Now who, that proposes to run my life want my blog updated?

Not Forgetting The Devil's Breath

I missed out Hyoscine also know as scopolamine administered via a small pin worn as a ring, kept on the inside of the palm.  A pad on the shoulders,  a pad on the back, to catch you by the elbow is all it takes to introduce the drug.  A small prick from the pin on that ring is all that you feel as you fall under the control of this drug.

It works better if you are first subjected to harassment, and is mentally weak.  Behavioural changes afterwards are very common among children exposed to this drug, the Devil's Breath.

Do I have enough justifications?  Not quite...

Afternotes, Shreds of Memory

A hard kill as opposed to a soft kill is, to turn off the lightings to the path where I jog and two motorized skateboards with steering come charging from behind with armed assailants wearing night vision goggles.  Small arms with a silencer.

Try again, a bazooka...

And again, gas, area wide coverage...
(This time, one of them stood confront and asked hypothetically "You think we wouldn't use gas?"  I said "Thank you." and passed him leeway, I wonder what happened to him.  There were rumors of mass suicide to demonstrate loyalty for a recently deceased king.)

And again...  And again...

Skin Diseases, Brain Trauma And Mind Control

Another way such agent actually "prescribe" mind control drugs to you is first to give you a skin disease.

A liquid contaminated with a protozoan or other parasites, a virus with a necrotizing bacteria that eats into you skin and introduce the virus infection, is poured onto your back or the back of you neck.  If you are unable to wash it off immediately, the disease takes root.  Then a doctor or TCM practitioner recommended by a "concerned" friend, relatives, an acquaintance or a shop assistant does more damage.

The doctor first implant rapport by giving you a drugged drink, through the nurse or the receptionist that causes fear and anxiety.  When the "love" drug takes hold and you feel anxious and unstable, the doctor comes forward to console you, the result of which is that you "fall in love" with him.  This drug is derived from the thorns from cactus or roses.  (I received one of this at the dentist, given to me by the dental assistance.)

The doctor, with placebos and contraindicated drugs that cause severe itching in patients with skin diseases, exacerbates the conditions.  He uses placebos and fall short of not giving the proper medications at all because on record he must be seen as providing proper treatment.  With your agreement that the itching is unbearable, that you need drugs to sleep, you are given mind control psychotic drugs.

You are caged.

At this point, they might decide against killing you as the drugs crush you into their full control.  A overdose of botulism injected and the rabies virus to cause a fear for water, hydrophobia, then an induced high fever together with a prescribed cold bath induces trauma.  The person is sprayed with freezing cold water.  It is just sadistic torture done on children, to induce brain trauma that opens the mind up to mind control later in life.  A key phrase spoken, the mere sight of "protected" persons lockups the mind and the person freezes.  Only with the release word(s) does the person recovers.  In between you are bombarded with hypnotic suggestions and commands that run you life for you.

I know, I know up close and personal...

Drowning while drugged, waterboarding as it is called is a very common way to induce brain trauma.

This is what happens when clandestine mind control methods spread into the hands of rogue governments, small kingdoms, and the mafias and indigenous gangs.  And when countries realise that their rich, academics, researchers and business decisions makers are so targeted, it is tit-for-tat and all hell break lose.

Happy New Year again...

Note:  Check the names of the medications on the web and read about what they are normally prescribed for.  Know the difference in packings for placebos and fake medications.  Check the bottles for tempering, for example if you see a white residue powder at the bottom of the bottle but the pill is of a different color, blue, your pills have been swapped.  You are probably given something contraindicated.

Check the names of the doctor and try to locate him on the web, if you find more than one person with the same name at two or more different locations, investigate further.  One of them could be an imposter.

Self-medicate given the wealth of information on the web is less dangerous.

Unless you have lived in the wild picking leaves, roots and mushrooms, I would advise against harvesting herbs from the local flora.  Herbivore you are not.

Most importantly resolve the issues that make you a target.  Otherwise, call your armada, earth will experience its first star war!

The Prince, Feared Or Loved?

Niccolò Machiavelli was Mona Lisa.  He dressed up as a woman to escape his pursuers, his friend was so impressed with him dressed up as a woman, he got him to dress up as a woman again and made a portrait.

Things happen during the Renaissance...

Botulism, Harassment, Murder And Mind Control

There is more than just botulism toxin dripped onto the back of your neck, it is orchestrated with consistent harassment.  If you stay in a flat, such agents will first create some pretexts and visit your flat to make a layout of your living space, then comes the incessant knocking on your ceiling.  They might even install video surveillance to track your movement in the flat to make sure that the knocking is always above you.   To add to the noise, the agents removes filters from your water drainage pipes so that the water drains loudly, and remove anti-knocking devices from water pipes around your flat so that they knock every time they turn the tap on then off.

A common pretext that is often used to gain access into your flat is to drop a piece of clothing from above your flat and wish to recover it, or to inspect for mosquitoes, or to check for gas leaks, or to check for faulty electrical wirings.  There can be many scenarios created as the imagination is capable of.

All  these are done to induce a state of anxiety in the target; the first step towards a mental breakdown and, as these agent would wish, a first step towards suicide.  It is not suicide but murder!

They test your mental state by shouting and making loud noise in your presence.  If you get startled easily, they have the feedback they needed...They would start an argument with you as you queue in the supermarket, knocking into you as you jog, as you walk, pretend to slip and spill food or water onto you when you dine outside, to gauge your mental state.

Workplace harassment is also part of the repertoire.  Agents installed at the workplace as contractors, uncooperative subordinates or contract workers, a student or parents of a student if you lecture and teach, a teacher that swap pages of your test answer sheets if you are a student, bullying by a fellow student, usually much older and way out of place, all these to continue with the harassment after you leave the flat.

If you live in a landed property, knocking comes next door and from bogus construction work just outside, on the street.

In particular, the agents swapped both the water and gas meters and install them at a nearby food stall in a hawker center, so that the utility bills explodes.

Swapping your medications or wrong medications in dosage or type is a common tactic to weaken the targets.  They do not kill with outright poisoning other than an induced stroke.  A soft kill is not ostentatious.  A soft kill is another unfortunate statistic that adds to the annual death rate and not the murdered count.  In particular, the one who ordered such agents might wish for one up on the annual suicide count.

They frame you for shoplifting by placing small items into your handbag and carrier from behind.  These sometimes involve agents pretending to be the enforcements or are actually with the police.  It was interesting to track where this path along the legal trail might lead...

My mind drifts to Nicholas Tesla in his final days.  He booked into a motel to get some rest because there were too much noise where he used to stay...

A concerted effort by an experienced team to commit murder.  The murdered targets can be schooling children, teenager, working adults and old persons.  It is unscrupulous in the first place to commit murder and they do get away with murders; it is not surprising that schooling children can be the targets.

I have seen enough to identify members of the team and pick out those in the know of such undertakings.  I know them.  This kill technique so common, it has a colloquial name "ghosts knock on the wall"; 鬼敲墙！

Happy holidays...

Note:  Personal note, Chinatown, 1987;  Hong Lim, 1988

Happy New Year 2017

And that's was the last year of peace in this region...

War! War! War!

Hey, I am not wishing for war.  I just know...

And so I ate some luncheon meat?  What mind control?  I will start your fire.

Temperature Particles Found

A cold spoon sticks to the bottom of a boiling pot.  It does not matter what material the spoon is made of, plastic, steel or porcelain, a force pulls on the spoon at the bottom of a metal pot with boiling water, over a stove.

Negative temperature particles on the cold spoon attract positive temperature particles on the boiling pot.

Try it; from that part in all of us, part Greek.

Note:  Hot pot, cold spoon.

Odd Give, Even Give Unequally...

It is interesting that,

$\cfrac{\pi^2}{6}=\cfrac{1}{1^2}+\cfrac{1}{2^2}+\cfrac{1}{3^2}+...$

and

$\cfrac{\pi^2}{8}=\cfrac{1}{1^2}+\cfrac{1}{3^2}+\cfrac{1}{5^2}+\cfrac{1}{7^2}+...$

but,

$\cfrac{\pi^2}{24}=\cfrac{1}{2^2}+\cfrac{1}{4^2}+\cfrac{1}{6^2}+\cfrac{1}{8^2}+...$

that odd and even numbers do not contribute equally to $\sum^{\infty}_1{\cfrac{1}{n^2}}$

$\cfrac{3}{24}=\cfrac{1}{8}\ne\cfrac{1}{24}$

Odd number contributes three times more than even numbers.

Why?

Towards Ionization!

Hei, knock on my door I'll give you war!  For real.

In the plot f(x) = ((x)^(1/3)-(2)^(1/3))/((2*x)^(1/3))-log((x/2)^(1/3)) and its derivative f(x),

the points of interest is beyond $x=1$, the gradient of $f(x)$ tends towards zero.  The corresponding energy levels piles up with higher values of $x$, ie $f(x+1)-f(x)$ narrows but only gradually compared to Rydberg formula.

No, ionization has nothing to do with particles in collision, coalescence and separation.

Teenage alcoholism and weapon grade encephalitis.  Alcohol actually provides temporary relief.  In these cases, alcohol clears the mind.

Just Like A Snake

Temperature particles provide continuous spectra in the ultraviolet, the infrared, and visible range ("Lemmings Over The Cliff..." dated 18 May 2016).  Temperature particles could also be responsible for smell and taste, that fresh vegetable and food taste bland after they have been stored in cool temperature.   The coolness of negative temperature particles neutralizes  positive temperature particle and takes the taste/smell away.

