a Christmas wish for a dialed portal.
Sunday, December 8, 2019
Lights in a jar
in remembrance of one
who advocates love
in an age of violence
when blood-lust was joy
Sparkles in a glass
in his eminence
by our hands abundance be
providence for us to give
love and care to share
Enlightenment in our hearts
darkness and shadows be gone
and forgiven
gentle and mercy to disarm
lawful be righteous and fair
Brightness aloft
our paths be guided
through trials and tribulations
resilience in failures
resolute in mis-course
Merry Christmas...
《Christmas 2019》
in remembrance of one
who advocates love
in an age of violence
when blood-lust was joy
Sparkles in a glass
in his eminence
by our hands abundance be
providence for us to give
love and care to share
Enlightenment in our hearts
darkness and shadows be gone
and forgiven
gentle and mercy to disarm
lawful be righteous and fair
Brightness aloft
our paths be guided
through trials and tribulations
resilience in failures
resolute in mis-course
Merry Christmas...
《Christmas 2019》
Pushing Sideways
Remember the pair of magnets that went sideways,
the force that pushes the magnets sideways must be due to the interactions of field lines perpendicular to the direction of attraction. One possibility is,
where at the ends of the magnets, the domains reversed and the field lines on one side run parallel to the end surface. The interactions of these field lines push the the magnets sideways. The diagram on the right shows a more definitive way to make such magnets. These are,
electromagnetic version of the same, where strong parallel field lines at the end of the them create a strong push side-way in a stater-rotor configuration.
Just thinking out loud.
the force that pushes the magnets sideways must be due to the interactions of field lines perpendicular to the direction of attraction. One possibility is,
where at the ends of the magnets, the domains reversed and the field lines on one side run parallel to the end surface. The interactions of these field lines push the the magnets sideways. The diagram on the right shows a more definitive way to make such magnets. These are,
electromagnetic version of the same, where strong parallel field lines at the end of the them create a strong push side-way in a stater-rotor configuration.
Just thinking out loud.
Saturday, December 7, 2019
Cosinusoidal Time
Consider the three time dimension spinning about the three space dimension,
The direction perpendicular to the \(t_c\) plane is the direction along that space dimension, \(x_c\). The expression for time along this axis is,
\(t_{\small{axis}}=-t_T sin \theta+t_g cos \theta=t(cos \theta-sin\theta)\)
\(t_{\small{axis}}=t\sqrt{2}cos(\theta+\cfrac{\pi}{4})\)
where we assume \(t_c=t_g=t_T=t\), that time is undifferentiated.
Is this formulation consistent with "the universe is a time particle" view; that a circular time wave passes through Earth at light speed, centered at the center of the universe?
Can the \(\sqrt{2}\) factor and the phase \(\small{\cfrac{\pi}{4}}\) be verified? What is the significance of these factors?
The direction perpendicular to the \(t_c\) plane is the direction along that space dimension, \(x_c\). The expression for time along this axis is,
\(t_{\small{axis}}=-t_T sin \theta+t_g cos \theta=t(cos \theta-sin\theta)\)
\(t_{\small{axis}}=t\sqrt{2}cos(\theta+\cfrac{\pi}{4})\)
where we assume \(t_c=t_g=t_T=t\), that time is undifferentiated.
Is this formulation consistent with "the universe is a time particle" view; that a circular time wave passes through Earth at light speed, centered at the center of the universe?
Can the \(\sqrt{2}\) factor and the phase \(\small{\cfrac{\pi}{4}}\) be verified? What is the significance of these factors?
If The Universe Is A Time Particle
If the universe is a time particle,
then a time wave around the center of the universe passes by Earth and imparts time onto the planet. A distorted particle in the path of such a wave travels through time. \
How do construct such a particle and be in such a particle to travel through time?
Since photons have been purported to be a wave with two orthogonal time component, spinning photons may just create a time field,
A distortion in this field can be introduced by displacing the light tube slightly,
Aligned in the direction of the time wave, it is then possible to move back and forth through time.
Yes, you sit inside the spinning tubes...
then a time wave around the center of the universe passes by Earth and imparts time onto the planet. A distorted particle in the path of such a wave travels through time. \
How do construct such a particle and be in such a particle to travel through time?
Since photons have been purported to be a wave with two orthogonal time component, spinning photons may just create a time field,
A distortion in this field can be introduced by displacing the light tube slightly,
Yes, you sit inside the spinning tubes...
Consistence About A Hartree
Hartree energy is defined as
\(E_H=2R_Hhc\)
From the post "Looking for Murder" dated 13 Oct 2018 it was proposed that,
\(R_H=\cfrac{1}{2\pi a_{\psi\,c}}\)
where \(R_H\) is the Rydberg constant
So,
\(E_H=2*\cfrac{hc}{2\pi a_{\psi\,c}}\)
but
\(f=\cfrac{c}{\lambda_{\psi\,c}}=\cfrac{c}{2\pi a_{\psi\,c}}\)
as
\(\lambda_{\psi\,c}=2\pi a_{\psi\,c}\)
we have,
\(E_H=2*hf_{\psi\,c}\) --- (*)
which only confirms that the prposed definition for \(R_H\) in the post "Looking for Murder" dated 13 Oct 2018 is dimensionally correct.
