Mr. Yogesh Vishwanath Chavan †Zero Energy Theory

Mr. Yogesh Vishwanath Chavan – Zero Energy Theory Free Online Research Papers Abstract This zero energy theory has great advantage over all other theories by its inner consistency due to which it can be used at microscopic as well as macroscopic level also. By using this theory, the attempt has been made in order to solve physics some mysterious unsolved problems like, dark matter, definition of space, expansion of the Universe and Gamma ray burst, origin of gravity, loss of mass and jets of black hole, curvature of space-time, equivalence between inertia and gravitational mass, difference in mass of proton and electron even having same value of charge, meaning of antimatter, the Universe with no anti matter etc. Thus, this theory can be used for unification of fundamental forces without need of any string i.e. no need of 11 dimensions, parallel universes, virtual particles like graviton etc. Keywords: Zero Energy, Uncertainty Principle, Gravity, Anti Neutrino, Anti matter. Introduction The energy of a body is its capacity for doing work. Almost all physics law is based upon this term – Energy. In fact, the basic physics law is â€Å"The total amount of energy in the Universe always remains constant and energy neither be created nor destroyed, but can be converted from one form to another form.† We all had accepted this law and are using over hundreds of years without thinking on following points. If energy neither be created nor destroyed, then what is the origin of all the galaxies (matter) in the Universe? In other words, Universe itself violates the above statement. If assume that, some fixed amount of matter was already available in the Universe which fly away apart in terms of galaxies during big bang explosion, then why only that fixed amount of matter was available to the Universe or who supply these fixed amount of matter to the Universe (God?) and why not supplying again? Thus, the statement of â€Å"total amount of energy in the Universe always remains constant† produces only unanswered mysterious questions. The available energy in the Universe is in the form of matter or radiations which are in terms of positive mass. If energy neither be created nor destroyed, then the total amount of energy in the Universe must be zero. Then, where is the negative energy? The only one way of escape from such mysterious questions is to modify above basic law as follow â€Å"The Universe must be continuously creating energy at every point such that the total amount of energy in the Universe should always remain equal to zero†. Thus from above statement, the energy can be created or destroyed in pairs of equal amount of positive and negative energy and for that there is no need of any supplier like God. The Universe is created from zero energy and is still in existence with summation of energy equal to zero. Or, The Universe is a Big Zero. But if it is created from zero energy, then which thing produces pairs of positive and negative energy continuously at every point. The answer is â€Å"Uncertainty Principle† which states that â€Å"Empty space cannot be exactly zero†; hence there should be formation of pairs of opposite sign of energy without violating energy balance as well as uncertainty principle. In other words, â€Å"Unc ertainty principle, itself is the God or Creator of the Universe.† But, if all galaxies are made from positive energy, where is the negative energy? Macroscopic Level (Gravity) Gravitational Constant (G) and Space as Negative Energy The gravitational force between two objects is written as: F = (G*M1*M2)/R2 (1) Where, M1 and M2 are the masses of the two objects, R is the distance between their centers and G is Gravitational Constant = 6.67*10-11 m3/ (Sec2*Kg). This equation is known as Law of Universal Gravitation. In above equation, M1, M2 and R2 are positive values. But, the gravitational force is always attractive force; hence R.H. side of above equation must be negative to represent attractive force. Therefore, gravitational constant (G) must be negative in sign. For e.g. in case of Coulomb’s electrostatic force law, the force between two charged particle is attractive because of multiplication of opposite sign of charge on the particles making negative sign for attractive electrostatic force. While in case of repulsive force, there is multiplication of same sign of charge on the particles making positive value for repulsive electrostatic force. As per law of gravitation, the sign of G must be negative; and dimensional analysis of constant ‘G’ shows that either (Kg/m3) or (Sec2) must be negative. But, Sec2 is a positive value, hence sign of (Kg/m3) must be negative. The gravitational force is an action at a distance force similar to electromagnetic force. In electromagnetic force (action at a distance force), ?0 (Permittivity of free space) and ?0 (Permeability of free space) are constants representing one of the property of space. Hence, G as a constant must have to represent the property of the space. In other words, the above dimensional analysis of ‘G’ shows that property of space representing (Kg/m3) must be negative and that property of space is nothing but critical density of vacuum (space). Or, Critical Density of Vacuum (Space) must be Negative i.e. â€Å"Empty Space must be filled of negative energy particles†. Or what we are saying about â€Å"Dark Matter/ Dark Energy is actually Negative Energy Particles†. Cosmic Microwave Background (CMB) and Mass of Neutrino In 1965, Penzias and Wilson announced the discovery of the Cosmic Microwave Background. They observed an excess flux at 7.35 cm wavelength equivalent to the radiation from a blackbody with a temperature of 2.725 degrees Kelvin. The temperature of the CMB is almost the same all over the sky. The effect of CMB is not because of adiabatic cooling of photons due to expansion of Universe from the big bang as per big bang theory, but it is continuous ongoing effect due to uncertainty principle. These CMB radiations or positive energy particles are opposite pair of dark energy in the Universe. Following simple calculation will show that mass of these CMB radiations is closer to the mass of neutrino, M (Approx. 