So, if you are able to isolate temperature particles, give positive temperature particles a good lick and varies its amount as you do so.  It is likely that a combination of positive and negative temperature particles in succession provide a unique favor to foodstuff.

Just like a snake tasting the air with its tongue is seeing in infrared.

Absorbed!

But a plot of (x^(1/3)-a^(1/3))/((a*x)^(1/3))-ln((x/a)^(1/3)) for a =1 to 7 is,

All absorption lines!  A plot of 1/a^2-1/x^2 for a =1 to 7 gives,

which indicates,

$h\ne\cfrac{c}{5\pi }a_{\psi\,c}m_{ \psi  }$

on the basis that Rydberg formula is correct.

What is $h$?

It Is Just Nice To Look At

Hei, from the previous post "A Deep Dark Secret" dated 25 Oct 2016,

$F=\cfrac { 2 }{ 5 } m_{ \psi  }a^{ 2 }_{ \psi  }\left( \cfrac { c }{ 2\pi a_{ \psi  } }  \right) ^{ 2 }\cfrac{1}{a_{\psi}}$

instead,

$F_{\rho}=\cfrac { 2 }{ 5 } m_{ \psi  }a^{ 2 }_{ \psi  }\left( \cfrac { c }{ 2\pi a_{ \psi  } }  \right) ^{ 2 }\cfrac{1}{a_{\psi}}$

since we are dealing with $\psi$ and it is force density all along, $m_{\psi}=m$ is mass density, a constant.  So,

$F_{\rho\,n_1}=\cfrac { 2 }{ 5 } m_{ \psi  }a^{ 2 }_{ \psi \,n1 }\left( \cfrac { c }{ 2\pi a_{ \psi n1 } }  \right) ^{ 2 }\cfrac{1}{a_{\psi\,n1}}$

then

$F_{\rho\,n_1}=\cfrac{1}{10\pi^2 }\cfrac{m_{ \psi  }c^2}{a_{\psi\,n1}}$

and similarly,

$F_{\rho\,n_2}=\cfrac{1}{10\pi^2 }\cfrac{m_{ \psi  }c^2}{a_{\psi\,n2}}$

In general,

$F_{\rho\,n}=\cfrac{1}{10\pi^2 }\cfrac{m_{ \psi  }c^2}{a_{\psi\,c}}.\cfrac{1}{\sqrt[3]{n}}$

since,

$\cfrac{1}{n}=\left(\cfrac{a_{\psi\,c}}{a_{\psi\,n}}\right)^3$

as such,

$W_{1\rightarrow 2}=\int^{a_{ \psi\,n2 }}_{a_{ \psi\,n1 }}{F_{\rho\,n}}\,da_{\psi}$

$W_{1\rightarrow 2}=\int^{n2}_{n1}{F_{\rho\,n}}\,d\left(\sqrt[3]{n}.a_{\psi\,c}\right)$

$W_{1\rightarrow 2}=\cfrac{m_{ \psi  }c^2}{10\pi^2 }\int^{n2}_{n1}{\cfrac{1}{\sqrt[3]{n}}}.d\sqrt[3]{n}$

and

$W_{1\rightarrow 2}=\cfrac{m_{ \psi  }c^2}{10\pi^2 }\left[ln(\sqrt[3]{n_2})-ln(\sqrt[3]{n_1})\right]=\cfrac{m_{ \psi  }c^2}{10\pi^2 }ln\left(\cfrac{\sqrt[3]{n_2}}{\sqrt[3]{n_1}}\right)$

$W_{1\rightarrow 2}=\cfrac{m_{ \psi  }c^2}{10\pi^2 }ln\left(\cfrac{a_{\psi\,n2}}{a_{\psi\,n1}}\right)$

Which is somehow very satisfying, irrespective of all and any maths and logic blunders.  It is so nice I took a second look.  Since,

$f_c=\cfrac{c}{2\pi a_{\psi\,c}}$

the expression suggests,

$h=\cfrac{c}{5\pi }a_{\psi\,c}m_{ \psi  }$

which would be a constant for a given particle.  Nice!

A plot of ln((n_2/n_1)^(1/3)) for n_1 = 1 to 7 is given below,

As $a_{\psi\,n1}\rightarrow a_{\psi\,n2}$, there is an release of energy as the particle grows bigger.  $W_{1\rightarrow 2}$ requires energy and reduces the amount of energy released.

The most energy required from a transition from $n_1=1$ results in a absorption line.

Big particle with less energy in $\psi$ is counter-intuitive.  Big particle has $\psi$ going around its circumference at lower frequency, given that $\psi$ has a constant speed, light speed.

But a big particle contains a small particle!?

A big particle has more extensive $\psi$ but its $\psi$ has less energy when we take reference at the circumference/surface of the particle.

$v=r\omega$

All inner $\psi$ have lower speed, when they do not slide along each other.  $\psi$ outside from the center of the particle need to move faster; the fastest of which is light speed.

A Deep Dark Secret

Consider the force holding $\psi$ in circular motion,

$L=I\omega$

$F=L\cfrac{\omega}{r}$

scale by $r$, as the further from the center the less $L$ has to turn,

$F=I\cfrac{\omega ^{ 2 }}{r}$

In the case of a point mass, $m$ in circular motion with velocity $v$, in circle of radius of $r$,

$F=mr^2\left(\cfrac{v}{r}\right)^2.\cfrac{1}{r}=m\cfrac{v^2}{r}$

where $I=mr^2$ and  $\omega =\cfrac{v}{r}$ in radian per second.  Which is as expected.

In this case of a sphere of $\psi$,

$F=\cfrac { 2 }{ 5 } m_{ \psi  }a^{ 2 }_{ \psi  }\left( \cfrac { c }{ 2\pi a_{ \psi  } }  \right) ^{ 2 }.\cfrac{1}{a_{ \psi  } }$

where $\omega=\cfrac{v}{2\pi a_{\psi}}$ in per second.

$m_{ \psi  }=\rho _{ \psi  }.\cfrac { 4 }{ 3 } \pi a^{ 3 }_{ \psi  }$

$F=\cfrac { 4 }{ 30\pi }  c^{ 2 }\rho _{ \psi  }.a^{ 2 }_{ \psi  }$

$W_{ 1\rightarrow 2 }=\int _{ a_{ \psi \,n1 } }^{ a_{ \psi \, n2 } }{ F } da_{ \psi  }$

If we assume that, $\rho _{ \psi  }\propto \psi$.

For,

$a_{ \psi \, n1 }>a_{ \psi \,\pi  }$ and $a_{ \psi \, n2 }>a_{ \psi \,\pi  }$

$\psi=\psi_{\pi}=constant$

We have,

$W_{ 1\rightarrow 2 }=\cfrac { 4 }{ 90\pi } c^{ 2 }\rho _{ \psi  }\left[ a^{ 3 }_{ \psi \, n_{ 2 } }-a^{ 3 }_{ \psi \, n_{ 1 } } \right]$

$W_{ 1\rightarrow 2 }=\cfrac { 4 }{ 90\pi  } c^{ 2 }\rho _{ \psi  }a^{ 3 }_{ \psi \, c }\left[ n_{ 2 }-n_{ 1 } \right]$

which is just as wrong as the other expressions derived previously.

For,

$a_{ \psi \, \, n1 }\lt a_{ \psi \, \, \pi  }$

$\, \, a_{ \psi \, \, n2 }\lt a_{ \psi \, \, \pi  }$

$W_{ 1\rightarrow 2 }=\int _{ a_{ \psi \, \, n1 } }^{ a_{ \psi \, \, n2 } }{ F } da_{ \psi  }=\cfrac { 4 }{ 30 \pi} c^{ 2 }\int _{ a_{ \psi \, \, n1 } }^{ a_{ \psi \, \, n2 } }{ \rho _{ \psi  }.a^{2 }_{ \psi  } } da_{ \psi  }$

From the post "Not Quite The Same Newtonian Field" dated 23 Nov 2014,

$\psi =-i{ 2{ mc^{ 2 } } }\, ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{ z })))+c$

$\psi =-{ 2{ mc^{ 2 } } }\, ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (a_{ \psi  })))$

under the assumption, $\rho _{ \psi  }\propto \psi$

$\rho _{ \psi  }=A. \psi$

We have,

$W_{ 1\rightarrow 2 }=A.\cfrac { 8 }{ 30 \pi } mc^{ 2 }\int _{ a_{ \psi \,  n1 } }^{ a_{ \psi \,  n2 } }{ -a^{ 2 }_{ \psi  } } ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (a_{ \psi  })))\,da_{ \psi  }$

At last, inevitably the ugly bride meets the in-laws ...

What is $m$?  This was a problem since $F_{\rho}$ or $F$, the force density was written down.  A force has to act on some mass.  If force density acts on mass density then,

$\rho_{\psi}=m$

then $\psi=f(\rho_{\psi})$, given $\rho_\psi$,

$\psi =-{ 2{ \rho_{\psi}c^{ 2 } } }\, ln(cosh(\cfrac { G }{ \sqrt { 2{ \rho_{\psi}c^{ 2 } } }  } (a_{ \psi  })))$

from which we solve for $\rho_{\psi}$ given $\psi$.  Which make sense because $W_{ 1\rightarrow 2 }$ is moving $\psi$ about, we must know the amount of $\psi$ in question to calculate $W_{ 1\rightarrow 2 }$.

