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

[Aethelwulf]

This all appears to be crazy talk to me! I never understand these formulas because I never know the values of the numbers as well as I have never been any good at math. I tried figuring out why I consistantly got the wrong answer on a previous post on thermal expansion but eventually just gave up... I was only off by a small amount but that little bit I am sure adds up when it needs to be accurate... I love physics but I need a giant cheat sheet just to get by lol. Well I am not sure what it is your doing on this topic but I am sure its smart guy stuff and it looks like ya got a good grip on it lol so ill leave it be!

 

Thank you. Physics isn't easy, I struggled with it for years and I still struggle with it today!

[Aethelwulf]

I missed out a factor of [math]c^2[/math], sorry about that!

 

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

Edited November 5, 2012 by Aethelwulf

[Aethelwulf]

I got a little muddled up in post 32, rather G takes on large values using

 

[math]\hbar c = GM^2[/math]

 

Not the quantization condition itself

 

[math]\hbar = \frac{GM^2}{c}[/math]

 

ref

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

[Aethelwulf]

So the other way to give the gravitoelectromagnetic equation

 

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

Edited November 5, 2012 by Aethelwulf

[Aethelwulf]

To gain some idea of how this would work, is by seeing [math]G[/math] as a conversion factor. The best way to describe this is on similar grounds to the usage of the conversion factor in Einstein's famous

 

[math]E = Mc^2[/math]

 

If it was not for the conversion factor [math]c^2[/math], the energy would be small, but it isn't, we gain a lot of energy from just a small piece of matter.

[Aethelwulf]

I was checking out all the units, I was right the first time around lol... there is no c^2 coefficient. Sorry again.

[Aethelwulf]

A few points to be made

 

First of all, these are not like your standard GEM equations you might see. I might also use the notation [math]E_g[/math] to denote a gravitational contribution of energy - this still has your usual units of energy so the subscript is nothing to be read as a different unit of energy. In my postulations, energy is still energy.

 

Later I will be going into new idea's and concepts. Many of these might involve new symbols and if there is a change in the units I will make this clear. We will explore what is called the strong Gravitational constant, [math]G_s[/math] and ask the question which a few physicists have before in the past, ''why is there a huge difference in the strong gravitational constant and the ordinary gravitational constant.'' I will be putting this down to the extremely large value of [math]G[/math] inside a particle to explain why there is a gravitational strong constant interaction at the fundamental level and the strength will drop off, something like [math]\frac{1}{r^2}[/math] the distance, eventually becoming weak at our measurement scale.

 

I will also be talking about new concepts, like the Stoney Mass, and other related topics to mainstream GEM approaches with different equations. This is all in the pipe line, I haven't finished any of it yet however.

Edited November 5, 2012 by Aethelwulf

[Aethelwulf]

So... as I said before,

 

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

 

Plays an important part in the idea of my tiny sphere's. The reasons behind this was because, whilst quantum field theory treats particles normally as pointlike particles, if you are dealing with a particle whose behaviour mimics a [math]\frac{1}{r}[/math] potential, then we are required to have a Poincare stress. However, we can overcome that problem if [math]G[/math] takes on a very large attractive value inside the sphere

 

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

 

Which would mean it would either cancel it out completely (which is not required) but could be greater than the coulomb repulsion which is generally believed by scientists capable of ripping a particle apart. The equation solves this by stating that

 

[math](\frac{4 \pi \epsilon GM^2}{r_s})^{\frac{1}{2}}[/math]

 

Plays the non-electromagnetic part which is believed to be required to solve this problem, as found in classical mechanics. Being a classical equation does not mean it is wrong however, but may mean that we may need to deal with a semi-classical model - doing so, we would invite wave functions; but we won't be doing this in this part.

 

I wish to discuss this [math]\frac{1}{r_s}[/math] potential with some math. Explain that the ''intrinsic (strong) gravitational constant'' will now be denoted as [math]G_I[/math], the ''[math]I[/math]'' for ''intrinsic'' and with that this means it takes the new description as

 

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

 

To implement a [math]\frac{1}{r_s}[/math] potential, we simply need to add a potential to our self-gravitoelectromagnetic energy equation. I provide this as

 

[math]E_{GEM} = \frac{1}{2}(\frac{4 \pi \epsilon G_I M^2}{r_s} + \frac{e^2}{r_s}) + (GM^2 + \frac{e^2}{4\pi \epsilon})r^{-1}_{s}[/math]

 

