Screaming For Experiments

Intuitively the first assignment was correct.

But for comparison sake, and other good wine,

we will kept the second version in mind.  Maybe,

In this case earth is a \(g^{-}\) particle in the new sign assignment.

Whether a particle exert an attractive or repulsive force depends on the boundary conditions applied to \(F\), the force density that results from a change in \(\psi\), \(\cfrac{\partial\psi}{\partial\,x}\) around the particle.   A negative \(F\) implies \(\psi\) extends further from the center and the mass of the particle is smaller.

What to do?

Find the particle pairs, find the smaller mass in the pair, spins/rotates, collide and sent it down a coil conductor and determine the type(s) of field/energy that manifests.  The less massive particle has a negative \(F\) and is assigned as a negative particle.

Collisions Vs Circular Motion

What if the energy extracted from the particle during collision is along the dimension at which the wave is at light speed, but the energy that manifests when the particle is in circular motion or spin by itself, is the oscillatory energy in an orthogonal dimension?

The oscillatory energy and \(KE\) at light speed of the wave are presented under different circumstances; circular motion vs collisions respectively.

What if in both cases, collisions and circular motion, the energy that present itself is always \(KE\) along the light speed dimension of the wave?  Oscillatory energy is not extracted at all.

The Origin Of Earth's Magnetic Field

Is earth a \(g^{+}\) or \(g^{-}\) particle?

Earth is a spinning \(g^{+}\) particle with a magnetic North pole.  A spinning \(g^{-}\) particle will generate an electric field pointing upwards, instead.  These are based on the new particle assignment in the post "Sunshine and Proton Beam Corrected" dated 01 Jul 2015.

If the field generated when the particle is in circular motion, is due to the oscillating energy component of the wave, then the first sign assignment was right.  Earth is then a spinning \(g^{-}\) particle.

In both cases, the spinning gravity particle generates a magnetic North pole.

Since Earth has negative gravitation potential energy, the origin sign assignment is more efficacious.  But our definition of a negative potential well, that zero gravitational potential is at a point infinitely far is arbitrary, and without sight of an opposite gravity particle.

\(\psi\) is pulsating, a field as the result of aligning \(\psi\) will also be oscillating.  Is earth magnetic field pulsing at \(7.489 Hz\)?  A field as the result of aligning the light speed dimension of the wave will not be oscillating.  For a point particle, \(\psi\) will be oscillating at high frequency (\(\sim PHz\)).

What is Schumann Resonance?

Sunshine and Proton Beam Corrected

This post corrects the post "Sunshine and Proton Beam" dated 23 May 2015.

Both these particles impart electric potential energy,

This electric potential is in \(\psi\) around the particles that exist in the \(t_c\) time dimension.  Which is an electron and which is a proton?

A bold guess is that electron that also carries heat and has low mass is the right particle and proton is the left particle, where \(t_g\) manifest itself perpendicular to two parallel, current carrying wires.

What's is the difference between existing in the \(t_c\) time dimension, as in the case of the electron, and being a wave along \(t_c\), as like a positive gravity particle, \(g^{+}\)?

They are not equivalent.

Energy in the wave can be dissipated and reduced, the particle still exist.  Energy that marks the particle existence (\(E=mc^2\)) cannot be dissipated, the particle disappears from existence if that happens.

The space dimensions are curled along the time dimensions \(t_c\).  When we align the two waves such that the time dimensions, \(t_g\) and \(t_T\) are parallel, we find that \(x\) for the case of electron is going down \(-x\) with respect to the proton's \(+x\) axis.  This is consistent with the fact that the charges have opposite force fields.

If this model is true, then electron give heat but not gravitational potential energy and proton gives gravitation potential energy but not heat; in addition to the electrical fields around them.  Energy along \(t_g\) or \(t_T\), that is the kinetic energy of the wave, is extracted by collisions.

We have a probelm!!!

If heat is lost as the result of a change in momentum along the time axis at which the wave is at light speed, ie \(t_T,\,\,v=c\), then we have made a mistake in particle sign assignment.  The focus is not on the dimension with oscillatory energy but the the dimension along which the wave is at light speed.

The new particle assignment is then,

If this is true then in circular motion, the energy component that manifest itself is also in the dimension of the wave at light speed.

What happens to the oscillatory energy component?  \(\psi\) defines the particle.  The existence of \(\psi\) makes it both a particle and a wave.  As long as the particle exist, \(\psi\) remains.

I love it!

But They Are All Here!?

Both electron and proton exist in the \(t_c\) time dimension and interact as charges.

Both negative and positive gravity particles exist in the \(t_g\)  time dimension and interact as gravity particles.

A negative gravity particle does not interact with a proton, simply because they exist in different realms.

But they are all here!?

Because of this non-interaction, we conclude that the forces are orthogonal and \(t_c\) is orthogonal to \(t_g\).  This give raise to the graphical representation of particle as the vertex of a cube.

\(F\) and \(\psi\) was derived independent of such a representation.

When the time dimensions are aligned, \(F\) is also rotated.  The direction of the space dimension gives the direction of positive \(F\).  We find that a negative particle exert a force that is opposite to a positive particle. There is no need to make \(F\) negative again.

Why is the mass of an electron smaller than a proton?  It might not be, that all negative particles have smaller mass compared to their positive counterpart.

They are still all here.

When the time dimensions are aligned, \(F\) is also rotated.  The direction of the space dimension gives the direction of positive \(F\).  \(F\) pointing in the opposite direction indicates a negative \(F\).  For \(F\) to be negative, \(\psi\) extends further into space resulting in a smaller particle mass.  For a negative particle, we define it to be exerting a negative force.  All negative particles will then have smaller particle mass compared to the corresponding positive particle.

The latter reasoning can be refuted by showing that not all negative particles have relatively smaller mass.  However, all negative particles having relatively smaller mass compared to their positive counterpart does not prove that the latter reasoning is right.

\(F\) has two constants of integration that are adjusted for boundary conditions.  If \(F(0)\lt 0\) then \(F\) shifts to the right resulting in greater \(\psi\) and so less particle mass.  A rotated \(F\) as the time dimensions align provides the boundary condition that \(F(0)\lt 0\).

???!

Right Answer Wrong Question

There is a problem, if the sign of the force exerted by the particle (in its force field) is accounted for by aligning the time dimensions...

The case for electron and proton is illustrated above.  When the time dimensions are aligned, the space dimensions \(x\), point in opposite directions.   Electrons exert a force opposite to that of protons.  They attract each other.

Then making the force density, \(F\), negative around an electron by extending \(\psi\) further beyond the origin (center of the particle), is double counting.  There is no need to make \(F\) negative as aligning the time dimensions accounts for the sign of \(F\) already.

If this is the case, what would account for the difference in mass of opposite particles?

An electron has smaller mass, its \(\psi\) will extend further beyond the origin compared to a proton, but why?

If we define a negative particle as one with a higher initial negative \(F\) as the result of \(\psi\) extending further in space. Then it follows that a negative particle has less mass and would always be in orbit around a more massive positive particle that spins almost in place when the two are bounded.

Or, do we argue that, by aligning the time dimensions, we find that \(F\) for a negative particle is negative and so indicates that \(\psi\) of this particle must extend further from the origin, as such resulting in a smaller particle mass?

What The Weak Fields?

In the posts "We Still Have A Problem" and "Less Mass But No Theoretical Mass" dated 23 Nov 2014, from the solutions for \(F\) and \(\psi\), we wrote down an expression for mass density of a particle with reference to the total energy released when the particle is annihilated.

\(m_{\rho\,particle} c^2=m_\rho c^2-\int^{x_a}_{0}{\psi}dx\)

This expression shows that the KE of the particle along space and real time \(t\),

\(m_{\rho\,particle} c^2\),

added to the total energy in \(\psi\),

\(m_{\rho\,particle} c^2+\int^{x_a}_{0}{\psi}dx\)

also manifested in space and time \(t\), is equal to

\(m_{\rho\,particle} c^2+\int^{x_a}_{0}{\psi}dx=m_\rho c^2\)

the KE of the particle along time \(t_c\), \(t_g\) or \(t_T\).  This is just a restatement of the conservation of energy across dimensions which we assume to be true.

The greater the extend of \(\psi\) in space, the smaller its mass in space and real time \(t\) is.