In the definition for Harttree Energy, this energy is approximately the electric potential energy of the hydrogen atom in its ground state and, by the virial theorem, approximately twice its ionization energy; the relationships are not exact because of the finite mass of the nucleus of the hydrogen atom and relativistic corrections.
\(a_{\psi}=14.77\,nm\) is a particle defined by \(\psi\) in circular motion; Hartree Energy sees an electron in orbit around a hydrogen nucleus.
This contention is not new and has been resolved by thinking that the electron is a wave going in orbit.
The posts presented here however, takes the view that the electron is made up of a \(\psi\) wave warped around a sphere. A \(\psi\) wave in circular motion. Enegry changes of this wave result in the spectral lines. The electron need not be in orbit around the hydrogen nucleus.
The size of a hydrogen atom is in the range of \(\times 10^{-15}\) meters, \(fm\),
but a consistent \(a_{\psi\,c}\), according to (*) is in the range of tenth of \(\times 10^{-9}\).
An electron at \(a_{\psi\,c}\) from the hydrogen nucleus cannot be at the ground state.
A particle, however, can have a force field around it up to \(\times10^{-9}\) meters.
Good night...
\(E_H=2R_Hhc\)
From the post "Looking for Murder" dated 13 Oct 2018 it was proposed that,
\(R_H=\cfrac{1}{2\pi a_{\psi\,c}}\)
where \(R_H\) is the Rydberg constant
So,
\(E_H=2*\cfrac{hc}{2\pi a_{\psi\,c}}\)
but
\(f=\cfrac{c}{\lambda_{\psi\,c}}=\cfrac{c}{2\pi a_{\psi\,c}}\)
as
\(\lambda_{\psi\,c}=2\pi a_{\psi\,c}\)
we have,
\(E_H=2*hf_{\psi\,c}\) --- (*)
which only confirms that the prposed definition for \(R_H\) in the post "Looking for Murder" dated 13 Oct 2018 is dimensionally correct.
In the definition for Harttree Energy, this energy is approximately the electric potential energy of the hydrogen atom in its ground state and, by the virial theorem, approximately twice its ionization energy; the relationships are not exact because of the finite mass of the nucleus of the hydrogen atom and relativistic corrections.
\(a_{\psi}=14.77\,nm\) is a particle defined by \(\psi\) in circular motion; Hartree Energy sees an electron in orbit around a hydrogen nucleus.
This contention is not new and has been resolved by thinking that the electron is a wave going in orbit.
The posts presented here however, takes the view that the electron is made up of a \(\psi\) wave warped around a sphere. A \(\psi\) wave in circular motion. Enegry changes of this wave result in the spectral lines. The electron need not be in orbit around the hydrogen nucleus.
The size of a hydrogen atom is in the range of \(\times 10^{-15}\) meters, \(fm\),
but a consistent \(a_{\psi\,c}\), according to (*) is in the range of tenth of \(\times 10^{-9}\).
An electron at \(a_{\psi\,c}\) from the hydrogen nucleus cannot be at the ground state.
A particle, however, can have a force field around it up to \(\times10^{-9}\) meters.
Good night...
For Completeness Sake, Freaking Out
From the post "What If The Particles Are Photons?" dated 12 Dec 2017,
\(f=\cfrac{c}{\lambda}=\cfrac{8^3}{18}\pi^2=280.74\,Hz\)
we can have,
\(f=3*280.74\,Hz=842.21\,Hz\)
But,
what is \(w\)? to realize such a 3D confinement portal. It is hoped that with
\(w\gt842.21\,Hz\) and
\(w\) being prime, we may have a prime dialed portal to walk into and walk out.
Just science friction.
\(f=\cfrac{c}{\lambda}=\cfrac{8^3}{18}\pi^2=280.74\,Hz\)
we can have,
\(f=3*280.74\,Hz=842.21\,Hz\)
But,
what is \(w\)? to realize such a 3D confinement portal. It is hoped that with
\(w\gt842.21\,Hz\) and
\(w\) being prime, we may have a prime dialed portal to walk into and walk out.
Just science friction.
Don't Freak Out
Yet another mistake...in the post "Freaking Out Entanglement" dated 14 Dec 2017,
\(P=\cfrac{8^3}{6}{\pi^2}∗f^2∗m_ac^2=1∗\cfrac{4}{3}\pi∗f^3∗m_ac^2\)
should be
\(P=\cfrac{8^3}{6}{\pi^2}∗f^2∗m_ac^2=1∗\cfrac{3}{4\pi}∗f^3∗m_ac^2\)
\(f=\cfrac{4*8^3}{18}\pi^3=3527.83 Hz\) Very dangerous! Do not look into the light, at all. No exposure in anyway.
\(P=\cfrac{8^3}{6}{\pi^2}∗f^2∗m_ac^2=1∗\cfrac{4}{3}\pi∗f^3∗m_ac^2\)
should be
\(P=\cfrac{8^3}{6}{\pi^2}∗f^2∗m_ac^2=1∗\cfrac{3}{4\pi}∗f^3∗m_ac^2\)
\(f=\cfrac{4*8^3}{18}\pi^3=3527.83 Hz\) Very dangerous! Do not look into the light, at all. No exposure in anyway.
and
\(f=64*\pi=201.06\,Hz\) is wrong!
