Due to Emergency the contents
of this page is shifted bottom.
The Earth will
be blown up by the scum, who call themselves as scientists. Society you are a stupid
cattle - because, despite the existence of this and similar web-sites, you did nothing
for the detention of terrorists, physicists from CERN and other scoundrels, sitting
in academies and promoting this blasphemous experiment.
September 11, 2010.
uuu uud udd ddd uus uds dds
uss dss
sss
These are triads, which can be made of up, down and strange
quarks. One of these triads is stable in isolated state; it is a proton, uud. All
other triads are more massive than proton (uud), as a result, they all decay. It
is known, that the triad udd, which is neutron, decays, if it is isolated. The cause
of decay is clear; - the mass of free neutron is bigger than the mass of free proton.
But the situation is different inside another environment:
1. inside the neutron-excessive nuclei a neutron decays at proton, electron, electron
antineutrino;
2. inside the stable nuclei the number of neutrons and protons do not change;
3. inside the neutron-deficit nuclei a proton decays at neutron, positron, electron
neutrino.
This shows us, that sometime a triad udd can be more stable than uud triad. Now
we can make a conclusion: stability of a quark triad depends on the properties of
a subnuclear condensate.
The question arises, - is there a subnuclear condensate, where some quark triads
be more stable as usual proton-neutron condensate, which are the nuclei of our Earth?
Till now physicists had received unstable nuclei, consisting from three types of
triads: protons (uud), neutrons (udd), and lambda-hyperons (uds).
Here are some examples of such hyper-isotopes:
hyper-hydrogen (pnl), (pnnl);
hyper-helium (ppnl), (ppnnl), (ppnnnl), (ppnnnnl), (ppnnnnnl), (ppnnll);
hyper-lithium (pppnnnl), (pppnnnnl),(pppnnnnnl), (pppnnnnnnl).
Till now physicists had received even an anti-hyper-hydrogen, which is also unstable.
But look at the beginning of this text, and try to understand, when unstable
free neutron becomes more stable than proton, if neutron is not already free, but
bonded in subnuclear condensate.
Try to answer, what will happen with the ordinary matter, if physicists will
create a stable condensate of lambda-hyperons, or of a mixture of neutrons and lambda-hyperons.
These subnuclear condensates have no limit of grows. Usual subnuclear condensate,
consisting of protons and neutrons, do, have the limit, which is uranium. Uranium
finishes the periodic table of stable elements. The existence or absence of the
limit is explained by the electrical charge of a condensate. Usual nuclei have an
electrical charge. Condensates, made of lambda-hyperons and/or neutrons, do not
have the electrical charge. As a result, they do not have Columbian barrier, which
do not let the ordinary nuclei to fuse.
The 9-th of November 2010 CERN plans to collide heavy ions. As a result, the
Solar system can be exploded into a nova star. If strangelets will capture ordinary
nucleons, ruin them and grow without decay on smaller strangelets, then the duration
of explosion will be about of 500-1000 days. If growing strangelets at some moment
would decay at smaller strangelets, then the duration of Earth explosion can be
much less.
There are some other microscopic condensates, which can be created at LHC, for
example, microscopic magnetic holes. Result of such experimental discovery will
be the same – the Earth’s explosion.
Heavy ion collision is a new dangerous step, planned by CERN physicists. They
will be started in two months. Politicians, businessmen, ordinary people believe
that “the best physicist” on the planet are correct. CERN propaganda had made its
criminal role; - courts, switched on by opponents of dangerous CERN’s experiments,
are failed.
So, people, be ready to be murdered. You did not believe us earlier, you do not
believe us now. You a lazy to ruin the CERN, you are lazy to strike in order to
save yourself. You are a crowd of silent cattle, which must be killed. I wish you
happy flight in icy comets to other star systems. Go and bring there a seeds of
new life. Do not fear to die, and fulfill your sacred mission. Fasten your safety
belts; - be ready to suicidal reproduction of biospheres.
September 7, 2010. From my
point of view, the most dangerous things, which can be made on proton-proton (p-p)
collisions at powerful colliders, are two different condensates: microscopic magnetic
holes and microscopic droplets of strange matter. In the bottom I’ll name both of
them by unifying term “droplets’.
