Where would one find a prevalent time field that accelerate all time particles to light speed in the time dimension?
Inside a time particle.
The universe is one big time particle.
Inside such a particle, the force on \(\psi\) is given by \(-\frac{\partial\psi}{\partial\,x}\). The value of this force and its gradient will locate us in the universe. We need two equations because we do not have an origin to start with. At the origin, time return to the space dimension.
Time is in flux, we are not going any where in space. The end of time is at the center of the universe. At the end, \(\psi\) get recycled on the surface of the same particle (or another particle) where \(v=c\) is perpendicular to the radial direction.
Since, space is orthogonal to time and vice versa, space and time are just orthogonal view of the same particle!
There is a time particle, is there a separate space particle? A space particle is our experience of a time particle as we travel circularly around the time particle. In the radial direction, we experience time. In a reciprocal way, a time particle is our experience of a space particle as we go around the space particle; along its radius, we experience space.
There is just space-time.
Which brings to mind the relationship between \(B\) and \(E\) fields. And how such fields are forcefully made once removed, that a \(g^{+}\) particle produces a \(T\) field that hold a \(T^{+}\) particle that in spin produces a \(E\) field which hold a \(p^{+}\). A spinning \(p^{+}\) produces a \(g\) field that in turn holds a \(g^{+}\) particle, to accommodate other fields, especially the temperature particle and temperature field.
If we are to be consistent, spinning space particle does not produce a time field, when spinning time particles produces a space field.
But a spinning \(B\) field produces an \(E\) field.
A spinning space field will produce a time field. And vice versa. What is a space field? The fountain of youth and Casimir effect where space is force out of a narrow gap comes to mind.
If spinning space particle produces a different field, then there is a possible new type of particle, un-thought of. If space is just gravity then time is just proton. Does an \(E\) field stop time?
One scenario leads us to two new possible particles in addition to the time particle, to form a tuple of three that is analogous to \((g^{+},\,T^{+},\,p^{+})\). Another scenario collapses time field into an electric field.
Walk through a charged parallel plate capacitor and walk though time...the Afghan artifacts!
Saturday, July 30, 2016
A Treasure What Worth?
If time particle establishing a time field is,
\(a_{\psi}=14.77\,nm\)
what is the significance of,
\(f_{\psi}=\cfrac{c}{2\pi a_\psi}=3230699.3\,GHz\)?
And,
\(T_{\psi}=\cfrac{1}{f_{\psi}}=3.09531*10^{-16}=0.309531\,fs\)?
And,
\(E=hf_{\psi}=6.626 069 57*10^{-34}*3230699.3*10^{9}=2.14068*10^{-18}\, J\)
Hartree energy is, \(E_H=4.359 744 34[19]*10^{-18}\)
So......???
Maybe \(a_{\psi}=14.77\,nm\) is just an electron.
\(a_{\psi}=14.77\,nm\)
what is the significance of,
\(f_{\psi}=\cfrac{c}{2\pi a_\psi}=3230699.3\,GHz\)?
And,
\(T_{\psi}=\cfrac{1}{f_{\psi}}=3.09531*10^{-16}=0.309531\,fs\)?
And,
\(E=hf_{\psi}=6.626 069 57*10^{-34}*3230699.3*10^{9}=2.14068*10^{-18}\, J\)
Hartree energy is, \(E_H=4.359 744 34[19]*10^{-18}\)
So......???
Maybe \(a_{\psi}=14.77\,nm\) is just an electron.
Friday, July 29, 2016
The Placid Field
I think,
\(a_{\psi\,74}=14.77\)
is the time particle, for given the extreme experimental conditions in the emission lamp of high temperature, high voltage and high collisions (high acceleration/deceleration) only one thing remain placid, the time field. The particle size remains low at \(n=74\).
In my dream, I won the No Bell prize again...
\(a_{\psi\,74}=14.77\)
is the time particle, for given the extreme experimental conditions in the emission lamp of high temperature, high voltage and high collisions (high acceleration/deceleration) only one thing remain placid, the time field. The particle size remains low at \(n=74\).
In my dream, I won the No Bell prize again...
Young Again
The presence of a fourth field and elements being made up of this fourth particle pair might explain the relative reactivity of inert gases. This suggests that inert gases do not have protons and electrons orbits in the outer layer of their atoms, instead a fourth type of positive and negative particles are present.
And we add another layer to the nucleus. Is there other evidence for the presence of a fourth wave? Electric, gravitational, temperature and...
In the time dimension the will be a corresponding electric, gravitational and temperature field each acting along one of the 3D time dimensional axis and in oscillation between space, which is now curled around each time dimension, and time, which is now linear.
It is easy to venture, a time field and a positive and a negative time particle conjugate pair. The implication of which,
\(v_t^2+v_g^2=c^2\)
that energy conservation across space and time, is subjected to a constant time field.
The good thing is, if time field exists, it can be made negative and I shall be young again. In reality...
time is driven to light speed and curls around each space dimension, so there must be a strong field and corresponding particle pair.
Sweet dreams...
And we add another layer to the nucleus. Is there other evidence for the presence of a fourth wave? Electric, gravitational, temperature and...
In the time dimension the will be a corresponding electric, gravitational and temperature field each acting along one of the 3D time dimensional axis and in oscillation between space, which is now curled around each time dimension, and time, which is now linear.