Sorry...
To Clear Things Up
I have made a mistake in the post "I Don't Know..." dated 12 Dec 2017, in the expression
\(P=\cfrac{8^3}{18}{\pi^2}∗f^2∗m_ac^2=1∗\cfrac{4}{3}\pi∗f^3∗m_ac^2\)
It should be
\(P=\cfrac{8^3}{18}{\pi^2}∗f^2∗m_ac^2=1∗\cfrac{3}{4\pi}∗f^3∗m_ac^2\)
where the \(RHS\) is divided by a spherical volume to obtain energy per unit volume. The expression leads to,
\(f=\cfrac{4*8^3}{3*18}\pi^3=1175.94 Hz\)
The previous result,
\(f=\cfrac{8^3}{24}\pi=67.02\,Hz\)
is not valid.
Sorry...
\(P=\cfrac{8^3}{18}{\pi^2}∗f^2∗m_ac^2=1∗\cfrac{4}{3}\pi∗f^3∗m_ac^2\)
It should be
\(P=\cfrac{8^3}{18}{\pi^2}∗f^2∗m_ac^2=1∗\cfrac{3}{4\pi}∗f^3∗m_ac^2\)
where the \(RHS\) is divided by a spherical volume to obtain energy per unit volume. The expression leads to,
\(f=\cfrac{4*8^3}{3*18}\pi^3=1175.94 Hz\)
The previous result,
\(f=\cfrac{8^3}{24}\pi=67.02\,Hz\)
is not valid.
Sorry...
Saturday, September 7, 2019
Torturing Fish
There is more oxygen in the air than in water, so a fish out of water is hyperventilating. It is not grasping for air.
A drop of oil on the outer gill, leaving the rest of the gills not coated, may prevent the fish from dying too soon. Other than oil, other options such as starch of liquid consistency, agar-agar with pineapple juice (that prevents the agar-agar from setting) can be experimented with. The point is to cover the gills and prevent hyperventilation.
Good night...z.z.z
A drop of oil on the outer gill, leaving the rest of the gills not coated, may prevent the fish from dying too soon. Other than oil, other options such as starch of liquid consistency, agar-agar with pineapple juice (that prevents the agar-agar from setting) can be experimented with. The point is to cover the gills and prevent hyperventilation.
Good night...z.z.z
The Old Model
The Old Model is to consider a chemical bond as a particle whose diameter is the bond length. As such, all resonance associated with a particle is applicable to the chemical bond. For example,
\(f_{res}=0.061\cfrac{c}{a_{\psi}}\)
\(a_{\psi}=\cfrac{1}{2}BL\)
\(f_{res}=0.122\cfrac{c}{BL}\)
The radius of such a particle is half the bond length, \(BL\).
\(f_{res}=0.061\cfrac{c}{a_{\psi}}\)
\(a_{\psi}=\cfrac{1}{2}BL\)
\(f_{res}=0.122\cfrac{c}{BL}\)
The radius of such a particle is half the bond length, \(BL\).
Wednesday, March 20, 2019
Ammonia Is Important
This post was missing; a cold stream of hydrogen gas mixed with a warm stream of nitrogen gas to produce ammonia without high pressure and temperature.
It is not a mix of cold nitrogen and hot hydrogen because I remember a jet of hydrogen gas from a high pressure cylinder smells like ammonia and that ammonia is found in dark underground where warm air mix in with the cold and damp.
This post came after the postulate that there is another class of chemistry involving temperature particles, where a temperature difference provides for the disparity in temperature charge, analogous the electric charge that fills oxidation states in normal chemistry.
I am not back; data is very expensive.
It is not a mix of cold nitrogen and hot hydrogen because I remember a jet of hydrogen gas from a high pressure cylinder smells like ammonia and that ammonia is found in dark underground where warm air mix in with the cold and damp.
This post came after the postulate that there is another class of chemistry involving temperature particles, where a temperature difference provides for the disparity in temperature charge, analogous the electric charge that fills oxidation states in normal chemistry.
I am not back; data is very expensive.
Thursday, January 10, 2019
Where It Isn't
One way to proof the validity of \(f_{osc}\) and the associated \(I_{osc}=q_e*f_{osc}\) where \(q_e\) is the electron charge, is to pass such a current, \(I_{osc}\), through a metal whose photoelectric threshold frequency cannot be obtained (usually where the expected threshold frequency is at the edge of the producible light spectrum) during the experiment to obtain the threshold frequency.
In another words, repeat the experiment to obtain \(f_{threshold}\) with \(I_{osc}\) following through the metal.
To find where there was none...
In another words, repeat the experiment to obtain \(f_{threshold}\) with \(I_{osc}\) following through the metal.
To find where there was none...
Wednesday, January 9, 2019
I Found The Rabbit Hole
This is being thick skin about why \(f_{osc}\) is not the photoelectric emission threshold frequency.
This line of best fit has gradient\(\rightarrow\infty\) as \(x\rightarrow\infty\). For any such line fitted closer to the origin, it has gradient \(\lt\infty\). "Threshold", the x-intercept of this line is not unique.
