Copper Antenna Bias

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

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

for copper,

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

What is this good for?

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

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

Electron Cloud Copper

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

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

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

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

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

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

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

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

where a length of $1\,m$ is used,

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

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

Most Penetrating X ray, Boron

With Boron $B$,

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

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

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

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

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

$V_{min}=0.1445\,V$

For a length of $d=10\,cm$,

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

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

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

$V_{min}=0.3710\,V$

For a length of $d=10\,cm$,

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

Boron requires less voltage.

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

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

$I_{res}=1.1974 A$

because the electric resistance of copper is very low at $16.78 nΩ·m$. 

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

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

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

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

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

Good night.

Probing X rays

There is a problem, if

$\Delta t=\cfrac{v}{c}-1$

when $v=0$

$\Delta t=-1$

when $v=c$

$\Delta t=0$

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

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

This requires $v$ to be nearer to $c$.

We should have used,

$\Delta t=\cfrac{v}{c}$

instead, taking $c$ as reference,

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

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

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

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

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

Quantifying Time Travel Backwards

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

For copper,

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

For Hydrogen $H_2$,

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

If $\Delta t=\cfrac{v}{c}-1$  and

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

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

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

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

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

$S_{p\,H_2}==0.2000007$

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

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

since,

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

but the high value of $c$ makes the increment insignificant.

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

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

Then,

 $S_p=\cfrac{S_{p\,1}+S_{p\,2}}{2}$

A series of $D$ values would narrows the variance in $S_{p}$.

Maybe then the expression,

$\Delta t=\cfrac{v}{c}-1$

$v=v_{boom}$

can be verified, after quantifying Earth's movement.

Probing Sideways

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

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

Probing The Nucleus

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



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

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

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

 Good morning.

Light Dispersion And Particles II

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

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

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

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

Then obviously,

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

X Ray Scope

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

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

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

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

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

Good night.

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

Nemo Of Many Sizes

With,

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

we have also,

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

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

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

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

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

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

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

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

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

Finding Nemo

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

still looking...

where's Nemo?

My guess is,

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

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

 $a_{\psi\,ne}=7.321\text{e-8}\,m$

Have a sunny day.

Light Dispersion And Particles

Another guess,

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

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

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

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

More X Rays

Since X ray refracts away from the normal,

$a_{\psi\,n}\gt a_{\psi\,s}$

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

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

The torus travels,

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

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

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

and

$\lambda_r=2n\pi a_{\psi\,n}$

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

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

Which leads to normal color lights...

Vita Rays

To set hydrogen gas $H_2$ into discharge,

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

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

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

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

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

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

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

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

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

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

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

For a electrode gap of $d=10\,cm$,

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

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

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

This is totally irresponsible.

Hubble, Huba, Huba

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

where $v_s$ is in the direction of Earth's movement.

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

When $v_s$ does not leave the medium boundaries parallel as they move in time,

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

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

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

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

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

Good night.

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

A Hot Pendulum

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

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

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

Maybe...

Glowing Pendulum

A transparent glow paint pendulum,

I don't know whether it will glow.

Color Spectrum And Particle Size

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

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

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

Cracking Balls

If this is the torus photon,

we can approximate,

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

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

We are not going to get $m$, because,

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

So,

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

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

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

or

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

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

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

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

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

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

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

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

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

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

$a_{\psi\,l}\rightarrow a_{\psi\,s}=a_{\psi\,c}$

$v\rightarrow c$

$f_{res}\rightarrow \sqrt{2}$

Given a simple string-ball pendulum,

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

$l=\cfrac{g}{8\pi^{2}}$

$g=9.806\,ms^{-1}$

$l=0.1242\,m$

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

Entangled LEDs

The frequency,

$f=280.73\,Hz$

 from the post "What If The Particles Are Photons?" dated 12 Dec 2017, considers a flat surface volume of,

$vol_{time}=T^3$

The frequency,

$f=67.02\,Hz$

from the post "I Don't Know..." dated 12 Dec 2017, considers a spherical surface volume of,

$Vol_{time}=\cfrac{4}{3}\pi T^3$

Both scenarios is along one space dimension.

The frequency,

$f=280.73\,Hz$

from the post "Freaking Out Entanglement" dated 14 Dec 2017, also considers a spherical surface volume of,

$Vol_{time}=\cfrac{4}{3}\pi T^3$

but is with all three space dimensions.

Does the LEDs acquire other colors, not designed for?

If $f=280.73\,Hz$ is true, then

Why is volume in space spherical, but volume in time is cubic?  Because the quantum in the time dimension is quantized in all three time dimensions in multiples of $mc^2$?  Maybe.

