RECONSTRUCTION OF THE GAS EQUATION
THE SOLUTION OF
THE BOLTZMANN CONSTANT
5
8.61022581810  e  k
2 107 137.036 e  k
BY
FRANCIS VIREN FERNANDES
DECEMBER 15, 2012
PV  nkT
If
n  1 particle
PV  kT
EE
I2 F
P 2 
R
A
V  R1  R2  R3
k  2 10 137.036 e
7
T  eV  ea
My Discoveries
ETHER IN DYNAMIC PULSATE MOTION
PLANCK ETHER MASS
mPl 
186 ETHER MASS
c
 Planck Mass  2.176450474×10 -8 kg
G
or  B 2  137.036     Plancklength 
2
I have discovered this correspondence in
the dynamic living pulsations of ether
B  1.3806680311036
WAVE-PARTICLE DUALITY
• Particle:
• Wave:
q  M R 10 
2
7
λ  2R 1
• Field:
M
1.859222909 1 0 9

 1.346611109 1 0 27 k g/m
R
1.380668038 1 0 36
ETHER TORUS[Ђ]
186ether velocity  Boltzmannradius  Planck ' sh
1.86 109  25812.8075 1.380668 1029  h
737 Aitheron x f = 186-ETHER [Ę]
ONE 186-PHOTON TORUS ◌
SOLID ANGLE=4π×10-7
TWIN MASS
The Fractal Structure of Light & Emanation of Matter
A pulsating ether sphere  rolling in a spinning pulsating ether torus Ђ
737 Wave-Maker
The energy of 737 equals the Planck’s constant
times one second
7.37 x 10-51 x c2 = h x 1.0
1.86 x 10-9 x VRk =7.37 x 10-51 x c2/1.38x10-29 x 1.0
So, the imposed condition is 1 second for the wavemaker to exist.
Coulomb’s Constant
K = c2 x 10-7
Velocity squared = v1 x v2
25812.8076 3.48181876510  K
5
The Boltzmann Constant
Superconductivity of SQUID & resistance
The ratio of the velocity of 186-ether to the
electrical resistance of a photon, c/e
(2 10 ) 137.036 
R
Qt
7
25812.8076m / s
  c / e  1.8711574691027 m / s.C
c
25812.8076  Boltzm ann' k
e
The Coulomb constant
25812.8076 x 34181.8779 m2/s2
f  2.5218363141041
1.859222909109 kg
f 
7.372496371051 kg
f 1.859222909109  4.6886558471032 kg
4.6886558471032 kg
length 
 34181.8779m
27
1.34661110910 kg / m eter
25812.8076m  (2 10 )137.036
7

t
E=mc2
25812.8076 3.48181876510  c
12
2
Ohm’s Law V=IR is F=ma
• OHM’S LAW Rewritten as F = ma
•
•
V=IR
• FERNANDES LAW
•
•
•
c
I F
2
a 10  F   I  
e
m m
7
(2 107 ) 137.036 
R
Qt
• Acceleration, a as volts equals current, I the root of force, F times
resistance, R.
•
• Resistance, R is current, I per photon mass about one 186-seed ether.
•
Voltage is acceleration
Consider 511 keV for an electron
eV
V
e
511000
24

3
.
1894

10
Volts
19
1.60217653 10
c 2 107 (2.998108 ) 2 107
24
m/s2
Acceleration 


3
.
1894

10
15
r
2.81794032510
q  M R10
2
(1.602176537 1 0
1 .8592229091 0
7
19 2
9
) 
1 .3806680311 0
36
1 0
7
CURRENT
Momentum = mv = Ft
1.859222909 x 10-9 kg x v = 25 x 3.20435306 x 10-20 s
v = 4.3087263 x 10-10 m/s
One coulomb of ether in kg = 1.859222909 x 10-9 kg x 6.24150948 x 1018 = 1.160435741 x 1010kg
Current is the momentum of one coulomb of ether,
Ether Current I = 5 amps = 1.160435741 x 1010kg x 4.3087263 x 10-10 m/s per one coulomb
Energy of ether drift, E = F r where r = 9.6064088 x 10-12m and F = I2
E = 25 x 9.6064088 x 10-12 = 2.4016022 x 10-10 J
E = m (cv) = 1.859222909 x 10-9 (2.99792458 x 108 x 4.3087263 x 10-10) = 2.4016022 x 10-10 J
Copper Current I = 5 amps = 2.672142823 x 10-27 x 6.24150948 x 1018 x 2.99792458 x 108 per
one Coulomb
The drift velocity of changed mass 2.672142823 x 10-27 kg of copper atoms
v = 7.34295219 x 10-4 m/s
E = m (cv) = 2.672142823 x 10-27 x (2.99792458 x 108 x 7.34295219 x 10-4) = 5.88235283 x 10-22 J
E = FR = 25 x 2.352941132 x10-23= 5.88235283 x 10-22 J
Remarks
The drift of ether and atoms of the conductor can be calculated by E = m (cv)
The mass of copper atoms changed under eVe stress from 1.05520602 x 10-25 kg to 2.672142823
x 10-27 kg
Maxwell thought that drift velocity was that of electrons. This has now been shown to be a false assumption.
NEWTON’S GRAVITATION CONSTANT DECONTRUCTED
G
G
B 2
c

