Home | Blog | Books | Links                                                                           Language: Danish Comments: post@finaltheories.com Deduction of the Theory | Mass and Energy | Evaluation of the Theory | Test of the Theory The Relativistic MassWe consider a body x·y·z at rest in relation to the zero-point field, with the rest mass m0 = ρ·x·y·z, where ρ is the density and x, y and z are the coordinates of the body in the respective directions, when the body is at rest in the zero-point field. We now impart a velocity v to the body - in the plus x direction. Because of the length contraction the volume becomes equal to                     x'·y·z = x(1  ̶  v2/c2)½·y·z. As the number of elementary particles, the body consists of doesn't change, the mass of the body remains unchanged. However, since the volume is smaller, the density will be greater. We denote the new density with ρ', which becomes equal to the rest mass divided by the new volume:                     ρ' = m0/[x'·y·z] = ρ·x·y·z/[x(1  ̶  v2/c2)½·y·z] = ρ/(1  ̶  v2/c2)½ = ρ(1  ̶  v2/c2)-½. If we insert the term m0/(x·y·z) for ρ, we find                     ρ'= m0/[x'·y·z] = m0(1 ̶ v2/c2)-½/[x·y·z]so                    ρ'x·y·z = m0(1 ̶ v2/c2)-½. If we set ρ'x·y·z equal to the mass m at the speed v, we find                     m = ρ'x·y·z = m0(1 ̶ v2/c2)-½, from which we find Einstein's relativistic mass:                     m = m0(1 ̶ v2/c2)-½. It should be noted that the mass m is the mass of a body of the original volume, x·y·z. The original body with the mass m0 will because of the speed v in the x-direction shrink by a factor (1 ̶ v2/c2)½, so it during the movement has the coordinates x'·y·z and therefore the mass                    m' = ρ'·x'·y·z = ρ(1  ̶  v2/c2)−½· x(1  ̶  v2/c2)½·y·z = ρ· x·y·z = m0 That is, the mass remains the same, but the density becomes larger due to the reduced volume. In connection with cyclotrons it is thus necessary to enhance the field strength, so the number of field lines that affects the body increases as the volume of the body decreases, if the orbit of the body shall be maintained. The Relativistic Mass Mass and Energy are Equivalent Quantities: E = mc2   Since the density and thereby the mass depends on the velocity, the force is defined as the change of impulse per unit time . The work the force Fs performs on a particle, which is moved a distance ds along the trajectory of the particle, is equal to the increase of the kinetic energy of the particle, dEk                     dEk = Fsds.   Using the chain rule, we find that , from which                     dEk = Fsds = vd(mv) = v2dm + mvdv. If we insert the expression for the relativistic mass ,we get . If we integrate from 0 to v and define that Ek = 0 for v = 0, we find the following expression for the kinetic energy: , where m0c2 is the rest energy of the particle.   The total energy, E, of the particle is then the sum of its rest energy and the kinetic energy, Ek:                     E = m0c2 + Ek = mc2. Thus, we find, that the total energy of a particle is its total mass m times the square of the velocity of light, c2, that is:                     E = mc2. The equation can also be written as                     mi = E/c2, where mi is called the inertial mass. It follows that the mass and energy are two sides of the same coin. As                     ε0μ0 = 1/c2, we get                     mi = ε0μ0E. where ε0 is the electric, and μ0 the magnetic constant. It can be seen that the mass, as well as the inertial mass, is a result of the electric and magnetic energy of the field. Since the SI unit for the mass is [V∙C∙F/m∙H/m], i.e.: Volts x Coulombs x Farad/meter x Henry/meter, there is much to suggest that the mass is of electromagnetic nature.The inertial mass mi = E/c2 (= m) acts like any other mass. This means that it exhibit inertia when exposed to an external force, as when, for example, it is influenced by a gravitational force, and the inertial mass possesses, as any other mass, both potential and kinetic energy. This applies whether the inertial mass stems from an increase of the energy of an existing mass, or whether it stems from electromagnetic radiation.That it applies to electromagnetic radiation can be seen from, that the energy of a photon according to the quantum theory equals E = h∙f, where h is Planck's constant and f is the frequency of the radiation. Since E = mc2 we find that                     E = h∙f = mc2 and hence that                     mi = h∙f  / c2. Mass and Energy are Equivalent Quantities To the Top Black Holes in the Euclidean Space   A black hole is a region of space in which the gravitational field is so strong that nothing - not even light - can escape from its pull. To calculate the escape velocity in a Euclidean space, we consider a heavy body with the mass M, which is located at the origin of a coordinate system. Another body with the inertial mass mi is starting in the distance r from the origin with the velocity v. If the body shall escape to infinity, it must have a sufficient kinetic energy ½miv2 to be able to counterbalance the gravitational potential energy GmiM/ r:                    miv2/2 = GmiM/ r,    where G is the gravitational constant.   