Quantum Of Existence

In the post "Like Doppler" dated 15 Jan 2018, energy discrepancy along the time dimension is compensated for by setting time to a negative value with respect to some instance $T$.  That a particle, with energy $mc^2$ along the time dimension, provides energy $\Delta E$,

$\Delta E=\Delta t*mc^2$

by coming into existence at a time $\Delta t$ before the reference $T$. 

This does not redefine energy but redefines time duration as the ratio of two energy terms.

$\Delta t=\cfrac{\Delta E}{mc^2}$

and that such a duration can be negative with respect to a reference instance $T$.

Time then, is the a span of existence in quantum of existence $mc^2$.

Time is either a dimensionless ratio of energy terms, or is also measured in units of energy; per $mc^2$.

As $\Delta t$ is measured in numbers of $mc^2$, $mc^2$ is the second, in which case, second $s$ is just another unit for energy; or time is dimensionless as it is a ratio of seconds.

All in all, we all have our personal second, $mc^2$,  that defines our existence.

...and time exists as long as there are two energy terms.


Note:  Remember the expression from defining one particle as a particle with $\theta=\theta_{\pi}$,

$2mc^2ln(cosh(\theta_{\pi}))=1$

from the post "Sonic Boom" dated 14 Oct 2017.

$mc^2=\cfrac{1}{2}*\cfrac{1}{2.4438}=\cfrac{1}{4.8876}$

This could be the discrepancy encountered previously that would account for the need for defining $\mu$ in the expression for light speed,

$c=\cfrac{1}{\sqrt{\varepsilon\mu}}$

This discrepancy results from the definition of time, the second ($s$).

Deja Vous Change

In view of the previous post "Like Doppler" dated 15 Jan 2018, the expression,

$\Delta t*v_s=\cfrac{v_{boom}}{c}*v_s=S$ --- (*)

from the post "Light Speed Back In Time" also dated 15 Jan 2018, might change to,

$\Delta t*v_s=\left(1-\cfrac{v_{boom}}{c}\right)*v_s=S$ --- (**)

The former expression (*) suggests an absolute change in $t$ and the latter expression (**) might suggest a relative change in $t$ with respect to the photon that remains at light speed $c$.  But consider this;

if

$1-\cfrac{v_{boom}}{c}=\cfrac{\Delta}{\text{refer.}}$

$\cfrac{v_{boom}}{c}=\Delta$

then,

$\text{refer.}=\cfrac{\cfrac{v_{boom}}{c}}{1-\cfrac{v_{boom}}{c}}=\cfrac{v_{boom}}{c-v_{boom}}$

Which reminds of the expressions,

$E=mc\int ^{c}_{v_{boom}}{ dv }$

quoted in the post "Like Doppler", originally from the post "No Poetry for Einstein" dated 06 Apr 2014;  and,

$E=mc\int ^{c}_{c-v_{boom}}{ dv }$

This implies that $\text{refer.}$ is with reference to the a photon at speed $c-v_{boom}$ instead of $c$.  ie the situation when $v_{boom}$ flips sign.  Given $v_{boom}$, numerically, it is possible that,

$c-v_{boom}$  and $c+v_{boom}$

that the torus collapse to a photon going in the opposite direction.

Expression (*) should be used, where $c$ is the reference.

Like Doppler

Another view of the torus photon's time trip backwards is,

baring in mind that the kinetic energy along the time dimension is $\color{red}{mc^2}$.  As we account for total energy across space and time dimensions,

$\Delta_{\small{forward}}= \cfrac{1}{2}m(c+v_{boom})^2+\color{red}{mc^2}-\left(\cfrac { 1 }{ 2 } mv_{ boom }^{ 2 }+\cfrac { 1 }{ 2 } mc^{ 2 }+\color{red}{mc^2}+\color{red}{mc^2}\right)$

$\Delta_{\small{forward}}=mcv-mc^2=mc(v-c)$

which is just the required energy to to accelerate a particle from ($v_i=v$) to ($v_f=c$) in the expression,

$E=mc\int ^{v_f}_{v_i}{ dv }$

from the post "No Poetry for Einstein" dated 06 Apr 2014.

