A New Realization Into The Second Order Dirac Operator And Planck Particles

[Aethelwulf]

There must be another way to describe the sphere's in my theory without being concerned with all these terms in this equation

 

[math]E_{GEM} = \frac{1}{2}(\frac{G_I M^2(\phi)}{r_s} + \frac{e^2}{4 \pi \epsilon r_s}) + (M\phi + k e^2)[/math]

 

But they were chosen for a good reason mind you. For instance, very early on, I said that particles may not be truly pointlike, but behave as though they have a potential [math]1/r[[/math], intrinsically-speaking of course. This is why in the term

 

[math]ke^2 = \frac{e}{4\pi \epsilon}r^{-1}[/math]

 

It was essential to have the [math]1/r = r^{-1}[/math] describing the potential inside of the sphere. The potential varies for instance as [math]1/r[/math] while the force (notably the strong gravitational constant) would vary as [math]1/r^2[/math]. The same can be said about the gravitational part [math]M\phi = \frac{G M^2(\phi)}{r_s}[/math] as it will also vary depending on [math]1/r[/math] inside of spheres which was going to lead me to try and explain that perhaps this is how different particle masses come about.

 

I think I might leave this part for now and go on to a new topic later.

[Aethelwulf]

How does the gravitational charge play a role in particles and antiparticles? To describe this, we need to use the alpha Dirac matrix.

 

 

 

 

We begin with a wave equation

 

[math]c \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial t} = 0[/math]

 

Concerning a wave function with the identity

 

[math]\Psi = e^{i(kx - \omega t)}[/math]

 

Pulling down the derivative with respect to [math]x[/math] and [math]t[/math] we have

 

[math]ikc - i\omega = 0[/math]

 

Keep in mind this is not hard to see since [math]\frac{\omega}{k} = c[/math]

 

Multipying by the quantum action we have

 

[math]i\hbar(kc - \omega) = 0[/math]

 

Using an equation

 

[math]\frac{\partial}{\partial t} \psi = -\alpha c \frac{\partial}{\partial x} \psi[/math]

 

Where alpha is the Dirac Matrix, using the same idea, pulling down derivatives and melding it into our previous derivation we have

 

[math]-i \hbar \omega = -\alpha -i k \hbar c[/math]

 

Using the equivalence of quantization method for charge [math]\hbar c = GM^2[/math] and simplifying we end up with

 

[math]\hbar \omega = \alpha k GM^2[/math]

 

Incidently, the angular frequency is a matrix also.

 

Dividing off the wave number on both sides yields the gravitational charge on the right handside, with the exception that it has the Dirac alpha matrix coefficient

 

[math]\frac{\hbar \omega}{k} = \alpha GM^2 [/math]

 

The alpha matrix is basically

 

[math]\alpha = \begin{pmatrix} 1 & 0 \\0 & -1 \end{pmatrix}[/math]

 

we have

 

[math]\frac{\hbar \omega}{k} = \begin{pmatrix} 1 & 0 \\0 & -1 \end{pmatrix} GM^2[/math]

 

The matrix alpha is in fact [math]+1[/math] for right moving waves and [math]-1[/math] for left moving waves (particles and antiparticles). If we hit the equation

 

[math]\frac{\hbar \omega}{k} = \begin{pmatrix} 1 & 0 \\0 & -1 \end{pmatrix} GM^2[/math]

 

With a column vector

 

[math]\displaystyle \binom{\frac{\partial \psi_R}{\partial x}}{\frac{\partial \psi_L}{\partial x}}[/math]

 

We get

 

[math]\frac{\hbar \omega}{k} \binom{\dot{\psi}_R}{\dot{\psi}_L} = - \begin{pmatrix} 1 & 0 \\0 & -1 \end{pmatrix}\displaystyle \binom{\frac{\partial \psi_R}{\partial x}}{\frac{\partial \psi_L}{\partial x}} GM^2[/math]

 

This is just a convienient trick to writing two linear equations into matrix notation for right movers and left movers.

