Reinventing The Laser

If the post "Blue Mussel Solar Cell" dated 4 Jun 2018 is successful,

In cases, where the basic particle generated in the colored glass tube is positive (red color glass for \(a_{\psi}\gt a_{\psi\,ne}\)), the applied voltage polarities will have to be reversed, such that the particles are always driven towards the emission hole.

If the system can be pulsed, and so squeezed (\(\theta=43.535^{o}\) from the post "Guns With No Recoil" dated 23 Sep 2017), we may have resonance and so reproduce the laser effect.

Goodnight...

The Basic Particle Phenomena

What has a jet pilot experiencing a psychedelic episode and a torus meant to power a set of VR goggles have in common?

 
Hint:  The canopy is tinted blue-purplish to provide shade.

Answer:  Basic particles!

Caution:  An overload of basic particles faints.

But which type, gravity, charge or temperature?

Have a nice trip.  A psychedelic one.

Blue Mussel Solar Cell

The question was how did the Chin terracotta and ancient Japanese pottery acquired their blue purplish finish.

And the tint of blue of high energy particles from posts "Glow Some More Glow",  "Two Color World" and "Animal And Plant Torture" all dated 29 Dec 2017.

The varied thickness of the fired glass plate introduces a charge density difference and directs the current flow in one specific way.

The closer the contacts on the glass the higher the voltage across the terminals.  Current increases with more collection length when the contacts run across the glass as parallel wires.

Goodnight...

Running High

The problem is, \(T_a\) changes constantly with the application of \(T_{ex}\),

In retrospect, the same happens in electrolysis where the electrolyte gets charged.

"Thermode" Are Hot And Cold Rod

Just as charges provide by the each electrode enable a redox half reaction,

temperature particles provided by each "thermode", one at each end of a temperature gradient established using a heating element, enable the chemical reaction hindered by a discrepancy in temperature charge.

The barrier prevents the resulting compounds from migrating to the opposite electrode/thermode and react in reverse.

In this case, the relative cold and hot end of the temperature gradient overcomes the energy level difference in the temperature particle clouds of the participating reagents and drive the chemical reaction in one direction.

The existence of temperature particles opens up a new possibility of temperature charge circuitry operating in an insulated thermostat confinement of constant temperature.

It is a hot day...

Enough Already?

Does the the accumulation of negative charge below zero charge temperature and the accumulation of positive charge above zero charge temperature cancels and give a reduced charge difference that do not cancels the energy level difference totally between temperature charge clouds (in an aqueous medium)?

Yes and no.

It is the amount of \(\psi\) provided via the temperature difference that cancels the energy level difference between temperature charge clouds between the reagent.  Negative charge are smaller and provide less \(\psi\) per particle, their polarity does not negate the presence of positive charges when they mix and coalesce.  \(\psi_{c}\) coalesce with a positive charge to result in a bigger positive charge with a higher amount of \(\psi\) embodied in a single particle.

In the range of temperature below zero charge temperature, since we are dealing with smaller negative charge, \(T_{ex}\) has to increase to provide the same amount of \(\psi\).

A body of \(\psi\) behave as a single particle.  An increasing embodiment of negative charges each remaining distinct as a particle increases the negative potential.  In this way, a body can be brought to a high negative potential.

Temperature \(\psi\) moves freely as we are in an environment of higher temperature potential; electric charge tends to distill into particles because we are in an environment of normally low electric potential.  Temperature behaves more like waves.

Have a nice day.

Note:  That would mean at high enough positive electric potential, positive charges can break into smaller negative charges and so switches the electric potential to negative.   And at high enough negative potential where the negative charges are pressed to coalesce, they merge into large positive particles and switches back to positive electric potential.  So, a large embodiment of charges fed with \(\psi\) (via a bombardment of basic particles) is naturally oscillatory.  Such an oscillatory system will emit EMWs and photons.

Too much of anything is a bad thing.

Scales and Feathers

Why does water at \(4^oC\) has the lowest density?  Temperature particle at close proximity interact as waves and attract each other, at \(4^oC\), the effective/resultant temperature charge in the containment of water is zero, the water molecules are not pulled together, they extend the furthest volume and so has the least density.  At temperature below \(4^oC\), negative temperature charge interacting as waves pull the molecules closer; at temperature above \(4^oC\), positive temperature charge also interacting as waves pull the molecules closer.  In both cases, the volume decreases and density increases.

If this is so, and...

if we have an analogous electric scale (not necessarily the voltage measure) which measures the amount of electric charge \(x_{T}\) in a containment, by varying this value we may have the equivalent of thermal expansion and consequently density changes due to the electric charge.

As thermal expansion and contraction are readily observable but not electric expansion and contraction, it may suggests that the outermost layer of an atom is positive temperature particles that acquire loosely held negative temperature particles.  This positive temperature particle layer is attracted to the electrons in orbits.

The interaction between atoms however, occurs at the lower layer just below the temperature particles where protons in paired orbits hold electrons in various modes (as illustrated in the posts "Just Rolling Along...Conducting Electricity" and "Rolling Inside" both dated 03 May 2016).

The cases of water and benzene, it is the layer of temperature particles below the electric charge layer that raise to prominent outside of the paired orbits as charged clouds that interacts with temperature particles in the environment directly.  These temperature charged clouds are orientated perpendicular to the plane of the paired orbits.  They effect physical properties such as density, thermal conductivity, thermal capacity, latent heat...etc, directly.

The resultant charge of the containment at various temperature depends on the constituent.  Water given the size of its temperature clouds has the lowest density at \(4^oC\).  A mixture of water and benzene has the lowest density at \(-22^oC\). 

