Get your own Free Homepage SUN ENERGY (archived as priority version)
Originated 10 June 1998, edited version.
First attempt. On the bottom picture in first line is the resting rod, on the second line is the same rod but which was accelerated to the velocity V.
Instead of picture:
m/2................R of string.............................m/2
m/2.............R' of string.....................m/2
Moving rod must be shorter, consequently one of the masses must pass the bigger distance, and have more velocity. But in the end of one cycle (t=2R/c) of acceleration, that mass must conjugate own velocity(V2) by changing photon with another mass which have in the end of one cycle (V1).
V2=V1+(R1-R)/t /1/
Accelerating machine must spend in the end of one cycle such energy: E=m*(V2)^2/4 - m*(V1)^2/4 /2/
And we receive capacity which goes not on the accelerating of body but on the warming of any body which is being accelerated. p=E/t=mV^3/(16*R). /3/
According to the principle of equivalence of masses we must conclude that any object making G is permanently in the state of acceleration. If we take the equation that has been received at the moment and use it for sun we'll receive:
p=G^3*m^4/(2*R^4*c^3). /4/
...Here can appear a question: "Why the rod must be heated, when it is in the state of rest relatively to the surface of gravitating object." The answers: 1."Because, relatively to the frame of reference of universe, the state of rest will be connected to the frame which flies into the object with the second cosmic speed" 2."Because, if we consider the point, moving from the state of rest from the region of universe, where the space-time is asymptotically flat, and taking into account the principle of equivalence of being in the state of accelerated motion or in gravity field, we may conclude that for the rest part of universe more natural state is the state where non accelerated frame of reference will fly into massive object with second cosmic speed and with acceleration equal g." On the other hand any body resting in the surface of massive object will be in the state of acceleration, relatively frame of reference of universe, or relatively probe point moving free in curved space-time in the same vicinity of massive object. Consequently free dropping body is in the natural state in which its inner state is unchangeable. Otherwise surface of object, so as any layer of it, is in the state of acceleration relatively any inertial frame of reference. In calculations the time was presumed to be equal to 2*r/c. It is just true for the case then acceleration and consequently velocity of stick is mach less the velocity of light. In different presumptions the results are quite different. It seems that in the process of conjugation of pairs of particles we see quantum frame changing, with quantum transition of particle pairs on higher level. Energy received by the pair of interacting particles in the process of conjugation of length after period of elementary interaction: e = m*v1^2/4 -m*v^2/4, /5/
where: v1=v+(r0-r)/t.
r = r0*sqrt(1-v^2/c^2) ~ r0*(1-v^2/(2*c^2)). /6/
v1 = v+r0*v^2/(2*c^2*t) /7/
Elementary power p = e/t; /8/
p = m/(4*t)*((v+r0*v^2/(2*c^2*t))^2-v^2) = m/(4*t)*(r0*v^3/(c^2*t)+r0^2*v^4/(4*c^4*t^2)). /9/
If v << c: p = m/(4*t)*r0*v^3/(c^2*t) =r0*m*v^3/(4*c^2*t^2). /10/
If v << c: t = 2*r0/c; p = m*v^3/(16*r0) /11/
v = a*t = a*2*r0/c . /12/
p = m*a^3*8*r0^3/(c^3*16*r0) = m*a^3*r0^2/(2*c^3). /13/
Let a = g =G*M/R^2. /14/
Then p = m*G^3*M^3*r0^2/(2*c^3*R^6). /15/
(Here p considered as dp, m - as dm, M - total mass of object.)
Second attempt. Let's consider object with constant density ro.