The time interval of the Earth explosion depends from several parameters: the number
of created droplets; the density of matter inside droplets; the velocity of particles
in plasma bubbles around the droplets.
Let density of droplet will be 300 times greater than nuclear density;
Let velocity of particles in the plasma bubbles will be v=c/10.
Then:
a. if there were created 100 000 000 000 droplets, then the Earth will be exploded
in 6 days;
b. if there were created 1000 000 droplets, then the Earth will be exploded in 272
days;
c. if there were created 1 droplet, then the Earth will be exploded in 75 years.
Result "c" can be rejected. Physicist will not stop if they will see one unusual
track.
Result "a" can also be rejected, because it is impossible to create 100 000 000
000 droplets in the very first days.
In the upper results we supposed that the droplets were made simultaneously.
Now we'll remade the program and let’s suppose that they are created continuously:
a. 10 droplets per second;
b. 1 droplet per second;
c. 1 droplet per 10 seconds;
d. 1 droplet per 100 seconds.
Then the Earth will be torn on fragments in
a. 98 days;
b. 175 days;
c. 311 days;
d. 552 days.
As we know LHC works with quite long stops. So, I think, that the option "d" is
the most reliable. As a result, I supposed that the total time of Earth explosion
into fragments is about 1000 days.
Several more interesting results:
1. The mass of Earth’s matter, ruined before the rest of it would start into cosmos
with velocity about 10 km/s, is about 10^{15} kg, - that is about one billionth
part of Earth. That also will be approximately the total mass of all droplets.
2. The Sun will be ruined in about a 1000 years. The Sun will undergo several successive
explosions as nova. By the way, that explains the rings of SN 1987A.
3. The energy output of matter’s collapse at droplets at the time of “Earth torn
on peaces” will be about the luminosity of the Sun. That means that neutrino flux
from the Earth at that time will be by (R/r)^{2} times bigger than today’s
neutrino flux from the Sun. (R – radius of Earths orbit; r – radius of the Earth).
The graph of heat power in the “continuously” case is a cubic parabola; consequently,
we are able to register the excessive flux of neutrino from the Earth already now!
If there is no such excessive flux, then I was wrong. But do not forget about future
ion-ion collisions; do not forget about future 2*7 TeV p-p collisions. Those are
other possibilities to kill the Earth, to kill the humankind.
CERN and all corrupted science are responsible for the probable future murder of
6 billion people and probable destruction of the Earth.
It was subscribed in the Universal Genetic Code of Eternal Living Universe that
we, the enemies of biosphere's suicidal reproduction must be undergone to ban, to
laughter and shame; otherwise the reproducing act would not occur. That is the law
of biosphere's reproduction. So, don’t fear to die and perform your sacred mission.
Fasten your safety belts; - be ready to biosphere’s suicidal reproduction!
Appendix: If you do not trust to my computed results, you can compose your own program
and test it. My program works by small steps, measured by time interval, for example,
dt=1000 seconds. At every step it computes the number of ruined protons by this
formula dN = dt v 4pr(i)^{2}k / 6, where v =
c/10, c – velocity of light; 4pr(i)^{2} – the
surface of a droplet at i-th step; r(i) – radius of a droplet at i–th step; k –
concentration of protons in a plasma bubble around a droplet, k = M_{Earth}/((4/3)pR_{Earth}^{3})
/ m_{proton}. The number “6” in a formula for dN is explained by the
fact that our space has 3 dimensions and by the fact that only 1/2 of particles
move in positive direction.
Radius of a droplet at the next step: r(i) = r(i - 1)((N + dN) / N)^{1/3}.
The total number of ruined protons at every step: N = N + dN.
Heat power at every step, if droplets are made continuously:
P(i) = P(i - 1) + dN m_{proton} c^{2} / dt / 2 * N_{dr};
if droplets were made simultaneously:
P(i) = dN * m_{proton}c^{2} / dt / 2 * N_{dr}.
In “simultaneously” case P-graph is square parabola; in “continuously” case P-graph
is cubic parabola;
N_{dr} – number of droplets created per step.
The program stops, when the total heat output, Q, becomes bigger than 3*10^{32}
Wt. This amount of energy is enough to send all matter of the Earth to infinity.