If this discrepancy is solely compensated for by one particle,  $mc^2$ along one timeline of existence,

$\Delta t=\cfrac{\Delta_{\small{forward}}}{mc^2}=\cfrac{v}{c}-1$

As $v\lt c$, $\Delta t\lt 0$,  a negative value indicates backward in time at the instance $T$ when the torus collapses.  In term of $\tau$,

$-\Delta \tau=\cfrac{1}{\Delta t}=\cfrac{1}{\cfrac{v}{c}-1}$

$\Delta \tau=\cfrac{1}{1-\cfrac{v}{c}}$

Both expressions are different from the one in the previous post "Light Speed Back In Time" dated 15 Jan 2018,

$\Delta t=\cfrac{v}{c}$

$\Delta \tau=\cfrac{c}{v}$

because previously the focus is on one particle along its timeline.  In this post, the total energy across both timelines are considered with the discrepancy compensated solely by one particle.  Both timelines are kept in view.

$v_{boom}$ is not the speed of the torus photon source.  It is assumed that $v_{boom}$ is in the same direction as $c$.  If $v_{boom}$ acquires a negative value.

$\Delta \tau=\cfrac{1}{1-\cfrac{-v}{c}}=\cfrac{1}{1+\cfrac{v}{c}}$

$v_{boom}$ can acquire a negative value when the photon ray is observed from the side moving at right angle to the line of sight.  To establish a common reference given two $\Delta \tau$ values, we use the geometric mean,

$\Delta \tau_{+}=\cfrac{1}{1-\cfrac{v}{c}}$

$\Delta \tau_{-}=\cfrac{1}{1+\cfrac{v}{c}}$

$\Delta \tau=\sqrt{\Delta \tau_{+}\Delta \tau_{-}}=\sqrt{\cfrac{1}{1+\cfrac{v}{c}}*\cfrac{1}{1-\cfrac{v}{c}}}$

This geometric mean is the ray $mc^2$, that appear previously in this derivation,

$\Delta t=\cfrac{\Delta_{\small{forward}}}{mc^2}$   or

$\Delta \tau=\cfrac{mc^2}{\Delta_{\small{forward}}}$

So,

$\Delta \tau_1=\cfrac{\Delta \tau_{+}}{\Delta \tau}=\cfrac{\sqrt{1-\cfrac{v^2}{c^2}}}{1-\cfrac{v}{c}}=\cfrac{mc^2}{\Delta_{\small{forward}}}=\cfrac{f_o}{f_1}$

$f_1=\cfrac{1-\cfrac{v}{c}}{\sqrt{1-\cfrac{v^2}{c^2}}}*f_o$

and

$f_2=\cfrac{1+\cfrac{v}{c}}{\sqrt{1-\cfrac{v^2}{c^2}}}*f_o$

which look like the transverse Doppler effect but has nothing to do with a moving source.  It is a ray moving in a medium that collapses the torus photons to $v_{boom}$ that in turn eject torus photons upon further collisions at $v_{boom}$.  A ray through a region of gas that emitted the torus photons.

OK??

Light Speed Back In Time

The problem with the torus photon travelling back in time,

is its first existence being destroyed at the instance it collapses and attain greater than light speed $c+v_{boom}$.

For conservation of energy across time and space dimensions to hold, the first existence with light speed $c$ in the time dimension is not destroyed when the particle travels back in time at the instance $T$ and creates a second timeline of existence with speed $v_{boom}$.

Another photon is created at the instance $T$.  This photon is sent back in time and is marked with a second timeline.

How is $-\Delta \tau$ related to $c+v_{boom}$?  Consider, per unit time,

where the areas are spacetime(s).  When the torus photon collapses and attains $c+v_{boom}$, given its existence (ie light speed $c$ in the time dimension), the total spacetime is,

$\Delta_i=\cfrac{1}{2}c(c+v_{boom})$

immediately after its passage back in time,

$\Delta_f=\cfrac{1}{2}c.v_{boom}$

The difference,

$\Delta=\Delta_f-\Delta_i=-\cfrac{1}{2}c^2$

if spacetime is like energy and must be conserved, this amount of spactime must be created in the past, during $\Delta \tau$ such that at instance $T$ the total spacetime is conserved.

$-\Delta\tau=\cfrac{\Delta_f-\Delta_i}{\Delta_f}$

$\Delta\tau=\cfrac{c}{v_{boom}}$

gives $\Delta\tau$ per unit time into the past the particle has to travel so as to create $\Delta$ and make up for the difference in spacetime up to the instance $T$ at speed $v_{boom}$.  The actual time period $\Delta t$ in the past is,

$\Delta t=\cfrac{1}{\text{per unit time}}=\cfrac{1}{\Delta \tau}=\cfrac{v_{boom}}{c}$

Which is total nonsense, crafted only for entertainment, but for a particle source at speed, $v_s$,

$\Delta t*v_s=\cfrac{v_{boom}}{c}*v_s=S$

where $S$ is the distance in front of the source (in the direction of its speed) that particles with speed $v_{boom}$ are detected.