36
1.380668031
10
2

c
1.859222909109
or
Ether  1.346611109 1027  6.371  106  8.579261033 kg
Force, F of earth mass, m oscillators F = m a
F  8.57926 1033  9.82
c2
F  m
r
F  8.42483332 1034 N
c2
F  8.42483332  10  m 
6.371106
m  5.9748 1024 kg
34
Force of entrained ether F = m a
Entrained Ether for Earth
The acceleration of entrained ether produces an acceleration of 9.8 m/s2
Radius of Earth 6.371 × 106 m
Magnetism
• MAGNETISM IDENTIFIED
The effect of magnetic flux [LHS] is caused by
current [RHS]
Ф = I x π x r x 10-7 x 137.036
UNIFICATION OF GRAVITY WITH ELECTROMAGNETISM
GEM
This is the unification of gravity with electromagnetism which is being searched
for because no one cared about the units of eV
MASS
Mass is an attribute of an oscillator measured differently by eV, R=I/m and F=ma.
These experimental mass measures differ by a rotational factor of
2 Pi x 10-7 x 137.036
Francis V. Fernandes (Born 1961)
Idiocy of energy units - It's too bad, but I have already apologized,
and there is nothing else I can do...
Richard P. Feynman (1961)
You Feynman recognized the error which physicists and
engineers memorized as truth; so here is my solution…
T  2
T  2
q
T  2
I
q  I T
ml
mg
2
q
2
I
Q=Ixt
This formula Q = I x t represents a physical
reality when squared
q  I t
2
2
2
m  r 107  F  t 2
r 10
F  m 2
t
F  m a
7
PV=kT
2
I
7
 R1  R2  R3  2 10 137.036 e  eV
2
R
2
R  R1  R3
h
I
 BA    R1  R3
2e
R
2
F
I
P 
A R1  R3
METHOD
• CV = velocity squared
• E = PV = kT = eVe = eae = hf = mc2
• m x r = e2
• Ф = I x π x r x 10-7 x 137.036
I know the formulas for volts, current and resistance
a = I x I/m
Time period of a pendulum
Particle volume of gas at 300kelvin
•
•
•
•
•
Vol = R1 x R2 x R3
Vol = 4.084572793 x 10-26 m3
R1 = 0.278731462 m
R2 = 0.557462906 m
R3 = 2.628722405 x 10-25 m
kT
ea  T  eV
5
8.61022581810  e  k
2 10 137.036 e  k
7
Energy, E=kT=PV
ea  T  eV
2 10 137.036 e  k
7
2 10 137.036 e  eV  kT
7
2 10 137.036 e  ea  kT
7
2 10 137.036 E  kT
7
E = eae = eVe
• At a temperature of say for example 300kelvin
and so
acceleration,a or volt,V=1.612224182x1017 m/s2
• E = kT = eae = 4.138530517x10-21 Joules
2 10 137.036 e  eV  kT
7
2 10 137.036 e  ea  k  300
7
Speed of hydrogen molecule mass, m
at 300K is 1927m/s
acceleration, a =1.612224182x1017 m/s2
2
2
v
v
a

r 3r
v  1927 m / s
q  m  r 10
2
7
r  7.6734889951012 m
angle x alpha-1 x e x ea
2 107 137.036 e  eV  kT
2 10 137.036 e  ea  k  300
7
Boltzmann, k  e  precession
T x e = ea x e = 300 x e = E = m(cv)2

19
E  3001.6021765310 J

17
E  4.8065295910 J

17
E  4.8065295910
17
E  4.8065295910
 (2 10 )137.036
 21
E  4.13853051710
J
7
J  PV  kT
R2 = 0.557462906 m
E  4.1385305171021 J  PV  kT
E  m c2  hf
f  6.2458304161012 Hz
1
t   1.6010681261013 s
f
m  4.6047362121038 kg
I
c
2.99792458108
27
R  