But does energy density has mass density?  Does energy has mass?

This path does not provide an easy answer to $W_{ 1\rightarrow 2 }$  as $L_{n1}\rightarrow L_{n2}$.  $W_{ 1\rightarrow 2 }$ might be similar to Rydberg formula,

$\cfrac{1}{\lambda}=R_{\small{H}}\left(\cfrac{1}{n^2_1}-\cfrac{1}{n^2_2}\right)$

A new passport and a ticket to nowhere...you asked for it!

Not Orbits!

There is no orbits, $\psi$ is held by $F_{\rho}$ the force density.  Moving $\psi$ away from the center of the particle act against $F_{\rho}$ or $\int{F_{\rho}}dr$.

There is no charge particle holding another oppositely charged particle, and no field between the oppositely charged particles, along which potential energy changes.

Adjusting for Bohr - Angular Momentum

The expression,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}-\cfrac { n_{ o2 }^{ 2 }-n_{ o1 }^{ 2 } }{ (n_{o1 }n_{ o2 })^{ 2 } }\right)$

from the post "Particles In Orbits" dated 18 Oct 2016, with $n_i=n_{oi}=i$,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}-\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{1 }n_{ 2 })^{ 2 } }\right)$

is ARBITRARY.  It is the result of comparing the leading constant to Planck relation $E=h.f$

The term,

$\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}$

is due to the difference in potential energy associated with the spin of $\psi$ with $n_1$ and $n_2$, at $a_{\psi\,n_1}$ and $a_{\psi\,n_2}$, respectively, and the term,

$\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{1 }n_{ 2 })^{ 2 } }$

is the change in energy as momentum changes due to a change in $n$, assuming that the particles after coalescence/separation are still at light speed.

Let's formulate the change in energy as per Bohr model due to a quantized change in momentum.  Since $n$ is an integer, the change in momentum is equally spaced and so if $L_{n1}$ is the momentum of the

$KE_{n1}=\cfrac{L_{n1}^2}{2m_1}$
$KE_{n2}=\cfrac{L_{n2}^2}{2m_2}$

the change in $KE$ is

$\Delta KE=KE_{n2}-KE_{n1}=-\Delta PE$

$\Delta PE=\cfrac{L_{n1}^2}{2m_1}-\cfrac{L_{n2}^2}{2m_2}$

since we know that the change in momentum is entirely due to a change in $n$.  For a particle spinning at light speed $c$,

$L_{ n2 }=\cfrac{2}{5}m_{ 2 }a^2_{ \psi \, \, n2 }.\cfrac{c}{2\pi a_{\psi\,\,n2}}=\cfrac{c}{5\pi}m_{ 2 }a_{ \psi \, \, n2 }$

$L_{ n1 }=\cfrac{2}{5}m_{ 1 }a^2_{ \psi \, \, n1 }.\cfrac{c}{2\pi a_{\psi\,\,n1}}=\cfrac{c}{5\pi}m_{ 1 }a_{ \psi \, \, n1 }$

where $I_{ni}=\cfrac{2}{5}m_i.a^2_{\psi\,\,ni}$ is the moment of inertia of a sphere of radius $a_{\psi\,\,ni}$.  So,

$\Delta PE=\cfrac { m_{ 1 }\left( a_{ \psi \, \, n1 }c \right) ^{ 2 } }{ 50 \pi^2} -\cfrac { m_{ 2 }\left( a_{ \psi \, \, n2 }c \right) ^{ 2 } }{ 50\pi^2 } =\cfrac { c^{ 2 } }{50\pi^2 } \left( m_{ 1 }a^{ 2 }_{ \psi \, \, n1 }-m_{ 2 }a^{ 2 }_{ \psi \, \, n2 } \right)$

As particle of radius $a_{ \psi \, \, n1 }$ is made up of $n_1$ basic particle of radius $a_{\psi\,c}$, $n=1$

$\cfrac { 1 }{ n_{ 1 } } =\left( \cfrac { a_{ \psi \, \, c } }{ a_{ \psi \, \, n1 } }  \right) ^{ 3 }$,   $a_{ \psi \, \, n2 }=\sqrt [ 3 ]{ n_{ 2 } } a_{ \psi \, \, c }$

and particle of radius $a_{ \psi \, \, n2 }$ is made up of $n_2$ basic particle of radius $a_{\psi\,c}$, $n=1$

$\cfrac { 1 }{ n_{ 2 } } =\left( \cfrac { a_{ \psi \, \, c } }{ a_{ \psi \, \, n2 } }  \right) ^{ 3 }$,   $a_{ \psi \, \, n1 }=\sqrt [ 3 ]{ n_{ 1 } } a_{ \psi \, \, c }$

Since we have assumed that the volume of the particles are conserved and their density is a constant,

$m_{ 1 }=n_{ 1 }m_{ c }$

$m_{ 2 }=n_{ 2 }m_{ c }$

So after substituting for $m_{i}$,

$\Delta PE=\cfrac { m_{ c }c^{ 2 } }{ 50\pi^2 } \left( n_{ 1 }a^{ 2 }_{ \psi \, \, n1 }-n_{ 2 }a^{ 2 }_{ \psi \, \, n2 } \right)$

And after substituting for $a_{\psi\,i}$,

$\Delta PE=\cfrac { m_{ c }c^{ 2 } }{50\pi^2 } \left( n_{ 1 }\left( \sqrt [ 3 ]{ n_{ 1 } } a_{ \psi \, \, c } \right) ^{ 2 }-n_{ 2 }\left( \sqrt [ 3 ]{ n_{ 2 } } a_{ \psi \, \, c } \right) ^{ 2 } \right)$

$\Delta PE=\cfrac { m_{ c }c^{ 2 } }{ 50\pi^2 } a^{ 2 }_{ \psi \, \, c }\left( n_{ 1 }^{ 5/3 }-n_{ 2 }^{ 5/3 } \right)$

$L_{ c }=\cfrac{2}{5}m_{ c }a^2_{ \psi \, c }.\cfrac{c}{2\pi a_{ \psi \, c }}$

$\Delta PE=\cfrac { L_c^{ 2 } }{ 2m_c } \left( n_{ 1 }^{ 5/3 }-n_{ 2 }^{ 5/3 } \right)$

What happened to $\cfrac{1}{n^2_1}-\cfrac{1}{n^2_2}$?  The above expression is the change in $KE$ required for the specified change in momentum, it is not the associated change in energy level of the particle in its field.

Consider again,

$L_{ n2 }=\cfrac{2}{5}m_{ 2 }a^2_{ \psi \, \, n2 }.\cfrac{c}{2\pi a_{\psi\,\,n2}}=\cfrac{c}{5\pi}m_{ 2 }a_{ \psi \, \, n2 }$

$\cfrac{L_{ n2 }}{L_{ n1 }}=\cfrac{m_2}{m_1}.\cfrac{a_{ \psi \, \, n2 }}{a_{ \psi \, \, n1 }}$

$\cfrac{L_{ n2 }}{L_{ n1 }}=\cfrac{n_2}{n_1}.\sqrt[3]{\cfrac{ n_2 }{ n_1 }}=\left(\cfrac{n_2}{n_1}\right)^{4/3}$

Which is not $\cfrac{L_{n1}}{L_{n2}}=\cfrac{n_1}{n_2}$ as in Bohr model of

$L=n\hbar$

And if we follow through the derivation for the energy difference between two energy levels, $n_1$ and $n_2$ we have,

$E_{\small{B}}=R_{\small{E}}\left(\cfrac{1}{\left(n_1^{4/3}\right)^2}-\cfrac{1}{\left(n_2^{4/3}\right)^2}\right)=R_{\small{E}}\left(\cfrac{1}{n_1^{2.667}}-\cfrac{1}{n_2^{2.667}}\right)$

this does not effect the first term, $\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}=\cfrac{1}{\sqrt[3]{n_1}}-\cfrac{1}{\sqrt[3]{n_2}}$ associated with the decrease in potential energy of the system in circular motion, as $\psi$ move from $n_1\rightarrow n_2$.

$E_{\small{B}}$ derived above is the potential energy change in the field ($E\propto\frac{1}{r}$) that accompanies a change in the angular momentum of $\psi$ as the energy level transition $n_1\rightarrow n_2$ occurs.  $E_{\small{B}}$ is recovered from the amount of potential energy associated with the system in circular motion, released.

$C=\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } -\cfrac { n_{ 2 }^{ 2.667 }-n_{ 1 }^{ 2.667 } }{ (n_{ 1 }n_{ 2 })^{ 2.667 } }$

Does a system in circular motion have potential energy $h.f$ in addition to the system $KE$ as quantified by its angular momentum?  The work done field against the force $-\frac{\partial\psi}{\partial r}$ as $\psi$ moves away from the center along a radial line, is strictly not $\propto\frac{1}{r^2}$ but, the Newtonian $F$,

$F\propto -\int{F_{\rho} dr}$

$F\propto -ln(cosh(r))$

$E\propto -\int{ln(cosh(r))} dr$

and things get very difficult.

A series plot of ((x)^(1/3)-(a)^(1/3))/(a*x)^(1/3)-((x)^(2.667)-(a)^(2.667))/(a*x)^2.667 for a =1 to 6 is given below,

where the plots for $n_1=2$ and $n_1=3$ is almost coincidental.

