So let's derive some understanding of spin for this gravimagnetic spin system I will lead us to. We may chose for simplicity that our pointers $$\psi(\uparrow, \downarrow)$$ may appear in a Hamiltonian of the form
$$\mathcal{H} = H_0 + \frac{P^2}{2M} + g\mathcal{O}P$$
Where $$H_0$$ is the unperturbed Hamiltonian. Here $$g$$ is a coupling on the observable denoted as $$\mathcal{O}$$ . This Hamiltonian however designates to one pointer, so we may choose to write two Hamiltonians if need be.
Usually it is taken that is our observable does not commute with the unperturbed Hamiltonian then we need not normally worry how it evolves during a measurement. This means then we may find a Hamiltonian given as:
$$\mathcal{H} = g\mathcal{O} P$$
In fact what we really have here is the Von Neumann Interaction. Since the observable and the momentum act in different Hilbert Spaces one can assume they commute [1].
Instead of measuring any joint observable simultaneously one can do this for each particle seperately and then multiply the results at the end of the trial, as the last paper I linked to explains.
The time evolution operator associated to such an observable can be given as
$$U(t) = exp [-igt\mathcal{O} P]$$
Indeed, one common approach to understanding pointer physics is by a coupling induced by a magnetic field along a certain spin directionality $$\nu(\vec{\sigma})$$ in fact, according to a derivation later on, we certainly measure an observable given as the form $$\hat{n} \cdot \vec{\sigma}$$ and then cleverly work out an angle between the two vectors.
Introducing a new equation, if an initial state were in fact a mixed state with a density matrix $$\rho$$ then we can express this as some pure states
$$\sum_a E_a \rho E_a$$
where $$E_a$$ is an operator. A linear map of the form
$$M:\rho \rightarrow \rho'$$
takes an initial density matrix to a final density matrix. We may make use of this later.
A mixed state $$\psi$$ with a probability $$\mathcal{P}$$ the expectation value is
$$< \hat{A} > = \sum_i \mathcal{P}_i < \psi_i | \hat{A}| \psi_i>$$
Thus an expectation value can be written with a trace of the density matrix as
$$Tr(\rho \hat{A}) = < A >$$
The unit density is given as $$Tr \rho = 1$$. The observable can be related to the density matrix as
$$\bar{\mathcal{O}} = Tr \rho \mathcal{O}$$
Thus an expectation of such an observable
$$\bar{\mathcal{O}} = \sum_i < \psi | \mathcal{O}| i > < i |\psi >$$
Where $$i$$ denotes the i'th state and the unit matrix reduces this to the expectation
$$< \psi|\mathcal{O}| \psi >$$
We can take this calculation in the basis in which $$\rho$$ was diagonal thus
$$\sum_i < i| \rho \mathcal{O}| i >$$
and we can expand our states by introducing $$j$$
$$\sum_{ij} < i |\rho| j > <j|\mathcal{O}|i >$$
To find the jth state for instance from the above expression, you may set $$i=j$$ solve to find probability for the jth state as
$$\sum_j \lambda_j <j| \mathcal{O}|j>$$
These jth and possible ith states could represent a spin subspace which can be given as $$a$$ and $$b$$. An example could be the following ket vector
$$\sum_{ab} \psi(a,b)|ab>$$
in terms of our observable, then solving for the subsystem $$a$$ and not $$b$$ then
$$\sum_{a'b'} <a'b'| \mathcal{O}| \psi(ab)| ab> $$
which if the observable does not act on (with?) $$b$$ then this reduced to the identity
$$Tr\mathcal{O} \rho$$
through atleast four steps. Thus $$a$$ can be either $$(\uparrow, \downarrow)$$ and $$b$$ can take on the observables $$(\downarrow, \uparrow)$$
This was to just give us some flavor of spin mechanics. From here I will be deriving a whole bunch of formulae from well known equations. My task is to find some way to express a relationship between the angle of two spin vectors with a magnetic moment present and finding a relationship to derive force along a certain axis of spin if one chosed, but they will be derived from gravitomagnetic field equations.
Deriving our Gravimagnetic Spin-Field Equations
The density of the gravitational field implies the relationship:
$$\frac{\nabla^2 \phi_{ij}}{4 \theta G_{ij}} = \box (g \phi)$$ [1]
with gravitational coupling in the form of $$g = \frac{\hbar c}{GM^2}$$ between two particles $$k \equiv (i,j)$$ which is defined in a set of interactions $$k \in \mathcal{I}$$.
