星疏寥散没明天
檐漏风鞭无落处
虐溅瀑洒泥塘潽
《夜狂雨》
Saturday, December 30, 2017
Through The Mind's Eye
If \(\psi\) is conservative, that it does not lose energy impart onto it; that the deformation from a spherical particle to a torus as the result of \(v_{max}=v_{boom}\), stores the kinetic energy and that this energy is completely recovered when the particle collapses back into a sphere.
\(E_{torus}-E_{absorbed}\rightarrow E_{sphere}+\cfrac{1}{2}m_av^2_{boom}\)
where \(-E_{absorbed}\) is the energy absorbed by the photon when it collapses.
If the torus is at light speed, \(v_{boom}\) will kick the particle beyond light speed. It travels back in time. As far as our perception is concerned, the particle appears first at a distance, with \(v_{boom}\) speed (as energy is conserved between space and the time dimension) causing other photon torus to be ejected. From there, the light emitted traces backwards to its point of origin.
If \(v_{boom}\) is slow, the light path can be seen growing at \(v_{boom}\) speed back to its origin. We are not observing the speed of light, but the emitted light (photon torus) from colliding particles at \(v_{boom}\) that returned from the time dimension, at a time before its emergence from the source. The path of light traces the movement of the source as Earth moved.
Given,
\(v^{ 2 }_{ { max } }=\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\} e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }\)
When
\( \psi _{ n }-\psi _{ max }\lt0\)
\( \psi _{ n }\lt\psi _{ max }\)
\(v^{ 2 }_{ { max } }\lt 0\)
Since we need the remaining particle to repel the ejected torus, \(a_{\psi\,n}=a_{\psi\,ne}\) is the best candidate with \(\psi_{max}=a_{\psi\,\pi}\). If not \(a_{\psi\,ne}\lt a_{\psi\,n}\lt a_{\psi\,\pi}\), will also eject a photon torus.
The trick is to have \(a_{\psi\,ne}\) at \(v_{boom}\) speed when they are formed. In a discharge tube, such particles in the Faraday Dark Space is above \(v_{boom}\) speed. Furthermore, to observe the arc outside of the discharge tube, \(v_{boom}\) is of the air at atmospheric pressure and not of the low pressure gas.
Why an arc outside of the discharge tube? The particles time traveled there. The apparent displacement in space is the result of Earth's movement. The path of light observed traces the discharge tube movement as Earth moved.
Time travel at \(c+v_{boom}\) speed returns to the space dimension with \(v_{boom}\) speed with work done in the time dimension. A particle exist because of its speed in the time dimension at \(c\). The total energy in the time dimension due to this speed is \(E=mc^2\) (post "No Poetry for Einstein" dated 6 Apr 2016). When the particle goes back in time, it starts off a second line of existence in space that requires this energy of \(E=mc^2\) in the time dimension. So, the work required in the time dimension is \(E=mc^2\). Leaving this quantum of energy behind, at particle at speed \(c+v_{boom}\) exit the time dimension with speed \(v_{boom}\).
Where \(v_{boom}\) is rate of delivery of kinetic energy of particles at light speed that would trigger a boom. Boom generates more photons that are visible as the particles return to the space dimension. And we see this boom light from afar tracing backwards to the source of the particle. It is not the speed of light in reverse.
The path of the boom light traces the relative movement of the source through time as Earth moved. The particle enters into the time dimension just as it is ejected at the source. It appears earlier in time, before the source moved into place due to Earth's motion. On switching off the discharge, the boom light appears to return to the source as it traces moment less and less earlier to the present. So, a pulse train that switches on and off periodically will provide a persistent beam of light from a point (in space) forward in time, to the source at present.
Like a light saber but you cannot fight with it.
\(E_{torus}-E_{absorbed}\rightarrow E_{sphere}+\cfrac{1}{2}m_av^2_{boom}\)
where \(-E_{absorbed}\) is the energy absorbed by the photon when it collapses.
If the torus is at light speed, \(v_{boom}\) will kick the particle beyond light speed. It travels back in time. As far as our perception is concerned, the particle appears first at a distance, with \(v_{boom}\) speed (as energy is conserved between space and the time dimension) causing other photon torus to be ejected. From there, the light emitted traces backwards to its point of origin.
If \(v_{boom}\) is slow, the light path can be seen growing at \(v_{boom}\) speed back to its origin. We are not observing the speed of light, but the emitted light (photon torus) from colliding particles at \(v_{boom}\) that returned from the time dimension, at a time before its emergence from the source. The path of light traces the movement of the source as Earth moved.
Given,
\(v^{ 2 }_{ { max } }=\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\} e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }\)
When
\( \psi _{ n }-\psi _{ max }\lt0\)
\( \psi _{ n }\lt\psi _{ max }\)
\(v^{ 2 }_{ { max } }\lt 0\)
Since we need the remaining particle to repel the ejected torus, \(a_{\psi\,n}=a_{\psi\,ne}\) is the best candidate with \(\psi_{max}=a_{\psi\,\pi}\). If not \(a_{\psi\,ne}\lt a_{\psi\,n}\lt a_{\psi\,\pi}\), will also eject a photon torus.
The trick is to have \(a_{\psi\,ne}\) at \(v_{boom}\) speed when they are formed. In a discharge tube, such particles in the Faraday Dark Space is above \(v_{boom}\) speed. Furthermore, to observe the arc outside of the discharge tube, \(v_{boom}\) is of the air at atmospheric pressure and not of the low pressure gas.
Why an arc outside of the discharge tube? The particles time traveled there. The apparent displacement in space is the result of Earth's movement. The path of light observed traces the discharge tube movement as Earth moved.
Time travel at \(c+v_{boom}\) speed returns to the space dimension with \(v_{boom}\) speed with work done in the time dimension. A particle exist because of its speed in the time dimension at \(c\). The total energy in the time dimension due to this speed is \(E=mc^2\) (post "No Poetry for Einstein" dated 6 Apr 2016). When the particle goes back in time, it starts off a second line of existence in space that requires this energy of \(E=mc^2\) in the time dimension. So, the work required in the time dimension is \(E=mc^2\). Leaving this quantum of energy behind, at particle at speed \(c+v_{boom}\) exit the time dimension with speed \(v_{boom}\).
Where \(v_{boom}\) is rate of delivery of kinetic energy of particles at light speed that would trigger a boom. Boom generates more photons that are visible as the particles return to the space dimension. And we see this boom light from afar tracing backwards to the source of the particle. It is not the speed of light in reverse.
The path of the boom light traces the relative movement of the source through time as Earth moved. The particle enters into the time dimension just as it is ejected at the source. It appears earlier in time, before the source moved into place due to Earth's motion. On switching off the discharge, the boom light appears to return to the source as it traces moment less and less earlier to the present. So, a pulse train that switches on and off periodically will provide a persistent beam of light from a point (in space) forward in time, to the source at present.
Like a light saber but you cannot fight with it.
Missing Post
The torus dipole is not the first dipole formed without two opposite charge, a deformed, egg shaped particle also present itself as a dipole (post "Obviously, The Cow Is Driving" dated 12 Nov 2017).
A post on an actual dipole disengaging but left only one particle is missing.
Can't continue along this thread if the post is missing.
A post on an actual dipole disengaging but left only one particle is missing.
Can't continue along this thread if the post is missing.
Friday, December 29, 2017
Animal And Plant Torture
Subject a squid or a prawn or any other marine life to a low negative electric field and see if they emit blue-violet colors.
Subject plant life to a similar negative electric field and observe their color change if any.
Don't electrocute anything, just a copper plate with a (low if in water) negative potential (DC) in the background and a camera in front.
What happened? Fried prawns? If so stir...
Subject plant life to a similar negative electric field and observe their color change if any.
Don't electrocute anything, just a copper plate with a (low if in water) negative potential (DC) in the background and a camera in front.
What happened? Fried prawns? If so stir...
Two Color World
If the torus is the photon,
it is ejected with zero velocity along the radial direction.
Only if the big particle and the photon is of the same charge, repulsive, does the photon gain velocity in the radial direction.
If the big particle is positive \(a_{\psi}\gt a_{\psi\,ne}\), then the photon has to be \(a_{\psi}\ge a_{\psi\,ne}\). Only after with some initial velocity, does the dipole spins and be self-propelling. (Post "Ladies and Gentlemen, We Have Light Speed!" 29 May 2014.)
A negative particle will not propel a ejected torus. The torus is likely to be reabsorbed.
...and we have a two color world; black and white.
How many shades of greys are in \(a_{\psi\,ne}\le a_{\psi\,\small{torus}}\le a_{\psi\,\pi}\) under normal conditions. Maybe \(a_{\psi\,\small{torus}}\gt a_{\psi\,ne}\) is in the infra-red. And ultra-violet is when \(a_{\psi\,\small{torus}}\lt a_{\psi\,ne}\) gains some initial speed by collisions or somehow.
...and we have colors.
it is ejected with zero velocity along the radial direction.
Only if the big particle and the photon is of the same charge, repulsive, does the photon gain velocity in the radial direction.
If the big particle is positive \(a_{\psi}\gt a_{\psi\,ne}\), then the photon has to be \(a_{\psi}\ge a_{\psi\,ne}\). Only after with some initial velocity, does the dipole spins and be self-propelling. (Post "Ladies and Gentlemen, We Have Light Speed!" 29 May 2014.)
A negative particle will not propel a ejected torus. The torus is likely to be reabsorbed.
...and we have a two color world; black and white.
How many shades of greys are in \(a_{\psi\,ne}\le a_{\psi\,\small{torus}}\le a_{\psi\,\pi}\) under normal conditions. Maybe \(a_{\psi\,\small{torus}}\gt a_{\psi\,ne}\) is in the infra-red. And ultra-violet is when \(a_{\psi\,\small{torus}}\lt a_{\psi\,ne}\) gains some initial speed by collisions or somehow.
...and we have colors.
Torus Photons
The torus particle emitted as a dipole could be the photon, except the common understanding of a photon is that it imparts energy and heats materials up; this photon however, on returning to its spherical shape absorbs energy.
Maybe, only initially does the photon collapse and absorb energy, the particle that results does the heating up afterwards. If the collapsed particle is of size \(a_{\psi\,c}\lt a_{\psi}\gt a_{\psi\,ne}\) then the material heats up as this negative particle coalesce into big positive particles. If the collapsed particle is \(a_{\psi}=a_{\psi\,ne}\), the material cools as energy is absorbed by the photon during its collapse from a torus. The neutral particles does not coalesce with big particle to form bigger particles. Without negative particles the material remains cool.
When a photon impacts a \(\psi_{max}\), it reduces \(\psi_{max}\) energy content as the photon collapses, the big particle grows into a bigger particle with lower energy content.
Photoelectric effect is kinetic, where the photons behave like particles delivers energy to the system at some input velocity and. ejects a particle at the same impact speed, \(v_{max}\).
The effects of additive color lights are on the color sensors in our eye. Energy is absorbed by the color photons and the particle embedded triggers a sense response for that color. Each color light produces a color particle.
Subtractive color is when the color photons are all absorbed but a narrow range of color photons are re-emitted. Re-emission is necessary because illumination is often not in the opposite direction to the view direction, unless a colored transparent slide with the illumination directly behind.
Maybe...
Note: Don't read this as if the gospels. These are just to note the possible inconsistencies if the torus is accepted as a photon. These notes are not proof of the photon being torus.
Maybe, only initially does the photon collapse and absorb energy, the particle that results does the heating up afterwards. If the collapsed particle is of size \(a_{\psi\,c}\lt a_{\psi}\gt a_{\psi\,ne}\) then the material heats up as this negative particle coalesce into big positive particles. If the collapsed particle is \(a_{\psi}=a_{\psi\,ne}\), the material cools as energy is absorbed by the photon during its collapse from a torus. The neutral particles does not coalesce with big particle to form bigger particles. Without negative particles the material remains cool.
When a photon impacts a \(\psi_{max}\), it reduces \(\psi_{max}\) energy content as the photon collapses, the big particle grows into a bigger particle with lower energy content.
Photoelectric effect is kinetic, where the photons behave like particles delivers energy to the system at some input velocity and. ejects a particle at the same impact speed, \(v_{max}\).
The effects of additive color lights are on the color sensors in our eye. Energy is absorbed by the color photons and the particle embedded triggers a sense response for that color. Each color light produces a color particle.
Subtractive color is when the color photons are all absorbed but a narrow range of color photons are re-emitted. Re-emission is necessary because illumination is often not in the opposite direction to the view direction, unless a colored transparent slide with the illumination directly behind.
Maybe...
Note: Don't read this as if the gospels. These are just to note the possible inconsistencies if the torus is accepted as a photon. These notes are not proof of the photon being torus.
What Donuts? Dipoles?
