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A Geometrical way to unite Green's theorem with the magnetic field

The work is defined proportionally to the metric dsd\mathbf{s} and the force as




W=∫CF⋅ds=∫CF⋅dxW = \int_{\mathcal{C}} \mathbf{F} \cdot d\mathbf{s} = \int_{\mathcal{C}} \mathbf{F} \cdot d\mathbf{x} (1).



∫mv ds=∫p⋅dq\int mv\ d\mathbf{s} = \int \mathbf{p} \cdot d\mathbf{q} (2).


differentiation gives us


∫dmvdtds=∫dpdt⋅dq=∫F⋅ds\int \frac{d mv}{dt} d\mathbf{s} = \int \frac{\mathbf{dp}}{dt} \cdot d\mathbf{q} =\int F \cdot d\mathbf{s} (3).



In previous work, I defined Greens formula (in non-vector form)



∮Fˉ⋅dRˉ=∫∫(∇ˉ×Fˉ)⋅n^dAˉ\oint \bar{F} \cdot d\bar{R} = \int \int (\bar{\nabla} \times \bar{F}) \cdot \hat{n} d\bar{A} (4).



Some notation will be changing now so the mathematics are a bit clearer, dRˉd \bar{R} has many different notations, some of them are (n^Aˉ)(\hat{n} \bar{A}) or even simply ds=dx2,−dy2d\mathbf{s} = \sqrt{dx^2, - dy^2} which is a metric term. If say n^ds\hat{n} d \mathbf{s} then the metric becomes (dy−dx)(dy - dx). In previous work, I showed that Greens theorem may be implemented into the Larmor total energy (of a spin-coupling) system to it's motion in R2\mathbf{R}^2. To give an extremely brief cover of that proof, we considered a force equation


∇×F=∂U(r)∂r⋅n^\nabla \times F = \frac{\partial U(r)}{\partial r} \cdot \hat{n} (5).


The UU term comes out of quantum theory, it is a potential energy term U=eVU = eV were ee is the charge and VV is the potential. In series of calculations, we arrived at a few important equations by noticing equivalent terms in the theory. The magnetic field relationship in the orbital momentum term is found as


B=1Mc2e1r∂V(r)∂rL\mathbf{B} = \frac{1}{Mc^2 e} \frac{1}{r} \frac{\partial V(r)}{\partial r}\mathbf{L} (6).


This is from a well-known derivation, from already well-established facts from the Larmor equation's theory. As it turned out, the component we were concerned with was ∂U(r)∂r\frac{\partial U(r)}{\partial r} by simply recognizing this term to be integral to the developing of Greens theorem,


∇⃗×F⃗⋅n^=∂V(r)∂r\vec{\nabla} \times \vec{F} \cdot \hat{n} = \frac{\partial V(r)}{\partial r} (7).


and simply distributing an area dot product with unit vector Aˉ⋅n^=dr\bar{A} \cdot \hat{n} = d\mathbf{r}


(∇⃗×F⃗)⋅ds=∂U(r)∂r(Aˉ⋅n^)=∮∂SF⃗⋅dRˉ(\vec{\nabla} \times \vec{F}) \cdot d\mathbf{s} = \frac{\partial U(r)}{\partial r} (\bar{A} \cdot \hat{n}) = \oint_{\partial \mathcal{S}} \vec{F} \cdot d\bar{R} (8).



Equation (6). has terms then it might take account for. We shall replace terms where necessary. Equation (6).



∮CB⋅dr=1eMc21rL[∂U(r)∂r(Aˉ⋅n^)]=1eMc21rL∮∂SF⃗⋅ds\oint_{\mathcal{C}}\mathbf{B} \cdot d\mathbf{r} = \frac{1}{e Mc^2} \frac{1}{r} \mathbf{L} [\frac{\partial U(r)}{\partial r} (\bar{A} \cdot \hat{n})] = \frac{1}{e Mc^2} \frac{1}{r} \mathbf{L} \oint_{\partial \mathcal{S}} \vec{F} \cdot d\mathbf{s} (9).


