On Modeling the electron loop and the origin of mass

Thanks to u/kmarinas86 inspiration for looking at the electron loops as a solenoid. I did take the challenge to try make some more rigorous notations about that model. Please take this model and try improve on it.

Consider modeling the electron loop that represented by a solenoid torus where we have a dense set of non interacting helices motion along a loop of radius $$R$$. Then we shall assume that the helix radius is $$r$$ and that $$R >> r$$. The idea is that mass is produced through this setup and assume that the speed of the current in the helix is $$v_h = c$$. Let $$v_* = v + v_l$$ be the sum of the translations speed $$v$$ and the helix speed $$v_l$$. The speed along the small helix loop is $$v_r$$. Then we have by Pythagoras,

$$v_r = c\sqrt{1-\frac{v_*^2}{c^2}} = c/\gamma$$

Now we make one turn inside the helix means $$2 \pi r = v_r T$$ hence,

$$T = \frac{2 \pi r}{v_r} =\frac{2 \pi r \gamma}{c}$$

And the length $$L_1$$ we make along the big loop while moving one turn in the inner loop is,

$$L_1 = v_l T = \frac{v_l 2\pi r \gamma}{c}.$$

So the number of turns in a loop is (with the length of the outer loop $$L_R=2\pi R$$),

$$N = \frac{L_R}{L_1} = \frac{2 \pi R}{ \frac{v_l 2\pi r v_l \gamma}{c}} = \frac{cR}{r v_l \gamma}.$$

And hence the loop density is

$$n = \frac{N}{L_R} = \frac{c}{2 \pi r v_l \gamma}.$$

Now the $$B$$ field for a solenoid is (we assume that we make a tube out of an ensemble of electrons hence $$I_{tot}$$ shall be multiplied by $$L_r = 2\pi r = 1 / \rho_r$$ But the charge density in view of the helix depends how many turns it take per unit length. I one turn we have $$cT$$ amount of charge, but we move only $$v_lT$$ along the bigger loop, which means that an initial charge $$e$$ will scale from $$e_*$$ like,

$$e = \frac{v_h T}{v_l T}e_* = c/v_le_*.$$

Or,

$$e_* = \frac{v_l}{c} e.$$

Then, for the theory of solenoids we can take the $$B$$ a field parallel to the loop and with value,

$$B = \mu_0 n I_{tot}.$$

This is the same as,

$$B = \mu_0 \frac{c}{2 \pi r v_l \gamma}\rho_R (e_* v_h) = \mu_0 \frac{c}{2 \pi r v_l \gamma} (\frac{v_l}{c} e) c \rho_R.$$

Or simplified

$$B = \frac{\mu_0 c e}{\gamma 2 \pi r} \rho_R.$$

The total magnetic energy per loop element is

$$E_m = \frac{1}{2}\frac{B^2}{\mu_0}A_r = \frac{\mu_0 c^2 e^2}{8 \pi^2 r^2 \gamma^2}A_r \rho_R^2 = \frac{\mu_0 c^2 e^2}{8 \pi \gamma^2} \rho_R^2$$

And for the full loop we get the energy,

$$E_M = \frac{\mu_0 c^2 e^2}{8 \pi \gamma^2}\rho_R$$

where we used,

$$A = \pi r^2.$$

The magnetic force is

$$F_B = B e_* v_r = \frac{\mu_0 c e}{2 \pi r \gamma} \rho_R \frac{v_l}{c} e \rho_r\rho_R \frac{c}{\gamma} = \frac{\mu_0 c e^2 v_l} {2 \pi r \gamma^2} \rho_r \rho_R^2.$$