After adjusting for $2\rightarrow 2.667$, the plots are essentially the same except that the overlapping plots are not $n_1=2$ and $n_1=5$ but $n_1=2$ and $n_1=3$.

It seems that $n_1=2$ and $n_1=3$ are degenerate, instead.

The important points are: a pair of degenerate plots ($n_1=2$ and $n_1=3$) very close together occurs naturally, that the plot for $n_1=1$ is an absorption line and the rest of the plots $n_1\ge 2$ are the emission background.

Good night.

Note: The potential energy of a ball in circular motion held by a spring, is the energy stored in the spring as the spring stretches to provide for a centripetal force.  It is clear in this case, that the $KE$ of the ball, (equivalently stated as angular momentum) is a separate issue from the potential energy stored in the spring.  The energy required to increase the ball's $KE_1\rightarrow KE_2$ is not the same as the energy required to stretch the spring further as the ball accelerates from $KE_1\rightarrow KE_2$.

It could be that $\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}$ defines the energy change needed at the perimeter, $a_{\psi\,\,n2}$ and $-\cfrac { n_{ 2 }^{ 2.667 }-n_{ 1 }^{ 2.667 } }{ (n_{ 1 }n_{ 2 })^{ 2.667 } }$ is the change in energy along a radial line from $n_1$ to $n_2$.

Note: Positive is emission line, and negative is absorption line

How Big To Be?

The emission spectrum is continuous with $n_2$ large.  As the total energy of the system increases with high electric field or high temperature, $n_1$ can take on smaller values.  The absorption lines with $n_1=1$ are distinctive against the background of continuous emission.  How small is $n_{1}$ to be?

$n_1=1$

In order to generate a continuous emission spectrum, how big is $n_2$ to be?

$n_2=n_{\small{large}}\gt\gt 77$

How to make $n_2$ big?  If $n_2$ is spinning about the center of the particle, will $n_2$ be big?  The centripetal force acting against a pinch force that pull $\psi$ away from the particle will allow more $\psi$ to attach itself to the particle and allows $n_2$ to increase.

Maybe...

Electrons May Not Be Involved!

With,

$C=\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } -\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$

Could it be that because the spectrum observations are conducted at high temperature,

$\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  }$

is due to temperature particles.  And,

$\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$

orbiting particles not necessarily electrons.

$C=C_{\small{T}}+C_{\small{B}}$

where $C_{\small{T}}$ is due to the energy density $\psi$ of interacting temperature particles and $C_{\small{B}}$ due to Bohr model not necessarily of electrons, with quantized momenta.

Temperature particles orbiting around electron orbits that in resonance produce the infrared spectrum and the ultraviolet spectrum was proposed previously "Lemmings Over The Cliff..." dated 18 May 2016.  Spectra lines in the visible spectrum will come from transitions between energy levels in the ultraviolet spectrum.

Electrons may not be involved!  Unless the observations are also conducted in the presence of a high electric field, electron are not affected.  If this is the case, temperature should be raised without the use of an electric field.  Within the confines of the experiment, there should be only one strong field affecting one type of particles.

Let electrons not be involved.  Maybe then spectra lines can be indicative of the existence of temperature particles in orbit around electron orbits as proposed in the post "Capturing $T^{+}$ Particles" dated 15 May 2016.

In which case, electron ionization energies has no direct relation with spectra lines (due to temperature particles), not even in the simplest model of Hydrogen atoms.

Have a nice day...

When Wrong Is Right

I know...

Emission spectrum is taken to be the complement of the absorption spectrum.

If all energy transitions have its complement then, they should all cancel.  Dark lines should not be visible and the background emission spectrum against which they appears must also be accounted for.  Ambient light or indirect illumination alone does not generate the background emission spectrum.

Emission spectrum in this blog refers to the backdrop upon which the dark absorption lines appears.

Redefining everything to be right...When wrong is right, what's left?  Right??

Eliminate all indirect illumination, does fine gaps appears in the background emission spectrum?

Wrong does not make left right, it just leaves behind more questions...  More questions are what's left!

What "It"?

Given,

$C=\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } -\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$

when $n_2\lt n_1$, for absorption lines, $n_1\leftrightarrow n_2$,

$D=\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }-\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } =-C$

And so we can expect the spectra line $n_{1\rightarrow 2}$ to cancel the reverse spectra line $n_{2\rightarrow 1}$.

For the sake of it....

The Total Picture

Why is emission that occurs at the same time as absorption lines almost continuous, whereas absorption lines are spaced far apart?

For each value of $n_2$, we have an absorption line.  All possible values of $n_2$ generates an absorption series.

With many $n_1=1$ plots corresponding to different allowable energy states, we have a number of absorption series that depict such plausible energy states.

Which is confusing because of the one added level of indirection.  For example,

$n_1=1$ for $n=1$

$n_1=1$ for $n=2$

and

$n_1=1$ for $n=3$

where the lowest energy level of a particle of two constituent basic particles is different (higher) than the lowest energy of a particle with three constituent basic particles.

What happened to $n_i=n_{oi}=i$?

This expression considers the lowest (first, $n_1=1$) energy level in a set of all particle sizes (all $n$).

Energy transitions occur for a particle given its number of constituent basic particle size $n$ with a particular first/lowest energy state.  The lowest energy level $n_1$, given $n$ is an absorption series given all values of  $n_2$.  Across all possible values of $n_1$ as particle $n$ size changes, we have different absorption series.

And the question was, why is it an absorption line?  Which leads to the question, are there fine gaps in the background emission spectrum?

Note:  In the plot above $n_1=1$ is indicative of the lowest energy level of a particular $n$.  Given all values of $n_2$, $n_1=1$ for a particular $n$ generates a absorption series.  When $n$ is higher $n_1=1$ is lower, with less energy.  Different $n$, generates different spectra series.

Shifty Spectra

Given equal probabilities that an emission or absorption occurs at all energy levels to all energy levels, why does not $-E_{1\rightarrow n}$, an emission line and $E_{n\rightarrow 1}$, where $n_2=n_1$, an absorption line, cancels?

When $n_2\lt n_1$, emission plots become absorption plots but the single absorption plot does not become an emission plot.  For the case of $n_1=1$ there cannot be a lower $n_2$, so there is no emission lines from $n_1$ to a lower energy level.  This does not mean that there is no emission lines  that would cancel the absorption lines from $n_1=1$.  A transition $E_{2\rightarrow 1}$ would generate a emission line that cancels the absorption line of $E_{1\rightarrow 2}$.

Do the forward transition and its the reversed transition cancels?  Only with numerical calculations can tell.

but we see that,

because the emission plots are not parallel for values of $n_2$ smaller than $n_2$ at the minimum point.  And in the following case its is not possible to tell graphically whether the forward transition and its reversed transition cancels,

Notice that $E_{2\rightarrow 3}$ is an emission line and $E_{3\rightarrow 2}$ is an absorption line.  All transitions to the left of the final energy state plot minimum point $n_{2\,\,min}$ have the opposite sign.  The corresponding reversed transition is at the minimum point and move downwards vertically to the final energy state.

It is also noted that for the plot $n_1=5$ another zero x-axis intercept occurs at $n_2=2$. And that for the plot $n_1=2$ another zero x-axis intercept occurs at $n_2=5$.

Both plots are coincidental.  This suggests that moving from $n_1=2$ to $n_2=5$ and in reverse, requires no net energy.  In moving from $n_1=$ to $n_2=5$, the loss in energy due to a decrease in $a_{\psi}$ at $n_2=5$ is made up for by the increase in orbital energy there, and vice versa.

It is possible that there is an energy gradient as the spectra line observations are being made.  When energy of the system is increasing, small particles tend to form and $n_1\gt n_2$.  When energy of the system is decreasing, big particles tend to form on separation after a collision and $n_1\lt n_2$.  In this way, $n_1\ne n_2$ and the emission lines are not coincidental with the absorption lines.  Emission and absorption that cancel do not occur with equal probability.

Which means, with no energy input to the system, the spectra line would disappear, but a different sets of lines reappears as the system cools.

Note: The difference between $\Delta E_{1\rightarrow 2}$ and $E_{1\rightarrow 2}$ is the overall change in energy state as demarcated by the definition of one single process and one energy state change of many in a more prolonged process.

In the case where separation follows coalescence,

$\Delta E_{1\rightarrow 2}\ne h.f$

instead there are both an emission line,

$E_{1\rightarrow n}$

and an absorption line,

$E_{n\rightarrow 2}$

As the small particle grows bigger its energy drops due to a larger $a_{\psi}$.

We All Have Our Say, Give And Take

We can do away with the notion of colliding atoms and think of the high energy experimental conditions as being conducive to the formation of small particles,

In the top most diagram, a small particle passes through a big particle with energy transitions $-E_{1\rightarrow n}$ and $E_{n\rightarrow 2}$ occurring, on entering the big particle and on leaving the big particle.  Both emission and absorption of energy occur in the same process.  The overall change in energy state is,

$\Delta E_{1\rightarrow 2}=E_{n\rightarrow 2}-E_{1\rightarrow n}$

but $-E_{1\rightarrow n}$ is energy emission and $E_{n\rightarrow 2}$ is energy absorption.

In the middle diagram, the small particle coalesce with the big particle and the lowest energy level attained is,

$n=n_{ large }+n_{ 1 }$

and the resulting bigger particle subsequently breaks into $n-n_2$ and $n_2$ particles where $n_2\ne n_1$.  This differ from the toppest case where $n\ne n_{ large }+n_{ 1 }$ but simply $n\gt n_1$.  The small particle passing through the big particle retains its distinctiveness because of its high momentum.  These two diagrams provide two scenarios as to what happened to the small particle inside the big particle.  In both cases a subsequent departure follows coalescence.