It is easy to venture, a time field and a positive and a negative time particle conjugate pair. The implication of which,
\(v_t^2+v_g^2=c^2\)
that energy conservation across space and time, is subjected to a constant time field.
The good thing is, if time field exists, it can be made negative and I shall be young again. In reality...
time is driven to light speed and curls around each space dimension, so there must be a strong field and corresponding particle pair.
Sweet dreams...
Bigger Field, Bigger Particles
Remember the unofficial data,
\(\cfrac{m_e}{m_p}=\cfrac{1}{74}\)
If,
\(a_{\psi\,74}=14.77\)
From the post "Sticky Particles Too...Many" dated 24 Jun 2016,
\(n\left(\cfrac{a_{\psi\,c}}{a_{\psi\,\pi}}\right)^3=n\left(\cfrac{0.7369}{\pi}\right)^3=1\)
\(\cfrac{n_1}{n_2}=\left(\cfrac{a_{\psi\,n_1}}{a_{\psi\,n_2}}\right)^3\)
\(a_{\psi\,85}=15.48\)
\(a_{\psi\,100}=16.32\)
and
\(a_{\psi\,166}=19.34\)
which are bigger than big particles of \(n=77\) when not subjected to high fields (zero field strength). In high fields, the respective basic particles coalesce into bigger particles, \(n\gt77\).
Which unfortunately invalidates, using spectra lines to identify elements. Only spectra lines emitted under the same field conditions are comparable, because \(a_{\psi}\) increases with increasing field strength.
And we walk further into the science fiction...
\(\cfrac{m_e}{m_p}=\cfrac{1}{74}\)
If,
\(a_{\psi\,74}=14.77\)
From the post "Sticky Particles Too...Many" dated 24 Jun 2016,
\(n\left(\cfrac{a_{\psi\,c}}{a_{\psi\,\pi}}\right)^3=n\left(\cfrac{0.7369}{\pi}\right)^3=1\)
\(\cfrac{n_1}{n_2}=\left(\cfrac{a_{\psi\,n_1}}{a_{\psi\,n_2}}\right)^3\)
\(a_{\psi\,85}=15.48\)
\(a_{\psi\,100}=16.32\)
and
\(a_{\psi\,166}=19.34\)
which are bigger than big particles of \(n=77\) when not subjected to high fields (zero field strength). In high fields, the respective basic particles coalesce into bigger particles, \(n\gt77\).
Which unfortunately invalidates, using spectra lines to identify elements. Only spectra lines emitted under the same field conditions are comparable, because \(a_{\psi}\) increases with increasing field strength.
And we walk further into the science fiction...
No Experimental Proof
Consider this,
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,c}))=\cfrac{1}{4}\)
\(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,c}=cosh^{-1}(e^{\frac{1}{4}})\)
and,
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,\pi}))=1\)
\(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,\pi}=cosh^{-1}(e)\)
So,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}}=\cfrac{cosh^{-1}(e)}{cosh^{-1}(e^{\frac{1}{4}})}\)
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}}=2.24921\)
From the post "Sizing Them Up" dated 13 Dec 2014,
\(a_\psi=19.34\,nm\)
\(a_\psi=16.32\,nm\)
\(a_\psi=15.48\, nm\)
\(a_\psi=14.77\,nm\)
of all the particles proposed none of them pair up to give a ratio of \(2.24921\). The extreme pair,
\(\cfrac{19.34}{14.77}=1.30941\)
Maybe \(a_{\psi\,c}\) readily coalesce into big \(a_{\psi\,\pi}\). Furthermore, if these are the positive particles and the last an electron, where are the other two conjugate negative particles; the negative temperature particle and the negative gravity particle. It could be that at higher temperature, the negative particles are drive off, which implies that at lower temperature, these particles will emit a faint spectra. Negative gravity particles may be more prominent under micro gravity; a new spectra line could be seen when the experiment is conducted in outer space. Under Earth's gravity, negative gravity particles will fall to the ground not affected by the electric or temperature gradients.
Still, the difference in \(a_{\psi}\) can be due the prevailing fields, temperature, electric and gravity, that the experiment is subjected to. The bigger the field the bigger the particle. Theoretically all particles with the same number of basic particles \(n\), should be of the same \(a_{\psi\,\pi}\).
But in consolation,
\(\cfrac{15.48}{14.77}\approx\cfrac{78}{74}=1.05\)
which is just \(\psi\) going around in circle...
For that matter, as we go boldly maybe four experimental values of \(a_{\psi}\) from the post "Sizing Them Up" suggests the presence of another type of field and its associated particle pair. The last value for \(a_{\psi}=14.77\) is not an electron, but a positive particle of a field relatively least in strength during the course of the experiment. The weakest field and so a particle of smallest \(a_{\psi}\).
What is this possible field?