\(T_{ex}\) from the post "Benzene Continued" dated 24 Apr 2018, is obtained by observing the volume of a mixture of reagent (in molar portion as dictated by the desired reaction equation).  It is the temperature at which the mixture expand greatly as temperature is lowered.  It was proposed that at this temperature, the charge on the temperature cloud is neutralized.

But the reagents are further apart with the observed increase in volume, why would a temperature difference between the reagents at \(T_{ex}\) prompt a chemical reaction that would otherwise requires high temperature and pressure?

The reagents are furthest apart at the temperature \(T_{ex}\), however, at the temperature DIFFERENCE of \(T_{ex}\) between the reagents, the energy difference between temperature clouds that have to merge in the resultant compound, which is the barrier to chemical reaction proposed here, is zero.

Should \(T_{ex}\) take reference with \(4^oC\), the temperature at which water, the aqueous medium in which the reaction take place is temperature charge neutral?  That one reagent is held at \(4^oC\) and the other reagent is at \(T_{ex}\) above (when its temperature cloud is larger) or below (when its temperature cloud is smaller) \(4^oC\).

\(T_{ex}\) alone, seem arbitrary.  Does a temperature difference of \(T_{ex}\) always contain the right amount of temperature charges to negate the energy level difference between the temperature clouds?

What happens when the reagents are not in an aqueous medium? 

What happens when \(T_{ex}\) spreads over a region where either reagent has an effective zero temperature charge when pure?  Does a mixture of positive and negative charges also negate the energy difference between temperature clouds?

Have a nice day.

Innate Hot And Cold, Latent

If two reagents mixed at the same temperature do not react, but when mixed at a temperature difference of \(T_{ex}\) reacted and turned dark blue,

we have canceled the temperature charge difference between the two reagents, and have also negated any heat change as the reaction proceeds.

It is providing the heat required first when the reaction is endothermic or removing heat first when the reaction is exothermic, so that the reaction can proceed without hindrance and without further energy change.

 Here lies the problem;  like electrical charges within a dielectric as it aligns when a capacitor is charged, the temperature charges within the temperature charge clouds of the resulting compounds do not increase the temperature potential of the containment linearly, correspondingly.  The reaction has an imbalance of energy (negative or positive) that was accounted for using latent energy when a significant amount shows up.  But where is this latent energy without a temperature charge cloud?

Furthermore, when the temperature charge clouds rearrange themselves as the molecules move, temperature changes and so does the thermal energy content of this containment of molecules.

So, if a reaction is difficult under normal conditions, try changing the relative temperatures of the reagents.  Reagents with a larger temperature charge clouds have lower temperature and will have to increase in temperature to match energy levels with reagents that have smaller temperature charge clouds.  And conversely, cool the reagents with smaller temperature clouds.

Have a nice day...

Benzene Continued

So, the reagent with an appropriate functional group is cooled and sprayed in the way of benzene in a warmer spray of mist.  At what specific temperatures depend on the interpretation of \(E\),

\(E=E_{ph}=\cfrac{n.hc}{2\pi a_{\psi}}\)

where \(a_{\psi}\approx 139\,pm\)  and  \(n=3\)

and the specific reagent.  The point is to mitigate the energy difference between the temperature particles clouds on the two reacting compounds.

What is \(E_{ph}\) on the Celsius temperature scale?

In experiments freezing water and benzene, at the temperature at approximately \(-22^oC\), a large expansion occurs in contrast with the gradual contraction found with temperature above  \(-22^oC\).  This might be due to the temperature particle clouds on water and benzene aligning themselves and both molecules form into a hyper-lattice.

The temperature clouds have to be attractive to from into a lattice.  Benzene freezes at \(5.5^oC\); water freezes at \(0^oC\).

What happens at \(-22^oC\)?

The temperature around a containment is due to temperature charge on its surface.  So it is the temperature charge around the mixture of water and benzene that give a temperature reading of \(-22^oC\).  Is this charge that mitigate the energy level difference between the two temperature particle clouds?  That the charge equivalent to \(-22^oC\) makes both temperature clouds on the reacting compounds neutral and is attractive.

If, we mix benzene and the reagent with a functional group to be substituted for an hydrogen on the benzene and lowers the temperature of the mixture in porous porous Vycor Glass till similar expansion occurs at say, \(T_{ex}\).  Does a temperature difference of \(T_{ex}\), the hotter spray being benzene, encourages the substitution reaction?

Water dissolves benzene, but a reagent that provides a functional group for substitution might react quickly.

Note:  \(T_{ex}\) can be negative, but the hot spray is benzene when the other reagent is expected to have a smaller temperature particle cloud.  Two running streams when the reagents are liquid will work as well as two sprays of mist.  The point of both set up is to encourage collisions between the reacting molecules.

The Other Torus

I was thinking,

but if the rings are due to temperature particles and,

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

given,

\(139*\cfrac { \sqrt { 3 }  }{ 2 } \quad \le \quad { a_{ \psi  } }\quad \le \quad 139\quad \, pm\)

and

\(2\pi a_{\psi}=n\lambda\)

with

\(n=3\)

\(f=\cfrac{n.c}{2\pi a_{\psi}}\)

What is the significance of \(f\)?

If a reagent with an needed functional group also has a temperature particle layer (just below the outer \(p^+\), \(e^-\) layer) at this frequency, does reaction occurs more readily?  Is the functional group accepted more readily?

For a lone particle, \(f\), because of

\(E=h.f\)

\(E=\cfrac{n.hc}{2\pi a_{\psi}}\)

is the energy of \(\psi\) that constitute the particle.  \(f_{ph}=f\) here, is then the energy of each of the two tori on the benzene.

How to manipulate \(f_{ph}\)?  Once again, it is a specific temperature.  At this temperature, all collisions between the reagents and benzene molecules result in an exchange of radicals, hopefully between a \(H\) from the benzene ring and the functional group from the reagent.  The difference in energy levels between the temperature \(\psi\) clouds on the benzene and the reagent with the needed functional group is the hindrance against chemical reactions.