Also based on the solution for \(F\) and \(\psi\), a negative force \(F\) requires that \(\psi\) extends further from the origin.  In this way, a particle with an attractive force field around it must have lower mass.  We concluded that,

\(m_{\rho\,e}\lt m_{\rho\,po}\)

an electron has less mass than a positive charge.   And in general, a negative particle has less mass than a positive particle.  So, when a negative particle in bounded by a positive particle at terminal velocity \(c\) (space is a thin fluid with viscosity), the negative particle will transcribe a bigger circular motion than the more massive positive particle.  In the case when the positive particle is much more massive, the negative particle orbits around the positive particle.  The positive particle is in a slow spin.

Since both particles are spinning, both oscillatory components in the negative and positive particle will manifest themselves.  In the case of temperature particles, \((T^{-},\,T^{+})\), they are an electric field and a gravitational field.  In the case of gravity particles, \((g^{-},\,g^{+})\), they are an electric field and a magnetic field \((B=T)\).  In the case of charges, \((e^{-},\,p^{+})\), they are a gravitational field and a magnetic field.

Although this energy oscillating along an orthogonal time dimension is aligned, the axes of the spin and rotation of the particles are not stationary, as a result, these fields generated are "weak" fields.

It is like a particle along a helical path generating a field given by the right hand screw rule, except this coiled path is rotating randomly.  The generated field around such motion is oscillatory and varying, and so a "weak" field.

The generated fields associated with the more massive spinning positive particles are more stable (more stable axis of rotation) and are comparatively stronger although still oscillatory.

Are these the strong and weak fields scientists talk about in particle physics?

Two Too Many

\(p^{+}\), \((g^{-},\,g{+})\), \((T^{-},\,T^{+})\) three types of protons!

Both \(g^{-}\) and \(T^{+}\) have energy oscillating in the \(t_c\) time dimension.  When the particle is in circular motion or spin, this component aligns in one direction and its characteristic as a positive charge manifest itself.

It is like a charge travelling down a coil displaying magnetic characteristics.

Both \(g^{-}\) and \(T^{+}\) go into circular motion in the presence of the respective opposite particle.  \((g^{-},\,g{+})\), \((T^{-},\,T^{+})\) then have characteristic of a positive charge.

The aligned electric field attracts an electron and set it in orbit around the nucleus.  All three particles are in coupled motion.

And that is how we have two new protons.

Sign Assignments Signing Off

In summary, the following are the particles with their sign assignments,

Have a nice day.

Still How Much Further Can Gravity Particles Go?

Let's look at radioactive decays where there are transitions between states of the same nucleus from a gravity particle perspective.  Much of this is taken from the page https://en.wikipedia.org/wiki/Radioactive_decay.

1.   Isomeric transition:

Excited nucleus releases a high-energy photon (gamma ray)  (A, Z)

Excited basic particles in the nucleus, \(p^{+}\), \(g^{-}\), \(T^{+}\) slow down, releasing gamma rays.

2.   Internal conversion:

Excited nucleus transfers energy to an orbital electron, which is subsequently ejected from the atom (A, Z)

An excited \(g^{+}\) particle breaks away from the proton pair \((g^{-},\,g^{+})\), temporarily.  The proton loses its charge and converts to a neutron.  A electron in orbit is freed.  After which the \(g^{+}\) particle is recaptured and the neutron reverts to the proton pair \((g^{-},\,g^{+})\).

At this point, every explanations are just concoctions that serve more to entertain than science.  None of this is vindicated.  Then again, no close enough observation is possible at the subatomic level.

Much of the existing explanation is worse speculation.

How Much Further Still Can Gravity Particles Go?

Let's look at different modes of beta decay from a gravity particles perspective.  Much of this is taken from the page https://en.wikipedia.org/wiki/Radioactive_decay.

1.   β− decay:

A nucleus emits an electron and an electron anti-neutrino (A, Z + 1)

Occurs in the nucleus, at a hydrogen particle, \((e^{-},\,g^{-},\,g^{+})\).  \(g^{+}\) is first ejected as the electron anti-neutrino, the particle loses its positive charge, the electron is freed and the photon that ejects \(g^{+}\) is captured. The proton pair \((g^{-},\,g^{+})\) forms up resulting in an increment to Z.

2.   Positron emission (β+ decay):

A nucleus emits a positron and an electron neutrino (A, Z − 1)

Occurs in the nucleus, at a hydrogen particle, \((e^{-},\,g^{-},\,g^{+})\).  \(g^{-}\) and \(g^{+}\) are both ejected.  \(g^{-}\) with its small oscillatory energy along \(t_c\) is detected as a positive electric charge.  \(g^{+}\) is also the electron neutrino.

3.   Electron capture:

A nucleus captures an orbiting electron and emits a neutrino; the daughter nucleus is left in an excited unstable state (A, Z − 1)

The nucleus captures a \(P_{g^+}\) photon of high energy, the photon slows emitting gamma rays in the process and excites an existing \(g^{+}\) particle.  The photon becomes a fast moving \(g^{+}\) particle.  Together with a \(g^{-}\), they form an extra proton pair \((g^{-},\,g^{+})\) that pull an electron from a low lying orbit.  However, this proton pair is unstable and break apart as the passing particle/photon moves on.  The excited \(g^{+}\) particle having gained enough energy as the photon approached, is emitted as a neutrino, as the photon passes.  The atomic number is reduced by one.

4.   Bound state beta decay:

A free neutron or nucleus beta decays to electron and anti-neutrino, but the electron is not emitted, as it is captured into an empty K-shell; the daughter nucleus is left in an excited and unstable state. This process is a minority of free neutron decays (0.0004%) due to the low energy of hydrogen ionization, and is suppressed except in ionized atoms that have K-shell vacancies. (A, Z + 1)

As with β− decay.  Subsequent fate of the electron does not involve any basic particle.

5.   Double beta decay:

A nucleus emits two electrons and two anti-neutrinos (A, Z + 2)

Two times β− decay.  Please refer to point 1.

6.   Double electron capture:

A nucleus absorbs two orbital electrons and emits two neutrinos – the daughter nucleus is left in an excited and unstable state (A, Z − 2)

Two times electron capture.  Please refer to point 3.

7.   Electron capture with positron emission:

A nucleus absorbs one orbital electron, emits one positron and two neutrinos (A, Z − 2)

As with electron capture in point 3, with the additional emission of a \((g^{-},\,g^{+})\) pair from the nucleus.  This could happen when the approaching photon is of high energy.

8.   Double positron emission:

A nucleus emits two positrons and two neutrinos (A, Z − 2)

Two times β+ decay.  Please refer to point 2.

Neutrinos and anti-neutrinos are the same \(g^{+}\) particles.

How Far Can Gravity Particles Go?

Let's look at radioactive decays with emission of nucleons from a gravity particles perspective.  Much of this is taken from the page https://en.wikipedia.org/wiki/Radioactive_decay.

1.   Alpha decay:

An alpha particle (A = 4, Z = 2) emitted from nucleus (A − 4, Z − 2)

A \(g^{-}\) particle is ejected from the nucleus by a \(P_{g^-}\) photon, which weakens the gravitational hold on the alpha particle.  The alpha particle is emitted and the photon recaptured as a \(g^{-}\) particle.

2.   Proton emission:

A proton ejected from nucleus (A − 1, Z − 1)

A \(p^{+}\), \((g_{-},\,g^{+})\) or \((T^{-},\,T^{+})\) group is ejected from the nucleus.

3.   Neutron emission:

A neutron ejected from nucleus (A − 1, Z)

A \(g^{-}\) or \(T^{+}\) particle is ejected but the ejecting photon is not recaptured.

4.   Double proton emission: 

Two protons ejected from nucleus simultaneously (A − 2, Z − 2)

Two \(p^{+}\), \((g_{-},\,g^{+})\) or \((T^{-},\,T^{+})\) groups are ejected from the nucleus.

5.   Spontaneous fission:

Nucleus disintegrates into two or more smaller nuclei and other particles

\(g^{-}\) particle(s) being ejected results in a weaken negative gravity potential at the nucleus.

6.   Cluster decay:

Nucleus emits a specific type of smaller nucleus (A1, Z1) which is larger than an alpha particle.  (A − A1, Z − Z1) + (A1, Z1)

\(g^{-}\) particle(s) being ejected results in a weaken negative gravity potential at the nucleus.

A photon recapture is key, when the nucleus does not change mass number and with the emission of gamma rays in this mode of radioactive decay.

Period Table Dux

What if the periodic table is wrong?

We have assumed that all elements are made up of electrons, protons and neutrons.

What if all particles, charge, temperature and gravity particles, are involved in building up the elements?

And there is a systematic way to go about it.