(note, dRˉd\bar{R} will now be replaced with it's more illuminating form as dsd\mathbf{s} the metric - we also pick up a term n^\hat{n} on the right hand side and
Unparseable latex formula:

\mathbf{B} \cdot \mathdf{r}

is just the current μj\mu j!) A current term should be expected since we are dealing with a curved trajectory, ideally, a near-perfect circle trajectory for simplicity.



Of course, this identity was much more rigorously derived in my other work, but this is a very efficient way to cut a long paper short. From this point, we can identify ΦB=B⋅Aˉ\Phi_{\mathbf{B}} = \mathbf{B} \cdot \bar{A} describes a magnetic flux: of course, since it implements Greens theorem on the right hand side, which meant we can rewrite everything further again by knowing ΦB=∮∂SA dr\Phi_{\mathbf{B}} = \oint_{\partial \mathcal{S}} \mathbf{A}\ dr



∮∂SA dr=1eMc21rijL⋅∮∂SF⃗⋅ds \oint_{\partial \mathcal{S}} \mathbf{A}\ dr = \frac{1}{e Mc^2} \frac {1}{r_{ij}} \mathbf{L} \cdot \oint_{\partial \mathcal{S}} \vec{F} \cdot d\mathbf{s} (10).




where ds=vdtd\mathbf{s} = v dt and A\mathbf{A} is the magnetic vector potential (not to be confused with area Aˉ\bar{A}. The last term looks a lot like equation (3). In fact, equation 3 is designed to describe the mechanical momentum between two points and the ijij-notation can describe the separation between two points sufficiently on the radius term. Equation 3 fully describes the mechanical work done which has units of energy. In the description of it having geometric consequence from Greens theorem, it would be a charge, following a classical path between two classical points on a two dimensional surface. It also has it's motion coupled with the angular momentum. By denoting the two points on the surface as (i,j)(i,j) attached to the radius means any distance traversed will not only involve an angle between ii and jj, but finding the distance traversed along a curved surface naturally can be found through trigonometry and calculating arc lengths.


Using the definitions for the metric term dsd\mathbf{s} that was given early, multipling the unit vector on both sides gives us



∮∂S dr(A⋅n^)=1eMc21rijL⋅∮∂SF⃗(dx,−dy) \oint_{\partial \mathcal{S}}\ dr (\mathbf{A} \cdot \hat{n}) = \frac{1}{e Mc^2} \frac {1}{r_{ij}} \mathbf{L} \cdot \oint_{\partial \mathcal{S}} \vec{F}(dx, - dy) (11).



*note! the use of ′′⋅′′''\cdot'' on ∮∂S\oint_{\partial \mathcal{S}} is to be taken as a multiplication sign, not a dot product.


Taking the curl on the force removes that ugly unit length term on the potential


∮∂SA dr=1eMc21rijL⋅∮∂S∇⃗×(Fx⃗dx,−Fy⃗dy) \oint_{\partial \mathcal{S}} \mathbf{A}\ dr = \frac{1}{e Mc^2} \frac {1}{r_{ij}} \mathbf{L} \cdot \oint_{\partial \mathcal{S}} \vec{\nabla} \times (\vec{F_x}dx, - \vec{F_y}dy) (12).



Using the left side of Greens theorem, we may have


∮(Fx⃗dx+Fy⃗dy)=∮C(Fy⃗,−Fx⃗)⋅(dy,−dx)=∮C(Fy⃗,−Fx⃗)⋅n^ ds\oint (\vec{F_x}dx + \vec{F_y}dy) = \oint_{\mathcal{C}} (\vec{F_y},- \vec{F_x}) \cdot (dy, -dx) = \oint_{\mathcal{C}} (\vec{F_y},- \vec{F_x}) \cdot \hat{n}\ d\mathbf{s} (13).