Now, a new relationship is given as
$$\frac{\eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi_{ij}}{4 \theta_{ij} k(\phi_{ij})} = \frac{\nabla^2 \phi_{ij}}{4\theta G_{ij}}$$
where $$k(\phi)$$ is a function derived from a metric equation and $$\eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi_{ij} = 4 \pi \rho k(\phi)$$.
The gravimagnetic field is given as
$$\frac{2 \vec{\omega} c}{\sqrt{G}} = \Phi$$
solving for $$G$$ yields
$$\frac{4 \vec{\omega}^2 c^2}{\Phi^2} = G$$
Since eq. [1] has dimensions of density, $$\frac{M}{\ell^3}$$, you can obtain a relationship as
$$4 \theta \frac{4 \vec{\omega}^2 c^2}{\Phi^2}\frac{\nabla^2 \phi_{ij}}{4\theta G_{ij}} = \nabla^2 \phi$$
Since $$\nabla^2 \phi = 4 \pi G \rho$$
In understanding the dimensions an equation can be given as
$$\4 \theta \frac{4 \vec{\omega}^2 c^2}{\Phi^2}\rho = \nabla^2 \phi$$
and we already know what $$ \frac{\nabla^2 \phi}{4\theta G_{ij}}$$ is, it is the density $$\box \phi$$ so we can begin to express these equations in relativistic terms.
Eq [1] can be gives as
$$\frac{\eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi}{4\theta k(\phi)} = \box \phi$$
which means the full General Relativistic version of
$$\box \phi = \frac{\nabla^2 \phi}{4 \theta G_{ij}}$$
is
$$\frac{\eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi}{4\theta k(\phi)} = T_{\mu \nu} \delta^{\mu \nu}$$
This means our final equation can in theory describe gravitomagnetic effects. In hindsight, the variables at work may describe the density of a particle due to gravimagnetic charges. Mass is afterall a charge as well and electromagnetic mass theories have existed for a long time. Charge in our current theory are the coefficients of the Lie Algebra's.
Indeed, we may even derive a different relationship
$$\frac{\eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi}{4\theta k(\phi)} = \frac{\nabla^2 \phi}{4 \theta G_{ij}}$$
Cancelling the $$4\theta$$ on both sides then rearranging yields
$$\eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi = \frac{\nabla^2 \phi}{ G_{ij}}k(\phi)$$
But a full interpretation or implications of this equation ellude me.
Now if we take the dot product the unit vector $$\hat{n}$$ with our angular term in eq.[4] (which is a process that calculates spin along a certain axis $$x^{i}$$), then to this multiply this with a column vector we shall gives as $$\begin{bmatrix} \alpha \\ \beta \end{bmatrix}$$ then we end up with the following equation:
$$\frac{\nabla^2 \phi_{ij}}{G_{ij}}((\hat{n} \cdot \vec{\sigma}_{ij}) \begin{bmatrix} \alpha \\ \beta \end{bmatrix}) = \frac{\nabla^2 \phi_{ij}}{G_{ij}} \vec{\theta}_{ij} \rightarrow T_{\mu \nu} \delta^{\mu \nu}$$
This single equation essentially describes the density of the gravitational field strength whilst the seperation between the particles, given an $$(i,j)$$-notation has a spin along a certain axis.
If we use a notation that expresses magnetic moments of the particles along the axes
$$\mu = \mu(\hat{n} \cdot \vec{\sigma}_{ij}) = \begin{bmatrix}\ \mu(n_3) & \mu(n_{-}) \\ \mu(n_{+}) & \mu(-n_3) \end{bmatrix}$$
Then plugging in this new definition, the equation becomes a gravimagnetic spin equation
$$\frac{\nabla^2 \phi_{ij}}{G_{ij}} \begin{bmatrix}\ \mu(n_3) & \mu(n_{-}) \\ \mu(n_{+}) & \mu(-n_3) \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \frac{\nabla^2 \phi_{ij}}{G_{ij}} \mu(\vec{\theta}_{ij}) = \rho$$
The magnetic part is the measure of the particles magnetic moments along the axes in question. If you wish to describe only one particles' gravimagnetic spin relationship just decompose the equation for $$i$$ and $$j$$ seperately.