The third diagram shows a simple coalescence without a subsequent separation.  Only one singular energy emission occurs as $n_1\rightarrow n_{\small{large}}+1$.

All three process occur simultaneously.  Emissions form the colored background against which dark absorption lines show up in contrast.

Have a nice day.

Note:  This discussion is of all $n_1$, not just $n_1=1$.  The case of $n_1=1$ generates an absorption series given the two processes involved; changing $a_{\psi}$ and changing orbital energy level as according to Bohr model.

Speculating While Others Sleep...

Continued from the previous post "Particles In Orbits" dated 18 Oct 2016,

$n_1=n_{o1}=1$

where the lowest energy level coincides with the smallest particle.  In general,

$n_i=n_{oi}=i$

if we argue that given light speed a constant,  angular momentum at light speed increases proportionally with the number of constituent basic particle $n$.  This makes $n_{oi}$ from Bohr model coincidental with $n_{i}$, the number of constituent basic particles.

Which brings us to the emission plots $n_1=2$ and $n_1=5$,

When experimental conditions are such that $n_i\ne n_{oi}$, these two plots will split and emerge as two separate emission lines.  A new line mysteriously appears.

When an appropriate external field is applied the angular momentum of the particle changes in the direction of the field, $n_i\ne n_{oi}$.

This is not Zeeman effect nor Stark effect.

But an indication of the two processes involved as particles coalesce and break.

Speculating while others sleep...

The problem is $n_2=80$ is not large enough for the various plots to be on a plateau (approaching an asymptote).  Since, spectra lines are observed only in high energy conditions and we postulate that high energy conditions enables $n_2\rightarrow large$, then

$R_{\small{H}}=\cfrac{1}{2\pi a_{\psi\,c}}$

is still on the chopping block, where

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{1}{\sqrt[3]{n_1}}-\cfrac { 1 }{ (n_{o1 })^{ 2 } }\right)$

$\Delta E_{n\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{1}{\sqrt[3]{n_1}}-\cfrac { 1 }{ (n_{o1 })^{ 2 } }\right)$

and

$\Delta E_{n\rightarrow 2}=h.f_{\psi\,c}\left(\cfrac{1}{\sqrt[3]{n_2}}-\cfrac { 1 }{ (n_{o2 })^{ 2 } }\right)$

And in a sleep deprived and convoluted way,

$\Delta E_{2\rightarrow 1}-\Delta E_{n\rightarrow 2}=h.f_{\psi\,c}\left(\cfrac{1}{\sqrt[3]{n_2}}-\cfrac{1}{\sqrt[3]{n_1}}+\cfrac { 1 }{ (n_{o2 })^{ 2 } }-\cfrac { 1 }{ (n_{o1 })^{2 } }\right)$

Which is really interesting...a transit from energy level $1$ to a higher level $n$ and a return to energy level $2$.  The negative sign here makes the absorption line an emission line.  This is how an absorption spectrum has a parallel emission spectrum!

A collision not of basic particles but atoms that contains the basic particles.  The high energy level state is achieved when the colliding atoms are close together on impact.  As the atoms part after the collision, the transition $\Delta E_{n\rightarrow 2}$ occurs.

Hmmm...

Particles In Orbits

The difference plot of the expression, $\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } -\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$

however, draws a sense of déjà vu...

The concepts leading to the expression $\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  }$ reverses the energy sign of conventional electron energy level transitions; $E_{1\rightarrow 2}$ is negative and the particle loses energy.  $E_{1\rightarrow 2}$ is emitted

This would make the plot for $n_1=1$ in the above graph, an absorption line.

For all values of $n_2$, only when $n_1=1$ is an absorption line.  The plots in black are emission lines against which we see the absorption line without direct illuminations.

Values of the plots in black below $y=0$ is ignored because $n_2\ge n_1$.

Rydberg constant is not murdered, instead a new process is given birth.

$\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  }$ occurs at the same time as $\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$ due to quantized energy levels in Bohr model theory.  The two process has reverse energy signs and hence married with a negative sign.  According to Bohr model $E_{1\rightarrow 2}$ is positive and the particle gains energy and transits to a higher energy level.

We have instead,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}-\cfrac { n_{ o2 }^{ 2 }-n_{ o1 }^{ 2 } }{ (n_{o1 }n_{ o2 })^{ 2 } }\right)$

where a change $n_1\rightarrow n_2$ is accompanied by a change in orbital energy level $n_{o1}\rightarrow n_{o2}$.

Only now are the particles in orbits.

Note:  $E_2=E_1+\Delta E_{1\rightarrow 2}$

Murder Yet Written

Of course,

$\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}\ne\cfrac{1}{n_1^2}-\cfrac{1}{n_2^2}=\cfrac{n_2^2-n_1^2}{(n_1n_2)^2}$

but are they parallel over the range of $n_1$ and $n_2$ in consideration.

A plot of $\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } -\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$ gives

where the values of the expression for $n_1=1$ is negative and the graphs of for $n_1=2$ and $n_1=5$ are coincidental.

A plot of the ratio $\cfrac { \cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  }  }{ \cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }  }$ gives,

The values of the ratio varies with both $n_2$ and $n_1$.  In the case of $n_1=1$, the ratio $\cfrac { \cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  }  }{ \cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }  }$ is close to $1$

Rydberg constant is not dead.  Yet...

Looking for Murder

From the previous post "" dated 13 Oct 2016,

$\Delta E_{1\rightarrow 2}=0.2063hf_{\psi\,c}$

and so we have,

$a_{\psi\,new}=0.2063a_{\psi}$

thus assuming that the first spectra line is of double intensity, ie from the pair $(n_2=2,n_1=1)$,

$a_\psi=0.2063*19.34=3.99\,nm$

$a_\psi=0.2063*16.32=3.37\,nm$

$a_\psi=0.2063*15.48=3.19\, nm$

$a_\psi=0.2063*14.77=3.05\,nm$

If, the first spectra line is the result of $(n_{\small{large}},1)$,

$a_\psi=0.762*19.34=14.74\,nm$

$a_\psi=0.762*16.32=12.44\,nm$

$a_\psi=0.762*15.48=11.80\, nm$

$a_\psi=0.762*14.77=11.25\,nm$

Which is just a bunch of numbers.  However, both cases point to the fact that the calculated $f_{\psi}$ from Planck's relation, $\Delta E=h.f_{\psi}$, is not the frequency $f_{\psi\,c}$ of $\psi$, to resonate the particle with.  $f_{\psi\,c}$ is defined as,

$f_{\psi\,c}=\cfrac{c}{\lambda_{\psi\,c}}=\cfrac{c}{2\pi a_{\psi\,c}}$

where there is one wavelength $m=1$ around the circular path of $\psi$ with radius $a_{\psi\,c}$ .

Depending on which spectra line is used to derive $f_{\psi}$, a factor of $0.2063$ or $0.762$ applies.  In the case where both the double intensity emission spectra line and the first absorption spectra line are used, with the appropriate factor, there can be only one value for $f_{\psi\,c}$.

Where do all these lead us?

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}\right)$

The murder of Rydberg constant?

$\cfrac{1}{\lambda_{2\rightarrow 1}}=R_{\small{H}}\left(\cfrac{1}{n_1^2}-\cfrac{1}{n_2^2}\right)$

Comparing the two expressions,

$R_{\small{H}}=\cfrac{1}{2\pi a_{\psi\,c}}$

which is a constant given $a_{\psi\,c}$.  Maybe...

Don't Worry, Be Creepy

It does not quite matter, given

$f(n)=\left(\cfrac{\sqrt[3]{n}-{1}}{\sqrt[3]{n}}\right)$

n	f(n)
70	0.7573572497
71	0.758501808
72	0.7596250716
73	0.7607277244
74	0.7618104192
75	0.7628737797
76	0.7639184021
77	0.7649448568
78	0.7659536896
79	0.7669454232

The value of $f(n)$ for the range $70\le n\le79$ average to

$f(n)=0.762$

from which we may estimate $f_{\psi\,c}$ from,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1}\sqrt[3]{n_2}}\right)$

where $n_1=1$

from the post "Sizing Them Up Again..." dated 4 Oct 2016.

But what is $n$ or $n_{\small{large}}$?

Better yet since the emission spectra line as the result of  $(n_2=2,\,n_1=1)$, (two basic particles, $n=1$ coalesce) that produces a pair of photons, $\Delta E_{1\rightarrow 2}$ has double the intensity and hence readily identifiable,

$\Delta E_{1\rightarrow 2}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{2}-\sqrt[3]{1}}{\sqrt[3]{1}\sqrt[3]{2}}\right)$

where $n_1=1$ and $n_2=2$.

$\Delta E_{1\rightarrow 2}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{2}-1}{\sqrt[3]{2}}\right)$

$\Delta E_{1\rightarrow 2}=0.2063hf_{\psi\,c}$

without worry about what $n_{\small{large}}$ might be.

This is different from Planck's relation $E=h.f$.  A factor of about $\frac{1}{5}$ creeps in.

The First Spectra Line

For the absorption spectrum where energy is measured with reference from a higher threshold, the highest energy level has the lowest drop from such a reference, we reflect the previous plot from the post "Emission Spectrum Simplified" dated 12 Oct 2016, about the axis $y=0$.