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,c}))=\cfrac{1}{4}\)
\(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,c}=cosh^{-1}(e^{\frac{1}{4}})\)
and,
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,\pi}))=1\)
\(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,\pi}=cosh^{-1}(e)\)
So,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}}=\cfrac{cosh^{-1}(e)}{cosh^{-1}(e^{\frac{1}{4}})}\)
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}}=2.24921\)
From the post "Sizing Them Up" dated 13 Dec 2014,
\(a_\psi=19.34\,nm\)
\(a_\psi=16.32\,nm\)
\(a_\psi=15.48\, nm\)
\(a_\psi=14.77\,nm\)
of all the particles proposed none of them pair up to give a ratio of \(2.24921\). The extreme pair,
\(\cfrac{19.34}{14.77}=1.30941\)
Maybe \(a_{\psi\,c}\) readily coalesce into big \(a_{\psi\,\pi}\). Furthermore, if these are the positive particles and the last an electron, where are the other two conjugate negative particles; the negative temperature particle and the negative gravity particle. It could be that at higher temperature, the negative particles are drive off, which implies that at lower temperature, these particles will emit a faint spectra. Negative gravity particles may be more prominent under micro gravity; a new spectra line could be seen when the experiment is conducted in outer space. Under Earth's gravity, negative gravity particles will fall to the ground not affected by the electric or temperature gradients.
Still, the difference in \(a_{\psi}\) can be due the prevailing fields, temperature, electric and gravity, that the experiment is subjected to. The bigger the field the bigger the particle. Theoretically all particles with the same number of basic particles \(n\), should be of the same \(a_{\psi\,\pi}\).
But in consolation,
\(\cfrac{15.48}{14.77}\approx\cfrac{78}{74}=1.05\)
which is just \(\psi\) going around in circle...
For that matter, as we go boldly maybe four experimental values of \(a_{\psi}\) from the post "Sizing Them Up" suggests the presence of another type of field and its associated particle pair. The last value for \(a_{\psi}=14.77\) is not an electron, but a positive particle of a field relatively least in strength during the course of the experiment. The weakest field and so a particle of smallest \(a_{\psi}\).
What is this possible field?
Wednesday, July 27, 2016
Stuck In Time
If a basic particle can be made big, really big,
What is this place inside the basic particle? If it is the time dimension, then the thin layer of \(\psi\) would have accelerated everything that passed through it to light speed.
What to do in the time dimension? A similar basic particle made big in the time dimension, by reciprocity will open into the space dimension.
What's the point of being in and out of the time dimension?
If basic particles coalesce into a big particle \(\psi\) can be thick enough. Does it burn or tear?
If one basic particle particle is made big in the space dimension its dual in the time dimension is also made big. There is already "a similar particle" the time dimension. What is needed is another particle in the space dimension along the direction of travel (along \(v\)) to step out of. This second particle is created at the same time as the first so that its distance in time is coincidental. Its location in space is variable. Not at all public transport unless all particle pairs are synchronized such that only one pair exists within a \(\Delta\) in space and time.
What is in the time dimension? If we ever get stuck.
What is this place inside the basic particle? If it is the time dimension, then the thin layer of \(\psi\) would have accelerated everything that passed through it to light speed.
What to do in the time dimension? A similar basic particle made big in the time dimension, by reciprocity will open into the space dimension.
What's the point of being in and out of the time dimension?
If basic particles coalesce into a big particle \(\psi\) can be thick enough. Does it burn or tear?
If one basic particle particle is made big in the space dimension its dual in the time dimension is also made big. There is already "a similar particle" the time dimension. What is needed is another particle in the space dimension along the direction of travel (along \(v\)) to step out of. This second particle is created at the same time as the first so that its distance in time is coincidental. Its location in space is variable. Not at all public transport unless all particle pairs are synchronized such that only one pair exists within a \(\Delta\) in space and time.
What is in the time dimension? If we ever get stuck.
Saturday, July 23, 2016
The Sun Says It All
A particle in space accelerates to light speed and enters into the time dimension. In the time dimension it accelerates to light speed and re-emerge from the time dimension into the space dimension. If this is the case, particles of lighter inertia will emerge first as they gain light speed first, particles of greater inertia follow afterwards. The color spectrum of these particles lends this portal from the time dimension its color.
Our Sun,
will go nova when it starts to spill heavier elements of multitude of colors. Not in an explosive manner but gradually. If any Sun spills particles from a common pool of particles in time. A new born sun, (a sun bomb), will have the same spectrum as the Sun.
The colors of a white dwarf, a nova or supernova, travels to us at light speed, the heavier particles in the time dimension are accelerated to light speed in the same way as in space,
But "time" in the time dimension is space. When two portals "space out" in time, they are of different age. When they are of different age in time, they are at a different place in space.
\(x\) is the time it takes for the heavier particles to accelerate to light speed in the time dimension. \(\Delta age\) is the distance it took for the heavier particle to accelerate to light speed in the time dimension. With a set of \((\Delta age,\,x)\) data, it is then possible to map the acceleration profile in the time dimension, \((x,\,t)\). (\(\Delta t\) plays a part as the initial condition when the spectrum of the heavier elements first reaches Earth at light speed; it allows for the calculation of \(x\), given \(c\)).
If the acceleration to light speed breaks up the big particles into basic particles then, the formations into different heavier elements in space will take specific conditions in space at the supernova. All particles that emerge from a portal are basic particles that reforms into different elements depending on local conditions. All particle that passes through from one dimension to the other are basic particles. The acceleration profile through the time dimension is not accessible.
This however does not rule out the possibility of local conditions, where big particle passes through to the time dimension and then returns to the space dimension.
Our Sun may just be one type of portal between space and time.
All creatures great and small...
Our Sun,
will go nova when it starts to spill heavier elements of multitude of colors. Not in an explosive manner but gradually. If any Sun spills particles from a common pool of particles in time. A new born sun, (a sun bomb), will have the same spectrum as the Sun.
