# A Thought Experiment. An Electron Orbiting close to the perimeter of ...

Discussion in 'Alternative Theories' started by Trapped, Jan 31, 2014.

1. ### TrappedBannedBanned

Messages:
1,058
I've been seeing loads of people post their ''pet theories'' so I have one of my own, but with some math with it. Using a completely geometrical vector picture of a trajectory in curvilinear geodesics of an electron which couples it's spin to the gravitational field of the black hole.

Particle Thought Experiment Following Geodesics

Concerning a particle like an electron bound by the gravitational curvature of a black hole, travelling around the perimeter of the event horizon, there are some conditions we must take into consideration.

First of all, there will be a spin-orbit coupling to the gravitational force. The gravitational feature of this configuration will not appear in the equations immediately, first we must describe the quantum relationship of our particle following this closed curve displacement. To do this, I will use vector notation. Then I took this energy to be the energy component in the Covariant Dirac Equation.

Before we get to the equation which describes the energy contributed by the spin of the particle and the angle of coupling to it's trajectory, we shall just quickly take a look at some equations that was derived in this work.

The first part we will concentrate on is

$\frac{\partial V(r)}{\partial r}$

$\nabla \times F = \begin{vmatrix}\hat{n}_1 & \hat{n}_2 & \hat{n}_3 \\ \partial_x & \partial_y & \partial_z \\F_x & F_y & 0 \end{vmatrix}$

You can write this as

$\nabla \times F = \frac{\partial F_y}{\partial x}\hat{n}_1 - \frac{\partial F_x}{\partial y}\hat{n}_2 + (\frac{\partial F_y}{\partial z} - \frac{\partial F_x}{\partial z})\hat{n}_3$

The first set of terms cancel out

$\nabla \times F = (\frac{\partial F_y}{\partial z} - \frac{\partial F_x}{\partial z})\hat{n}_3$

A unit vector squared just comes to unity, so if you multiply a unit vector of both sides we get

$\nabla \times F \cdot \hat{n} = (\frac{\partial F_y}{\partial z} - \frac{\partial F_x}{\partial z})$

Now, a force equation can be given as

$F = \frac{\partial V(r)}{\partial r} \hat{n}$

Again, if one multiplies the unit vector on both sides we get

$F \cdot \hat{n} = \frac{\partial V(r)}{\partial r}$

This leads us to some of our calculations we will use to describe our thought experiment.

We have a unit force related to the inverse curl operator acting on the electromagnetic potential $A$

$(\frac{1}{eMc^2})\frac{1}{r} F \cdot \hat{n}\ J = \frac{1}{eMc^2}\frac{1}{r}\frac{\partial V(r)}{\partial r}\ J =(\nabla \times)^{-1} \mathbf{A}_{\mu}$ 1.

where

$\mathbf{A}_{\mu}$ 2.

Is a four potential

so that

$F \cdot\hat{n} = \frac{\partial V(r)}{\partial r}$ 3.

is related to the quantity

$(\nabla \times)^{-1} \mathbf{A}_{\mu}$ 4.

(which if my memory serves right) shouldn't be much of a surprise because the inverse of the curl involves the potential of a system, so that the spin-coupling energy can be written as

$\Delta Mc^2 = \frac{2 \mu}{\hbar Mc^2 e} r^{-1} (\nabla \times)^{-1} \mathbf{A}_{\mu} (J \cdot S)$

where $J$ is the angular momentum of the black hole and $S$ is the spin of the electron, which produces $|J||S|cos \theta$ which acts as the angle between the two quantities.

Using purely vector notation, this equation is not only about a force inside the model gravi-spin-orbit coupling due to frame dragging effects (the unification of gravitational and magnetic forces), but is about the orbit itself, a closed curve displacement. Specifically, the perimeter of the closed curve in which the electron is moving in, in vector notation.

Even though, the equation is written in terms of a coupling between the magnetic force with the spin of the particle, gravitomagnetism refers to the kinetic effects of the gravitational field, which is an analogy to the magnetic effects of moving electric charges.

An electron accelerating round the perimeter of a black hole would experience, similar analogies.