Now the electrical potential, assuming a solid infinitely thin tube, consists of terms like,

$$\frac{k e^2}{|x-y|}\rho_r^2\rho_R^2$$

And we note that we can use the expansion assuming $$r_1=r_2$$,

$$\frac{1}{|x-y|} = P_0(\cos(\theta))/r + P_1(\cos(\theta))/r + \ldots$$

And sloppily assume symmetry so that all but the first term remains when we integrate it, e.g. on a loop and hence one of the $$\rho_r$$ is integrated away. Hence,

$$V_1 = \frac{ke^2}{r}\rho_r\rho_R^2$$

The electrical energy between the loops for one turn if they are tight means that the potential is the same as between a point and an infinite line. This is $$\pi/L_*$$ where $$L_*$$ is the shortest distance between to consecutive parts in the helix (actually not we have the second turn, the third turn and so on as well to handle, but we keep it simple for now), that is from some geometry in the picture using congurence,

$$\frac{L_r}{L_h} = \frac{L_*}{L_1}.$$

Hence,

$$L_* = L_1\frac{L_r}{L_h} = L_1 \frac{v_r T}{v_h T} = L_1 \frac{v_r}{v_h}= L_1 \frac{c/\gamma}{c} = L_1\gamma.$$

And we get,

$$V_{2} = \frac{k \pi (e \rho_r \rho_R)^2}{L_*} = \frac{\gamma k \pi e^2}{L_1}\rho_r^2\rho_R^2 = \frac{\gamma k \pi e^2}{2\pi r \gamma v_l / c}\rho_r^2\rho_R^2 = \frac{k c e^2}{2rv_l}\rho_r^2\rho_R^2.$$

We see that both $$V_1$$ and $$V_2$$ are radial symmetric And also $$F_B$$ (and the potential it belongs to) so in all we can try ti find the optimum. As the $$V_1$$ and $$V_2$$ singularities means they dominate repelling at small radius and the magnetic force is attractive and dominates for large distances. Hence we must have an optimum. The force balance are,

$$F_B = -\frac{\partial}{\partial r} (V_1 + V_2).$$

Writing this out means,

$$\frac{\mu_0 c e^2 v_l} {2 \pi r \gamma^2} \rho_r \rho_R^2 = \frac{2 ke^2}{r^2}\rho_r\rho_R^2 + \frac{3 k c e^2}{2r^2v_l}\rho_r^2\rho_R^2.$$

This is the same as to solve,

$$\frac{\mu_0 c } {2 \pi \gamma^2} = \frac{2 k }{(rv_l)} + \frac{3 k c}{2 \pi (rv_l) (rv_l)}.$$

We see that we can solve assuming $$x=rv_l$$

$$\frac{\mu_0 c } {2 \pi \gamma^2} = \frac{2 k }{x} + \frac{3 k c}{2\pi x^2}.$$

Or multiplying the hole setup with $$x^2$$,

$$x^2 \frac{\mu_0 c } {2 \pi \gamma^2} - x \frac{2 k }{x} - \frac{3 k c}{2 \pi} = 0$$

Use $$\mu_0\epsilon_0 = 1/c^2$$ to get

$$x^2 \frac{\mu_0 c } {2 \pi \gamma^2} - x \frac{2 \mu_0 c^2 }{4\pi} - \frac{\mu_0 3 c^3}{8 \pi^2} = 0$$

Divide by $$\mu_0 c/(2\pi \gamma^2)$$ yields,

$$x^2 - \gamma^2 xc - \gamma^2 \frac{3 c^2}{4 \pi} = 0$$

Now dive by $$c^2$$ produces,

$$(x/c)^2 - \gamma^2 (x/c) - \gamma^2 \frac{3}{4 \pi} = 0$$

Put $$y=x/c$$ and we have,

$$y^2 - \gamma^2 y - \gamma^2 \frac{3}{4 \pi} = 0$$

Put

$$\beta = \gamma^2,$$

and finally we have

$$y^2 - y \beta - \frac{3}{4 \pi}\beta = 0$$

And with solution (the other lead to a forbidden negative value),

$$\frac{rv_l}{c} = y = \frac{\beta + \sqrt{\beta^2 + \beta \frac{3}{\pi}}}{2}$$

Now think about having a spiral in the hand. Then you can start moving the whole solid spiral, that's the speed $$v$$ we had in the beginning. So in rest we should not have any current, e.g. in rest $$v = - v_l$$ and then $$\gamma$$ is 1. But this means,

$$y = \frac{rv_l}{c} = C_{\beta},$$

with r.h.s. A constant defined by the solution to the equation above.