The colors of a white dwarf, a nova or supernova, travels to us at light speed, the heavier particles in the time dimension are accelerated to light speed in the same way as in space,
But "time" in the time dimension is space. When two portals "space out" in time, they are of different age. When they are of different age in time, they are at a different place in space.
\(x\) is the time it takes for the heavier particles to accelerate to light speed in the time dimension. \(\Delta age\) is the distance it took for the heavier particle to accelerate to light speed in the time dimension. With a set of \((\Delta age,\,x)\) data, it is then possible to map the acceleration profile in the time dimension, \((x,\,t)\). (\(\Delta t\) plays a part as the initial condition when the spectrum of the heavier elements first reaches Earth at light speed; it allows for the calculation of \(x\), given \(c\)).
If the acceleration to light speed breaks up the big particles into basic particles then, the formations into different heavier elements in space will take specific conditions in space at the supernova. All particles that emerge from a portal are basic particles that reforms into different elements depending on local conditions. All particle that passes through from one dimension to the other are basic particles. The acceleration profile through the time dimension is not accessible.
This however does not rule out the possibility of local conditions, where big particle passes through to the time dimension and then returns to the space dimension.
Our Sun may just be one type of portal between space and time.
All creatures great and small...
Friday, July 22, 2016
Pop Here
Is the direction of \(v=c\) in space preserved as \(\psi\) emerge from the time dimensions?
How does \(\psi\) re-emerge, if at all? Through a parallel "time particle" in a parallel "time world"? A particle in space pops into the time dimension and pops out of light speed, travels in time and pops back to the space dimension when it reaches light speed in the time dimension. For every portal in space, there is a dual in time. And so for every particle in space there is a dual particle in time; a time particle. The dual are two manifestations of the same entity, in space and in time. And so energy changes to the particle must be in both space and time,
\(E=2*\cfrac{1}{2}mc^2\)
And Einstein laugh from his grave,
\(E=mc^2\)
The direction of \(v\) should be preserved because no force acts on the particle other than along \(v\). That's how aliens beam themselves to inner Earth first, avoiding all up top.
How does \(\psi\) re-emerge, if at all? Through a parallel "time particle" in a parallel "time world"? A particle in space pops into the time dimension and pops out of light speed, travels in time and pops back to the space dimension when it reaches light speed in the time dimension. For every portal in space, there is a dual in time. And so for every particle in space there is a dual particle in time; a time particle. The dual are two manifestations of the same entity, in space and in time. And so energy changes to the particle must be in both space and time,
\(E=2*\cfrac{1}{2}mc^2\)
And Einstein laugh from his grave,
\(E=mc^2\)
The direction of \(v\) should be preserved because no force acts on the particle other than along \(v\). That's how aliens beam themselves to inner Earth first, avoiding all up top.
Stable Four
For a basic particle,
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,c}))=\cfrac{1}{4}\)
then it takes four basic particles to balance the charge of one big particle with,
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,\pi}))=1\)
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,c}))=\cfrac{1}{4}\)
then it takes four basic particles to balance the charge of one big particle with,
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,\pi}))=1\)
This might explain the stable four electron configuration, but a charge with one-fourth the charge of an electron is no longer an electron.
In the case of paired orbits, the orbiting particles are at a phase angle of \(\theta_p=\cfrac{\pi}{4}\) when they are on the same plane, two at the intersections of the orbits and when their orbits bring two of them on a perpendicular plane. In the last case, a third circle can be drawn perpendicular to the two orbits, that intersects all the revolving particles. All particles are at the same speed.
In the case of four orbitals around a single nucleus, it is possible that all particles passes through one of the two intersections of their orbits (taken in pairs) and are \(\theta_p=\cfrac{\pi}{4}\) with each other. At any time during their revolutions, a circle can be drawn through all particles and they rest at \(\theta_p=\cfrac{\pi}{4}\) phase to each other.
Maybe. The revolving particles might not even be in fixed circular orbits but are always part of a circular orbit.
So, a stable four "electron" configuration and a stable eight "electron" configuration are for different reasons (c/f "Stable Eight" dated 08 Jul 2016).
And it rained in July...
Thursday, July 21, 2016
Earth Shield Frequency
From the post "A Shield" dated 27 May 2016,
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{gradient|_{\pi}}{m}}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2sech^2(\pi)}{m}}=\cfrac{sech(\pi)}{2\pi}\cfrac{G}{\sqrt{m}}\)
\(G=\cfrac{\pi \sqrt { 2{ mc^{ 2 } } } }{a_e}\)
\(a_E=6371e3m\)
\(m_E=5.972e24 kg\)
\(G=\)pi*sqrt(2*5.972*(299792458)^2*10^(24))/(6371e3)
\(G=5.109e14\)
If instead we use half the gradient at \(\theta_{\psi\,\pi}=3.135009\)
\(gradient|_{\pi}=sech^2(3.135009)\)
without the constant \(G\).
Approximating the tangent with a line with gradient \(tan(\beta)\),
\(tan(\beta)=sech^2(3.135009)\)
\(\cfrac{1}{2}\beta=\cfrac{1}{2}tan^{-1}(sech^2(3.135009))\)
\(tan(\cfrac{1}{2}\beta)=0.00377\)
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2*0.00377}{m}}\)
\(f_{res}=0.009772\cfrac{G}{\sqrt{m}}\)
\(f_{res}=2.0430 Hz\)
which is low....
