Arc EffectDo you know, than the Sun reminds the empty ball? This paradoxical conclusion I had made in the May 2000. I was shocked by the results of logical deduction and computations. The density and pressure in stars grow only to the certain level. This is the level, where the thermal velocities of molecules become equal to their orbital velocities. If we go dipper of this level the density and pressure must drop. The stars become almost empty at their centers!This logical deduction led to the conclusion, which was a heresy. Such heretical conclusion must be tested by different methods. In a couple of years, in summer of 2002, I had found the second method of this problem solution. I began to discuss this problem in Internet. This discussion had convinced me in my correctness. I am very thankful to all participants of discussion. Especially to Tiber (nickname), who tried to refute the Arc Effect, and who made several interesting deductive steps in order to defend the standard approach. In my bottom proof I will use his steps, to prove the Arc Effect. This proving method can be named the analytical proof. In autumn of 2003 I had received another proof, which can be named the mathematical proof, and which is applicable not only to gas, but to any medium of interacting particles. In 2003 I had made the computer program, which solves the systems of differential equations, which could not be solved by usual way. This program had given me the series of interesting results, first of all,  the expanded theorem of virial. Analyses of these results can be regarded as program proof of Arc Effect. There is one more proof,  historical. I found out that one of the first persons, who had said about the hollow planets and stars was famous Leonard Euler. His name can be found in math and physics books such frequently as the name of Einstein, or even more. Consequently, the rating of wisdom of Euler is comparable to rating of Einstein. Great physicists did not accept the conclusion of Euler about the hollow planes and stars. This paradox is more paradoxical than Special Relativity of Einstein. That is why those great physicists are weaklings and got to the camp of misguided. But Euler was correct. And the new equation of Hydrostatic equilibrium is correct, because it meets the conclusions of Euler. Hydrostatic law. Standard approach.How to compute the pressure in the depth of a planet or a star? These objects have the spherically symmetrical fields. Let's at first investigate the field, with parallel force lines. Let's divide the object on the series of parallel layers with the thickness dr. If in one layer there is already the pressure p, then in the bottom layer will be the pressure p+dp. In order to find dp, it is necessary to divide the weight dF of the upper layer by its surface S. dp =  dF / S =  dM g / S =  rg dV / S =  rg dr. dp/dr =  rg. (1) The sign minus occurs because dr directed up, and the weight dF directed down. If we use the concentration n of particles of mass m then dp/dr =  nmg, (1,a) or through the potential energy dP in gravity field dp/dr =  n dP/dr. (1,b) Thus we had received several forms of the hydrostatic law, which is used not only to the field with parallel force lines, but to all objects, including planets and stars. Analytical proof of the Arc Effect.Potential and kinetic energies (P, K) of a particle of gas, moving in the gravity field of the object, constantly changes at the time from one collision to another collision with other particles, but the sum of these energies remains the same. The selection of zeropoint of potential energy can be arbitrary. Then we can suppose that potential energy at the moment just after the collision is equal to its kinetic energy but with opposite sign. Then at every moment till the next collision the sum of these energies will also be equal to zero. P + K = 0. (2) Between the collisions the particles move by ellipselike trajectories and we can divide the kinetic energy on two items K = K_{u} + K_{w}, (3) mv^{2}/2 = mu^{2}/2 + mw^{2}/2, (4) where: u  the radial velocity of a particle; w  tangential velocity of a particle; K_{u}, K_{w}  radial and tangential parts of the kinetic energy. It is clear that the radial part of kinetic energy is responsible for the radial increment of pressure. Combining (2) and (3) we'll have P + K_{u} + K_{w} = 0. (5) K_{u} =  P  K_{w} =  P  L^{2}/(2mr^{2}). (6) The radial kinetic energy can be changed by the effective potential energy W with the opposite sign. W = P + L^{2}/(2mr^{2}). (7) The change of potential energy is responsible for the change of the radial kinetic energy. The change of the radial kinetic energy is responsible for the radial increment of pressure. Consequently, now we can change the equation (1,b) which is valid for the homogeneous field, substituting in it the value P by W, and denoting some values by indexes _{L}. dp_{L}/dr =  n_{L} dW_{L}/dr. (8) Index shows us that this increment is partial, which is made only by n_{L} particles, having the angular moment L. Let's substitute (7) into (8) dp_{L}/dr =  n_{L} (dP/dr  L^{2}/(mr^{3})). (9) Knowing that dP/dr = mg, (10) and L^{2}/(mr^{3}) = m^{2}w^{2}r^{2}/(mr^{3})= mw^{2}/r, (11) we'll have the equation dp_{L}/dr =  n_{L}(mg  mw^{2}/r). (12) The pressure is additive, consequently dp/dr = dp_{1}/dr + dp_{2}/dr + .. + dp_{i}/dr dp/dr =  mg(n_{1} + n_{2} + … + n_{i}) + (m/r)*(n_{1}w_{1} + n_{2}w_{2} + … + n_{i}w_{i}) dp/dr = mng + (m/r) _{0}Integral^{w.max} w^{2}dn. (13) The value (1/n) _{0}Integral^{w.max} w^{2}dn is the average value of a square of tangent velocity [w^{2}]. Consequently dp/dr = mng + mn[w^{2}]/r. (14) If the distribution of velocities is isotropic, then [w^{2}] = (2/3)[v^{2}]. (15) And dp/dr =  mng + (2/3)mn[v^{2}]/r, (16) or dp/dr =  r(g  (2/3)v^{2}/r). (17) Using the generally known relations p=nkT, r=nm, (3/2)kT=mv^{2}/2, we can write dp/dr =  rg + 2p/r. Thus we have proved that the hydrostatic law for real gas in the spherically symmetrical field must have the Arc Effect item 2p/r. The classical hydrostatic law must be changed by the new one, if we work with spherically symmetrical field. Mathematical proof of the Arc Effect.Equilibrium in the spherically symmetrical problem without field.Did you see the flat fan? It can be gathered into a rod. You also can imagine the volume fan, which also can be gathered into a rod. Let's suppose that we gathered the Sun in a rod. Thus we can sum the forces of weight of all particles, and their vector sum now will not be equal to zero, because all forces are parallel now. We transformed the threedimensional Sun into onedimensional. In the further text we will use the word FORCE and we'll understand that it is the sum of radial forces, gathered together by turn. In order to underline this unusual method of force summing, I'll write the word FORCE in capital letters. Under gathering the center of the object will coincide with the zeropoint of the axis r. The surface of the object of radius R will coincide with the point R on axis r. Let's suppose that the axis r goes in positive direction and also in negative direction. It is obvious, that the object will accelerate into negative direction of axis r, if the sum of FORCES directed inward, will be greater then the sum of FORCES directed out of the object. The object will accelerate into positive direction of axis r, if the sum of FORCES directed inward, will be smaller then the sum of FORCES directed out of the object. The object will not accelerate along axis r, if the sum of FORCES directed inward and out of the object will be equal to zero. These reasoning can be repeated the same, relatively to the any internal spherical shell of the object. Let some homogeneous medium, for example, water, gas, dust is placed into a spherical vessel and squeezed by the vessel to the certain pressure p. There is no any outside field. According to this section the medium does not create own gravity field. Consequently, the pressure everywhere will be the same. Let's investigate the thin layer between two spheres with radii r and r+dr. The centers of these spheres coincide with the center of the vessel. Internal and external layers apply two FORCES to our investigated layer. The surface of external sphere is bigger then the surface of internal sphere by the value dS. The sum of these two FORCES is not equal to zero. It is directed inward and we will write it with the minus sign. The origin of this FORCE is clear it begins from the stretched shell of a vessel. That is why this FORCE can be named as POTENTIAL FORCE. df_{pot} =  pdS. There are no any additional FORCES. Consequently, the layer must accelerate into negative direction of axis r. But acceleration is impossible in the equilibrium state. Recalling that according to quantum mechanics, particles continue to move even under the zero temperature, and taking into account that external sphere is bigger than the internal sphere by the value dS, we come to conclusion, that KINETIC FORCE compensates the POTENTIAL FORCE. KINETIC FORCES directed upward and we will take them with the plus sign. df_{kin} = pdS. The equilibrium equation for thin layer will have the form: df_{pot}
+ df_{kin} = 0. Equilibrium in the spherically symmetrical problem with gravity field.If the object keeps itself by the gravity forces and if it does not squeezed by the outer shell, then the equilibrium equation will not have the item dfpot, but will receive instead the GRAVITY FORCE item df_{grav}. Besides the pressures in this case inside and outside the layer are not equal now, and the equation will receive now one more item, PRESSURE GRADIENT FORCE, df_{grad}. df_{grav}
+ df_{grad} + df_{kin}
= 0. Dividing by S and dr, we'll have the new hydrostatic law dp/dr =  rg + 2p/r. Spherically symmetrical object with gravity field, enclosed by shell with pressure p_{0}.Let's imagine the elastic spherical shell situated concentrically inside the gaseous object with gas inside and outside the shell. Let's blow out the gas, situated outside the shell. Let temperature of the shell remain the same. Then the equation for pressure gradient will be the same as in the above section. It will be valid at the interval lasting from r=0 to r=R_{shell}. But if we situate the whole object inside the shell and then begin to squeeze the object, or begin to heat the gas, then we'll receive another equation. df_{grav}
+ df_{grad} + df_{kin}
+ df_{pot }= 0. dp/dr =  rg + 2(pp_{0})/r. Thus, the mathematical proof give us the new hydrostatic law in more general form, which is applicable not only for gaseous object, but for the object composed by any isotropic medium, consisting of mutually interacting particles. The object can be squeezed by the shell with pressure p_{0}. In the real stars the pressure of emitted light can play the role of the shell. Program proof of the Arc Effect.