The spin of an electron is $\frac{1}{2}$ and when being described in a gravitational field $g_{\mu \nu}$ it given by the covariant Dirac equation

$[i \gamma^{\mu}(x)\mathcal{D}_{\mu} - mc^2]\Psi(x) = 0$

The coupling to the gravitational field uses the connection $\mathcal{D}_{\mu} = \nabla_{\mu} + \Gamma_{\mu}(x), \Gamma_{\mu}(x)$

Placing our particle energy into the Covariant Dirac Equation, then it can couple to gravity in such way:

$[i \gamma^{\mu}(x)\mathcal{D}_{\mu} - \frac{2 \mu}{\hbar Mc^2 e} r^{-1} (\nabla \times)^{-1} \mathbf{A}_{\mu} (J \cdot S)]\Psi(x) = 0$

The simplified version of this equation is

$[i \gamma^{\mu}(x)\mathcal{D}_{\mu} - \frac{1}{r}\frac{2 \mu}{\hbar Mc^2 e} F \cdot \hat{n} (J \cdot S)]\Psi(x) = 0$

or even

$[i \gamma^{\mu}(x)\mathcal{D}_{\mu} - \frac{1}{r}\frac{2 \mu}{\hbar Mc^2 e} \frac{\partial V(r)}{\partial r}(J \cdot S)]\Psi(x) = 0$

Not only does this describe the spin connection to the gravitational field, but also of the magnetic field of the electron.

Last edited: Jan 31, 2014
2. ### Google AdSenseGuest Advertisement

to hide all adverts.
3. ### TrappedBannedBanned

Messages:
1,058
There is a really nice way to picture my theoretical model using the math presented. We shall just take a quick look at the magnetic components of the equation

$(\frac{1}{eMc^2})\frac{1}{r} F \cdot \hat{n}\ J = \frac{1}{eMc^2}\frac{1}{r}\frac{\partial V(r)}{\partial r}\ J =(\nabla \times)^{-1} \mathbf{A}_{\mu}$

Now, I ask the question, what is it that allows the gravitational feature of this physics to allow a coupling of the equation above to the dynamics of our spinning black hole? We have been arguing cases for fields which distort spacetime and result in electrons following a tight orbit in our hypothetical experiment. The key word here was... rotating. It needs to allow a frame dragging process and to obtain this, we first understand that for there to be frame-dragging, there must be a local Coriolis force field surrounding the gravitationally-dragged particle. According to Sciarma who has worked on matter, inertia and other gravitationally-related subjects, says you obtain the coupling of gravimagnetic fields through a cross product, similar to what I am about to perform and solve for you

$v_{B} \times (\nabla \times)^{-1} \mathbf{A}_{\mu} = \frac{2(\omega \times v)}{\sqrt{G}}$

This equation allows us to see the coupling where $v_B$ is the rotational velocity of the black hole and $(\omega \times \mathcal{v})$ is the Coriolis acceleration experienced by gravimagnetic system.

4. ### Google AdSenseGuest Advertisement

to hide all adverts.
5. ### TrappedBannedBanned

Messages:
1,058
Perhaps more interesting than that, is if you multiply through $\sqrt{G}M$ which is an equivalent expression for the charge of the system, $\frac{e}{\sqrt{4 \pi \epsilon}}$

$F = \sqrt{G}M(v_{B} \times (\nabla \times)^{-1} \mathbf{A}_{\mu})$

It's purely gravitational four force on the particle is dictated by

$F_{\mu} = \Gamma^{\lambda}_{\mu \nu} u^{\mu} p^{\nu}$

This force seems to be completely gravimagnetic in nature.

6. ### Google AdSenseGuest Advertisement

to hide all adverts.
7. ### TrappedBannedBanned

Messages:
1,058
An even nicer mathematical situation can be investigated. We know the magnetic components have been defined as

$(\frac{1}{eMc^2})\frac{1}{r} F \cdot \hat{n}\ J = \frac{1}{eMc^2}\frac{1}{r}\frac{\partial V(r)}{\partial r}\ J =(\nabla \times)^{-1} \mathbf{A}_{\mu}$

You can also rewrite the magnetic part as

$B = -\frac{v \times E}{c^2}$

Let's set this equal to the last expression

$-\frac{v_e \times E}{c^2} = (\nabla \times)^{-1} \mathbf{A}_{\mu}$

The electric field is radial and the momentum of the electron is $p = M_e c$. If we substitute this into the equation and also change the order of cross products, we have a new equation again

$-\frac{r \times p}{Mc^2}|\frac{E}{r}| = (\nabla \times)^{-1} \mathbf{A}_{\mu}$

To retain what we began with, you simply replace the electric field with the gradient of the electric potential and you get back all that neat stuff we began with.