The first spectra line is not the line with double intensity, but

$\Delta E_{1\rightarrow n=large}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_{\small{large}}}-\sqrt[3]{1}}{\sqrt[3]{1}\sqrt[3]{n_{\small{large}}}}\right)=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_{\small{large}}}-1}{\sqrt[3]{n_{\small{large}}}}\right)$ --- (*)

To see what is $\left(\cfrac{\sqrt[3]{n_{large}}-{1}}{\sqrt[3]{n_{large}}}\right)$, we plot

As $n_{\small{large}}\rightarrow 77$, the increment in$\left(\cfrac{\sqrt[3]{n_{large}}-{1}}{\sqrt[3]{n_{large}}}\right)$ decreases but the plot is not asymptotic towards a steady value around $n=70\approx 80$.  (When $n\rightarrow \infty$, we have an asymptote towards $y=1$.)

Given that $n$ takes on integer values ($n+1$ being the number of constituent basic particles the big particle has before its breakage into a basic particle ($n=1$) and another big particle, $n$), and that expression (*) at high consecutive values of $n$ has very close values, spectra lines involving high values of $n$ will seem to split into numerous close lines.

This could be the explanation for split spectra lines.  The above is an emission spectra line plot, NOT an absorption spectra line plot.

To be sure that the first spectra line is given by (*), we look at other coalescence/breakage possibilities.  When big particles of $n+1$, $n+2$ and $n+3$ constituent basic particles break into pairs of $(n,\,1)$, $(n,\,2)$ and $(n,\,3)$  particles respectively,

Since the particle pairs $(n=3,\,n+3)$ are at closer energy levels than $(n=1,\,n+1)$, the transition from $n+3$ to $n=3$ requires less energy than the transition fro $n+1$ to $n=1$.  As such breakage into a small particle of higher $n$ involve lower absorption energy.  Such lines will be higher up in the absorption spectrum plot when this plot is reflected about $y=0$.

After considering other possible particle pairs, $(n=i,\,n+i)$, the first spectra line is still given by (*).

But how large is $n_{\small{large}}$? $n_{\small{large}}=77$? $n_{\small{large}}=74$?...

Emission Spectrum Simplified

Unfortunately in this scheme of things, we have $\psi$ balls of various sizes, $n$ in collisions.

When 2 particles of size $n=1$ coalesce 2 photons $E_{1\rightarrow2}$ are emitted, and when 2 particles of sizes $n=1$ and $n=2$ coalesce 2 different photons $E_{1\rightarrow 3}$ and $E_{2\rightarrow 3}$ are emitted.

This is not the simple scheme where emissions as the result of hops from energy level to energy level has equal intensity, but in the example above, $E_{1\rightarrow 2}$ has twice the intensity of $E_{1\rightarrow 3}$ and $E_{2\rightarrow 3}$.

The plot below shows ((x+1)^(1/3)-x^(1/3))/(((x+1)*x)^1/3), ((x+2)^(1/3)-x^(1/3))/(((x+2)*x)^1/3) and ((x+3)^(1/3)-x^(1/3))/(((x+3)*x)^1/3).

where particle of size $n=1$, $n=2$ and $n=3$ coalesce with particle of size $n=x$.

We see that the highest energy transition occurs with $E_{1\rightarrow n=large}$, when a basic particle $a_{\psi\,c}$ (ie. $n=1$) coalesces with a large particle $n\rightarrow 77$.  Two photons are released $E_{1\rightarrow n=large}$ and $E_{large\rightarrow large+1}$.

Emission occurs in bands as $n=x$ increases and such bands narrows with increasing $n=x$.

In the graph above, the top most three horizontal lines maroon, red and blue correspond to $x=1$.  The next band of maroon, red and blue correspond to $x=2$.  Each band progressively narrows as $x$ increases.

As $x$ increases, all graph approaches asymptotically to zero, ie as $x$ increases all emissions due to the coalescence of particles of various sizes, $n$ approaches zero.

Good night.

Sizing Them Up Again...

With this new picture of particles coalescence and disintegration, we will calculate $a_{\psi\,n=1}$ again...

From the post "No Nucleus Needed" dated 03 Oct 2016,

$\Delta E_{2\rightarrow 1}=E_{n_1}-E_{n_2}=h\cfrac{c}{2\pi a_{\psi\,n_1}}-h\cfrac{c}{2\pi a_{\psi\,n_2}}$

where positive values for $\Delta E_{n}$ is energy absorbed.  And from the post "Speculating About Spectra Series" dated 29 Sep 2016,

$a_{\psi\,n}=\sqrt[3]{n}.a_{\psi\,c}$ --- (*)

where  $n=1,\,2,\,3,..77$,

we have,

$\Delta E_{2\rightarrow 1}=h\cfrac{c}{2\pi }\left(\cfrac{1}{a_{\psi\,n_1}}-\cfrac{1}{a_{\psi\,n_2}}\right)=\hbar.c\left(\cfrac{a_{\psi\,n_2}-a_{\psi\,n_1}}{a_{\psi\,n_1}.a_{\psi\,n_2}}\right)$

Substitute (*) in,

$\Delta E_{2\rightarrow 1}=\hbar\cfrac{c}{a_{\psi\,c}}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1}\sqrt[3]{n_2}}\right)$

since,

$2\pi a_{\psi\,c}=\lambda_{\psi\,c}$  with $m=1$

assuming that there is $m=1$ wavelength along the circular path,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1}\sqrt[3]{n_2}}\right)$

where  $f_{\psi\,c}=\cfrac{c}{\lambda_{\psi\,c}}$.

Each possible value of $\Delta E_{2\rightarrow 1}$ due to a pair $\left(n_2,\,\,n_1\right)$ of particles before and after a collision, corresponds to one spectra line.  As more possible tuples of $\left(n_2,\,\,n_1\right)$ are made available given the set of experimental conditions, more spectra lines appear.

Now for experimental data from the web...

Where Small is Highest

The smaller the particle the higher $E_n$ is.  Big particle absorb energy to break into smaller particles.  When small particles coalesce energy is released as a photon that we observe as the emission spectrum.

Big particles are more susceptible to collision impacts and break into smaller particles, absorbing part of the energy of impact on breaking up.  Energy is absorbed as a photon, from which we obtain the absorption spectrum.

But $E_n$ is not $\psi$.  $E_n$ is the result of $\psi$ on a circular path at light speed $c$.

So,

$\lambda_\psi\ne\lambda_n$

and the photon packets emitted or absorbed,

$h.f_{vis}=h.\left(f_{n_1}-f_{n_2}\right)$

$\lambda_{vis}=\cfrac{c}{f_{vis}}=\cfrac{c}{f_{n_1}-f_{n_2}}$

where $f_{vis}$ and $\lambda_{vis}$ are obtained from the experimental spectrum(s) observed.

Obviously,

$\lambda_{vis}\ne\lambda_n$  and

$\lambda_{vis}\ne \lambda_{\psi}$

But,

$\lambda_{\psi\,n}=\lambda_{n}$,  $m=1$

only when $m=1$ that there is one wavelength around the circular path of radius $a_{\psi}$.  The factor $2\pi$ appears as the wavelength, $\lambda_{\psi}$ is wrapped around a circular path.

And since the energy transitions as particles coalesce and disintegrate is the reverse of electron energy level transitions, we do not have an issue with negative energy.

And in the rest state of no collisions, we have $a_{\psi\,n=1}$ where $a_{\psi}$ is the smallest at the highest energy possible,

$E_{n=1}=E_{max}$

$a_{\psi}=a_{\psi\,1}=a_{\psi\,c}$

So, paradoxically $\psi$ with $a_{\psi\,c}$ has the highest energy but the smallest size.

No Nucleus Needed

$\psi$ wraps around a sphere.  A certain amount of energy is associated with $\psi$ motion around the sphere.  If $h$ amount of energy is in the system, going once per second around the sphere, then travelling at light speed $c$ around a radius of $a_{\psi\,n}$, the amount of potential energy associated with this motion is,

$E_n=h.\cfrac{c}{2\pi a_{\psi\,n}}$

but,

$2\pi a_{\psi\,n}=m\lambda_m$

where $m$ is the number of wavelength, $\lambda_m$ of $\psi$ around the circle.

With $m=1$,

$E_n=h.\cfrac{c}{\lambda_m}=h.f$

which is Planck's relation given $m=1$.  When $\psi$ is a half wave in time and a half wave in space space,

$m=\cfrac{1}{2}$

$E_n=h.\cfrac{2c}{\lambda_m}=2h.f=h^{'}.f$

with a mysterious factor of $2$.

When $m=2, \,3...$

$E_n=h.\cfrac{c}{m\lambda_m}=\cfrac{h}{m}.f=h_m.f$

$h_m$ changes with $m$ because $c=f\lambda$.

And,

$h_{m_1}.f\gt h_{m_2}.f$

when $m_2\gt m_1$.

When there is more wavelengths around the circular path, stored energy decreases, which suggest another way $\psi$ can absorb and release packets of energy; by changing $m$.  Also, given $a_{\psi\,n}$ at fixed value, $f$ increases with $m$ but $h_m$ decreases, so, $E_n=h_m.f$ remains a constant.  $f$ increases with decreasing $a_{\psi}$, with increasing $E_n$.  Catastrophe when the number of particles with small $a_{\psi}$ is also large.

