Precession Under Zero Gravity

$F_{gB}$ can explain precession.  The diagram below shows a disc spinning about a tilted axis.

Mass at the disc edge with velocity $v_1$ behaves analogously like a moving negative charge generating a $B$ field that is in the same direction as the gravity equivalent $g_{B}$.

This field cuts the interior of the disc again where the disc is at velocity $v_2$.  The interaction of this velocity, $v_2$ and the gravitational field, $g_{B}$ generates a force, $F_{gB}$ pointing towards the axis of rotation of the disc, below the disc.

$F_{gB}=\pi v_2\times g_{B}$ per unit mass

Mass further into the interior of the disc beyond $v_2$ generates $g_{B}$ that is in the opposite direction to those due to the outer edge of the disc at the point $v_2$ but its value is lower due to lower velocity.  $g_B$ at $v_2$ is still downwards as we sum the effects of all mass along a radial line on the disc.

Similarly, at the edge diagonally opposite, an $g_B$ field intersect the interior of the disc downwards at a local where the velocity is $v_2$.  A similar force, $F_{gB}$ develops, pointing towards the axis of rotation but above the disc.

These pair of $F_{gB}$ form a torque that levels the disc until both $F_{gB}$ are parallel to the surface of the disc and generates no torque.  This happens when the surface is perpendicular to the axis of rotation.

When the disc is spinning at an incline, any tendency to topple generates a velocity component $v$ in red as shown below.

This velocity component together with $g_B$ creates a horizontal force $F_{gB}$ (out of the paper) that cause the disc to precess in an anti-clockwise direction.  Irrespective of the direction of spin of the disc, $g_B$ is always downwards, the tendency to fall will always result in a anti-clockwise precession.  When $v$ is reversed, upwards, the precession will be clockwise.

This process does not involve earth's gravity.  This means that under zero gravity, precession will still occur, following the right hand rule to give the direction of the precession.

Let's denote $g_B(r)$ as the variation of $g_B$ along a radial line on the disc,

$g_B(r)=\int^{r_o}_{r}{g_B}d\,r-\int^{r}_{0}{g_B}d\,r$

where $r_o$ is the radius of the disc. And the total force across a radial line on the disc,

$F_{gB}=\int^{r_o}_{0}{\pi vcos(\theta).g_B(r)} d\,r$  per unit mass

where $v$ is the drop velocity that varies along radius of the disc,

$v=\sqrt{r^2+L^2}.\omega$

where $\omega$ is the drop angular velocity.

The most obvious formulation for $g_B$ due to mass $m$, at distance $x$ from this mass travelling at $v$ is,

$g_B=\left\{\cfrac{\mu_gm}{6\pi}\right\}^{ 1/2 }v\cfrac { 1 }{ x^{ 3/2 } }$

where $\mu_g$ is a constant of proportionality.  This is however without verification.  It is derived from

$B_c=\left( \cfrac { \mu_omc^{ 2 } }{ 6\pi  }  \right) ^{ 1/2 }\cfrac { 1 }{ x^{ 3/2 } }$

published in the post "nλ But Not As Neil Bohr" by analogy.  $\mu_g$ might be used again in $\psi$ wave associated with gravity.  A EMW exchange energy between the B field and E field as it travels at light speed.  A GW (Gravitational Wave) should also oscillate between two types of energies stored in corresponding conservative force fields.  The existence of GW, formulated in the post "Gravity Wave and Schumann Resonance", predicted the exist of $g_B$, the orthogonal conservative force field that, together with $g$ make such energy exchange possible and thus propagation as a wave possible.

The problem is, if the gravitational particles are to be taken literally then in a wave it travels at light speed $c$.  At this speed, $B_c$ was derived and by analogy $g_B$ was written down.  But, an ordinary mass like an elemental mass on a flat disc does not travel at light speed when spinning.  An electron can achieve $c$ in orbit and even as conduction electrons over short distances can achieve $c$ when subjected to an E field.  But is $g_B$ valid for $v\lt c$?

From the posts "How Time Flies" and "Coriolis Force And My Left Foot", $\psi$ exists in three-dimensional space and one orthogonal time dimension, is at light speed in one of the space dimension and in the time dimension.  If all dimensions are equivalent, then a similar formulation of $\psi$ in two time dimensions at the equivalent of light speed in these time dimensions, would mean that even stationary in 3D space the associated particle will be radiating wave (has a $\psi$ around it.)  And \fo (g_B\) or $B$ to exist without the particle moving at all, would require a third time dimension.

But a $B$ field, supposedly do not exist when the charge is stationary, that a static charge on a insulator does not generate a $B$ field.

$\psi$ is an energy wave that required three orthogonal components, $v=c$, $E$ and $B$.  It exist in three dimensional space, oscillating between two dimensions and traveling down a third.  A forth dimension is time as part of the velocity term, $v$.

It will takes four dimensions to manifest an energy wave.  Two of such dimensions exchange energy between themselves (ie. oscillations) and a third dimension perpendicular to the energy oscillations, on which the wave travels and a fourth time dimension upon which the wave existence and velocity is based.

But the electrostatic force and gravity, both conservative force fields, already exist with stationary mass and charge.  This means such stationary fields requires only motion in two dimensions to manifest themselves. One space and one time dimension.  In the case of electrostatic force and gravity, one axis is the respective time axis ($t_c$ and $t_g$ respectively) and one other space dimension.  It is the association with a time dimension that gives the space dimension its conservative properties.  Conservation means the same across time in total, be it momentum or energy.

The associated particle can exchange energy between the conservative field and kinetic energy along the space dimension.

If all energy phenomenon are due to motion along dimensional axes than all energies are equivalent to

$E=\cfrac{1}{2}mv^2$

the kinetic energy along such axis, where $m$ is the mass of the associated particle, and $v$, its velocity along that axis.  In the case where the axis is also the time axis on which $v$ the velocity is derived,

$E=mv^2$

from the post "No Poetry for Einstein".  From this, we can actually generalize $v$ as the rate of change along one dimension with respect to another.  Together with the idea that all dimensions are equivalent, then there can be other physical phenomenon yet undiscovered amid the five dimensions (3 space + 2 time).  Moreover, existence itself has potential energy $E=mc^2$ which is kinetic energy along its time axis.  This energy is released when existence itself is destroyed, when the particle stops on the time axis.

Still, is gravitational particle for real and requires light speed?  If it is required to be at light speed in at least two dimensions (one space and one time dimension) but it is naturally not, and so it is weak.  The charge particle responsible for electrostatic forces is already at light speed along $t_c$ naturally and easily achieve light speed under an electric potential, and so it is strong.