10-39 Kg). From Wien’s Law, a blackbody (the Universe) at temperature T has a maximum wavelength of ?max (m)=0.0029/T. Using this value in Planck’s relation for the CMB radiations, we find E=M*C2=h*f=(h*C)/?max (2) M*C2=(h*C*T)/0.0029 Solving for the Mass (M), we have M=(h*T)/(C*0.0029) =(6.62*10-34*2.725)/(3*108*0.0029) =2.0754 * 10-39 Kg The continuous creation of pairs of energy of opposite sign, due to uncertainty principle solves the horizon problem (the extreme uniformity of CMB radiations in different regions of space). Also, it solves the flatness problem. Flatness problem means the density of the universe is pretty close to the critical density. Zero Energy and No Boundary Condition of the Universe As per thermodynamics, zero energy means no volume condition i.e. no boundary condition. From above theory, the Universe must be started from zero energy or no boundary condition and is still expanding in zero energy or no boundary condition by converting zero energy into pair of positive and negative energy particles due to uncertainty principle. Or, The Universe is expanding in No Boundary Condition. In other words, the Universe not started with big bang explosion, but the matters in the Universe were created with continuous increase in the size of the Universe. The Universe is expanding at the speed equal to the speed of light, from Hubble’s Law. After every one second, the increase in the radius of the Universe must be equal to 3*108 m, or, dR/dt=C=3*108 m/sec. The density of the Universe always remains constant equal to the critical density of vacuum. Let us calculate the matter in the Universe after one second from the birth of the Universe due to uncertainty principle. R=Radius of the Universe after 1 Second=3*108 m, Therefore, M=Matter in the Universe after 1 Second=Rho(Critical)*{4*Pi*R3/3} {Where, Rho(Critical)=Critical Density of Vacuum=7.94*10-27 Kg/m3 for H=Hubble’s constant=65 Km/Sec/Mpc}. = 7.94*10-27*4*(22/7)*(3*108)3/3 =0.9 Kg From above, during early stage, the Universe was not as hot as per big bang theory; but, it was too cold. The gravity effect was very negligible to halt the expansion of the Universe. Hence, the Universe expanded with no barrier for its expansion. With further expansion of the Universe, the gravity effect becomes considerable due to tremendous increase in the matter. From above formula, it can be shown that just after 1000 years; the matter in the Universe was around 2.8* 1031 Kg or 14 times the mass of the sun. It was spread around the Universe everywhere in the radius of approx. 9.48*1018 m (1000*3.16*107*3*108 m) instead of collecting at single point; it was segregated making clusters of matter which finally converts into galaxies. Thus, this theory also solves the structure problem i.e. the existence of large scale structure, like walls of cluster of galaxies and voids. According to Hubble’s law, the age of the Universe is equal to 1/H which is 15 Billion years old. Hence, the size of the Universe (R), is- R=C*t=3*108*15*109*3.16*107=0.142*1027 m Where, 1Year=3.16*107 Sec. The total matter (M) in the Universe after 15 billion years is, M=7.94*10-27*4*(22/7)*(0.142*1027)3/3 M=9.52*1052 Kg (4) Same amount of dark matter exists in the form of space, so that total energy in the Universe becomes equal to zero. The above calculated value of total mass in the Universe is quiet acceptable due to existence of more than billion of galaxies in the Universe and more than billion of stars in a galaxy. Let us calculate how much amount of matter; the Universe must be creating per second at this moment. dV=Increase in Volume of the Universe per second=4*(22/7)*{(R2)3-(R1)3}/3. Where, R1=0.14*1027m and R2-R1=3*108 m ? dV=4*(22/7)*(R2-R1)*{(R2)2+R2*R1+(R1)2}/3=4*(22/7)*3*108*3*(0.142*1027)2/3 R2?R1 =76*1060 m3 ? Increase in Matter of the Universe per Second=7.94*10-27*76*1060 = 0.60*1036 Kg (5) From above it is seen that, really, â€Å"Size does Matter†; i.e., larger the size of the Universe, greater is the creation of matter per second. This large amount of matter (in terms of radiations) at the edge of the Universe may be the source of Gamma rays burst which occurs sixteen times a month and to be roughly uniformly distributed in all direction across the sky. Disappearance of the Sun, Zero Energy and Gravity What will happen, if the sun disappears suddenly? According to Newton’s Classical Theory, the planets will start to move away from the sun instantaneously due to absence of gravitational force. While according to A. Einstein’s General Theory of Relativity, there is a universal limit equal to speed of light for transformation of information from one