$E_n$ is independent of $m$.  $h$ is defined as one wavelength going around once per second along the circular path, then a corresponding increase in the number of wavelengths, $m$ increases the frequency by $m$ but this increase cancels with the decrease in $h$ by the reciprocal of $m$.

With these issues with $m$ aside, when two particles coalesce,

$a_{\psi\,n_1}\rightarrow a_{\psi\,n_2}$ 

the bigger particle has less potential energy associated with its $\psi$ in circular motion,  The amount of energy released due to this drop in potential is.

$\Delta E=E_{n_1}-E_{n_2}=h\cfrac{c}{2\pi a_{\psi\,n_1}}-h\cfrac{c}{2\pi a_{\psi\,n_2}}$

$\Delta E=hf_{n_1}-hf_{n_2}=h\left(f_{n_1}-f_{n_2}\right)$

$f_{n_1}\gt f_{n_2}$

using $h$ to denote all cases of $m$, as discussed above, we have an expression similar to the discrete energy level transitions expression for electrons around a nucleus,  In this case, however it is just $\psi$ alone going around a circular path and energy level decreases with increasing $a_{\psi\,n}$

No nucleus needed.

Spectra Ghost At The Rear

Happy Birthday to me...03 Oct 1968.

If spectra lines reflect the collision and subsequent coalescence of particles, then the observation that more spectra lines appear at lower experiment setup temperature, can be explained by lower collisions rate at lower temperature with less momentum which allows bigger particles to form.  Bigger particles have higher $a_{\psi\,n}$ which appear as a higher spectra line.  At higher temperature, big particles are broken up in high velocity impacts before further coalescence forms bigger particle.

So, at low temperature more spectra lines due to bigger particles appear.

Have a nice day.

The Quantum

From the previous post "Speculating About Spectra Series" dated 29 Sep 16,

$a_{\psi\,n}=\sqrt[3]{n}.a_{\psi\,c}$

which suggests energy transition,

$a_{\psi\,n_1}\rightarrow a_{\psi\,n_2}$

$n_1\rightarrow n_2$

occurs as small particles coalesce into bigger particles and when a big particle breaks into smaller particles.

The second case explains the occurrence of two or more simultaneous energy transitions, as a bigger particle breaks into two or more smaller particles.

$a_{ \psi \, 6 }\begin{matrix} \nearrow  \\ \rightarrow  \\ \searrow  \end{matrix}\begin{matrix} a_{ \psi \, 1 } \\  \\ a_{ \psi \, 2 } \\  \\ a_{ \psi \, 3 } \end{matrix}$

$n_{ 6 }\begin{matrix} \nearrow  \\ \rightarrow  \\ \searrow  \end{matrix}\begin{matrix} n_{ 1 } \\  \\ n_{ 2 } \\  \\ n_{ 3 } \end{matrix}$

In the case above, a particle made up of $n=6$ basic particles breaks into particles of size $n=1$, $n=2$ and $n=3$.  $n$ denoting arbitrary energy levels before has now a physical interpretation; it is the number of constituent basic particles.

I have found the quantum!  $a_{\psi\,c}$ is the quantum.

Speculating About Spectra Series

From the post "Sizing Them Up" dated 13 Dec 2014,

$a_\psi=19.34\,nm$

$a_\psi=16.32\,nm$

$a_\psi=15.48\, nm$

$a_\psi=14.77\,nm$

If,

$a_\psi=a_{\psi\,c}=19.34\,nm$

$a_\psi=a_{\psi\,c}=16.32\,nm$

$a_\psi=a_{\psi\,c}=15.48\, nm$

$a_\psi=a_{\psi\,c}=14.77\,nm$

in which case we have,

$n.\cfrac{4}{3}\pi \left(a_{\psi\,c}\right)^3=\cfrac{4}{3}\pi \left(a_{\psi\,n}\right)^3$

where $1\le n\le 77$ and

$a_{\psi\,c}=a_{\psi\,n=1}$

That is to say, $n$ small particles of  radius $a_{\psi\,c}$ coalesce into a bigger particle of $a_{\psi\,n}$.

$a_{\psi\,n}=\sqrt[3]{n}.a_{\psi\,c}$

where $n=1,\,2,\,3,..77$

If each of this particle is responsible for a spectra line then, a spectra series due to one type of particle will line up nicely on a $y=\sqrt[3]{n}$ plot with a common scaling factor.  An the maximum number of stable lines due to stable particle, observable or otherwise is $77$ or $78$.  Unstable particles that grows beyond the plateau on the $\psi$ vs $r$ graph where $\psi$ pinch off with decreasing force will also result in faint spectra line.

Just speculating.

Bubbles And Balls

This may be how a concrete slab fell outside of a 12 storey unoccupied warehouse.  When $\psi$ of a particle growing big at resonance on the floor increases suddenly, the increasing $\psi$ sends the floor confined within the particle back into the past.  A spatial shift that accompanies the temporal shift moves the concrete slab out of the building and it fell.

At the instant when the concrete slab fell, there is no hole on the floor as the slab comes from the future.  The whole appears when the experiment is being conducted.

When $\psi$ of the particle grown big (slowly increasing $\psi$) is allowed to collapse suddenly, rapidly decreasing $\psi$ sends the slab into the future instead.  A hole appears immediately but the slab will fall some time in the future, at a location determined by the movements of Earth and the time shift.

The bubble is to be use in specific ways and not to be use in specific ways...And the ball is to be used in specific ways and not to be used in specific ways...

Don't Walk Into The Bubble, AK!

When a shield around a confinement sudden collapses the decreasing $\psi$ sends its content into the future.

Earth is revolving around the Sun.  The temporal differential results in a spatial shift.  If the shield envelops a whole city then, the city is shifted in space as the result of a shift in time because earth is in motion.

The collisions between the area outside the shield and the area shifted in time and space within the shield will be a catastrophe.

In a similar way, if $\psi$ is allowed to build up engulfing a body, the body will be sent back in time.  When the spatial shift is significant, there will be destruction inside the shield outside the area of overlap.

Strengthen the $\psi$ shell after its build up and let the shield weaken without collapsing the $\psi$ shell.  This way, the temporal force that results from changing $\psi$ with time is minimal.

Don't walk into the bubble, AK!

Particles Big And Small

The vector sum of the force due to two quarter charges ($\cfrac{1}{4}q$) is,

$\sqrt{\left(\cfrac{1}{4}\right)^2+\left(\cfrac{1}{4}\right)^2}F=0.35355F$

which might suggest a single charge of magnitude $\approx \cfrac{1}{3}q$.  $F$ is the force due to a charge $q$.

Which brings us to the suggestion that a quarter charge Helium $He$, masquerading as $H$ that manifest a charge of one third the normal charge due to the presence of two quarter charges.  These quarter charge Helium $He$ atoms might form where big particles ($a_{\psi\,\pi}$) break into small particles ($a_{\psi\,c}$), in a particle collider, in abundance.

And in general, a new zoo of stable elements made from small particles.

More And Smaller

If,

$a_{\psi\,74}=14.77$

and from the post "Sticky Particles Too...Many" dated 24 Jun 2016,

$\cfrac{n_1}{n_2}=\left(\cfrac{a_{\psi\,n_1}}{a_{\psi\,n_2}}\right)^3$

$a_{\psi\,c}=a_{\psi\,1}=3.52\,\,nm$

This is the size of the smallest particle possible, breaking off from the tip of a sonic cone in a sonic boom or when big particles disintegrate in high speed collisions.

This particle has $\small{\cfrac{1}{4}}$ the charge magnitude (gravity, electric or temperature) but $\sqrt [ 3]{\cfrac{1}{74}}=0.2382$ the radius of a big particle.

But the classical electron radius is quoted at,

$r_e=2.817 940 3267[27] e-15\,\,m$

What is this value $a_{\psi\,1}$, a million times bigger?  Then again, the visible spectrum is $390\le\lambda_{vis}\le 700\,\,nm$; can we expect particles that are photons to be that much smaller that their wavelengths?

$\lambda_{\psi}$ or $\lambda_{n}$ is not $\lambda_{vis}$!

Could it be that $r_e$ measured under the formulation for point particles, is actually the size of the void where $v=c$ along a radial line, at the perimeter of this void?

Which suggests that interactions between particles allow overlaps in $\psi$ and that such interactions (mechanical/gravitational, electrical, ie field interactions) is limited down to $r_e$, below which $\psi$ does not exist.

$r_e$ is the size of the hole in all particles that changes with exterior conditions acting on $\psi$.  Pushing $\psi$ at the start of the plateau in the $\psi$ vs $x$ graph, at resonance frequency, increases $r_e$ and the over all size of the particle.

Maybe...

Alien Intervention

Yes, I know, Madam.

No, the thing to do to resolve this is not to leave this planet.  You would wish.  As your colleague would say: "Why don't you tell him."  There is no need for that either!

However, lacking data to validate or invalidate,

$\cfrac{m_e}{m_p}=\cfrac{1}{74}$

the solution might just to seek help from outer space.  Warden!  Where's help when you need it?

A particle charge of $\cfrac{1}{4}$ the normal charge (gravitational, electric or temperature) however, is consistent with some experimental observations of weak fields ((gravitational and electric only).

Water meter, gas meter and all things to measure...  Peace comes with a prize.  Alien intervention is indeed needed here, else it would be a rampage... through science.

Have a nice day.