As the functional group reagent is often smaller than benzene (low \(\lambda\), high \(f\), high \(E\)), it is the energy of temperature particle clouds on benzene that has to increase to facilitate a substitution.  Conversely, the reagent that provides the functional group is lowered in temperature and made to collide with benzene.

Heat up benzene and spray it at the reagent which remains at a lower temperature.

If \(f_{ph}\) can be raised by increasing the temperature of benzene, then at the specific temperature when \(E=h.f_{ph}\) is equal to the energy level of the temperature clouds on the reagent, the energy barrier is surmounted and reactions proceeds.

But what charge is the benzene \(\psi\) torus?  Under normal conditions, benzene rejects all reagents and is chemically stable.  So, the benzene torus is positive and repulses positive temperature clouds on other possible reagent including other benzene rings.

Collisions bring the reactants close enough for \(\psi\) to interact as waves and be attractive, but that still requires compatible energy levels as \(\psi\) clouds merge and subsequently separate.

Which bring us to endothermic and exothermic reactions....because electrical charges and energy levels do not change temperature charges and energy levels.  Electric and temperature are two orthogonal types of fields that do not affect each other.

Have a nice day...

Time After Time

If you can make lightning strike again at the same spot,

It does not matter whether the charges are positive or negative as long as they cause a change in oxidation state of the Vanadium ions.  The ions in circular motion, in a constant magnetic field, separate into layers.  On discharge, a reversion in oxidation states releases the charge and is tapped from two circulating ion layers to drive an external load.

Do charges flow from a pool of water at \(80^oC\)?  Micro-lightning?

Monolithic Sword

If the the whole sword, blade and grip, is forged from a single piece of smelt,

\(n=512\)

\(l_s=0.795\,m\)

and the number of folds the smelt is subjected to,

\(n_b=9\)

If you are on the right path, the smelt first seems sticky and will stick to the hammer as you rise the hammer for the next strike to elongate the smelt.

Collecting Collapsed Torus

Collecting collapsed torus,

the particles will move towards the positive rail, into the collection channel.

Solar Cell Skin

If X ray generated by the Sun from Hydrogen, \(H\) and Helium, \(He\) are captured on solar cells and plant life,

	Z	density (g/m^3)	density/Z (g/m^3)	v boom (m/s)	x ray penetration (m)
Hydrogen, H	1	8.99E-005	8.9900E-005	3.0884E-01	5.8195E-05
Helium, He	2	1.79E-004	8.9250E-005	3.0661E-01	5.7775E-05

where X ray peneration is calculated by taking reference with copper, \(Cu\),

\(S_{p\,H_2}=\cfrac{v_{boom\,H_2}}{v_{boom\,Cu}}*S_{p\,Cu}\)

\(S_{p\,Cu}=0.2\,m\)

\(S_{p\,H}=\cfrac{3.0884E-01}{1061.42}*0.2=5.8195E-05\,m\)

The X ray torus photons collapse on the surface but penetrate only by \(50\,\text{to}\,60\,\mu m\) into the surface; the collection channel that collects the particles cannot be deeper than these depths, with the orientation of the collecting device considered.

If the particles collected are from X rays...

诗赋多词
言当行达
霸溃纹理
钢腐不持
滴水破溅
漫洪没顶

《淹败》

Note: Tidal floods and a broken dam are two different things.

Gravity Sword

A sword that resonate at \(7.489\,Hz\),

given that the speed of sound in steel is \(5000\,ms^{-1}\).

Unfortunately...

\(f.\lambda=v\)

\(l_s=\cfrac{\lambda}{2}=\cfrac{v}{2f}\)

\(l_s=\cfrac{5000}{2*7.489}\)

\(l_s=333.82\,m\)

over three hundred and thirty meters!

If however, we fold the molten smelt repeatedly during the forging process,  such that many half wavelengths spread over the length of the sword we set at,

\(l_s=0.72\,m\)

\(l_s=n.\cfrac{\lambda}{2}=n.\cfrac{v_n}{2f}\)

where \(n\) is the number of folds on the sword, and \(v_n\) is the speed of sound along the sword,

\(n=l_s*\cfrac{2f}{v_n}\)

we assume further that the speed reduced by a factor of \(\cfrac{1}{n^2}\)

\(v_n=\cfrac{v}{n^2}\)

\(n=\cfrac{v_n}{2fl_s}\)

\(n=\cfrac{333.82}{0.72}\)

In practice, the number of folds possible is binary,

\(n_b=\left\lceil\cfrac{log(n)}{log(2)}\right\rceil=\left\lceil 8.856 \right\rceil=9\)

ie,

\(n_p=2^9=512\)

This would be,

\(p=\cfrac{n_p}{n}=\cfrac{512}{333.82}*{0.72}=1.104\)

times the intended \(l_s=0.72\).  So, with nine folds of the smelt, we turn it such that the folds run perpendicularly along the sword and elongate the smelt to a length of,

\(l_b=p*0.72=0.795\)

and file this blade down to \(0.72\).  This way the sword at \(0.72\,m\) will have the intended \(n\) number of folds along its length.

The speed of sound along the blade is an approximated at \(5000\,ms^{-1}\) and we have assumed that the folds on the blade slow the speed of sound along it by a factor of \(\cfrac{1}{n^2}\).

If both these assumptions are true, this sword should be light as a feather.

Tale Spin

\(\nabla.A\ne\nabla.(-A)\),

refers to a point at the periphery of the spinning field, \(P\) and not at its center \(C\).  The flow of \(E\) at the center is reversed when a negative sign is applied.