Inevitable energy losses can be neglected if T < 1000 days. If T > 1000, then the
true value of T will be the more, the more it’s computed value.
Note 1. It was supposed that every proton, p, colliding with microscopic magnetic
hole, NS, decays and the droplet becomes bigger. NS_{n} + p = NS_{n+1}
+ e^{+}, where: e^{+} - positron; subscripts _{n} and
_{n+1} are the number of x-bosons in the magnetic hole; x-boson is a smallest
magnetic excitation in ferromagnetic Savidy vacuum, ns. As a result of proton capture
the region around magnetic hole will have an excess of neutrons. That will lead
to beta decays and to flux of electronic antineutrino from that region.
Note 2. It was supposed that every proton, p(uud), and every neutron, n(udd), colliding
with droplet of strange matter, N(uds), decays and droplet becomes bigger. N(uds)
+ p(uud) = (N+1)(uds) + K^{+}, where: K^{+} - positively charged
kaon. (N) and (N+1) are the number of lambla-hyperons in a growing strangelet. Lambla-hyperon
is a strange nucleon (uds) in a strangelet N(uds). A strangelet can capture neutrons
too: N(uds) + n(udd) = (N+1)(uds) + K^{0}, where: K^{0} - neutral
kaon.
K^{+} and K^{0} decay. As a result an additional flux of different
sorts of neutrino and antineutrino can be registered.
Every capture of nucleon by a droplet is accompanied by heat output of about half
of rest energy of nucleon. That means that the binding energy of a droplet is extremely
high; that specific energy output of nucleon collapse onto droplet is hundred times
more than the specific energy output in uranium or hydrogen bombs. But the time
interval of collaptical explosion is quite big, because, in order to explode the
Earth into peaces, the droplets must successively grow from 10^{-23} kg
to 10^{15} kg / N_{dr}.
March 19, 2010. Today night
the criminals from CERN accelerated protons to the record energy 3.5 TeV per beam.
At the regions of collisions, probably, the rays were on skew lines (two lines that
do not intersect but are not parallel). It is not excluded that there were accidental
collisions of protons.
If the collapse was switched, then most
probably tomorrow morning all people will start to cosmos.
According to plan the first collisions
at 3.5 TeV must be fulfilled the 30-th of March.
March 10, 2010. By the way,
it seems, unstable condensates were already been created at Tevetron. That follows
from the multimuon events (up to 8 per droplet and antidroplet) described at the
end of 2008 by CDF collaboration.
From the other hand, some authors say that "strange condensate" is stable if it
has more than 10 lambdas, 10(uds).
...This portrait is dedicated to CERN decision to start collider work the 20-th
of February with energies 3.5 TeV per proton.
Rolf, Adolf, Collider, Death.
...
...
Let’s speak about “Hawking radiation”.
Bell paradox. If two rockets, connected by a string,..
...
...These formulae give Unruh/Hawking temperatures at the depth
R, which is equal equal to Compton radius of a particle R_{c}...
...
... This is not my picture. Perhaps its author is my good friend, supporter.
Citizens! Do not think that some good Uncle Ivan would come and destroy the CERN
and its doomsday machine, LHC. Think yourself for your salvation. And not just think,
but act. I have no flight experience, no money for the fighter and for a nuclear
warhead to bomb the LHC. If you will not stop the CERN now, the CERN will kill all
of you in the first quarter of this year, with the probability of about 50%.
...
Proton-proton collision, which leads to creation of stable excitation of physical
vacuum, can look like this inelastic collision.
Directions of both protons were changed...
...
... In the ordinary vacuum different virtual particles constantly appear and
disappear. In the bottom figure we can see the appearance of electro-positron pair
in the point A and their annihilation in the point B.
...
In order to transform virtual pair into real pair we must spent energy. On the
next figure we can see the x-boson, consisting from two fermions, particle and antiparticle.
If the energy of p-p collision at collider is sufficient we can create an exited
region in vacuum containing some quantity of such x-bosons. They are oriented in
the same direction and have the superconductive and ferromagnetic properties.
If proton enters into the field of such exited vacuum it can be ruined onto x-boson
and positron...
