And They Danced

Since,

\(\cfrac{d\,KE}{d\,t}=\cfrac{1}{2}m\cfrac{dv^2}{d\,t}=mv\cfrac{d\,v}{d\,t}=v.F\)

We consider again an electron in circular orbit around a hydrogen nucleus,

\(\cfrac{q^2}{4\pi \varepsilon_o r_e^2}=v.\cfrac{m_ev^2}{r_e}\) --- (1)

assuming that the electron is in light speed \(v=c\), and not \(v=2\pi c\) (not yet and not necessary),

\(\cfrac{q^2}{4\pi \varepsilon_o r_e}=m_ec^3\)

So,

\(r_e=\cfrac{q^2}{4\pi \varepsilon_o.m_ec^3}\)

where \(m_e\) is the electron mass, \(q\) is the electron charge.

We have a lot of a doubts upon the valid value for \(\varepsilon_o\), however, in calculations for ratios such as \(\cfrac{r_p}{r_e}\) or \(\cfrac{m_p}{m_e}\) where \(\varepsilon_o\) cancels, we can ignore the value for the constant.

With this mind, the ratio of \(\theta_\psi\) of a big particle (\(n=77\)), to a basic particle is,

\(r_{\small{Q}}=\cfrac{\theta_{\psi}}{\theta_{\psi\,c}}=\cfrac{a_{\psi}}{a_{\psi\,c}}=\cfrac{3.135009}{0.7369}=4.25432\)

But both particles have the same inertia \(\cfrac{1}{c}\), under the action of the force field, in the time dimensions.  Should they have the same inertia in the space dimension?

When expression (1) is written down, the hydrogen nucleus is stationary and the electron with reference to the nucleus as a center, orbits at a radius \(r\).  But the nucleus under the attraction of the electron should respond physically too.

To consider the motion of the nucleus together with the circular motion of the electron...we have a dance.

and more obvious,

In both cases,

\(\cfrac{q^2}{4\pi \varepsilon_o r_p^2}=v.\cfrac{m_pv^2}{r_p}\)

\(r_p=\cfrac{q^2}{4\pi \varepsilon_o.m_pc^3}\)

And so,

\(\cfrac{r_p}{r_e}=\cfrac{m_e}{m_p}\)

This expression suggests that \(r_e\gt r_p\).  In which case, we have instead,

Both of which are just guesses.  One scenario is just as awkward as the other because they are equivalent, one electron is \(\pi\) rad out of phase compared to the other scenario.

Since, \(f=\cfrac{c}{2\pi r}\) as we have assumed that they are both at light speed,

\(\cfrac{r_p}{r_e}=\cfrac{m_e}{m_p}=\cfrac{f_e}{f_p}\)

If  \(f_p\gt f_e\) then it is even more likely that the black colored proton crosses the orbit of the electron and be outside of the wider orbit, as the left part of the diagram shows.  One the other hand, if \(f_e\gt f_p\) then the red colored electron can also cross the black orbit be outside of the proton orbit.  When this happens the attraction force acting as the centripetal force must do work against the changing momentum of the particle in an orbit bent away from its line of action, and force the orbit inward.  The particle slows, the orbital radius (inter-particle distance) decreases...and the electron eventually falls into the nucleus.

And the whole universe collapses.

If the inter-particle distance increases, at near light speed the particles will break attraction and not be in orbit around each other.

When \(f_p\ne f_e\), no matter where the starting positions of the particles are, one of the particle relative to the other will move closer to the orbit intersections at \(A\) or \(B\),

When either one of the two particles is outside the confine of the other's orbit, either the inter-particle distance decreases at light speed and the system collapses or the inter-particle distance increases at near light speed and the system breaks apart.

Could the particles move in perpendicular planes?

No, a second force will be needed for one particle to be in orbit around a perpendicular plane.  Without such a force, the motions of both particles will be in the same plane.

\(f_p=f_e\) is not the only possible conclusion,  Suppose,

\(v_p\ne v_e\)

that the particles are not of the same speed in spin.  In that case,

\(\cfrac{r_p}{r_e}=\cfrac{m_e v_e^3}{m_p v_p^3}\)

\(\cfrac{f_e r_e}{f_p r_p}=\cfrac{v_e}{v_p}\)

and so,

\(\cfrac{m_e f_e^3}{m_p f_p^3}=\cfrac{r_p^4}{r_e^4}\) --- (*)

This expression still cannot guarantee that one particle does not move closer to the orbital intersections and crosses outside.

This is the conventional solution,

where the proton is considered much more massive that it is in a slow spin about an axis not in its interior, with the electron in wide orbit around it.  Expression (*) still applies.  The particles does not spin as if there is a rod holding them, but they interact with each other through their fields.

This does not indicate in any way, why the mass of a proton is more than the mass of an electron.

The wanted conclusion \(m_e=m_p\) takes a delay.  And so they danced...

Lost A Planet

I am missing this equation,

\(\lambda^2-\lambda-1=0\)

Where is it???

Big Mushroom

And if you turn a "CHL Bood" given by,

\(x=-\cfrac{|F_{drag}|}{n.m_b\omega^2}sin(\cfrac{\pi}{2}-s)*cos(s)\)

and

\(y=\cfrac{|F_{drag}|}{n.m_b\omega^2}sin(\cfrac{\pi}{2}-s)\)

upwards, we have a mushroom cloud,

The blast drove \(T^{-}\), negative temperature particles to terminal speed, the drag force they experience sheds the the big particles into smaller groups of \(n\) basic particles and send them off to a trajectory defined  by the parametric equations above.  The negative temperature particles are driven off first in a blast, because of their lighter inertia.  They cause moisture in the atmosphere to condense into clouds.  The positive temperature particles that follow infuse into the clouds and create a mix of heat and steam.

The top widest canopy is likely to correspond to \(n=1\).

A small "bood" for man, a big KaBoom for mankind.

Note: It is possible to disregard the direction of \(x\) and flip \(x\) around,

\(x=\cfrac{|F_{drag}|}{n.m_b\omega^2}sin(\cfrac{\pi}{2}-s)*cos(s)\)

\(y=\cfrac{|F_{drag}|}{n.m_b\omega^2}sin(\cfrac{\pi}{2}-s)\)

\(x=y*cos(s)\)

Since,

\(sin(\cfrac{\pi}{2}-s)=cos(s)=\cfrac{y}{\cfrac{|F_{drag}|}{n.m_b\omega^2}}\)

\(\cfrac{|F_{drag}|}{n.m_b\omega^2}x=y^2\)

\(y=\sqrt{\cfrac{|F_{drag}|}{n.m_b\omega^2}}\sqrt{x}\)

in this case the quadratic is parameterized by \(\cfrac{1}{\sqrt{n}}\).

A profile is different from an equation.  In a profile the direction of the slope of the shape presented with reference to \(y=0\) or \(x=0\) is important.  An equation shows the relationship between the variables involved.

No Sustained "Bood"

From the post "Just When You Think c Is The Last Constant" dated 26 Jun 2016, we have,

\(c=\sqrt { \cfrac { S_{ n }(32\pi ^{ 4 }-1) }{ 64\pi .\theta _{ \psi  }ln(cosh(\theta _{ \psi  }))tanh(\theta _{ \psi  }) } *\left( \cfrac { \theta _{ \psi  } }{ 0.7369 }  \right) ^{ 3 } } \)

when we replace \(\cfrac{77}{2}=S_n\).

When \(S_n\) is variable,

\(v=\sqrt { \cfrac { S_{ n }(32\pi ^{ 4 }-1) }{ 64\pi .\theta _{ \psi  }ln(cosh(\theta _{ \psi  }))tanh(\theta _{ \psi  }) } *\left( \cfrac { \theta _{ \psi  } }{ 0.7369 }  \right) ^{ 3 } } \) --- (1)

and

\(S_n=39-\cfrac{0.5S}{\omega_o}v\) --- (2)

where \(S\) is a constant.  From the post "An Egg With Bood..." dated 12 Jul 2016,

\(\omega^2=\cfrac{F_{drag}}{\delta m}\cfrac{sin(\theta)}{r}\)

at the front tip of the particle, \(r\rightarrow 0\) and \(\theta\rightarrow 0\).  If we assume

\(\lim\limits_{r,\,\theta\rightarrow 0}{\cfrac{sin(\theta)}{r}}=1\)

at the tip of the particle,

\(\omega^2_t=\cfrac{F_{drag}}{\delta m}\)

Since, \(F_{drag}\propto v^2\)

\(\omega^2_t=T_t.v^2\) --- (3)

where \(T_t\) is a constant.  The particle spin, \(\omega_t\) decreases with decreasing \(v\).

When \(v\) increases under the action of an attractive field \(F.c\), \(S_n\) according to expression (2) decreases, which through expression (1) decreases \(v\).  As \(v\) decreases \(\omega_t\) decreases via (3).  But under the action of the field, \(v\) increases again.  It seems that the cycle repeats but this is not a sustained oscillation.  Energy in the spin of the particle is not being exchanged for translational kinetic energy in \(v\) and vice versa.  \(S_n\) presents a lower values because the face that presents a higher number of basic particles towards the attractive field, is being turn away by spin.  As the particle spins, \(S_n\) represents the average number of basic particles on one side of the big particle for which they constitute.

This is just a short burst of EB, before \(\omega_t\) settles to a constant value.

Substitute (3) into (1)

\( \omega _{ t }=\sqrt { \cfrac { S_{ n }T_{t}(32\pi ^{ 4 }-1) }{ 64\pi .\theta _{ \psi  }ln(cosh(\theta _{ \psi  }))tanh(\theta _{ \psi  }) } *\left( \cfrac { \theta _{ \psi  } }{ 0.7369 }  \right) ^{ 3 } } \)

given \(T_t\) which is a constant, \(\omega_t\) is determined.  The particle can spin with constant \(\omega_t\) and travel at a speed lower than \(c_{39}\).

All that is presented here is under the assumption that \(c\), light speed is a constant.

They All Shed...

If pieces of \(\psi\) were to fly off the particle each of some elemental mass \(\delta m=n.m_b\), in \(n\) integer multiple of basic particles, \(m_b\). Then given,

\(y_g=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)\)

\(\because\) \(\delta m=n.m_b\),

\(y_g=\cfrac{|F_{drag}|}{n.m_b\omega^2}sin(\theta)\)

where \(m_b\) is the inertia of the basic particle in the field.