This is a line to fit V-I characteristics of a gas discharge tube at the onset of glow discharge,
where \(x=log(I)\). The drop in voltage here is attributed to negative charges created.
This should not be confused with,
\(K_{max}=hf-\Phi\)
where \(K_{max}\) is the maximum kinetic energy of ejected electrons and \(\Phi=hf_o\) the Work Function defined by \(f_o\), the threshold frequency. This relation answers the question: "What happens to the excess energy of the photon beyond \(\Phi\) after impact?"
Using sodium, \(Na\)
\(f_o=4.39*10^{14}\,Hz\)
If we reverse and find \(a_{\psi}\) using
\(f_{osc}=c\sqrt{\cfrac{2\pi}{a_{\psi}}}\) --- (*)
with a change in notation,
\(f_{hole}=c\sqrt{\cfrac{2\pi}{a_{\psi\,hole}}}\)
\(a_{\psi\,hole}=2\pi\left(\cfrac{c}{f_{hole}}\right)^2=2.93\text{e-}12\,m\)
What is this small hole?
Could it be that \(f_{osc}\) of sodium, opens up this small hole and subsequent photons that falls into it, is split into two parts,
\(K_{max}=hf-\Phi\)
\(\Phi=hf_{hole}\), \(f\) is the frequency of the impacting photons and \(K_{max}\) is the maximum kinetic energy of the ejected particles. \(f_{osc}\) of sodium is given by,
\(f_{osc}=c\sqrt{\cfrac{2\pi}{a_{\psi\,Na}}}\)
where \(a_{\psi\,Na}\) is the size of sodium \(\psi\) cloud.
\(f_{osc}\) opens up a hole denoted by \(a_{\psi\,hole}\), and all photons at greater size than \(a_{\psi\,hole}\) (lower frequency than \(f_{hole}\), as indicated by (*)) are rejected by the hole. If this is the role of \(f_{osc}\), then using a narrow bandwidth of photons at \(f_{osc}\) and illuminating the metal with photons at bigger size than \(a_{hole}\) will cause the metal to appear very bright and without the emission of charged particles. All illuminating photons is rejected by the hole opened by \(f_{osc}\).
Photons of size smaller, \(a_{\psi}\lt a_{\psi\,hole}\) falls into the hole. When do such photons return after falling into the hole? What dictates the size of the hole? Hole of different sizes could explain the seemingly different processes that occurs simultaneously during photoelectric emission where two \(K_{max}\) values were noted. The post related to this is "More Fodo Effects" dated 04 Jun 2014; two related diagrams are shown below,
Impurities could explain the two \(K_{max}s\).
\(f_{hole}\) is the threshold frequency. Given \(f_{\psi}\), \(a_{\psi}\) can be derived given,
\(2\pi a_{\psi}=n\lambda=n\cfrac{c}{f_{\psi}}\)
\(f_{\psi}=\cfrac{c}{2\pi a_{\psi}}\), with \(n=1\)
but an impacting photon interact through,
\(f_{\psi}=c\sqrt{\cfrac{2\pi}{a_{\psi}}}\)
when \(f_{\psi}\gt f_{hole}\), \(a_{\psi}\lt a_{\psi\,hole}\) and the photon goes through the hole, and its energy is split via,
\(K_{max}=hf-\Phi\)
\(\Phi=hf_{hole}\) and \(K_{max}\) is the maximum kinetic energy of a emitted particle.
\(f_{osc}\) that triggers \(f_{hole}\) is lower than \(f_{hole}\). \(f_{hole}\) alone does not cause photoelectric emission. \(f_{osc}\), a frequency below the threshold, has first to create the hole, via a mode of oscillation with resonance at,
\(f_{osc}=c\sqrt{\cfrac{2\pi}{a_{\psi}}}\)
where \(a_{\psi}\) is the size of the \(\psi\) ball.
This hole is in effect, a region of negative energy that minus off energy of subsequent impacting photons of higher frequencies. Photons of lower frequencies are too big to fit into the hole. And only the resonance frequency, \(f_{osc}\) opens the hole in the metal.
Besides being convoluted, this explanation does not hold \(a_{\psi\,hole}\) at all, apart from the exertion that \(f_{osc}\) caused it. It however, allows for many size holes that can account for more than one \(K_{max}\) and allows \(f_{osc}\) to be different from the threshold frequency. But charges are still created within the metal at \(f_{osc}\) and, higher frequency photons interacting with the holes created can be outside the metal, ie emitted, when they return from the time dimension.
How does \(f_{osc}\) create charges? \(f_{osc}\) set the \(\psi\) cloud of the atom into resonance and the atom loses its orbiting valence electrons. And, how a photon of higher energy \(f_{\psi}\gt f_{hole}\) ejects a particle of kinetic energy \(K_{max}\) after filling in the hole? The hole requires \(E=hf_{hole}\), what remains of the impacting photons with higher energy after making good for the hole is like a torus photon (posts "What Donuts? Dipoles?" and "Torus Photons" dated 29 Dec 2017).
\(K_{max}=hf_{\psi}-hf_{hole}\)
When the torus photon collapses however, the resulting particle may not have a full unit charge. It may be a partial charge of size smaller than \(a_{\psi\,c}\).
Is \(f_{osc}\) valid?
