ÄËß ÐÅØÅÍÈß ÏÐÎÁËÅÌÛ ÊÐÀÒÊÎÑÐÎ×ÍÎÃÎ ÏÐÎÃÍÎÇÀ ÇÅÌËÅÒÐßÑÅÍÈÉ
È ÓÌÅÍÜØÅÍÈß ÑÅÉÑÌÈ×ÅÑÊÎÃÎ ÐÈÑÊÀ ÄËß ÍÀÑÅËÅÍÈß ÑÒÐÀÍ ÍÅÎÁÕÎÄÈÌÎ
ÐÅØÈÒÜ ÍÈÆÅÑËÅÄÓÞÙÈÅ ÂÎÏÐÎÑÛ.
1. Earthquakes and other natural disasters short-term prediction.
2. Appropriateness of connection of start point resonance interaction of interplanetary magnetic and gravitational fields having the
geomagnetic field with the earthquake start point.
3. Dependence of tectonic earthquake magnitude from sizes and types of seismogenous geological faults.
4. Theoretical calculation for earthquake forcasting and other natural disastersfor whole territory of the world on unlimitted time forward.
5. Electromagnetic model of earthquake centre beginning and developing.
6. Geocurrent mechanism formation and electrical field in the ground.
7. Interaction of dipolar planetary fields and its influence on earthquake cetnre developing.
8. Connection of strong earthquakes with flood deformation of the earth crust.
9. Connection appropriates of emitting and observed seismic waves, which are begins in centre during the earthquake preparing.
10. Buildings and constructions seismic stability increasing.
NATURAL DISASTERS’ ACCURATE PREDICTION
CHAPTER 1. Earth-current’s and Earth’s electric field’s formation mechanism.
1.1. The electric conductivity.
According to the classic theory Earth’s inner layers are placed under high pressure and temperature, under these circumstances the metals are in
ionization state.
In our model we shall consider the Earth as a volume, in which ions and electrons are spread with different densities, ions stay in a stationary
state, while electrons wander freely in the volume of the Earth.
The model, we have chosen, looks like P. Drude’s model. In this, the electrons of a solid body are considered as an ideal gas, to which the kinetic
theory of gasses applicable is. In this theory molecules and atoms of a solid body are considered as elastic balls and there is no account taken of
the interaction between electrons and ions in the time between collisions. In the science this approach is known under the name of “free,
independent electrons’ approach”.
The electron collisions occur quickly with their reflection from ions as from stiff balls, there is no account taken of the interaction between
electrons. The probability distribution in a time unit ~1/ò, where ò is the relaxation time or free path travel time.
Based on these suppositions of P. Drude we can define now the substance’s electric conductivity σ.
Electrons move at the velocity  . The equivalent
current  ,
where e is the electron’s charge and V is the substance’s volume.
(1)
where N is the quantity of electrons,  , v is the drift
speed,  .
Let’s suppose that an outside force  is influencing the electron.
Then, based on the probability theory and Newton’s laws, we can describe
the motion of electrons:
 , (2)
from which follows that . (3)
Let’s suppose that  , so, for ò < t < T we have
 . (4)
where  m is the mass of the electron. At t < T,
 where 
For metals at the room temperature we have ò ~ 10-14 sec., n ~ 10-22 cm-3,
so  .
Here  where z is the valence, A is the molecular mass, π is
the density and NA is the Avogadro constant. ò = T-1/2, T is the
temperature, n ~ T-1/2, that is, nò = const, so σ is little changed at the rise of temperature.
1.2. The dynamic conductivity
Let’s suppose that the substance is inside of an alternating electric field. Let’s examine a harmonically alterating field:
(5)
 from where we have:
(6)
where  is the dynamic conductivity.
Further we have
(7)
 , from where
 , (8)
where tg δ = ωò,  . At ωò >> 1 we have
 , from where
 , (9)
At n = 1022 cm-3, e ≈ 10-10 CGS units and m ≈ 10-27 g we get
Let’s suppose that ω = 1015 ω0,
then  so
 , (10)
here ωt >> 1. In the case of ωt << 1, we have |σ(ω)| ≈ σ, then
(11)
The alternating in the time current and electric field cause a magnetic field that can be calculated from the equations of Maxwell:
(12)
Then  , where r is the distance of a conductor with current
and  is the magnetic field strength, which is produced by the
current and the field:
(13)
(14)
from where
(15)
At σ ≈ 1015 CGS, c = 1010 cm/sec and ω ≈ 1015 ω0 we have:
(16)
In the presence of only the current we have: 
then  from where
(17)
The natural electric field of the Earth:
|E| ≈ 10-3 CGS (18)
and
|B| ≈ 102/r CGS. (19)
1.3. The geo-electric field
In a non-uniformly heated substance arises an electric field, directed against the gradient of temperature – the effect of Zeebeck:
(20)
There is no current in the state of equilibrium.
fig. 1. the effect of Zeebeck.