Similarly,

\(x_g=\cfrac{|F_{drag}|}{n.m_b\omega^2}sin(\theta)cos(\theta)\)

Pieces of elemental mass will follow the trajectory defined by \((x_g,\,\,\,y_g)\) as they leave the front tip of the big particle.

However to plot this trajectory correctly as discussed in the post "Egg Shaped Egg" dated 12 Jul 2016,

\(x=-\cfrac{|F_{drag}|}{n.m_b\omega^2}sin(\cfrac{\pi}{2}-s)*cos(s)\)

The negative sign flips \(x\).  The argument to the term \(cos(\theta)\) still changes from zero to \(\pi/2\) as we draw the tip of the distortion first.

\(y=\cfrac{|F_{drag}|}{n.m_b\omega^2}sin(\cfrac{\pi}{2}-s)\)

A parametric plot of \(x\) and \(y\) with various \(\cfrac{1}{n}\) is,

The outermost envelope corresponds to \(n=1\).

This is how a big particle shred itself of basic particles.

If a sonic shock wave is due to gravity particles shedding under high drag force and Cherenkov radiation is due to charge particles shedding under high drag force.  Then there is another, "CHL Bood" effect when temperature particles shed under high drag force.

Another No Bell Prize!  And this time for "Bood".

Note:  "Bood" is otherwise a sonic boom.  "CHL Bood" is a thermal boom.

Nothing To Do...With An Egg

Actually the parametric is just,

\(x=1-sin(\cfrac{\pi}{2}-s)*cos(s)\)

\(y=sin(\cfrac{\pi}{2}-s)\)

\(0\le s\le\pi/2\)

Since, \(sin(\cfrac{\pi}{2}-x)=cos(x)\)

\(x=1-y^2\)

\(y=\pm\sqrt{1-x}\)

A plot of (1-x)^(1/2) and -(1-x)^(1/2) gives,

which looks more like a wave front than egg.  And a blimp,

a bunch of pile up waves.  This however suggests that there is something wrong with the expression,

\(x_g=rcos(\theta_1)=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)cos(\theta_1)\)

from the post "Egg Shaped Egg" dated 12 Jul 2106.

\(x_g\) here, starts with a maximum value, when \(r=0\) and \(\theta_1=0\) and end up with \(x_g=0\) when \(\theta_1=\pi/2\).

If we considers the effects of \(F_{D}=F_{drag}cos(\theta)\).  If the particle is at terminal speed, \(F.c=constant\), \(F\) is a constant, then \(F_{D}\) is fully compensated by the internal forces of particle, after it deformed.  If the particle deforms in the same way in all directions, the length along the line of action of the force \(F_{drag}sin(\theta)\) is \(r\), so the length along \(F_D\) is,

\(\cfrac{F_{drag}sin(\theta)}{r}=\cfrac{F_{drag}cos(\theta)}{x_g}\)

\(x_g=\cfrac{rcos(\theta)}{sin(\theta)}\)

But \(r=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)\)

\(x_g=\cfrac{|F_{drag}|}{\delta m.\omega^2}cos(\theta)\)

which is still a circle!  What happened?  The statement " If the particle deforms in the same way in all directions..." changes the particle back into a circle/sphere.  If we instead allow the changes along the surface of particle to dictate the change along \(x_g\), where like the point at the tip of the particle, an elemental mass on the surface if moved by a centripetal force tangentially by an infinitesimal distance.

The displace/distortion tangential to the surface due to \(F_{D}=F_{drag}cos(\theta)\) is,

\(\cfrac{F_{drag}sin(\theta)}{r}=\cfrac{F_{drag}cos(\theta)}{x_t}\)

\(x_t=\cfrac{rcos(\theta)}{sin(\theta)}\)

the projection of this length along the horizontal axis is,

\(x_g=x_tcos(\pi/2-\theta)=\cfrac{rcos(\theta)}{sin(\theta)}.sin(\theta)\)

Since,  \(r=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)\)

So,

\(x_g=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)cos(\theta)\)

which is the same as before.  And the egg remains.

An Egg With \(B\)ood...

\(F^{*}_c=F_{drag}sin(\theta)=\delta m.r.\omega^2\)

\(\omega^2=\cfrac{F_{drag}}{\delta m}\cfrac{sin(\theta)}{r}\)

where \({F_{drag}}\) is a force not the distance from the origin to the elemental mass, \(\delta m\) on the surface of the particle.  The particle will deform such that it rotates as a whole at a constant angular velocity.

\(\omega^2=A.{F_{drag}}\)

 \(A\) is a constant.

If we define that when the particle spins at \(\omega_o\) , it will present an \(S_n=38.5\), and that \(\omega\) is proportional to \(S_n\),

\(\cfrac{\omega}{\omega_o}=\cfrac{S_n}{38.5}\)

And we consider that the particle starts at \(S_n=39\) when \(\omega=0\), spinning decreases \(S_n\) and at \(\omega=\omega_o\),  \(S_n=38.5\), we have,

\(S_n=39-0.5\cfrac{\omega}{\omega_o}\)

\(S_n=39-\cfrac{0.5}{\omega_o}\sqrt{AF_{drag}}\)

as \(F_{drag}\propto v^2\)

\(S_n=39-\cfrac{0.5S}{\omega_o}v\)

where \(S\) is a constant.

where \(\omega\le\omega_o\).  And we define \(\omega_o\) to be the the maximum angular velocity attained by the particle as \(B\) oscillates, assuming that the particle does not continue to increase in spin when \(S_n=38.5\).

Are we ready to make \(B\) spin?  It is Bu...ood as in wood.

Fat Butt Egg

We take another look at the forces on the particle under the field force \(F\) and drag force \(F_{drag}\).

\(F_{drag}=F^{*}_D+F^{*}_c\)

\(F^{*}_D\) is not the resolution of the resultant force \(F+F_{drag}\).

\(F^{*}_D\) the component of \(F_{drag}\) along \(F\), reduces the front force, \(F\) and provide less kinetic energy change in he forward direction.  At the same time, \(F^{*}_c\) tends to rotate the particle and impart rotational kinetic energy.  In this way, energy change from the original \(F.c\) is split between rotation and forward translation.

It would seem that \(F^{*}_c\) is rotates the particle in opposite directions.  \(F^{*}_c\) in the rear rotates upwards and \(F^{*}_c\) in the front rotates downwards.  This is actually consistent with a failure scenario when the particle fracture.  A bullet, for example, on impact flares up (opens up) at the rear but hold firm at the front (a shiny impact spot).

 \(F^{*}_c\) in the rear and \(F^{*}_c\) in the front may not act to rotate the particle in a common rotational sense.  The particle is twisted when they rotate in clockwise and anticlockwise direction.

But \(F_{drag}\), as it move through to the rear of the particle, is decreased by the internal forces that holds \(\psi\) together.  The rotational forces at the front of the particle dictates the rotation of the particle.

 However,

\((F-F^{*}_{D})_{front}=(F-F^{*}_{D})_{rear}\) --- (*)

when the deformed particle has one velocity.

A difference in the front and rear horizontal forces will deform the particle, until (*) holds.  As \(F^{*}_D\) in the rear is less than \(F^{*}_D\) in the front, the particle compresses to the front and the rear is flattened.  Such a compression adds another force to the equation (*), on an elemental mass \(\delta m\) on the rear surface of the particle.

\((F-F^{*}_{D})_{rear}-F_{material}=(F-F^{*}_{D})_{front}\)

where the right hand side of the equation is of an elemental mass at the tip of the particle, \(F_{material}\) is the force due to material property (compression, elongation).

 \(F^{*}_c\) in the rear wobbles the particle and direct the tip of particle away from the horizontal rotational axis.  It tends to stretch the particle in the upward direction away from the horizontal rotational axis.  \(F^{*}_c\) in front compresses the particle in the upward direction towards the horizontal axis.  When the particle is rotating,

\(-F^{*}_c+F_{material}=F^{*}_c\)

\(F_{material}=2F^{*}_c\)

The internal force reverses \(F^{*}_{c}\) at the rear when the whole deformed particle is rotating in one direction; the sense of the rotation being dictated by the force component at the front of the particle.

If you don't want you missile to wobble make sure it is free to deform at the rear so that \(F^{*}_c\) cancels with internal force completely.  A soft ass.

In this case, \(F^{*}_c\) in front is solely due the drag force acting through the \(C.G\) of the particle.  If the particle first suffers compression under \(F^{*}_{c}\) then,

\(F^{*m}_{c}=F^{*}_{c}-F_{material}\)

we replace \(F^{*}_{c}\) at the front, with the force \(F^{*m}_{c}\) to account for a lower rotation.

If you want your missile not to rotate, it should compress at the front.  A soft front and all in all a soft missile.  It is after all such material deformations, that the shape of the missile is to fit an aerodynamics profile.

What happen to the end of the particle?  The particle is similarly deformed at the rear.  The rear is compress towards the tip to account for loss in \(F_{drag}\) passing through the particle.  And it is stretched in both vertical directions, up and down, away from the horizontal rotational axis, the result of which is that internal force reverse \(F^{*}_{c}\) at the rear and the particle spins in one direction (as the front).

When is an egg a missile?  South Korean protests over missile deployment.

Note: \(F_{drag}\) is virtual due to entanglement that shares energy of the particle. \(F_{drag}\) is defined as if the kinetic energy of the particle is changed directly and immediately.  The \(C.G\) defines the position of the particle.  Moving the \(C.G\) moves the particle.  If the the virtual force does not act through the \(C.G\) then relatively, the particle has moved for no apparent reason.  Alternatively, if this force does not act through the \(C.G\) it will generate a torque that imparts additional rotational kinetic energy.  All energy change should be part of the total energy change that defines this virtual force.  Otherwise there must be another force.  A force with displacement imparts energy into a system.  A virtual force imparts energy directly into a system without displacement.

Egg Shaped Egg

From the previous post "Lay Another Egg..." dated 12 Jul 2016,

\(|F^{*}_{c}|=|F_{drag}|sin(\theta)=\delta m.r.\omega^2\)

where \(\delta m\) is an elemental mass on the surface of the egg. \(r\) is the radial distance from the axis along the direction of travel and \(\theta\) the angle at the \(C.G\) between \(F_{drag}\) and the rotational axis.

\(r=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)\)

This is not a polar plot, \(r\) is the distance from the horizontal rotational axis.