Since $v_{boom}$ is known, by varying $v_s$ and measuring $S$ this is a new way to determine $c$ from the gradient of the plot of $S$ vs $v_s$.

$S$ is the distance beyond which the collapsed particle does not penetrate.  Given the spread of velocities around $v_{boom}$, a subject material placed around $S$ will have collapsed particles embedded in it as the particles travel back from the future.

Time Travel Into A Wall

X rays are refracted away from the normal at the refracting surface.  It is possible that X rays are the results of torus photons and collapsing at the refracting surface send them into the past with greater than light speed, $c+v_{boom}$.  This means a X ray setup need to be orientated with respect to earth's movement (Eastward) such that the future position is into the material to be examined.

If time travel into the past is the reason why X rays are penetrating, then in the westward direction, X rays will fail to penetrate as the future position is behind the X ray source away from the subject material refracting surface.

Conversely, in the eastward direction, X rays are most penetrating.  In a vertical position, X rays exposures are intensified in the eastward direction.  X ray source seems to be more intense eastward.

and the danger of being stuck in a wall on return from light speed is very real...

Why else should X rays be penetrating?

Sizing Particles Up

Knowing that the particles are passing at velocity $v$ in the Faraday Dark Space of a discharge tube,

it would be possible to flash X_ray through an aperture at an appropriate frequency, $f$ to obtain a still scattering pattern of each type of particles, $a_{\psi\,c}$,  $a_{\psi\,ne}$ and $a_{\psi\,\pi}$.

Given some reference, their size ($a_{\psi\,c}$,  $a_{\psi\,ne}$ and $a_{\psi\,\pi}$) can be determined.

At this point,  although the argument for the existence of $a_{\psi\,c}$ is plausible, particles need not have quantized sizes.  Particle size need not be in integer multiples of $a_{\psi\,c}$.  The requirement,

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

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

need not be true.

Spectrum Of Charge

When basic particles $a_{\psi\,c}$ are colliding with molecules of summed atomic numbers $Z$, the energy input is divided by $Z$ number of particles around the nucleus,

$\cfrac{density}{Z}$

When the molecules is moving, there are $Z$ number of particles around its nucleus at the required velocity, $v_{boom}$,

$density*Z$

Along another thread of thought,

when a particle can accumulate charge and this charge can flip polarity, then there is no need for sign assignment because each particle is its opposite charge.  There is also no need to set $B=T$, the magnetic field is the temperature field when temperature particles are interacting as waves, to provide for symmetry among the particles proposed.

The magnetic field is just the $E$ field rotating and fields of similar nature exist for gravity and temperature.

This resolve the issue of the opposite charges being anti-matter to each other.

There is no opposite charge, a particle acquires charge as its size changes. It is neutral at a specific value of $\small{1.499*a_{\psi\,c}}$.  Below this value the particle has a negative charge; above this value the charge is positive.  The opposite charge particle pairs identified earlier are complement of each other.  They are complement particles.

It is possible to align previously $p^{+}$ and $e^{-}$ particles such that both their $t_{T}$ and $t_{g}$ axes are opposite to each other and so cancels.  This suggests that the complement pairs are matter-anti matter pair.

But which is matter and which is anti-matter, and why??

The structure/symmetry of this model is now held together by complement, matter-anti matter pairs instead of opposite charge pairs.

绵雨风瑟瑟
点点如针冰
南洋耐冬寒
更盼明年春

《要过年啦！》

Projecting 3D Image

From the post "Phase Shift, No Normal" dated 07 Jan 2018,

$\angle P=\cfrac{cf+cf' }{\lambda_i}{sin(\theta_i)}*\pi$

where

$cf'=\lambda _{ r }\left[ tan(\theta _{ r })-{ 2 }{ cosec(2\theta _{ r }) }+{ cosec(\theta _{ r }) } \right]$

and

$cf=\lambda _{ i }\left[ tan(\theta _{ i })-{ 2 }{ cosec(2\theta _{ i }) }+{ cosec(\theta _{ i }) } \right]$

Specifically, if we consider,

$\cfrac { \lambda _{ i } }{ \lambda _{ r } } =\cfrac { sin(\theta _{ i }) }{ sin(\theta _{ r }) } =1.5$

$\theta _{ r }=sin^{-1}\left(\cfrac { sin(\theta _{ i }) }{ 1.5 }\right)$

and plot $\angle P$ with respect to $\theta_i$ only, we have,

When we zoomed in $-\small{\cfrac{\pi}{2}}\le \theta_i\le\small{\cfrac{\pi}{2}}$,

$cf'$ is sinusoidal.  The phase different changes from zero up to $\approx 1.25\pi$.