Which as you may have noticed is identical to the last term in the equation above. Any time a curl is operated on this equation becomes a divergent equation. Already, that is featured in the equation we recently showed



∮∂SA dr=1eMc21rijL⋅∮∂S∇⃗×(Fx⃗dx,−Fy⃗dy) \oint_{\partial \mathcal{S}} \mathbf{A}\ dr = \frac{1}{e Mc^2} \frac {1}{r_{ij}} \mathbf{L} \cdot \oint_{\partial \mathcal{S}} \vec{\nabla} \times (\vec{F_x}dx, - \vec{F_y}dy) (14).



Doing this gives you the right side of Greens theorem as the last component in the equation above




∮∂SA dr=1eMc21rijL⋅∮∂S(∂Fx∂x−∂Fy∂y)dAˉ \oint_{\partial \mathcal{S}} \mathbf{A}\ dr = \frac{1}{e Mc^2} \frac {1}{r_{ij}} \mathbf{L} \cdot \oint_{\partial \mathcal{S}} (\frac{\partial F_x}{\partial x} - \frac{\partial F_y}{\partial y}) d\bar{A}



∵ ∮C(Fx⃗,−Fy⃗)⋅n^ ds=∫∫D(∇⋅(Fx⃗,−Fy⃗)Aˉ=∮C(∂Fx∂x−∂Fy∂y)dAˉ\oint_{\mathcal{C}} (\vec{F_x}, -\vec{F_y}) \cdot \hat{n}\ d\mathbf{s} = \int \int_D (\nabla \cdot (\vec{F_x}, -\vec{F_y})\bar{A} = \oint_{\mathcal{C}} (\frac{\partial F_x}{\partial x} - \frac{\partial F_y}{\partial y}) d\bar{A}



To finish, let's take one last look at equation (9).




∮CB⋅dr=1eMc21rL[∂U(r)∂r(Aˉ⋅n^)]=1eMc21rL∮∂SF⃗⋅ds\oint_{\mathcal{C}}\mathbf{B} \cdot d\mathbf{r} = \frac{1}{e Mc^2} \frac{1}{r} \mathbf{L} [\frac{\partial U(r)}{\partial r} (\bar{A} \cdot \hat{n})] = \frac{1}{e Mc^2} \frac{1}{r} \mathbf{L} \oint_{\partial \mathcal{S}} \vec{F} \cdot d\mathbf{s} (9).


Simplifying the equation, we get


Unparseable latex formula:

(\mu j \cdot) \hat{n} = \frac{1}{e Mc^2} \frac{1}{r} \mathbf{L} [\frac{\partial U(r)}{\partial r} d\mathbf{r}]}




where μ\mu is the magnetic vacuum constant.


The magnetic field can also be given as


B=−v×Ec2\mathbf{B} = - \frac{v \times \mathbf{E}}{c^2}



Express the electric field as the gradient of the electric potential, while keeping in mind that the electric field is radial then we substitute this in and changing the order of the cross products




B=r×pMc2∣Er∣\mathbf{B} = \frac{r \times p}{Mc^2}|\frac{E}{r}|


where EE is the energy.




Knowing this then, we quickly return to this equation




Unparseable latex formula:

(\mu j) \cdot \hat{n} = \frac{1}{e Mc^2} \frac{1}{r} \mathbf{L} [\frac{\partial U(r)}{\partial r} d\mathbf{r}]}





=B×dr= \mathbf{B} \times d\mathbf{r}



because


dr=A⋅n^d\mathbf{r} =\mathbf{A} \cdot \hat{n}




You can see how you can construct that by decomposing the angular momentum into it's full form and using partial notation we have




(μj)⋅n^=r×peMc2[∂Urdr]r(\mu j) \cdot \hat{n} = \frac{r \times p}{e Mc^2} \frac{[\partial U_r d\mathbf{r}]}{r}


=[∂Urdr]rMc2r⋅pe=EErr⋅pe= \frac{[\partial U_r d\mathbf{r}] r}{Mc^2 r} \cdot \frac{p}{e} = \frac{E}{E} \frac{r}{r} \cdot \frac{p}{e}