Note: The previous version of this post has mistaken \(f_{osc}\gt f_{o}\) or \(f_{hole}\) that was wrong.
This line of best fit has gradient\(\rightarrow\infty\) as \(x\rightarrow\infty\). For any such line fitted closer to the origin, it has gradient \(\lt\infty\). "Threshold", the x-intercept of this line is not unique.
This is a line to fit V-I characteristics of a gas discharge tube at the onset of glow discharge,
where \(x=log(I)\). The drop in voltage here is attributed to negative charges created.
This should not be confused with,
\(K_{max}=hf-\Phi\)
where \(K_{max}\) is the maximum kinetic energy of ejected electrons and \(\Phi=hf_o\) the Work Function defined by \(f_o\), the threshold frequency. This relation answers the question: "What happens to the excess energy of the photon beyond \(\Phi\) after impact?"
Using sodium, \(Na\)
\(f_o=4.39*10^{14}\,Hz\)
If we reverse and find \(a_{\psi}\) using
\(f_{osc}=c\sqrt{\cfrac{2\pi}{a_{\psi}}}\) --- (*)
with a change in notation,
\(f_{hole}=c\sqrt{\cfrac{2\pi}{a_{\psi\,hole}}}\)
\(a_{\psi\,hole}=2\pi\left(\cfrac{c}{f_{hole}}\right)^2=2.93\text{e-}12\,m\)
What is this small hole?
Could it be that \(f_{osc}\) of sodium, opens up this small hole and subsequent photons that falls into it, is split into two parts,
\(K_{max}=hf-\Phi\)
\(\Phi=hf_{hole}\), \(f\) is the frequency of the impacting photons and \(K_{max}\) is the maximum kinetic energy of the ejected particles. \(f_{osc}\) of sodium is given by,
\(f_{osc}=c\sqrt{\cfrac{2\pi}{a_{\psi\,Na}}}\)
where \(a_{\psi\,Na}\) is the size of sodium \(\psi\) cloud.
\(f_{osc}\) opens up a hole denoted by \(a_{\psi\,hole}\), and all photons at greater size than \(a_{\psi\,hole}\) (lower frequency than \(f_{hole}\), as indicated by (*)) are rejected by the hole. If this is the role of \(f_{osc}\), then using a narrow bandwidth of photons at \(f_{osc}\) and illuminating the metal with photons at bigger size than \(a_{hole}\) will cause the metal to appear very bright and without the emission of charged particles. All illuminating photons is rejected by the hole opened by \(f_{osc}\).
Photons of size smaller, \(a_{\psi}\lt a_{\psi\,hole}\) falls into the hole. When do such photons return after falling into the hole? What dictates the size of the hole? Hole of different sizes could explain the seemingly different processes that occurs simultaneously during photoelectric emission where two \(K_{max}\) values were noted. The post related to this is "More Fodo Effects" dated 04 Jun 2014; two related diagrams are shown below,
Impurities could explain the two \(K_{max}s\).
\(f_{hole}\) is the threshold frequency. Given \(f_{\psi}\), \(a_{\psi}\) can be derived given,
\(2\pi a_{\psi}=n\lambda=n\cfrac{c}{f_{\psi}}\)
\(f_{\psi}=\cfrac{c}{2\pi a_{\psi}}\), with \(n=1\)
but an impacting photon interact through,
\(f_{\psi}=c\sqrt{\cfrac{2\pi}{a_{\psi}}}\)
when \(f_{\psi}\gt f_{hole}\), \(a_{\psi}\lt a_{\psi\,hole}\) and the photon goes through the hole, and its energy is split via,
\(K_{max}=hf-\Phi\)
\(\Phi=hf_{hole}\) and \(K_{max}\) is the maximum kinetic energy of a emitted particle.
\(f_{osc}\) that triggers \(f_{hole}\) is lower than \(f_{hole}\). \(f_{hole}\) alone does not cause photoelectric emission. \(f_{osc}\), a frequency below the threshold, has first to create the hole, via a mode of oscillation with resonance at,
\(f_{osc}=c\sqrt{\cfrac{2\pi}{a_{\psi}}}\)
where \(a_{\psi}\) is the size of the \(\psi\) ball.
This hole is in effect, a region of negative energy that minus off energy of subsequent impacting photons of higher frequencies. Photons of lower frequencies are too big to fit into the hole. And only the resonance frequency, \(f_{osc}\) opens the hole in the metal.
Besides being convoluted, this explanation does not hold \(a_{\psi\,hole}\) at all, apart from the exertion that \(f_{osc}\) caused it. It however, allows for many size holes that can account for more than one \(K_{max}\) and allows \(f_{osc}\) to be different from the threshold frequency. But charges are still created within the metal at \(f_{osc}\) and, higher frequency photons interacting with the holes created can be outside the metal, ie emitted, when they return from the time dimension.
How does \(f_{osc}\) create charges? \(f_{osc}\) set the \(\psi\) cloud of the atom into resonance and the atom loses its orbiting valence electrons. And, how a photon of higher energy \(f_{\psi}\gt f_{hole}\) ejects a particle of kinetic energy \(K_{max}\) after filling in the hole? The hole requires \(E=hf_{hole}\), what remains of the impacting photons with higher energy after making good for the hole is like a torus photon (posts "What Donuts? Dipoles?" and "Torus Photons" dated 29 Dec 2017).