Being able to quantify phase here, allows for manipulating phase in optical signals easier.

By varying phase, it is possible to project a pseudo-3D image out from a flat screen.

Tungsten Stronger

Consider Tungsten Carbide, $WC$,  $Z=74+6$, density $15.6\,gcm^{-3}$, molar mass  $195.85\,gmol^{-1}$

$v_{boom}=3.4354*\cfrac{density}{Z}$

$v_{boom}=3.4354*\cfrac{15.6*10^{3}}{74+6}=669.90\,ms^{-1}$

and,

$T_{boom}=v^2_{rms}*\cfrac{Molar\,mass}{3*8.3144}$

$T_{boom}=669.90^2*\cfrac{195.85*10^{-3}}{3*8.3144}=3523.67\,K=3250.52\,^oC$

also,

$T_{p}=v^2_{rms}*\cfrac{Molar\,mass}{2*8.3144}$

$T_{p}=669.90^2*\cfrac{195.85*10^{-3}}{2*8.3144}=5285.51\,K=5012.35\,^oC$

which is a very high temperature, but might just anneal tungsten doped with carbon to even greater strength.

Unfortunately, Osmium Carbide, $OsC$ does not occur naturally.

Phase Shift, No Normal

There's nothing wrong with Snell's Law,

This diagram is constructed by touching a particle of radius $\lambda_i$ on a medium boundary, with a ray $AO$ through its center that intersect the boundary at point $O$.  The particle shrinks to size $\lambda_r$ at the point of contact $C$.  The line through its center to point $O$ is the new wave front.

Consider $CO$,

$\lambda_i tan(\theta_i)=OC=\cfrac{\lambda_r}{tan(\theta_r)}$

$\cfrac{\lambda_i }{\lambda_r}=\cfrac{1}{tan(\theta_i)tan(\theta_r)}$

$\cfrac{\lambda_i }{\lambda_r}=\cfrac{cos(\theta_i)cos(\theta_r)}{sin(\theta_i)sin(\theta_r)}$

or

$\cfrac{\lambda_i }{\lambda_r}=\cfrac{cos(\theta_i-\theta_r)+cos(\theta_i-\theta_r)}{cos(\theta_i-\theta_r)-cos(\theta_i-\theta_r)}$

The right diagram should be,

What happens on one wave front happens on the next and the two green triangles on the right shares the same hypotenuse but are displaced.

$OC=O'C'=\cfrac{2\lambda_i}{sin(\theta_i)}=\cfrac{2\lambda_r}{sin(\theta_r)}$

and so,

$\cfrac{\lambda_i}{\lambda_r}=\cfrac{sin(\theta_i)}{sin(\theta_r)}$

The particle collapse to a smaller radius at the contact point.  The wave fronts are disjointed and so are the incident and refracted rays.  The is no normal that meets both incident and refracted rays, as the wave fronts are not continuous.

A discontinuous wave front suggests a phase change with refraction.

Centers of the particles are at $a$ and $a'$

$oa=\cfrac{\lambda_i}{cos(\theta_i)}$

so,

$of=\cfrac{oa-\lambda_i}{sin(\theta_i)}$

$cf=oc-of=\lambda_i tan(\theta_i)-\cfrac{\cfrac{\lambda_i}{cos(\theta_i)}-\lambda_i}{sin(\theta_i)}$

$cf=\lambda_i \left(tan(\theta_i)-\cfrac{\cfrac{1}{cos(\theta_i)}-1}{sin(\theta_i)}\right)=\lambda_i \left[tan(\theta_i)-{2}{cosec(2\theta_i)}+{cosec(\theta_i)}\right]$

Similarly the distance from the contact point $c$ to where the wave front meets the boundary at $f'$ is,

$cf'=\lambda_r tan(\theta_r)-\cfrac{\cfrac{\lambda_r}{cos(\theta_r)}-\lambda_r}{sin(\theta_r)}$

$cf'=\lambda_r \left(tan(\theta_r)-\cfrac{\cfrac{1}{cos(\theta_r)}-1}{sin(\theta_r)}\right)=\lambda_r \left[tan(\theta_r)-{2}{cosec(2\theta_r)}+{cosec(\theta_r)}\right]$

and since between two consecutive wave fronts along the medium boundary is a phase difference of $2\pi$, so the phase difference between the incident and refracted wave is,

$\angle P=\cfrac{cf+cf'}{OR}*2\pi=\cfrac{cf+cf' }{2\lambda_i}*{2\pi}{sin(\theta_i)}=\cfrac{cf+cf' }{\lambda_i}{sin(\theta_i)}*\pi$

or

$\angle P=\cfrac{cf+cf' }{\lambda_r}{sin(\theta_r)}*\pi$

And we have the phase difference after refraction!