Any massive object: the Sun, the Earth, a star, a planet, a nuclei of galaxies
and quasars are empty inside. This emptiness is filled with rarefied gas
and radiation at high temperatures. There are two causes leading to formation
of the emptiness. The first cause is the pressure of photon gas. The second
cause is the Arc Effect discovered in 2000. On the bottom figures you can
see the dependence of some physical values from the distance to the center
of the objects. We'll investigate the objects of different masses. The graphs
were received with the help of a computer program, which integrates the
system of differential equations. I made it in the year 2003 and developed
this year (2007). On the right side I'll place the graphs, received according
to classical equations, on the left side I'll place the graphs received
according to my equations, accounting the Arc Effect.
The Sun-like object, M = 1M_{Sun};
R = 1R_{Sun}; m_{particle }= 1.5 a.u.m.
Our model
Standard model
T_{center} = 7.66 millions K.
Total pressure p_{max} =8.4*10^{8}atm at
r=0.56R
Density r_{max}=2.5 t/m^{3}
at r=0.632R
T_{max}=18 millions K.
p_{max} =7.96*10^{9}atm at center.
r_{max}=7.8t/m^{3}
Green - concentration of particles; red - gas pressure; yellow
- radiation pressure; orange - total pressure; white - temperature;
blue - acceleration g. Graphs were normalized to the scale of window
and surface gravity acceleration in both cases are equal.
Object with M = 10M_{Sun};
R = 50R_{Sun}; m_{particle }= 1.5 a.u.m.
Our model
Standard model ruined
T_{center} = 1.2 millions K.
p_{max} =1.2*10^{4}atm at r=0,6R.
r_{max}=0,19 kg/m^{3}
at r=0,717R
T_{max}=1.77 millions K.
p_{max} =31519atm at center.
r_{max}=0,2kg/m^{3}
at r=0,617R
As we can see the radiation had naked the center in both cases.
The pressure and particles concentration is not sufficient for nuclear
fusion in such stars. Mainstream astrophysicists slander us. Or
try to hide the radiation pressure under their carpets.
Object with M = 70M_{Sun};
R = 500R_{Sun}; m_{particle }= 1.5 a.u.m.
Our model
Standard model become as ours
T_{center} = 363 thousands K.
p_{max} =45atm at r=0.74R.
r_{max}=1.86 g/m^3 at
r=0.88R.
T_{center} = 391 thousands K.
p_{max} = 58atm at center.
r_{max}=1.81 g/m^{3}
at r=0,87R.
Pay attention at the values. Nuclear reactions are impossible
for giant stars.
Let's diminish the radius, for example at ten times (object with
M = 70M_{Sun}; R = 50R_{Sun}; m_{particle}=1.5
a.u.m.). We'll have almost the same graphs but the temperature will
rise ten times, common pressure will rise in 10^{4} times,
because it now dependent mostly of radiation. And the density will
rise in 10^{3} times. That is also insufficient. Energy
sources of standard astronomy disappeared.
But we have no need to take that to heart. We know the other
sources of star's energy. And we need to know the real non-falsified
internal structure of stars. Let’s now look at the objects, smaller
then our Sun.
White dwarf, Object with M = 1M_{Sun}; R = 0.01R_{Sun};
m_{particle }= 2a.u.m. Density of condensation of gas into
liquid (degenerated gas) = 2*10^{9} kg/m^{3}.
T_{center} = 920 millions K.
p_{max} = 6.9*10^{16}atm r=0,581R.
The liquid state with r_{max}=2*10^{9}
kg/m^{3
}stretches from 0.83R to 0.52R.
T_{center} = 1249 millions K.
p_{max} =2.1*10^{17}atm at center.
The liquid state with r_{max}=2*10^{9}
kg/m^{3}
stretches from r=0,81R to center.
Pay attention on the temperatures! But this is not so extraordinary
if we mention that surface temperatures of some white dwarfs are 10000
K, 60000 K, and higher. (The difference in heights of green graphs is
the result of program normalization. In fact we have launched the same
density of condensation of gas into liquid.)
Jupiter-like object: M=0.00095M_{Sun};
R = 0.1025R_{Sun};
m_{particle }= 4a.u.m. - was taken arbitrarily.