We will start with the temperature particle pair, \((T^{-},\,T^{+})\); a hydrogen nucleus of some sort.  There is then \((g^{-},\,g^{+})\), also some sort of hydrogen nucleus and \((e^{-},\,p^{+})\), the hydrogen.

There are three types of hydrogen, protium, deuterium and tritium.  Besides tabled parameters such as Nuclear Spin, Magnetic Moments, Nuclear Quadrupole Moment, radioactiveness, and mass, high thermal conductivity of natural hydrogen (protium) and helium are noted.  It is also noted that only deuterium has a non zero Nuclear Quadrupole Moment and a Nuclear Spin of one.

Even as this blog is filled with bold speculations this warning is due; the rest of this post is FICTION!

Strictly speaking, only \(p^{+}\) contribute to the atomic number.   But both \(g^{-}\) and \(T^{+}\) have energy oscillating along \(t_c\), in the presence of a respective opposite particle, the spin of the particle is restrained and the vector along \(t_c\) points predominantly in one direction.  So, the pairs, \((g^{-},\,g^{+})\) and \((T^{-},\,T^{+})\) acquire a positive electric potential and behave like a positive charge, the proton, but oscillating.

So,  \(p^{+}\), \((g^{-},\,g^{+})\) and \((T^{-},\,T^{+})\) all contribute to the atomic number.

All particles contribute to the non integer mass number, ie they all contribute mass.  \(p^{+}\), \(g^{-}\) and \(T^{+}\) contribute to the integer mass number; these particles has positive electric potentials.  (From the post "Less Mass But No Theoretical Mass" date 23 Nov 2014.  Positive particle have more mass.)

A half spin is due to the oscillating energy along \(t_c\) of the wave/particle from which even mass number has integer spin and odd mass number has half spin.

In view of their high thermal conductivity as gases, the normal hydrogen, protium is \((e^{-},\,T^{-},\,T^{+})\) besides \((e^{-},\,p^{+})\).  The particle \((e^{-},\,g^{-},\,g^{+})\) will be radioactive.

Helium is

\((2e^{-},\,p^{+},T^{-},\,T^{+},\,g^{-},\,g^{-})\) or

\((2e^{-},\,2T^{-},\,2T^{+},\,g^{-},\,g^{-})\) or

\((2e^{-},\,2p^{+},\,g^{-},\,g^{-})\) .

The first is less likely because, then it is possible for \(He^{+}\) to be

\((e^{-},\,p^{+},T^{-},\,T^{+},\,g^{-},\,g^{-})\)

whereas in reality \(He\) is very stable, a noble gas.

\((2e^{-},\,2p^{+},\,g^{-},\,g^{-})\) will not be very conductive of heat, although

\((2e^{-},\,p^{+},T^{-},\,T^{+},\,T^{+},\,g^{-})\)

is possible even when the negative gravitational potential that holds the nucleus together is small.

Deuterium is \((e^{-},\,p^{+},\,T^{+})\) for integer spin.  But \((e^{-},\,g^{-},\,g^{+},\,T^{+})\) will be radioactive. And \((e^{-},\,T^{-},\,T^{+},\,T^{+})^{+}\) heat conductive.

Radioactive tritium is \((e^{-},\,(g^{-},\,g^{+}),\,(e^{-},\,g^{-},\,g^{+}),\,g^{-})\)  where the pair \((g^{-},\,g^{+})\) serves as a proton and the three \(g^{-}\)s contribute to the mass number.

The particle within tritium, \((e^{-},\,g^{-},\,g^{+})\), a hydrogen, behave like a neutron in the nucleus.  It is radioactive and can decay by ejecting the pair \(e^{-}\) and \(g^{+}\) and retaining \(g^{-}\) or, by retaining \(e^{-}\) and ejecting both \(g^{-}\) and \(g^{+}\).

\(n\to e^{-}+\,g^{-}+\,g^{+}\)

\(e^{-}\) is an electron, \(g^{+}\) is a small packet of gravity energy with oscillating energy in \(t_T\).   \(g^{+}\) could be the electron anti-neutrino.   In the nucleus, \((g^{-},\,g^{+})\) a proton is converted to \(g^{-}\), a neutron.  And \(g^{-}\) is a small packet of gravity energy with oscillating energy in \(t_c\) that could be detected as a small positive charge, a positron.  In this case, \(g^{+}\) is also the electron neutrino.

This hydrogen \((e^{-},\,g^{-},\,g^{+})\), residing in the nucleus, could be the root of beta decay.

Tritium decays by ejecting an electron and a \(g^{+}\) from \((e^{-},\,g^{-},\,g^{+})\) within its nucleus .  This could happen via a collision with a \(P_{g^{+}}\) photon that ejects the \(g^{+}\) particle first.  The particle loses its positive charge, and the electron is also ejected.

The resulting atom \((e^{-},\,(g^{-},\,g^{+}),\,g^{-},\,g^{-})\), is highly unstable, \(^3H\).  It acquires a \(g^{+}\) again to become \(^3_2He^{1+}\), \((e^{-},\,(g^{-},\,g^{+}),\,(g^{-},\,g^{+}),\,g^{-})^{1+}\), where the two \((g^{-},\,g^{+})\) behave like protons.  The three particles \(g^{-}\) in the makeup of this atom make its atomic mass 3.

Why not lose an electron directly?  Such an electron lost outside of the nucleus will be ionization.  Radioactivity occurs in the nucleus through \(g^{-}\) and \(g^{+}\) gravity particles.

In the posts "Hic, Hic, Hic, I Cna Fly..." and "Guess What? Radioactivity Defined" both dated 26 Jun 2015, radioactivity is attributed to gravity particles.  In particular, the hydrogen, \((e^{-},\,g^{-},\,g^{+})\) can reside within the nucleus and undergo beta decay.  Furthermore, when a nucleus has a \(g^{-}\) particle, the gain of a \(g^{+}\) particle converts the neutron (\(g^{-}\) by itself) to a proton, (\(g^{-},\,g^{+}\)).  The loss of a \(g^{+}\) particle from the pair (\(g^{-},\,g^{+}\)) converts the proton to a neutron.

The loss of  \(g^{-}\) particles from a nucleus weakens the negative gravitational potential that holds the nucleus together, as a result an alpha particle, \(^4_2He\) may break away from the nucleus.

There is no gamma rays being emitted in this model.  Both electron neutrino and electron anti-neutrino are the same \(g^{+}\) particle.  The first issue can be dealt with in the following way.  If the photon that ejected the positive gravity particle \(g^{+}\), is slowed down and subsequently captured in the later stage as a \(g^{+}\) particle.  This implies that the photon that ejected the particle in the first place, is just another particle of the same nature, accelerated to light speed.   Then the energy emitted as the photon slows can account for the gamma rays emitted.  At the same time this scenario provides for a mechanism to recapture \(g^{+}\).

This post does not have the coverage to claim consistency.   We need to look at all types of radioactive decays to have greater confidence in gravity particles being responsible for radioactivity.

It is Fourier Afterall

We know that the two oscillatory dimension components in a particle wave are complementary, given a constant total energy.  Only one of the two dimensions is needed to fully describe the oscillation.

So we need only two dimensions to describe the wave.  In addition, we can swap the dimension in which the wave is at light speed with the oscillating dimension and the wave equation remains the same.  We will differentiate between space and time dimensions, and we can reduce our descriptions to three unique orthogonal pairs,

\((t,x)\), \((t,t)\) and \((x,x)\).  All three are sinusoidal with speed \(c\) along an orthogonal dimension.

And thanks to Fourier Transforms, the sum of any combinations, of any numbers of these three can be described analytically using integrable functions.

This is provided we describe waves as,

\(\cfrac{\partial^2\psi}{\partial\,t^2}=c^2\cfrac{\partial^2\psi}{\partial\,x^2}\) --- (*)

It is not wavelets but good old Fourier, that fully describe physics.

However, as we have noted previously, as we swap a time dimension for a space dimension, the wave becomes a standing wave wrap around a center; a spherical wave.  A wave wrapped into a particle.  This information is lost when we think space and time dimensions are interchangeable as we develop the equivalent view above.    When a space dimension is swapped for a time dimension, the wave presents itself differently in space.  And since each time dimensions has characteristic energy, electric, temperature or gravity, all time dimensions must be differentiated to retained their characteristics.

As such the wave equation (*) by itself is inadequate.  The description of a particle is supplemented with information of which dimension carries oscillating energy, which time dimension the particle exist in and, which dimension it is at light speed.

Have a nice day.