The ejected particle,
\(v^{ 2 }_{ { max } }=\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\} e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }\)
\(v^{ 2 }_{ { max } }\lt0\)
\(v_{ { max } }=i\sqrt{\left|\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\} e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }\right|}\)
from the post "Peanuts And Peanut Butter" dated 09 Dec 2017, is a torus shaped particle.
Does it remain a torus of total \(\psi=\psi_n\) or does it expand continuously and disperse into a mist of \(\psi\). Does the torus collapse into a spherical particle?
If this entity disperses into a mist then, all particles in this mist tend to positive charge as the mist coalesce with the particles, and all bigger particles are positive (except neutral at \(a_{\psi\,ne}\)).
If this particle collapses into a spherical particle, it will be with the absorption of energy, as \(\psi\) then goes around a smaller radius \(a_{\psi}\), at a higher frequency.
Both positive and negative fields are present around the torus and the field lines wrap around the particle through the hole in the middle.
This could be the smallest magnetic particle with both poles, or an electric dipole or a gravity dipole.
When the torus collapses, only one particle is left. The single particle can be positive or negative depending on its size. This is not a dipole formed from two opposite charges.
The absorption line from this entity when it collapses will not fit any of the absorption series. The formation of this torus may result in a separate emission of energy, in which case, the absorption line will not be seen. If the torus is ejected with less energy absorbed than when a spherical particle is ejected, ie. without a separate emission of energy, then the absorption line as the torus collapses can be seen. Also, the lower energy absorption line due to the impact that created the torus, can also be seen.
How big does the torus get?
Good night.
Note; What happen to the impact particle? The theory goes up to the point of impact in time, when \(\psi_n\) is ejected, \(\psi_d=\psi_n-\psi_{max}\) is oscillating in the big particle. What happens to the impact particle afterwards? Never say. Unknown.
It could coalesce with the big particle and the process emits a photon.
\(v^{ 2 }_{ { max } }=\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\} e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }\)
\(v^{ 2 }_{ { max } }\lt0\)
\(v_{ { max } }=i\sqrt{\left|\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\} e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }\right|}\)
from the post "Peanuts And Peanut Butter" dated 09 Dec 2017, is a torus shaped particle.
Does it remain a torus of total \(\psi=\psi_n\) or does it expand continuously and disperse into a mist of \(\psi\). Does the torus collapse into a spherical particle?
If this entity disperses into a mist then, all particles in this mist tend to positive charge as the mist coalesce with the particles, and all bigger particles are positive (except neutral at \(a_{\psi\,ne}\)).
If this particle collapses into a spherical particle, it will be with the absorption of energy, as \(\psi\) then goes around a smaller radius \(a_{\psi}\), at a higher frequency.
Both positive and negative fields are present around the torus and the field lines wrap around the particle through the hole in the middle.
This could be the smallest magnetic particle with both poles, or an electric dipole or a gravity dipole.
When the torus collapses, only one particle is left. The single particle can be positive or negative depending on its size. This is not a dipole formed from two opposite charges.
The absorption line from this entity when it collapses will not fit any of the absorption series. The formation of this torus may result in a separate emission of energy, in which case, the absorption line will not be seen. If the torus is ejected with less energy absorbed than when a spherical particle is ejected, ie. without a separate emission of energy, then the absorption line as the torus collapses can be seen. Also, the lower energy absorption line due to the impact that created the torus, can also be seen.
How big does the torus get?
Good night.
Note; What happen to the impact particle? The theory goes up to the point of impact in time, when \(\psi_n\) is ejected, \(\psi_d=\psi_n-\psi_{max}\) is oscillating in the big particle. What happens to the impact particle afterwards? Never say. Unknown.
It could coalesce with the big particle and the process emits a photon.
Not Involved Reaching Zero
If the post "To Infinity But Slightly Less..." dated 28 Dec 2017 is correct, that the extent of the fluid in the thermometer is due to temperature charge particles extending in the temperature field by \(r\) to \(V=0\). The temperature particles carry the fluid with them as they reach for \(V=0\). Then,
all thermometers, no matter the fluid type, will have the same extend at steady state, given the same temperature charge \(q_s\), provided that all fluids are free to move in the bores of the thermometers.
Essentially no material properties of the fluid are involve here. Not drag, not viscosity and not thermal expansion.
The fluid acts like a dye to mark the measurement, the imbued temperature particles gauge the temperature field and carry the dye along.
all thermometers, no matter the fluid type, will have the same extend at steady state, given the same temperature charge \(q_s\), provided that all fluids are free to move in the bores of the thermometers.
Essentially no material properties of the fluid are involve here. Not drag, not viscosity and not thermal expansion.
The fluid acts like a dye to mark the measurement, the imbued temperature particles gauge the temperature field and carry the dye along.
Glow Some More Glow
This explains the repeating patterns in the positive column,
Both discharges due to negative particles (blue) and positive particles (red) occurs in the column. Interval \(a\) and interval \(b\) are similar, where negative particle coalesce into neutral then positive particles. Interval \(a\) is longer than \(b\) because \(b\) occurs in the relatively more positive region closer to the anode and the discharge region at the cathode creates more negative charges than all subsequent blue regions.
The background red in the positive column is due to positive particles repelled by the anode and is returning to the cathode. Beyond \(v_{boom}\) speed the positive particles do not cause the low pressure gas in the tube to discharge. On collision with the cathode, the cathode glows.
Some negative particles that reach the anode causes it to glow with a blue tint.
Merry Christmas...
Note: A red glow is consistent with a larger sized positive particles that have lower energy. A blue glow is consistent with a smaller sized negative particles that have higher energy.
Both discharges due to negative particles (blue) and positive particles (red) occurs in the column. Interval \(a\) and interval \(b\) are similar, where negative particle coalesce into neutral then positive particles. Interval \(a\) is longer than \(b\) because \(b\) occurs in the relatively more positive region closer to the anode and the discharge region at the cathode creates more negative charges than all subsequent blue regions.
The background red in the positive column is due to positive particles repelled by the anode and is returning to the cathode. Beyond \(v_{boom}\) speed the positive particles do not cause the low pressure gas in the tube to discharge. On collision with the cathode, the cathode glows.
Some negative particles that reach the anode causes it to glow with a blue tint.
Merry Christmas...
Note: A red glow is consistent with a larger sized positive particles that have lower energy. A blue glow is consistent with a smaller sized negative particles that have higher energy.
Discharge Tube Color
This is a low pressure gas discharge,
The positive column is due to \(a_{\psi}\gt a_{\psi\,ne}\) positive particles
The Crookes' Dark Space is due to the time needed by the newly ejected \(a_{\psi\,c}\) to attain \(v_{boom}\) speed of the gas concealed. The glow beyond this dark space is due to discharge caused by the negative particles at \(v_{boom}\). More negative particles are generated during discharge. When the particles has speed beyond \(v_{boom}\) discharge stops.
At the same time \(a_{\psi\,c}\) coalesce into bigger particles.
The Faraday Dark Space is due to \(a_{\psi\,ne}\) not accelerated by any electric field and race through with a constant velocity. Their speed is higher than \(v_{boom}\) and does not cause discharge.
At the same time \(a_{\psi\,ne}\) coalesce into bigger particles.
Beyond Faraday Dark Space, \(a_{\psi}\gt a_{\psi\,ne}\), the particles are positive and is retarded by the positive anode (and the negative cathode). When their speeds reduce to \(v_{boom}\), they cause the gas to discharge (red region). Below \(v_{boom}\) speed discharge stops and a dark striation develops. This is just like the Faraday Dark Space but it developed under a positive field and is much shorter.
\(a_{\psi\,c}\) particles generated during the last discharge accelerate under the positive field, also cause discharge when they attain \(v_{boom}\) speed (bright region within the red region). But discharge stops when the particles are beyond \(v_{boom}\) speed. A dark striation develops, and \(a_{\psi\,c}\) particles ejected during the last discharge coalesce into positive particles, which are decelerated by electric field in the tube. And another discharge occurs at the end of the dark furrow when the positive particles have slowed to \(v_{boom}\) speed.
First discharge in the positive column (middle left in the diagram above) is due to positive particles. Beyond this bright strip, the positive particle are at speed lower than \(v_{boom}\) and does not cause the concealed gas to discharge. Negative \(a_{\psi\,c}\) particles generated during the discharge are the cause of subsequent discharges. These negative particles also coalesce with the positive particles and form bigger positive particles. For this reason their number reduces, and the subsequent discharges fade gradually, til they reach the anode. The positive particles are reversed near the anode and travels back to the cathode. On their way, they cause discharge when their speed reaches \(v_{boom}\) in the reverse direction (red region). Before attaining \(v_{boom}\) speed, a dark region occurs just before the anode.
Some negative particles will collide with the anode at high speed and causes it to glow (bright/blue region). In the same manner, some positive charge reaches the cathode and cause that to glow (red color). Since positive particles are the result of negative particles coalescence, they reach the cathode later and the glow at the cathode occurs later than the glow at the anode, as the discharge tube lights up from the cathode side.
The negative glow can be extended by reducing the voltage at the anode. The Faraday Dark Space correspondingly reduces, as the negative particle takes longer to go beyond \(v_{boom}\). The glow at the cathode, attributed to positive particles is also reduced as the number of positive particles in the tube is reduced. The tube lights up with one color
Positive particles exist as the result of coalesce without the need for ionization of the low pressure gas.
If discharge is due to one type of particles, then there should be only one color.
The positive column is due to \(a_{\psi}\gt a_{\psi\,ne}\) positive particles
The Crookes' Dark Space is due to the time needed by the newly ejected \(a_{\psi\,c}\) to attain \(v_{boom}\) speed of the gas concealed. The glow beyond this dark space is due to discharge caused by the negative particles at \(v_{boom}\). More negative particles are generated during discharge. When the particles has speed beyond \(v_{boom}\) discharge stops.
At the same time \(a_{\psi\,c}\) coalesce into bigger particles.
The Faraday Dark Space is due to \(a_{\psi\,ne}\) not accelerated by any electric field and race through with a constant velocity. Their speed is higher than \(v_{boom}\) and does not cause discharge.
At the same time \(a_{\psi\,ne}\) coalesce into bigger particles.
Beyond Faraday Dark Space, \(a_{\psi}\gt a_{\psi\,ne}\), the particles are positive and is retarded by the positive anode (and the negative cathode). When their speeds reduce to \(v_{boom}\), they cause the gas to discharge (red region). Below \(v_{boom}\) speed discharge stops and a dark striation develops. This is just like the Faraday Dark Space but it developed under a positive field and is much shorter.
\(a_{\psi\,c}\) particles generated during the last discharge accelerate under the positive field, also cause discharge when they attain \(v_{boom}\) speed (bright region within the red region). But discharge stops when the particles are beyond \(v_{boom}\) speed. A dark striation develops, and \(a_{\psi\,c}\) particles ejected during the last discharge coalesce into positive particles, which are decelerated by electric field in the tube. And another discharge occurs at the end of the dark furrow when the positive particles have slowed to \(v_{boom}\) speed.
First discharge in the positive column (middle left in the diagram above) is due to positive particles. Beyond this bright strip, the positive particle are at speed lower than \(v_{boom}\) and does not cause the concealed gas to discharge. Negative \(a_{\psi\,c}\) particles generated during the discharge are the cause of subsequent discharges. These negative particles also coalesce with the positive particles and form bigger positive particles. For this reason their number reduces, and the subsequent discharges fade gradually, til they reach the anode. The positive particles are reversed near the anode and travels back to the cathode. On their way, they cause discharge when their speed reaches \(v_{boom}\) in the reverse direction (red region). Before attaining \(v_{boom}\) speed, a dark region occurs just before the anode.
Some negative particles will collide with the anode at high speed and causes it to glow (bright/blue region). In the same manner, some positive charge reaches the cathode and cause that to glow (red color). Since positive particles are the result of negative particles coalescence, they reach the cathode later and the glow at the cathode occurs later than the glow at the anode, as the discharge tube lights up from the cathode side.
The negative glow can be extended by reducing the voltage at the anode. The Faraday Dark Space correspondingly reduces, as the negative particle takes longer to go beyond \(v_{boom}\). The glow at the cathode, attributed to positive particles is also reduced as the number of positive particles in the tube is reduced. The tube lights up with one color
Positive particles exist as the result of coalesce without the need for ionization of the low pressure gas.
If discharge is due to one type of particles, then there should be only one color.
Thursday, December 28, 2017
To Infinity But Slightly Less...
Maybe this is what happens inside a thermometer where the temperature charged material flows with the temperature field until \(V=0\),
from the previous post "Only If You Started Somewhere" dated 27 Dec 2017. In this case, \(r\ne\infty\) at \(V=k\Delta T=0\).
Note: Mercury has a thermal expansion of \(60.4\,\mu m\, (m^{-1}k^{-1})\) at \(25\,^oC\), that means a thermometer of one meter will expand by \(60.4 \,\mu m\) per Kelvin increase in temperature.
from the previous post "Only If You Started Somewhere" dated 27 Dec 2017. In this case, \(r\ne\infty\) at \(V=k\Delta T=0\).