\(y_g=r\)

The distance along the rotational axis of \(\delta m\) from the \(C.G\) is given by,

\(x_g=rcos(\theta_1)=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)cos(\theta_1)\)

This may be wrong Pls refer to "Nothing To Do...With An Egg" dated 13 Jul 2016

We plot the parametric pair,

\(x=1-sin(\cfrac{\pi}{2}-s)*cos(s)\) and

\(y=sin(\cfrac{\pi}{2}-s)\)

\(0\le s\le\pi/2\)

Implemented in the plotting software, \(s\) is measured with reference to the \(y=0\) axis.  In the intended graph, \(\theta\) starts at \(s=\pi/2\) at the front tip and ends with \(s=0\) at the half circle defined by an axis that divides the particle into front and back.  \(\theta_1\) is always measured with reference to the \(x=0\) axis and start from \(0\) to \(\pi\) and varies linearly with \(s\). So, \(\theta\rightarrow (\pi/2-s)\) and \(\theta_1\rightarrow s\).  Up to this point, the  front tip will be at the origin where \(s=0\), to invert the plot \(x\rightarrow 1-x\).

As spin, \(\omega\), increases, the front tip sharpens.  As drag force, \(|F_{drag}|\), increases, the front tip flattens.  The ratio,

\(R_D=\cfrac{|F_{drag}|}{\omega^2}\)

a drag-spin ratio determines the shape of the particle.

The front tip is always flat as \(\delta m\) moves tangentially away under the action of \(F_c\) a centripetal force.  The front top does not sharpen with a discontinuity in gradient.

What happen to the back of the particle?

Lay Another Egg...

From the post "What Change KE?" dated 30 Jun 2016,

\(\cfrac{d\,KE}{d\,t}=constant\)

for a particle in a field.

\(\cfrac{1}{2}m.v\cfrac{dv}{dt}=\cfrac{1}{2}v.m\cfrac{dv}{dt}=constant\)

\(\cfrac{1}{2}v.F=constant\)

\(v.F=B\)  where \(B\) is a constant.

The Newtonian force \(F\) is inversely proportional to \(v\).  But the drag force due to massive entanglement, sharing energy is proportional to \(v^2\).

\(F_{drag}=A.v^2\)

where \(A\) is a constant of proportionality.  For a particle driven in a field,

As such, the resultant force on the furthest point in the direction of travel on the particle is,

\(F_c=F_{drag}-F\)

\(F_c=Av^2-\cfrac{B}{v}\)

where \(A\) and \(B\) are both constants.

\(F_c\) act as a centripetal force moving \(\psi\) away from the furthest point in the direction of travel on the particle.  The particle has symmetry along \(F\).  This movement of \(\psi\) is in a plane perpendicular to the direction of travel, intersecting \(F_c\) at the tip of the particle.  The result is to distort \(\psi\) from a sphere into an egg shape.  This distortion by \(F_c\) is along an infinitesimal circular arc centered at the \(C.G\) of the particle as the forces involved passes through the \(C.G\) of the particle.

This egg has no spin yet...

But more importantly, \(F\) increases with higher concentration of \(\psi\) at its rear.  \(B\) increases with increased \(\psi\).  The one with a bigger buttock gets a bigger kick in the rear.

Distortion stops when \(F_c=0\), and \(B=B^{'}\)

\(F_c=Av^2-\cfrac{B^{'}}{v}\)

\(v^3=\cfrac{B^{'}}{A}\)

\(v=\sqrt[3]{\cfrac{B^{'}}{A}}\) --- (*)

It is expected that \(F\) is the force that causes the most distortion on the particle.  On any other point on the surface of the particle, the force components involved are less than \(F\) and causes less distortion to \(\psi\).

On a distorted egg,

\(F_{drag}\) is a virtual force that always act through the \(C.G\).  The force \(F\) due to the field is the same.  These forces are of the same magnitude, but at a point not at the front tip of the particle, \(F_{drag}\) they are not co-linear.

\(\overset{\rightarrow }{F}+\overset{\rightarrow }{F_{ drag }}=\overset{\rightarrow }{F_{ R }}\)

The resultant force, \(F_{\small{R}}\) that develops can be resolved into two directions.

\(\overset{\rightarrow }{ F_{ R }}=\overset{\rightarrow }{F^{ * }_{ D }}+\overset{\rightarrow }{F^{ * }_{ c }}\)

The normal component of \(F^{*}_{D}\) to the surface pulls the particle outwards and is countered by internal forces in the particle that holds \(\psi\) together.  The tangential component of \(F^{*}_{D}\) resistance the movement of \(\psi\) from the front tip of the particle.

\(F^{*}_c\) spins the particle perpendicular to the axis along its velocity.

So, when not spinning, the particle elongates from the front tip and its surface collapses inwards.  When the particle starts to spin, it elongates less and bulges outwards.

The bulge behind the particle causes \(\psi\) to increase and so \(F.c\) increases.  But at maximum distortion, the drag force and the force in the field is equal, at the front tip of the particle.

\(F_{drag}=F\)

When the field is attractive, spinning causes the particle to lower velocity, then the decrease in \(F_{drag}\) (\(\propto v^2\)) and the increase in \(F\) (\(\propto \frac{1}{v}\)) will cause the particle to increase in speed again, along \(F\).

When the field is repulsive, spinning causes the particle to increase velocity, then the increase in \(F_{drag}\) (\(\propto v^2\)) and the decrease in \(F\) (\(\propto \frac{1}{v}\)) will increase \(F_c\), the force that move \(\psi\) away from the front tip of the particle and causes the particle to distort further and spin faster.  The particle does not destroy itself because the maximum linear speed that spinning can bring the particle to is at \(c_{38.5}\).  There is no wave with an oscillating \(B\) when the field is repulsive.

What happened to expression (*), that seems to suggest another speed limit.  (*) allows \(B^{'}\) to be found.  \(B\) changes to \(B^{'}\) as the result of the particle spinning and distorts \(\psi\).  The speed attained with spin is fixed; the maximum velocity of the particle is a constant at \(c_{38.5}\) when the field is repulsive.

There is still no explicit mechanism by which \(B\) oscillates when \(EB\) is in an attractive field.  \(B\) may not be sinusoidal.  But \(c_{38.5}\) is another candidate for light speed when the field is repulsive.

\(EB\) remains extraterrestrial and unreachable.

ISA: Don't Spin Around To Time Travel Yet

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.

Don't Spin Around To Time Travel Yet

From the previous post "Increasing Light Speed" dated 10 Jul 2016,

\(c_{measured}=c_{ave}=\cfrac{2}{\pi}c_{max}+c_o\)

The first question is:  Why would \(EB\) be dipping in and out of light speed \(c_o\)?  The second question is:  What is \(c_{max}\)?  From the post "Just When You Think c Is The Last Constant" dated 26 Jun 2016, we have,

when we use \(\cfrac{77}{2}=38.5\)

\(c=\sqrt { \cfrac { 38.5*(32\pi ^{ 4 }-1) }{ 64*3.135009*\pi .ln(cosh(3.135009))tanh(3.135009) }  *\left( \cfrac {3.135009}{ 0.7369 }  \right) ^{ 3 }}\)

\(c_{38.5}=77.5871223\)

when \(\left\lceil\cfrac{77}{2}\right\rceil=38\)

\(c=\sqrt { \cfrac { 38*(32\pi ^{ 4 }-1) }{ 64*3.135009*\pi .ln(cosh(3.135009))tanh(3.135009) }  *\left( \cfrac {3.135009}{ 0.7369 }  \right) ^{ 3 }} \)

\(c_{38}=77.0816633\)

both without the adjustment to \(c_{adj}\).

If \(c_{max}=c_{38.5}-c_{38}=0.505458945\)

and,

\(c_o=\cfrac{c_{38}+c_{38.5}}{2}\)

this assumes that \(B\) is symmetrical about \(c_o\) and that its value increases in the same way as it decrease; \(B\) is odd about point \((0,c_o)\) and is mirrored about \(x=\pi/2\) in the graph in "Increasing Light Speed" dated 10 Jul 2016.  Then,

\(c_{measured}=c_{ave}=\cfrac{2}{\pi}(c_{38.5}-c_{38})+\cfrac{c_{38}+c_{38.5}}{2}\)

\(c_{ave}=\cfrac{2}{\pi}*0.505458945+77.0816633=77.656177\)

and when we adjust for this value,

\(c_adj=c_{ave}.\cfrac { 2ln(cosh(3.135009)) }{ 4\pi\times10^{-7}  }\)

\(c_{adj}=302032247\)

This is not closer to the defined value of \(c=299792458\) than the adjusted values of \(c_{38}\) at \(299797757\).  But it does suggest a mechanism by which \(EB\) acquire velocity above light speed and present itself as \(B\), by spinning.

The factor \(S_n=\cfrac{77}{2}=38.5\) as the average of \(77\) basic particles on one side of the big particle is applicable only when the particle is in spin.  When not in spin but under a repulsive field that pushes the big particle away, the effective number of basic particle facing the repulsive force is \(S_n=\left\lceil\cfrac{77}{2}\right\rceil=38\), as the particle turns away from the repulsive force as it is propelled forward.

Why would \(EB\) drop speed after attaining a maximum speed?  Why would \(EB\) return from the time dimension?  If \(EB\) returns to the space dimension because it attained the equivalent light speed in the time dimension, the field that propels the particle must act both in the space and time dimension simultaneously.  That would make space and time the same dimension not orthogonal.

\(B\) may not be the force in time we are looking for.  Don't spin around to time travel yet.

Increasing Light Speed

Please refer to the later post "No Sustained "Bood" dated 14 Jul 2016.  No oscillations.
In the post "Energy Accounting With Fourier" dated 08 Jul 2016, it was proposed that \(E\) itself has speed \(c\) that varies in a sinusoidal, 

\(v=c_{max}sin(t)+c_o\)

When \(E\) is above light speed in lapses into the time dimension.  \(E\) is Fourier transformed 

\(E\overset{Fourier}{\longrightarrow }B\)

to \(B\).

If this is the case, the average light speed, \(c_{ave}\)

\(c_{ave}=c_{max}.\cfrac{1}{\pi}\int_{0}^{\pi}{sin(x)}+c_o\)

\(c_{ave}=\cfrac{2}{\pi}c_{max}+c_o\)

It is likely that when we measure light speed by observing \(E\) or \(B\),

\(c_{measured}=c_{ave}=\cfrac{2}{\pi}c_{max}+c_o\) 

we are measuring \(c_{ave}\) and not \(c_o\).  If \(c_{max}\) increases linearly with \(E\) then it might be possible to obtain \(c_o\), the true light speed, by extrapolating from varies values of \(c_{max}\) to the c-intercept when \(E=0\).