So we can place a magnetic moment coefficient to the equation we just derived
$$\frac{\nabla^2 \phi_{ij}}{G_{ij}}(\mu(\hat{n} \cdot \vec{\sigma}_{ij}) \begin{bmatrix} \alpha \\ \beta \end{bmatrix}) = \frac{\nabla^2 \phi_{ij}}{G_{ij}} \mu(\vec{\theta}_{ij}) \rightarrow T_{\mu \nu} \delta^{\mu \nu}$$
We usually say that
$$(\mu(\hat{n} \cdot \vec{\sigma}_{ij}) \begin{bmatrix} \alpha \\ \beta \end{bmatrix}) = \mu(\theta_{ij})$$
would calculate the angle between two spin vectors and would look like
$$\frac{1+ cos\theta_{ij}}{2}$$
so what you really have
$$(\mu(\hat{n} \cdot \vec{\sigma}_{ij}) \begin{bmatrix} \alpha \\ \beta \end{bmatrix}) = \mu(\frac{1+ cos\theta_{ij}}{2})$$
Now I am going to derive a force equation along a certain spin axis with a magnetic moment present. The force between two particles can be given as
$$F_{ij} = -\frac{\partial V(r_ij)}{\partial r_{ij}}\hat{n}_{ij}$$
This specific equation will be useful as it contains a unit vector $$\hat{n}_{ij}$$ where $$r_{ij}$$ calculates the distance between two particles $$i$$ and $$j$$ respectively.
I therefore present a new form of this equation as I came to the realization that squaring everything would yield
$$-\frac{\partial^2 V^2 (r_{ij})^2}{\partial^2 r^{2}_{ij}} \mu(\hat{n} \cdot \vec{\sigma}_{ij})^2 = -\frac{\partial^2 V^2 (r_{ij})^2}{\partial^2 r^{2}_{ij}} \begin{bmatrix}\ \mu(n_3) & \mu(n_{-}) \\ \mu(n_{+}) & \mu(-n_3) \end{bmatrix}^2$$
Thus we have derived our force equation compensation for a magnetic moment along the spin axis.
Final Thoughts
It has been comforting to come across a paper which seems to take the idea of a quantum Coriolis Field seriously http://arxiv.org/ftp/arxiv/papers/1009/1009.3788.pdf - of course Motz however never actually called it a quantum coriolis field, but you certainly infer that from his calculations in his paper, ''On the quantization of mass.''
The Coriolis effect for a rotating shell of matter generates inside of itself a field called the Coriolis Force.
It will be taken for now to assume that particles are not truely pointlike particles, that this seemed to have been the route we have taken because it was ''easier'' to deal with.
It goes to say, even for a particle, there must be a small ''twisting effect'' of the gravitational field which could be deemed negligable by General Relativistic effects.
As small as frame-dragging would be for particles, the idea of gravitomagnetic forces for a particle as a rotating sphere may have interesting relationships.
For instance, gravitmagnetism allows bodies to exchange energy in the form of coupling external gravitomagnetic fields niether would they ever undergo a direct collision, though normally in QM we never tend to think of shoving two particle into the same location, in fact the more you try and do this the more energy you require.
It seems that the more I read on this subject, there is some evidence that gravitomagnetic forces might be noticable around water molecules Gravitomagnetism . If this can be extended to particles, I wonder the true implications. In the work so far however, I have entertained the idea of treating mass like a quantization of charge. Charge is simply the coefficients of the Lie Algebra's on the theory.
I want to see mass and charge as ''being the dimensions'' spoke about contained inside of a particle presumably begin to treat particles with a classical radius $$\frac{e^2}{mc^2}$$. I don't like the idea of thinking particle's as having no dimensions but still possessing charge and mass, especially since your usual standard definition of mass requires some volume to define the density of an object. In fact I propose that there is unique limit on experimentation and what information we can get from particles. With current technology, mostly by studying particle collisions we are left believing that particles are pointlike objects. I just argue they are so incredibly small, yet still being spheres, that there sizes in experiments look like they are as if they are pointlike. It puts me in mind of the Weyl Limit, where Neutrino's may be considered massless - they have such a ridiculously small mass anyway that you may treat them like massless radiation. Along the same lines, particles are not pointlike, but are so very small that even in experiments today they seem like they act like pointlike particles.
Indeed, in these latter equations I derived, I made sure that a magnetic moment was in there... magnetic moments simply cannot be generated by a pointlike system since a magnetic field can only be generated by an element of electrical current which implies in some dimensions atleast. Three perhaps, or am I too greedy?