Note: Mercury has a thermal expansion of \(60.4\,\mu m\, (m^{-1}k^{-1})\) at \(25\,^oC\), that means a thermometer of one meter will expand by \(60.4 \,\mu m\) per Kelvin increase in temperature.
Measuring \(T_k\)
From the previous post, "Only If You Started Somewhere" dated 27 Dec 2017,
in the case of,
when \(\Delta T=0\) ie. \(T=T_g\), \(r\) is a measure for \(T_k\) up to \(V=0\). \(r\) need not be at infinity, just
\(\left.V=0\right|_{r=r_o}\)
when ever \(V=0\) at \(r_o\).
in the case of,
\(\left.V=0\right|_{r=r_o}\)
when ever \(V=0\) at \(r_o\).
Wednesday, December 27, 2017
Only If You Started Somewhere
From the post "Match Making Kinetic Theory And Temperature Particles" dated 27 Dec 2017:
It should be,
\(kT_g=\cfrac{q_g}{\varepsilon_g *r_g}\)
where \(kT_g\) is in Joules per particle, and \(\cfrac{q_g}{\varepsilon_g *r_g}\) is potential (\(J\)) due to temperature charges on one particle.
Only if one started somewhere somehow, anywhere anyhow...
\(T_k=r*\cfrac{T}{T_g}\)
where \(r=\cfrac{q_g}{\varepsilon_g k\Delta T}\) or
\(T_k=r_g*\cfrac{T}{\Delta T}\)
where \(r_g=\cfrac{q_g}{\varepsilon_g kT_g}\)
(T_k\) is made up of \(\cfrac{T}{\Delta T}\) up to the length \(r_g\) of a field projected by the total temperature charge per particle, up to a potential of \(V=kT_g\). Where \(T_g\) is the temperature of the gas when the temperature particles it contains, fill the gas evenly at a volume of \(V_m\), radius \(r=3\). \(T\) is the measured temperature at the surface, and \(\Delta T\) is the difference between the measured temperature \(T\) and \(T_g\).
If the temperature charge particles are free from the gas particles, \(T=T_g\) where the measured temperature is the same as the temperature of the gas just below the inner surface of the containment. \(T\) is independent of \(T_k\).
If the temperature charge particles are bounded to the gas particles and moves to the interior of the gas as the gas particles moves, then \(T\ne T_g\) and \(T_k\) is related as above.
At high temperature and high pressure it is likely that \(T\) is independent of \(T_k\) as more temperature particles are free from the gas particle. \(T_g\) then measures the temperature charge content of the gas, as in the case of a conductor imbued with electric charges.
\(T_g\equiv\cfrac{\rho_{e\,area}}{\varepsilon}=D\)
Have a nice day.
It should be,
\(kT_g=\cfrac{q_g}{\varepsilon_g *r_g}\)
where \(kT_g\) is in Joules per particle, and \(\cfrac{q_g}{\varepsilon_g *r_g}\) is potential (\(J\)) due to temperature charges on one particle.
Only if one started somewhere somehow, anywhere anyhow...
\(T_k=r*\cfrac{T}{T_g}\)
where \(r=\cfrac{q_g}{\varepsilon_g k\Delta T}\) or
\(T_k=r_g*\cfrac{T}{\Delta T}\)
where \(r_g=\cfrac{q_g}{\varepsilon_g kT_g}\)
(T_k\) is made up of \(\cfrac{T}{\Delta T}\) up to the length \(r_g\) of a field projected by the total temperature charge per particle, up to a potential of \(V=kT_g\). Where \(T_g\) is the temperature of the gas when the temperature particles it contains, fill the gas evenly at a volume of \(V_m\), radius \(r=3\). \(T\) is the measured temperature at the surface, and \(\Delta T\) is the difference between the measured temperature \(T\) and \(T_g\).
If the temperature charge particles are free from the gas particles, \(T=T_g\) where the measured temperature is the same as the temperature of the gas just below the inner surface of the containment. \(T\) is independent of \(T_k\).
If the temperature charge particles are bounded to the gas particles and moves to the interior of the gas as the gas particles moves, then \(T\ne T_g\) and \(T_k\) is related as above.
At high temperature and high pressure it is likely that \(T\) is independent of \(T_k\) as more temperature particles are free from the gas particle. \(T_g\) then measures the temperature charge content of the gas, as in the case of a conductor imbued with electric charges.
\(T_g\equiv\cfrac{\rho_{e\,area}}{\varepsilon}=D\)
Have a nice day.
Temperature Potential
What is,
\(T_k=r*\cfrac{T}{T_g}\)
\(r=\cfrac{q_g}{\varepsilon_gk\Delta T}\)
or
\(\Delta T=T-T_g=\cfrac{q_g}{\varepsilon_gk*r}\)
using \(V_m\), from the post "Otherwise An Idiot..." dated 28 Dec 2017. Does it make sense?
\(T_g=\cfrac{\rho_c}{\varepsilon_c}\ne\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\)
only for \(P\rightarrow \infty\) is
\(\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\)
Or should we consider instead,
\(T_k=r_g*\cfrac{T}{\Delta T}\)
\(r_g=\cfrac{q_g}{\varepsilon_gk T_g}\)
What is
\(T_g=\cfrac{q_g}{\varepsilon_gk *r_g}\)
??
Have a nice day?
\(T_k=r*\cfrac{T}{T_g}\)
\(r=\cfrac{q_g}{\varepsilon_gk\Delta T}\)
or
\(\Delta T=T-T_g=\cfrac{q_g}{\varepsilon_gk*r}\)
using \(V_m\), from the post "Otherwise An Idiot..." dated 28 Dec 2017. Does it make sense?
\(T_g=\cfrac{\rho_c}{\varepsilon_c}\ne\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\)
only for \(P\rightarrow \infty\) is
\(\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\)
Or should we consider instead,
\(T_k=r_g*\cfrac{T}{\Delta T}\)
\(r_g=\cfrac{q_g}{\varepsilon_gk T_g}\)
What is
\(T_g=\cfrac{q_g}{\varepsilon_gk *r_g}\)
??
Have a nice day?
The Price Of Three
...an idiot because,
\(T_k=r*\cfrac{T}{Tg}\)
where \(r=\cfrac{q_g}{\varepsilon_g k \Delta T}\)
when
\(T_g=\cfrac{\rho_g}{\varepsilon_g}=\cfrac{\rho_c}{\varepsilon_c}=T\)
as measured on the inner wall of the containment.
\(\cfrac{T}{T_g}=1\)
and so,
\(T_k=r\)
but,
\(\Delta T=T_g-T=0\)
and so,
\(T_k=\cfrac{q_g}{\varepsilon_g k \Delta T}\rightarrow \infty\)
What went wrong?
\(T=\cfrac{\rho_g}{\varepsilon_g}=\cfrac{\rho_c}{\varepsilon_c}\)
but
\(\cfrac{\rho_c}{\varepsilon_c}\ne\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\)
in the volume \(V_m\), the temperature charge did not all move to the surface. If they did, the surface density will be numerically the same as volume density. What then is the point of \(V_m\) or \(r=3\)?
\(T=\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_g}{\varepsilon_g}=\cfrac{rho_v}{\varepsilon_g}\)
as \(P\rightarrow \infty\)
such that \(T\) now measures the temperature of the volume \(V_m\) with a certain volume temperature charge density and not just its surface temperature charge.
It seems that pressure must still be high, and we can be off by a factor as much as \(3\). The change in size \(V\rightarrow V_m\) simply removes the factor \(\small{\cfrac{1}{3}}\) from the expression,
\(r=\cfrac{q_g}{3\varepsilon_g k \Delta T}\)
to obtain,
\(r=\cfrac{q_g}{\varepsilon_g k \Delta T}\)
the assumption of high pressure remains. The price of this \(3\) is a big volume.
Have a nice day.
\(T_k=r*\cfrac{T}{Tg}\)
where \(r=\cfrac{q_g}{\varepsilon_g k \Delta T}\)
when
\(T_g=\cfrac{\rho_g}{\varepsilon_g}=\cfrac{\rho_c}{\varepsilon_c}=T\)
as measured on the inner wall of the containment.
\(\cfrac{T}{T_g}=1\)
and so,
\(T_k=r\)
but,
\(\Delta T=T_g-T=0\)
and so,
\(T_k=\cfrac{q_g}{\varepsilon_g k \Delta T}\rightarrow \infty\)
What went wrong?
\(T=\cfrac{\rho_g}{\varepsilon_g}=\cfrac{\rho_c}{\varepsilon_c}\)
but
\(\cfrac{\rho_c}{\varepsilon_c}\ne\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\)
in the volume \(V_m\), the temperature charge did not all move to the surface. If they did, the surface density will be numerically the same as volume density. What then is the point of \(V_m\) or \(r=3\)?
\(T=\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_g}{\varepsilon_g}=\cfrac{rho_v}{\varepsilon_g}\)
as \(P\rightarrow \infty\)
such that \(T\) now measures the temperature of the volume \(V_m\) with a certain volume temperature charge density and not just its surface temperature charge.
It seems that pressure must still be high, and we can be off by a factor as much as \(3\). The change in size \(V\rightarrow V_m\) simply removes the factor \(\small{\cfrac{1}{3}}\) from the expression,
\(r=\cfrac{q_g}{3\varepsilon_g k \Delta T}\)
to obtain,
\(r=\cfrac{q_g}{\varepsilon_g k \Delta T}\)
the assumption of high pressure remains. The price of this \(3\) is a big volume.
Have a nice day.
Otherwise An Idiot...
From the previous post "Match Making Kinetic Theory And Temperature Particles" dated 28 Dec 2017, volume \(V\) is not explicit in the final expression for \(T_k\),
\(T_k=r*\cfrac{T}{T_g}\)
where \(r=\cfrac{q_g}{3\varepsilon_g k \Delta T}\)
It went into hiding underneath "density",
\(\rho_n=\cfrac{nN_A}{V}\)
as per unit volume, where its surface area, \(4\pi*1^2\) is three times its volume \(\small{\cfrac{4}{3}\pi* 1^3}\) so much so that the surface charge density when all the temperature charge is pushed by an infinite pressure to the surface is three time smaller than the volume charge density when all the temperature charge is evenly distributed throughout its volume.
However, if we consider a volume of radius \(r=3\) then,
\(V=\cfrac{4}{3}*\pi 3^3=4\pi*3^2=surface\,area\)
then both surface charge density and volume charge density will be the same. There is then no need for infinite pressure, nor a very high pressure approximation.
This \(V\)olume with \(r=3\) shall be called the Measurement Volume, \(V_m\)
Using this volume \(V_m\),
\(T_g=\cfrac{\rho_{\small{g\,P\rightarrow \infty}}}{\varepsilon_g}=\cfrac{\rho_n*q_g}{\varepsilon_g}\)
where,
\(\rho_n=\cfrac{nN_a}{V_m}=\cfrac{Q}{V_m}\)
is the number of temperature charges per volume evenly distributed, and \(q_g\) is the charge per gas particle, and \(Q\) is the total number of temperature charge in the gas of volume \(V_m\).
\(T_k=r*\cfrac{T}{T_g}\)
where \(r=\cfrac{q_g}{\varepsilon_g k \Delta T}\)
exactly, irrespective of the pressure of the gas, and its approximation to infinite value.
Or is it...?
\(T_k=r*\cfrac{T}{T_g}\)
where \(r=\cfrac{q_g}{3\varepsilon_g k \Delta T}\)
It went into hiding underneath "density",
\(\rho_n=\cfrac{nN_A}{V}\)
as per unit volume, where its surface area, \(4\pi*1^2\) is three times its volume \(\small{\cfrac{4}{3}\pi* 1^3}\) so much so that the surface charge density when all the temperature charge is pushed by an infinite pressure to the surface is three time smaller than the volume charge density when all the temperature charge is evenly distributed throughout its volume.
However, if we consider a volume of radius \(r=3\) then,
\(V=\cfrac{4}{3}*\pi 3^3=4\pi*3^2=surface\,area\)
then both surface charge density and volume charge density will be the same. There is then no need for infinite pressure, nor a very high pressure approximation.
This \(V\)olume with \(r=3\) shall be called the Measurement Volume, \(V_m\)
Using this volume \(V_m\),
\(T_g=\cfrac{\rho_{\small{g\,P\rightarrow \infty}}}{\varepsilon_g}=\cfrac{\rho_n*q_g}{\varepsilon_g}\)
where,
\(\rho_n=\cfrac{nN_a}{V_m}=\cfrac{Q}{V_m}\)
is the number of temperature charges per volume evenly distributed, and \(q_g\) is the charge per gas particle, and \(Q\) is the total number of temperature charge in the gas of volume \(V_m\).