For each value of v, there is a critical value of r, so a particle with velocity v is only able to escape to infinity, if                    r ≥ 2GM / v2.   When the velocity is equal to the velocity of light c, we get the radius of a black hole with the mass M  from which nothing, not even light, can escape,                    rSchwarzschild =2GM / c2.   The value of the radius of a black hole is called the Schwarzschild's radius. Black Holes in the Euclidean Space To the Top Gravitational Redshift and BlueshiftLight and other electromagnetic radiation, which originates from a source located in a strong gravitational field, will have a longer wavelength than radiation emitted from a source situated in a region with a weaker gravitational field. As, the long-waved end of the visible electromagnetic spectrum, is red, the extension of the wavelength of the radiation is called a redshift. Blueshift is, on the other hand, a shortening of the wavelength of the emitted radiation, or an increase of the frequency of the radiation. The name originates from the fact that the shorter end of the visible spectrum is blue or violet. Redshift and blueshift can be derived by considering the energy outside and inside a gravitational field. We first consider the energy of the electromagnetic radiation when it is situated in an area that is not affected by the gravitational field. If a photon has the velocity c and the inertial mass mi, its total energy will be equal to                     E = mic2. If, the photon moves into a gravitational field with a gravitational potential equal to                     GmiM/ r, where G is the gravitational constant, M is the mass of the body, and r the distance to the center of mass of the heavy body, the total energy will change a bit. We denote the new energy as E', from which                     E' = mic2 + GmiM/ r.Since the energy of the photon is a function of the distance r on its way through the gravitational field, the energy E' can be written as According to the quantum theory the energy of a quantum of radiation is equal to Planck's constant h  times the frequency f                     E = h ∙ f .   From this, we can find the frequency in a gravitational field f ' expressed by the initial frequency f, so . We then find that the redshift or blueshift equals where .The sign of z determines whether it is a redshift or a blueshift. Gravitational Redshift and Blueshift Clocks Go Slower in a Gravitational Field than OutsideSince the frequency of a quantum of electromagnetic radiation is changing under the influence of a gravitational field, it can be expected that a gravitational field also affect mechanical clocks. If we use a stable oscillator as a clock, and let its frequency represent a time unit, the frequency will as we have seen go slower when the oscillator is affected by a gravitational field. This means that the clock runs more slowly in a gravitational field than in the free space. As previously, the energy of an oscillator in the empty space can be expressed as E = mic2, and as E', when the photon is in a gravitational field. This gives us the following expression for E', ,where GmiM / r is the gravitational potential. As, a quantum of radiation contains the energy                    E = h ∙ f ,where f is the frequency of the emitted radiation, and h is the Planck constant, the relationship between the frequencies can be written as .Since, the frequency is changed, and an oscillation represents a time unit, the clock will be slower. We then find the following connection between the time in a gravitational field t' and time in the zero-point field t, .    It is seen, that the clocks go slower, when they are in a gravitational field. Clocks Go Slower in a Gravitational Field than Outside Energy and Mass are Deflected in a Gravitational Field   The existence of closed universes and black holes in a flat Euclidean space depends on that mass as well as energy is deflected in a gravitational field. This can be shown in the light of the fact that the inertial mass of a body mi, is equal to the energy the body contains. If the growth of energy amounts to E, the inertial mass increases with mi = E/c2. This means that if there is an increase of inertial mass, there is a corresponding increase of the gravitational mass.   