Mercury And Dark Light

Mercury, $Hg$, $Z=80$, density $13.534\,gcm^{-3}$, molar mass $200.592\,gmol^{-1}$

$v_{boom}=3.4354*\cfrac{density}{Z}$

$v_{boom}=3.4354*\cfrac{13534}{80}=581.18$

and

$T_{boom}=v^2_{rms}*\cfrac{Molar\,mass}{3*8.3144}$

$T_{boom}=581.18^2*\cfrac{200.592*10^{-3}}{3*8.3144}=2716.37\,K$  or  $2443.22\,^oC$

also,

$T_{p}=v^2_{rms}*\cfrac{Molar\,mass}{2*8.3144}$

$T_{p}=581.18^2*\cfrac{200.592*10^{-3}}{2*8.3144}=4074.55\,K$  or  $3801.40\,^oC$

Other than,

$c+v_{boom}=c+581.18$

there's nothing new here.  How far back in time does a mere $581.18\,ms^{-1}$ send us?

It is possible to stir up $v_{boom}$ as in a vortex; since $v_{boom}$ is the target velocity, a cylindrical drum will work better.

My post on mercury well (H.G. Wells) is missing!  Bunch of Nazis!

Light Bending Light

Another ray can bend such a ray,

as the particles collapse and gain greater speed after the collisions, the rays bend away from the respective normals at the point of collision.  The rays act as an intervening medium in each others path.  Each ray provides the energy needed for the collapse, but the ray does not refract again when it emerge from each other.

Does the rays deflect further from the collisions?  No, ever seen two rays disperse each other?

The rays trigger the collapse in each other.

Displaced Normal

This might be the right diagram, but which is the normal?

This is not normal.  If the photon torus is at light speed $c$, the collapsed spherical particle is at $c+v_{boom}$.  It is faster than the speed of light in the second medium, unless the second medium slows it down sufficiently.  The diagram shows the refracted ray bending away from the normal, as its speed is higher.

Which is normal? It does not matter, the point of incident however, is at the contact point of the two particles.

Dark Light.

Wrong Snell's Law Again

If instead,



$(\cfrac{\lambda_i}{sin(\theta_i)}+a)tan(\theta_i)=OC=\cfrac{\lambda_r}{tan(\cfrac{1}{2}\theta_r)}$

$\cfrac{\lambda_i}{\lambda_r}(\cfrac{1}{sin(\theta_i)}+\cfrac{a}{\lambda_i})tan(\theta_i)=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}$

If $\cfrac{a}{\lambda_i}\approx 1$

$\cfrac{\lambda_i}{\lambda_r}(sec(\theta_i)+tan(\theta_i))=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}$

$\cfrac{\lambda_i}{\lambda_r}(tan(\cfrac{1}{2}(\theta_i+\cfrac{\pi}{2})))=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}$

$\cfrac{\lambda_i}{\lambda_r}=\cfrac{1+cos(\theta_r)}{sin(\theta_r)}\cfrac{1+cos(\theta_i+\cfrac{\pi}{2})}{sin(\theta_i+\cfrac{\pi}{2})}$

$\cfrac{\lambda_i}{\lambda_r}=\cfrac{1+cos(\theta_r)}{sin(\theta_r)}\cfrac{1-sin(\theta_i)}{cos(\theta_i)}$

or

$\cfrac{\lambda_i}{\lambda_r}=\cfrac{cot(\cfrac{1}{2}(\theta_i+\cfrac{\pi}{2}))}{tan(\cfrac{1}{2}\theta_r)}$

which is still not Snell's Law.

Wrong Snell's Law

The previous post "Not Snell's Law" dated 03 Jan 2018 is wrong.



Notice the swap $\lambda_i\leftrightarrow a$.

$\cfrac{a*sin(\theta_i)+\lambda_i}{cos(\theta_i)}=OC=\cfrac{\lambda_r}{tan(\cfrac{1}{2}\theta_r)}$

$\cfrac{\lambda_i}{\lambda_r}\cfrac{\cfrac{a}{\lambda_i}*sin(\theta_i)+1}{cos(\theta_i)}=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}$

If $\cfrac{a}{\lambda_i}\approx 1$ ie a torus with a small hole,

$\cfrac{\lambda_i}{\lambda_r}\cfrac{sin(\theta_i)+1}{cos(\theta_i)}=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}$

which is the same as before. Wrong!