Let's take arbitrarily the density of condensation of gas into liquid
equal to 2000 kg/m^{3 }for our equations, and 6000 kg/m^{3}
for standard model.
T_{center} = 183 thousands K.
p_{max} = 6.4*10^{6}atm at r=0.57R.
The liquid state with r_{max}=2000
kg/m^{3
}stretches from 0.76R to 0.54R.
T_{center} = 444 thousands K.
p_{max} = 6.4*10^{7}atm at center.
The liquid state with r_{max}=6000
kg/m^{3
}stretches from 0.25R to center.
The Earth-like object: M = M_{Earth}; R = R_{Earth};
m_{particle }= 20a.u.m. - was taken arbitrarily.
T_{center} = 25 thousands K.
p_{max} = 6.6*10^{5}atm at r=0.62R.
The liquid/solid state with r_{max}=6000
kg/m^{3
}stretches from 0.97R to 0.47R.
T_{center} = 33 thousands K.
p_{max} = 1.9*10^{6}atm at center.
The liquid/solid state with r_{max}=6000
kg/m^{3
}stretches from 0.95R to center.
The difference in heights of green graphs is the result of program
normalization. In fact we have launched the same density of condensation
of gas into liquid/solid, r=6000 kg/m^{3}.
The received values of temperature surpass the values, proposed
in literature, somewhere about 5 times. The central temperature
can be decreased by decreasing of the coefficient of temperature
gradient. In all upper cases we have used the coefficient 2/5, which
is reasonable for gaseous medium, but is controversial for liquid/solid
medium. Let's diminish it to 1/10 and try again.
T_{center} = 7.9 thousands K.
p_{max} = 7*10^{5}atm at r=0.59R.
The liquid/solid state with r_{max}=6000
kg/m^{3
}stretches from 0.97R to 0.17R.
T_{center} = 8.2 thousands K.
p_{max} = 1.9*10^{6}atm at center.
The liquid/solid state with r_{max}=6000
kg/m^{3
}stretches from 0.97R to center.
Let's analyze the graphs in our model. The drop of pressure begins
approximately at r=0.6R, but the temperature is already quite high there.
That means that there must exist the point of transition of substance
from solid state into liquid. It is known from seismology that the border
between the core and mantle is situated at r=0.54R. That border strongly
reflects P and S waves and refracts the P-waves. Our graphs of pressure
and temperature allow us to divide the horizontal graph of density on
two regions: solid and liquid. But standard model graphs do not obey
us to do that, - the continuing growth of pressure rises the temperature
of melting. At last, lets analyze the point of transition from liquid
state into gas. It is the boundary between the external and internal
core. Standard geology says that external core is liquid and internal
core is solid. We say: the external core is liquid; the internal core
is gaseous. We can see there the crack at the density graph. But it
is known from the thermodynamics that at high temperature there is no
distinct boundary between liquid and gas. There must be a smooth transition.
Consequently, I'll paint by hand the region of liquid phase in dark-blue
color. I'll smooth the crack and paint the gas phase into yellow color.
The external core (dark-blue) is liquid and keeps the S-waves
out; the internal core (yellow) is gaseous and keeps both waves
out.
Used equations.
If we do not account
the light's pressure and mass the following systems is valid.
With Arc Effect.
Standard-like equations.
dp/dr = - gr
+ 2p/r.
dT/dr = - (2/5)mg/k.
dM/dr = 4pr^{2}r.
g = GM/r^{2}.
p=nkT.
r=nm.
dp/dr = - gr.
dT/dr = - (2/5)mg/k.
dM/dr = 4pr^{2}r.
g = GM/r^{2}.
p=nkT.
r=nm.
If we account the light's pressure
and mass the following systems is valid.
If we take into account the density of condensation of gas into liquid/solid,
the p=nkT or n=p/(kT) is changed into:
n = p/(kT)
if p/(kT) smaller than n_{max}.
n = n_{max}.
if p/(kT) biger than n_{max}.
The temperature gradient
equation.
In the case of Earth we tried two variants for temperature gradient:
dT/dr = - (2/5)mg/k and dT/dr = - (1/10)mg/k.