I Don't Know

What are we actually measuring when we measure temperature?

When using a thermometer...

possibly an equilibrium of all particles evenly distributed between the target of measurement and the thermometer.  All these particles contain KE in a 2D plane.  When confined to a long tube, all such particles contributes to the extend of the remaining free dimension.  More vibration energy results in a greater length along the tube.

Using thermal Imaging...

temperature photons where the time dimension along which the wave that makes up the temperature particles have light speed, are replaced with a space dimension, \(x_2\).  These photons are enable to eject temperature particles analogous to photoelectric effect.  With reference to thermal imaging, these photons are in the infra red spectrum.  But how do we know in the first place, we are capturing temperature related information consistently?

Using our hands...

oscillating energy along \(t_c\) of all particles affects chemical changes.   As long as the particles are enable to affect biochemical changes that we register as heat, we will detect them as hot.

More Temperature Particles Speculations

Hot air raises, because of collisions with negative temperature particles.  The oscillatory energy in \(t_g\) imparts gravitational potential.  Hot air raises not because it is less dense.

Hot air is also positively charge, because of collisions with positive temperature particles.  The oscillatory energy in \(t_c\) imparts an electric potential.  It is through this electric potential that positive temperature particles in collisions with valence electrons increase the rate of chemical reactions.

The positive charge surrounding a flame is not entirely due to positive ions, otherwise a flame in an \(E\) field will widen equally to both sides of the parallel plates extending the \(E\) field.  In experimental observations, the tip of the flame bends towards the negative plate.  Furthermore, a positive electric potential emanates from a flame.  Combustion by itself should generate equal positive and negative charges, the flame as a whole should be charge neutral.

Our common experience of heat should be both/either temperature particles.  Heat flow is diffusion from higher particle concentration to lower particle concentration.  It is possible that a positive temperature particles pairs up with a negative temperature particle in a solid conductor, like the hydrogen atom.  This paired entity should be more stable and so less "heat like".  When the temperature particles are separated in a solid, they form the opposite poles of a magnet.

In a flame, temperature particles remain separated without pairing.  Under an \(E\) field, the pronounced separation magnify its magnetic property.  Oscillations in the flame observed when it is subjected to an \(E\) field is due to this magnified magnetic property and ambient EMW.

All charged up and floating...magnetic too.

What's The Smug?

A heavy, dense mass will be a negative gravity particle conductor.

A radioactive mass that decays from a neutron to a proton (loss mass) can be a positive gravity particle conductor.

Just as you might think my insanity is not complete...

If a fan feels cooling because \(T^{-}\) particles gain greater momentum before the fan under the action of gravity \(g\), where in part the variation in \(g\), \(\cfrac{\partial\,g}{\partial t}\) is due to the fan oscillating from side to side that throws the \(T^{-}\) particles into a Coriolis path and that \(g\) is a standing wave with a low frequency along the flow of \(T^{-}\).

\(T^{+}\) can be retarded by the addition of an \(E\) field pointing backwards.  The change in \(E\), \(\cfrac{\partial\,E}{\partial t}\) needed is provided for by the side to side rotation of the fan.  The air current before the fan will even be more cooling.

The underlying principle of accelerating temperature particles can be easily tested by reversing the rotation of the fan blades but still pushing air forward.  The fan will then be more cooling behind, where there is no perceivable air flow, than in front, in the path of the air current.  (It is assumed here that, \(T^{+}\) particles with energy oscillating along \(t_c\), are able to excite valence electrons upon collisions and so affect the rate of chemical reactions.  These are the particles we feel as heat.)

Air conditioning on the cheap.

Transient Anti-Matter

In this model when electron and proton collide, there can be gravitation energy, heat and EMW being released, but they are not matter/anti-matter pair.

If anti-matters are of negative time (\(time\,\,speed=-c\)), ie that they travel along the time dimensions in the negative direction, then their existence is only in transient as they would relatively zoom past anything with positive time speed at twice the speed of light (\(2c\)).

Have a nice day.

No Hay Problema, Amigos

We have a problem;  does opposite particles flow in the opposite direction, if they can even exist in the same conductor?

Positive current is thought to flow in the opposite direction to electron flow in a electric conductor.  This positive current cannot be a flow of free protons!  Such a flow of positive charges is admissible only conceptually.

We do not have a problem after all.

Then we have this,

The same magnetic coil does not drive the particles of different polarity apart, or do they?  In this case the particles are not in circular motion, they behave as packets of energies in the \(t_T\) time dimension with characteristic of that time dimension, ie temperature particles.  And would separate in a \(B\) field generated by the coil.

When the particles are in circular motion along a conductor coil, their oscillating components are aligned along the long axis of the coil and interacts strongly with external forces of a similar nature.   If the aligned oscillating components have energy along the \(t_c\) time dimension, they behave like positive charges.

Particle Flow

How to create gravity photons?

From collisions of gravity particles onto a hard surface, like electrons on lead blocks for X-rays...

Where to get gravity particles??

Ejected from radioactive material using gravity photons.

And round and round we go...

Fortunately by analogy, if there is such a thing as gravity particle conductor for positive gravity particles and negative gravity particles,

We have a flow of gravity particles in the respective gravity particle conductors.  In a analogous way, for electron flow and proton flow, with their respective conductor coils,

where \(\cfrac{\partial\,g}{\partial\,t}\) is generated using a temperature particle torus as proposed in the post "Spinning Temperature Particles" dated 23 Jun 2015.  And theoretically we have,

using temperature particle conductors.  The top schematic may be at odd with magnetic induction heating.  In this case, it is proposed that heat flow (flow of positive temperature particles) is generated directly.   In the case of a varying magnetic field, electron flow is induced in the conductor coil and collisions of such electrons generate heat.

 What would be a positive/negative gravity particle conductor?

PhotoGravitational

In the post dated 26 Jun 2015, "Guess What? Radioactivity Defined", it was postulated that the emission and absorption of gravity particles are the cause of radioactivity.  More accurately, the emission and absorption of gravity particles manifest radioactivity.  The cause, are gravity photons that eject gravity particles from a radioactive material.

The arrival of a gravity photon above the threshold frequency triggers the emission of a gravity particle.   This is consistent with a Poisson distribution (k=1) for radioactive decay (Exponential decay), where

\(\cfrac{d\,N}{dt}=-\lambda N\)

\(N(t)=N_oe^{-\lambda t}\)

where \(N=N_o\) at \(t=0\)

\(\lambda\) give the average number of gravity photons arriving per second, which can be a very small number (x10^-5).  We also have, the mean lifetime \(\tau\),

\(\tau=\cfrac{1}{\lambda}\)

and the half life, given by,

\(t_{{1}/{2}}=\tau ln(2)=\cfrac{ln(2)}{\lambda}\)

And we have "photogravitational effect" that is radioactive decay.  This has implications in radioactive clean up, if radioactivity can be triggered so.

Gravity Photons Ejecting Gravity Particles.

Provided that the artificial gravity devices suggested in previous posts can provide an opposing gravitational force, we may verify,

\(h_{tg}.f=\psi_{tg}+\Phi_{tg}\)

where \(\Phi_{tg}\) is the work density function for work done against an opposing gravity field.

Photons beyond the threshold frequency, \(f\) will be able to eject gravity particles from the radioactive material, both positive and negative particles alike.  Such photons exist in \(t_g\) time dimension, in order to interact with gravity particles.

Two types of photons for two types of gravity particles.

Guess What? Radioactivity Defined

Given that a positron decay involves a proton being converted to a neutron (apparent gain in mass) while a small positive charge (positive electric potential) is released.

It is likely that,

the particle on the right with energy oscillating in the \(t_c\) time dimension is a negative gravity particle and the particle on left with energy oscillating in the \(t_T\) time dimension is a positive gravity particle.

Beta emissions allows for both proton to neutron and neutron to proton change and is not indicative here.  But we see that, when a positive gravity particle is accelerated to light speed, \(x\to c\) and \(t_g\to0\), it is an electron.  High energy positive gravity particles could be the beta particles detected during radioactive decays.  On the other hand, when a negative gravity particle is accelerated, it becomes like a temperature particle with energy oscillating in \(t_c\).

When a positive gravity particle leave the atom, the binding energy at the nucleus due to negative gravity particles remains intact.  There is however an apparent loss in mass.

Just a guess.

Hic, Hic, Hic, I Cna Fly...

If gravity particles are particles that give us weight,

we are in serious trouble with our definition of mass based on gravity (circular momentum in a magnetic field is different).  These particle are in pairs, one type acting opposite to the other.  One impart gravity potential, makes an object lighter; the other impart opposite potential energy (negative), makes the object heavier.