This might explain why \(c\) seems to increase with increasing \(E\).

To \(f\) and Forward

From the post "What Time?" dated 08 Jul 2016,

\(\cfrac { d^{ n } }{ dt^{ n } } \left\{ t^{ n -1}f(t) \right\} \overset { Fourier }{ \longrightarrow  } (-1)^{(n-1)}(i2\pi f)\cfrac { d^{ n-1 } }{ df^{ n-1 } } \left\{ F(f) \right\}\)

with light speed \(t=\cfrac{x}{c}\),

\(dt=\cfrac{1}{c}dx\)

\(c^n\cfrac{d^n}{dx^n}\left(\left(\cfrac{x}{c}\right)^{(n-1)}f(\cfrac{x}{c})\right)\overset{Fourier}{\longrightarrow }(-1)^{(n-1)}(i2\pi f)\cfrac{d^{n-1}}{df^{n-1}}\left(F(f)\right)\)

What if \(f(\cfrac{x}{c})=\cfrac{c}{x}=\cfrac{1}{t}=f\)?  Then it is possible to obtain, \(F(f)\) for \(\cfrac{1}{t}\) in

\(\cfrac{1}{t}\overset{Fourier}{\longrightarrow }F(f)\)

Let \(n=1\), and \(f(\cfrac{x}{c})=\cfrac{c}{x}\),

\(c\cfrac { d }{ dx } \left( \cfrac { c } { x } \right) \overset { Fourier }{ \longrightarrow  } (i2\pi f)F(f)\)

\(-\left( \cfrac { c } { x } \right)^2 \overset { Fourier }{ \longrightarrow  } (i2\pi f)F(f)\)

If \(f(t)=f_t\) and \(f_t\overset{Fourier}{\longrightarrow }F(f)\) then

\(f_t^2\overset{Fourier}{\longrightarrow }F(f)\star F(f)\)

where \(\star\) denotes convolution and we differentiate \(f\) in the frequency domain from \(f_t\), the function in the time domain.  So,

\(i2\pi f. F(f)=-F(f)\star F(f)\)  --- (1)

When \(n=1\), and \(f(\cfrac{x}{c})=(\cfrac{c}{x})^2\),  \(F_1(f)=F(t)\star F(f)\)

\(c\cfrac { d }{ dx } \left( \cfrac { c } { x } \right)^2 \overset { Fourier }{ \longrightarrow  } (i2\pi f)F_1(f)\)

\(-2\left( \cfrac { c } { x } \right)^3 \overset { Fourier }{ \longrightarrow  } (i2\pi f)F_1(f)\)

\( (i2\pi f)\left(F(f)\star F(f)\right)=-2F(t)\star F(f)\star F(f)\) --- (2)

Consider,  \(n=2\) and \(f(\cfrac{x}{c})=\left(\cfrac{c}{x}\right)^2\),

\(c^2\cfrac{d^2}{d\,x^2}\left(\cfrac{c}{x}\right)\overset{Fourier}{\longrightarrow }-(i2\pi f).\cfrac{d}{df}\left(F(f)\right)\)

\((-1)(-2).\cfrac{c^3}{x^3}\overset{Fourier}{\longrightarrow }-(i2\pi f).\cfrac{d}{df}\left(F(f)\right)\)

which is,

\(2f^3\overset{Fourier}{\longrightarrow }-(i2\pi f).\cfrac{d}{df}\left(F(f)\right)\)

Since,

\(f_t^3\overset{Fourier}{\longrightarrow }(F(f)\star F(f))\star F(f)\)

where \(\star\) denotes convolution and we differentiate \(f\) in the frequency domain from \(f_t\), function in the time domain.

\(-(i2\pi f).\cfrac{d}{df}\left(F(f)\star F(f)\right)=2F(f)\star F(f)\star F(f)\) --- (3)

Comparing (2) and (3),

\(\cfrac{d}{df}\left(F(f)\star F(f)\right)=F(f)\star F(f)\)

substitute (1) in,

\(\cfrac{d}{df}\left(i2\pi f. F(f)\right)=i2\pi f. F(f)\)

One possible solution is,

\(i2\pi f. F(f)=e^{f}\)

and \(F(f)=\cfrac{1}{i2\pi f}e^{f}\)

So,

\(\cfrac{1}{t}\overset{Fourier}{\longrightarrow }\cfrac{1}{i2\pi f}e^{f}\) --- (*)

But what does (*) mean?

A plot of y=e^x/x and y=x,

And how do check such an expression?  The Duality property of Fourier tranform \(F[F(t)]=f(-f)\).

Note:  The change from \(dt\) to \(dx\) via \(c.t=x\) is not necessary.

\(\cfrac{1}{t}=f\) is a function that is different from \(\cfrac{1}{\Delta t}=\cfrac{1}{T}=f\) the definition for frequency.  The later is divided by an time interval \(\Delta t=T\), the period of an oscillation.  \(\cfrac{1}{t}\) is the reciprocal of the passage of time with respect to some origin \(t=0\).  \(\Delta t=T\) is without such an origin.  \(t\) is meaningless without first defining the origin, \(\Delta t\) finds a reference point when it is applied after it has been given a value.

Phonetically Yours

Remember

\(B=-i\cfrac{\partial\,E}{\partial\,x}\)

and \(i\) that bends \(B\) to be normal to \(x\) the radial distance from a center, effectively making \(B\) go round in a circle.  \(B\) is tangential to the circle with radius \(x\).

Why does this work?  After making \(B\) into a circle by applying \(i\),

\(\cfrac{\partial}{\partial\,x}\equiv\cfrac{1}{c}.\cfrac{\partial}{\partial\,t}\)

\(\because\)  \(t.c=x\) at light speed, \(c\)

which is a time derivative with the constant \(\cfrac{1}{c}\).  We applied a time derivative to \(E\) after bending it into a circle by applying \(i\).

The good news is, \(B\) is in the time dimension and affects a force in the space dimension by effecting time.  We may have \(B\) to affect time directly in the time dimension.  The bad news is, \(B\) is in the time dimension, to be affected by \(B\) we have to join \(B\) and go around in circles.  In which case, Lorentz's force is more important than it is.

Can the time force needed to manipulate time be in \(B\)?  Thick magnetic coating on the hull of a timecraft might provide the carrier containment or bounce needed when subjected to a \(B\) field to travel through time.  Which reminds me of the color of the UFO in the video "UFO Over Vasquez Rocks" found on Youtube.  Lodestone paint maybe.  Or just wrap yourselves in a big silver coil.

\(\psi\) phonetically sounds like the word "shit" in a local asian dialect.  Or "stuck" in another language.

Note: In the post "Magnetic Field In General, HuYaa" dated 13 Oct 2014,

\(B=-i\cfrac{\partial E}{\partial x^{'}}\)

\(x^{'}\) is tangential along \(B\).  In the expression above, \(x\) is along \(E\).  Both expression are equivalent.  In the case of,

\(B=-i\cfrac{\partial E}{\partial x^{'}}\)

the perspective is with the dipole/particle observing the orientation of \(B\) with reference to \(E\).  In the case of,

\(\cfrac{\partial B}{\partial\,t}=-i\cfrac{\partial E}{\partial x^{'}}\)

the perspective is at a fixed location in space, away from the dipole/particle, observing \(B\) and \(E\) at the fixed location.

Loop Wire Antenna

From the post "Energy Accounting With Fourier" dated 08 Jul 2016, extra energy is needed for \(B\) in circular motion.  The extra energy needed reduces the total power in \(B\).  When \(B\) returns from the time dimension, where is this energy when a radiating wire is not in a circular loop (\(E\) not circular).

Is the power \(\cfrac{1}{4}P_{total}\) the radiated power?

Yes! (but no, read further...)  Because in a circular loop wire antenna, the gap in the loop at the connections is radiating power.  This gap is the part of the wire loop that deviates from a circle.  When \(B\) is transformed back to \(E\), \(E\) is supposedly circular, a complex sinusoidal, \(A(cos(i2\pi f t)+isin(i2\pi f t))\) because \(B\) is a complex sinusoidal.  That part of \(E\) that encounters a break in the loop is radiated into space.

When the antenna is a straight wire, one of the components, (\(Ecos(i2\pi f t\) or \(Esin(i2\pi f t)\)) of the circular \(E\) is radiated.  The actual power fraction is \(P_f=\cfrac{1}{2}\left(\cfrac{1}{2}\right)^2=\cfrac{1}{8}\).

And no!  The initial power fraction of \(P_f=\cfrac{1}{4}\) is loss bending the \(E\) into a circular wave.  This part of the power is not recovered as long as \(E\) remains circular.

In the case of the gap in a circular loop,

\(P_{str}=\left(\cfrac{\theta_{ant}}{2\pi}\right)^2\) a guess

where \(\theta_{ant}\) is the angle subtended by the gap at the center of the loop.  The power fraction \(P_f\) that applies to a straight wire antenna does not appear here because the applied \(E\) is already circular.  No energy is needed to made \(B\) circular.

An Intuitive Bull

Why Fourier Transform moves us to the time domain?

Circles, lots and lots of circles...

When we derive \(F_{time}\), we are bending to form a circle around which time travels by applying constrains and conditions.  Fourier transform represents everything from \(t=\cfrac{x}{c}\) in circles, \(e^{-i2\pi f t}\).  Both paths leads to Rome, but Fourier transform does not conserve energy.  Fourier coefficients are not constrained by,

\(a^2_{t1}+a^2_{t2}+...a^2_{tn}+...=A.E_s(b_{s1},b_{s2},...b_{sn}...)\)

where \(a_{tn}\), \(n=1,2,3..\) are coefficients of the transform in time and \(b_{sn}\) are the coefficients/parameters in space.  \(E_s(...b_{sn}...)\) is a function that provides the total energy of the system in space.  \(A\) is a constant, for the system, ie. for the set of equations that provides \(b_{sn}\).

Fortunately, the energy change in moving between two valid points (energy states) in a conservative system is irrespective of the path taken from one point to the other.  Since energy is conserved between space and time, the discrepancy will always be \(\cfrac{1}{4}\) the total energy in space, as shown in one instance in the post "Energy Accounting With Fourier" dated 08 Jul 2016.

We naturally derive physical parameters in per unit time as the "force in a field" was mistaken as the Newtonian force and not power, the change in energy per second.