Strictly speaking,

\(\nabla.A\)

does not represent the effect of spinning \(E\) at point \(P\).  Neither does,

\(-\cfrac{\partial E}{\partial t}\)

nor the more familiar,

\(-\cfrac{\partial B}{\partial t}\)

represents the spinning end of a field \(B\) centered at \(C\).  It is the negative rate of change with time of \(B\) at a fixed arbitrary point \(P\).

Lunch!

Sphere, Cylinder And Egg

Remember this,

if we line them up and send them down a line we have,

\(\psi\) wraps around into a sphere manifest itself as circular waves with centers along the line of travel, forming a cylinder.

At this point, we do not know the direction of this wave (right hand rule or left hand) but we known form symmetry, by rotating the line in any plane containing it by \(180^o\) that this direction reverses.  So, the sense of the wave (clockwise vs anticlockwise) reversed with the follow of the particles along the straight path.

But given one direction of flow, the clockwise or anticlockwise sense remains arbitrary.

Only upon the application of Lenz's Law based on conservation of energy, that the particle is not self propelling like a photon, that the right hand rule for the flow of positive particles is true.

And we are still open as to the nature of this field, cylindrical along the line of travel.

Egg, refers to the deformed shape of the particles as they are propelled in a field.

Have a nice day.

Data Crunch And Rotor Speed

If, the sonic boom at the region on the blade closer to where they connect to the rotor mast of a helicopter, is responsible for its lift,

then, at higher altitudes where the air is thin and water has a lower density value (humility) and a reduced boom velocity, a decrease in rotor speed may just maintain lifts.

Intuitively, the normal reaction to a lost in lift is to increase rotor speed.

However, it is not altitude per se that requires a reduced rotor speed, in cold regions a reduction in water density may also require a similar reduction in rotor speed to take flight.

Good night...

Dust And Water Droplets

Fluorescence tubes are rare, but,

With charges flowing towards the left inside the discharge tube, maybe a flow of water around the tube and a ring of light dust particles around the tube in the direction of the \(g\) field can be seen.  A tiny drop of ink might indicate the flow direction of water more clearly.

Water smear and dust tend to stay on the tube when it is on, but these phenomena are attributed to static electricity and not the presence of a \(g\) field proposed here.

Maybe...

Field Party

The fields around a discharge tube may show the way,

Do these fields exist?

Neutral Particles, Neutral Temperature And KaBoom

The flow into a point is the reverse of the flow out of the point by introducing a negative sign only if there is high symmetry like in the case at the center of a sphere where the radius indicates equal vectors.  When the vectors are not equal reversing the flow by introducing a negative sign generates a new flow into the point, not the negative of the flow initially.

If charge is a continuum, when does it end?  How big can a positive charge gets given the right conditions?

What happens when a neutral charge is spinning?

If the manifestation of any one field depended on the relative phase between two, as charge sweeps from negative to zero, to positive, then when the charge is zero... well everything goes zero as if nothing is in spin.

Still there would be three types of neutral particles; electric, gravity and temperature.

Neutral temperature is not at zero temperature as defined on the Kelvin scale.  Neutral particles are not at \(a_{\psi}=0\) but are big at \(a_{\psi\,ne}\).

What if there is a field as a neutral particle spins?  This field is the field at zero phase between the two fields that together manifest the observed field.  It is one field in isolation.  And each type of neutral particle (electric, gravity and temperature) presents one field.

It is possible, that each field instead has three orthogonal components, two of which cancels when two fields are orthogonal, the remaining component presents itself when the particle is in spin.  As charge changes from positive to negative, a phase shift from \(90^o\) to \(-90^o\) occurs (relatively) between the two fields and a different component remains as another set of two components cancel.  In this paradoxical way a neutral particle in spin generates the highest possible fields compared to charged particles.

A neutral particle suggests that all the wave energy is in oscillations.  This is still a wave-particle?  It no longer exert a field in space stationary but it has size \(a_{\psi\,ne}\) and provides the strongest field in spin.  It is the highest negative charge possible.

As long as spins generates a field,

the layered atom model still holds.

Maybe all spinning particles in a nucleus are neutral that, generates the strongest field in spin possible.  And isotopes are of off neutral particles in spin that have a charge, generate a weaker fields in spin and can be bigger (positive) or smaller (negative) than a nucleus of neutral particles.

This makes \(a_{\psi\,c}\) very dangerous as \(a_{\psi\,ne}+a_{\psi\,c}\) makes a neutral charge positive.  A change in charge switches the type of field generated in spin and rearranges the layers inside the nucleus and may cause the disintegration of the atom.

KaBoom!

No Sign Assignment Needed

A hiatus because the model for particles,

and all its variations is not consistent with,

because,

\(t_g\ne -t_c\)

\(t_c\ne -t_T\)

\(t_T\ne -t_g\)

and etc.

Luckily, charge continuum, that charge varies continuously from negative to zero, to positive values  does not implies that the various fields (electric, gravity and temperature) are a entangled mess of opposing fields.  It implies that,

\(\nabla\cdot E=-\nabla\cdot g\)

\(\nabla\cdot T=-\nabla\cdot E\)

and,

\(\nabla\cdot g=-\nabla\cdot T\)

where \(E\), \(g\) and \(T\) are electric, gravity and temperature fields.

If the fields are a mess of opposing fields then, one possible conclusion is that there is only one electric field that produces a magnetic field on spin and gravity is just electric particles interacting at close proximity as waves that attract each other.  Similarly, temperature fields are waves interacting at close distance, and interactions at greater distance manifest as the magnetic fields.

Distance apart being relative to the size of the particle in consideration, on Earth's surface, we are interacting with Earth as waves.