point to another point. Hence, the planets will not start to move away from the sun at same time, but the disappearance of the sun will produce the gravitational waves (Space-time waves) travelling at the speed of light and when these waves will reach towards the planets, they will start to move away from the sun. But, both these theories are on wrong track. Because disappearance of the sun means all energy of the sun vanishes suddenly which will create zero energy density region having volume equal to the volume of the sun. Now, as proved earlier, space is nothing but negative energy; hence the energy density around the su n is negative. Therefore, this zero energy density will exert force on negative energy particles having direction away from the center of the sun. We know that, Force=Mass*Acceleration. But, here, the mass is negative; hence these negative energy particles must have to accelerate towards the center of the sun (-Ve Acceleration). When this effect will reach towards the planets with speed of the light as per general theory of the relativity, the planets will experience attractive force proportional to their mass because of acceleration of space which is nothing but the gravity. The overall effect will be the movement of the planets towards the center of the sun instead of moving away from the sun. This acceleration must be more than the gravitational acceleration produced due to the presence of the sun, because gravitational acceleration produced by the sun is only due to annihilation of the positive mass of the sun with negative mass of the space. From above the gravity can be define d as below: â€Å"Gravitational force is an action at a distance force which is due to creation of zero energy density region around the positive mass because of annihilation of this positive mass with negative mass of the space.† From above analysis, two points come ahead viz. – The source object in order to produce gravitational effect on the other objects must have to lose its mass continuously by making annihilation with negative mass of the space. The object which experiences the gravitational force does not receive any amount of energy from the source object. Hence, for making displacement against the gravitational pull, it must have to convert its available energy into kinetic energy equivalent to the gravitational force multiplied by the displacement made by the object. For e.g., the ball releases from certain height, on the earth, should have to convert its available rest energy into kinetic energy because the ball does not receive any type of energy from the earth. Hence, the rest energy of the ball is not always remains constant, but should have to decrease by converting into kinetic energy. Let us derive the new energy balance equation for both the cases i.e. free fall of the object under gravity and the object thrown in upward direction against gravity of the earth between points A and B (RA>RB). Case 1: In case of a body, falling freely under gravity, there is a continuous conversion of its rest energy into kinetic energy such that at any point in its path, the total energy of the body always remains constant equals to the rest energy of the body. Thus, for freely falling body, (Rest Energy)A=(Reduced Rest Energy)B+(K.E.)B (6) From above formula, let us calculate, the velocity V, at which all the rest energy of the ball converts into the kinetic energy. Therefore, (Reduced Rest Energy)B become equals to zero i.e. (Rest Energy)A = (K.E.)B. From special theory of relativity, (Rest Energy)A=(M0*C2)=(Kinetic Energy)B={M0*C2*[(1/sqrt(1-(V2/C2)))-1]} ? V={?3/ 2}*C= 0.866*C = 2.598*108 m/s (7) As per special theory of relativity, the particle like photon should have only kinetic energy and no rest energy because they are travelling at the speed of light. Hence, at the above derived velocity, the entire kinetic energy of the ball should have to convert into photons where energy of each photon depends upon the temperature of the ball at this velocity, V. Suppose any charged particle (say electron, -Ve charge) falls freely under the influence of very strong gravitational force, then, the entire rest energy of the electron must have to convert into kinetic energy at above derived velocity. But, the electron will never converts into photons because of law of conservation of charge. The charge on the particle prevents conversion of its available energy into photons or in terms of radiations. Charge helps to conserve available energy. If rest energy of particle does not convert into K.E., then charge on particle must have to conserve this rest energy or prevent to convert this re st energy into photons. In other words, Charge is responsible for Rest Energy or Conservation of Energy in Charged Particles. Hence, existence of matter (Galaxies, stars, planets and finally living cells like human being) is due to charge without which entire universe must be filled of only radiations. Any neutral particle if carry some amount of energy, all its stored energy should have to convert into K.E. as there is no such a thing like charge which will prevent conversion of this stored energy into the form of K.E. and then into the form of photons. Neutral particle should have to carry only Kinetic Energy. Therefore, free neutrons are not stable, because neutrons do not carry any charge, hence all its stored energy must have to convert into Kinetic Energy, while in nucleus of