1
.
871157469

10
ohm s
19
m q 1.6021765310
I  m  R  4.6047362121038 kg 1.8711574691027 ohm s
I  8.6161865541011 Am p
R2 = 0.557462906 m
38
m  4.60473621210 kg
q  m  R2 10
2
7
R2  0.557462906m
R1 = 0.278731462 m
38
m  4.60473621210 kg
q  m  R2 10
2
7
R2  0.557462906
m
q  2m  R3 10
2
7
R2
R1 
 0.278731453
m
2
R3 = 2.628722405 x 10-25 m
Ro = 3.053023765 x 10-21 m
Volume = 4.084572793 x 10-26 m3
Volum e
 Area
R2
 26
4.08457279310
 7.32707641026 m 2
Area, A 
0.557462906
 2.067837171015
MagneticField , B  
A 7.32707641026
B  2.8221811871010 Am p/ m eter
DONE AWAY WITH SIN COS i j k
R3
&
Ro
I
B
R
8.6161862341011
 21
R 

3
.
053023765

10
m
10
2.82218118710
R3  R  2 107 137.036  2.6287224051025 m
Drift velocity E/B = v
F I 2 (8.6161862341011 ) 2
Electricfield , E  

q
q
1.602176531019
E  0.046336133N / C
E
0.046336133
12
v 

1
.
641855357

10
m/s
10
B 2.82218118710
R3 2.62872240410 25
12
v


1
.
641855434

10
m/s
13
t
1.60106812710
Drift velocity squared cv
q  m  R3
2
14
m  9.76508447310 kg
Energy, E  e  300kelvin  eea
E  m cv  4.8065294131017 J
Squid Velocity of 186-ether propels hydrogen
gas molecules – the reason for Brownian motion
186ether velocity  Boltzmannradius  Planck ' sh
9
1.86 10  25812.8075 1.380668 10
29
h
v
25812.8076
17
2
a 

1
.
612224182

10
m
/
s
t 1.6010681271013
v2 v2
a

r 3r
v  1927 m / s
q  m  r 10
2
7
12
r  7.67348899510 m
VELOCITY SQUARED OF LIGHT SPEED C2
25812.8076 X 3.481818765 X 1012
m  4.604736212
 1038 kg
q 2  m  r  107
R2  0.557462906
m
v2
0.557462906
12

3
.
481818761

10
m / s[ FTL]
13
1.601068132
 10
v1  25812.8076



t
1.601068132
 1013
  4.132806365 109 m
mc  h
m  5.347985402
 1034 kg
E  m c2  4.806529576
 1017 J
E  eT  eea  e  300J
T x e = ea x e

17
E
4.8065295910
a 2 
19 2
e
(1.6021765310 )
a  1.87245284410 m / s
T  ea  eV  300kelvin
21
2
Analysis of Volume = R1 x R2 x R3
AT 300 KELVIN
A = R1x R3
R1 = 0.278731462 m
2m
Used to calibrate
voltmeters
B A is the magnetic
flux quantum
Pressure = Current
squared per area A
R2 = 0.557462906 m
R3 = 2.6287224 X 10-25 m
m
m*
The photon mass
measured due to
737-frequency
The FTL component
velocity of speed of
light squared
Measured in
magnetism
The source of
magnetism
PV=kT
2
I
 R1  R2  R3  2 107 137.036 e  e  a
R1  R3
F
 V  2 107 137.036 E
A
Spectroscopy
The mass of a proton, m2 is changed
from 1.672622216 x 10^-27kg to
3.634114179 x 10^-29kg when
0.068amps of current is passing thru
water during electrolysis
The energy as measured for a proton E = F R = I^2 R
The radius of a proton q^2 = m R 10^7 and so
R = 1.5347 x 10^-11m
E = 0.068^2 x 1.5347 x 10^-11 = 7.0964528 x 10^-14 J
E = m1c^2 and so m1 = 7.895868606 x 10^-31kg
Originally before electricity was passed the energy of
the proton E = mc^2
E = 1.672622216 x 10^-27 x c^2 = 1.503275 x 10^-10 J
E = I^2 R
1.503275 x 10^-10 = 0.068^2 x R
R = 3.251029 x 10^-8 m
Q^2 = 7.895868606 x 10^-31 x 3.251029 x 10^-8
In any interaction its always
I1 x I2 or m1 x m2 or r1 x r2
So m1 x m2 = 7.895868606 x 10^31kg x 1.672622216 x 10^-27
Square root of these masses =
3.634114179 x 10^-29kg [ R = I /
m]
Q^2 = 3.634114179 x 10^-29 x
7.063536 x 10^-10
This is the oscillator that yields
the clock frequency
The radius of the proton changes
from R = 1.5347 x 10^-11m to
7.063536 x 10^-10m
I
 1.871157469 10 27 amp / kg
m
0.068