If such particles can be introduced and remove from a body, like colored photons, temperature particles and charge particles, objects can be made lighter or heavier by introducing such gravity particles, which might be misinterpreted as changes in mass.  In particular, isotopes; when a particle carrying negative gravitational potential is ejected from the nucleus, the nucleus is seen to have gain in mass, although it is its gravitational potential that increased.  When a particle carrying positive gravitational potential is ejected from the nucleus, the nucleus is seen to have loss mass, although it is its gravitational potential that decreased.  The ejected particles when stopped as the result of collisions, release their component energy as heat (energy oscillating in \(t_T\) or charge potential (energy oscillating in \(t_c\)).  This could be radioactivity that Madame Curie first discovered.  The changes in mass as radioactive isotopes decay can be re-interpreted as residing particles being ejected from the nucleus.

The gravity particle carrying positive gravitational potential energy can be injected around a nucleus to make them more massive.  The nucleus is now a heavy isotope.  When such gravity particles are subsequently ejected, the nucleus's mass decreases, and the packet of energy released can be detected upon collisions as heat or charge potential.  And so, a body can be made radioactive by injecting positive gravity particles.

It is possible that the negative gravitational particles provides the gravitational attraction at the nucleus that binds the nucleus together.  When one such particle is ejected, the nucleus breaks apart, in particular an alpha particle can break away from the nucleus.  Positive gravity particles orbit around the negative gravity particles at the nucleus.  When positive gravity particles are ejected, the nucleus is seen to have loss apparent mass, likely in the form of a neutron decay to a proton.  When negative gravity particles are ejected, the nucleus is seen to have gain apparent mass, likely in the form of a proton decay to a neutron.  In both cases energy are expelled, irregardless of the apparent opposite change in mass (gain mass vs loss mass).

Without experimental proof, we should also inject a pinch of salt.  But still, Higgs Boson, my left foot, gravity particles more likely.

Gravity particles provide for the attraction force that forms the nucleus and account for charge, energy and mass changes in radioactive decays.  In fact the emission and absorption of gravity particles are the cause of radioactivity.

The Pumpkin

The pumpkin is generated using Scilab 5.5.1,

function Myparam3d_2()

    my_handle             = scf(100001);
    clf(my_handle,"reset");
    my_plot_desc          = "";
    my_handle.figure_name = my_plot_desc;
    xtitle(my_plot_desc,"x","y","z");

    //flower????
     t = 0:0.01:(2.0*%pi);
     x = (3.0+5*cos(t*20)).*cos(t);
     y = (3.0+5*cos(t.*20)).*sin(t);
     z = 10*0.02*sin(t.*20);

     p=get("hdl"); //get handle on current entity (here the polyline entity)
     p.foreground=5;  
     param3d(x,y,z,45,60,"",[2,0]);
   
endfunction

Myparam3d_2();
clear Myparam3d_2

There or about.  You may need to use the right mouse button to move the perspective of the graphic output to get the proper view.

Ultraviolet Appears White

The first implication of photons being ejected from colored object is that, blue light will illuminate everything.  Even red objects reflect red photons, with higher energy, when illuminated in blue light. Red objects appear magenta in blue light.  But blue objects do not eject blue photons in red light, and appears dark.  Green objects appear cyan colored in blue light but do not eject any photons in red light and appear dark.

Cyan which is the combination of blue and green, appear green in green light but light cyan in blue light.  Under both blue and green light it appears cyan.  This is so because blue light also eject green photons.

Blue light has higher frequency than green light and red light has the lowest frequency in the visible spectrum.  And

\(h_p.f=\psi_p+\Phi_p\)

photons are ejected as long as

\(h_p.f\ge\psi_p\)

and so, \(\Phi_p\ge0\).

Ultraviolet would then appears white, as it is able to eject all color photons.

What is Color?

From the previous post it was proposed that the temperature particles are trapped in the vicinity of an atom/nucleus where the \(\psi\)s are equal.

Photons with its \(\psi_p\) can also be trapped in the vicinity of an atom/nucleus where its \(\psi_p\)s are equal to the \(\psi_{nucleus}\) around the nucleus.

This is color!

When such photons are illuminated by a board spectrum light, they are freed from the nucleus and are ejected, by a process similar to the photoelectric effect.  The board spectrum illumination contains frequencies higher than the photons' threshold frequencies,

\(h_p.f=\psi_p+\Phi_p\)

where \(h_p\) is the equivalent Planck's constant, \(f\), the frequency of the illumination, \(\psi_p\), energy density of the photon in orbit and \(\Phi_p\), the work function (energy density) represents the amount of energy acquired by the photons when they are ejected.

So, red photons in orbit around the nucleus will make the material appear red when they are ejected.

And so there be colors.

A Weaker \(S\) Pole!

This is how the temperature particles are separated,

How do I know which is which, gravity (\(t_g\)) is prevalent around us, whereas electric field (\(t_c\)) is much weaker, in transient.  \(\psi\) in \(t_g\) will be comparatively low in a higher ambiance of gravity around us, this I associate with the weaker \(S\) pole.

No harm dreaming out loud.

Furthermore, if Curie temperature is analogous to threshold frequency then we have,

\(h_TT_{cu}=\psi_T\)

where \(h_T\) is a normalized Planck's constant, \(T_{cu}\) is Curie temperature and \(\psi_T\) is energy density.

Energy density is not the same as energy.  Energy density is more appropriate here as we are dealing with infinitely small point particles.

If we allow a \(B\) field to replace the stopping voltage,

\(h_TT=\psi_T+\Phi_T\)

where \(T\ge T_{cu}\) and the work function \(\Phi_T\ge0\).

Just like photoelectric effect.

In this case, the temperature particle of \(\psi_T\) resides in the vicinity of the atom where,

\(\psi_{atom}=\psi_T\)

where \(\psi_{atom}\) is the energy density around the atom, ie the energy densities of the two entities matches up.

Where the loci of \(\psi_{atom}=constant\) is circular, then the temperature particle is in circular orbit around the atom, along a particular locus of \(\psi_{atom}\).

If this is insanity, mine comes in a set menu.  A buffet of madness.

Spinning Temperature Particles

The following diagram shows temperature particles in a magnetic field torus,

The particles are driven in opposite directions in the torus.  One produces a gravitational field and the other, an electric field in the perpendicular direction given by the right hand screw rule.

If we are to pulse the gravitational field at Earth's gravitation frequency of 7.489 Hz, we have an anti-gravitational device.

When we are able to isolate one of the pair,

we have a device that produces a gravitational field or electric field only.

Hot very hot.

Spin Left Spin Right

In a 3 dimensional space, each spacial axis can acquire its own spin.  From the previous post "Negative Energy, Screw It", one spin differentiate a photon from EMW.  There is then two other degrees of freedom of spin left.

One of these remaining spins accounts for the left or right handedness of the wave in space.

And we have one other spin left...

Negative Energy, Screw It

There is no need to define negative energy nor negative time,

just opposing spins, given the right hand screw rule.  The sign in the expression,

\(i\cfrac{\partial\,\psi_{t_g}}{\partial\,x_1}\)

for the diagram on the left changes to,

\(-i\cfrac{\partial\,\psi_{t_g}}{\partial\,x_1}\)

with opposite spin, as the diagram on the right.

In both cases, \(x_1\) serve as the radius of the cross section circle perpendicular to \(x_2\), the direction of travel of the particle.  \(i\) rotates \(x_1\), \(\pi/2\) anti-clockwise into the perpendicular direction and the expression denotes the change in \(\psi_{t_g}\) in the direction perpendicular to \(x_1\), in the plane containing the circle that defines the spin.

The diagram on the left with a anti-clockwise spin is a photon, as we know them.

Conventions, Left And Right, And Negative Time!

What is the difference between the left and right diagram?

In both cases, the particle is the circular motion anti-clockwise around the circle.  \(t_g\) is perpendicular to the plane of the circle and is in the direction given by the right hand screw rule.  \(t_c\) sweeps outwards on the left and \(t_c\) sweeps inwards on the right.

Looking along \(t_c\), opposite to its direction,  \(t_c\) and \(x\) swapped position on the right diagram, which means given an anti-clockwise circulation, \(t_g\) lags \(x\).  For the diagram on the left, \(t_g\) leads \(x\), if \(t_c\) is to point inwards, when the the particle is rotating anti-clockwise.