In fact this equation is approximate, - it does not contain the radiation
part. The computer modeling had showed me that inclusion of radiative
part into the temperature gradient equation almost does not change the
graphs. From my point of view the coefficient 2/5 is
valid for the ideal gas object in thermodynamical equilibrium. I would
name the equation dT/dr = - (2/5)mg/k as the equation
of gravitational temperature gradient. This gradient is intrinsic
and does not lead to heat transfer from hotter region to cooler region.
The heat transfer will begin in the case dT/dr = - ((2/5)+a)mg/k,
where a is a coefficient of temperature gradient, responsible
for heat transfer, and which value is determined by the thermal conductivity
of the gas. I would name the last equation as the equation of total
temperature gradient. The computer modeling had showed me that
decreasing of the coefficient 2/5 leads to inclination
of graphs to the center of the object, but the increasing of the coefficient
leads to inclination of graphs out from the center. If the coefficient
becomes very small or equal to zero, the system can not be integrated.
But when we launch the condensation with some arbitrary n_{max},
the system becomes integrable again. The cause is clear, - the condensation
closes the exponential precipice, appearing at the center of the object
in the isothermal case.
The pressure gradient equation.
I saw two types of equation of hydrostatic equilibrium.
dp/dr = - gr, for ideal gas object. d(p+p')/dr = - g(r+r'), for ideal
gas object filled with radiation.
From my point of view these equations lead to errors because they
do not account the Arc Effect. These equations are correct for the object
filled with non-colliding particles, for example, for star congestions
filled be stars, for photon gas enclosed in some vessel, for very rarefied
gaseous object with particles having the length of free pass, greater
than the object's radius.
If we account the collisions between particles, we'll have the following
equations:
dp/dr = - gr + 2p/r, for ideal gas object. dp/dr = - gr + 2(p-p_{0})/r,
for ideal gas object, enclosed by the shell with pressure
p_{0}. d(p+p')/dr = - g(r+r') + 2p/r, for
ideal gas object filled with radiation. d(p+p')/dr = - g(r+r') + 2(p-p_{0})/r,
for ideal gas object filled with radiation, enclosed by the shell with
pressure p_{0}.
Pay attention that the last item in these equations does not contain
the radiative part of pressure, p'. The cause is clear,
- the light does not obey the Arc Effect, because the light beam do
not change the direction of another light beam at their cross-point;
the photons can exert pressure upon matter screen, but photons do not
exert pressure upon other photons.
The computer modeling had showed me, that the usage of the symmetrical
equation, containing p' in the last item, leads to
appearing of a hole in the center of the object. The more the temperature
of the object is the more the hole.
You can try my computer program Arc Effect - 2007. It is uploaded
at "narod/ru" and at
"webcenter", exe-files,
70kb.
March, 2007.
Text from May 2000.
Notes from March 2007 I'll paint in yellow.
Any massive object: the Sun, the Earth, a star, a planet, a nuclei
of galaxies and quasars are empty inside. This emptiness is filled with
rarefied gas and radiation at high temperatures. In average,
at constant radius, the more massive object
is, the more temperature is inside of its depths...(cut).
On the other hand, the further a star on a scale of evolution, is the
thinner its shell. The rupture of a star shell leads to the explosion
of nova or supernova. (...) To understand
these paradoxical conclusions let's try to analyze the dependence of
a pressure from the distance to the center of an object.
Let’s assume at first, that the massive object is a giant gas ball.
(...) Let’s break a star into a series
of concentric layers. The gravity force pulls each layer to the center
of the object; and its weight dP is transferred to a bottom layer. Having
divided the weight by the area of sphere between these spherical layers
we shall receive an increment of a pressure, which is added to the whole
pressure, transmitted from the top layer to bottom: dp = dP / S. In
order that the object be in balance it is necessary, that the pressure
of gas and radiation in the bottom layer be greater than the pressure
of gas and radiation in the top layer by the quantity: dp = p_{i+1}
- p_{i}. The expression dp = dP / S is the equation of hydrostatic
equilibrium. It can be rewritten in the form:
dp = dm * g / (4pr^{2})
= -r * g * dr,
or shortly: dp/dr = -rg.