8. ### TrappedBannedBanned

Messages:
1,058
$v_{B} \times (\nabla \times)^{-1} \mathbf{A}_{\mu} = \frac{2(\omega \times v)}{\sqrt{G}}$

(1)

$[i \gamma^{\mu}(x)\mathcal{D}_{\mu} - \frac{2 \mu}{\hbar Mc^2 e} r^{-1} (\nabla \times)^{-1} \mathbf{A}_{\mu} (J \cdot S)]\Psi(x) = 0$

(2)

Moving part of it to the right

$(i \gamma^{\mu}(x)\mathcal{D}_{\mu})\psi = (\frac{2 \mu}{\hbar Mc^2 e} r^{-1} (\nabla \times)^{-1} \mathbf{A}_{\mu} (J \cdot S))\psi$

(eq. 3)

Let $i \gamma^{\mu}(x)\mathcal{D}_{\mu}$ be simplified into some single notation, say, $\mathcal_{D}_{\mu \dagger}$ so that we may rewrite later using a variational mathematical approach. I don't want to confuse variational notation when I attempt to write it out. Probably not tonight no less.

Just to quickly finish, we have an equation which is about the curvature of spacetime on the LHS, the same geodesic action which is related to the trajectory of our electron, whilst on the right, we have a geometrical interpretation as well involving the magnetic field as the same electron moves in the EM field which as we all know, is how charge comes about. Both are fields, created by a system moving through a gravitational warped field, in a trajectory which couples to the magnetic field to the spin of the particle.

Using the concise language of mathematics, this is really a vector bundle $V$ on a Riemannian manifold $\mathcal{M}$. You do complicated things like involving ''connections'' as they are called in GR, the connection $\Gamma$ defines the gravitational field and this uses the mathematical object called the Covariant Derivative... if one ever get's to study how these objects involve themselves in the original field equations of Einstein, one appreciates how beautiful nature really is.

The curvature forms on the left hand side of our equation (3) are dictating how the gravitational force and the magnetic force couples through the cross product given in equation (1).

9. ### Steve100O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔OValued Senior Member

Messages:
2,346
Is Trapped Reiku?

Messages:
1,058
.... nvm

11. ### TrappedBannedBanned

Messages:
1,058
I haven't been writing it out properly but it can be easily fixed. I forgot the order of how to write the inverse of the operator, it should have been

$(\frac{1}{eMc^2})\frac{1}{r} F \cdot \hat{n}\ J = \frac{1}{eMc^2}\frac{1}{r}\frac{\partial V(r)}{\partial r}\ J = \nabla \times A_\mu$

You can also rewrite the magnetic part as, this is true

$B = -\frac{v \times E}{c^2}$

again, setting it equal to the last expression

$-\frac{v_e \times E}{c^2} = \nabla \times A$

$-\frac{r \times p}{Mc^2}|\frac{E}{r}| = \nabla \times A_{\mu}$

so that one can get the EM potential

$-\frac{r \times p}{Mc^2}|\frac{E}{r}| = \nabla \times A_{\mu}$

$-(\nabla \times)^{-1} \frac{r \times p}{Mc^2}|\frac{E}{r}| = A_{\mu}$

and

$B = -\frac{2 \omega}{\sqrt{G}}$

For a gravitational feature of magnetism

That's was quite an error which was carried through so you'll need to forgive the mistake and go through the equations replacing it with the correct way to write the inverse of the curl operator correctly. That also means I got the cross product wrong in the coupling term

$v_{B} \times (\nabla \times \mathbf{A}_{\mu}) = \frac{2(\omega \times v)}{\sqrt{G}}$

Last edited: Feb 6, 2014
12. ### originHeading towards oblivionValued Senior Member

Messages:
11,154
Nothing to see here. These are just random equations that trapped did not understand and then he tried to 'glue' together with junior highschool algebra. Move along and try not to step in the pile of crap...