Do we have a problem here?  Yes.  In the post "Whacko and the Free Photons" dated 30 May 2014,  EMW obeys Lenz's Law, photons fly off by themselves.

Given the direction of \(t_c\), the diagram on the left is a photon set free from its source, and the diagram on the right represents an EMW that collapses when power to the source is switched off.

We have a problem with negative time; \(t_c\,\, vs\,\,-t_c\).

It is possible to rotate in an anti-clockwise sense, but have \(t_c\) pointing in the opposite direction by moving in the \(-x\) direction for both cases.  But the direction of \(t_c\) has real physical implications.  \(t_c\) inwards is a EMW and \(t_c\) outwards is a photon.   We can travel in the \(-x\) direction by reversing time;  \(-t_T\).

This suggests that reversing one of \(t_c\), \(t_g\), or \(t_T\) reverses all time simultaneously.  EMW then, is, relatively, in negative time.  And so, has NEGATIVE energy!

All these notion of negative time and negative energy are avoided by instead adapting spins in clockwise or anti-clockwise sense;  we can travel in the \(-x\) direction by simply reversing the sense of the circular motion;  clockwise spin vs anti-clockwise spin.

Irrespective of which representation is valid, in all cases, EMWs and photons are, relatively, in opposite time of each other, positive energy vs negative energy, clockwise vs anti-clockwise spin and one being folded out into the other.

Can we choose which to produce, EMW or photons.  Ooops, but first thing first, given any conservative field,

\(v^2+v^2_{tx}=c^2\)

where \(v\) is velocity in space and \(v_{tx}\) is velocity in the \(t_x\) time dimension.  We can, by increasing the field, accelerate the particle such that,

 \(v\to c\)

\(v_{tx}\to0\)

We can theoretically, by accelerating a basic particle to light speed with a spin, ie

to create a wave in space.  Is this a Heat Wave?

It seems that EMW being different from photons complicates matters, slightly.  We don't get photons/EMW in a straight forward manner.  Given the left diagram and the post "Whacko and the Free Photons" dated 30 May 2014, an anti-clockwise spin produces a photon and an clockwise spin produces an EMW.

\(B\) Not \(T\)? Where Are All These Particles?

If we review the basic particle repeated below,

we find that these particles are still unique without denoting which time dimension do the particles exist in.  So we can reduce the diagrams to,

In this case, a charge is a particle with electric potential energy oscillating in the orthogonal dimension.  Similarly, gravity and temperature particle each has energy oscillating in the respective time dimensions.  In this formulation, collisions releases energy in the time dimension along which the wave is at light speed \(c\); an electron gives heat upon collision but a proton provides gravity potential energy upon collision.

How in circular motion, these particle manifest the energy in the orthogonal time dimension is not clear.  In the case of an electron, we see that as \(x\) is rotated in a circle, anticlockwise, \(t_c\) points toward the center of the circular path and sweeps out the area of the circle, \(t_T\) is always perpendicular to the circle and move on the circumference of the circle, pointing in the direction given by the right hand screw rule.  There is no field perpendicular to the circular path.

 Previously, when a particle with oscillating energy in the \(t_T\) dimension but exist in the \(t_c\) dimension (ie. an electron), moves in circular motion and \(x\) goes round in a circle, we see that \(t_T\) is perpendicular to the circular path with a changing \(\psi\) along it.  So there is a force field of  \(t_T\) nature in this perpendicular direction.

The new reduced view does not provide any indication of a force perpendicular to path of the particle when it goes into circular motion.   However in the case of an electron, the rotating \(t_c\) that sweeps out a circle is consistent with the view of a rotating dipole.  In this case, the perpendicular force field created when the particle goes into circular motion is then not related to \(t_T\).  \(B\) is not \(T\).

We may conclude for consistency, \(g_B\) exists as the result of a rotating \(t_g\) and \(T_B\) exists as the result of a rotating \(t_T\).  All these are consistent with the dipole formulation.  Each force has a complementary perpendicular force when the particle is in circular motion.

In both scenarios, we are dealing with the same set of particles.  The identity of individual particles are however, different.  In the top diagram, charges are in the middle pair.  In the latter diagram, charges are at the top and bottom on the right.  Which is which?

The key is that \(B\) field interacts with moving electrons NOT stationary electrons.  In the reduced view, \(B\), an rotating \(t_c\) will interact with oscillating \(t_c\) (both electrons and protons) irregardless of whether they are in motion!

If \(B\) is of the nature \(T\), all particles with \(t_T\) component will be effected by \(B\).  It will also interact with moving gravity particles with light speed along \(t_T\).

Another key question is, can \(B\) separate temperature particles.  In the reduced view, only moving and oscillating \(t_c\) components are effected by \(B\), and so only one of the temperature particle is affected by \(B\).

The reduced view is not consistent.  In the original perspective, a particle exists in a particular time frame, with characteristics of that time dimension.  It is wave with component potential energy of other orthogonal dimensions.  An explicit view, indicating the time dimension of existence, is needed to differentiate waves with speed in space but in different time dimensions.  A wave with speed \(c\) in \(t_c\) is different from wave with the same speed \(c\), but in the \(t_g\) time dimension.  In the former, the wave carries electric potential energy, and the latter carries gravitational potential energy.

\(B\) can still be \(T\).

And where are all these particles in a material?  Are they orbiting in a nucleus-satellite formation like electrons and protons.

Sticky Magnetic Force

If particles are responsible for magnetism, then \(\psi\) is responsible for its force field and naturally, the force density, \(F_m\) between the magnet and a piece of magnetic material stuck on it, is,

\(F_m=\cfrac{\partial\,\psi}{\partial x}|_{x=0}\)

the change in \(\psi\) across the contact boundaries of the two objects, at \(x=0\).  This is not a good formulation because contact are made over an extended area and both objects are not point particles.

This is counter intuitive, because a strongly held material will also be strongly magnetic.  It would seem then, that the drop in \(\psi\) across the material boundary should be small.  In fact, \(\psi\) permeates through the material to a greater extent than the drop across the material boundary; although it provides for a greater force the drop is still comparatively small.  In this way, a strongly held nail will attracts more nails than a weakly held nail.  It would seem that this change in \(\psi\) across the boundary is a fraction of (proportional to) the amount of \(\psi\) that permeated through the boundary.

It is expected that \(\psi\) decreases with \(x\) and the force is attractive.  When \(\psi\) increases in the presence of another field, the force is repulsive.  In this way, the above expression is without a negative sign.  It is the force on another body, not the force on the body exerting \(\psi\).

Let's see how far can \(\psi\) go...

Temperature, Magnetism, Are They The Same?

Hysteresis, remanence both suggest something is retained in the material of the magnet.  If Curie temperature is a measure of the amount of opposite particles that would cancel the magnetic potential, just as opposite charges cancels an electric potential...

The following diagram shows how the magnetic potential on a ferromagnetic material can be destroyed without the application of heat directly.

Intuitively, I am just turning the \(S\) pole into a \(N\) pole and in between the magnetic potential goes to zero.

We may find that the temperature required to neutralize the magnetic pole is much lower than Curie temperature.  The duration of the exposure will definitely play a part, given temperature.

This actually raises the question of quantum.  All particles are in quantum of \(\psi\), the existence of temperature particles implies that temperature is also in quantum.  Since, one of the temperature particle pair has charge potential, it will pop electrons from metal.  The other particle, provides gravitational potential and make light things float in air.  Collisions extract energy from the oscillating component of the wave/particle, destroying the wave/particle wholly or partially in the process.

If temperature is analogous to frequency, then Curie temperature might just be analogous to threshold frequency of photoelectric phenomenon.  In which case, the temperature required to neutralize the magnetic pole is still Curie temperature and time duration does not play a role when temperature is less than Curie temperature.

How then are such temperature particles held within the material?  Like an electron in orbit?

Are we bouncing the temperature particles off the material or are we neutralizing their potentials with the opposite particles?

And the pin drops.

What Is Permanent?

A permanent magnet is then of a material that retains, separates and localizes temperature particles.

We have already material polarized with opposite charges at opposing ends.  This is a charge magnet.

We can then have material that retains, separates and localizes gravity particles.  They would then be gravity magnets.

Hey! This is my dream.

Why not gravity magnet in the first place?  Because \(B\) field is produced by a flow of electrons.  A flow of electrons also generates heat.  The rest is speculative.

Ultra Strong Electromagnet

If temperature particles are responsible for magnetism, then a constant supply of such particles will create the strongest magnet.