The trouble is, given,

\(\nabla\cdot E=-\nabla\cdot g\)

and,

\(\nabla\cdot T=-\nabla\cdot E\)

we have,

\(\nabla\cdot T=-(-\nabla\cdot g)\)

\(\nabla\cdot T=\nabla\cdot g\)

which contradicts the last statement,

\(\nabla\cdot g=-\nabla\cdot T\)

Taken together, \(\nabla\cdot T=0\) and since all assignments are arbitrary, so is \(\nabla\cdot E=0\) and \(\nabla\cdot g=0\).

Unless,

\(-\nabla\cdot T\ne \nabla\cdot -T \)

and that,

\(\nabla\cdot E=\nabla\cdot (-g)\)

similarly,

\(\nabla\cdot T=\nabla\cdot (-E)\)

and,

\(\nabla\cdot g=\nabla\cdot (-T)\)

that the back (negative) end of a vector spinning is not the negative of the front (positive) end of the same vector spinning.

With this in mind,

where a negative charge particle spins (blue lines) to generate a temperature field, but a positive charge particle spins (red lines) to generate a gravity field.  A reversed spin generates the same field in the reversed direction.

Do we collapse every atom in the universe this way?

Inside, Outside, Small And Big

We have,

and

When we consider the direction of the \(E\) field outside of the particle, deformed at high velocity, flowing electrons generate photons and flowing protons generate electromagnetic waves.

When we consider the direction of the \(E\) field inside of the particle, deformed at high velocity, flowing electrons generate electromagnetic waves and flowing protons generate photons.

Which is which?

Do electrons or protons flow in a electric current?

In could be argued that when charge carriers collide with the conductor lattice that smaller particles are generated and photons are emitted by the flow of smaller electron.

But this leaves the question: do electrons or protons flow in a copper wire?

银树冰枝折腰断
一剪寒梅更耐冬
天地苍白冷彻时
朵朵醉晕赤热心

《红梅》

守诚得信迎新年
忠义阔怀安国邦
风适日丽昭太平
花木欣荣大地春

《2018 戊戌。狗年》

Copper Antenna Bias

From the post "Most Penetrating X ray, Boron" dated 20 Jan 2018,

\(V_{min}=\cfrac{1}{2}\cfrac{M}{qN_A}\left(3.4354*\cfrac{density}{Z}\right)^2\)

for copper,

\(V_{min}=\cfrac{1}{2}*\cfrac{63.546}{6.0221*10^{23}*1.6021*10^{-19}*10^{3}}*\left(3.4354*\cfrac{8960}{29}\right)^2\)

What is this good for?

If basic particles are created at the end of \(0.371\,V\) per meter of copper, then as an antenna, coupled to a signal source at a DC bias of \(V=0.371\,V\), that end will have an accumulation of charged particles.  Is it more effective as an antenna for both transmitting and receiving?

Note:  Bare copper is expected to oxidize quickly under the same condition.  Maybe the copper can be coated with a suitable resins to prevent oxidation.

Electron Cloud Copper

From the previous post "Most Penetrating X ray, Boron" dated 20 Jan 2018,

\(A_x=1.1974*16.78*10^{-9}*\cfrac{0.1}{3.710}=5.41\text{e-10}\)

If we assume that this is due to a particle of radius \(a_{\psi}\) we have,

\(a^2_{\psi}=\cfrac{A_x}{\pi}=1.7239\text{e-10}\,m\)

\(a_{\psi}=1.313\text{e-5}\,m\)

This could be the size of the electron cloud of a copper atom.

This is wrong because a length \(0.1\,m\) was used, it should be,

\(A_x=1.1974*16.78*10^{-9}*\cfrac{1}{0.371}=5.41\text{e-8}\)

where a length of \(1\,m\) is used,

\(a_{\psi}=1.313\text{e-4}\,m\)

This is the effective size of the electric field around a copper atom and not the radius of the electron orbits around the nucleus, in the electron cloud.

Most Penetrating X ray, Boron

With Boron \(B\),

\(v_{boom}=3.4354*\cfrac{density}{Z}=3.4354*\cfrac{2340}{5}=1607.80\,ms^{-1}\)

\(S_{p\,B}=\cfrac{1607.8}{1061.42}*0.2=0.303\,m\)

This is the most penetrating X ray using Boron \(B\), at \(30\,cm\).

\(V_{min}=\cfrac{1}{2}\cfrac{M}{qN_A}\left(3.4354*\cfrac{density}{Z}\right)^2\)

\(V_{min}=\cfrac{1}{2}*\cfrac{10.806}{6.0221*10^{23}*1.6021*10^{-19}*10^{3}}*\left(3.4354*\cfrac{2340}{5}\right)^2\)

\(V_{min}=0.1445\,V\)

For a length of \(d=10\,cm\),

\(v_{min}=\cfrac{0.1445}{0.1}=1.445\,V\)

Compare this to copper \(Cu\) of density \(8960\,kgm^{-3}\),

\(V_{min}=\cfrac{1}{2}*\cfrac{63.546}{6.0221*10^{23}*1.6021*10^{-19}*10^{3}}*\left(3.4354*\cfrac{8960}{29}\right)^2\)

\(V_{min}=0.3710\,V\)

For a length of \(d=10\,cm\),

\(v_{min}=\cfrac{0.371}{0.1}=3.71\,V\)

Boron requires less voltage.

Both \(v_{min}\) are resonance value, only at exactly these values are X rays emitted.

The value of \(V_{min}\) for copper however, does not compare with the post "X Ray, Inner Electron Cloud And Just As Shocking TWO" dated 15 Oct 2017, where,

\(I_{res}=1.1974 A\)

because the electric resistance of copper is very low at \(16.78 nΩ·m\). 