References
http://theory.physics.helsinki.fi/~plasma/luennot05/summary_0502.pdf
http://www.citebase.org/fulltext?format=application/pdf&identifier=oai:arXiv.org:quant-ph/0501072
Motz ''On the quantization of mass''
Sciarma ''On the Origin of Inertia''
$$\mathcal{H} = H_0 + \frac{P^2}{2M} + g\mathcal{O}P$$
Where $$H_0$$ is the unperturbed Hamiltonian. Here $$g$$ is a coupling on the observable denoted as $$\mathcal{O}$$ . This Hamiltonian however designates to one pointer, so we may choose to write two Hamiltonians if need be.
Usually it is taken that is our observable does not commute with the unperturbed Hamiltonian then we need not normally worry how it evolves during a measurement. This means then we may find a Hamiltonian given as:
$$\mathcal{H} = g\mathcal{O} P$$
In fact what we really have here is the Von Neumann Interaction. Since the observable and the momentum act in different Hilbert Spaces one can assume they commute [1].
Instead of measuring any joint observable simultaneously one can do this for each particle seperately and then multiply the results at the end of the trial, as the last paper I linked to explains.
The time evolution operator associated to such an observable can be given as
$$U(t) = exp [-igt\mathcal{O} P]$$
Indeed, one common approach to understanding pointer physics is by a coupling induced by a magnetic field along a certain spin directionality $$\nu(\vec{\sigma})$$ in fact, according to a derivation later on, we certainly measure an observable given as the form $$\hat{n} \cdot \vec{\sigma}$$ and then cleverly work out an angle between the two vectors.
Introducing a new equation, if an initial state were in fact a mixed state with a density matrix $$\rho$$ then we can express this as some pure states
$$\sum_a E_a \rho E_a$$
where $$E_a$$ is an operator. A linear map of the form
$$M:\rho \rightarrow \rho'$$
takes an initial density matrix to a final density matrix. We may make use of this later.
A mixed state $$\psi$$ with a probability $$\mathcal{P}$$ the expectation value is
$$< \hat{A} > = \sum_i \mathcal{P}_i < \psi_i | \hat{A}| \psi_i>$$
Thus an expectation value can be written with a trace of the density matrix as
$$Tr(\rho \hat{A}) = < A >$$
The unit density is given as $$Tr \rho = 1$$. The observable can be related to the density matrix as
$$\bar{\mathcal{O}} = Tr \rho \mathcal{O}$$
Thus an expectation of such an observable
$$\bar{\mathcal{O}} = \sum_i < \psi | \mathcal{O}| i > < i |\psi >$$
Where $$i$$ denotes the i'th state and the unit matrix reduces this to the expectation
$$< \psi|\mathcal{O}| \psi >$$
We can take this calculation in the basis in which $$\rho$$ was diagonal thus
$$\sum_i < i| \rho \mathcal{O}| i >$$
and we can expand our states by introducing $$j$$
$$\sum_{ij} < i |\rho| j > <j|\mathcal{O}|i >$$
To find the jth state for instance from the above expression, you may set $$i=j$$ solve to find probability for the jth state as
$$\sum_j \lambda_j <j| \mathcal{O}|j>$$
These jth and possible ith states could represent a spin subspace which can be given as $$a$$ and $$b$$. An example could be the following ket vector
$$\sum_{ab} \psi(a,b)|ab>$$
in terms of our observable, then solving for the subsystem $$a$$ and not $$b$$ then
$$\sum_{a'b'} <a'b'| \mathcal{O}| \psi(ab)| ab> $$
which if the observable does not act on (with?) $$b$$ then this reduced to the identity
$$Tr\mathcal{O} \rho$$
through atleast four steps. Thus $$a$$ can be either $$(\uparrow, \downarrow)$$ and $$b$$ can take on the observables $$(\downarrow, \uparrow)$$
This was to just give us some flavor of spin mechanics. From here I will be deriving a whole bunch of formulae from well known equations. My task is to find some way to express a relationship between the angle of two spin vectors with a magnetic moment present and finding a relationship to derive force along a certain axis of spin if one chosed, but they will be derived from gravitomagnetic field equations.
Deriving our Gravimagnetic Spin-Field Equations
The density of the gravitational field implies the relationship:
$$\frac{\nabla^2 \phi_{ij}}{4 \theta G_{ij}} = \box (g \phi)$$ [1]
with gravitational coupling in the form of $$g = \frac{\hbar c}{GM^2}$$ between two particles $$k \equiv (i,j)$$ which is defined in a set of interactions $$k \in \mathcal{I}$$.
