we derived not long ago

    \(i \gamma^{\mu} \nabla_{\mu} \psi = r^{-1} \psi\)

    In other work I showed that the zitter radius is given as an inverse value of the curvature \(r^{-1}\) and that applying the anomalous magnetic value \(g=2\) on such an equation as the one above would indeed also yield the squared reciprocal

    \(i \gamma^{\mu} \nabla_{\mu}\ g\ \psi = r^{-2} \psi\)

    But this will have little interest for this particular excerpt. These excerpts are to give some deeper meanings to the terminology I have been using. The Schwinger quantization method states for a stationary particle

    \(m\dot{V} = e_1(\mathbf{E} + \frac{1}{c}\ v \times \mathbf{B}) + g_2(\mathbf{B} - \frac{1}{c}\ v \times \mathbf{E})\)

    (see reference 4.)

    Where \(e_1,g_2\) are the electric and magnetic charges. Schwinger states that if \(\mathbf{r}\) is the vector position of the mass relative to the fixed body then

    \(\mathbf{E} = e_2 \frac{\mathbf{r}}{r^3}\)

    \(\mathbf{B} = g_2 \frac{\mathbf{r}}{r^3}\)

    Introducing those fields into the equation of motion, Schwinger finds that the total conserved angular momentum component consists of two terms!

    \(J = \mathbf{r} \times mV - (e_1g_2 - e_2g_1)\frac{1}{c} \frac{\mathbf{r}}{r}\)

    From this one can find a quantization of the charges as shown in the beginning of the first paper

    \(\hbar = \frac{(e_1g_2 - e_2g_1)}{c}\)

    Which is analogous to

    \(\hbar = \frac{Gm^2}{c}\)

    where \(Gm^2\) is the gravitational charge.

    Motz ended up showing an interesting equation of motion with that all important expression \(m\ddot{r}\) we used a lot when studying the torque properties of systems in my first paper

    \(m\ddot{\mathbf{r}} = f(\mathbf{r}, \dot{\mathbf{r}} \nabla r^{-1} - 4Gm^2 \frac{\dot{\mathbf{r}}}{c}[\frac{\dot{\mathbf{r}}}{c} \nabla r^{-1}]\)

    The product of the Coriolis force and the Centrifugal force had been shown before to be equal to \(m\ddot{\mathbf{r}}\) and could even be given in Newtons second law for co-rotating frames in a Euler-Langrange equation, and the Coriolis force is a radial force, thus the radial equation for the centrifugal part was

    \(m\ddot{\mathbf{r}} = mr\dot{\theta}^2 - \frac{dU}{dr}\)

    Interestingly, the Centrifugal force is equivalent to the static magnetic field (see references 1.) and can be taken in a simple limit where the motion is defined on a plane perpendicular to \(\omega\) - this part of the force is a radial force which can be written as
    \(m\omega^2 \ddot{\mathbf{r}}\). The net force of both combined Coriolis and Centrifugal fields was given as

    \(m\omega \times (\omega \times \mathbf{r}) + 2m \omega \times \mathbf{v}_{\mathrm{r}}\)

    This means we can rewrite the net force equation

    \(F_{net} = m\omega^2 \ddot{\mathbf{r}} + 2m \omega \times \mathbf{v}_{\mathrm{r}} = m\omega(\omega\ddot{\mathbf{r}} + 2 \mathbf{v}_{\mathrm{r}})\)

    Ref.

    1. A very rare and difficult to find paper by Motz Downloads/Il%20Nuovo%20Cimento%20B%20Volume%2012%20issue%202%201972%20[doi%2010.1007%252Fbf02822633]%20L.%20Motz%20--%20Gauge%20invariance%20and%20the%20quantization%20of%20mass%20(of%20gravitational%20charge).pdf

    2. http://www.gravityresearchfoundation.org/pdf/awarded/1971/motz.pdf

    3. http://benthamopen.com/toaaj/articles/V004/SI0064TOAAJ/151TOAAJ.pdf

    4. This equation was of interest for a number of things, but what sprung out was \(v \times \mathbf{B}\) which earlier in my work I calculated directly

    \(v \times \mathbf{B} = \frac{2(\omega \times v)}{\sqrt{G}}\)

    which the RHS was identified by Motz as the gravimagnetic field. So naturally the Schwinger equation a two-part equation, describing the gravimagnetic field and the gravielectric field \(v \times \mathbf{E}\)
     
  9. James R Just this guy, you know? Staff Member

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    Reiku?
     
  10. origin Heading towards oblivion Valued Senior Member

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    Gravitational charge??? Oh yeah, reiku for sure...
     
  11. Anew Life isn't a question. Banned

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    Just stopping by to mention that I like the calculous relevance above. seems calculous is and had been used for pulling land gratis of whaleing, initially; mechanical science and outerspace relativism is what it is.

    is the United States of America

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    somewhat of a whale pulled ?nation.

    Please Register or Log in to view the hidden image!

     
    Last edited: Dec 10, 2014
  12. origin Heading towards oblivion Valued Senior Member

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    Why did you feel the need to create a sock puppet? Was it just out of habit?
     
  13. origin Heading towards oblivion Valued Senior Member

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    Did you stop taking your meds reiku?
     
  14. origin Heading towards oblivion Valued Senior Member

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