It is possible that under an attractive the particle turns and a higher number of basic particles face the attractive force, resulting in an greater attractive force,
when \(\left\lfloor\cfrac{77}{2}\right\rfloor=39\)
\(c=\sqrt { \cfrac { 39*(32\pi ^{ 4 }-1) }{ 64*3.135009*\pi .ln(cosh(3.135009))tanh(3.135009) } *\left( \cfrac {3.135009}{ 0.7369 } \right) ^{ 3 }} \)
\(c_{39}=78.0893095\)
which gives,
\(c_{adj}=c_{39}.\cfrac { 2ln(cosh(3.135009)) }{ 4\pi\times10^{-7} }\)
\(c_{adj}=303716848\)
Of all the various values for \(c\), which value applies depended on whether an attractive or repulsive force is used in the experiment and whether the particle is set to spin.
Would a particle under an attractive set into spin? Yes. In this case, its speed reduces to \(c_{38.5}\).
Does this spin results in an \(B\) field? If it does than travel to the time dimension to manifest \(B\) is unnecessary. Fourier transform is applicable because of circular motion:- the spin.
In the previous post "Don't Spin Around To Time Travel Yet " spin results in an higher effective \(S_n=38.5\) compared to when it is not in spin, \(S_n=38\) and so spin results in higher speed. In this case, under an attractive force, spin results in the same \(S_n=38.5\) but this is comparatively lower than when the particle is not in spin, \(S_n=39\). The result here is a lower speed, \(c_{38.5}\lt c_{39}\), when the particle spins.
If spin produces the \(B\) field, for it to oscillate, spin has to be stopped. As spin decreases to zero (effectively zero) the particle acquires higher speed to be once again at \(c_{39}\) and the cycle repeated.
Why and how would spin stop?
In this, the particle acquires higher speed\(c_{39}\) with spin and in the other case, the particle has lower speed \(c_{39}\) when in spin. This asymmetry suggests that in one case, the situation is not cyclic. In one of these cases, the particle continues to spin but the system find an equilibrium among the forces; \(B\) does not oscillate. In the other, the particle's spin reduces to near zero, \(B\) disappears (not completely but near zero), \(S_n\) returns to the value when the particle is not in spin and the cycle repeats. \(EB\) in this case, has an oscillating \(B\) field. \(E\) field present itself as though the traveling particle is a line of charges along the direction of the particle's velocity in both cases.
Which is which depends on the mechanism for spin. How does the \(EB\) particle acquire its spin?
Note: ISA is acronym for I Say Again.