Now, a new relationship is given as
$$\frac{\eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi_{ij}}{4 \theta_{ij} k(\phi_{ij})} = \frac{\nabla^2 \phi_{ij}}{4\theta G_{ij}}$$
where $$k(\phi)$$ is a function derived from a metric equation and $$\eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi_{ij} = 4 \pi \rho k(\phi)$$.
The gravimagnetic field is given as
$$\frac{2 \vec{\omega} c}{\sqrt{G}} = \Phi$$
solving for $$G$$ yields
$$\frac{4 \vec{\omega}^2 c^2}{\Phi^2} = G$$
Since eq. [1] has dimensions of density, $$\frac{M}{\ell^3}$$, you can obtain a relationship as
$$4 \theta \frac{4 \vec{\omega}^2 c^2}{\Phi^2}\frac{\nabla^2 \phi_{ij}}{4\theta G_{ij}} = \nabla^2 \phi$$
Since $$\nabla^2 \phi = 4 \pi G \rho$$
In understanding the dimensions an equation can be given as
$$\4 \theta \frac{4 \vec{\omega}^2 c^2}{\Phi^2}\rho = \nabla^2 \phi$$
and we already know what $$ \frac{\nabla^2 \phi}{4\theta G_{ij}}$$ is, it is the density $$\box \phi$$ so we can begin to express these equations in relativistic terms.
Eq [1] can be gives as
$$\frac{\eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi}{4\theta k(\phi)} = \box \phi$$
which means the full General Relativistic version of
$$\box \phi = \frac{\nabla^2 \phi}{4 \theta G_{ij}}$$
is
$$\frac{\eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi}{4\theta k(\phi)} = T_{\mu \nu} \delta^{\mu \nu}$$
This means our final equation can in theory describe gravitomagnetic effects. In hindsight, the variables at work may describe the density of a particle due to gravimagnetic charges. Mass is afterall a charge as well and electromagnetic mass theories have existed for a long time. Charge in our current theory are the coefficients of the Lie Algebra's.
Indeed, we may even derive a different relationship
$$\frac{\eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi}{4\theta k(\phi)} = \frac{\nabla^2 \phi}{4 \theta G_{ij}}$$
Cancelling the $$4\theta$$ on both sides then rearranging yields
$$\eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi = \frac{\nabla^2 \phi}{ G_{ij}}k(\phi)$$
But a full interpretation or implications of this equation ellude me.
Now if we take the dot product the unit vector $$\hat{n}$$ with our angular term in eq.[4] (which is a process that calculates spin along a certain axis $$x^{i}$$), then to this multiply this with a column vector we shall gives as $$\begin{bmatrix} \alpha \\ \beta \end{bmatrix}$$ then we end up with the following equation:
$$\frac{\nabla^2 \phi_{ij}}{G_{ij}}((\hat{n} \cdot \vec{\sigma}_{ij}) \begin{bmatrix} \alpha \\ \beta \end{bmatrix}) = \frac{\nabla^2 \phi_{ij}}{G_{ij}} \vec{\theta}_{ij} \rightarrow T_{\mu \nu} \delta^{\mu \nu}$$
This single equation essentially describes the density of the gravitational field strength whilst the seperation between the particles, given an $$(i,j)$$-notation has a spin along a certain axis.
If we use a notation that expresses magnetic moments of the particles along the axes
$$\mu = \mu(\hat{n} \cdot \vec{\sigma}_{ij}) = \begin{bmatrix}\ \mu(n_3) & \mu(n_{-}) \\ \mu(n_{+}) & \mu(-n_3) \end{bmatrix}$$
Then plugging in this new definition, the equation becomes a gravimagnetic spin equation
$$\frac{\nabla^2 \phi_{ij}}{G_{ij}} \begin{bmatrix}\ \mu(n_3) & \mu(n_{-}) \\ \mu(n_{+}) & \mu(-n_3) \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \frac{\nabla^2 \phi_{ij}}{G_{ij}} \mu(\vec{\theta}_{ij}) = \rho$$
The magnetic part is the measure of the particles magnetic moments along the axes in question. If you wish to describe only one particles' gravimagnetic spin relationship just decompose the equation for $$i$$ and $$j$$ seperately.