Given a voltage of \(3.71\,V\), the current need not be at \(I_{res}=1.1974\,A\) unless the cross sectional area of the wire is at,

\(V=3.71=1.1974*16.78*10^{-9}*\cfrac{0.1}{A_x}\)

\(A_x=1.1974*16.78*10^{-9}*\cfrac{0.1}{3.71}=5.41\text{e-10}\,m^2\)

The electron cloud at resonance and colliding particles at \(v_{boom}\) are different phenomenon.  It may not be correct to associate them.

Then again \(v_{min}\) for a solid has never been verified.  How much free space is there, within a solid, to accommodate basic particles?

Good night.

Probing X rays

There is a problem, if

\(\Delta t=\cfrac{v}{c}-1\)

when \(v=0\)

\(\Delta t=-1\)

when \(v=c\)

\(\Delta t=0\)

However near the equator, earth's motion is about \(460\,ms^{-1}\).  In one second, earth has moved \(460\,m\).  This means collapsed photon returning from one second in the future is \(460\,m\) away from the present position.  The argument here for X ray penetration is that the photons materialize inside the object just in front of the X ray source, through a space of about \(0.2\,m\).  This requires,

\(\Delta t=\cfrac{0.2}{460}=4.35\text{e-4}\,s\)

This requires \(v\) to be nearer to \(c\).

We should have used,

\(\Delta t=\cfrac{v}{c}\)

instead, taking \(c\) as reference,

\(\cfrac{S_{p\,Cu}}{S_{p\,H_2}}=\cfrac{\cfrac{v_{boom\,Cu}}{c}}{\cfrac{v_{boom\,H_2}}{c}}\)

\(S_{p\,H_2}=\cfrac{v_{boom\,H_2}}{v_{boom\,Cu}}*S_{p\,Cu}\)

\(S_{p\,H_2}=\cfrac{0.154}{1061.42}*0.2=2.902\text{e-5}\,m\)

X ray using Hydrogen gas (\(H_2\)), penetrates only \(30\,\mu m\).  Either a very thin anode of less than thirty micron is used followed immediately by the object elements, or another element should be used to generate X ray in the posts "Probing The Nucleus" and "Probing Sideways" both dated 19 Jan 2018.

Better yet, without an anode at all.  A thin object film placed at \(S_p\) (from the post "Quantifying Time Travel Backwards" dated 20 Jan 2018), without an intervening anode.

Quantifying Time Travel Backwards

If torus photons find their way into an object via time travel backward in time, then the penetrating power of X rays is the photons' ability to bring surface features out of the object along the direction of incidence.  The object's material absorption coefficient applies only on the trip out of the object.

For copper,

\(v_{boom\,Cu}=3.4354*\cfrac{8960}{29}=1061.42\,ms^{-1}\)

For Hydrogen \(H_2\),

\(v_{boom\,H_2}=3.4354*\cfrac{0.08988}{1*2}=0.154\,ms^{-1}\)

If \(\Delta t=\cfrac{v}{c}-1\)  and

\(\cfrac{S_{p\,Cu}}{S_{p\,H_2}}=\cfrac{\cfrac{v_{boom\,Cu}}{c}-1}{\cfrac{v_{boom\,H_2}}{c}-1}\)

where \(S_{p\,Cu}\) is the penetrating depth due X rays from copper...

\(S_{p\,H_2}=S_{p\,Cu}*\cfrac{\cfrac{v_{boom\,H_2}}{c}-1}{\cfrac{v_{boom\,Cu}}{c}-1}\)

We estimate \(S_{p\,Cu}\approx 20\,cm\),

\(S_{p\,H_2}=0.2*\cfrac{\cfrac{0.154}{299792458}-1}{\cfrac{1061.42}{299792458}-1}\)

\(S_{p\,H_2}==0.2000007\)

X ray due to \(H_2\) at \(v_{boom}\) is only slightly more penetrating than X ray due to \(Cu\) at \(v_{boom}\).  An increase in value for \(S_p\) because the term,

\(\left|\cfrac{v_{boom\,H_2}}{c}-1\right|\gt\left|\cfrac{v_{boom\,Cu}}{c}-1\right|\)

since,

\(v_{boom\,H_2}\lt v _{boom\,Cu}\lt c\)

but the high value of \(c\) makes the increment insignificant.

It is best to measure \(S_{p}\) directly.  In the setup below,

\(S_p\) is varied as the detector, at a fixed distance \(D\) from the metal plates with surface features sandwiched in the middle, detects a clear image of the interior of the plates.  If X ray penetration is due to a spread of torus photons around \(v_{boom}\), the detector will see the hidden surface features in the metal plates, over a range of values \(S_{p}\).  Increasing the detector distance \(D\) narrows this range.  The plates are sufficiently thin to allow the interior features to be detected by X ray.

Then,

 \(S_p=\cfrac{S_{p\,1}+S_{p\,2}}{2}\)

A series of \(D\) values would narrows the variance in \(S_{p}\).

Maybe then the expression,

\(\Delta t=\cfrac{v}{c}-1\)

\(v=v_{boom}\)

can be verified, after quantifying Earth's movement.

Probing Sideways

This is the setup from the post "Probing The Nucleus" dated 19 Jan 2018, sideways,

where the target element is lowered to accommodate Earth's drop in trajectory, in the eastward direction, on a level surface.

Probing The Nucleus

The element (\(H_2\)  or  \(He\)) in the discharge tube produces the smallest particle, \(a_{\psi\,c}\),



The target nucleus is stable, possibly Tungsten (\(Z\)) or Osmium (\(Os\)).  The injected particle returns from a future position inside the target element.

If the setup is at the nano-scale, the injected particle may find itself inside the nucleus of the target element and so provide information from inside the nucleus.

The target element maybe unstable and so provide information of many nuclear processes.

 Good morning.

Light Dispersion And Particles II

The dispersion of white light in the post "Light Dispersion And Particles" dated 17 Jan 2018 is wrong,

because \(a_{\psi\,ne}\) cannot be located within the spread \(a_{\psi\,c}\le a_{\psi\,ne}\le a_{\psi\,\pi}\). 