\(T_k=r*\cfrac{T}{T_g}\)
where \(r=\cfrac{q_g}{\varepsilon_g k \Delta T}\)
exactly, irrespective of the pressure of the gas, and its approximation to infinite value.
Or is it...?
Match Making Kinetic Theory And Temperature Particles
How do Kinetic Theory of gasses and temperature particles coexist?
Consider a confinement of particles,
Given the surface density of temperature particle on the inner surface of the containment, collisions of the gas on the containment inner wall result in an exchange of temperature particles. The gas carries a certain amount of temperature particles and on the wall there is a certain amount of temperature particle.
The pressure of the gas dictates the efficiency of this temperature particle exchange.
There are more temperature particles with the gas than on the wall of the containment. Posted previously, temperature particles are caught in orbit around electron orbits (post "Not To Be taken Too Seriously, Please" dated 15 May 2016).
If the gas was a solid in thermal contact with the containment,
\(\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_g}{\varepsilon_g}\) --- (**)
where \(\rho\)'s are the respective temperature charge density, and \(\varepsilon\)'s are the equivalents of dielectric constants.
But the gas is not in full contact with the containment wall. The gas is in contact with the inner wall due to its pressure. Given infinite pressure, the frequency of collision is so great that it is as if all the gas particles are in contact with the inner wall.
\(\cfrac{nN_a}{A}*q_g=\rho_{\small{g,\,P\rightarrow \infty}}\)
\(n\) the number of moles of gas, \(N_A\) Avogadro's number, \(A\) area of the inner wall, \(q_g\) the number of temperature particles caught in electron orbits per particle of gas. And \(\rho_{\small{g\,P\rightarrow \infty}}\), the surface charge density of the gas when its pressure is infinite, hypothetically.
\(Q=nN_a*q_g\)
where \(Q\) is the total amount of temperature particles in the gas. \(q_{g}\) is the temperature of the gas particle. \(nN_A\) is the total number of gas particles in the containment.
When pressure is zero, hypothetically,
\(\rho_{\small{g,\,P=0}}=0\)
the charge density presented by the gas is also zero. It is as if the gas is not in contact with the inner wall.
What charge density is between zero pressure and infinite pressure?
We shall try,
\(\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\cfrac{P}{1+P}\) --- (+)
\(\cfrac{\rho_c}{\varepsilon_c}({1+P})=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}{P}\)
Consider a confinement of particles,
Given the surface density of temperature particle on the inner surface of the containment, collisions of the gas on the containment inner wall result in an exchange of temperature particles. The gas carries a certain amount of temperature particles and on the wall there is a certain amount of temperature particle.
The pressure of the gas dictates the efficiency of this temperature particle exchange.
There are more temperature particles with the gas than on the wall of the containment. Posted previously, temperature particles are caught in orbit around electron orbits (post "Not To Be taken Too Seriously, Please" dated 15 May 2016).
If the gas was a solid in thermal contact with the containment,
\(\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_g}{\varepsilon_g}\) --- (**)
where \(\rho\)'s are the respective temperature charge density, and \(\varepsilon\)'s are the equivalents of dielectric constants.
But the gas is not in full contact with the containment wall. The gas is in contact with the inner wall due to its pressure. Given infinite pressure, the frequency of collision is so great that it is as if all the gas particles are in contact with the inner wall.
\(\cfrac{nN_a}{A}*q_g=\rho_{\small{g,\,P\rightarrow \infty}}\)
\(n\) the number of moles of gas, \(N_A\) Avogadro's number, \(A\) area of the inner wall, \(q_g\) the number of temperature particles caught in electron orbits per particle of gas. And \(\rho_{\small{g\,P\rightarrow \infty}}\), the surface charge density of the gas when its pressure is infinite, hypothetically.
\(Q=nN_a*q_g\)
where \(Q\) is the total amount of temperature particles in the gas. \(q_{g}\) is the temperature of the gas particle. \(nN_A\) is the total number of gas particles in the containment.
When pressure is zero, hypothetically,
\(\rho_{\small{g,\,P=0}}=0\)
the charge density presented by the gas is also zero. It is as if the gas is not in contact with the inner wall.
What charge density is between zero pressure and infinite pressure?
We shall try,
\(\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\cfrac{P}{1+P}\) --- (+)
\(\cfrac{\rho_c}{\varepsilon_c}({1+P})=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}{P}\)
\(\cfrac{\rho_c}{\varepsilon_c}+\cfrac{\rho_c}{\varepsilon_c}P=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}{P}\)
we have,
\(T=\Delta T*P\) --- (*)
where \(\Delta T=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}-\cfrac{\rho_c}{\varepsilon_c}\) and
\(T=\cfrac{\rho_c}{\varepsilon_c}\)
this is temperature measured.
But \(P=\cfrac{nRT_k}{V}=\rho_nkT_k\)
where \(\rho_n=\cfrac{nN_A}{V}\) is the number density of the gas in the volume \(V\).
\(T=\Delta T*\rho_n*kT_k\)
where \(T\) is the temperature of the gas as measured, and \(T_k\) is the temperature of the gas calculated from Kinetic Theory of gases.
There is of course no reason for the expression
\(X=\cfrac{P}{1+P}\)
except that,
\(\left.X\right|_{P\rightarrow \infty}=1\)
\(\left.X\right|_{P=0}=0\)
and \(P\) is linear after expansion, in the expression (*).
Does \(T=\Delta T*\rho_n*kT_k\) make sense?
Let \(T_g=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\)
with
\(T=\cfrac{\rho_c}{\varepsilon_c}\)
\(T(1+\rho_n*kT_k)=T_g*\rho_n*kT_k\)
\(T=T_g*\cfrac{\rho_n*kT_k}{1+\rho_n*kT_k}\)
so, \(T_k\rightarrow \infty\)
\(T=T_g\)
which is the same as (**), as if the gas is a solid at full thermal contact with the containment inner wall.
What is \(T_g=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\)?
Consider,
\(\cfrac{n}{\cfrac{4}{3}\pi(1)^3}\)
and
\(\cfrac{n}{4\pi(1)^2}\)
for \(\cfrac{n}{\cfrac{4}{3}\pi(1)^3}\rightarrow\cfrac{n}{4\pi(1)^2} \)
we have to introduce a factor \(\cfrac{1}{3}\). ie,
\(3\rho_A=\rho_n\)
as \(P\rightarrow \infty\)
So,
\(T=\Delta T*\rho_n*kT_k=\Delta T*3\rho_A*kT_k\)
Since,
\(T_g=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}=\cfrac{\rho_A*q_g}{\varepsilon_g}\)
\(T=3\Delta T*T_g*\cfrac{\varepsilon_gkT_k}{q_g}\)
\(\cfrac{T}{T_g}=\Delta T*3\varepsilon_gk*\cfrac{T_k}{q_g}\)
where \(\Delta T=T_g-T\) is the result of assuming,
\(\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\cfrac{P}{1+P}\)
It is the result of how temperature is measured.
\(T_g\) is hypothetical but a constant given a volume of gas with a given amount of temperature particles.
\(3\varepsilon_gk\) is a constant, if \(\varepsilon_g\) exist.
If the gas is a solid, \(T_g\) is just the particle temperature of the gas given \(q_g\). \(q_g\) like electric charges on a conductor spread over the surface of the conductor. But \(T_k\) will not make sense, unless the temperature particles behave like a gas in the solid.
\(T_k = \cfrac{T}{T_g}*\cfrac{q_g}{3\varepsilon_gk\Delta T}\)
where \(q_g\) exert a field in space \(r\) and \(\Delta T\) is the temperature potential across the gas-containment boundary. Let,
\(\Delta T=\cfrac{q_g}{3\varepsilon_gk*r}\)
\(r\) is the distance moved to provide for a potential change of \(\Delta T\) in the presence of \(q_g\).
\({T_g}\)
is temperature, if the gas is a solid imbue with \(q_g\) amount of temperature particles per gas particle; a total amount of \(Q\) temperature charges.
So,
\(T_k=r*\cfrac{T}{T_g}\)
Kinetic temperature is the moment of the measured temperature per temperature as a solid.
What is \(3\varepsilon_gk\), where \(3\) has a unit of per meter?
Have nice day?
When Big Particle Collide
\(E=hf=h\cfrac{c}{\lambda}=h\cfrac{c}{2\pi a_{\psi}}\)
The bigger \(a_{\psi}\) has lesser energy, \(E\).
If the energy transition \(2n_e\rightarrow n_e + n_e\) results in the absorption line, then \(a_{\psi\,14.77}\) has also grow to \(2n_e\), even though unaffected by the high temperature high and voltage. It has grow big because of the high collision rate.
How did the time particle grow bigger by collisions?
From the post "No Solution But Exit Velocity Anyway" dated 14 Jul 2015,
\(v^{ 2 }_{ { max } }=\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\} e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }\)
When is \(v_{max}=0\)? When
\(\psi_n=\psi_{max}\)
two big particles collide and coalesce.
To suggest that \(\psi_{max}\) changes with temperature would be wrong, as only temperature particles are affected by the temperature field force. Big temperature particle are also more stable when the temperature field is high, for the difference in field potential that give raise to a pinch force, that pulls \(\psi\) away is less. A high surrounding potential will result in a negative (inward) pinch force that compresses \(\psi\).
For,
\((e^{ 2\psi _{ max } }-1)=0\)
\(\psi_{max}=0\)
this is a null answer, just as \(\psi_n=0\).
Note: The placid field remains.
The bigger \(a_{\psi}\) has lesser energy, \(E\).
If the energy transition \(2n_e\rightarrow n_e + n_e\) results in the absorption line, then \(a_{\psi\,14.77}\) has also grow to \(2n_e\), even though unaffected by the high temperature high and voltage. It has grow big because of the high collision rate.
How did the time particle grow bigger by collisions?
From the post "No Solution But Exit Velocity Anyway" dated 14 Jul 2015,
When is \(v_{max}=0\)? When
\(\psi_n=\psi_{max}\)
two big particles collide and coalesce.
To suggest that \(\psi_{max}\) changes with temperature would be wrong, as only temperature particles are affected by the temperature field force. Big temperature particle are also more stable when the temperature field is high, for the difference in field potential that give raise to a pinch force, that pulls \(\psi\) away is less. A high surrounding potential will result in a negative (inward) pinch force that compresses \(\psi\).
For,
\((e^{ 2\psi _{ max } }-1)=0\)
\(\psi_{max}=0\)
this is a null answer, just as \(\psi_n=0\).
Note: The placid field remains.
They Are All Photons
This is another interpretation of the set of data from the post "Sizing Them Up" dated 3 Dec 2014,
where \(a_{\psi}=16.32\,nm\) is the value of \(a_{\psi\,ne}\).
But if,
\((n_2=2n_e)\rightarrow (n_1=n_e)+n_e\)
gives one photon \(a_{\psi\,ne}\), then
\((n_3=3n_e)\rightarrow (n_2=2n_e)+n_e\)
also gives one photon \(a_{\psi\,ne}\). And both,
\((n_4=4n_e)\rightarrow (n_2=2n_e)+2n_e\)
\((n_2=3n_e)\rightarrow (n_1=2n_e)+n_e\)
give photon \(a_{\psi\,2ne}\).
What if they are all photons liberated, and we have,
where the smallest photon \(a_{\psi\,ne}\) has the highest energy.
\(a_{\psi}\) remains undetermined...
where \(a_{\psi}=16.32\,nm\) is the value of \(a_{\psi\,ne}\).
But if,
\((n_2=2n_e)\rightarrow (n_1=n_e)+n_e\)
gives one photon \(a_{\psi\,ne}\), then
\((n_3=3n_e)\rightarrow (n_2=2n_e)+n_e\)
also gives one photon \(a_{\psi\,ne}\). And both,
\((n_4=4n_e)\rightarrow (n_2=2n_e)+2n_e\)
\((n_2=3n_e)\rightarrow (n_1=2n_e)+n_e\)
give photon \(a_{\psi\,2ne}\).
What if they are all photons liberated, and we have,
where the smallest photon \(a_{\psi\,ne}\) has the highest energy.
\(a_{\psi}\) remains undetermined...
Tuesday, December 26, 2017
Misinterpreting More Data
If for the energy state transition \((n_2=2n_e)\rightarrow (n_1=n_e)\)
\(2n_e\rightarrow n_e + n_e\,\, (photon)\)
one of the resultant the particles \(a_{\psi\,ne}\) is the photon emitted, from the post "Good Morning Big Particles Split" dated 24 Dec 2017,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}=1.5\)
\(a_{\psi\,\pi}=1.5*a_{\psi\,ne}\)
and from the post "No Experimental Proof" dated 29 Jul 2016,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}}=\cfrac{cosh^{-1}(e)}{cosh^{-1}(e^{1/4})}=2.24921\)
we have,
\(a_{\psi\,c}=\cfrac{1.5}{2.24921}*a_{\psi\,ne}\)
and from the post "Sizing Them Up" dated 3 Dec 2014,
The first row of the table gives the values of \(a_{\psi\,ne}\), \(a_{\psi\,\pi}\) and, \(a_{\psi\,c}\) for temperature particle. This particle is identified solely base on the value of \(f_{ne}=2466067.5\,GHz\) resemblance to microwave heat frequency a million times smaller at \(2.4\,GHz\).