Not Snell's Law

If photon collapsing at the medium boundary is the cause of refraction,

where the torus photon collapses at the contact point $C$ and the photons stack up to give a wave front spacing of $\lambda$.  Bearing in mind that,

$2\pi a_{\psi}=\lambda$

Consider the line segment $OC$,

$\cfrac{\lambda_i}{sin(\theta_i)}=OC=\cfrac{\lambda_r}{sin(\theta_r)}$

and so,

$\cfrac{sin(\theta_i)}{sin(\theta_r)}=\cfrac{\lambda_i}{\lambda_r}$

And the beam does not change color, $f$ is a constant,

$\cfrac{sin(\theta_i)}{sin(\theta_r)}=\cfrac{f\lambda_i}{f\lambda_r}=\cfrac{v_i}{v_r}$

where $\lambda_i$ measures the thickness of the torus and $\lambda_r$ measures the diameter of the collapsed spherical photon.

Maybe...

But the details are not,

Also consider the line segment $OC$, and not considering the wave fronts of the incident and refracted waves,

$\cfrac{\lambda_i*sin(\theta_i)+a}{cos(\theta_i)}=OC=\cfrac{\lambda_r}{tan(\cfrac{1}{2}\theta_r)}$

$\cfrac{\lambda_i}{\lambda_r}\cfrac{sin(\theta_i)+\cfrac{a}{\lambda_i}}{cos(\theta_i)}=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}$

If we approximate  $\cfrac{a}{\lambda_i}\approx 1$ (ie. torus with a small hole) and let $90^o-\theta_{i\,c}=\theta_i$,

$\cfrac{\lambda_i}{\lambda_r}\cfrac{sin(90^o-\theta_{i\,c})+1}{cos(90^o-\theta_{i\,c})}=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}$

$\cfrac{\lambda_i}{\lambda_r}\cfrac{cos(\theta_{i\,c})+1}{sin(\theta_{i\,c})}=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}$

$\cfrac{\lambda_i}{\lambda_r}=\cfrac{tan(\small{\cfrac{1}{2}}\theta_{i\,c})}{tan(\small{\cfrac{1}{2}\theta_r})}$

Do we have a problem here?  Yes, this is not Snell's Law.

Image From The Future

What if all lighting are discharge phenomena, where torus photons are ejected and gained light speed but collapsed later in their path and are projected into the past, somewhere/anywhere before the source?

The gray beam is due to photons returning from the future.  And when the experiment setup is rotated,

the gray beam remains parallel to the direction of Earth's motion.

This beam generates a virtual image of the source.  On the opposite side of the source we see two images, one refracted and another from the future.

Silicon Steel

If of steel the annealing temperature is $T_{boom}$ of the dopant $Fe_3C$ then other additives with their corresponding $T_{p}$ as annealing temperature may result in steel of varying desirable properties.

For example, silicon $Si$, as an additive.  $Fe_3Si$, $Z=3*26+14$, density $7220\,kgm^{-3}$, molar mass $195.6205 g\,mol^{-1}$,

$v_{boom}=3.4354*\cfrac{density}{Z}$

$v_{boom}=3.4354*\cfrac{7220}{3*26+14}=269.60\,ms^{-1}$

And,

$T_{boom}=v^2_{rms}*\cfrac{Molar\,mass}{3*8.3144}$

$T_{boom}=269.60^2*\cfrac{195.6205*10^{-3}}{3*8.3144}=570.03\,K$  or  $296.89\,^oC$

and also,

$T_{p}=v^2_{rms}*\cfrac{Molar\,mass}{2*8.3144}$

$T_{p}=269.60^2*\cfrac{195.6205*10^{-3}}{2*8.3144}=855.05\,K$  or  $581.90\,^oC$

If steel uses $Si$ as an additive, does heat treatment at $581.90\,^oC$ causes the material to strengthen as in annealing of steel?

Basic Particles And Steel

Why does annealing steel leads to a change in its ductility and hardness?

$a_{\psi\,c}$ generated at boom temperature, $T_{p}$ coalesce with $a_{\psi}\lt a_{\psi\,ne}$and $a_{\psi\,\pi}$ to form greater charges.  Both of which results in stronger bonds as paired orbits share an electron.