\(K_{max}=hf_{\psi}-hf_{hole}\)
When the torus photon collapses however, the resulting particle may not have a full unit charge. It may be a partial charge of size smaller than \(a_{\psi\,c}\).
Is \(f_{osc}\) valid?
Note: The previous version of this post has mistaken \(f_{osc}\gt f_{o}\) or \(f_{hole}\) that was wrong.
Tuesday, January 8, 2019
Pulling Crystals
This was what I thought of pulling silicon crystal,
and the artifacts that it creates. This is what I thought that would reduce such ringing artifacts,
\(\theta\) for diamond cubic crystals will be,
\(OA=OC=\cfrac{3}{4}\sqrt{\cfrac{2}{3}}a\)
\(OH=\cfrac{1}{4}\sqrt{\cfrac{2}{3}}a\)
\(\theta=arcsin(\cfrac{OH}{OC})=arcsin(\cfrac{1}{3})=19.47^o\).
where \(C\) is at the first layer and is bonded to the next layer of silicon is at \(O\). The next pull does not form up the bond along \(OA\), but it is at \(19.47^o\) from \(O\). Bonds similar to \(CO\) form up else where in the crystal. After four steps, because \(OH=\cfrac{1}{4}AH\), the atom needed at \(A\) falls into place. At each step, a sheet of atoms is shifted into place.
Good night...
and the artifacts that it creates. This is what I thought that would reduce such ringing artifacts,
\(\theta\) for diamond cubic crystals will be,
\(OA=OC=\cfrac{3}{4}\sqrt{\cfrac{2}{3}}a\)
\(OH=\cfrac{1}{4}\sqrt{\cfrac{2}{3}}a\)
\(\theta=arcsin(\cfrac{OH}{OC})=arcsin(\cfrac{1}{3})=19.47^o\).
where \(C\) is at the first layer and is bonded to the next layer of silicon is at \(O\). The next pull does not form up the bond along \(OA\), but it is at \(19.47^o\) from \(O\). Bonds similar to \(CO\) form up else where in the crystal. After four steps, because \(OH=\cfrac{1}{4}AH\), the atom needed at \(A\) falls into place. At each step, a sheet of atoms is shifted into place.
Good night...
Townsend Discharge In Solids
Cont'd from the post "Hollow Metal" dated 29 Dec 2018, if the matter is a gas, \(I_{osc}=q_e*f_{osc}\) might be Townsend discharge, where \(q_e=1.602176565\text{e-19}\,C\)
A table of \(I_{osc}\) for covalent bond is given below.
Apart from the fact that \(I_{osc}\approx10^{-6}\,\text{to}\,10^{-5}\), there is no reason to believe that \(I_{osc}\) triggers Townsend discharge.
However, if \(I_{osc}\) does trigger a discharge, then the negative charges produced during resonance will reduce the measured voltage and increase the current. The changes in current and voltage are continued only if the resonance is sustained. This means, the current remains at \(I_{osc}\). Superimposed on this is the effect of created charges that reduces the voltage and increases the current. The applied voltage is also constant. Townsend discharge then does not occur unless \(I_{osc}\) is achieved first with a specific applied voltage (\(V_{osc}\)) and maintained at this specific voltage. As long as there are free charges inside the tube, the tube is conductive. If this is the case, a discharge tube when subjected to a higher voltage that produces a higher current, \(I\gt I_{osc}\), will not glow or arc as there are no free charges in the tube. This would refute the ionization-avalanche mechanism as the explanation for discharge because this mechanism should also occur at higher voltage. Does \(V_{osc}\) exist for sustained discharge? And that all discharge must pass through \(V_{osc}\)?
Does this happen in a metal like \(Cu\) where a current \(I_{osc}=1.025\text{e-05}\,A\) creates more charge carriers?
Maybe...
Note: no data means no data but its does not mean no information, not including them will be wrong.
A table of \(I_{osc}\) for covalent bond is given below.