(1)
where: r is a density on a distance r
from the center of an object; g is an acceleration of free fall for
given r; dm is a mass of a layer.
dm = - r * S * dr, (2)
(When we move from outside to the center, dr is negative,
consequently, dm and dp are positive.)
Alas, the equation (1) is only fair for a homogeneous
gravitational field, and for the object with non-colliding
particles. It is not applicable to a spherically symmetric gravitating
object. This mistake leads to a grandiose growth of pressure and density
inside of planets and stars. So as the equation of hydrostatic equilibrium
for spherically symmetric objects does not exist, or I did not met it
earlier, it is necessary to deduce it here, proceeding from elementary
principles. (That was my first proof of Ark Effect,
logical proof. Several years later I have found another proves:
analytical, mathematical, historical, computer modeling).
So, on page the description with the help
of figures 2-6 it was shown, that the formula (1) should not include
the acceleration of a free fall g, but it must include
the value v_{0}^{2}/r
- v_{t}^{2}/r.
Thermal velocity v_{t}
coincides with the most probable velocity v_{pr}.
If we use the average squared velocity v, than:
v_{0}^{2}/r
- v_{t}^{2}/r
= v_{0}^{2}/r
- (2/3)v^{2}/r.
Let's name this value the absolute (or internal) acceleration of free
fall, g_{a}.
where: v_{0} – orbital velocity of a particle for a given
r; v_{t} – average thermal velocity of a particle; g_{0}
- Newtonian acceleration of free fall, or external acceleration of free
fall, acceleration of free fall in vacuum.
For Newtonian acceleration of free fall we can write:
g_{0} = g_{0}M/r^{2},
(4)
where: g_{0} - gravitational constant,
renamed here as external gravitational constant, or gravitational constant
in vacuum. The similar expression can be written down for absolute (internal)
acceleration of free fall, acceleration of free
fall in medium, without friction:
Outside of the object, or far from the object,
or in the vacuum cavity inside the object,
the probe body will be drawn to the object with the force:
F = g_{0}Mm/r^{2},
(8)
The element of the object itself with the same
regional density, which has the same temperature as the object,
will be drawn to a part of the object, situated under a probe body,
with the force:
F = g_{a}Mm/r^{2}
= gg_{0}Mm/r^{2}, (9)
Let's compare expressions (8) and (9) to the Coulomb law for vacuum
(10) and for medium (11):
F = (1/(4pe_{0})) *
Qq/r^{2}, (10)
F = (1/(4pee_{0}))
* Qq/r^{2} = (1/(4pe_{a})) *
Qq/r^{2}. (11)
As we can see, the role of permeability of vacuum, or the role of
the electrical constant e_{0} plays
the factor g_{0}; the role of relative
permeability e plays the factor
g, the role absolute permeability of medium
e_{a} plays g_{a}.
To these strange analogies it is possible to add the following ones
from electrodynamics:
em = (c/v)^{2}; or
if m = 1, then: e =
(c/v)^{2}, (12)
And its analogue, the equation (7):
g = g_{a} /
g_{0} = 1 - (v_{t}/v_{0})^{2}.
Consequently, g_{a} is a gravitational
constant, applicable to a body, situated in a spherical layer of object,
where the molecules of the layer have the average thermal velocity v_{t},
and with the same regional density. The
average squared velocity can be defined from the equation: mv^{2}/2
= 3/2*kT where: m
- weight of a molecule of air; T - the temperature
of gas (...). The value v_{t}
can be find from equation v_{t}^{2}
= (2/3)v^{2}.
And g_{0} is the gravitational
constant in vacuum, which is applicable to a probe body far outside
of attracting object.
At last, g is a relative gravitational
permeability of a medium.
The equation
g = g_{a}
/ g_{0}
= 1 - (v_{t}/v_{0})^{2}
for the object with density, which is not equal to the regional density
of the object, will have the coefficient
r_{object}/r_{probe}.
But main profit of this approach is in the demonstration of symmetries
between the Newton and Coulomb laws. In fact such record of the formulae
unites the attraction force in vacuum and repulsion force of Archimedes.
Stop correcting.
I don't correct the further 2000-text. I did
not solve the riddle of my excellent formula G' = 1/exp(a
+ 1/a)
yet.