Unless the refracted size of \(a_{\psi\,\pi}\), denoted as \(\left(a_{\psi\,\pi}\right)_r\),

 \(\left(a_{\psi\,\pi}\right)_r\lt a_{\psi\,ne}\)

Then obviously,

where the refracted rays are correctly denoted as \(\left(a_{\psi}\right)_r\).

X Ray Scope

A torus photon being a dipole travelling in the direction opposite to its \(E\) field through its center may be focused using a positive temperature particle current carrying, tapped coiling with an \(E\) field in the direction of sight,

The focus is at the apex of the cone formed by the tapped coil.  Since the photons does not collapse at the surface of the coils as they are guided to the focus, only the detector at the focus need to adjust for \(\theta_s\) mentioned in the post "Hubble, Huba, Huba" dated 17 Jan 2018.

As board-spectrum X ray is expected to penetrate differently, \(-\Delta t\) varies for different bands of X rays.  There cannot be one adjustment to \(\theta_s\) across the spectrum.  A layered detector, that enables a composite image along a range of \(\theta_s\) using post image processing is shown below,

All collapsed photons from a narrow band of X rays will impact along the red line defined by \(\theta_s\) across fine layers of detection medium.  A composite image of that band of X rays is formed by picking all the pixels that lie on the red line.

\(\theta_s\) is defined by the movement of the detector during \(\Delta t\) after a torus photon collapses at the detector.  \(\Delta t\) is the duration the collapsed photon travels back in time from a future position.

Good night.

Note: An example of a positive temperature particle carrying current, hot water.

Nemo Of Many Sizes

With,

\(a_{\psi\,ne}=7.321\text{e-8}\,m\)

we have also,

\(a_{\psi\,c}=4.884\text{e-8}\,m\)

\(a_{\psi\,\pi}=10.98\text{e-8}\,m\)

Which match none of the values from the post "Sizing Them Up" dated 13 Dec 2014,

\(a_{\psi}=19.34\,nm\)

\(a_{\psi}=16.32\,nm\)

\(a_{\psi}=15.48\,nm\)

\(a_{\psi}=14.77\,nm\)

What happened?  The hydrogen spectrum is not the sun spectrum.  Particles from the sun is different from particles from hydrogen on Earth.  A notion that would make a universal \(a_{\psi\,c}\) size impossible.

Is there a universal \(a_{\psi\,c}\) value?

Finding Nemo

Finding \(a_{\psi\,ne}\) in light dispersion...

still looking...

where's Nemo?

My guess is,

\(\lambda_{ne}=460\,nm\),  which is also the solar irradiance peak.

 \(a_{\psi\,ne}=\cfrac{\lambda_{ne}}{2\pi}=\cfrac{460*10^{-9}}{2\pi}=7.321\text{e-8}\,m\)

 \(a_{\psi\,ne}=7.321\text{e-8}\,m\)

Have a sunny day.

Light Dispersion And Particles

Another guess,

which reinforce the idea that \(a_{\psi\,ne}\) is invisible or transparent.

\(a_{\psi\,ne}\) shatters at the medium boundary.  Some particles recombine and give \(a_{\psi}\ge a_{\psi\,ne}\) up till \(a_{\psi}\le a_{\psi\,\pi}\)

Since, particles of size \(a_{\psi}\ge a_{\psi\,ne}\) require recombination to form, they are expected to be of lower number density (lights of lower intensity).

...anymore guesses and physics is a TV game show...

More X Rays

Since X ray refracts away from the normal,

\(a_{\psi\,n}\gt a_{\psi\,s}\)

So, the approximation that the kinetic energy \(\small{\cfrac{1}{2}mv^2_{boom}}\) is stored solely along \(a_{\psi\,l}\) does not hold.  Part of energy extending \(a_{\psi\,l}\) leads to the shortening of \(a_{\psi\,n}\) to \(a_{\psi\,s}\).

But as a whole, the torus can still be modeled as a totally elastic body characterized by the equivalent of a spring constant \(k\).  \(a_{\psi\,s}\) is the bore of the spring contracting as the spring lengthens.

The torus travels,

perpendicular to the plane of the torus, in the direction opposite to the field (\(E\)) of the dipole.  This was derive from the model of photon as a dipole.

So, from \(a_{\psi\,s}\) and \(a_{\psi\,n}\) the respective the incident and refractive \(\lambda\)s are derived,

\(\lambda_i=2n\pi a_{\psi\,s}\)

and

\(\lambda_r=2n\pi a_{\psi\,n}\)

where \(n\) is the number of wavelengths along \(2\pi a_{\psi}\).  Most of the time, \(n=1\) and is not expected to change across the refracting medium boundary.

Is X ray visible after collapse?  That depends on \(a_{\psi\,n}\), the size of \(a_{\psi}\) after collapse.  Is \(a_{\psi\,n}\) visible?

Which leads to normal color lights...

Vita Rays

To set hydrogen gas \(H_2\) into discharge,

\(V_{min}=\cfrac{1}{2}\cfrac{M}{qN_A}\left(3.4354*{density}*{Z}\right)^2\)

\(M\)is molar mass and \(N_A\), Avogadro's number (post "Mental Discharge" dated 26 Dec 2017) from which,

\(V_{min}=\cfrac{1}{2}*\cfrac{2.016}{6.0221*10^{23}*1.6021*10^{-19}*10^3}*\left(3.4354*{0.0899}*{2}\right)^2\)

\(V_{min}=3.98*10^{-9}\,V\)

with a gap width of \(d=12\,\mu m\)

\(v_{min}=\cfrac{V_{min}}{d}=3.32*10^{-4}\,V\)

To set carbon \(C\), \(Z=6\) of density \(1346\,kgm^{-3}\) into discharge,

\(V_{min}=\cfrac{1}{2}\cfrac{M}{qN_A}\left(3.4354*\cfrac{density}{Z}\right)^2\)

In this case, particles move and collide into carbon atoms.