The insulator blocks both heat and electric current.  The key point is the heat source.  At a particular ambient temperature, increasing \(B\) will increase \(N\), the magnetic field through the magnetic material incrementally (a slight gradient positive slope) as the material saturates magnetically.

A heat source will increase \(N\) further.

Speculatively yours.

Tropical Santa Claus

But a \(N\) pole is not hot?

Neither is the \(S\) pole hot.

\(p_{tT}(t_g,x,t_c)\) particles carry gravitational potential energy.  At the end where such particles reside, the material will also have higher gravitational potential energy.  This would mean, as a whole, the CG of the material will shift away from this end.

\(p_{tT}(t_c,x,t_g)\) particles carry electric potential energy.  At this end, the material will be as if positively charged.

Paradoxically, neither types of particle carries \(t_T\) potential energy that would set up a temperature gradient and cause a flow of heat.  (This is not true.  The wave as a whole carries temperature potential energy.  But in this case, the component characteristics of the wave show through instead.)

At this point we are in danger of playing with words.  Fiction all this may be, but physics is not a play of words, neither is logic.

Maybe charges should be particles that have energy oscillating in the \(t_c\) dimensions, instead of existing in the \(t_c\) dimension.  And so, temperature particle have energy oscillating in \(t_T\) and not the wave itself existing in \(t_T\).  In either way, the six particles pair up into three sets nicely all the same.

But, we started with the notion that charges are charges because they exist in charge time, \(t_c\) and since they are along the same time line they are able to interact so.  At the same time however, a charge provides an electric potential around it.  And so, a temperature particle must also provide a temperature energy potential.  The problem is, in 3D space, \(t_c\), \(t_g\) and \(t_T\) are not differentiated; all particles interact in general time, \(t\).

As a whole, the particle waves manifest the characteristic of the time dimension in which they exist.  So a charge is a particle that exist in charge time, \(t_c\) and provides an electric potential.  However, the wave component properties can manifest themselves under special circumstances.  When the wave is destroyed for example, KE at light speed along that particular dimension is release as energy of that dimension.  The oscillating energy is also release with characteristic of that dimension.  These happens irrespective of the time dimension in which the wave exist.

It was proposed previously, that circular motion is also a special circumstance, under which the oscillating energy present itself with its time dimension characteristics.  For example, a proton in circular motion creates a gravitational field, since its oscillating energy is of \(t_g\) time dimension.  Under the same notion (post "Positively, Glass Rub With Silk" dated 21 Jun 2015), it was proposed then, that \(B\) field is an aligned \(T\) field.  (Temperature can destroy a \(B\) field.)

It would seem that being trapped in a magnetic material also constitute a special circumstance under which the component characteristics of the wave manifest themselves.   This assertion would be consistent with the shift in CG and the slight positive charge potential at the \(N\) pole of the magnet.

Is a magnet slightly hotter than its environment?  If the ability to retain heat is related proportionately to the magnet's ability to retain its magnetic property, this then is consistent with temperature particles being responsible for magnetism.  And that \(B=T\).

This is one sweaty Santa Claus playing with words!

Magnetic Monopoles, Very Hot

If \(B\) is synonymous with \(T\), this is the way to make magnets,

Temperature particles created at the heat source is separated by the \(B\) field coil.  Particles repelled by the \(S\) pole of the \(B\) field, that would cancel a \(N\) pole, embed into the magnetic material and create a \(S\) pole.  Similarly, particles repelled by the \(N\) pole of the \(B\) field, that would cancel a \(S\) pole, embed into the magnetic material and create a \(N\) pole.

This is why a magnetic material somehow retains the magnetic polarity it is subjected to under high heat.

If such temperature/magnetic particles can exist and flow independently, then we have another pair of particles just as electron and proton, capable of being manipulated in an equivalent circuitry.

These are magnetic monopoles!  It is likely that the particle that creates a \(N\) pole, which is perceived to be stronger, has energy oscillating in \(t_c\) and the particle that creates a \(S\) pole has energy oscillating in \(t_g\).

Only more experiments will tell...  Have a hot magnetic day.

Positively, Glass Rub With Silk

Amber is an organic resin, glass can be pulled into flexible fibers.  Glass is a better choice.  In order to discourage the formation of \(H_2\) it would be useful to pump UVC 274.3 nm light into the fiber made with a less dense clagging around a central glass cylindrical core to confine the light.

The glass fiber will then conduct protons.

If optical fibers came from some UFO wreckage,  I think there has been a sweet misunderstanding.

A protonic circuit do not heat up, but gravity effects such a circuit.  Just as there are electrical sensors/circuits to detect temperature changes, in a analogous way, there can be a protonic sensors/circuits to detect changes in gravity.

Gravity will effect the resistance to proton flow in a proton conductor, just as temperature increases the resistance to electron flow in a electric conductor.  Does resistance increase or decrease with gravity, in the case of proton flow?  Likely to increase.  In both cases then, there is an increase in energy potential along the orthogonal time axis that results in an increase in resistance to particle flow.  In the case of electrons, that was energy oscillating between \(t_T\) and \(x\), and in the case of protons, energy oscillating between \(t_g\) and \(x\).

Both proton and electron flow in a coil produces \(B\) fields, albeit in the opposing sense; protons being just opposite charge to electrons.

What if  we are wrong!

If \(B\) field produced in the case of circulating electrons is actually energy along \(t_T\), that is to say the magnetic field are basically ordered/aligned temperature gradients.  Then circulating protons will produce gravitational fields.  In each case then, the energy in the orthogonal time axes aligns and manifest itself as the particle goes into circular motion.

The notion that \(B\) fields are actually aligned temperature gradients can be tested.  If we have very fine heated particles in circular motion, it would be as if small \(B\) field vectors are going in circular motion.  Does this produce an electric field through the center of the circle, perpendicular to the plane containing the circular motion.

We may have just found the heat particle pairs, postulated previously,

These are particles corrected from previously where energy oscillates between two space dimensions.  Instead, these particles are waves that exist in the \(t_T\) time dimension, has light speed along another time dimension \(t_g\) or \(t_c\), and energy that oscillating between the remaining time dimension and one other space dimension.  \(p_{tT}(t_g,x,t_c)\) in circular motion will produce an electric field and \(p_{tT}(t_c,x,t_g)\) in circular motion will produce a gravitational field.  In both cases, these fields are perpendicular to the plane containing the circular motion.

There is no \(T\) but aligned or misaligned \(B\).

Aligned \(T\) is \(B\).

Misaligned \(B\) is \(T\)

Or... 

There is no \(B\) but aligned or misaligned \(T\).

Aligned \(T\) is \(B\).

Misaligned \(B\) is \(T\)

The problem is polarity, the \(N\) and \(S\) poles of a magnet.  May be the three time dimensions proposed should be \(t_B\), \(t_c\) and \(t_g\); instead of \(t_T\), \(t_B\)?

In which case, we have immediately, two monopoles independent of each other and quite capable of acting alone.  If \(B\) is flow, then there must be a "through" end and a "from" end.  In which case the "through" end cannot be separated from the "from" end.  And so, when there is a \(S\) pole there is a \(N\) pole.

Does magnetic field confines heat flow between a hot and cold object?

The implicit notion here is that when these particles are in circular motion their energies oscillating in the orthogonal dimensions manifest themselves.  That the \(B\) field, for example, whatever its nature may be, is part of the wave.

In the case of an electric potential, we know that the presence of a lot of charges (of the oppose sense) can effectively nullifies the potential.  A high positive potential is cancelled  by adding electrons.  The \(B\) field can also be cancelled by high temperature (Curie temperature).  We however, create temperature particles in pairs and have never separated these pairs experimentally.

If \(B\) is just \(T\), then a magnetic field will serve to separate the temperature particle pair.  One of the pair cancels the \(N\) pole of a magnet and is pushed away by the \(S\) pole.   The other of the pair cancels the \(S\) pole of the magnet and is similarly pushed away by the \(N\) pole.

We are back to question of magnetic monopoles, but this time they are temperature particles.

Only experiments can tell... whether you get the Nobel Prize or not.  Thank you Madame Curie.

Proton Conductor

Protons carry both electric potential and gravitational potential.  It has no potential energy in the \(t_T\) time dimension and so does not heat up the circuits as electrons do.

A proton beam will increase the gravitational potential of the target and causes it to fall towards a massive body.  In effect, it is a tractor beam.

To complete this picture, we need the missing piece, a proton conductor.

What is a proton conductor??

Akan Datang...