The rest of the rows, if they are due to other particles and not part of a spectral series for temperature particles, gives the relevant values for the other particles, yet identified specifically.
It is likely that this is just the temperature particle and its spectral series.
Have a nice day.
\(2n_e\rightarrow n_e + n_e\,\, (photon)\)
one of the resultant the particles \(a_{\psi\,ne}\) is the photon emitted, from the post "Good Morning Big Particles Split" dated 24 Dec 2017,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}=1.5\)
\(a_{\psi\,\pi}=1.5*a_{\psi\,ne}\)
and from the post "No Experimental Proof" dated 29 Jul 2016,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}}=\cfrac{cosh^{-1}(e)}{cosh^{-1}(e^{1/4})}=2.24921\)
we have,
\(a_{\psi\,c}=\cfrac{1.5}{2.24921}*a_{\psi\,ne}\)
and from the post "Sizing Them Up" dated 3 Dec 2014,
| \(a_{\psi\,ne}\) (nm) | \(\lambda_{ne}\) (nm) | \(f_{ne}\) (GHz) | \(a_{\psi\,\pi}\) (nm) | \(\lambda_{\pi}\) (nm) | \(f_{\pi}\) (GHz) | \(a_{\psi\,c}\) (nm) | \(\lambda_{c}\) (nm) | \(f_{c}\) (GHz) |
| 19.35 | 121.57 | 2466067.5 | 29.02 | 182.35 | 1644045.0 | 12.90 | 81.07 | 3697802.5 |
| 16.32 | 102.57 | 2922728.6 | 24.49 | 153.86 | 1948485.7 | 10.89 | 68.41 | 4382553.6 |
| 15.48 | 97.25 | 3082568.8 | 23.22 | 145.88 | 2055045.9 | 10.32 | 64.86 | 4622229.7 |
| 14.77 | 92.79 | 3230699.3 | 22.15 | 139.19 | 2153799.5 | 9.85 | 61.88 | 4844347.4 |
The first row of the table gives the values of \(a_{\psi\,ne}\), \(a_{\psi\,\pi}\) and, \(a_{\psi\,c}\) for temperature particle. This particle is identified solely base on the value of \(f_{ne}=2466067.5\,GHz\) resemblance to microwave heat frequency a million times smaller at \(2.4\,GHz\).
The rest of the rows, if they are due to other particles and not part of a spectral series for temperature particles, gives the relevant values for the other particles, yet identified specifically.
It is likely that this is just the temperature particle and its spectral series.
Have a nice day.
Mental Discharge
From the post "Checking Discharge Voltage" dated 23 Dec 2017,
\(V_{min}=\cfrac{1}{2}\cfrac{m_i}{q}\left(3.4354*\cfrac{density}{Z}\right)^2\)
\(m_i\) is the mass of one particle.
\(V_{min}=\cfrac{1}{2}\cfrac{M}{qN_A}\left(3.4354*\cfrac{density}{Z}\right)^2\)
\(M\) is molar mass and \(N_A\), Avogadro's number.
This set of data assumes that the noble gas atom with a charge of \(q\) is accelerated under the voltage \(V_{min}\) across a gap of \(d=1\,m\) to achieve the speed \(v_{boom}\).
This set of values for \(V_{min}\) is too low, a typical discharge voltage value for Argon, for a gap of \(12\,\mu m\) is \(137\,V\).
If we take the per atom view, instead of per particle view above,
\(V_{min}=\cfrac{1}{2}\cfrac{m_i}{q}\left(3.4354*{density}*{Z}\right)^2\)
\(V_{min}=\cfrac{1}{2}\cfrac{M}{qN_A}\left(3.4354*{density}*{Z}\right)^2\)
where we have taken that an atom with \(Z\) number of particles (of one type), with a charge of \(q\), is accelerated under \(V_{min}\) for a meter (\(d=1\,m\)) to achieve \(v_{boom}\) speed.
The last column is,
\(V=\cfrac{V_{min}}{d}\)
\(d=12\,\mu m\)
to obtain the value of the minimum discharge voltage \(V\) for a gap of \(12\,\mu m\). The minimum discharge voltage for a gap distance of \(12\,\mu m\) for Argon is \(182\,V\). This is higher than the quoted value of \(137\,V\). (A higher voltage is needed for a gap smaller than one meter, because the particle must reach \(v_{boom}\) before reaching the opposite electrode.)
Notice that to accelerate Helium \(He\) to the correct \(v_{boom}\) speed within a similar gap, the applied voltage is in the mini-voltage (mV) range. And for Neon \(Ne\), a few volts. Hydrogen gas, interestingly is discharged at \(\approx0.33\,mV\), which is at the cellular voltage level.
Good day!
Note: Vmin=0.5*(3.4354*density/Z)^2*M/(6.0221*10^23*1.6021*10^(-19)*10^3) for the first table.
Vmin=0.5*(3.4354*density*Z)^2*M/(6.0221*10^23*1.6021*10^(-19)*10^3) for the second table.
Note: Hydrogen gas was added to the last table on 29 Dec 2017.
\(V_{min}=\cfrac{1}{2}\cfrac{m_i}{q}\left(3.4354*\cfrac{density}{Z}\right)^2\)
\(m_i\) is the mass of one particle.
\(V_{min}=\cfrac{1}{2}\cfrac{M}{qN_A}\left(3.4354*\cfrac{density}{Z}\right)^2\)
\(M\) is molar mass and \(N_A\), Avogadro's number.
| Atomic No. | Name | Symbol | Mass(g/mol) | Density(g/L) | v_boom(m/s) | Vmin(V) |
| 2 | Helium | He | 4.002602 | 0.1785 | 0.3066 | 1.9501E-09 |
| 10 | Neon | Ne | 20.1797 | 0.8999 | 0.3092 | 9.9952E-09 |
| 18 | Argon | Ar | 39.948 | 1.7837 | 0.3404 | 2.3993E-08 |
| 36 | Krypton | Kr | 83.798 | 3.733 | 0.3562 | 5.5110E-08 |
| 54 | Xenon | Xe | 131.293 | 5.887 | 0.3745 | 9.5440E-08 |
| 86 | Radon | Rn | 222 | 9.73 | 0.3887 | 1.7381E-07 |
This set of data assumes that the noble gas atom with a charge of \(q\) is accelerated under the voltage \(V_{min}\) across a gap of \(d=1\,m\) to achieve the speed \(v_{boom}\).
This set of values for \(V_{min}\) is too low, a typical discharge voltage value for Argon, for a gap of \(12\,\mu m\) is \(137\,V\).
If we take the per atom view, instead of per particle view above,
\(V_{min}=\cfrac{1}{2}\cfrac{m_i}{q}\left(3.4354*{density}*{Z}\right)^2\)
\(V_{min}=\cfrac{1}{2}\cfrac{M}{qN_A}\left(3.4354*{density}*{Z}\right)^2\)
where we have taken that an atom with \(Z\) number of particles (of one type), with a charge of \(q\), is accelerated under \(V_{min}\) for a meter (\(d=1\,m\)) to achieve \(v_{boom}\) speed.
| Atomic No. | Name | Symbol | Mass \(gmol^{-1}\) | Density \(gL^{-1}\) | v_boom \(ms^{-1}\) | Vmin \(V\) | Vmin \(\div12\mu m\) |
| 1 | Hydrogen | H2 | 2.016 | 0.0899 | 0.1544 | 3.9844E-09 | 3.32E-04 |
| 2 | Helium | He | 4.002602 | 0.1785 | 0.3066 | 3.120E-08 | 2.60E-03 |
| 10 | Neon | Ne | 20.1797 | 0.8999 | 0.3092 | 9.995E-05 | 8.33E+00 |
| 18 | Argon | Ar | 39.948 | 1.661 | 0.3170 | 2.184E-03 | 1.82E+02 |
| 36 | Krypton | Kr | 83.798 | 3.733 | 0.3562 | 9.256E-02 | 7.71E+03 |
| 54 | Xenon | Xe | 131.293 | 5.887 | 0.3745 | 8.115E-01 | 6.76E+04 |
| 86 | Radon | Rn | 222 | 9.73 | 0.3887 | 9.507E+00 | 7.92E+05 |
The last column is,
\(V=\cfrac{V_{min}}{d}\)
\(d=12\,\mu m\)
to obtain the value of the minimum discharge voltage \(V\) for a gap of \(12\,\mu m\). The minimum discharge voltage for a gap distance of \(12\,\mu m\) for Argon is \(182\,V\). This is higher than the quoted value of \(137\,V\). (A higher voltage is needed for a gap smaller than one meter, because the particle must reach \(v_{boom}\) before reaching the opposite electrode.)
Notice that to accelerate Helium \(He\) to the correct \(v_{boom}\) speed within a similar gap, the applied voltage is in the mini-voltage (mV) range. And for Neon \(Ne\), a few volts. Hydrogen gas, interestingly is discharged at \(\approx0.33\,mV\), which is at the cellular voltage level.
Good day!
Note: Vmin=0.5*(3.4354*density/Z)^2*M/(6.0221*10^23*1.6021*10^(-19)*10^3) for the first table.
Vmin=0.5*(3.4354*density*Z)^2*M/(6.0221*10^23*1.6021*10^(-19)*10^3) for the second table.
Note: Hydrogen gas was added to the last table on 29 Dec 2017.
Charge Bodies Courtship
Maybe this is how electrons reside in a charged body,
The two like bodies are repulsive this way,
and two unlike bodies are attractive this way,
And when two negatively charged bodies are brought closer,
they attract each other and stick.
Maybe, only electrons are mobile,
a negatively charged body is the result of surface property. If this is possible, two of these will attract each other, initially. When brought closer, the surface charge flip and they repel as in the very first case. Normally, two negative charges are more attracted to each other than one positive and one negative charge pair, (\(-1.49824\) vs \(1\)).
If this is so, applying another dielectric (eg. alcohol vapor or other volatile oil) onto the bodies will change its charge attribute without discharge and recharge.
Good night...
and two unlike bodies are attractive this way,
And when two negatively charged bodies are brought closer,
they attract each other and stick.
a negatively charged body is the result of surface property. If this is possible, two of these will attract each other, initially. When brought closer, the surface charge flip and they repel as in the very first case. Normally, two negative charges are more attracted to each other than one positive and one negative charge pair, (\(-1.49824\) vs \(1\)).
If this is so, applying another dielectric (eg. alcohol vapor or other volatile oil) onto the bodies will change its charge attribute without discharge and recharge.
Good night...
Monday, December 25, 2017
Big Body VS Wave-Particles
What about the century old law "Like charges repel and unlike charges attract"?
Neutral particle, \(a_{\psi}=a_{\psi\,ne}\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,ne} }=\left.3 \cfrac { \partial V\, }{ \partial \, x } \right|_{x=a_{\psi\,\pi}}\)
Negative particle, \(a_{ \psi \,c }\lt a_{\psi}\lt a_{\psi\,ne}\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } -1.49824\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
Positive particle, \(a_{\psi\,ne}\lt a_{\psi} \lt a_{ \psi \,\pi }\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,\pi } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
Between the same charge, the \(\small{\cfrac { \partial T\, }{ \partial \, x }}\) are always in the opposite directions for a test particle placed in the middle the two particles. A positive \(\small{\cfrac { \partial T\, }{ \partial \, x }}\) increases \(KE\) as the test point moves away from the particle. The particles are driven apart as they move away from the test particle.
This component is in the same direction in the interaction between particle of opposite charges. A test placed in the middle is pulled closer to one of the particle and the particles move closer.
Between neutral and other charges, because of the singular \(\cfrac { \partial T\, }{ \partial \, x }\) component (only one direction as the case for opposite charges), it is always attractive. This was explained at a macro level with induced charges or surface charge polarization.
\(\psi\) is in circular motion. The change in \(T\) along its path is manifested as a force along the radial line, along the direction as the centripetal force. An increase in \(T\) is aided by a force in the same direction as the centripetal force, inwards. A decrease in \(T\) is a force opposite to the centripetal force. This force is outwards along the radial line. This reversal in sign is already accounted for in the derivation for \(\psi\) and \(\cfrac{dq}{dx}\).
A positive change in \(V\) will drive both particles closer as a test charge placed in the middle will experience a drop in potential if the particles move closer. A negative change in \(V\) will drive the interacting particles apart, as a test charge placed in the middle of these particles will experience a drop in potential if the particles moves apart. The particles, however do not move without a change in \(T\).