Edited December 4, 2012 by Aethelwulf

[Aethelwulf]

Incidently none of this is relativistic. For a moment I was worried about my dimensions

 

[math]\hbar \omega = \alpha k GM^2[/math]

 

But since rearranging

 

[math]\frac{\hbar \omega}{GM^2} = \alpha k[/math]

 

And since [math]\hbar c = GM^2[/math] then simplifying yields

 

[math]\frac{\omega}{c} = \alpha k[/math]

 

This is true because of the relation

 

[math]\frac{\omega}{k} = c[/math]

 

Thus

 

[math]\omega = kc[/math]

 

then

 

[math]\frac{\omega}{c} = \alpha k[/math]

 

You can also tell it has the right dimensions by knowing that the wave number has dimensions of [math]\ell^{-1}[/math], meaning that (from my most earliest equations in the first post that

 

[math]E = \frac{GM^2}{\ell}[/math]

 

Anyway, as I stated, this part

 

[math]\hbar \omega = \alpha k GM^2[/math]

 

Is not relativistic, to make it relativistic, we would need to add a mass term with the beta matrix

 

[math]\hbar \omega = \alpha k GM^2 + \beta mc^2[/math]

 

This means that even though

 

[math]\hbar \omega = \alpha k GM^2[/math]

 

it is dealing with a gravitational charge squared, it is not actually talking about a system with a mass which is a curious outcome. To do that you need to add the full form of

 

[math]\hbar \omega = \alpha k GM^2 + \beta mc^2[/math]

 

Only now can you begin to write the equation out relativistically. I will expand on this tomorrow as there is no doubt plenty to write about it.

Edited December 4, 2012 by Aethelwulf

[Aethelwulf]

I will note however so I don't forget tomorrow, that

 

[math]\hbar \omega = \alpha k GM^2 + \beta Mc^2[/math]

 

Can be rewritten also as a Hamiltonian (or a Dirac Hamiltonian as)

 

[math]\mathcal{H} = i GM^2 \alpha \cdot \vec{\nabla} + \beta Mc^2[/math]

[Aethelwulf]

So let's revisit the equation

 

[math]\hbar \omega = \alpha k GM^2 + \beta mc^2[/math]

 

Let us replace the energy [math]\hbar \omega[/math] with it's respective quantum mechanical operator and also inserting a wave function

 

[math]i\hbar \frac{\partial}{\partial t} \psi = (\alpha k GM^2 + \beta mc^2)\psi[/math]

 

So we have our Dirac-like equation in this form now.

 

What is odd, is that in the beginning of investigating the relationship [math]GM^2 = \hbar c[/math] is that I associated [math]GM^2[/math] with inertial systems strictly. The normal Dirac Equation looks something like

 

[math]i \hbar \dot{\psi} = -i\hbar \partial_x \cdot \alpha + mc^2 \beta[/math]

 

If you wanted to talk about massless systems, you have an equation simply reduced to (In the normal Dirac form)

 

[math]E = \alpha pc[/math]

 

So as far as I can see

 

[math]i\hbar \dot{\psi} = (\alpha k GM^2 + \beta mc^2)\psi[/math]

 

Is incompatible in describing massless systems. What a mass term does however is that it couples particles and antiparticles together

 

[math]i\hbar \dot{\psi}_R = \alpha k GM^2 \psi_R + \beta mc^2 \psi_L[/math]

 

[math]i\hbar \dot{\psi}_L = \alpha k GM^2 \psi_L + \beta mc^2 \psi_R[/math]

 

You can just write it in this form

 

[math]i\hbar \dot{\psi} = \alpha k GM^2 \psi + \beta mc^2 \psi[/math]

 

Keeping in mind the left is columns and the right is made of matrices. It is actually the matrix beta which interchanges psi-left to psi-right particles, which is how they become coupled through the mass term for particles like electrons (fermions). The solutions are also Lorentz Covariant.

 

For massless systems, there is the limit such that [math]M \rightarrow 0[/math], the second limit one may think about then is for slowly moving systems ie. non-relativistic.

 

At this limit you rewrite the equations as

 

[math]i\hbar \dot{\psi}_R = \beta mc^2 \psi_L[/math]

 

[math]i\hbar \dot{\psi}_L = \beta mc^2 \psi_R[/math]

 

But here is the catch, particles are never truly at rest - when physicists treat particles at rest, they set the momentum to zero which is a wrong way to apply the physics, because particles are never truly at zero.