\(\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o}=F.c=\cfrac{\partial\,KE}{\partial\,t}\ne F_{newton}\) --- (*)

The notion of flux was abandoned a long time ago.

The unit meter in \(c\) from the (*), unlike the "per unit time" mentioned above,  remains unaccounted for.  This could be the reason for the confusion between intensity and power.

You know the bull is going round and round when what it unloads forms into a circle.

And The Bull Drives On

Why the time derivative?

The answer was in

\(\cfrac { d^{ n } }{ dt^{ n } } \left\{ f(t) \right\} \overset { Fourier }{ \longrightarrow  } (i2\pi f)^{ n }.F(f)\)

when \(n=1\).

\(\cfrac { d }{ dt } \left\{ f(t) \right\} \overset { Fourier }{ \longrightarrow  } i2\pi .f .F(f)\)

The scaling factor \(i2\pi\) is the imaginary clock face staring right back at you.

The factor \(f\) justifies windowing a function of interest in the frequency domain (or equivalently the time domain) and repeat that window for a periodic variation and then apply the transform.  It is also the reason why in the frequency domain we always obtain results in per unit time (per second), over one time period, \(f=\cfrac{1}{T}\).

And first thing first.  Although in retrospect, nothing stopped us from making up periodic intervals just because we can obtain the appropriate answer.  And the bull drives on.

What Time?

The time dimension is no longer represented by second (s) but by frequency (s^-1); a count or measure in the first time derivative.

What is then the relationship between Fourier transform and the time derivative?

Inverse Fourier transform as defined, transforming from the time domain to the frequency domain,

\(F_{ f}^{-1}[F(f)]=f(t)=\int_{-\infty}^{\infty} { F(f).e^{ i2\pi ft } } df\)

The transform of the nth derivative in time of \(f(t)\) is,

\(\cfrac { d^{ n } }{ dt^{ n } } \left\{ f(t) \right\} \overset { Fourier }{ \longrightarrow  } (i2\pi f)^{ n }.F(f)\)

and the transform to the (n-1)th derivative in frequency of \(F(f)\) is,

\( t^{ n-1 }f(t)\overset { Fourier }{ \longrightarrow  } \cfrac { 1 }{ (-i2\pi f)^{ n-1 } } \cfrac { d^{ n-1 } }{ df^{ n-1 } } \left\{ F(f) \right\} \)

Combining,

\( \cfrac { d^{ n } }{ dt^{ n } } \left\{ t^{ n-1 }f(t) \right\} \overset { Fourier }{ \longrightarrow  } (i2\pi f)^{ n }.\cfrac { 1 }{ (-i2\pi f)^{ n -1} } \cfrac { d^{ n-1 } }{ df^{ n-1 } } \left\{ F(f) \right\} \)

\(\cfrac { d^{ n } }{ dt^{ n } } \left\{ t^{ n -1}f(t) \right\} \overset { Fourier }{ \longrightarrow  } (-1)^{(n-1)}(i2\pi f)\cfrac { d^{ n-1 } }{ df^{ n-1 } } \left\{ F(f) \right\} \)

with light speed \(c\),

\(t=\cfrac{x}{c}\)

\(dt=\cfrac{1}{c}dx\)

\(c^n\cfrac{d^n}{dx^n}\left(\left(\cfrac{x}{c}\right)^{(n-1)}f(\cfrac{x}{c})\right)\overset{Fourier}{\longrightarrow }(-1)^{(n-1)}(i2\pi f)\cfrac{d^{n-1}}{df^{n-1}}\left(F(f)\right)\)

If \(n=2\) and \(f\left(\cfrac{x}{c}\right)=\cfrac{x}{c}\)

\(c^2\cfrac{d^2}{d\,x^2}\left(\cfrac{x}{c}\right)^2\overset{Fourier}{\longrightarrow }-(i2\pi f).\cfrac{d}{df}\left(F(f)\right)\)

\(2\overset{Fourier}{\longrightarrow }-(i2\pi f)\cfrac{d}{df}\left(F(f)\right)=2\delta(f)\)

This implies,

\(\cfrac{d}{df}\left(F(f)\right)=-\cfrac{1}{i\pi f}\delta(f)\)

\(F(f)=-\cfrac{1}{i\pi}\int{\cfrac{1}{ f}\delta(f)}d\,f\)

\(F(f)=-\cfrac{1}{i\pi}\int{\delta(f)}d\,ln(f)\)

If \(y=ln(f)\) then \(e^{y}=f\)

\(dy=d\,ln(f)\)

\(F(f)=-\cfrac{1}{i\pi}\int{\delta(e^{y})}d\,y\)

We could have,

\(\left(\delta(e^{y})+A\right)^{'}=\delta^{'}(e^{y})=-\delta(\left(e^{y}\right)^{'})=-\delta(e^y)\) --- (*)

where \(A\) is a constant,

\(F(f)=\cfrac{1}{i\pi}\int{\left(\delta(e^{y})+A\right)^{'}}d\,y\)

\(F(f)=\cfrac{1}{i\pi}\delta(e^{y})+\cfrac{A}{i\pi}\)

\(F(f)=\cfrac{1}{i\pi}\delta(f)+\cfrac{A}{i\pi}\)

since, \(f\gt0\) but \(\delta(f)=0\) for \(f\ne0\)

\(F(f)=\cfrac{A}{i\pi}\) a constant

for \(f\gt0\).

Since \(A\) from (*) is arbitrary, we let \(A=i\pi\)

\(F(f)=1\)

More correctly, \(2\delta(x)\rightarrow \delta(x)\) because the delta function cannot be scaled.

\(F(f)=\cfrac{A}{i2\pi}\) a constant

for \(f\gt0\).

This shows,

\(t\overset{Fourier}{\longrightarrow }F(f)=\cfrac{A}{i2\pi}\)

\(F(f)\) in the frequency domain is measured in units of \(2\pi i\), the perimeter of a circular clock of radius \(i\).  \(F(f)\) is the number of cycles around an imaginary clock.

Which is not the answer we want.  Why do we take the time derivative of \(F\) obtained from the space dimension with considerations for the time dimension to obtain an equivalent expression for \(F\) in the time dimension.

\(\cfrac{d\,F_{space}}{d\,t}=F_{time}\)

still?  A notion to be proven, not stated, but phrased as a question.

Defining Light Speed

For the fun of it, given Fourier,

\(F_{ f }[f(x)]=F(f)=\int_{-\infty}^{\infty} { f(x).e^{ -i2\pi fx } } dx\)

as

\(f.\lambda=c\)

\(f.x=\cfrac{c}{\lambda}x\)

If we denote inertia in time as

\(m_t=\cfrac{1}{c}\) and

\(n_\lambda=\cfrac{x}{\lambda}\)

the number of wavelengths in a distance \(x\).  The expression,

\( f.x = \cfrac { n_{ \lambda  } }{ m_{ t } }  \)

where \(\cfrac{n_\lambda}{m_t} \) is the number of wavelength per unit inertia in the time dimension.

\(f\) stretches \(x\) into spacetime, \(fx\).  When measured in units of wavelength, \(\lambda\), the expansion of spacetime is weighted by its inertia in time, \(m_t=\cfrac{1}{c}\).

Furthermore, if the time and space dimensions are equivalent, that \(f.x\) is a square, then

\(m^2=c.n_\lambda\)

\(c=\cfrac{m^2}{n_\lambda}\)

or

\(n_\lambda=\cfrac{m^2}{c}\)

where \(m^2\) is a complete square of a rational number of our choosing.  If we define \(c\) to be a complete square too,

\(c=b^2\)

where \(b\) is rational, then

\(n_\lambda=\cfrac{m^2}{b^2}\)

\(n_{\lambda}\) is always rational, provided spacetime, \(fx\) stretches uniformly, equally in both \(f\) and \(x\) directions, that both time and space dimensions are equivalent.

\(c=299792458\) is a complete square of rational number \(\cfrac{17314515817660047984}{10^{15}}\)

approximately...

Energy Accounting With Fourier

When,

\(G=\cfrac { \sqrt { 2c }  }{ a_{ \psi  } } \theta _{ \psi  }\)

and

\( G=\cfrac { 1 }{ \sqrt { 2 }  } \left( \cfrac { 1 }{ c }  \right) ^{ 3/2 }.\cfrac { 1 }{ tanh(\theta _{ \psi  }) } \)

An illustrative plot of 1/tanh(x) and 2x is shown below,

where both positive and negative \(G\) are admissible because of the \(\sqrt{2}\) factor in both equations.  This provides the option to make \(F_{\rho}\) positive or negative and both.

And we have another particle zoo.

What if \(E\overset{Fourier}{\longrightarrow }B\)?  That the \(B\) field is the Fourier transform of an \(E\) field and \(E\) field is the Inverse Fourier Transform of \(B\).  Or that they are of one entity that is oscillating around light speed.  Above light speed it present itself as a \(B\) field in space, below light speed it is an \(E\) field in space.  An entity we shall call \(EB\).  At the threshold of \(v=c\), we apply the Fourier Transform when we move above light speed \(F_f[E]=B\) and the Inverse Fourier Transform \(F_f^{-1}[B]=E\)when we move below light speed.

When \(E=cos(2\pi f_o x)\),

\(F_f[E]=B=\cfrac{1}{2}\{\delta(f+f_o)+\delta(f-f_o)\}\) --- (*)

where \(f\) is defined as part of \(e^{-i2\pi f x}\), a complex sinusoidal in the product of space and time, spacetime.

This suspicion arises from the postulate that in the time domain relevant parameters are periodic and is defined over a per unit time interval (per second, \(Hz\)).

Expression (*) introduces negative frequency \(-f_o\) and complex wave part \(isin(2\pi fx)\).  And a factor of one half that divides equally the magnitude (coefficient) of \(E\) between the positive and negative frequency, \(f_o\) and \(-f_o\).

The average power of the wave, over one period is with a factor \(P_f=\cfrac{1}{2}.1+\cfrac{1}{2}.\left(\cfrac{1}{4}+\cfrac{1}{4}\right)=\cfrac{3}{4}\)

The average value of \(E\) or \(B\) is with a factor \(D_{f}=\sqrt{\cfrac{3}{4}}=0.8660\).  And \(\cfrac{1}{D_f}=1.15470\)

\(D_f\) or \(\cfrac{1}{D_f}\) might amplify the discrepancies in \(\varepsilon_o\) and \(\mu_o\); calculated vs experimental values.