\(\cfrac { \partial \, T }{ \partial \, x }\) decides whether the particle move closer or apart, ie are attractive or repulsive
\(\cfrac { \partial \, V }{ \partial \, x }\) determines the change in \(V\) as the particle move relative to each other. It is likely that,
\(\cfrac { \partial V\, }{ \partial \, x }=0\)
because the force \(\cfrac{d\psi}{dx}=0\) on the plateau just beyond \(a_{\psi\,\pi}\). If,
\(\cfrac { \partial V\, }{ \partial \, x }=0\)
then there is no change in \(V\).
For \(\cfrac { \partial \, V }{ \partial \, x }\) positive, particles moving closer experience a drop in \(V\). For \(\cfrac { \partial \, V }{ \partial \, x }\) negative, particles moving apart experience a drop in \(V\). And vice-versa.
Furthermore, it does not matter which force comes from which particle acting on the test particle in the middle of the particles. It is possible to swap ownership of the forces as long the same forces still acts on the test particle.
In the case of two interacting negative charge particles, the two negative forces can swap ownership and be equivalent to the interaction of two positive particles with two positive forces. This is because,
\(\cfrac { \partial T\, }{ \partial \, x }\)
changes the energy (a scalar) of the test particle directly,
In a vector force, a pair of forces in tension is different from a pair of forces in compression.
What happen to \(a_{\psi\,c}\) particle coalescing? It is awkward here because we are using wave equation results to explain charged big body interactions. In big body interaction we can use test particle or test point between them and deduce the movement of the charge bodies from the movement of the test particle. Wave however, superimpose and interact directly.
As waves, the negative particles act on each other directly and force ownership swap cannot happen as the forces are vectors. Negative particles are attractive to each other because of the negative
\(\cfrac { \partial \, T }{ \partial \, x }\)
component. A positive
\(\cfrac { \partial \, T }{ \partial \, x }\)
As particles coalesce, \(a_{\psi\,ne}\) is neutral and beyond \(a_{\psi\,ne}\) positive.
"Like charges repel and unlike charges attract" is not true at the quantum level where waves dictate interactions. The use of a test particle or test point between the bodies accounts for the presence of an intervening medium between the charged bodies.
Have a nice day.
Neutral particle, \(a_{\psi}=a_{\psi\,ne}\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,ne} }=\left.3 \cfrac { \partial V\, }{ \partial \, x } \right|_{x=a_{\psi\,\pi}}\)
Negative particle, \(a_{ \psi \,c }\lt a_{\psi}\lt a_{\psi\,ne}\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } -1.49824\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
Positive particle, \(a_{\psi\,ne}\lt a_{\psi} \lt a_{ \psi \,\pi }\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,\pi } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
Between the same charge, the \(\small{\cfrac { \partial T\, }{ \partial \, x }}\) are always in the opposite directions for a test particle placed in the middle the two particles. A positive \(\small{\cfrac { \partial T\, }{ \partial \, x }}\) increases \(KE\) as the test point moves away from the particle. The particles are driven apart as they move away from the test particle.
This component is in the same direction in the interaction between particle of opposite charges. A test placed in the middle is pulled closer to one of the particle and the particles move closer.
Between neutral and other charges, because of the singular \(\cfrac { \partial T\, }{ \partial \, x }\) component (only one direction as the case for opposite charges), it is always attractive. This was explained at a macro level with induced charges or surface charge polarization.
\(\psi\) is in circular motion. The change in \(T\) along its path is manifested as a force along the radial line, along the direction as the centripetal force. An increase in \(T\) is aided by a force in the same direction as the centripetal force, inwards. A decrease in \(T\) is a force opposite to the centripetal force. This force is outwards along the radial line. This reversal in sign is already accounted for in the derivation for \(\psi\) and \(\cfrac{dq}{dx}\).
A positive change in \(V\) will drive both particles closer as a test charge placed in the middle will experience a drop in potential if the particles move closer. A negative change in \(V\) will drive the interacting particles apart, as a test charge placed in the middle of these particles will experience a drop in potential if the particles moves apart. The particles, however do not move without a change in \(T\).
\(\cfrac { \partial \, T }{ \partial \, x }\) decides whether the particle move closer or apart, ie are attractive or repulsive
\(\cfrac { \partial \, V }{ \partial \, x }\) determines the change in \(V\) as the particle move relative to each other. It is likely that,
\(\cfrac { \partial V\, }{ \partial \, x }=0\)
because the force \(\cfrac{d\psi}{dx}=0\) on the plateau just beyond \(a_{\psi\,\pi}\). If,
\(\cfrac { \partial V\, }{ \partial \, x }=0\)
then there is no change in \(V\).
For \(\cfrac { \partial \, V }{ \partial \, x }\) positive, particles moving closer experience a drop in \(V\). For \(\cfrac { \partial \, V }{ \partial \, x }\) negative, particles moving apart experience a drop in \(V\). And vice-versa.
Furthermore, it does not matter which force comes from which particle acting on the test particle in the middle of the particles. It is possible to swap ownership of the forces as long the same forces still acts on the test particle.
In the case of two interacting negative charge particles, the two negative forces can swap ownership and be equivalent to the interaction of two positive particles with two positive forces. This is because,
\(\cfrac { \partial T\, }{ \partial \, x }\)
changes the energy (a scalar) of the test particle directly,
In a vector force, a pair of forces in tension is different from a pair of forces in compression.
What happen to \(a_{\psi\,c}\) particle coalescing? It is awkward here because we are using wave equation results to explain charged big body interactions. In big body interaction we can use test particle or test point between them and deduce the movement of the charge bodies from the movement of the test particle. Wave however, superimpose and interact directly.
As waves, the negative particles act on each other directly and force ownership swap cannot happen as the forces are vectors. Negative particles are attractive to each other because of the negative
\(\cfrac { \partial \, T }{ \partial \, x }\)
component. A positive
\(\cfrac { \partial \, T }{ \partial \, x }\)
component drives the particles apart.
"Like charges repel and unlike charges attract" is not true at the quantum level where waves dictate interactions. The use of a test particle or test point between the bodies accounts for the presence of an intervening medium between the charged bodies.
Have a nice day.
Sunday, December 24, 2017
Flying Toaster Really?
Maybe, \(a_{\psi\,ne}\) are gravity particles that are attracted to negative particles as if by virtual of their mass. \(a_{\psi\,c}\) up to \(a_{\psi\,ne}\) are negative charge particles, and beyond \(a_{\psi\,ne}\) to \(a_{\psi\,\pi}\) are positive charge particles. Temperature are fragments of \(\psi\) free.
In this case, temperature can be made into particles of any type, positive, neutral and negative. And conversely, temperature cannot exist independent of particles. Gravity do not exist, as attractive forces are all electric in nature.
The existence of temperature as fragments of \(\psi\) increases all particle size, making them eventually all positive.
This postulate is without the need for, and motivation to find, gravity particles and temperature particles. Time is just a delusion.
Is the toaster really flying?
Note: Only between \(a_{\psi\,ne}\) is the interaction neutral. Between positive charge and \(a_{\psi\,ne}\) the force is still repulsive, as the
\(3 \cfrac { \partial V\, }{ \partial \, x }\)
component in
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,ne} }=\left.3 \cfrac { \partial V\, }{ \partial \, x } \right|_{x=a_{\psi\,\pi}}\)
and
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,\pi } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
moves in opposite directions, and the
\(+\cfrac { \partial \, T }{ \partial \, x }\)
component increases the distance between them.
In this case, temperature can be made into particles of any type, positive, neutral and negative. And conversely, temperature cannot exist independent of particles. Gravity do not exist, as attractive forces are all electric in nature.
The existence of temperature as fragments of \(\psi\) increases all particle size, making them eventually all positive.
This postulate is without the need for, and motivation to find, gravity particles and temperature particles. Time is just a delusion.
Is the toaster really flying?
Note: Only between \(a_{\psi\,ne}\) is the interaction neutral. Between positive charge and \(a_{\psi\,ne}\) the force is still repulsive, as the
\(3 \cfrac { \partial V\, }{ \partial \, x }\)
component in
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,ne} }=\left.3 \cfrac { \partial V\, }{ \partial \, x } \right|_{x=a_{\psi\,\pi}}\)
and
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,\pi } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
moves in opposite directions, and the
\(+\cfrac { \partial \, T }{ \partial \, x }\)
component increases the distance between them.
Charge Mystery
It is paradoxical that \(a_{\psi}\) grows bigger with higher energy conditions but a bigger particle has lower energy (\(\lambda=2\pi a_{\psi}\), \(E=hf\)). And that it is easier to pluck \(\psi\) from a bigger \(a_{\psi}\).
\(a_{\psi\,c}\) has the highest energy per particle and can be generated via collisions at \(v_{boom}\) velocities.
We should generate \(a_{\psi\,c}\) without high temperature by using \(v_{boom}\), and then subject the confinement of particles to, in turn, high voltage, high gravity and high temperature to enable \(a_{\psi\,2ne}\) of the corresponding charge nature. Maybe three different spectral series will be generated, and we have evidence for particles that respond to an electric field, a gravity field and a temperature field exclusively.
\(\psi\) does not innately have charge attributes in the theoretical treatment here, electric charge particles, gravity particles and temperature particles as \(\psi\) particles behave the same on the absorption spectra. It is still possible that all three types of particles give the same set of spectral lines.
The charge attributes remain a mystery.
\(a_{\psi\,c}\) has the highest energy per particle and can be generated via collisions at \(v_{boom}\) velocities.
We should generate \(a_{\psi\,c}\) without high temperature by using \(v_{boom}\), and then subject the confinement of particles to, in turn, high voltage, high gravity and high temperature to enable \(a_{\psi\,2ne}\) of the corresponding charge nature. Maybe three different spectral series will be generated, and we have evidence for particles that respond to an electric field, a gravity field and a temperature field exclusively.
\(\psi\) does not innately have charge attributes in the theoretical treatment here, electric charge particles, gravity particles and temperature particles as \(\psi\) particles behave the same on the absorption spectra. It is still possible that all three types of particles give the same set of spectral lines.
The charge attributes remain a mystery.
Back To Spectral Series
Could it be that,
and we are back to the spectral series interpretation of the data presented in the post "Sizing Them Up" dated 3 Dec 2014.
The spectral lines then, are not supportive evidence for the existence of three types of particles, electric charge, gravity and temperature; and the existence of time particles. Although the value \(f=2466067.5\,GHz\) is still indicative of temperature.
Maybe the property of electric charge, gravity and temperature depends on the size of \(a_{\psi}\). Which would lead to the conclusion that charge property changes as \(a_{\psi}\) grows.
Maybe, there are three types of particles simply because there is an electric field, a gravity field and a temperature field; and if we consider ourselves travelers in a time field, a time particle (at least one). Only one big time particle in existence does not rule out the possibility that we can use \(\psi\) of the time charge nature
Maybe...maybe...
and we are back to the spectral series interpretation of the data presented in the post "Sizing Them Up" dated 3 Dec 2014.
The spectral lines then, are not supportive evidence for the existence of three types of particles, electric charge, gravity and temperature; and the existence of time particles. Although the value \(f=2466067.5\,GHz\) is still indicative of temperature.
Maybe the property of electric charge, gravity and temperature depends on the size of \(a_{\psi}\). Which would lead to the conclusion that charge property changes as \(a_{\psi}\) grows.
Maybe, there are three types of particles simply because there is an electric field, a gravity field and a temperature field; and if we consider ourselves travelers in a time field, a time particle (at least one). Only one big time particle in existence does not rule out the possibility that we can use \(\psi\) of the time charge nature
Maybe...maybe...
Good Morning Big Particles Split
From the post "Small Negative, Big Positive" dated 24 Dec 2014,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}\lt 1.5\)
at the limit,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}=1.5\)
obviously,
\(a_{\psi\,ne}=\cfrac{2}{3}a_{\psi\,\pi}\)
and
\(2a_{\psi\,ne}=\cfrac{4}{3}a_{\psi\,\pi}\gt a_{\psi\,\pi}\)
Since, \(2a_{\psi\,ne}\gt a_{\psi\,\pi}\) particles of this size do no occur normally. These particles occur only under higher energy conditions; high acceleration, high temperature and/or high voltage.
So, \(n_2=2n_e\) to \(n_1=n_e\) energy transition is the first to be seen as high energy conditions are applied, although it is not necessarily the lowest absorption line (least \(\lambda\), highest energy). When \(3n_e\) size particles are possible given higher energy conditions, transitions from \(n_3=3n_e\) to \(n_2=2n_e\), \(n_3=3n_e\) to \(n_1=n_e\) are possible. It is likely that these lines sandwich the first line between them.
These transitions do not reverse because the resultant particles are positively charge or neutral.
Since, \(n_e\) particles are neutral, high collision rate under high energy conditions will force them to fuse and the reverse transition \(n_1=n_e\) to \(n_2=2n_e\) emits a spectral line. In this way, the first line fades when higher energy conditions are applied. The appearance of other absorption lines on the spectrum, sees the fading of the first line and other lines involving \(n_1=n_e\), ie the first series.