If we instead do not half the gradient at \(\theta_{\psi\,\pi}=3.135009\),
\(tan(\beta)=sech^2(3.135009)=0.00754\)
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2*0.00754}{m}}\)
\(f_{res}=0.01382\cfrac{G}{\sqrt{m}}\)
\(f_{res}=2.8893 Hz\)
If we take the gradient, without \(G\) at the origin instead,
\(sech^2(0)=1\)
\(\cfrac{1}{2}\beta=0.39270\)
\(tan(\cfrac{1}{2}\beta)=0.41421\)
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2*0.41421}{m}}\)
\(f_{res}=0.10243\cfrac{G}{\sqrt{m}}\)
\(f_{res}=21.41 Hz\)
What happen when \(\psi\) is pushed along its profile? Maybe \(a_{\psi}\) increases...chicken! Big chicken!
Or,
\(tan(\beta)=sech^2(0)=1\)
\(\cfrac{1}{2}tan(\beta)=0.5\)
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2*0.5}{m}}\)
\(f_{res}=0.11254\cfrac{G}{\sqrt{m}}\)
\(f_{res}=23.53 Hz\)
We are comparing \(\psi\) to the energy curve of a SHM system, \(U=\cfrac{1}{2}kx^2\) the gradient of which \(U^{'}=kx\).
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{gradient|_{\pi}}{m}}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2sech^2(\pi)}{m}}=\cfrac{sech(\pi)}{2\pi}\cfrac{G}{\sqrt{m}}\)
\(G=\cfrac{\pi \sqrt { 2{ mc^{ 2 } } } }{a_e}\)
\(a_E=6371e3m\)
\(m_E=5.972e24 kg\)
\(G=\)pi*sqrt(2*5.972*(299792458)^2*10^(24))/(6371e3)
\(G=5.109e14\)
If instead we use half the gradient at \(\theta_{\psi\,\pi}=3.135009\)
\(gradient|_{\pi}=sech^2(3.135009)\)
without the constant \(G\).
Approximating the tangent with a line with gradient \(tan(\beta)\),
\(tan(\beta)=sech^2(3.135009)\)
\(\cfrac{1}{2}\beta=\cfrac{1}{2}tan^{-1}(sech^2(3.135009))\)
\(tan(\cfrac{1}{2}\beta)=0.00377\)
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2*0.00377}{m}}\)
\(f_{res}=0.009772\cfrac{G}{\sqrt{m}}\)
\(f_{res}=2.0430 Hz\)
which is low....
If we instead do not half the gradient at \(\theta_{\psi\,\pi}=3.135009\),
\(tan(\beta)=sech^2(3.135009)=0.00754\)
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2*0.00754}{m}}\)
\(f_{res}=0.01382\cfrac{G}{\sqrt{m}}\)
\(f_{res}=2.8893 Hz\)
If we take the gradient, without \(G\) at the origin instead,
\(sech^2(0)=1\)
\(\cfrac{1}{2}\beta=0.39270\)
\(tan(\cfrac{1}{2}\beta)=0.41421\)
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2*0.41421}{m}}\)
\(f_{res}=21.41 Hz\)
What happen when \(\psi\) is pushed along its profile? Maybe \(a_{\psi}\) increases...chicken! Big chicken!
Or,
\(tan(\beta)=sech^2(0)=1\)
\(\cfrac{1}{2}tan(\beta)=0.5\)
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2*0.5}{m}}\)
\(f_{res}=0.11254\cfrac{G}{\sqrt{m}}\)
\(f_{res}=23.53 Hz\)
We are comparing \(\psi\) to the energy curve of a SHM system, \(U=\cfrac{1}{2}kx^2\) the gradient of which \(U^{'}=kx\).
Wednesday, July 20, 2016
Second Guessing Again
I was wrong. From the post "Pound To Rescue Permittivity" dated 30 May 2016,
\(\cfrac{q}{\varepsilon_o}=4\pi a^2_{\psi}m.2c^2ln(cosh(\pi))\)
where we generalized to
\(\cfrac{q}{\varepsilon_o}=4\pi a^2_{\psi}m.2c^2ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi}))\)
but for the case of a basic particle, from the post "New Discrepancies And Hollow Earth" dated 23 Jun 2016,
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,c}))=\cfrac{1}{4}\)
and a big particle,
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,\pi}))=1\)
The charge of a basic particle and a big particle differ by a factor of \(\cfrac{1}{4}\).
So, in the post "And They Danced" dated 18 Jul 2016,
\(\cfrac{1}{4}\cfrac{q^2}{4\pi \varepsilon_o r_e^2}=v.\cfrac{m_ev^2}{r_e}\)
the new factor \(\cfrac{1}{4}\) still cancels in the ratios of mass and orbital radii derived here,
\(\cfrac{r_p}{r_e}=\cfrac{m_ev_e^3}{m_pv_p^3}\)
But as we rearrange,
\(\cfrac{q^2}{4\pi \varepsilon_o r_e^2}=v.\cfrac{4m_ev^2}{r_e}\)
\(m_e\rightarrow 4m_e\)
because we have assumed that the electric charge on a electron in orbit is the same as the electric charge of a free electron. If the experimenter has also made the same assumption, this would introduce a factor of four to the quoted experiment value of the mass of electron. The quoted mass of an electron is,
\(m_e=\cfrac{1}{4}m_e^{'}\)
where \(m_e^{'}\) is the value for electron mass used here.