If we cannot measure negative frequency \(f_o\) and its power then,

The average power of the wave is with a factor \(P_f=\cfrac{1}{2}.1+\cfrac{1}{2}.\cfrac{1}{4}=\cfrac{5}{8}\)

Which gives \(D_{f}=\sqrt{P_f}=\sqrt{\cfrac{5}{8}}=0.79057\) and \(\cfrac{1}{D_f}=1.26491\).

However if instead we allow,

\(E=cos(2\pi f_o x)+isin(2\pi f_o x)\) --- (**)

where

\(F_f[isin(2\pi f_o x)]=i.\cfrac{1}{2}i\{\delta(f+f_o)-\delta(f-f_o)\}=\cfrac{1}{2}\{\delta(f-f_o)-\delta(f+f_o)\}\)

then,

\(F_f[E]=B=\delta(f-f_o)\) --- (*)

\(B\) field is at \(f_o\) with no complex part, and there is no extra factor due to the transform, \(P_f=D_f=1\).

Does this mean a complex wave part must exist, either in space or in time for conservation of energy between space and time, through the transform?  No.

Expression (**) implies that \(E\) is also circular.

This means extra energy is required to set \(B\) into circular motion.  When \(E\) is already in circular motion, no extra energy is needed for \(B\) to coil around the space dimension.  Circular motion itself stores potential energy, \(P_{cir}\).  This extra energy required results in a discrepancy factor \(P_f\ne1\).  When energy is required,

\(P_{cir}=1-P_f=\cfrac{1}{4}\)

one fourth of the total energy is required.  \(P_{cir}\) could be the reason why Fourier Transform was not admitted.  Without accounting for \(P_{cir}\), Fourier Transform will result in energy discrepancy.

Negative frequency, \(-f_o\) would suggest that \(B\) is travelling back in time.  How did that happen?

Note: Fourier transform is defined as,

\(F_{ f }[f(x)]=F(f)=\int_{-\infty}^{\infty} { f(x).e^{ -i2\pi fx } } dx\)

and the Inverse Fourier transform is,

\(F_{ f}^{-1}[F(f)]=f(x)=\int_{-\infty}^{\infty} { F(f).e^{ i2\pi fx } } df\)

where all parameters are explicit.

\(P_f=\cfrac{1}{2}\).(power in \(E\))\(+\cfrac{1}{2}\).(power in \(B\))

With reference to power in \(E\) part of the wave (ie. taken to be \(1\)), power in \(B\) is reduced by a factor of \(\left(\cfrac{1}{2}\right)^{2}=\cfrac{1}{4}\) because its magnitude was reduced by \(\cfrac{1}{2}\) when it splits into positive and negative frequency.  Both positive and negative frequency contributes equally to the power of the wave, \(\cfrac{1}{4}+\cfrac{1}{4}\)

UFO Shrinking Trick

Cont'd from the post "Stable Eight" dated 08 Jul 2016...

In fact all distances \(\Delta a\) that results in a \(\pi\) phase shift, ie.

\(\Delta a=m.2\pi a_{\psi}\)

where \(m=1,3,5..\) is odd, will also result in field confinement.

Given \(n\), the number of complete wavelength(s) along \(2\pi a_{\psi}\) the perimeter of \(\psi\), if \(n\ne\cfrac{1}{2}\), but \(n=1\) minimum we have,

\(2\pi a_{\psi}=\lambda_{\psi}\)

then

\(\pi a_{\psi}=\cfrac{1}{2}\lambda_{\psi}\)

However, if the minimum is \(n=2\),

\(\cfrac{\pi}{2} a_{\psi}=\cfrac{1}{2}\lambda_{\psi}\)

which is impossible given \(m=1\) as,

\(\cfrac{\pi}{2} a_{\psi}\lt 2a_{\psi}\)

as the minimum distance for \(\Delta a\) is \(2a_{\psi}\) in order for \(\psi\) of the particles not to touch.

Since, stable nuclei of eight exist, \(n\le 1\) for \(m=1\); at least for the particles in the stable group of eight.  \(m\) on the other hand can be large.

Which points to disassociating/reducing a confined field by increasing the frequency of the particles in the stable group.

When \(m\) is large, \(n\) can be large, increasing the frequency of \(\psi\) will visibly shrink the nuclei and the body as a whole.

Is this how UFOs do their disappearance act?  "UFO Over Vasquez Rocks " https://www.youtube.com/watch?v=h0TpWceUZnQ

Good night.

Stable Eight

\(F_{\rho}.c\) is sinusoidal and travels outwards with velocity \(c\).  It is possible that a combination of different \(F_{\rho}.c\) with different phases relative to each other be bounded in space.  Instead of its resultant field reaching to infinity in space, this group of particles has significant field only within a distinct boundary.

This might explain the relative stability of certain nuclei and elements.

With this in mind, consider,

\(f\lambda_{\psi}=c\)

\(2\pi a_{\psi}=\cfrac{1}{2}\lambda_{\psi}\)

\(\small{n=\cfrac{1}{2}}\).  \(\psi\) is a half wave that spends half its time in the time dimension and half its time in the space dimension from the posts "Not All Integer But Half Has A Place" dated 12 Dec 2014 and "Half A Wave Really? Exicting! Illuminated!" dated 12 Dec 2014.

For a \(\pi\) phase difference or a length of \(\small{\cfrac{1}{2}\lambda}\) such that the fields from two particles cancel, the two particles will have to be \(\Delta a\) apart,

\(\Delta a = 2\pi a_{\psi}\)

center to center.  In this arrangement, the resultant field cancels along the line joining the two particles, beyond the two particles.  Their resultant field is concentrated in a plain perpendicular to and bisecting the line joining the two centers of the particles.

Two such pair of particles, where their field planes intersect and are \(\Delta a\) apart line to line.  The resultant fields are then restricted to between the parallel lines that bisect the lines joining the particle pair center to center.

Then, when we have another group of four particles where their resultant fields are similarly restricted, and aligned where their field strips overlap at \(\Delta a\) apart, the field strips cancels beyond the two groups.  The resultant field is essentially within the square between the two groups of four particles.

When the field is so confined, the particles as a group of eight does not interact with particles outside of its field.  They are relatively stable.

Inert elements are stable; eight big particles orbiting as such a group in a nuclei are stable.

Group of two and four are also relatively stable given their restricted fields.

Yet it is a sad day to have reach this far...

Free Lunch

If both basic particles and big particles have the same inertia to a change in energy in the time dimension, big particles are just as easy to move along the energy number line as basic particles.

There is an opportunity here for gain without effort.

Free lunch is being served big.

And \(G\) Came Last

From previously, "Pinch, Pull And Let Go" dated 08 Jul 2016,

\(F.c=\cfrac{dKE}{dt}\)

\(KE\) is as defined in the space dimension \(\cfrac{1}{2}mv^2\)

\(F.c=\cfrac{1}{2}m\cfrac{dv^2}{dt}\)

with \(v=c\)

\(F.c=m.c\)

with \(m=\cfrac{1}{c}\)

\(F.c=\cfrac{1}{c}.c=1\)

However, force density \(F\) was derived as,

\(F=i\sqrt{2mc^2}.G.tanh\left(\cfrac{G}{\sqrt{2mc^2}}(x-x_z)\right)\)

where we make \(x_z=0\).  When \(m=\cfrac{1}{c}\),

\(F=i\sqrt{2c}.G.tanh\left(\cfrac{G}{\sqrt{2c}}(x)\right)\)

Is it possible logically that,

\(F.c=\sqrt{2c}.G.tanh(\theta_{\psi}).c=1\)

from which,

\(G=\left(\cfrac{1}{c}\right)^{3/2}.\cfrac{1}{\sqrt{2}.tanh(\theta_{\psi})}\)

where \(\theta_{\psi}=\cfrac{G}{\sqrt{2c}}(a_{\psi})\)

\(\cfrac{1}{c}\) is the inertia of a particle in the time dimension.

\(G\) is not the gravitational constant, but the constant of proportionality in the derivation for \(F_{\rho}\).

Is \(\cfrac{1}{c}\) the inertia of a basic particle or a big particle of \(n\) constituent basic particles?

\(\cfrac{1}{c}\) was derived with the consideration of \(F_{\rho}\), both basic particles and big particles with \(n\) constituent basic particle shares the same \(F_{\rho}\) expression.  Both basic particles and big particles have the same inertia \(\cfrac{1}{c}\).  When \(F_{\rho}\) is scaled by \(m\),

\(F_{\rho}\to m.F_{\rho}\)

then inertia,

\(\cfrac{1}{c}\to \cfrac{m}{c}\)

\(m.F_{\rho}\) is due to \(m\) distinct basic particles or \(m\) distinct big particles.

\(\psi\) is energy density without material mass.  Inertia in the time dimension is the resistance to change in energy as expressed as,

\(\Delta E=\cfrac{d\,KE}{dt}\)

Both basic particles and big particles although with different \(a_{\psi}\) are made of the same \(\psi\).  They both have the same resistance to a change in energy in the time dimension, as such they both have the same inertia.  With \(m\) distinct particles, each will present the same resistance, as such there is \(m\,\,\,times\) the individual inertia to a change in energy.

Good night.

Pinch, Pull And Let Go

How to make a hollow particle?  When \(a_{\psi}\lt a_{\psi\,\pi}\) set off a perturbation in \(\psi\) towards the center of the particle.  Pinch, pull and let go.

When is \(a_{\psi}\lt a_{\psi\,\pi}\)?  When \(n\lt 77\), when the particle is smaller.

We can direct \(\psi\) only if \(\dot{x}\) retains its direction when the particle returns from the time dimension.  We point ourselves in space at Jupiter, travel to light speed in space and exit into the time dimension, travel in time by the comparative age of Earth and Jupiter (Jupiter is older by 50000 years; a guess); travel back in time by 50000 years and pops back into the space dimension, in the same direction.  The gain is in the higher speed limit in the time dimension of \(\small{4\sqrt{}2.\pi^2c}\) and possibly a short travel distance of 50000 years in time.

If these details are true, only places of about the same age as the starting point is worth travelling to this way.

But this enunciation does bring out a point; the need to travel through the distance in time between the starting and destination points.

Which lead us to "time particles".

Not only do the entangled particles in space be hollow, they have to exist in the time dimension as one big particle.  One big time particle.  Only then, there is entanglement.

Which bring us back to the Durian constant...