Let's now return to a fine structure constant
a, which is the running constant of the electromagnetic interactions,
and to the early received gravitational constant, which in the normalized
units is dimensionless, denoted there by G'; and it is the gravity fine
structure constant:
G' = 1/exp(a + 1/a),
(13)
The transition to SI units can be carried out under the formula:
G = (G'e^{3}/(4sqr(a)e_{0}^{3/2}m_{el}m_{pr}^{2}))^{2/3}.
(13 a)
From this formula G is equal to 6.671479888E-11 N*m^{2}/kg^{2},
that is in the perfect consent with the experimental value of G. According
to CODATA-1999 the experimental value of G lays between 6.663E-11 and
6.683E-11 in SI units. According to CODATA before 1999 the experimental
value of G was: (6.67259E-11 + /- 0.00085E-11) N*m^{2}/kg^{2}.
There is a question, why the accuracy of G has fallen? There is the
second question, why for normalized units the formula 1/G ' = exp(a
+ 1/a) works, instead of, for example,
the easier one: 1/G ' = exp(1/a)?
The answer consists in the following:
The quantity a can be regarded as the velocity
of an electron in an atom of hydrogen, measured in the shares of the
velocity of light.
The quantity (1 - a^{2}) – is the
relativistic amendment, which can be represented as (1-(v/c)^{2}).
The ratio of quantities exp(1/a) and exp(a
+ 1/a) with good accuracy is equal to (1-
a^{2}) or to (1-(v/c)^{2}).
Electron can be represented as a cloud, as an atmosphere of an atom,
for which the gravigeometrical paradox works too.
That is why, the quantity G in old denotation, or
g_{?} in new denotation is also the
running constant. The expressions exp(1/a)
and exp(a+1/a)
just also must give 1/g_{a}' and
1/g_{0}'. But what correspondences
between them are? Preliminary and, while what not the final analysis,
gives:
It is obvious, that g_{0} should
be greater than g_{a}.
Hence, g_{0}' should be greater,
than g_{a}'.
Hence, 1/g_{0}'should be less,
than 1/g_{a}'.
(1/a) is less, than (a+1/a),
and they both are more than 1.
Consequently: exp(1/a) is less, than exp(a+1/a).
At last we conclude, that:
1/g_{0}' = exp(1/a),
or g_{0}' = 1/exp(1/a),
(14)
1/g_{a}' = exp(a+1/a),
or g_{a}' = 1/exp(a+1/a),
(15)
In the expression (18) the relative gravitational permeability of
a medium can be received through a constant a with good precision. On
the other hand, we can check up this result applying the formula (7):
g=1-(v_{t}/v_{0})^{2}.
In order to do this we must know the value of the orbital velocity:
v_{0}=Sqr(g_{0}M/r) and the
value of average thermal velocity of molecules of the air near the surface
of the Earth v_{t} = Sqr(3kT/(29*a.u.m.)), where a.u.m.
– atomic unit of mass, 29 - molar weight of the air; and let T be equal
to 300 K. Then:
v_{0}=7910 m/s; v_{t} = 508 m/s; G = 0.995876
This result does not absolutely coincide with the result of the formula
(18) for several reasons:
1. The temperature 300 K is not very much approached to the real
value of temperature. The check has shown, that for planets and stars
we can receive rather good results.
2. The reason of discrepancy is contained also in that, that the
sole value G, used in physics and astronomy now, actually in one cases
"works" as g_{a}, and in the others
- as g_{0}. So, between planets and
stars the value g_{0} works, but
in the ground laboratories at pumped out air the value
g_{0} works, and at presence of the
air the value g_{a} works. Last two
items are correct, if we measure the force between the Earth and the
probe body. But actually, the mass of the Earth is unknown, and we measure
G between two probe bodies, between which the additive item
g_{0} works. All this confuses a
situation at all.
Despite of these difficulties, it is necessary to move forward in
the study of gravitation. This will unite the gravitation with electromagnetism.
It, at the end, will simplify the understanding of gravitation. The
experiment with the Gravity-Thermal Coil is extremely necessary. This
experiment will show us the structure of the depths of planets and stars.
This experiment will allow us to solve many energetic, ecological and
technical problems.