The published result of the ratio of electron mass to proton mass is,
\(\cfrac{m_e}{m_p}=\cfrac{1}{1836.152 673 89}\)
If we replace \(c\) with \(2\pi c\) as the electron is in circular motion, the expression
\(\cfrac{q^2}{4\pi \varepsilon_o r_e}=m_ec^3\)
from the post "And They Danced" dated 18 Jul 2016, becomes
\(\cfrac{q^2}{4\pi \varepsilon_o r_e}=m_e(2\pi c)^3\)
When we associated the extra factor with \(m_e\),
\(\cfrac{q^2}{4\pi \varepsilon_o r_e}=(2\pi)^3m_e.c^3\)
\(m_e\rightarrow (2\pi)^3m_e\)
The missing factor of \(2\pi\) accounts for circular motion. This factor cancels as we take the ratio, but if the experiment to obtain the mass of an electron is based on linear speed, the factor of \(2\pi\) for circular motion will reduce mass by a factor of \((2\pi)^3\) here.
\(m_e=(2\pi)^3m_e^{'}\)
Since the proton as the nucleus is in spin, and we consider its rotational kinetic energy with the use of moment of inertia, \(I\)
\(I=\cfrac{1}{2}\left(\cfrac{2}{5}m\right)r^2\omega^2=\cfrac{1}{2}\left(\cfrac{2}{5}m\right)v^2\)
\(m\rightarrow \cfrac{2}{5}m\)
we introduced the factor \(\cfrac{2}{5}\) to the mass of the proton, \(m_p\). The actual mass should be
\(m_p=\cfrac{5}{2}m_p^{'}\)
where \(m_p^{'}\) is the value for proton mass used here.
And if we accept the results from the post "And They Danced" dated 18 Jul 2016,
\(\cfrac{m_e f_e^3}{m_p f_p^3}=\cfrac{r_p^4}{r_e^4}\)
and,
\(\cfrac{m_e }{m_p }=\cfrac{1}{77}\)
with the following adjustments,
\(m_e\rightarrow \cfrac{1}{4}m_e\)
due to the assumption that the particles have the same charge magnitude. And
\(m_p\rightarrow \cfrac{5}{2}m_p\)
due to the use of mass instead of moment of inertia as the proton is in spin.
And also, for linear speed as opposite to circular motion,
\(m_e\rightarrow (2\pi)^3m_e\)
we have,
\(\cfrac{m_e}{m_p}\rightarrow \cfrac{\cfrac{1}{4}m_e (2\pi)^3}{\cfrac{5}{2}m_p }=\cfrac{1}{77}\)
where \(I_e\) and \(I_p\) are moment of inertia of the electron and proton respectively.
\(\cfrac{q}{\varepsilon_o}=4\pi a^2_{\psi}m.2c^2ln(cosh(\pi))\)
where we generalized to
\(\cfrac{q}{\varepsilon_o}=4\pi a^2_{\psi}m.2c^2ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi}))\)
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,c}))=\cfrac{1}{4}\)
and a big particle,
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,\pi}))=1\)
The charge of a basic particle and a big particle differ by a factor of \(\cfrac{1}{4}\).
So, in the post "And They Danced" dated 18 Jul 2016,
\(\cfrac{1}{4}\cfrac{q^2}{4\pi \varepsilon_o r_e^2}=v.\cfrac{m_ev^2}{r_e}\)
the new factor \(\cfrac{1}{4}\) still cancels in the ratios of mass and orbital radii derived here,
\(\cfrac{r_p}{r_e}=\cfrac{m_ev_e^3}{m_pv_p^3}\)
But as we rearrange,
\(\cfrac{q^2}{4\pi \varepsilon_o r_e^2}=v.\cfrac{4m_ev^2}{r_e}\)
\(m_e\rightarrow 4m_e\)
because we have assumed that the electric charge on a electron in orbit is the same as the electric charge of a free electron. If the experimenter has also made the same assumption, this would introduce a factor of four to the quoted experiment value of the mass of electron. The quoted mass of an electron is,
\(m_e=\cfrac{1}{4}m_e^{'}\)
where \(m_e^{'}\) is the value for electron mass used here.
The published result of the ratio of electron mass to proton mass is,
\(\cfrac{m_e}{m_p}=\cfrac{1}{1836.152 673 89}\)
If we replace \(c\) with \(2\pi c\) as the electron is in circular motion, the expression
\(\cfrac{q^2}{4\pi \varepsilon_o r_e}=m_ec^3\)
from the post "And They Danced" dated 18 Jul 2016, becomes
\(\cfrac{q^2}{4\pi \varepsilon_o r_e}=m_e(2\pi c)^3\)
When we associated the extra factor with \(m_e\),
\(\cfrac{q^2}{4\pi \varepsilon_o r_e}=(2\pi)^3m_e.c^3\)
\(m_e\rightarrow (2\pi)^3m_e\)
The missing factor of \(2\pi\) accounts for circular motion. This factor cancels as we take the ratio, but if the experiment to obtain the mass of an electron is based on linear speed, the factor of \(2\pi\) for circular motion will reduce mass by a factor of \((2\pi)^3\) here.