\(V_{min}=\cfrac{1}{2}*\cfrac{12.0096}{6.0221*10^{23}*1.6021*10^{-19}*10^3}*\left(3.4354*\cfrac{1346}{6}\right)^2\)

\(V_{min}=3.70*10^{-2}\,V\)

For a electrode gap of \(d=10\,cm\),

\(v_{min}=\cfrac{V_{min}}{d}=0.370\,V\)

Discharge tube of these elements, carbon and hydrogen gas produces torus photons that penetrate into matters,

Are such photons rejuvenating?  Or do they promote cell damage instead?

This is totally irresponsible.

Hubble, Huba, Huba

In the case of X-ray, Snell's Law still applies when the movement of the medium through time (as Earth move) is such that the medium boundaries before and after the movement are parallel.

where \(v_s\) is in the direction of Earth's movement.

Because \(\lambda_r\gt\lambda_i\), x ray blends away from the normal.

When \(v_s\) does not leave the medium boundaries parallel as they move in time,

the incident and reflected rays does not obey Snell's Law.

The angle of refraction is rotated as the medium boundary is rotated (\(\theta_s\)) with respect to the old position of the medium boundary back in time.

The photon upon its collapse, travels back in time, inside the medium (in this case) at a position its contact point at the boundary will be in the future.

In the diagram below \(S=v_s*\Delta t\) is the movement of the satellite in time \(\Delta t\).  The major axis of the satellite is rotated by \(\theta_s\) as a result of the movement.

\(\theta_s\) must be added to all calculated refraction angle, glancing or otherwise.  All surface normals are adjusted for this \(\theta_s\) that is added to the calculated refracted angle as the result of photons travelling back in time upon collapse on the surface of the refracting medium.

Good night.

Note:  Displacement does not alter orientation, rotation changes orientation.

A Hot Pendulum

If the length \(12.42\,cm\) prove to be to short for a sustained swing,

\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{g}{4l}}=\cfrac{1}{2}*\cfrac{1}{2\pi}\sqrt{\cfrac{g}{l}}=\cfrac{\sqrt{2}}{2}\)

a length of \(4*l=49.68\,cm\) at half the required frequency might swing a bit longer.

Maybe...

Glowing Pendulum

A transparent glow paint pendulum,

I don't know whether it will glow.

Color Spectrum And Particle Size

If we make a guess at the color spectrum as due to particles of increasing size, and align the blue region with small \(a_{\psi\,c}\) particles,

The ultraviolet region ends abruptly with the smallest \(a_{\psi\,c}\) but the infra-red region extends indefinitely with increasing \(a_{\psi\,\pi}\) that can exist in  high intensity fields (high temperature, high acceleration, high voltage).

And neutral \(a_{\psi\,ne}\) is without a charge may just be invisible.

Cracking Balls

If this is the torus photon,

we can approximate,

\(E=hf=h*\cfrac{2\pi a_{\psi\,l}}{v}=\cfrac{1}{2}mv^2_{boom}\)

that all energy due to \(v_{boom}\) is stored in \(a_{\psi\,l}\) and the initial energy before impact and also the energy of the photon upon collapse remains unchanged in \(a_{\psi\,s}\).

We are not going to get \(m\), because,

\(h=2\pi a_{\psi\,l}mc\)

So,

\(2\pi a_{\psi\,l}mc*\cfrac{2\pi a_{\psi\,l}}{v}=\cfrac{1}{2}mv^2_{boom}\)

\(v^2_{boom}=2(2\pi a_{\psi\,l})^2\cfrac{c}{v}\)

\(v_{boom}=\sqrt{\cfrac{2c}{v}}*2\pi a_{\psi\,l}\)

or

\(a_{\psi,l}=\cfrac{v_{boom}}{2\pi\sqrt{2}}*\sqrt{\cfrac{v}{c}}\)

We may also define a \(k_{\psi}\) such that,

\(\cfrac{1}{2}ka^2_{\psi\,l}=\cfrac{1}{2}mv^2_{boom}=h*\cfrac{2\pi a_{\psi\,l}}{v}=(2\pi a_{\psi\,l})^2m*\cfrac{c}{v}\)

\(k=8\pi^2\cfrac{c}{v}m\)

as if the photon is a spring extended to accommodate and store the energy in \(v_{boom}\).  Naturally,

\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{k}{m}}=\cfrac{1}{2\pi}\sqrt{8\pi^2\cfrac{c}{v}}=\sqrt{\cfrac{2c}{v}}\)

This means a collision frequency at \(f_{res}\) will stretch a torus photon along its larger radius.  Given \(E_p\) as the energy per photon, a power setting of,

\(P_{res}=f_{res}E_{p}\, Js^{-1}\)

will create a halo around the ray impacting a blocking medium.  Since \(E_p\) is very small, \(P_{res}\) is low, the effect of which is small.

\(f_{res}\) is the frequency to resonate a torus photon and may also be the frequency to grow a basic particle, because if the torus collapses,

\(a_{\psi\,l}\rightarrow a_{\psi\,s}=a_{\psi\,c}\)

\(v\rightarrow c\)

\(f_{res}\rightarrow \sqrt{2}\)

Given a simple string-ball pendulum,

\(f=\cfrac{1}{2\pi}\sqrt{\cfrac{g}{l}}=\sqrt{2}\)

\(l=\cfrac{g}{8\pi^{2}}\)

\(g=9.806\,ms^{-1}\)

\(l=0.1242\,m\)

Do the balls heat up, start to glow, become positively charged and explode?