The average velocity of electrons in the point x (fig. 1) is defined as
 , where 
from where  ,  ,
 ,  ,
cv is the specific heat capacity, KB is the Boltzmann constant, the field of
the Earth ≈ 10-3 CGS.
1.4. The geo-current
According to the kinetic equation of Lorentz for electron gas in a solid body we have:
(21)
in which  is the function of the electricity distribution, N0 is the
density of distribution of dispersers, R is the radius of dispersers, θ is the distribution angle, measured with respect to the normal to
the surface of disperser, v and v´ are respectively the velocities of the electron before and after the dispersion.
Let’s suppose, that the electron gas is at a constant field E along the x-axis and there is a gradient in the same direction.
For slowly alternating and weak fields,  VT is the velocity of the thermal
motion,  , L is the free path and
 is any dynamic value. Let’s find the solution of the equation of Lorentz in the form
 , in which  is a correction
of f0(v), where f0(v) is the distribution of Maxwell:
 in which  . After the transformation the equation
of Lorentz results in:  , where  ,
 . From here  . For the given current we have
 . Apparently  = 0, then we get
 and, converting to spherical coordinates,
we shall have:
 , (22)
 where  From here
(23)
For the density of the thermal current we have: 
from where
(24)
At T = const we get jx = σE, where  is the thermal conductivity.
Then  , where  .
In the case of E = 0, according to the effect of Zeebeck, the thermal field will be 
, from where jx = 0, so
(25)
Let’s observe the process of the heat distribution in the direction x (fig. 2):
fig. 2. the heat distribution in the direction x.
[qx(x) – qx(x+dx)] dt ds = ρc ds dT dx, (26)
in which qx(x) is the value of the heat density in the point x, ds is a surface element, placed perpendicularly to the direction of the heat
propagation, ρ is the density of the substance, c is the thermal capacity.
From (26) we have:  from where  then
 .
From (20) we get:  , then  .
From the equation of Maxwell , where η is the charge density, we get  , from where
(27)
From (25) we have  , then  .
Let’s suppose L ≈ 10-8 cm, ρ ≈ 1 g/cm3, then
(28)
where η is the spatial charge density in the place, where occurs the alteration of the temperature in the time, and
for the ground. In a volume V = 10km·100km·100km = 1020 cm3 we shall have η = -1016 CGS units. Let’s notice
that the charge of the Earth ≈ -1015 CGS units.
Let’s observe the equation of the charge collisions:  , then
 .
The heat, that the Earth receives, causes alternation of T, so  , that means, that the thermal electric
field E of Zeebeck is changed, that causes the spatial charge η, if  .
In the case of  the temperature alters linearly, that agrees with a homogenous substance, that is
heated on its borders by a constant source. Since everywhere, besides at the borders,  ,
then the charge arises at the borders of the
substance, that is, at the break of the gradient. In this case the surface charge density  .
Supposing the surface of the Earth S ≈ 1018 cm2, then  . Let’s suppose
 , then η = -1011 CGS units.
As is well known, there is an atmospheric current  ,
which transfers the electric charges from the Earth surface to the atmosphere. The charge of the Earth is restored by lightnings and magnetic fields.
The total Earth-current reaches its maximum at 7PM GMT and its minimum at 4AM GMT. The value of the Earth-current ≈ 1800A ≈ 1012
CGS units.
By the same reason as in the atmosphere there are caused lightnings inside the Earth, which are responsible for natural disasters.
Since the Earth revolves in its orbit around the Sun, there is  caused, that causes charges in the upper layers of the Earth.
In this way is compensating the Sun the loss of charge by the Earth, and, there is the atmosphere current caused as a result of Sun activity.
Since the heat follows the law (25), then we shall have:  with the creation of an enlargement’s field,
that is, as a result of the heat motion inside the Earth, the heat-wave causes quick enlargement of bodies, and if, e.g., the magma is enlarging more
quickly than the crust, then there is an eruption of volcano caused.
So, the heating by the Sun causes an electric field and charge. The dipole moment will have the following value: d ? Q 2RA ? 1015 ?109 ? 1024 CGS
units. This dipole moment rotates with the period T = 24 hours. So  , R is the distance from the
dipole, R >> RA. If R is an interplanetary distance, then
(29)
is the field of the Earth. On planets at R ≈ 1015 cm we shall have |E| ≈ 10-22 CGS units, and the field of the Earth
|E| ≈ 10-3 CGS units. (30)
1.5. The heat waves.
As is well known, the temperature on the Earth’s surface has daily and yearly periodicity. Let’s resolve the problem of
the distribution of periodic heat oscillations in the ground (0 < x < ∞).