| atomic no. | symbol | name | Covalent (single bond) pm | f_osc (10^14)Hz | _hf eV | q_e*f_osc A |
| 1 | H | hydrogen | 38 | 1.2190 | 0.504 | 1.953E-05 |
| 2 | He | helium | 32 | 1.3284 | 0.549 | 2.128E-05 |
| 3 | Li | lithium | 134 | 0.6492 | 0.268 | 1.040E-05 |
| 4 | Be | beryllium | 90 | 0.7921 | 0.328 | 1.269E-05 |
| 5 | B | boron | 82 | 0.8299 | 0.343 | 1.330E-05 |
| 6 | C | carbon | 77 | 0.8564 | 0.354 | 1.372E-05 |
| 7 | N | nitrogen | 75 | 0.8677 | 0.359 | 1.390E-05 |
| 8 | O | oxygen | 73 | 0.8795 | 0.364 | 1.409E-05 |
| 9 | F | fluorine | 71 | 0.8918 | 0.369 | 1.429E-05 |
| 10 | Ne | neon | 69 | 0.9047 | 0.374 | 1.449E-05 |
| 11 | Na | sodium | 154 | 0.6056 | 0.250 | 9.702E-06 |
| 12 | Mg | magnesium | 130 | 0.6591 | 0.273 | 1.056E-05 |
| 13 | Al | aluminium | 118 | 0.6918 | 0.286 | 1.108E-05 |
| 14 | Si | silicon | 111 | 0.7133 | 0.295 | 1.143E-05 |
| 15 | P | phosphorus | 106 | 0.7299 | 0.302 | 1.169E-05 |
| 16 | S | sulfur | 102 | 0.7441 | 0.308 | 1.192E-05 |
| 17 | Cl | chlorine | 99 | 0.7553 | 0.312 | 1.210E-05 |
| 18 | Ar | argon | 97 | 0.7630 | 0.316 | 1.222E-05 |
| 19 | K | potassium | 196 | 0.5368 | 0.222 | 8.600E-06 |
| 20 | Ca | calcium | 174 | 0.5697 | 0.236 | 9.127E-06 |
| 21 | Sc | scandium | 144 | 0.6262 | 0.259 | 1.003E-05 |
| 22 | Ti | titanium | 136 | 0.6444 | 0.266 | 1.032E-05 |
| 23 | V | vanadium | 125 | 0.6721 | 0.278 | 1.077E-05 |
| 24 | Cr | chromium | 127 | 0.6668 | 0.276 | 1.068E-05 |
| 25 | Mn | manganese | 139 | 0.6374 | 0.264 | 1.021E-05 |
| 26 | Fe | iron | 125 | 0.6721 | 0.278 | 1.077E-05 |
| 27 | Co | cobalt | 126 | 0.6695 | 0.277 | 1.073E-05 |
| 28 | Ni | nickel | 121 | 0.6832 | 0.283 | 1.095E-05 |
| 29 | Cu | copper | 138 | 0.6397 | 0.265 | 1.025E-05 |
| 30 | Zn | zinc | 131 | 0.6566 | 0.272 | 1.052E-05 |
| 31 | Ga | gallium | 126 | 0.6695 | 0.277 | 1.073E-05 |
| 32 | Ge | germanium | 122 | 0.6803 | 0.281 | 1.090E-05 |
| 33 | As | arsenic | 119 | 0.6889 | 0.285 | 1.104E-05 |
| 34 | Se | selenium | 116 | 0.6977 | 0.289 | 1.118E-05 |
| 35 | Br | bromine | 114 | 0.7038 | 0.291 | 1.128E-05 |
| 36 | Kr | krypton | 110 | 0.7165 | 0.296 | 1.148E-05 |
| 37 | Rb | rubidium | 211 | 0.5173 | 0.214 | 8.289E-06 |
| 38 | Sr | strontium | 192 | 0.5423 | 0.224 | 8.689E-06 |
| 39 | Y | yttrium | 162 | 0.5904 | 0.244 | 9.459E-06 |
| 40 | Zr | zirconium | 148 | 0.6177 | 0.255 | 9.897E-06 |
| 41 | Nb | niobium | 137 | 0.6420 | 0.266 | 1.029E-05 |
| 42 | Mo | molybdenum | 145 | 0.6241 | 0.258 | 9.999E-06 |
| 43 | Tc | technetium | 156 | 0.6017 | 0.249 | 9.640E-06 |
| 44 | Ru | ruthenium | 126 | 0.6695 | 0.277 | 1.073E-05 |
| 45 | Rh | rhodium | 135 | 0.6468 | 0.267 | 1.036E-05 |
| 46 | Pd | palladium | 131 | 0.6566 | 0.272 | 1.052E-05 |
| 47 | Ag | silver | 153 | 0.6075 | 0.251 | 9.734E-06 |
| 48 | Cd | cadmium | 148 | 0.6177 | 0.255 | 9.897E-06 |
| 49 | In | indium | 144 | 0.6262 | 0.259 | 1.003E-05 |
| 50 | Sn | tin | 141 | 0.6329 | 0.262 | 1.014E-05 |
| 51 | Sb | antimony | 138 | 0.6397 | 0.265 | 1.025E-05 |
| 52 | Te | tellurium | 135 | 0.6468 | 0.267 | 1.036E-05 |
| 53 | I | iodine | 133 | 0.6516 | 0.269 | 1.044E-05 |
| 54 | Xe | xenon | 130 | 0.6591 | 0.273 | 1.056E-05 |
| 55 | Cs | caesium | 225 | 0.5010 | 0.207 | 8.027E-06 |
| 56 | Ba | barium | 198 | 0.5340 | 0.221 | 8.556E-06 |
| 57 | La | lanthanum | 169 | 0.5781 | 0.239 | 9.261E-06 |
| 58 | Ce | cerium | no data | no data | no data | no data |
| 59 | Pr | praseodymium | no data | no data | no data | no data |
| 60 | Nd | neodymium | no data | no data | no data | no data |
| 61 | Pm | promethium | no data | no data | no data | no data |
| 62 | Sm | samarium | no data | no data | no data | no data |
| 63 | Eu | europium | no data | no data | no data | no data |
| 64 | Gd | gadolinium | no data | no data | no data | no data |
| 65 | Tb | terbium | no data | no data | no data | no data |
| 66 | Dy | dysprosium | no data | no data | no data | no data |
| 67 | Ho | holmium | no data | no data | no data | no data |
| 68 | Er | erbium | no data | no data | no data | no data |
| 69 | Tm | thulium | no data | no data | no data | no data |
| 70 | Yb | ytterbium | no data | no data | no data | no data |
| 71 | Lu | lutetium | 160 | 0.5941 | 0.246 | 9.518E-06 |
| 72 | Hf | hafnium | 150 | 0.6136 | 0.254 | 9.830E-06 |
| 73 | Ta | tantalum | 138 | 0.6397 | 0.265 | 1.025E-05 |