\(m_e=(2\pi)^3m_e^{'}\)
Since the proton as the nucleus is in spin, and we consider its rotational kinetic energy with the use of moment of inertia, \(I\)
\(I=\cfrac{1}{2}\left(\cfrac{2}{5}m\right)r^2\omega^2=\cfrac{1}{2}\left(\cfrac{2}{5}m\right)v^2\)
\(m\rightarrow \cfrac{2}{5}m\)
we introduced the factor \(\cfrac{2}{5}\) to the mass of the proton, \(m_p\). The actual mass should be
\(m_p=\cfrac{5}{2}m_p^{'}\)
where \(m_p^{'}\) is the value for proton mass used here.
\(\cfrac{m_e f_e^3}{m_p f_p^3}=\cfrac{r_p^4}{r_e^4}\)
and,
\(\cfrac{m_e }{m_p }=\cfrac{1}{77}\)
with the following adjustments,
\(m_e\rightarrow \cfrac{1}{4}m_e\)
due to the assumption that the particles have the same charge magnitude. And
\(m_p\rightarrow \cfrac{5}{2}m_p\)
due to the use of mass instead of moment of inertia as the proton is in spin.
And also, for linear speed as opposite to circular motion,
\(m_e\rightarrow (2\pi)^3m_e\)
\(\cfrac{m_e}{m_p}\rightarrow \cfrac{\cfrac{1}{4}m_e (2\pi)^3}{\cfrac{5}{2}m_p }=\cfrac{1}{77}\)
\(\cfrac{m_e }{m_p }=\cfrac{1}{77*(2\pi)^3*\cfrac{2}{5}*\cfrac{1}{4}}\)
or
\(\cfrac{m_p }{m_e }={77*(2\pi)^3*\cfrac{2}{5}*\cfrac{1}{4}}=1909.98\)
and if we have,
\(\cfrac{m_e }{m_p }=\cfrac{1}{74}\)
\(\cfrac{m_p }{m_e }={74*(2\pi)^3*\cfrac{2}{5}*\cfrac{1}{4}}=1835.57\)
and if,
\(\cfrac{m_e }{m_p }=\cfrac{1}{75}\)
\(\cfrac{m_p }{m_e }={75*(2\pi)^3*\cfrac{2}{5}*\cfrac{1}{4}}=1860.37\)
and if,
\(\cfrac{m_e }{m_p }=\cfrac{1}{76}\)
\(\cfrac{m_p }{m_e }={76*(2\pi)^3*\cfrac{2}{5}*\cfrac{1}{4}}=1885.18\)
Which seems to adjust for the experimental value of the ratio of proton to electron mass, but the introduction of moment of inertia adds complications to the original expression,
\(\cfrac{m_e f_e^3}{m_p f_p^3}=\cfrac{r_p^4}{r_e^4}\rightarrow\cfrac{I_e f_e^3}{I_p f_p^3}=\cfrac{r_p^4}{r_e^4} \)
where \(I_e\) and \(I_p\) are moment of inertia of the electron and proton respectively.
Lunch time!
Tuesday, July 19, 2016
No Show Still
\(\cfrac{1}{75}\), \(\cfrac{1}{76}\), \(\cfrac{1}{74}\)
and it is unofficial data from rare noble gases, which does not explain why \(m_e\lt\lt m_p\).
Maybe lunch will help.
and it is unofficial data from rare noble gases, which does not explain why \(m_e\lt\lt m_p\).
Maybe lunch will help.
Monday, July 18, 2016
\(r_p\) Not Published
The good thing about,
\(\cfrac{m_e f_e^3}{m_p f_p^3}=\cfrac{r_p^4}{r_e^4}\)
from the previous post "And They Danced" dated 18 Jul 2016, is that the presence of \(\cfrac{f_e^3}{f_p^3}\) allows for the possibility of,
\(\cfrac{1}{77}\lt\cfrac{m_e}{m_p}\le1\)
where an electron in orbit around a nucleus is a basic particle or a group of basic particles of number \(n\lt77\). Such particles coalesce into big particles \(n=77\), on being ejected from the nucleus. If this is the case, an electron beam from cathode ray tube will initially convergence and would need time to coalesce before it has momentum to turn a wind wheel or paddle wheel.
But \(r_p\), the radius of the hydrogen nucleus orbit/spin is not available for a hydrogen nucleus. The common notion is that the hydrogen nucleus is stationary without spin and the electron in orbit around it.
No spin, No show.
\(\cfrac{m_e f_e^3}{m_p f_p^3}=\cfrac{r_p^4}{r_e^4}\)
from the previous post "And They Danced" dated 18 Jul 2016, is that the presence of \(\cfrac{f_e^3}{f_p^3}\) allows for the possibility of,
\(\cfrac{1}{77}\lt\cfrac{m_e}{m_p}\le1\)
where an electron in orbit around a nucleus is a basic particle or a group of basic particles of number \(n\lt77\). Such particles coalesce into big particles \(n=77\), on being ejected from the nucleus. If this is the case, an electron beam from cathode ray tube will initially convergence and would need time to coalesce before it has momentum to turn a wind wheel or paddle wheel.
But \(r_p\), the radius of the hydrogen nucleus orbit/spin is not available for a hydrogen nucleus. The common notion is that the hydrogen nucleus is stationary without spin and the electron in orbit around it.
No spin, No show.
Their Whistles Make The Difference
Continuing from the previous post "And They Dance" dated 18 Jul 2016.