The bounded solution of the equation of heat conductivity  at 0 < x < ∞) and t > -∞.
Always supposing u(0,t) = u × Acosωt we have  , in which
 , 
is the phase shift of the heat oscillations (the Fourier’s law).
From here, comparing u(x,t) = A(x)cos[-ω(t-δ)] with the equation (28), we get  ,
that is  ,
from where  , where  . Then
 , from where we have  ,
 .
From here follows that t = δ + 104 n = 103 x + 104 n. At x = 10cm t ≈ 104 n.
There follow the maximums of the charge density at n = 1, 2, 3… during this time.
 . From equation (20) we have
 . (31)
A(x) ≈ 300K, then E ≈ 10-3 CGS units, that agrees well with the Earth’s field,
a ≈ 4×10-3 cm2/sec, the higher the frequency, the stronger
the field.
In a geocentric coordinate system the planets revolve with different frequencies. The Sun and last planets have nearly the same frequencies,
so their influence is practically equal. For example, for Mars (30)  CGS units,
in which R ≈ 1013 cm, that is (228 – 150) •106 km.
This is the field, caused on Mars by the dipole moment of the Earth.
The Sun causes on a planet the charge density:
 . (32)
It is evident, that, if  , then on the surface of the planet
 and, from there 
then  . From here we have:  .
At T = 100T0, R = 109 R0 we have:  ,
in which a2 = 10-3 a02, ω = 10-4 ω0,
 .
Let’s suppose that r = R = 1013 cm, that is 78•106 km. Then |E| ≈ |H| ≈ K•10-17.
A corresponding current is caused by the field:
J = 10-15 E ≈ 10-2 K CGS units. The atmospheric current ≈ 10-7 CGS units. The Sun causes the current
 .
The current falls to zero when the field comes into equilibrium.
The field caused by the Sun: |E| ≈ 10-3 CGS units. It agrees with the reality.
1.6. The formation mechanism of the geo-current and geo-electric field.
It is known, that the conductivity of conductors is defined by the equation  ,
in which m is the electron’s mass, n is the concentration of the electrons, e is the electron’s charge and ò is the electron’s free path time.
We have, that n ≈ 1013 cm-3, ò ≈ 10-14 sec, e = 4.8•10-10 CGS units,
m = 10-27 g, so  .
The inside layers of the Earth consist of a conducting substance, moreover, starting from the depth more than 100 km the substance of the Earth
is in ionization state and shows conducting properties.
At the depth of about 100 km we have:  CGS units.
Let’s suppose that the conducting substance is inside of the electric field. Then the inside current in this conductor (before it comes
into equilibrium, if it is a finite conductor) is defined by Ohm’s differential equation, in which we shall suppose E = const, that is,
we suppose that the conductor is infinite.
 . (33)
For the “real” conductors σ ≈ 1015 CGS units, and for the conductors on the Earth’s surface
σ ≈ 105 - 106 CGS units.
Then we shall have:  CGS units for the conducting substance and
 CGS units for the outside layers of the Earth’s surface.
As is well known, in a non-uniformly heated substance an electric field is formed, directed against the gradient of the temperature
(the Zeebeck effect).
In the case of the gradient of the temperature is directed along the x-axis only, we shall have
 , (34)
where KB is the Boltzmann constant, T is the temperature dependent on x. For metals at the room temperature
 . (35)
Since  does not depend on the properties of the substance, there is only necessary that the substance
has free electrons.
For the Earth let’s suppose Q equal to its value for metals. Such a supposition is acceptable, because the upper layer of the Earth is negatively
charged, that is provided by the difference of the conductivity depending on the depth and the presence of the charging geo-current.
From the electron theory of Lorentz for the heat current density in the x direction at a stable temperature gradient (25) can be obtained:
 , where L is the free path distance of the electron, k is the heat conduction factor.
Researching the heat propagation process in the x direction, we get  , where c is the specific thermal
capacity, ρ is the substance density.
Using the equation of Maxwell and supposing the dielectric constant ≈ 1, from the equation system we get:
 , η is the spatial charge density,
 , (36)
 .  , substituting the expression for k we get
 . Supposing L = 10-8 cm we get ρ = 1g/cm3.
η is the spatial charge density in the place where there is a temperature alteration in the time
 .
For the ground we have  , then  , from where
 .
Supposing that 1 day = 24 hours, |ΔT| ≈ 10 °K, we get  .
At T = 300 °K we get  , so ηground ≈ 10-8 CGS units.