atom, the charge on proton prevents neutron to convert its available rest energy into kinetic energy. Case 2: Consider a ball thrown in upward direction against gravitational field of the earth from point B to point A. As per classical theory, for throwing a ball in upward direction, we have to supply kinetic energy to the ball, which then converts into potential energy, when ball after achieving certain height comes to rest. First of all, the concept of supplying kinetic energy to the ball looks like strange idea, because there is only one way of supplying kinetic energy to any system i.e. nothing but photons. Here, you are supplying only kinetic energy and no rest energy to the ball and for achieving this, you have to transfer photons from your body through hand to the ball without any loss of photons to the surrounding. Is it possible? No. In case of inertia also, when we apply the force on an object (Inertial Mass) in order to move it with certain velocity, we are not transferring any type of kinetic energy to the object by application of force. In all these cases (inertia mass as well a s gravitational mass) the object converts some amount of its available rest energy into the kinetic energy which depends upon the value of force. Thus, this theory also achieves â€Å"Equivalence between Gravitational Mass and Inertial Mass† from the point of view that â€Å"Due to Application of Force in both cases, the object converts some amount of its rest energy into kinetic energy which depends upon the value of the force†. Therefore, (Rest Energy)A=(Reduced Rest Energy)B+(K.E.)B (8) The equations (6) and (8) both are same which shows consistency in this theory. The theory also eliminates the need of potential energy which is required in case of classical theory, where to determine potential energy at any point some datum reference point is required like surface of the earth. Why experiment have not been conducted to know the difference of rest mass (or rest energy) at point A and at point B, when the object will be thrown in upward direction. If experiment is carried out, you should get same amount of rest mass at both points as per this theory; while as per classical theory, the object at point A should have rest mass when it is at rest plus the mass equivalent to supplied kinetic energy at point B. Casimir Effect and Zero Energy It is a small attractive force that acts between two close parallel uncharged conducting plates. Its existence was first predicted by the Dutch physicist Hendrick Casimir in 1948 and confirmed experimentally by Steven Lamoreaux, in 1996. According to modern physics, one can interpret these so called vacuum fluctuations, as pairs of particles and anti particles that suddenly appear together, move apart, and then come back together again, and annihilate each other. But, it is not true. In fact, this effect can be considered as gravitational force between these plates. The negative energy particles (space) between these plates annihilates with positive mass of these plates creating zero energy density between them. While surrounding energy density remains same due to lot of negative energy particles around the plates. This zero energy density is nothing but gravitational pull as proved earlier. Hence, these plates attract towards each other. Therefore, as per modern physics, this effect cannot be linked with the possibility of faster than light (FLT). Loss of Mass of Black Holes and Jets of Black Holes From new definition of gravity, higher the mass of the object, not only there will be more destroy of space (negative energy) giving rise to strong gravitational field. There is same amount of destroy of mass of the object. In other words, for massive objects like black hole, due to above gravity effect, the continuous loss in mass of the black hole should take place. Also, for any object falling freely in strong gravitational field of the black hole, it’s all rest energy must be converted into kinetic energy by gaining velocity, V equals to {(?3/ 2)*C}. The temperature of the object at this velocity must be very high; hence its kinetic energy should have to convert into high energy photons like Gamma rays or X rays making jets around the black holes. Gravitational Red Shift of the Spectral Lines and Loss of Energy It is observed that the wavelength of sodium light coming from the sun (strong gravitational field) is greater than that from the sodium lamp on the earth which gives slight displacement in spectrum of light towards the red end called as gravitational red shift of spectral lines. It is considered as the effect of time dilation (a clock lying in a gravitational field appears to run slow) given in general theory of relativity. But, the time dilation should occur only in strong gravitational field of the sun and is very less in case of gravity of the earth. Therefore, even if it is consider that red shift is due to time dilation on the sun, the light after reaching on the earth from the sun, the red shift in spectral line should have to disappear due to very less time dilation on the earth. Thus, gravitational red shift is not due to effect of time dilation. The photon due to its mass (in terms of kinetic energy) experiences force due to gravity effect produced by the sun. In order to m ove away from the surface of the sun, the photon has to do work against gravity of the sun. Therefore, the energy balance equation can be written as: K.E. of photon at surface of the sun=K.E. of photon at surface of the earth+Work done by photon. This work done must be converted into less energy photons (different wavelength than original photons) which make loss in energy of the main photon. Hence the photon which receives at surface of the earth has energy less than that produced on the sun by amount equal to work done by photon. Therefore, we observed the gravitational red shift of the spectral lines. Curvature of Space-Time and Bending of Light According to Einstein’s General theory of Relativity, a body of high mass bends the space-time continuum around itself. The continuum becomes curved. In May 1919, during total solar eclipse, the deviations in the position of the stars which appears to be lying in the neighborhood of the sun disc were observed in gravitational field of the sun due to the deflection or bending of the rays of light coming from stars. Hence, confirming this prediction of the General theory of Relativity. In fact, there is no such a thing happening like curvature of space-time continuum in strong gravitational field. There is nothing like singularity in case of black holes or concepts like worm holes, the terms appears due to curvature of space-time continuum. The correct reason for the bending of light in strong gravitational field can be explained as below: The photon contains energy in the form of kinetic energy, (K.E.=m*C2); hence, the energy or mass of the photon experiences strong gravitational force. We know from Newton’s first law of motion, force is that external agent which changes the state of rest or of uniform motion of a body along a straight line. By applying a force we can change (1) the magnitude or (2) the direction or (3) both magnitude and direction of the velocity of the body. In case of photon, it is travelling at the speed of light; hence, it is not possible to reduce its velocity as per special theory of relativity. Therefore, the effect of gravitational force is to change the direction of photon from rectilinear motion. The following simple calculation will show that the bending of light do not depends upon energy of the light, but entirely depends upon its distance from the sun and mass of the sun. Thus, for all type of photons from higher energy to lower energy photons, the bending effect remain same whic h is considered by Einstein as curvature of space-time. Consider a ray of light which is travelling in a rectilinear direction at a distance of 109 m from center of the sun and parallel to the surface of the sun. Then, as per second kinematical equation and from Newton’s law of gravitation, S=Displacement of photon towards the sun=(u*t+0.5*g*t2), where, u=Initial velocity of the photon in the direction of displacement=0; g= Gravitational strength of the sun or acceleration of photon=-(G*M)/R2, G=Gravitational Constant=6.67*10-11 m3/(Kg*Sec2), M=Mass of the sun=2*1030 Kg, and R=Distance of photon from the sun=109 m. Let us calculate this displacement of the photon for time of travel equal to 1 Sec. Note that, the above eq. is independent of energy or matter of photon. S={0*1-[(0.5*6.67*10-11*2*1030*12) / (109)2]}=-66.7 m Figure 1: Bending of light The Projectile motion of light under strong gravity From above figure.1, for one second of travel of photon or for travel of photon at a distance equal to 3*108 m, the photon is displaced towards the sun about 66.7 m. Thus, the effect of bending of light in strong gravitational field of the sun is not due to any curvature of space-time, but it is the case of the projectile motion under the influence of gravity. It is similar to the parabolic path made by the ball on the earth which is thrown in upward direction at an angle to the surface of the earth. Therefore, there is nothing like concept of curvature of space-time in case of defining gravity. Microscopic Level (Matter and Anti-Matter) The Difference Between Mass of Electron and Proton In case of electrostatic force, we can use same equation (6) or (8) from point of view of application of force (i.e. no supply of kinetic energy from external medium). Here, the kinetic energy of the electron will be integral of electrostatic energy between radius during the origin of the electron (i.e. from the neutron) and the orbital radius of the electron around nucleus of atom (For H2 atom, the orbital radius of the electron at ground state is 5.29*10-11 m). The equation can be written as below: (Reduced Rest Energy)=(Rest Energy)-(Kinetic Energy) (9) But, Kinetic Energy = ? (Fe*dR) = ? {q2/ (4*?*?0*R2)}*dR = ? {e/ R2}*dR e= {q2/ (4*?*?_0)} ? Kinetic Energy = {(e/R1)-(e/R2)} = {(e*(R2-R1))/(R2*R1)} Where, R1=Radius during origin of the electron and R2= Orbital radius of the electron in H2 atom If the R.H. side of the above equation (Reduced Rest Energy) becomes equal to zero, then we can calculate the radius during origin or birth of the electron. By rearranging terms, we get R1 = {(e*R2)/((R2*M0*C2)+e)} = {(2.567*9*10-29*5.29*10-11)/((5.29*10-11*9.1095*10-31*9*1016)+(2.567*9*10-29))} ? R1 = 2.8178*10-15 m (10) Where, M0=Rest mass of the electron=9.1095*10-31 Kg This is closer to the radius of the neutron which shows that the birth of electron occurs at periphery of neutron. The above calculation shows that â€Å"Rest Mass of Electron i.e. 9.1095*10-31 Kg is due to Birth of Electron takes place at Periphery of Neutron i.e. apprx. at 2.82*10-15 m†. Note that the value of R1