Good morning.
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}\lt 1.5\)
at the limit,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}=1.5\)
obviously,
\(a_{\psi\,ne}=\cfrac{2}{3}a_{\psi\,\pi}\)
and
\(2a_{\psi\,ne}=\cfrac{4}{3}a_{\psi\,\pi}\gt a_{\psi\,\pi}\)
Since, \(2a_{\psi\,ne}\gt a_{\psi\,\pi}\) particles of this size do no occur normally. These particles occur only under higher energy conditions; high acceleration, high temperature and/or high voltage.
So, \(n_2=2n_e\) to \(n_1=n_e\) energy transition is the first to be seen as high energy conditions are applied, although it is not necessarily the lowest absorption line (least \(\lambda\), highest energy). When \(3n_e\) size particles are possible given higher energy conditions, transitions from \(n_3=3n_e\) to \(n_2=2n_e\), \(n_3=3n_e\) to \(n_1=n_e\) are possible. It is likely that these lines sandwich the first line between them.
These transitions do not reverse because the resultant particles are positively charge or neutral.
Since, \(n_e\) particles are neutral, high collision rate under high energy conditions will force them to fuse and the reverse transition \(n_1=n_e\) to \(n_2=2n_e\) emits a spectral line. In this way, the first line fades when higher energy conditions are applied. The appearance of other absorption lines on the spectrum, sees the fading of the first line and other lines involving \(n_1=n_e\), ie the first series.
Good morning.
Reinterpreting Spectral Lines
From the post "Peanuts And Peanut Butter" dated 09 Dec 2017, where \(\psi\) can disperse in a circular fashion around the point of impact.
\(v^{ 2 }_{ { max } }\lt0\)
and so,
\(v_{ { max } }=i\sqrt{\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\} e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }}\)
What happens when \(a_{\psi\,c}\) collides with one of these rings?
What happens to fragments of \(\psi\) less than \(a_{\psi\,c}\)?
If \({a_{\psi\,ne}}=182.29\,nm\) exist and that \(a_{\psi}\) changes polarity beyond \({a_{\psi\,ne}}\) then all emission spectral series ends at \(n=n_e\) ie, no energy transition between energy states \(n_1\gt n_e\) and \(n_2\gt n_e\). Energy transitions between particles of \(n\ge n_e\) do not occur as these particles are repulsive. A particle of \(n\ge n_e\) can still receive particle of size \(n\lt n_e\), because these particles are still attractive.
An absorption line occurs when a big particle splits into smaller particles of higher energy. But if the split involves negative particles \(n\lt n_e\), these particles will collide again and coalesce with an equal emission of energy. Both absorption and emission will occur at the same time within same confinement of particles. The effects will cancel and neither an absorption nor an emission line will be seen. Absorption spectral lines can be seen only if subsequent collisions and coalescence do not occur. This is when a big particle splits into particles of \(n\ge n_e\). The first absorption line is when \(n=2n_e\) splits into two particles of \(n=ne\). Two particles of \(n=n_e\) will not coalesce and emit a spectral line as they are repulsive (at the limiting border line between positive and negative). This is the lowest absorption spectral line visible.
We have to reinterpret the data presented in the post "Sizing Them Up" dated 3 Dec 2014. The data are energy transitions from \(n_2=2n_e\) to \(n_1=n_e\) and not of the particle \((n+1=2)\rightarrow (n=1)\) where the photon emitted is the particle itself.
These are data for the photon absorbed when particles of each type of size \(n=2n_e\) split into two particles \(n=n_e\). No corresponding emission line occurs because the resultant particles does not coalesce; there is no energy transition back from \(n_1=n_e\) to \(n_2=2n_e\).
What is the value of \(n_e\)? Is \(2*n_e\) possible given that \(a_{\psi\,\pi}\), \(n=77\).
Why would radiating at \(2.4660675\,Hz\) that encourages the breakup of particles, heat up the matter? If \(a_{\psi}=19.34\,nm\) is not the temperature particle, then it should not resonate to \(2.4660675\,Hz\). It is still possible that the resultant smaller particles at \(n=n_e\) absorb particles of \(n\lt n_e\) and become positive. These particles will not split again until \(n=2n_e\). The result is an increase in the number of positive particles. It is still possible to conclude that \(a_{\psi}=19.34\,nm\) is due temperature particles.
All similar radiations, \(\cfrac{f}{10^{6}}\) will increase the respective number of positive particles.
Negative particles are generated when big particle split into particles of \(n\lt n_e\), the smallest of which is \(a_{\psi\,c}\).
So, a stream of negative particles will first converge and coalesce, when they are of size \(n=n_e\) they are neutral. When these coalesce further with particles of \(n\lt n_e\) they become positive, and the stream of particles begins to diverge.
So, a stream of negative electric charges turns neutral then positive. When the stream is neutral, all deflection mechanisms using electric fields and magnetic coils will fail.
And phosphorous luminescence due to positive charges on a CRT screen comes to the rescue...There is no need for electrons knocking out electrons to create positive ions for phosphorous luminescence; the electrons turned positive.
This blog is definitely not for the faint heart-ed!
Note: Negative time is associated with negative temperature particle; cold. We look for negative time particles in \(v_{boom}\), especially, \(2n_e\rightarrow n_e\) and then \(n_e\rightarrow (n_c=1)+(n_e-1)\) that reduce positive particles and generates negative particles.
So, to generate \(T^{-}\), first, radiation at \(2.4660675\,Hz\) breaks the \(n=2n_e\) positive particles up, then a stream of particles at \(v_{boom}\) pops \(a_{\psi\,c}\) from the neutral \(a_{\psi\,ne}\). Both \(a_{\psi\,c}\) and the resultant particle are negative.
If time particles exist, neutral time particle stream stops time, negative time particle stream reverses time and positive time particle stream carries us into the future.
\(TRT\): Time Ray Tunnel
\(v^{ 2 }_{ { max } }\lt0\)
and so,
\(v_{ { max } }=i\sqrt{\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\} e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }}\)
What happens when \(a_{\psi\,c}\) collides with one of these rings?
What happens to fragments of \(\psi\) less than \(a_{\psi\,c}\)?
If \({a_{\psi\,ne}}=182.29\,nm\) exist and that \(a_{\psi}\) changes polarity beyond \({a_{\psi\,ne}}\) then all emission spectral series ends at \(n=n_e\) ie, no energy transition between energy states \(n_1\gt n_e\) and \(n_2\gt n_e\). Energy transitions between particles of \(n\ge n_e\) do not occur as these particles are repulsive. A particle of \(n\ge n_e\) can still receive particle of size \(n\lt n_e\), because these particles are still attractive.
An absorption line occurs when a big particle splits into smaller particles of higher energy. But if the split involves negative particles \(n\lt n_e\), these particles will collide again and coalesce with an equal emission of energy. Both absorption and emission will occur at the same time within same confinement of particles. The effects will cancel and neither an absorption nor an emission line will be seen. Absorption spectral lines can be seen only if subsequent collisions and coalescence do not occur. This is when a big particle splits into particles of \(n\ge n_e\). The first absorption line is when \(n=2n_e\) splits into two particles of \(n=ne\). Two particles of \(n=n_e\) will not coalesce and emit a spectral line as they are repulsive (at the limiting border line between positive and negative). This is the lowest absorption spectral line visible.
We have to reinterpret the data presented in the post "Sizing Them Up" dated 3 Dec 2014. The data are energy transitions from \(n_2=2n_e\) to \(n_1=n_e\) and not of the particle \((n+1=2)\rightarrow (n=1)\) where the photon emitted is the particle itself.
| \(a_{\psi}\) (nm) | \(f\) (GHz) | \(\lambda\)(nm) | |||
| 19.34 | 2466067.5 | 121.57 | |||
| 16.32 | 2922728.6 | 102.57 | |||
| 15.48 | 3082568.8 | 97.25 | |||
| 14.77 | 3230699.3 | 92.79 |
These are data for the photon absorbed when particles of each type of size \(n=2n_e\) split into two particles \(n=n_e\). No corresponding emission line occurs because the resultant particles does not coalesce; there is no energy transition back from \(n_1=n_e\) to \(n_2=2n_e\).
What is the value of \(n_e\)? Is \(2*n_e\) possible given that \(a_{\psi\,\pi}\), \(n=77\).
Why would radiating at \(2.4660675\,Hz\) that encourages the breakup of particles, heat up the matter? If \(a_{\psi}=19.34\,nm\) is not the temperature particle, then it should not resonate to \(2.4660675\,Hz\). It is still possible that the resultant smaller particles at \(n=n_e\) absorb particles of \(n\lt n_e\) and become positive. These particles will not split again until \(n=2n_e\). The result is an increase in the number of positive particles. It is still possible to conclude that \(a_{\psi}=19.34\,nm\) is due temperature particles.
All similar radiations, \(\cfrac{f}{10^{6}}\) will increase the respective number of positive particles.
Negative particles are generated when big particle split into particles of \(n\lt n_e\), the smallest of which is \(a_{\psi\,c}\).
So, a stream of negative particles will first converge and coalesce, when they are of size \(n=n_e\) they are neutral. When these coalesce further with particles of \(n\lt n_e\) they become positive, and the stream of particles begins to diverge.
So, a stream of negative electric charges turns neutral then positive. When the stream is neutral, all deflection mechanisms using electric fields and magnetic coils will fail.
And phosphorous luminescence due to positive charges on a CRT screen comes to the rescue...There is no need for electrons knocking out electrons to create positive ions for phosphorous luminescence; the electrons turned positive.
This blog is definitely not for the faint heart-ed!
Note: Negative time is associated with negative temperature particle; cold. We look for negative time particles in \(v_{boom}\), especially, \(2n_e\rightarrow n_e\) and then \(n_e\rightarrow (n_c=1)+(n_e-1)\) that reduce positive particles and generates negative particles.
So, to generate \(T^{-}\), first, radiation at \(2.4660675\,Hz\) breaks the \(n=2n_e\) positive particles up, then a stream of particles at \(v_{boom}\) pops \(a_{\psi\,c}\) from the neutral \(a_{\psi\,ne}\). Both \(a_{\psi\,c}\) and the resultant particle are negative.
If time particles exist, neutral time particle stream stops time, negative time particle stream reverses time and positive time particle stream carries us into the future.
\(TRT\): Time Ray Tunnel
One Of A Kind
From the post "Small Negative, Big Positive" dated 24 Dec 2014,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}\lt 1.5\)
at the limit,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}=1.5\)
In the case of temperature particle, \(a_{\psi\,\pi}=273.43\,nm\)
\({a_{\psi\,ne}}=\cfrac{273.43}{1.5}=182.29\,nm\)
This particle is neutral.
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,ne} }=\left.3 \cfrac { \partial V\, }{ \partial \, x } \right|_{x=a_{\psi\,\pi}}\)
when two of its kind interact. But this particle is attracted by, \(a_{ \psi \,c }\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } -1.49824\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
and repelled by, \(a_{ \psi \,\pi }\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,\pi } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
One of a kind...
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}\lt 1.5\)
at the limit,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}=1.5\)
In the case of temperature particle, \(a_{\psi\,\pi}=273.43\,nm\)
\({a_{\psi\,ne}}=\cfrac{273.43}{1.5}=182.29\,nm\)
This particle is neutral.
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,ne} }=\left.3 \cfrac { \partial V\, }{ \partial \, x } \right|_{x=a_{\psi\,\pi}}\)
when two of its kind interact. But this particle is attracted by, \(a_{ \psi \,c }\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } -1.49824\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
and repelled by, \(a_{ \psi \,\pi }\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,\pi } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
One of a kind...
Cooling Sonic Boom
If \(a_{\psi\,c}\) are negative, \(v_{boom}\) effects are also negative, as two smaller particles (both of higher energy) result from the collision, and energy is drawn from the surrounding. At least one of the resultant particle is negative, a \(a_{\psi\,c}\).
This might explain the condensations in a sonic boom.
Goodnight again...
This might explain the condensations in a sonic boom.
Goodnight again...
Small Negative, Big Positive
From the post "No Experimental Proof" dated 29 Jul 2016,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}}=\cfrac{cosh^{-1}(e)}{cosh^{-1}(e^{1/4})}\)
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}}=2.24921\)
From the post "Deep Blue Deeper" dated 01 Jun 2016,
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi } } =3k-2\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi}}\)
where,
\(\cfrac{\partial\psi}{\partial x}=\cfrac{\partial V}{\partial x}+\cfrac{\partial T}{\partial x}=k\)
a constant and, \(x\le a_{\psi\,\pi}\).
and the post "Why A Positron And Deep Blue..." dated 01 Jun 2016,
\(\cfrac { dq }{ dx } =\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi}}\)
But we know that \(a_{\psi\,c}\) is at a higher frequency than \(a_{\psi\,\pi}\). And intuitively,
\(\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,\pi}}\lt \left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,c}}\)
\(\because a_{\psi\,c}\lt a_{\psi\,\pi} \) and both particles are waves wrap around a center at light speed.
