A More Accurate Spin Orbit Equation, A Coupling Involving Eccentricity

[Dubbelosix]

Quite a bit of a calculus goes into the derivation, of eccentricity, but it wasn't my derivation. My contribution is its direct application to a spin orbit equation, two final forms came as

[math]B = k_B \frac{4 \pi^2 r e}{2 h^2c^2} \cdot \frac{\partial U(r)}{\partial r} \cdot (\frac{mk}{L^2} [1 + \sqrt{1 + \frac{2L^2E}{mk}} \cos (\theta - \theta_0)])J[/math]

where [math]B[/math] is the gravimagnetic field, we identify torsion as encoded in this as being part of the central potential [math] \frac{\partial U(r)}{\partial r}[/math], its full form is

[math]\Omega = \frac{1}{2mc^2} \frac{\partial U(r)}{\partial t}[/math]

and 

[math]k = Gm^2[/math]

and

[math]\epsilon = \frac{2L^2E}{mk}[/math]

Is the eccentricity. In a later post Ill show how you derive the full result from calculus, but its ugly and complicated so I'm not doing that today. The main point is we have a correction term on the spin orbit equation, the deviation of the orbit described by the eccentricity. It shows how much it will deviate from a perfect circle so has real world applications, even inside of the interior of atoms. 

The alternative formula is

[math]B = \frac{1}{2e\ mc^2} \cdot \frac{\partial U(r)}{\partial r} \cdot (\frac{mk}{L^2} [1 + \sqrt{1 + \frac{2L^2E}{mk}} \cos (\theta - \theta_0)])J[/math]

And [math]J[/math] is the total angular momentum. You get the first equation by plugging in the inverse Bohr mass

[math]\frac{1}{m} \equiv \frac{mv^2}{m^2v^2} = k_B \frac{4 \pi^2 e^2 r}{h^2}[/math]

into the standard spin orbit equation, which is, 

[math]B = \frac{1}{2e\ mc^2} \frac{1}{r} \frac{\partial U}{\partial r}J[/math] 

Ironically, I found that it didn't matter whether I used Bohrs inverse mass or inverse radius, they both produced the same equation, so they are not too much different animals. 

 

 

Edited July 13, 2021 by Dubbelosix

[Dubbelosix]

Coming from the main headbanger here? I mean, I understand you hate being told when your own theories are a load of bull. But you wouldn't have any insight into knowing whether a theory is wrong or right, or just plain speculative, based on the baloney you post. 

[Dubbelosix]

Anyway, minds to ignore the old bag above. In this post Ill explain quickly the calculus used to derive the eccentricity. Its all standard, its not something I derived myself, whoever did, was doing this stuff at high level university stuff, it was the insight I had where we could use it to describe the radius of curvature. Hopefully I put it here in an understandable way, my only contribution, as its a bit complicated. We start off with a Langrangian

[math]\mathcal{L} = \frac{1}{2} m\dot{r}^2 + \frac{L^2}{2mr^2} - \frac{Gm^2}{r}[/math]

We are able to define the shape of the orbit from

[math]\dot{r} = \frac{dr}{d\theta} \dot{\theta} =  \frac{dR}{d\theta} \frac{L}{mr^2}[/math]

The angle is defined by a succinct integral equation as

[math]\int_{R_0}^{R} =\frac{dR}{R^2\sqrt{\frac{2mE}{L^2} - \frac{1}{r^2} + \frac{2mk}{L^2R}}} = \int_{\theta_0}^{\theta} d\theta = \theta - \theta_0[/math]

The r-integration is followed by making

[math]u = \frac{1}{r}[/math]

and 

[math]du = \frac{dR}{R^2}[/math]

This gives

[math]-\int_{u_0}^{u} \frac{du}{\sqrt{\frac{2mE}{L^2} - \frac{2mk}{L^2} + \frac{2mk}{L^2} u - u^2}} = -\int_{u_0}^{u} \frac{du}{\sqrt{\frac{m^2k^2}{L^4}(1 +  \frac{2L^2E}{mk^2} - (u - \frac{2mk}{L^2})2}}[/math]

So pretty complicated stuff tracking these variables, a lot more complicated than Id mess around with normally outside of mathlab. Next we can make from this, 

[math]u - \frac{mk}{L^2} = \frac{mk}{L^2} \sqrt{1 + \frac{2L^2E}{mk^2}}\ cos\ \alpha[/math]

and also

[math]du = - \frac{mk}{L^2} \sqrt{1 + \frac{2L^2E}{mk^2}}\ sin \alpha\ d\alpha[/math]

So that the final orbit of curvature can be taken, with eccentricity as

[math]u = \frac{1}{r} = \frac{mk}{L^2}[1 + \sqrt{1 + \frac{2L^2E}{mk^2}}\ cos (\theta - \theta_0)][/math]

[Dubbelosix]

Again, take the standard orbit equation, and go right ahead by replacing one such inverse term of radius for this equation and we get back the modified spin orbit equation as

[math]B = \frac{1}{2e\ mc^2} \cdot \frac{\partial U(r)}{\partial r} \cdot (\frac{mk}{L^2} [1 + \sqrt{1 + \frac{2L^2E}{mk}} \cos (\theta - \theta_0)])J[/math]

[Dubbelosix]

"This is a JOKE" sayeth scienceforums jester. Yet here I am presenting it in fullon physics exchange. Not saying one side will favour my investigation than another, only that I'm confident enough to do so. Plus, the main reason why I'm reposting this, the site doesn't hold latex anymore? My equations never display anymore.

 

https://physics.stackexchange.com/questions/653386/on-the-bohr-model-and-kepler-orbits-allowing-eccentricity-in-the-full-poincare-g

Edited July 25, 2021 by Dubbelosix

[Dubbelosix]

I've took the liberty of writing the latex out on naked scientist forums, so it can be seen in its fuller beauty.

https://www.thenakedscientists.com/forum/index.php?topic=82903.0

Edited August 20, 2021 by Dubbelosix

[Dubbelosix]

On 7/28/2021 at 8:01 PM, JeffreysTubes8 said:

No, that's not why this is a joke, it's unfortunately because of your math skills.

My math skills? I know now why people want you banned. Even Ocean won't stick up for you because of the nonsense you post.

Edited August 20, 2021 by Dubbelosix

[Vmedvil2]

On 8/23/2021 at 12:54 AM, JeffreysTubes8 said:

He does stick up for me even though he knows I'm not copacetic, "borderline" as I may be moderation doesn't ban me for a couple threads piqued the power the be here's curiosity, mainly displays in mathematics in just two threads predominantly being Historical Codex and extreme Algebra challenge where I solved it without, unconventionally, completing the square.

Believe it or not they want those 32 coordinates. They saw some of the proof in blue marker before I took it down.

https://www.quantamagazine.org/how-big-data-carried-graph-theory-into-new-dimensions-20210819/