An increase of the inertial mass will then result in a corresponding increase in the potential energy in a gravitational field, ghmi = ghE/c2, where g is the gravitational acceleration, mi is the inertial mass and h is the height. This applies to both energy and material bodies. Fig. 32. The deflection of electromagnetic radiation in a gravitational field. Using the inertial mass of the photons, we can calculate the deflection of electromagnetic radiation in a gravitational field, which stems from a massive body. An electromagnetic radiation from a distant object will describe a characteristic hyperbolic path under the influence of a central force, when it passes the gravitational field of a celestial body. The deflection of the photons is shown in the figure where the deflection for the sake of understanding is much exaggerated. We designate the minimum distance between the radiation and the centre of mass of the celestial body as r, the angle of the asymptote as φ and the eccentricity of the hyperbola as e. Here, the connection between φ and e is equal to                     cosφ = 1/e, where e = c/a.The deflection angle of the radiation, α, is shown in the figure, and is                     α = π - 2φ. From the law of conservation of energy and angular momentum, we find the total energy E as a function of r: ,   where μ = miM /(mi + M) ≈ mi is the reduced mass of the particle, ½∙μ(dr/dt)2 is the radial kinetic energy, L2/(2μr2) is centrifugal potential, and GmiM /r  is the gravitational potential. A solution, to the radial energy equation, is                     r = p /(e∙cosθ +1),   where  p = L2/GmiMμ.    The total energy E is most easily found at the point C, where θ = 1800  and thus                     r = p /(1 - e) and dr/dt = 0. If these terms are inserted in the energy equation, we get . If we insert the expression for p and set μ ≈ mi , we find . For an energy quantum with mass mi , the total energy E and the angular momentum L, with respect to the centre of the celestial body, the eccentricity is thus equal to . The constants of motion E and L can be easily obtained at the point C. If we at this point call the velocity v, and replace the reduced mass with mi , we get , where, the angular momentum is equal to                     L = mirv. As, the velocity of the photons, is equal to the speed of light c - and thus, unless we are dealing with a black hole, much larger than the escape velocity ve = (2GM /r)½ at the point C - we can ignore the change of the velocity of the photons and assume that the gravitational field only changes the direction of the velocity. If we set v = c, where c is the velocity of the photons in the “free space”, the eccentricity can be written as .   As anticipated, the masses of the photons cancel out. The value of the eccentricity, determines whether the electromagnetic radiation can escape from the gravitational field of a celestial body, or whether it will be captured. If e ≥ 1 will the mass of the celestial body, not be large enough to hold on to the light, which thereby will describe a hyperbola - or a parabola when e =1.   It is seen that e =1 when                     r = 2GM / c2.    This value of the radius is called the Schwarzschild's radius, and if the radius is less than this value, then  e <1, and the electromagnetic radiation will describe an ellipse. The radiation thus will be unable to escape the gravitational field of the object. Here, the object can be a black hole or an entire universe. If it is a universe, it is referred to as a closed universe. If we look at the cases, where e is much larger than 1, and thus  2GM / r much smaller than c2, the eccentricity can be simplified to                     e = c2r / (GM). The angle of deflection of the radiation α can, by using the connections                        α = π − 2φ, cosφ = 1/e  ande = c2r/(GM), be written as .   Since x = GM /(c2r) << 1, cos-1x can be developed in a Taylor series: . As we only consider the first term within the brackets, the angle of deflection becomes .      Light passing along a body of mass M will then be subjected to an angular deflection, α, which also entails a reduction of the gravitational potential. Here G is the gravitational constant, M is the mass of the celestial body, c is the velocity of light, and r is the distance between the light and the centre of mass of the celestial body. Energy and Mass are Deflected in a Gravitational Field To the Top                                                              © J. Balslev 2010