The point was,
\(r_p\lt r_e\)
then,
\(\cfrac{m_e f_e^3}{m_p f_p^3}=\cfrac{r_p^4}{r_e^4}\lt\cfrac{r_p}{r_e}\lt 1\)
and if we attribute the factor \(\cfrac{ f_e^3}{f_p^3}\) solely to \(m_e\),
\(\cfrac{m_e.\cfrac{f_e^3}{f_p^3}}{m_p}\lt 1\)
as, \(f_p\gt0\) and \(f_e\gt0\)
\(\cfrac{m_e}{m_p}\lt \cfrac{f_p^3}{f_e^3}\)
This might be the reason why, \(m_e\) is made small (relatively, \(m_p\) made big) as \(f_e\gt f_p\) with the electron at light speed. Whereas, if \(m_e=m_p\) everything collapses to 1 and all expressions are satisfied.
This does not imply \(m_e=m_p\). \(r_e\) and \(r_p\) are the orbital radii of the electron and hydrogen nucleus respectively.
And so they dance...
The point was,
\(r_p\lt r_e\)
then,
\(\cfrac{m_e f_e^3}{m_p f_p^3}=\cfrac{r_p^4}{r_e^4}\lt\cfrac{r_p}{r_e}\lt 1\)
and if we attribute the factor \(\cfrac{ f_e^3}{f_p^3}\) solely to \(m_e\),
\(\cfrac{m_e.\cfrac{f_e^3}{f_p^3}}{m_p}\lt 1\)
as, \(f_p\gt0\) and \(f_e\gt0\)
\(\cfrac{m_e}{m_p}\lt \cfrac{f_p^3}{f_e^3}\)
This might be the reason why, \(m_e\) is made small (relatively, \(m_p\) made big) as \(f_e\gt f_p\) with the electron at light speed. Whereas, if \(m_e=m_p\) everything collapses to 1 and all expressions are satisfied.
This does not imply \(m_e=m_p\). \(r_e\) and \(r_p\) are the orbital radii of the electron and hydrogen nucleus respectively.
And so they dance...
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),
\(\cfrac{q^2}{4\pi \varepsilon_o r_e}=m_ec^3\)
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...
\(\cfrac{d\,KE}{d\,t}=\cfrac{1}{2}m\cfrac{dv^2}{d\,t}=mv\cfrac{d\,v}{d\,t}=v.F\)
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.
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...
Sunday, July 17, 2016
Thursday, July 14, 2016
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.
\(x=-\cfrac{|F_{drag}|}{n.m_b\omega^2}sin(\cfrac{\pi}{2}-s)*cos(s)\)
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.
\(c=\sqrt { \cfrac { S_{ n }(32\pi ^{ 4 }-1) }{ 64\pi .\theta _{ \psi }ln(cosh(\theta _{ \psi }))tanh(\theta _{ \psi }) } *\left( \cfrac { \theta _{ \psi } }{ 0.7369 } \right) ^{ 3 } } \)
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.
Wednesday, July 13, 2016
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.
\(y_g=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)\)
\(\because\) \(\delta m=n.m_b\),
\(y_g=\cfrac{|F_{drag}|}{n.m_b\omega^2}sin(\theta)\)
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.
\(x=1-sin(\cfrac{\pi}{2}-s)*cos(s)\)
\(y=sin(\cfrac{\pi}{2}-s)\)
\(0\le s\le\pi/2\)
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,
\(\cfrac{F_{drag}sin(\theta)}{r}=\cfrac{F_{drag}cos(\theta)}{x_t}\)
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.
Tuesday, July 12, 2016
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.
\(\omega^2=\cfrac{F_{drag}}{\delta m}\cfrac{sin(\theta)}{r}\)
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.
\(\omega^2=A.{F_{drag}}\)
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.
\(F_{drag}=F^{*}_D+F^{*}_c\)
\(F^{*}_D\) 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?
\(|F^{*}_{c}|=|F_{drag}|sin(\theta)=\delta m.r.\omega^2\)
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.
\(\cfrac{d\,KE}{d\,t}=constant\)
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.
Monday, July 11, 2016
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.
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.
\(c_{measured}=c_{ave}=\cfrac{2}{\pi}c_{max}+c_o\)
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.
Sunday, July 10, 2016
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,
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
\(\cfrac { d^{ n } }{ dt^{ n } } \left\{ t^{ n -1}f(t) \right\} \overset { Fourier }{ \longrightarrow } (-1)^{(n-1)}(i2\pi f)\cfrac { d^{ n-1 } }{ df^{ n-1 } } \left\{ F(f) \right\}\)
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,
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)
\(\cfrac{d}{df}\left(F(f)\star F(f)\right)=F(f)\star F(f)\)
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.
Saturday, July 9, 2016
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.
\(B=-i\cfrac{\partial\,E}{\partial\,x}\)
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.
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.
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.
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 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.
Friday, July 8, 2016
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.
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
\(F_{ f }[f(x)]=F(f)=\int_{-\infty}^{\infty} { f(x).e^{ -i2\pi fx } } dx\)
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, \(\cfrac{1}{4}+\cfrac{1}{4}\)
\(G=\cfrac { \sqrt { 2c } }{ a_{ \psi } } \theta _{ \psi }\)
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, \(\cfrac{1}{4}+\cfrac{1}{4}\)