If applying the Durian constant is correct, then this big time particle contains \(\cfrac{4}{3}\pi(2c)^3\) number of basic particles.  One big lump of entanglement.

Why?  My guess is it takes an equivalent distance of \(2c\) to reach light speed under the action of the field in the time dimension.  At \(2c\) from the center of a particle, a test particle \(\small{\cfrac{1}{c}}\) returns to the space dimension.

Unfortunately,

Mass \(m\) is the proportional constant in the space dimension \(x\).  If a point of unit mass move through \(x\).  A point mass of \(m=2\) moves through by \(\cfrac{1}{2}\).  In a very strained but analogous way, in the time dimension at light speed \(c\) (\(F.c\),energy is flowing out of the particle at light speed), a time particle moves through time by,

\(n_c=\cfrac{c}{\left(\cfrac{1}{c}\right)}\)

number of time particle of inertia \(\cfrac{1}{c}\) per second.  But since we are in the time dimension,

\(n_{c\,t}=\cfrac{d}{dt}\left\{n_c\right\}\)

we differentiate \(n_c\) in time again to obtain \(n_{c,t}\) the number of particles per unit time in the time dimension.

\(n_{c\,t}=2c\)

And so in the time dimension a sphere of entangled particle is,

\(A_D=\cfrac{4}{3}\pi (2c)^3\)

Why do we consider a windows of "per second" over a distance of \(c\) in time?  It is for the same reason we differentiate \(n_c\) in time again to obtain \(n_{c,t}\).  In the post "A Mass In Time And In Mind" dated 04 Jul 2016, the expression

\(F.c=\cfrac{dKE}{dt}\)

was differentiated with respective to time to arrive at,

\(\cfrac{dF}{dt}=\cfrac{1}{c}\cfrac{d^2KE}{dt^2}\)

the force in time.

In the time dimension displacement is measured in time, in order to obtain a dimensionless count in number, we differentiate once with time.  In order to obtain a per unit time count, (count per second), we differentiate again with respective to time.  So a dimensionless count is consider over a per second window.  The time dimension is the equivalent of the frequency domain where relevant parameters are in per unit time.

Intuitively speaking...

Hollow Particle, Driving Miss \(\psi\)

A perturbation in \(\psi\) will normally oscillate through the center of the particle,

when it does not gain light speed along its trip to the center of the particle.

When \(\psi\) does gain light speed at \(x=a_{\psi\,ex}\), the particle is hollow in the inside.

\(\psi=0\) for \(x\lt a_{\psi\,ex}\).

At the limiting case, when \(a_{\psi\,ex}=0\),  \(a_{\psi\,\pi}=a_{\psi\,c}\).

For a particle to start entangling,  it is either hollow or made hollow, or is a basic particle.

And a hollow vessel makes noise.

If Earth is hollow, travelling at light speed towards the center bring us up to the surface, or to another planet.  The problem is when is where?  A leg on Mars, an arm on Venus...

How to direct \(\psi\) through the time dimension?

Particle \(FM\) Radio

It was proposed in the post "Sticky Particles..." dated 23 Jun 2016, that at \(a_{\psi\,c}\) \(\psi\) returns to the time dimension.  It exits from time dimension and return to the space dimension on the outer surface of the particle where \(\dot{x}=c\)...

If \(\psi\) carries information then entanglement as understood, can occur if \(\psi\) is returned to the surface of another particle in close proximity in time.

This however is non specific, not two particles are specifically entangled.

Since we are dealing with \(a_{\psi}\lt a_{\psi\,\pi}\),

\(q|_{a_\psi}=2\dot{x}F_{\rho}|_{a_\psi}\)

and the exit point \(a_{\psi\,ex}\) at the interior of the particle,

\(\cfrac{1}{4\pi a_{\psi\,ex}^2}\cfrac{q|_{a_\psi\,ex}}{\varepsilon_o}=-F_{\rho}|_{a_{\psi\,ex}}.\dot{x}\)

where \(F_{\rho}|_{a_{\psi\,ex}}\) explicitly points in the negative \(x\) direction.

And the rate of change of \(F_{\rho}\), \(\cfrac{d F_{\rho}}{dt}\),

\(\cfrac{d F_{\rho}}{dt}=-\cfrac{dF_{\rho}}{dt}.\dot{x}+\cfrac{d\dot{x}}{dt}F_{\rho}\)

For the time being,

\(F_{\rho}=F_{\rho}|_{a_{\psi\,ex}}\)

\(\cfrac{dF_{\rho}}{dt}=0\)

\(\cfrac{d F_{\rho}}{dt}=-\cfrac{d\dot{x}}{dt}F_{\rho}|_{a_{\psi\,ex}}\)

We can change \(\dot{x}\) through the spin of the particle.  \(\dot{x}\) is always relative to space around the particle.

\(\cfrac{d F_{\rho}}{dt}\) acts to counter the spin at the surface of the particle and if and when \(\psi\) returns to the space dimension on the surface of another particle, induces a opposite spin on that particle.

From the post "A Mass In Time And In Mind" dated 04 Jul 2016,

\(\cfrac{dF_{\rho}}{dt}=\cfrac{1}{c}\cfrac{d^2KE}{dt^2}\)

\(\cfrac{dF_{\rho}}{dt}\) is a force on a mass \(\cfrac{1}{c}\) in the time dimension  (equivalent to \(m\) in the space dimension).

Entanglement here is made specific by creating two or more particles at the same time and thus be in close proximity in time.

If we modulate \(\dot{x}\) then we have a particle \(FM\) radio, because changing \(\dot{x}\) changes the frequency of \(\psi\).  If spin is difficult to detect then,

\(-\cfrac{d\dot{x}}{dt}F_{\rho}|_{a_{\psi\,ex}}=\cfrac{1}{c}\cfrac{d^2KE}{dt^2}\)

offers a direct way for detection as the second order time derivative of kinetic energy, \(KE\).

Note: Claude E. Shannon, Information is "entropy".  Information is energy need to wait...

Magic Booted Off The Bang Wagon

From the posts "Sticky Particles Too...Many" dated 24 Jun 2016, etc,

\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,c}))=\cfrac{1}{4}\)

\(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,c}=0.7369\)

When \(n=77\),

\(\theta_{\psi}=\cfrac{G}{\sqrt{2mc^2}}a_{\psi\,\pi}\lt\pi\)

The pinch force in the post "Big Particle Exists" dated 28 Jun 2016 holds the particles with \(a_{\psi}=a_{\psi\,77}\) with greater magnitude when they are slightly displaced further apart center to center.

When \(n=76\), this pinch force is even higher as the particles are displaced slightly, as \(F_{\rho}\) increases monotonously with radial distance \(r\).

Does this contribute to a macroscopic view that when the number of nucleons is \(2\), \(8\), \(20\) or \(28\) the nucleus is strongly bounded?

Not directly.  There is no reason why \(2\) nucleons in composite would make the constituent basic particles prefer a group of \(n=77\).  But, \(n=77\) would make all number of nucleons of this number of constituent basic particles more strongly bounded.  Furthermore, \(n=77\) being less than \(n=78\) would make such isotopes lighter in all magnitudes of charge, \(q|_{a_\psi}\), (gravitational mass, electric charge and temperature).

Do the quantum fractions, \(\cfrac{1}{77}\) and \(\cfrac{1}{78}\) exist? For that matter, \(\cfrac{1}{38}\) and \(\cfrac{1}{39}\) when we consider the interaction of only the exposed number of the constituent basic particles in each direction on half the surface of the big particle, \(a_{\psi\,\pi}\) together with the matter of attractive or repulsive interactions.

And kicked off the bang wagon I was.

Note:  The least repulsive force and the maximum attractive force that result from the constituent basic particle redistributing or the big particle \(a_{\psi\,\pi}\) rotating away from a repulsive force and rotating towards an attractive force, do not result in an integer effective number of \(n=39\), or \(n=38\) either.

These numbers are guides not absolute truth.

Vectoring DragonBall

Only if there is a determined and fixed direction of spin \(\omega\) along the top containment half sphere does orientating the half sphere produce a vectoring effect.

Vectoring points the rate of change in kinetic energy, \(\cfrac{d\,KE}{dt}\) in a specific direction.

Now where are you going?

An UFO With Seven Dragon Balls...

If you ever seen an UFO with a plasma propulsion supporting it,

it is likely that the plasma ball is spinning in a up-down direction instead of horizontally.  This direction creates lifts that provide for anti-gravity.

In a similar way, a missile that proposes to generate lift and velocity with a ball of fire, is best also spinning the ball of fire in a plane parallel to its direction of travel.  Changing the axis of this plane of spin, changes the direction of travel.  Vectoring.

It is the velocity along a circular path of the plasma in the interior of the containment of the fire ball that is \(v_s\), the spin velocity developed in the previous posts "Spinning Plasma" and "Spinning Plasma" dated 05 and 04 Jul 2016, respectively.

Unlike a jet engine, the spinning dragon ball is contained not open ended.  No hot material is lost from the fire ball.

KO!

Spinning Plasma

When the dragon ball is spinning at circular velocity \(c\), the power output from the particle is,

\(v_s=\dot{x}=c\)

\((\dot{x}-v_s).F=0\)

zero.  This is the limiting case, when the power output from the particle is tangential to the surface of the sphere, perpendicular to a radial line.  It is as if power is being shaved from the surface of the sphere by an observer standing in relative speed \(\dot{x}-v_s\) just above the surface.

Power output slows and remains around the particle could explain the increase in energy detected around spinning plasma.  Such power is directly proportional to the spin velocity \(v_s\) of the plasma, on the equatorial plane of the particle,

\(P_{rad}=(\dot{x}-v_s).F\)

By conservation of energy,

\(P_{\bot}=v_sF\)

which is consistent with the limiting case of \(v_s\to c\),

\(P_{bot}\to P=c.F\)

with \(\dot{x}=c\)

where the power output is tangential to a radial line.  Furthermore,

\(\overset {\rightarrow }{ \cfrac{ P_{\bot} }{v_s}}+\overset {\rightarrow}{\cfrac{P_{rad}}{(c-v_s)}}=\overset{\rightarrow }{\cfrac{P}{c}}\)

where the power vector \(\overset{\rightarrow }{\cfrac{P}{\dot{x}}}\) takes the direction of \(\dot{x}\), in view of the fact that,

\(F.\dot{x}=\cfrac{d\,KE}{dt}\)

\(F=\cfrac{1}{\dot{x}}\cfrac{d\,KE}{dt}\)

\(F\) is not a force, but a power vector.