\(H^{+}\) Again

\(H_2\) is easy enough by electrolysis, we then obtain \(H^{+}\) using UVC at 274.3 nm.  A layer of \(H^{+}\) around a sphere will cause light to bend around an object centered at the sphere rendering it invisible.

It is expected that within the \(H^{+}\) enclosure, \(n(r)\) is progressively less affected towards the center of the sphere.  This way, the \(n(r)\) increases towards the center of the sphere.

Which might explain the lighting discharge and UV lights of uncloaking UFO captured on film; and the cloud cover when the UFO is cloaked (\(2H_2+0_2\to2H_2O\)).

On yes, wearing such a sphere over your head will given you one big head and suffocates you immediately.

An so, I have my personal UFO.

We See You Spidey!

No, a screen in front of an object is not going to make the object invisible.  To be invisible, all light rays from around the object must bend around to avoid the object and emerge in the same direction on the opposite side.  Consider a sphere of decreasing refractive index along the radial line towards the center,

Light rays from one side of the sphere are bent by decreasing refractive index and emerge parallel to the incidence ray on the opposite side.  Rays can still be blocked from within such a sphere, but what if instead \(n(r)\) is made less dense than the surrounding and then made to increase progressively toward the center of the sphere.

In this case an object at the center of such a sphere can avoid the light rays round it.   But first \(n(r=r_o)\) must be made less then \(1\).  How can that be done?

We know from Snell's Law,

\(\cfrac{\psi_{n\,i}}{\psi_{n\,r}}=\cfrac{sin\,\theta}{sin\,\alpha}=n\)

If \(\psi_{n\,r}\) were to increase at the point of incidence,

\({\psi_{n\,r}}\gt{\psi_{n\,i}}\)

then it would be as if,

\(\cfrac{\psi_{n\,i}}{\psi_{n\,r}}=n\lt1\)

There after the ray is refracted progressively by smaller refractive angle as \(n(r)\) increases.  In effect the ray bends concave around the center of the sphere.

How then to increase \({\psi_{n\,r}}\) at the surface of the sphere?

Spider Web Man

We are familiar with the situation on the right, a reflective mirror with a silver coated surface.  But is the diagram on the left complementary?

If electrons slow the type of photons we perceive in our vision and in extreme cases able to bounce such photons away, a veil of electrons in front of an object might mask it from sight.

Spider webs with high electric static on it might explain why they are effectively invisible.  Dress up in spider web and we may turn invisible.

By a very thin line...

Even Light Can Bend

From the post "Success Is In The Way You Handle The Question" dated 13 Jun 2014,

\(n=1+\cfrac { C }{ \lambda^{ 4/3 }_{o} }\)

A sample plot with \(C=1\), n=1+1/x^(4/3) is given below,

It is possible to manipulate the expression \(n(\lambda)\),

\(n=1+\cfrac { C }{ \lambda ^{ 4/3 }_{ o } } \)

\( n^{ 2 }=1+\cfrac { C^{ 2 } }{ \lambda ^{ 8/3 }_{ o } } +2\cfrac { C }{ \lambda ^{ 4/3 }_{ o } } \)

Consider,

\( 2\cfrac { C }{ \lambda _{ o }^{ 4/3 } } =\cfrac { 1 }{ (\lambda _{ o }^{ 2 })^{ 2/3 } } =2\cfrac { C }{ (\lambda ^{ 2 }-D_{ 1 })^{ 2/3 } } =2\cfrac { C(\lambda ^{ 2 }-D_{ 1 })^{ 1/3 } }{ \lambda ^{ 2 }-D_{ 1 } } \)

when we consider the Taylor expansion of ,

\((\lambda^2-D_1)^{1/3}\) about \(\lambda=0\),

\((\lambda^2-D_1)^{1/3}=(-1)^{1/3}D_1^{1/3}-\cfrac{(-1)^{1/3}}{3D_1^{2/3}}\lambda^2-\cfrac{(-1)^{1/3}}{9D_1^{5/3}}\lambda^4+...\)

Ignoring higher terms of \(\lambda^4\),

\( 2\cfrac { C }{ \lambda _{ o }^{ 4/3 } } =\cfrac { B_1\lambda^2 }{ \lambda ^{ 2 }-D_{ 1 } }+\cfrac{A_1}{\lambda ^{ 2 }-D_{ 1 }}\)

Consider,

\( \cfrac { C^{ 2 } }{ \lambda ^{ 8/3 }_{ o } } =\cfrac { C^{ 2 } }{ (\lambda _{ o }^{ 2 })^{ 4/3 } } =\cfrac { C^{ 2 } }{ (\lambda ^{ 2 }-D_{ 2 })^{ 4/3 } } =\cfrac { C^{ 2 }(\lambda ^{ 2 }-D_{ 2 })^{ -1/3 } }{ \lambda ^{ 2 }-D_{ 2 } }\)

The Taylor expansion of

\((\lambda^2-D_2)^{-1/3}\) about \(\lambda=0\),

\((\lambda^2-D_2)^{-1/3}=-(-1)^{2/3}D_2^{-1/3}-\cfrac{(-1)^{2/3}}{3D_2^{4/3}}\lambda^2-2\cfrac{(-1)^{2/3}}{9D_2^{7/3}}\lambda^4+...\)

Ignoring higher terms of \(\lambda^4\),

\( \cfrac { C^{ 2 } }{ \lambda ^{ 8/3 }_{ o } } =\cfrac { B_2\lambda^2 }{ \lambda ^{ 2 }-D_{ 2} }-\cfrac{A_2}{\lambda ^{ 2 }-D_{ 2 }}\)

When,

\( D_{ 1 }=C_{ 1 }\)

\( D_{ 2 }=C_{ 2 }\)

and if we consider only the real parts, we obtain,

\( n^{ 2 }=1+\cfrac { B_{ 1 }\lambda^2 }{ \lambda ^{ 2 }-C_{ 1 } } +\cfrac { B_{ 2 }\lambda^2 }{ \lambda _{ o }^{ 2 }-C_{ 2 } }+Err\)

where \(Err\) is an error term.

If we let,

\(Err=\cfrac { B_{ 3}\lambda^2 }{ \lambda ^{ 2 }-C_{3 } } =\cfrac{A_1}{\lambda ^{ 2 }-C_{ 1 }}-\cfrac{A_2}{\lambda ^{ 2 }-C_{ 2 }}+expansion\,\,error\)

because,

\(\cfrac { A_{ 1 } }{ \lambda ^{ 2 }-C_{ 1 } } -\cfrac { A_{ 2 } }{ \lambda ^{ 2 }-C_{ 2 } } =\)

\( \cfrac { A_{ 1 }(\lambda ^{ 2 }-C_{ 2 })-A_{ 2 }(\lambda ^{ 2 }-C_{ 1 }) }{ (\lambda ^{ 2 }-C_{ 1 })(\lambda ^{ 2 }-C_{ 2 }) } =\cfrac { (A_{ 1 }-A_{ 2 })\lambda ^{ 2 }-(A_1C_{2 }-A_2C_{ 1 }) }{ \lambda ^{ 4 }-(C_{ 1 }+C_{ 2 })\lambda ^{ 2 }+C_{ 1 }C_{ 2 } } \)

Ignoring \(\lambda^4\),

\(\cfrac { \cfrac { (A_{ 2 }-A_{ 1 }) }{ (C_{ 1 }+C_{ 2 }) } \lambda ^{ 2 }-\cfrac { A_2C_{1}-A_1C_{ 2 } }{ C_{ 1 }+C_{ 2 } }  }{ \lambda ^{ 2 }-\cfrac { C_{ 1 }C_{ 2 } }{ C_{ 1 }+C_{ 2 } }  } \approx \cfrac { B_{ 3 }\lambda ^{ 2 } }{ \lambda ^{ 2 }-C_{ 3 } } \)

which is valid when \((A_2C_1-A_1C_2)/(C_1+C_2)\) is small compared to \(n^2\) and \(B_3\gt\gt A_2C_1-A_1C_2\).

Quite arbitrarily,

\( n^{ 2 }=1+\cfrac { B_{ 1 } \lambda^2}{ \lambda ^{ 2 }-C_{ 1 } } +\cfrac { B_{ 2 }\lambda^2 }{ \lambda ^{ 2 }-C_{ 2 } }+\cfrac { B_{ 3}\lambda^2 }{ \lambda ^{ 2 }-C_{3 } }\)

which is Sellmeier experimental fit for \(n\).

Even light can bend; given a dose of insanity everything else bends.  Obviously insanity is in \(Err\).