DR. Nobody
Registered Member
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 $$m \ne n$$.
$$Gm^2 = \frac{\pm \alpha \hbar \hat{n}}{4 \pi \epsilon_0 \mu_o c}$$
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 $$m \ne n$$.
somewhat of a whale pulled ?nation.