But,

\(\overset {\rightarrow }{ \cfrac{ P_{\bot} }{v_s}}=\overset {\rightarrow}{\cfrac{P_{rad}}{(c-v_s)}}=\overset{\rightarrow }{\cfrac{P}{c}}=F\)

\(\left|\overset {\rightarrow }{ F}\right|+\left|\overset {\rightarrow }{ F}\right|=\sqrt{2}\left|\overset {\rightarrow }{ F}\right|\)

since the total output power is a constant, we have instead,

\(\cfrac{1}{\sqrt{2}}\left|\overset {\rightarrow }{ F}\right|+\cfrac{1}{\sqrt{2}}\left|\overset {\rightarrow }{ F}\right|=\left|\overset {\rightarrow }{ F}\right|\)

both expressions suggest that there can be a reduction by a factor of \(\frac{1}{\sqrt{2}}\) as the power vectors sum vectorially, in the time dimension. Where did this lost energy go?  Into the space dimension from which we harness.  So, as we manipulate mass in the time dimension \(\frac{1}{c}\), by manipulating light speed \(\dot{x}=c\) in the space dimension, we can gain energy in the space dimension.  However, \(c\) is a constant, as the example illustrates, \(c\) is manipulated by adding an orthogonal component to \(c\).  In this case the spin velocity \(v_s\).

The maximum lost of power in the time dimension is at \(\cfrac{1}{\sqrt{2}}P\) when \(v_s=\cfrac{c}{2}\), and so the maximum power available to the space dimension is \(\left(1-\frac{1}{\sqrt{2}}\right)P\) when \(v_s=\cfrac{c}{2}\).

If there is ever a bullshit alert this will surely sets it off loudly.

Good night.

Light Speed Spin And Other Tales

The way to harness energy from a dragon ball is to slow it down in a vortex of the appropriate field,

This way the energy remains around the particle for long enough time to be put to good use.  Since the energy from the particle is the same as the vortex containment field, the system is self-sustaining.

All that is needed is a little spin, near light speed.

Singularity Is A Question

\(F_{\rho}\) points inwards and accelerate \(\psi\).  \(\psi\) returns to the time dimension at \(c\), so it is possible to create a thin shell of \(\psi\), very dense that is at light speed a very short distance from the surface of the spherical shell.

\(\psi\) is the equivalent of \(B\) or \(E\) per unit volume.

\(\psi\) was derived with the assumption, \(c\) is the terminal speed limit and a constant.  When we derive an expression for \(c\) in the post "Just When You Think \(c\) Is The Last Constant" dated 26 Jun 2016 and find that it is a constant for a given \(n\).  That is just consistent with the stated assumption.

A manufactured coincident.

Why should \(\dot{x}=c\) in \(q|_{a_\psi}=2\dot{x}F_{\rho}|_{a_{\psi}}\)?  Why should changes on the surface of the sun travel to Earth at the speed of light \(c\)?

Is light perception the way we detect the force field (temperature field) around a temperature particle?

When,

\(F.\dot{x}=\cfrac{d\,KE}{dt}\)

is it possible that \(\psi\) be at \(\dot{x}\lt c\) such that the power from a big particle of many \(n\) is reduced and harnessed?

A Mass In Time And In Mind

From the post "Into A Pile Of Deep Shit" dated 09 Jun 2016,

\(\left( 1-\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, t_{ c } } =2\cfrac { \partial \, T }{ \partial \, t_{ c } }\)

For \(\cfrac { \partial \, x }{ \partial \, t_{ c } } \ne 0\), multiply both sides by \(\cfrac{ \partial \, t_{ c } }  { \partial \, x }\)

\(\left( 1-\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, t_{ c } } \cfrac { \partial \, t_{ c } }{ \partial \, x } =2\cfrac { \partial \, T }{ \partial \, t_{ c } } \cfrac { \partial \, t_{ c } }{ \partial \, x }\) -- (*)

The reason (*) is still possible is because, the wave \(\psi\) is stationary in space, that \(\cfrac{\partial\,\psi}{\partial\,t}|_{x=x_o}=0\).

A wave \(\psi\) in circular motion, radiating energy is analogous to circular motion in mechanics.  In circular motion,

in turning \(v\) into a circular path a centripetal force directed to the center the circle acts on the body.  As \(\Delta \theta\to0\) in the diagram, \(F_{m}\) is directed at the center of the circle.  If the body encounters resistance along its circular path but remains in circular motion, a resistance force pointing away from the center of the circle acts on the body.  Work done against this resistance force is the energy needed to overcome the resistance along the circular path.  This force acts on the mass of the body, \(m\) and is a Newtonian force.

In the case of \(\psi\), in circular motion around the center of a particle, a similar "force" develops as \(\psi\) encounters resistance along its path.  "Work done" against this resistance is radiated outward away from the center of the particle along a radial line.  This force acts on \(\psi\) (not on mass \(m\) but on energy density, \(\psi\)) and results in a rate of change of kinetic energy,

\(F.c=\cfrac{dKE}{dt}\)

where \(\dot{x}=c\).

\(F=\cfrac{1}{c}\cfrac{dKE}{dt}\)

Image \(\psi\) along an energy number line mark consecutively with increasing energy magnitude.  \(\cfrac{dKE}{dt}\) is the equivalent of velocity along the energy number line. \(\cfrac{1}{c}\) is the equivalent of mass.  \(F\) is then the momentum of energy density, \(\psi\).  \(\psi\) is a point of energy denoted as \(\cfrac{1}{c}\) just as a point mass is notated as mass density, \(m\).

We can have,

\(F.c=ma.c=mc.a=\cfrac{dKE}{dt}\)

but \(F\) does not act on \(m\) but on \(\psi\).  If the equivalent mass is \(mc\) then \(mc.a=\cfrac{dKE}{dt}\) would make \(a\), a velocity.  However, \(m\) is a space concept of inertia and does not exist in time.  However we can view the time dimension from the perspective of space and obtain,

\(E=mc^2\)

where \(c\) is the speed limit in the time dimension or

\(E=32\pi^4c^2\)

where \(c\) is the speed limit defined in the space dimension.

\(32\pi^4v_x^2+v_t^2=32\pi^2c^2\)

both of which deals with the time dimension at the terminal speed \(c\) in the time dimension from the perspective of the space dimension.

\(m\) marks a point in space, \(x\) along a space number line.  \(\psi\) marks a point of energy in time along a time number line.  In the time dimension, the first rate of change is the second rate of change in space; Acceleration (in space) is the analogue of velocity (in time) and the change of energy, not over time and stationary in space is the analogue of the change of time (position) along the time number line.

Energy \(\cfrac{1}{c}\) in the time dimension is the equivalent of mass \(m\) in the space dimension.  In the space dimension,

\(F=ma\)

\(m\) is the inertia, resistance along a space number line.

In the time dimension,

\(F.c=\cfrac{dKE}{dt}\)

\(\cfrac{dF}{dt}.c=\cfrac{d^2KE}{dt^2}\)

assuming \(c\) is a constant.

\(\cfrac{dF}{dt}=\cfrac{1}{c}\cfrac{d^2KE}{dt^2}\)

A force in the time dimension is the rate of change of \(F\) acting on \(\psi\).

The first rate of change in time is the second rate of change in space; the change of position in time is the first rate of change in space (change in velocity; change in \(KE\)).  Inertia in the time dimension is \(\cfrac{1}{c}\).

This is just consistent coincident.

Why \(\mu_o\)?

If we have mistaken,

\(\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o}=F_{newton}\)

then effectively

\(\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o.c}=F_{newton}\)

that

\(\varepsilon_o\rightarrow \varepsilon_o^{*}.c\)

in which case,

\({\varepsilon_o^{*}.c}=\cfrac{1}{2c^2ln(cosh(\theta_{\psi}))}\)

\({\varepsilon_o^{*}}=\cfrac{1}{2c^3ln(cosh(\theta_{\psi}))}\)

\(c=\cfrac{1}{\sqrt[3]{2ln(cosh(\theta_{\psi})).\varepsilon_o^{*}}}=\cfrac{1}{\sqrt{\mu_o\varepsilon_o^{*}}}=\cfrac{1}{\sqrt{\varepsilon_o2ln(cosh(\theta_{\psi}))}}\) --- (*)

\(\mu_o\) in the denominator is first cube rooted then squared then divided by \(2ln(cosh(\theta_{\psi}))\).  The far right hand side, is the actual measured term, and has the factor \(2ln(cosh(\theta_{\psi}))\).  The mistaken theoretical term involving \(\varepsilon_o^{*}\) does not, so \(\mu_o\) inserted into the expression is divided by \(2ln(cosh(\theta_{\psi}))\).

With the result,

\({\varepsilon_o}=\cfrac{1}{2c^2ln(cosh(\theta_{\psi}))}\) 

when \(\varepsilon_o\rightarrow \varepsilon_o^{*}.c\), we have

\({\varepsilon_o^{*}.c}=\cfrac{1}{2c^2ln(cosh(\theta_{\psi}))}\) 

Multiplied by \(\mu_o\),

\({\mu_o\varepsilon_o^{*}.c}=\mu_o.\cfrac{1}{2c^2ln(cosh(\theta_{\psi}))}.c\)

if this were to be

\(\mu_o\varepsilon_o=\cfrac{1}{c^2}\)

\(\mu_o.\cfrac{1}{2c^2ln(cosh(\theta_{\psi}))}.c=\cfrac{1}{c^2}\)

\(\mu_o=\cfrac{2ln(cosh(\theta_{\psi}))}{c}\)

such that we move the measured results to the mistaken theoretical expression of \(\cfrac{1}{c^2}\).

So when we apply information in (*) to the \(\mu_o\) above,

\(\mu_o=\cfrac{1}{2ln(cosh(\theta_{\psi}))}\left(\sqrt[3]{\cfrac{2ln(cosh(\theta_{\psi}))}{c}}\right)^2\)

\(\mu_o=\left(\cfrac{1}{c\sqrt{2ln(cosh(\theta_{\psi}))}}\right)^{2/3}\)

when \(\theta_{\psi}=3.135009\),

\(\mu_o=1.315500e-6\)

The defined value of \(\mu_o\) is \(\mu_o=1.256637e-6\)

Which is interesting.  \(\mu_o\) may have originated in the mistake in \(c\) in,

\(\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o}=F.c\ne F_{newton}\)