Notice, these dimensionless factors attached to the momentum divided by the charge? In Gaussian units Er=ℏc=GM2E r = \hbar c = GM^2. This might be important because the fine structure may in fact be the dimensionless quantity, conserved as a coupling constant in this theory.
(edited 9 years ago)

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Reply 1
Avatar for Graviphoton
Graviphoton
OP
And to be honest, it probably is the electromagnetic dimensionless fine structure constant. If so, we finally reach some new conserved quantities in an equation

(μj)⋅n^=αmve(\mu j) \cdot \hat{n} = \alpha \frac{mv}{e}

Since our system is travelling a trajectory of a closed boundary (surface) there should be a gyroradius equation which can be obtained here as given above. The unit length n^\hat{n} appears from the area which makes up drd\mathbf{r}.


To finalize my point, the term mve \frac{mv}{e} is a conserved quantity through α\alpha which comprises the term (μj)⋅n^(\mu j) \cdot \hat{n}.
(edited 9 years ago)
Reply 2
Avatar for Graviphoton
Graviphoton
OP
So in short, what do we learn?


It says the left hand side of Greens Theorem becomes a part of the components which make the magnetic field whenthe right hand side is related by simply a curl on the force vector in a model in which the metric space is coupled to not only the motions direction but a fundamental intrinsic relationship to it's spin on the metric.
Original post by Graviphoton
...


Fabulous to see you again G. It is a shame you have chosen to hide this magnificent piece of gobbledegook in the Creative Corner.
Reply 4
Avatar for Graviphoton
Graviphoton
OP
Gobblegook?

It's pure calculus my friend. If you want discuss it, then do it. Your ad hominems don't mean much until you show a working knowledge of what is being discussed.
Original post by Graviphoton
Gobblegook?

It's pure calculus my friend. If you want discuss it, then do it. Your ad hominems don't mean much until you show a working knowledge of what is being discussed.


Remind me of your qualifications again? Oh that's right you can't as you don't have any.
Original post by Mr M
Remind me of your qualifications again? Oh that's right you can't as you don't have any.


I remember this guy. Also very glad to see him back; his threads were amazing.
Reply 7
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Graviphoton
OP
Original post by Mr M
Remind me of your qualifications again? Oh that's right you can't as you don't have any.




You know what the title is? It's a geometric interpretation of Green's theorem which happens to appear in a magnetic term which when considered with spin and orbit motion, couples.

My qualifications, ... I had a two year time studying graphic design, about a year doing classical mechanics at college (before I dropped out) to take more time assessing my life. In between that time, I still study quantum mechanics (as a hobby) but now take up chess as a fuller-time hobby, in which I play grandmasters from all over the country.


For some reason, you want me to question my life? My life is great, I have someone I love and a life to look forward for. You on the other hand, seem to find more direction in life in attacking people... with that... good luck. It won't bother me.
Original post by Graviphoton
You know what the title is? It's a geometric interpretation of Green's theorem which happens to appear in a magnetic term which when considered with spin and orbit motion, couples.

My qualifications, ... I had a two year time studying graphic design, about a year doing classical mechanics at college (before I dropped out) to take more time assessing my life. In between that time, I still study quantum mechanics (as a hobby) but now take up chess as a fuller-time hobby, in which I play grandmasters from all over the country.


For some reason, you want me to question my life? My life is great, I have someone I love and a life to look forward for. You on the other hand, seem to find more direction in life in attacking people... with that... good luck. It won't bother me.


You always make me smile. I particularly enjoyed this thread where the simplest question tied you up in knots.
Reply 9
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Graviphoton
OP
Original post by Mr M
You always make me smile. I particularly enjoyed this thread where the simplest question tied you up in knots.



You were trying to make an asinine joke... hence the word trying. For some reason you think you need a certificate to prove to everyone you're smart.

Did you know Green actually wrote his theorem without any previous education in mathematics? What was his qualifications then?