So what happens when the spiraling are close to the speed of light, e.g. $$v_l = c-\epsilon/2$$. then $$\gamma is large  and C_{\beta} = \gamma^2$$

$$1/\gamma_{v_l} \approx \frac{\sqrt{\epsilon}}{c}.$$

This means (in the rest frame $$\gamma_v=1$$),

$$\frac{\gamma_{v_l} r v_l}{ c} \approx \frac{r c}{\sqrt{\epsilon}} = 1,$$

or,

$$r = \sqrt{\epsilon}$$

And we see by moving closer to the speed of light we end up possibly having an infinitely small radius. But it can't be too small. we have in the rest frame,

$$B = \frac{\mu_0 c e}{2 \pi r} \rho_R.$$

We could assume that there is a limit at how large the magnetic field can go and in a sense it get stuck there.

In the calculation of the energy (in our lab system) we have the area. In this system, $$v_r$$ go to $$c$$ because $$\gamma_v=1$$ e.g. it is relativistic and around the circle the radius get contracted to a point on the circle so in stead of $$\pi r^2$$ is is plausable that we should have

$$A = 2 \pi r \cdot o.$$

where $$o = 1 \quad [m]$$. But then the correction of the energy at rest means,

$$E_M = o\frac{\mu_0 c^2 e^2}{4 \pi r}\rho_R = o\frac{\mu_0 e^2 }{4\pi r}\rho_R c^2 = m_e c^2,$$

with

$$m_e = o\frac{\mu_0 e^2}{4 \pi r}\rho_R$$

This means that if $$B$$ reaches a limit we have the same dependance in the mass nad hence this mass term is also quantasized and independet on the size of the solenoid.

Now consider the angular quantization condition that the following expression is quantasized,

$$m v R$$

As this expression is independent of $$\gamma$$ we can concider this in any refference frame.

$$m_e v R$$

Especially in the frame where the only motion if from the helical motion (and itis not moving as a solid). So,

$$m_e v R = m_e c (r v / c) R / r = m_e c (1 / (2 \pi R)) R / r = m_e c / (2 \pi r)$$

And in the helical frame $$x$$ is fixed so that we are at the limit of the B field this expression is constant, which we call $$\hbar_*$$, So what happens is essentiall that when you start to increase the original speed in the lab frame, there is a compensation to maintain the helical magnetic field at the limit and as a consequence we get the angular quantization in GUTCP.

$$m v R = \hbar_*.$$

Now introducing a net current and move it with velocity $$v$$. If we assume that translating the system or if we move the velocity is relatively the same. Hence,

$$vr = v.$$

Then for small velocities,

$$m_evR =m_e c (r v /c) R/r = mc \rho_R R/r = \frac{m_ec}{2 \pi r} = h_*$$

As the units $$h$$ indicate that it is independent of $$v$$ we should have this expression for all $$v$$ and as we must reach the maximum $$B$$ field $$r$$ is fixed as well meaning that $$h_*$$ is constand and we can take $$h_*=h$$. Now we know that

$$m_e = C/r_*$$

Where $$r_*$$ is the radius close to the speed of light essentially what we have is

$$r_* = D_\epsilon r$$

In the scaling. But $$r$$ is fixed as well, it's just undefined. What we had before was a condition for $$r_*$$ with this new condition we are able to fixate both $$\epsilon$$ and $$r$$. Now why are they both fixed? The reason are that you can interchange $$r$$ and $$v_l$$ via $$rv_l = c \beta$$ But because the condition for the $$B_{max}$$ involves an $$r_*$$ that means that $$v_l$$ is fixed and hence also $$r_*$$ But $$r_*$$ is a function of $$r,v_l$$ hence $$r$$ is fixed and also $$\epsilon$$. In all it should be possible to calculate $$h$$ from this observation and it looks like $$B_{max}$$ decides both $$m_e$$ and $$h$$. Now finally due to the observation that $$B$$ can be gotten from a boost of a electrostatic setup one speculate that the limit for the $$E$$ field and the limit for the $$B$$ field could be calculated through this setup and one could speculate that this decides the fundamental charge $$e$$. To analyze this we could boost with the speed $$v_l$$ then using

$$B_{max} = \frac{\gamma(v_l)v_lE_{max}}{c^2}$$

we get $$E_{max}$$ and it should be the true $$E_{max}$$ for space. So this means that given a random space with two limits for the fields will decide the fundamental constants $$h,e,m_e$$. And the number of free parameters are 2.