| 74 | W | tungsten | 146 | 0.6219 | 0.257 | 9.964E-06 |
| 75 | Re | rhenium | 159 | 0.5960 | 0.246 | 9.548E-06 |
| 76 | Os | osmium | 128 | 0.6642 | 0.275 | 1.064E-05 |
| 77 | Ir | iridium | 137 | 0.6420 | 0.266 | 1.029E-05 |
| 78 | Pt | platinum | 128 | 0.6642 | 0.275 | 1.064E-05 |
| 79 | Au | gold | 144 | 0.6262 | 0.259 | 1.003E-05 |
| 80 | Hg | mercury | 149 | 0.6156 | 0.255 | 9.863E-06 |
| 81 | Tl | thallium | 148 | 0.6177 | 0.255 | 9.897E-06 |
| 82 | Pb | lead | 147 | 0.6198 | 0.256 | 9.930E-06 |
| 83 | Bi | bismuth | 146 | 0.6219 | 0.257 | 9.964E-06 |
| 84 | Po | polonium | no data | no data | no data | no data |
| 85 | At | astatine | no data | no data | no data | no data |
| 86 | Rn | radon | 145 | 0.6241 | 0.258 | 9.999E-06 |
| 87 | Fr | francium | no data | no data | no data | no data |
| 88 | Ra | radium | no data | no data | no data | no data |
| 89 | Ac | actinium | no data | no data | no data | no data |
| 90 | Th | thorium | no data | no data | no data | no data |
| 91 | Pa | protactinium | no data | no data | no data | no data |
| 92 | U | uranium | no data | no data | no data | no data |
| 93 | Np | neptunium | no data | no data | no data | no data |
| 94 | Pu | plutonium | no data | no data | no data | no data |
| 95 | Am | americium | no data | no data | no data | no data |
| 96 | Cm | curium | no data | no data | no data | no data |
| 97 | Bk | berkelium | no data | no data | no data | no data |
| 98 | Cf | californium | no data | no data | no data | no data |
| 99 | Es | einsteinium | no data | no data | no data | no data |
| 100 | Fm | fermium | no data | no data | no data | no data |
| 101 | Md | mendelevium | no data | no data | no data | no data |
| 102 | No | nobelium | no data | no data | no data | no data |
| 103 | Lr | lawrencium | no data | no data | no data | no data |
| 104 | Rf | rutherfordium | no data | no data | no data | no data |
| 105 | Db | dubnium | no data | no data | no data | no data |
| 106 | Sg | seaborgium | no data | no data | no data | no data |
| 107 | Bh | bohrium | no data | no data | no data | no data |
| 108 | Hs | hassium | no data | no data | no data | no data |
| 109 | Mt | meitnerium | no data | no data | no data | no data |
| 110 | Ds | darmstadtium | no data | no data | no data | no data |
| 111 | Rg | roentgenium | no data | no data | no data | no data |
| 112 | Cn | copernicium | no data | no data | no data | no data |
| 113 | Nh | nihonium | no data | no data | no data | no data |
| 114 | Fl | flerovium | no data | no data | no data | no data |
| 115 | Mc | moscovium | no data | no data | no data | no data |
| 116 | Lv | livermorium | no data | no data | no data | no data |
| 117 | Ts | tennessine | no data | no data | no data | no data |
| 118 | Og | oganesson | no data | no data | no data | no data |
Apart from the fact that \(I_{osc}\approx10^{-6}\,\text{to}\,10^{-5}\), there is no reason to believe that \(I_{osc}\) triggers Townsend discharge.
However, if \(I_{osc}\) does trigger a discharge, then the negative charges produced during resonance will reduce the measured voltage and increase the current. The changes in current and voltage are continued only if the resonance is sustained. This means, the current remains at \(I_{osc}\). Superimposed on this is the effect of created charges that reduces the voltage and increases the current. The applied voltage is also constant. Townsend discharge then does not occur unless \(I_{osc}\) is achieved first with a specific applied voltage (\(V_{osc}\)) and maintained at this specific voltage. As long as there are free charges inside the tube, the tube is conductive. If this is the case, a discharge tube when subjected to a higher voltage that produces a higher current, \(I\gt I_{osc}\), will not glow or arc as there are no free charges in the tube. This would refute the ionization-avalanche mechanism as the explanation for discharge because this mechanism should also occur at higher voltage. Does \(V_{osc}\) exist for sustained discharge? And that all discharge must pass through \(V_{osc}\)?
Does this happen in a metal like \(Cu\) where a current \(I_{osc}=1.025\text{e-05}\,A\) creates more charge carriers?
Maybe...
Note: no data means no data but its does not mean no information, not including them will be wrong.