Time Paradox Not

In one example of time paradox, a scientist creates a portal that looks into the past one minute, and he shoots himself, at the image he sees, in the past.

This is not a paradox, it is likely the forces that delay the image he sees, as he takes aim, also delays the bullet that has to travel through the portal in reverse.  When the bullet emerge from the portal on the other side it is in the present and the scientist has walked away to take aim on the other side.  He misses all the time!  Forces acting this way prevent a positive feedback loop.

What is technically improbable is by no means limiting in theory.

Have nice day.

Of Dragon's And Mon's

If,

"re" is " 鲁"，

"pu" is "国"，

and

"lic" is "历", in Cantonese.

But I have also come to know that,

"weapon" is "尾巴" in Mandarin.

Given the first two above,

"weapon" is "武功" in Cantonese.

this means it is possible to convert Cantonese to Mandarin by notching out certain frequencies or shifting frequencies in Cantonese.

Cantonese and Mandarin is the same language.  Physiological and possibly anatomical differences between the ancient species that spoke them, heard the languages differently and so sound them off differently.

Have a nice day.

Note:  Plato's Republic and 孔子 - 鲁国历 penned by 乐欬 (ngok6 koi3), if "Plato" is Cantonese in ancient phonetics with "pla" pronounced as "glo" and "to" pronounced as "ko".  卜商 is also possible but unlikely because the "n" sound is missing.

高柴! gou1 caai2 where "p" is a "g", "a" is a "o" as above and "t" is a "j".

秋月恨离愁
更恨无星伴
云抹海线白
一等又明天

《孤月》

Other Dimensions?

There is no need for a separate space time particle, all particles have a space and time component.  These particles, free to move in a 3D volume, generate three space dimensions.  Around each space dimension, a time dimension wraps tightly in a close spiral.

Does the volume contains the three space dimensions, or do the three space dimensions extend the volume?

The volume and the three space dimensions both describe the same entity.

If the time dimension, as one of the two orthogonal dimensions about which energy oscillates in a particle, is available us, then the other space dimension in the same oscillation should also be accessible.  And as we measure $E$, $g$ and $T$ field around the particle, we access this space dimension.  Conceptually, the energy in these fields push a marker along a linear 1D scale de-marking a value indicative of the energy level in that field.  When such a scale is located physically around a particle, we obtain an indication of the field around that particle, along the direction of the scale.

Is it possible that other dimensions exist, yet undetected?  This is a different question from whether
other energies (energy pair) exist.  Each such energy has an associated particle, a force field and an orthogonal oscillating energy that manifest itself when the particle goes into spin.  This new particle is still in the old space time dimension framework.  Whereas a new dimension would imply a new space time framework, not necessarily with new energy pair.  (Such energy pairs are needed only for wave/particles to exist.  Is free energy without a particle embodiment possible?)

Have a nice day.

ABC Con Gatcha

For all triples $(a,\, b,\, c)$ of coprime positive integers, with $a + b = c$,

we see that for $\varepsilon\gt 0$ where

$c\gt rad(abc)^{1+\varepsilon}$

with the radical of a positive integer $n$, the radical of  $n$, denoted $rad(n)$, is the product of the distinct prime factors of $n$;

is the small blue circle smaller than the red circle of radius $a+b=c$.

The diagram does not count the valid number of triples for which the blue circle is small, but if we consider all valid triples of co-prime and scale each corresponding green circle $\small{abc}$ to a uniform size, and allow all with smaller blue circle to sieve through the red circle, then obviously, by the  pigeonhole principle, since the green circles are bigger and the blue circle can be as big as the green circle, only some blue circles will fall through the red circle at the center.

Is the number of blue circles that fell through finite?

We note that $abc\gt \sqrt{ab^2+a^2b}\gt c \gt rad(abc)^{1+\varepsilon}$, in other words, the blue circle is the smallest and has radius less than $\sqrt{abc}$.

Using Euler's totient function,

$\varphi(ab)=\varphi(a)\varphi(b)$

where $a$ and $b$ are co-prime.  We may,

$\varphi(abc)=\varphi(a)\varphi(b)\varphi(c)$

where $a$, $b$ and $c$ are co-prime.  As,  $n\gt\varphi(n)$

$abc\gt\varphi(abc)$

and

$abc\gt c$

So, we have either,

$abc\gt\varphi(abc)\gt c\gt\varphi(c)$

or

$abc\gt c\gt \varphi(abc)\gt\varphi(c)$

because,

$\varphi(abc)=\varphi(a)\varphi(b)\varphi(c)\gt\varphi(c)$

When,

$abc\gt\varphi(abc)\gt c\gt rad(abc)^{1+\varepsilon}$

Since, $abc$ is not prime, $\varphi(abc)$ is finite; that the number of positive integer not greater than $abc$, and relatively prime with $abc$ is countable, the number of blue circles that fell through is also countable.  If $abc$ is prime (which is impossible as it has three factors) than $\varphi(abc)=abc-1$, which is as large as $abc$.

When

$abc\gt c\gt \varphi(abc)\gt rad(abc)^{1+\varepsilon}$

which is the same as the first case, the position of $c$ is irrelevant.  In both cases, however,  the count of terms $\varphi(abc)$ is greater than the product of terms $rad(abc)^{1+\varepsilon}$, which can be true (in the limiting case of equality $\varphi(abc)=rad(abc)^{1+\varepsilon}$ ) only for the first two prime numbers $1$ and $2$.

When

$abc\gt c\gt rad(abc)^{1+\varepsilon}\gt \varphi(abc)\gt\varphi(c)$

we have,

$c\gt rad(abc)^{1+\varepsilon}\gt \varphi(c)$ --- (*)

where $rad(abc)^{1+\varepsilon}$ is squeeze between $c$ ($c\gt a$ and $c\gt b$) and $\varphi(c)$ which makes $rad(abc)^{1+\varepsilon}$ countable.

For expression (*) to be possible, when $c$ is large but not divisible by large powers of prime then, as $rad(abc)^{1+\varepsilon}$ exists in the range from $c$ to $\varphi(c)$, this implies that $a$ and $b$ must be of high powers of primes, because $a$, $b$ and $c$ are co-prime and factor $a$ and $b$ must contribute less to $rad(abc)$ than $c$, on condition that $rad(abc)\lt c$ is true for the triple $(a,\, b,\, c)$.

This is not to be taken too seriously...

When Time Passes in $c$?

Just as observing $B$ spinning around a dipole, at the dipole, yields

$B=-i\cfrac{\partial E}{\partial x}$

and observing $B$ away from the dipole, at a location fixed in space as the dipole passes,

$\cfrac{\partial B}{\partial t}=-i\cfrac{\partial E}{\partial x}$

as in the post "Magnetic Field In General, HuYaa" dated 13 Oct 2014.

For time, $\tau$ at the spacetime particle,

$\tau=f(x)$

at a point away from the spacetime particle,

$\cfrac{\partial \tau}{\partial t}=f(x)$

as the particle passes.  This provides intuitively, an explanation for the time derivative needed, after the expression for time has been derived at the spacetime particle and we needed an expression observing the spacetime particle from afar away from the particle, in a second perspective.

Time at light speed is particles carrying the time field passing by us at light speed.

Which lead us back to the notion of time travel by isolation; a solid block of lead with 42 cm thick wall, where inside the walled cell, time stood still.

And the oldest cat is 41 yrs...poor thing.

The End Of Time, SpaceTime Particle

Where would one find a prevalent time field that accelerate all time particles to light speed in the time dimension?

Inside a time particle.

The universe is one big time particle.

Inside such a particle, the force on $\psi$ is given by $-\frac{\partial\psi}{\partial\,x}$.  The value of this force and its gradient will locate us in the universe.  We need two equations because we do not have an origin to start with.  At the origin, time return to the space dimension.

Time is in flux, we are not going any where in space.  The end of time is at the center of the universe. At the end, $\psi$ get recycled on the surface of the same particle (or another particle) where $v=c$ is perpendicular to the radial direction.

Since, space is orthogonal to time and vice versa, space and time are just orthogonal view of the same particle!

There is a time particle, is there a separate space particle?  A space particle is our experience of a time particle as we travel circularly around the time particle.  In the radial direction, we experience time.  In a reciprocal way, a time particle is our experience of a space particle as we go around the space particle; along its radius, we experience space.

There is just space-time.

Which brings to mind the relationship between $B$ and $E$ fields.  And how such fields are forcefully made once removed, that a $g^{+}$ particle produces a $T$ field that hold a $T^{+}$ particle that in spin produces a $E$ field which hold a $p^{+}$.  A spinning $p^{+}$ produces a $g$ field that in turn holds a $g^{+}$ particle, to accommodate other fields, especially the temperature particle and temperature field.

If we are to be consistent, spinning space particle does not produce a time field, when spinning time particles produces a space field.

But a spinning $B$ field produces an $E$ field.

A spinning space field will produce a time field.  And vice versa.  What is a space field?  The fountain of youth and Casimir effect where space is force out of a narrow gap comes to mind.

If spinning space particle produces a different field, then there is a possible new type of particle, un-thought of.  If space is just gravity then time is just proton.  Does an $E$ field stop time?

One scenario leads us to two new possible particles in addition to the time particle, to form a tuple of three that is analogous to $(g^{+},\,T^{+},\,p^{+})$.  Another scenario collapses time field into an electric field.

Walk through a charged parallel plate capacitor and walk though time...the Afghan artifacts!