Now he is respected as one of the greatest mathematicians that ever lived. So don't come in here bud, try and raise an eye-brow with stuff that is irrelevant to the OP and quite frankly, none of your buisiness.
(edited 9 years ago)
Original post by Graviphoton
(I am sure a mod will also see it my way that you are simply trying to derail this thread.)


Improbable. I can't see why TSR would want to provide a safe haven for someone who has been banned from Forums all over the internet.
Reply 11
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Graviphoton
OP
Original post by Mr M
Improbable. I can't see why TSR would want to provide a safe haven for someone who has been banned from Forums all over the internet.




You seem to know so much about me.... not.

I only know of two forums that have effectively banned me and that was because of corrupt moderation.

Say, why don't you actually tackle the OP? You don't show any working knowledge, but you do show a need to attack me. Perhaps this was more personal than I realized. You're probably some obnoxious troll that has encountered me on another forum and thus, taking this chance to personally attack me, with things that are blatantly off-topic.
Reply 12
Avatar for Graviphoton
Graviphoton
OP
Original post by Mr M
That's not strictly true is it? Your further education college course did not have the title 'classical mechanics'.





Oh so you do know me... in person? You're right, the class what not called classical mechanics, but it was largely to do with Newtonian mechanics, which is classical mechanics.


So... who are you?
Original post by Graviphoton
So... who are you?


Just a fan.
Reply 14
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Graviphoton
OP
Original post by Mr M
Just a fan.


No you're a stalker with the sole intention to derail and undermine my work.

Again, good luck with that. You say you're mathematics teacher, but you act like you don't even understand the work.
Reply 15
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Cubic
8
Original post by Graviphoton
Your ad hominems


At least you spelt it right this time. :smile:
Reply 16
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Graviphoton
OP
Original post by Mr M
....
Good luck. I think the first time you were permanently banned I expressed a hope that you would eventually grow up and starting acting your age. I can only repeat that again."



You're calling me by a name I haven't used in about 10 years. Things change you know.

And yes, I did create multiple accounts at that site... because I was unfairly banned from the very beginning.

Don't start sharing half information here, if you want to get into a discussion about sciforums, I won't hesitate, I have plenty to say about the ''practices'' of mods at that site!
Hi
Saw "Geometrical" in the latest thread title so came to see what was what


Can I suggest that you reference other people's work when posting it - a link to the site or a simple citation will do

Also - I am wondering what your purpose was in posting - did you need help understanding ( I would suggest physics study help) or were you just marvelling at the work (tbh the physics band would probably be the one's to share your interest anyway)
Reply 18
Avatar for Graviphoton
Graviphoton
OP
Original post by TenOfThem
Hi
Saw "Geometrical" in the latest thread title so came to see what was what


Can I suggest that you reference other people's work when posting it - a link to the site or a simple citation will do

Also - I am wondering what your purpose was in posting - did you need help understanding ( I would suggest physics study help) or were you just marvelling at the work (tbh the physics band would probably be the one's to share your interest anyway)



I just wanted some reflection who might understand the implications of the theory... for instance, it's resemblance to the Larmor energy equation is striking that a further theory that describes the magnetic field with the Green term will involve a coupling of L to S where S denotes the spin vector.

That probably doesn't mean much to you or many people and even to some others it sounds wack, but I have offered a vigorous proof using pure calculus. Elsewhere, someone found a simpler tensor form of the equation I derived and said it was fully covariant. So it seems to have possibilities of having real world applications.
Original post by Graviphoton
I just wanted some reflection who might understand the implications of the theory... for instance, it's resemblance to the Larmor energy equation is striking that a further theory that describes the magnetic field with the Green term will involve a coupling of L to S where S denotes the spin vector.

That probably doesn't mean much to you or many people and even to some others it sounds wack, but I have offered a vigorous proof using pure calculus. Elsewhere, someone found a simpler tensor form of the equation I derived and said it was fully covariant. So it seems to have possibilities of having real world applications.


Hmmmm - I am still left wondering why "creative corner" was your forum of choice for this

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