For the volume of a 10 km deep earth layer V = 10 km • 10000 km • 10000 km = 1024 cm3 we get
|ηground| ≈ 1016 CGS units everywhere on the Earth.
Let’s notice that the charge of the Earth ≈ 1015 CGS units.
The heat, transferred by the Sun to the Earth, causes an alteration of the temperature with a period 24 hours (365 days). The resulted thermal
electric field causes a spatial charge η if  .
When   ,
that is, the temperature is linearly altered, that agrees with the heat distribution in a homogenous substance.
Then  everywhere besides on the boundaries, where the second derivative has a break. In this case
 has sense for the surface charge density.
For S ≈ 1018 cm2 (the Earth’s surface) we have:  ,
then |ηground| ≈ 108 CGS units, that is seven powers less than the Earth’s
charge, from where follows that the inside layers are heterogeneous and  .
In connection with the daily revolution of the Earth there are arisen a geo-electric field and a charge, which is leaving continuously the
Earth and returning back to the Earth. The above Earth lightnings are considered as the supporting mechanism of the Earth’s charge, but beside it
there is the charge support mechanism by the Sun heating.
The dependence of the temperature on the time and depth is given by the equations of mathematical physics:
 , (37)
where ω is the frequency of the temperature oscillations on the surface,  .
 is the temperature dependent on the depth.
 is the phase shift of thermal oscillations. At the depth 10 m (103 cm)
the delay is δ ≈ 106 sec ≈ 102 hours ≈ 4 days.
Coming from the above defined thermo-electric field’s expression and from the solution for heat oscillations we have:
 . (38)
The higher the frequency of heat oscillations the stronger the field.
As it was shown above, even at the depth 10 m the delay of heat oscillations is ~ 4 days, so the thermo-electric field is continuously supported
and there is no possibility to eliminate the temperature gradient by the heat exchange.
For the Earth we have: ω ≈ 10-4 sec-1, a2 ≈ 4•10-3 cm2/sec,
T0 ≈ 300 K, that is E ≈ 10-3 CGS units, that agrees with the electric field in the
proximity of the Earth.
From here we can suppose, that the supporting mechanism of the electric field of the Earth is the heating of the Earth by the Sun.
Taking into account the temporal changes of the field E we have:
 .
For the Earth we have:
 . (39)
These are the daily oscillations of the Earth’s field’s strength taking no account of the atmosphere current and the Earth’s residual charge.
It is known that the Earth’s current reaches its maximum at 7PM and its minimum at 4AM GMT.
So, E ≈ 10-3 cos (10-3 t), that is, the field oscillations with the frequency ≈ 10-3 sec-1
and the period ≈ 24 hours agree with the data from the observation, that the current alters with the period ≈ 15 hours.
|E| has a 12 hours period, |j| ≈ |E|. It is known, that jatmosphere ≈ 10-7 CGS units,
and the conductivity of the air on the surface σatmosphere ≈ 10-4 CGS units.
1.7. The planetary dipole fields’ interaction
We have found above, that the electric field, caused by Sun-heating has the expression:
 , where T0 is the average temperature on the surface of
the planet and ω is the rotation frequency of the planet on its own axis.
Let’s have: T0 = 300 K,  , a2 = 4•10-3 a0,
 , then  CGS units. The charge density
 , which is allied with this, will be:
  . From here we have:
 , where φ0 is a constant number. Then
 or
 ,  .
Let’s the surface is lighted by the Sun:  , then for the dipole moment of the planet we get:
 .
An alterating dipole moment causes electric and magnetic fields on distances longer than the planet’s radius:
 , where R is the
distance from a planet to another one: R >> Rop.
For the other planet, in its system, the dipole moment of an interacting with it planet will alter with a frequency, equal to its rotation frequency
on its own axis. If ω1 is the rotation frequency of the other planet, then
 , from where d ≈ 1022d0 CGS units.
Substituting the known values:  ,  ,
c2 ≈ 1021 (cm/sec)2, we get  .
Following the above-obtained result, the electric field of the interacting planet causes a current density on the other planet:
 in inside conducting layers and
 in the surface layer, where  .
Let’s notice, that the atmospheric current density of the Earth is jatmosphere ≈ 10-7 CGS units.
The dipole moment of the first planet, caused by the Sun, acts on the other (second) planet on the surface point of the second planet, from where
is the first one visible, in the way, that causes a current j = 10-5j0 in inside layers, that can lead to an abrupt discharge of the planetary
interlayer capacitor.
It is known, that the earthquakes are accompanied by an increase of Earth currents.  .
Let’s observe some possible planet positions (fig.3):
fig.3
At θ = 90° - β and òò multiplied we have resonance, so, if the angle between the Sun and the planet, visible from the Earth, is a right angle, then,
probably, an earthquake can occur in this place.
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