So we can place a magnetic moment coefficient to the equation we just derived
$$\frac{\nabla^2 \phi_{ij}}{G_{ij}}(\mu(\hat{n} \cdot \vec{\sigma}_{ij}) \begin{bmatrix} \alpha \\ \beta \end{bmatrix}) = \frac{\nabla^2 \phi_{ij}}{G_{ij}} \mu(\vec{\theta}_{ij}) \rightarrow T_{\mu \nu} \delta^{\mu \nu}$$
We usually say that
$$(\mu(\hat{n} \cdot \vec{\sigma}_{ij}) \begin{bmatrix} \alpha \\ \beta \end{bmatrix}) = \mu(\theta_{ij})$$
would calculate the angle between two spin vectors and would look like
$$\frac{1+ cos\theta_{ij}}{2}$$
so what you really have
$$(\mu(\hat{n} \cdot \vec{\sigma}_{ij}) \begin{bmatrix} \alpha \\ \beta \end{bmatrix}) = \mu(\frac{1+ cos\theta_{ij}}{2})$$
Now I am going to derive a force equation along a certain spin axis with a magnetic moment present. The force between two particles can be given as
$$F_{ij} = -\frac{\partial V(r_ij)}{\partial r_{ij}}\hat{n}_{ij}$$
This specific equation will be useful as it contains a unit vector $$\hat{n}_{ij}$$ where $$r_{ij}$$ calculates the distance between two particles $$i$$ and $$j$$ respectively.
I therefore present a new form of this equation as I came to the realization that squaring everything would yield
$$-\frac{\partial^2 V^2 (r_{ij})^2}{\partial^2 r^{2}_{ij}} \mu(\hat{n} \cdot \vec{\sigma}_{ij})^2 = -\frac{\partial^2 V^2 (r_{ij})^2}{\partial^2 r^{2}_{ij}} \begin{bmatrix}\ \mu(n_3) & \mu(n_{-}) \\ \mu(n_{+}) & \mu(-n_3) \end{bmatrix}^2$$
Thus we have derived our force equation compensation for a magnetic moment along the spin axis.
Final Thoughts
It has been comforting to come across a paper which seems to take the idea of a quantum Coriolis Field seriously http://arxiv.org/ftp/arxiv/papers/1009/1009.3788.pdf - of course Motz however never actually called it a quantum coriolis field, but you certainly infer that from his calculations in his paper, ''On the quantization of mass.''
The Coriolis effect for a rotating shell of matter generates inside of itself a field called the Coriolis Force.
It will be taken for now to assume that particles are not truely pointlike particles, that this seemed to have been the route we have taken because it was ''easier'' to deal with.
It goes to say, even for a particle, there must be a small ''twisting effect'' of the gravitational field which could be deemed negligable by General Relativistic effects.
As small as frame-dragging would be for particles, the idea of gravitomagnetic forces for a particle as a rotating sphere may have interesting relationships.
For instance, gravitmagnetism allows bodies to exchange energy in the form of coupling external gravitomagnetic fields niether would they ever undergo a direct collision, though normally in QM we never tend to think of shoving two particle into the same location, in fact the more you try and do this the more energy you require.
It seems that the more I read on this subject, there is some evidence that gravitomagnetic forces might be noticable around water molecules Gravitomagnetism . If this can be extended to particles, I wonder the true implications. In the work so far however, I have entertained the idea of treating mass like a quantization of charge. Charge is simply the coefficients of the Lie Algebra's on the theory.
I want to see mass and charge as ''being the dimensions'' spoke about contained inside of a particle presumably begin to treat particles with a classical radius $$\frac{e^2}{mc^2}$$. I don't like the idea of thinking particle's as having no dimensions but still possessing charge and mass, especially since your usual standard definition of mass requires some volume to define the density of an object. In fact I propose that there is unique limit on experimentation and what information we can get from particles. With current technology, mostly by studying particle collisions we are left believing that particles are pointlike objects. I just argue they are so incredibly small, yet still being spheres, that there sizes in experiments look like they are as if they are pointlike. It puts me in mind of the Weyl Limit, where Neutrino's may be considered massless - they have such a ridiculously small mass anyway that you may treat them like massless radiation. Along the same lines, particles are not pointlike, but are so very small that even in experiments today they seem like they act like pointlike particles.
Indeed, in these latter equations I derived, I made sure that a magnetic moment was in there... magnetic moments simply cannot be generated by a pointlike system since a magnetic field can only be generated by an element of electrical current which implies in some dimensions atleast. Three perhaps, or am I too greedy?
References
http://theory.physics.helsinki.fi/~plasma/luennot05/summary_0502.pdf
http://www.citebase.org/fulltext?format=application/pdf&identifier=oai:arXiv.org:quant-ph/0501072
Motz ''On the quantization of mass''
Sciarma ''On the Origin of Inertia''
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