If \(\cfrac { \partial \, T }{ \partial \, x }\propto \cfrac{1}{a_{\psi}}\), as in the case of gravitational force under earth's surface (a force being the rate of change of energy with distance), that,
\(\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,c}}=\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}} \left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,\pi}}\)
\(\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,c}}=2.24912 \left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,\pi}}\)
So,
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } } =3k-2\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,c}}\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } } =3k-2*2.24912\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,\pi}}\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } } =3\left.\cfrac{\partial V}{\partial x}\right|_{a_{\psi\,\pi}}-1.49824\left.\cfrac{\partial T}{\partial x}\right|_{a_{\psi\,\pi}}\)
where \(\left.\cfrac{\partial\psi}{\partial x}\right|_{a_{\psi\,\pi}}=\left.\cfrac{\partial V}{\partial x}\right|_{a_{\psi\,\pi}}+\left.\cfrac{\partial T}{\partial x}\right|_{a_{\psi\,\pi}}=k\)
since,
\(\cfrac{\partial\psi}{\partial x}=\cfrac{\partial V}{\partial x}+\cfrac{\partial T}{\partial x}=k\)
is true for values of \(x\le a_{\psi\,\pi}\). So,
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } -1.49824\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
but,
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,\pi } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
\(\cfrac { dq }{ dx }\) is also a force; from the two expressions, the resultant interaction of \(a_{\psi\,c}\) and \(a_{\psi\,\pi}\) is attractive, as the common component in the two sums,
\(3 \cfrac { \partial V\, }{ \partial \, x } \)
cancels and the leading coefficients of the term
\(\cfrac { \partial \, T }{ \partial \, x }\)
sums to a negative value (\(-0.49824\)) and results in a decrease in \(T\) (kinetic energy \(KE\)) along \(x\).
In the case when \(a_{\psi\,\pi}\) interacts with \(a_{\psi\,\pi}\) both particle experience positive change in kinetic energy in the direction of \(x\) and they move further apart. The potential component still cancels, as both particles change \(PE\) in the opposite directions. The interaction of two \(a_{\psi\,c}\) particles however, is still attractive due to the negative coefficient to the \(KE\) term,
\(\cfrac { \partial \, T }{ \partial \, x }\)
So, \(a_{\psi\,c}\) particles will coalesce to size \(n=n_e\), where
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}=ratio_e\)
until the substitution into the expression,
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } } =3k-2\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,c}}\)
does not result in a negative coefficient to the term,
\(\cfrac { \partial \, T }{ \partial \, x }\)
ie,
\(2*ratio_e\lt 3\)
\(ratio_e\lt 1.5\)
or
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}\lt 1.5\)
Only after attaining the size \(n_e\), \(a_{\psi\,ne}\), do the particles repel each other. There is not just one opposite charge particle but \(a_{\psi\,c}\) till \(a_{\psi\,ne}\) are all opposite charge to \(a_{\psi\,\pi}\).
If \(a_{\psi\,c}\) is the complement negative particle, from the post "Sizing Them Up" dated 3 Dec 2014,
where the spectra lines are due to basic particles \(a_{\psi\,c}\), \(n=1\). And the size of big particles \({a_{\psi\,\pi}}\) are given by,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}}=2.24921\)
and
\(2\pi a_{\psi}=\lambda\)
It was expected that \(a_{\psi\,c}\) are negative particles. But the value of \(2466067\,Hz\) suggests integer reduced micro wave frequency that agitates \(T^{+}\) particles. Based on this, \(a_{\psi}=19.34\,nm\) is a \(T^{+}\) particle. What gives?
Simple; an input of energy at reduce resonance frequency (by an integer divisor) imparts energy onto the particle; \(a_{\psi\,c}\), a negative temperature particle, grows into \(a_{\psi\,\pi}\), a positive temperature particle; the matter heats up.
Goodnight...
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}}=\cfrac{cosh^{-1}(e)}{cosh^{-1}(e^{1/4})}\)
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}}=2.24921\)
From the post "Deep Blue Deeper" dated 01 Jun 2016,
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi } } =3k-2\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi}}\)
where,
\(\cfrac{\partial\psi}{\partial x}=\cfrac{\partial V}{\partial x}+\cfrac{\partial T}{\partial x}=k\)
a constant and, \(x\le a_{\psi\,\pi}\).
and the post "Why A Positron And Deep Blue..." dated 01 Jun 2016,
\(\cfrac { dq }{ dx } =\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi}}\)
But we know that \(a_{\psi\,c}\) is at a higher frequency than \(a_{\psi\,\pi}\). And intuitively,
\(\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,\pi}}\lt \left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,c}}\)
\(\because a_{\psi\,c}\lt a_{\psi\,\pi} \) and both particles are waves wrap around a center at light speed.
If \(\cfrac { \partial \, T }{ \partial \, x }\propto \cfrac{1}{a_{\psi}}\), as in the case of gravitational force under earth's surface (a force being the rate of change of energy with distance), that,
\(\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,c}}=\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}} \left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,\pi}}\)
\(\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,c}}=2.24912 \left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,\pi}}\)
So,
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } } =3k-2\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,c}}\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } } =3k-2*2.24912\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,\pi}}\)
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } } =3\left.\cfrac{\partial V}{\partial x}\right|_{a_{\psi\,\pi}}-1.49824\left.\cfrac{\partial T}{\partial x}\right|_{a_{\psi\,\pi}}\)
where \(\left.\cfrac{\partial\psi}{\partial x}\right|_{a_{\psi\,\pi}}=\left.\cfrac{\partial V}{\partial x}\right|_{a_{\psi\,\pi}}+\left.\cfrac{\partial T}{\partial x}\right|_{a_{\psi\,\pi}}=k\)
since,
\(\cfrac{\partial\psi}{\partial x}=\cfrac{\partial V}{\partial x}+\cfrac{\partial T}{\partial x}=k\)
is true for values of \(x\le a_{\psi\,\pi}\). So,
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } -1.49824\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
but,
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,\pi } }=\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi\,\pi}}\)
\(\cfrac { dq }{ dx }\) is also a force; from the two expressions, the resultant interaction of \(a_{\psi\,c}\) and \(a_{\psi\,\pi}\) is attractive, as the common component in the two sums,
\(3 \cfrac { \partial V\, }{ \partial \, x } \)
cancels and the leading coefficients of the term
\(\cfrac { \partial \, T }{ \partial \, x }\)
sums to a negative value (\(-0.49824\)) and results in a decrease in \(T\) (kinetic energy \(KE\)) along \(x\).
In the case when \(a_{\psi\,\pi}\) interacts with \(a_{\psi\,\pi}\) both particle experience positive change in kinetic energy in the direction of \(x\) and they move further apart. The potential component still cancels, as both particles change \(PE\) in the opposite directions. The interaction of two \(a_{\psi\,c}\) particles however, is still attractive due to the negative coefficient to the \(KE\) term,
\(\cfrac { \partial \, T }{ \partial \, x }\)
So, \(a_{\psi\,c}\) particles will coalesce to size \(n=n_e\), where
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}=ratio_e\)
until the substitution into the expression,
\(\left.\cfrac { dq }{ dx }\right|_{ a_{ \psi \,c } } =3k-2\left.\cfrac { \partial \, T }{ \partial \, x }\right|_{a_{\psi\,c}}\)
does not result in a negative coefficient to the term,
\(\cfrac { \partial \, T }{ \partial \, x }\)
ie,
\(2*ratio_e\lt 3\)
\(ratio_e\lt 1.5\)
or
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,ne}}\lt 1.5\)
Only after attaining the size \(n_e\), \(a_{\psi\,ne}\), do the particles repel each other. There is not just one opposite charge particle but \(a_{\psi\,c}\) till \(a_{\psi\,ne}\) are all opposite charge to \(a_{\psi\,\pi}\).
If \(a_{\psi\,c}\) is the complement negative particle, from the post "Sizing Them Up" dated 3 Dec 2014,
| \(a_{\psi_c}\) (nm) | \(f_c\) (GHz) | \(\lambda_c\)(nm) | \(a_{\psi\,\pi}\)(nm) | \(f_{\pi}\)(GHz) | \(\lambda_{\pi}\)(nm) | |||
| 19.34 | 2466067.5 | 121.57 |
|
|
| |||
| 16.32 | 2922728.6 | 102.57 |
|
|
| |||
| 15.48 | 3082568.8 | 97.25 |
|
|
| |||
| 14.77 | 3230699.3 | 92.79 |
|
|
| |||
where the spectra lines are due to basic particles \(a_{\psi\,c}\), \(n=1\). And the size of big particles \({a_{\psi\,\pi}}\) are given by,
\(\cfrac{a_{\psi\,\pi}}{a_{\psi\,c}}=2.24921\)
and
\(2\pi a_{\psi}=\lambda\)
It was expected that \(a_{\psi\,c}\) are negative particles. But the value of \(2466067\,Hz\) suggests integer reduced micro wave frequency that agitates \(T^{+}\) particles. Based on this, \(a_{\psi}=19.34\,nm\) is a \(T^{+}\) particle. What gives?
Simple; an input of energy at reduce resonance frequency (by an integer divisor) imparts energy onto the particle; \(a_{\psi\,c}\), a negative temperature particle, grows into \(a_{\psi\,\pi}\), a positive temperature particle; the matter heats up.
Goodnight...
Saturday, December 23, 2017
Checking Discharge Voltage
From the post "Neon Light Boom" dated 22 Dec 2017,
\(V_{min}=A\left(\cfrac{P}{T}\right)^2\)
and \(A=\cfrac{3.4354^2}{2}\cfrac{m_i^3}{qZ^2k^2}\)
but
\(\cfrac{P}{T}=\cfrac{n}{V_{ol}}R=\cfrac{M}{m_iN_AV_{ol}}R=\cfrac{density}{m_i}k\)
\(V_{min}=\cfrac{3.4354^2}{2}\cfrac{m_i^3}{qZ^2k^2}\left(\cfrac{density}{m_i}k\right)^2\)
\(V_{min}=\cfrac{3.4354^2}{2}\cfrac{m_i}{qZ^2}density^2\)
\(V_{min}=\cfrac{1}{2}\cfrac{m_i}{q}\left(3.4354*\cfrac{density}{Z}\right)^2\) ---(*)
as \(v_{boom}=3.4354*\cfrac{density}{Z}\)
\(qV_{min}=\cfrac{1}{2}{m_i}v_{boom}^2\)
which just says the voltage applied converts to \(KE\) at \(v_{boom}\), provided that the gap length between the electrodes are sufficiently wide. This implies that the expression for \(V_{min}\) is dimensional-ly consistent.
However, if \(q\rightarrow \cfrac{1}{4}q\) then,
\(\cfrac{1}{4}qV_{min}=\cfrac{1}{2}{m_c}v_{boom}^2\)
and (*) becomes,
\(V_{min\,c}=2\cfrac{m_c}{q}\left(3.4354*\cfrac{density}{Z}\right)^2\) ---(**)
\(m_c\), however, is not available.
How does this tally with experimental values of \(V_{min}\)?
\(V_{min}=A\left(\cfrac{P}{T}\right)^2\)
and \(A=\cfrac{3.4354^2}{2}\cfrac{m_i^3}{qZ^2k^2}\)
but
\(\cfrac{P}{T}=\cfrac{n}{V_{ol}}R=\cfrac{M}{m_iN_AV_{ol}}R=\cfrac{density}{m_i}k\)
\(V_{min}=\cfrac{3.4354^2}{2}\cfrac{m_i^3}{qZ^2k^2}\left(\cfrac{density}{m_i}k\right)^2\)
\(V_{min}=\cfrac{3.4354^2}{2}\cfrac{m_i}{qZ^2}density^2\)
\(V_{min}=\cfrac{1}{2}\cfrac{m_i}{q}\left(3.4354*\cfrac{density}{Z}\right)^2\) ---(*)
as \(v_{boom}=3.4354*\cfrac{density}{Z}\)
\(qV_{min}=\cfrac{1}{2}{m_i}v_{boom}^2\)
which just says the voltage applied converts to \(KE\) at \(v_{boom}\), provided that the gap length between the electrodes are sufficiently wide. This implies that the expression for \(V_{min}\) is dimensional-ly consistent.
However, if \(q\rightarrow \cfrac{1}{4}q\) then,
\(\cfrac{1}{4}qV_{min}=\cfrac{1}{2}{m_c}v_{boom}^2\)
and (*) becomes,
\(V_{min\,c}=2\cfrac{m_c}{q}\left(3.4354*\cfrac{density}{Z}\right)^2\) ---(**)
\(m_c\), however, is not available.
How does this tally with experimental values of \(V_{min}\)?