We can see, by putting our potential [math]r^{-1}_{s}[/math] outside of the parenthesis, we are dealing with the squared gravitational charge [math]GM^2[/math] and the squared electric charge divided by [math](4\pi \epsilon)[/math]. The equation can be simplified to

 

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

 

Our potential depends on the radius then of our particle in question. It is speculated, that [math]G_I[/math] (the strong intrinsic gravitational constant) is very strong inside our particle, and is even stronger the ordinary strong gravitational constant usually denoted as [math]\Gamma[/math]. Thus [math](\Gamma < G_I)[/math]. It is also speculated, that while [math]G_I[/math] is strong inside the potential [math]r^{-1}_{s}[/math] of freedom, it drops off as your usual law [math]r^{-2}[/math] outside of the particle, so very close on the fundamental level, [math]G_I[/math] is actually the source of [math]\Gamma[/math] solving two problematic questions:

 

1) which is more fundamental, [math]G[/math] or [math]\Gamma[/math]

 

...the answer is neither, it is actually the strong instrinsic [math]G_I[/math]

 

2) Why is the strong gravitational constant so different to the ordinary [math]G[/math]?

 

... The answer is because the strong intrinsic gravitational constant is the source of the strong gravitational constant.

 

 

I shall leave it at here for now.

Edited November 6, 2012 by Aethelwulf

[Aethelwulf]

Of course... our radius potential [math]r^{-1}[/math] is just [math]r^{-1}_{s}[/math] so I will change this.

[Aethelwulf]

Just a few points on the self-GEM energy equation

 

[math]E_{GEM} = \frac{1}{2}(\frac{4 \pi \epsilon G_I M^2}{r_s} + \frac{e^2}{r_s}) + (GM^2 + \frac{e^2}{4\pi \epsilon})r^{-1}_{s}[/math]

 

One must keep an open eye that [math]GM^2[/math] is not playing the same role as the gravitational charge in the numerator of [math]\frac{4 \pi \epsilon G_I M^2}{r_s}[/math]. [math]G_I[/math] has a different guise. Because of this, one could, if they wished, re-represent the GEM self energy equation as

 

[math]E_{GEM} = \frac{1}{2}(\frac{4 \pi \epsilon G_I M^2}{r_s} + \frac{e^2}{r_s}) + (\hbar c + \frac{e^2}{4\pi \epsilon})r^{-1}_{s}[/math]

 

Since [math]\hbar c = GM^2[/math]. But I don't know, I find it less appeasing because it does not represent the squared gravitational charge as you would find directly from Motz' work.

Edited November 6, 2012 by Aethelwulf

[Aethelwulf]

I have noticed, I haven't made myself clear on dimensions... I have been a bad boy. Throughout a lot of this work, I have actually switched on a few occasions from SI units and cgs-Gaussian from time to time... I certainly haven't mean't to cause a confusion. It's just that I have in my head a certain formula. One example is that on a few occasions I have noted in the past that (perhaps not here) that

 

[math]e = \sqrt{G}M[/math]

 

this is true if in context of my work mostly here that

 

[math]\epsilon = 1[/math]

 

Notice how in the self energy equation

 

[math]E = \frac{1}{2} \frac{4\pi \epsilon GM^2}{r_s}[/math]

 

This is only a true equation if actually

 

[math]e^2 = 4 \pi \epsilon GM^2[/math]

 

but in your standard units, I derived earlier that

 

[math]Er = GM^2[/math]

 

so that contradicts

 

[math]Er_s = 4 \pi \epsilon GM^2[/math]

 

Unless epsilon was dimensionless. This doesn't exactly make my equations wrong, but when I haven't clearly stated this, it can be troublesome for someone who might be following this. For instance,

 

[math]Er_s = 4 \pi \epsilon GM^2[/math]

 

Would be correct if [math]\epsilon[/math] was in fact a relative permittivity which is actually dimensionless. http://www.ebyte.it/library/educards/sidimensions/SiDimensionsAlfaList.html

 

So whilst I haven't explained any of this, I carried on, not really thinking about it. So if it has caused any troubles, I am sorry about that... in the previous works where I haven't stated this, you will need to do some nip and tuck. Unfortunately, I can't edit most of these posts now, so I will continue with this highlight.