M = ro*V. /16/
g = M*G/R^2 = ro*V*G/R^2 =4/3*pi*ro*R*G /17/
The element of mass is in the sphere with thickness dR.
dm = ro*S*dR = ro*4*pi*R^2*dR. /18/
For elementary power we use dm instead of m/2 in /11/.
dp = m*v^3/(16*R) = v^3/(8*R)* ro*4*pi*R^2*dR = v^3*ro*pi*R/2*dR. /19/
To receive total power we take the integral I.
p = Idp = Iv^3*ro*pi*R/2*dR = Ia^3*t^3*ro*pi*R/2*dR = Ia^3*8*R^3*ro*pi*R/2*c^3*dR = 4*pi*ro/c^3*Ia^3*R^4*dR = 4*pi*ro/c^3*I(4/3*pi*ro*R*G) ^3*R^4*dR = 4*pi*ro/c^3*(4/3*pi*ro*G) ^3*IR^7*dR = 4*pi*ro/c^3*(4/3*pi*ro*G) ^3*IR^7*dR 4*pi*ro^4/c^3*(4/3*pi*G) ^3*IR^7*dR = 4*pi*ro^4/c^3*(4/3*pi*G) ^3*R^8/8 = pi*M^4/(c^3*V^4)*(4/3*pi*G) ^3*R^8/2 = 3/8*M^4*G^3/(R^4*c^3) /20/
This result differs from /4/ by coefficient 3/4. So we have more explicit equation fore total power received by object with constant density. p = 3/8*M^4*G^3/(R^4*c^3) /25/
Third attempt. In the next step it is necessary to use variable density. Let's say, density is proportional to depth to the star (R0-R).
ro = ro0*(1-R/R0), /26/
where ro0 is density in the center of a body, R0 is radius of a body.
g = G*M/R^2 = G/R^2*Idm = G/R^2*I4*pi*R^2*ro*dR = 4*pi*G*ro0*R*(1/3-R/(4*R0)). /27/
To find ro0 its necessary to use:
M = Idm = pi*ro0*R0^3/3. /28/
ro0 = 3*M/(pi*R0^3). /29/
g = 12*G*M*R/R0^3*(1/3-R/(4*R0)). /30/
p = Iv^3/(8*R)*dm = 4*pi*12^3*G^3*M^3*ro0/(R0^9*c^3)*I(1/3-R/(4*R0))^3*R^7*(1-R/R0)*dR = 4*pi*12^3*G^3*M^3*ro0/(R0^9*c^3)*R0^8*0.000038 = 0.8*G^3*M^4/(c^3*R0^4). /31/
So we have for total power:
two points rode: p =1/2*M^4*G^3/(c^3*R^4),
spherical body of constant density: p(c) =3/8*M^4*G^3/(c^3*R^4),
spherical body of linear variable density: p(v) = 4/5*M^4*G^3/(c^3*R^4).
For sun we have M = 1.989E+30 kg, G = 6,672E-11, R = 6.96E+8 m. p(v) = 5.88E+29(watt), p(observable) = 3.86E+26(watt). We see that result is much more bigger then observable. But it shows that capacity changes only a little then we examine objects of different density.
Fourth attempt. On the next step we must consider the time of interaction (conjugation). It seems that it is not equal to the time that light took to pass the radius of the object, but depends of temperature of vacuum (CMBR), or some characteristic frequency connected to N - unique quantum number.
Power for spherical object with constant density. Begin from /10/: p = r0*m*v^3/(4*c^2*t^2). Let R = r0, dM = m/2, dp = p, dM = 4*pi*R^2*ro*dR, v = g*t = 4/3*G*pi*R*ro*t. dp = R*4*pi*R^2*ro*(4/3*G*pi*R*ro)^3*t/(2*c^2)*dR = R^6*G^3*4^4*pi^4*ro^4*t/(2*c^2*3^3)*dR. /32/
p = G^3*4^4*pi^4*ro^4*t/(2*c^2*3^3)*I(R^6)dR = 3/14*G^3*M^4*t/(c^2*R^5). /33/
It is impossible to make numerical calculation in /33/, because we don't know the time of elementary conjugation. But we can receive another equation, which also have that time. We can write: p = n*E/t, /34/
where n - quantity of particles in the massive object, E - elementary energy given to particle, which proposed to be equal to energy of quantum of space, t - time of conjugation. Then: n = M/m, E = h*nu, nu = 1/t.
m - average mass of particle in object of mass M. Here is also proposed that energy of quantum of space defines similarly to light's one.
p = M*h/(m*t^2). /35/
t = sqrt(M*h/(m*p)). /36/
Constituting /36/ in /33/, we'll receive: p = k*G^2*M^3*h^(1/3)/(c^(4/3)*R^(10/3)*m^(1/3)), /37/
where k = (3/14)^(2/3) = 0.35809, numerical coefficient in the case of constant density.