From the post "Steel Fishing" dated 02 Jan 2018, if annealing temperature is equal to $v_p$ of iron carbide $Fe_3C$, it is bad news for the posts "Collecting Charge Current",  "Collecting Charge Current II" and "Sun Particle Capture" and "Torus In Entanglement", where it was proposed that very finely grind $TiO_2$ in a suspension has a changed density that will be in boom under sunlight.

Iron carbide $Fe_3C$ in steel, unless clustering has retained its density, the validity of $T_{p}$ as the annealing temperature of steel at and beyond $0.8\%$ of carbon, suggests that density cannot be changed.

Can density be changed by forming a suspension of very finely grind powder?

Yes, the needed boom interactions will be at the interface between the fluid and the powder clusters.  Within a cluster, which may contain millions of molecules, the density is still of $TiO_2$ pure.  In the same way, within steel, $Fe_3C$ cluster responds to $T_{p}$ as if a pure material with unchanged density.

Both can lah...

Steel Fishing

Iron Carbide $Fe_3C$, $Z=3*26+6$, density $7640\,kgm^{-3}$, molar mass $179.54\,gmol^{-1}$,

$v_{rms}=3.4354*\cfrac{density}{Z}$

$v_{rms}=3.4354*\cfrac{7640}{3*26+6}=312.46\,ms^{-1}$

and

$T_{boom}=v^2_{rms}*\cfrac{Molar\,mass}{3*8.3144}$

$T_{boom}=312.46^2*\cfrac{179.54*10^{-3}}{3*8.3144}=702.75\,K$  or  $429.25\,^oC$

It would seems that $T_{boom}$ for $Fe_3C$ has nothing to do with annealing of steel nor its critical temperature.

$T_{p}$ however,

$T_{p}=v^2_{rms}*\cfrac{Molar\,mass}{2*8.3144}$

$T_{p}=312.46^2*\cfrac{179.54*10^{-3}}{2*8.3144}=1054.12\,K$  or  $780.97\,^oC$

It could be that clusters of $Fe_3C$ with the effective density of $7640\,kgm^{-3}$ are broken up in steel at $780.97\,^oC$ and that gives it strength.

$v_{p}$ is more relevant here because, instead of multiple collisions to gain the required kinetic energy input to set off a boom effect, in this non-homogeneous medium, only one collision on the quasi-nucleus of $Fe_3C$ can be expected to deliver energy at the rate required.  Multiple collisions on the quasi-nucleus do not add up to $v_{boom}$.

Maybe, $T_{boom}=429.25\,^oC$ is just as effective annealing, it is simply not experimented with before.

Just when $T_{boom}$ was deemed more relevant (post "RMS Than Most Probable" dated 1 Jan 2018).

Note: For annealing steel is heated up $20-50^oC$ above critical temperature at $723^oC$.

Hot Tube Turned

When torus at light speed collapses back into a spherical particle it attains speed $c+v_{boom}$; this drive it into the past.  (post "Through The Mind's Eye" dated 30 Dec 2017.).  This happens all the time in a fluorescence tube.  The charges are in the position where the tube will be in the future and may end up outside the tube.  They are not visible as they do not cause a discharge of the air outside the tube.  Set the tube at the discharge voltage of air, the tube will not light up but a specific path outside the tube will.

The escaped particles are negative charged and eventually coalesce into positive charges and heats up the room.  But first they charge outside of the glass tubes negatively.

The tube is re-orientated to accommodate the particles that escape,

the tube itself need not be redesigned but only made longer, considering how far the particles have traveled from the future and how much Earth has moved then, Eastward.

To see the particles outside of the tube, discharge the tube at the air discharge voltage.  Assuming that the discharge is due to nitrogen gas $N_2$ in the air at $78\%$, $Z=7*2$, density $1.165\,gcm^{-3}$ and molar mass $28.01286\,gmol^{-1}$

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

$V_{min}=\cfrac{1}{2}\cfrac{28.01286*10^{-3}}{1.602176565\text{e-}19*6.02214129e23}\left(3.4354*{1.165*0.78}*{2*7}\right)^2$

$V_{min}=2.7728\text{e-}4\,V$,

electrode gap $d=1\,m$

And for a gap of $12\,\mu m$,

$V_{min}=23.11\,V$

Goggles required!

RMS Than Most Probable

It seems that $v_{rms}$ is more relevant than $v_p$,

multiple impact generates the neeeded $KE$ input to the system that averages (root-mean-squared) to $v_{boom}$.

In cases where $v_{boom}$ is generated directly, for example a concussion charge down a sealed tube full of water or ice, the exact speed of the explosive wave front equals to $v_{boom}$ of the material used.  This would be an invention that makes mining, excavations, demolitions and other construction explosions cheap.

Happy New Year 2017