m
m  3.63411424 10 29 kg

Transient mass
The Planck energy involves 2Pi x 137.036
E = m x 2Pi x 137.036 x c2 x 10-7 = 2.81225386 x
10-16 J
E=hf
f = 4.24422646 x 1017 Hz
t = 1/f = 2.35614196 x 10-18s
18000 / 2.35614196 x 10-18= 7.639607603 x 1021
Thus the time for hydrogen to evolve at the
cathode is inversely proportional to the frequency
of the changed mass of a proton under external
current stress during electrolysis of water.
Mass Length Time period
This experiment is conclusive evidence that …..
• Our measure of mass is at light speed. Mass is
already at light speed.
• The frequency of the oscillator changes drastically
on insults
• Frequency change means mass and radial length
change
• I am able to crack these issues because voltage is
acceleration, current squared force and
resistance always constant at 1.87 x 10^27
amp/kg or velocity / elementary charge
Magnetism
• h/2e = I x Pi x R x 137.036
• Ф = 0.068 x Pi x 7.0635359 x 10-17 x 137.0359991
I have deconstructed the magnetic flux quantum.
Gravitational force
F = mg = 0.008kg x 9.8 = 0.0784 N
F = I2 = 0.392 A x 0.2 A = 0.0784 N
• So, current of 0.392 A is required to flow thru a 0.50 m
conductor placed 90 degrees to the magnetic field, of
strength 0.40 T for a levitation effect.
• Let us consider current generated by the magnetic field, B
of 0.4 T on 0.5m length of the conductor.
•
I = 0.4 T x 0.5 m = 0.2 A
• F = I2 = 0.392 A x 0.2 A = 0.0784 N
The Rydberg Mass
13.6 eV the first ionization electron volt
energy of an electron for the hydrogen atom
corresponds to a Rydberg photon and not a
511 keV electron
13.60569223 x e = 2.179872077 x 10-18 J
E = mc2 and so m= 2.425434789 x 10-35 kg
The Reason for Spectra
2.425434789 x 10-35 x c x λ = h
inverse of lambda is the Rydberg number
Radius
Energy
R= 1.5347 x 10-11 m
E = F x R = I2 x R
E = mc2 where m is external photon
Resistance  
I
m rydberg
c
  1.8711574691027 m / sC
e
m rydberg  m proton m externalphoton
Changed measured mass
1.
2.
3.
4.
Maxwell drift of changed copper mass
Electrolysis of water
Spectrum of Hydrogen
A magnetic field which contributes to gravitational
force of a rod. I have shown how a plane has lift in a
similar manner.
THERE IS ONE MASS [ 7.37x10-51kg x f ] MEASURED
DIFFERENTLY AS IT MOVES ALONG 4 PATHS NAMELY
R1 R2 R3 R0
Twin Mass –Ether & Matter
Point mass 737-wavemaker must exist
Twin Mass Structure
Ether Mass and associated charge
7.372496 x 10-51 kg Oscillator x frequency , f = Photon mass
c = f x λ
Photon mass x # = 1.859222909 x 10-9 kg
(Matter)
(Elementary charge / oxidation state)
1.859222909 x 10-9 kg Torus x # = Ether Mass
(186-ether)
etherm ass
 radiusofphoton
27
1.34661110910 kg / m
STRUCTURE OF SPACE
1. Twin mass = Ether and Matter
2. Etheric sea = Ether and Ether
3. Void
= Soul
Entropy Units! ????
H
S
T  m ol
eV e T  e
S
 

m ol T m ol T
eVe
S
 n  e  m ol1
eV  m ol
Entropy
Slowing down of photon speed is my idea of generating
free energy from the structure of space
Entropy is the energy required to raise the energy of one
mole of substance from zero to 298 kelvin
I would redefine entropy as the energy input to slow down
speed of a photon from light speed c to drift velocity v
E = mc2 = m x 25812.8076 x 3.481818 x 1012
E = mxcxv