Power for spherical object with linear variable density.
ro = ro0*(1-R/R0), g = G*M/R^2, M = I(ro*4*pi*R^2*dR) = ro0*4*pi*R^3*(1/3-R/(4*R0)).
ro0 = 3*M0/(pi*R0^3).
M = 12*G*M0*R^3*(1/3-R/(4*R0)).
g = 12*G*M0/R0^3*R*(1/3-R/(4*R0)).
dp = g^3*t*ro*4*pi*R^3/(2*c^2)*dR.
p = 12^3*G^3*M0^3*t*ro0*2*pi/(R0^9*c^2)*I((1/3-R/(4*R0))^3*(1-R/R0)*R^6*dR) = 12^3*G3*M0^3*t*ro0*2*pi/(R0^2*c^2)*(0.00005637) = 0.5844*G^3*M0^4*t/(R0^5*c^2). /38/
Constituting /36/ in /38/ we'll have: p = k1*G^2*M0^3*h^(1/3)/(c^(4/3)*R0^(10/3)*m^(1/3)), /38/
where k1 = 0.5844^(2/3) = 0.699, numerical coefficient in the case of linear to center density.
Numerical values for sun. M = 2*10^30, R =6,96*10^8, G = 6,67*10^(-11), h = 6,63*10^(-34), m(H) = 1,67*10^(-27), m(He) =4*m(H)
Sun consists of 90% of H and 10% of He, consequently we must take average mass of particle:
m = mH*1.3.
p under linear density: 2,74*10^27. /2.99*10^27 in the case of only H/
p under const. density:1,40*10^27. /1.53*10^27 in the case of only H/
p observable = 3,86*10^26, (+ flow of neutrino + sun wind + accumulation of nuclear power).
Fifth attempt. Frequently appearing in modern physics value N here is the quantity of layers situated on every imaginable plane. These layers constitute 4-dimensional space lattice. In every plane (bundle of lines, frame of reference) there are N planes (lines, frames), moving relatively each other and rotating relatively each other. Linear and angular velocity is defined through N. If we divide velocity of light on N we'll have quanta of velocity: v = c/N /39/
Energy connected with this quanta velocity: E = m*v^2/2. /40/
Frequency: nu = c^2*m/(h*N^2) /41/
This frequency is the smallest possible frequency. From the other side we know that Hubble constant is the smallest possible frequency and can substitute it in order to define N.
N = sqrt(c^2*m/(H*h)). /42/
H =65 +/- 8 km/(sec*Mpc) = 65E6/3.1E22 = 2.1E(-18) (1/sec).
N = sqrt(1.08E41) = 3.29E20
If we multiply number N by H, we also receive frequency:
N*H = 3.29E20*2,1E(-18) = 691(1/sec). /43/
Suppose that period, defined by this frequency, is the time of conjugation of distance between particles in accelerated object.
T = 1/(N*H) = 0.00145(sec) /44/
(I can not explain, but further coincidences make me use sqr(2), in expressions concerning the time. May be the coordinate, degenerated in time coordinate for macroscopic observer, forms the angle pi/4 to N layers in 4-dimensional space.)
t = T*sqr(2) = 0.00205(sec) /45/
Now we constitute /45/ in /38/ and receive capacity of the sun.
p = 0.5844*G^3*M0^4*t/(R0^5*c^2) = 3.79E+26
p(observable) = 3.86E+26
p(strange coincidence) = M*c^2/N/sqr(2) = 3.85E+26
As for first p, I think that in reality we must take not observable radius, but a bit smaller, because density of the sun is decreased asymptotically near its surface. The last p is very strange and is not correct if this formulae was used for planets. But further coincidences in using N = 3.29E+20 gave me much more confidence in correctness of last expression if it is used to steady stars. Other coincidences with usage of this N lead to revision of space, time and mass units.
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