 

This would actually be seen in light of a non-vanishing mass on the (so-called) momentum term of my equation [math]\alpha k GM^2 \psi[/math]. There is no true vanishing term of momentum for systems. The mass in this term cannot truly vanish either if [math]\beta mc^2 \psi[/math] is allowed to be non-zero.

 

This all comes to the inconsistency of relativity theory, it is completely classical meaning it does not take into account uncertainty. Particles are never truly at rest and their momentum is never truly zero, but in my case you can have zero mass which means this term [math]\alpha k GM^2 \psi[/math] is inconsistent when describing massless systems... or is it?

 

Craig at this site pointed out that photons might have a mass, but I explained it would be tremendously small, a proposed value is about [math]10^{-51}[/math]. Does treating photons with mass solve more problems than what it creates in theory? If a photon does have a mass, the squared mass term in [math]\alpha k GM^2 \psi[/math] would have to be small - very small in fact! I could be just implementing bad physics to be honest, when early on replacing [math]\hbar c[/math] for [math]GM^2[/math], even when they are completely equivalent.

Edited December 8, 2012 by Aethelwulf

[Aethelwulf]

Some might however treat the photon with its ''mass,'' given as [math]E/c^2[/math].

 

In this case, a photons ''so-called'' mass is

 

[math]\frac{E}{c^2} = M[/math]

 

In this case, we may alter the equations to satisfy

 

[math]i\hbar \frac{\partial}{\partial t} \psi = (\alpha k G (\frac{E}{c^2})^2 + \beta mc^2)\psi[/math]

 

Which may actually be a solution out of this problem.

Edited December 8, 2012 by Aethelwulf

[Aethelwulf]

This actually comes back to the so-called ''electromagnetic mass theories,'' which involves radiation pressure and inertia. Einstein concluded a body of mass gives up energy in the form of [math]E/c^2 = M[/math]. It was, as wiki states and I quote

 

'' In 1900, Henri Poincaré associated electromagnetic radiation energy with a "fictitious fluid" having momentum and mass.''

 

''Friedrich Hasenöhrl showed in 1904, that electromagnetic cavity radiation contributes the "apparent mass" [math]m_{0}=\tfrac{4}{3}\tfrac{E_{em}}{c^{2}}[/math] to the cavity's mass. He argued that this implies mass dependence on temperature as well.''

Edited December 8, 2012 by Aethelwulf

[Aethelwulf]

So, suppose for a second that in the equation

 

[math]i\hbar \frac{\partial}{\partial t} \psi = (\alpha k G (\frac{E}{c^2})^2 + \beta mc^2)\psi[/math]

 

The part [math]\frac{E}{c^2}[/math] is attributed to the ''contribution of mass,'' in that case we simply go back to my equations which describe that contribution, this one specifically

 

[math]M = \frac{1}{2} \frac{GM^2}{r_s c^2}[/math]

 

Therefore we would have

 

[math]i\hbar \frac{\partial}{\partial t} \psi = (\alpha k G (\frac{GM^2}{r_s c^2}) + \beta mc^2)\psi[/math]

Edited December 8, 2012 by Aethelwulf

[Aethelwulf]

Using variational calculus, you can also find the Langrangian to

 

[math]i\hbar \frac{\partial}{\partial t} \psi + \alpha k G (\frac{GM^2}{r_s c^2} - \beta mc^2)\psi = 0[/math]

 

We variate this with [math]\psi^{\dagger}[/math]

 

We will have the Dirac Langrangian

 

[math]\psi{\dagger}(i\hbar \frac{\partial}{\partial t} \psi + \alpha k G (\frac{GM^2}{r_s c^2}\psi - \beta mc^2\psi) = \mathcal{L}[/math]

Edited December 8, 2012 by Aethelwulf

[Aethelwulf]

... deleted

Edited August 29, 2013 by Aethelwulf

[Aethelwulf]

You can use gamma matrices to simplify this even further but I won't this time around.

[Aethelwulf]

....

Edited August 29, 2013 by Aethelwulf