Edited November 6, 2012 by Aethelwulf

[Aethelwulf]

Let me provide you with another example, the self GEM energy equation

 

[math]E_{GEM} = \frac{1}{2}(\frac{4 \pi \epsilon G_I M^2}{r_s} + \frac{e^2}{r_s}) + (GM^2 + \frac{e^2}{4\pi \epsilon})r^{-1}_{s}[/math]

 

I have been in my head, speaking about this equation as if the permittivity was in fact the relative dimensionless case. If I had presented it correctly and hadn't jumped around without defining myself properly, the way to properly write this equation is

 

[math]E_{GEM} = \frac{1}{2}(\frac{4 \pi \epsilon_{r}(\omega) G_I M^2}{r_s} + \frac{e^2}{r_s}) + (GM^2 + \frac{e^2}{4\pi \epsilon})r^{-1}_{s}[/math]

 

Where [math]\epsilon_{r}(\omega) = \frac{\epsilon (\omega)}{\epsilon_0}[/math]

 

That way, [math]e^2 = GM^2[/math] and [math]Er_s = GM^2[/math]. Just about dimensional consistency.

Edited November 6, 2012 by Aethelwulf

[Aethelwulf]

The potential

 

[math]U(r_s) = (GM^2 + \frac{e^2}{4\pi \epsilon})r^{-1}_{s}[/math]

 

is consistent with the non-relative permittivity.

 

You can see this because

 

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

 

where [math]\frac{1}{4 \pi \epsilon}[/math] is the Coulomb constant, but has the dimensions fit to satisfy the potential energy.

Edited November 6, 2012 by Aethelwulf

[Aethelwulf]

In fact I jumping too much, this isn't what I wanted to discuss next. I will get to what I planned soon.

Edited November 6, 2012 by Aethelwulf

[Aethelwulf]

ref http://fqxi.org/data/essay-contest-files/Hestenes_Electron_time_essa.pdf?phpMyAdmin=0c371ccdae9b5ff3071bae814fb4f9e9

[Aethelwulf]

I keep running into different problems. I can't have

 

[math]E_{GEM} = \frac{1}{2}(\frac{4 \pi \epsilon G_I M^2}{r_s} + \frac{e^2}{r_s}) + (GM^2 + \frac{e^2}{4\pi \epsilon})r^{-1}_{s}[/math]

 

and claim it solves the Poincare stress problem because

 

[math]\frac{1}{2}(\frac{4 \pi \epsilon G_I M^2}{r_s})[/math]

 

is not entirely non-electromagnetic. Of course, even if our epsilon was the dimensionless permittivity, it still is an electrical storage of energy. To fix the problem, I need to restate it as

 

[math]E_{GEM} = \frac{1}{2}(\frac{G_I M^2}{r_s} + \frac{e^2}{4 \pi \epsilon r_s}) + (GM^2 + \frac{e^2}{4\pi \epsilon})r^{-1}_{s}[/math]

 

The new problem from this is that the first part [math]\frac{1}{2}(\frac{G_I M^2}{r_s} + \frac{e^2}{4 \pi \epsilon r_s})[/math] is practically identical to potential part [math](GM^2 + \frac{e^2}{4\pi \epsilon})r^{-1}_{s}[/math] - the only real difference is the strong gravitational constant inside the particle. So I don't know... I am finding it quite difficult right now to collect the essential ingredients to describe this... So I have tried a new way.

 

Noticing that [math](GM^2)r^{-1}[/math] is simply the potential [math]M\phi[/math], then we may rewrite the gravitational part as this and interpret it as a coupling of the mass term on the field [math]\phi[/math] - this field may play the role of a gravitational gradient potential field. Knowing this [math](4\pi \epsilon r_s)^{-1}[/math] part, we may simply denote that as [math]k[/math]. Replacing all these statements together into the equation we have a new looking potential part,

 

[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 isn't changing the dynamics. We still have our gravitational part [math]M\phi[/math] and we have our electric part [math]k e^2[/math]. This way we can avoid the potential looking too much like our first set of terms. Also, if mass is coupled to the potential [math]\phi[/math], then mass depends on the potential itself which is an interesting proposition.

Edited November 7, 2012 by Aethelwulf

[Aethelwulf]

So let's just take a few steps back and rewrite some of these equations... properly.

 

The gravitostatic equation of contribution of energy to mass would be

 

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

 

This keeps as the gravitational analogue of

 

[math]E_{EM} = \frac{1}{2} \frac{e^2}{4 \pi \epsilon R_{classical}}[/math]

 

The contribution of mass in my equation is found then as

 

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

 

where [math]GM^2[/math] is the squared gravitational charge. Now the confusion of me skipping between two different unit systems have been cleared up.

 

Now going back to a similar process to Wein (1900), the attraction of the gravitational field can be understood as

 

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

 

And clears that part up.

Edited November 7, 2012 by Aethelwulf