Is there antimatter-antimass

Big Bang HTL 4, textbook

202 Solutions b: With a proton spacing of 0.2 · 10 –12 m, the tunnel probability is 9 · 10 –7 ≈ 10 –6. Only around every millionth proton can therefore tunnel through. Or, to put it the other way round: the probability that a proton will tunnel at this distance is one millionth. c: Because the tunnel probability covers many orders of magnitude, the y-axis must be represented logarithmically. Fig. 14: Graphic representation of Table 17.4 1g of plutonium has 6.023 · 10 23 ________ 238 atoms. Decay per second ∆ N = N (0) ln2 ___ T 1/2 t = 2.53 10 21 ln2 _____________ 86.4 365 24 3600 nuclei = 6.4 10 11 nuclei. This results in a power of 5.48 · 6.4 · 10 11 MeV ____ s = 3.5 · 10 12 MeV ____ s = 0.57W. 1.8kg of plutonium is required for 1kW power. A gas consisting of deuterium ions is to be regarded as a monatomic gas. The kinetic energy results as E kin = 3 __ 2 k T and the temperature as T = 2 E kin _____ 3 k = 2 10 3 1.6 10 −19 ___________ 3 1.38 · 10 −23 K ≈ 8 · 10 9 K. However, in practice the gas only needs to have a temperature of around 10 8 K. The particles have a temperature distribution (Maxwell distribution) around the mean temperature. As a result, some particles have a lower, others a higher temperature, namely greater than 10 10 K. U = 1 _____ 4 π ε 0 Q Q ′ _____ r = 1 _____ 4 π ε 0 (46e) · (46e) ________ 2 · RR = R 0 · A 1/3 = 1.4 · 10 -15 3 √ ___ 119 m = 6.9 · 10 -15 m. U = 46 2 · (1.6 · 10 −19) 2 __________________ 4 π 8.9 10 −12 2 6.9 10 −15 J = 3.5 10 –11 J = 220MeV 17 Particle physics and standard model T 1/2 = ln2 / λ and therefore λ = In2 / T 1/2. Therefore the law of decay can also be written down as follows: N (t) = N 0 · e - In2 ___ T 1/2 t After the mean lifetime τ, the number of particles has decreased to N 0 / e. The following applies: N 0 / e = N 0 · e - ln2 ___ T 1/2 t ⇒ ln (1 __ e) = - ln2 ___ T 1/2 ⇒ τ = T 1/2 ___ ln2 at the weak force the interaction in the foreground. Nevertheless, one can say that the weak force holds the u-quark, electron and anti-electron neutrino together in the d-quark. But it is so weak that it decays after around 15 minutes, and with it the neutron. Not according to the current state of knowledge! The electron could only decay into particles that have a lower or no mass, i.e. neutrinos, photons or the hypothetical gravitons. However, none of these particles are charged. Therefore the principle of the conservation of the electric charge would be violated. The strengths of the individual forces change with the energy. The values ​​given in Tab. 17.7 are only correct for low energies. In the context of SUSY (which is also a suggestion for GUT), all forces should be the same at around 10 17 GeV (see figure). Such energies prevailed shortly after the Big Bang. F17 F18 F19 F18 F19 F20 F21 Fig. 15 Fig. 16a shows only one of the possibilities. Anti-particles are represented as going back in time. It interacts with its own field. Sleptonen, Zino, Photino, Higgsino and Gravitino If neutrinos were to fly at the speed of light, no time would pass for them from our point of view, because for v = c tb = tr √ _____ 1 - v 2 __ c 2 = tr √ ____ 1 - 1 = 0 apply. But if they don't age, their condition would be frozen. However, this would also prevent them from transforming, because that presupposes the flow of time. It is alluded to that neutrinos, if they were to travel at faster than light speeds, could possibly move back in time, which would turn the principle of cause and effect on its head. a: The sun has an output of 3.9 · 10 26 W or J / s. An incredible 3.9 · 10 26 J / 4.2 · 10 –12 J = 9.3 · 10 37 reactions must therefore take place per second. Therefore, 1.86 · 10 38 neutrinos are produced in the sun per second. b: The surface of a sphere is calculated with O = 4 r 2 π. Because we want the surface in square centimeters, we convert the distance immediately beforehand: 150 billion meters = 1.5 · 10 11 m = 1.5 · 10 13 cm. This gives a spherical surface of about 2.8 · 10 27 cm 2. In total, 1.86 · 10 38 fly through the entire surface of the sphere and therefore 1.86 · 10 38 neutrinos / 2.8 · 10 27 cm 2 ≈ 6.6 · 10 10 neutrinos / cm 2. The decay is ambiguous because there are three different possibilities. For example, does the muon consist of one electron and two neutrinos or two electrons and one positron and two neutrinos? That is why one has taken the point of view that the muon is elementary. A proton consists of two up quarks and one down quark. The mass of a proton should therefore be 12 MeV / c 2 (see Tab. 17.4, p. 155). In fact, it is around 941 MeV / c 2! The largest part of the mass of protons - and by the way also of neutrons - is formed by field energy. No! There are mutliple reasons for this. One of them is that the neutron consists of three quarks that are charged (with the color charges red, green and blue). The antineutron consists of three antiquarks (with the color charges yellow, purple and magenta; see Fig. 17.28 right), whose charge differs from that of the “neutron quarks”. a: Assume the equation ∆ E · ∆ t ≥ h __ 13. In order for a virtual particle to emerge from nothing, at least the energy ∆ E = mc 2 is necessary. If you start at the top and solve for ∆ t, you get ∆ t ≥ h _____ 13mc 2. The upper limit to the speed a particle can reach is c. From c = ∆ s __ ∆ t follows ∆ s = c · ∆ t ≤ h ____ 13mc. b: From ∆ s ≤ h ____ 13 mc follows m ≥ h ____ 13 ∆ sc. This implies m ≥ 6.6 · 10 −34 __________ 13 · 10 −18 · 3 · 10 8 kg = 1.7 · 10 −25 kg. The mass of these virtual bosons is around 100 times that of the protons! No. Antiparticles have exactly the same mass as their corresponding particles (m e + = m e -, m p = m _ p…). If antimatter were to have antimass, particles and antiparticles would complement each other to produce zero energy. In fact, when particles and antiparticles are annihilated, the rest mass of both particles is released as energy. In many modern acceleration systems, particles are shot at corresponding antiparticles. Due to the large kinetic energy of the particles (around a thousand GeV), the rest energy released during the collision (about 1 MeV for electrons, 2 GeV for protons) only accounts for a small part of the total energy released. a) c) d) e) allowed; b) forbidden: lepton number not included This is the so-called Lagrangian (the word is pronounced like “Lagroschien” with a slightly squeezed o). This is the mathematical formulation of the standard model, and it is extremely complex. If you write it down, it takes up several pages. You can read about it here, for example: http://einstein-schrodinger.com/Standard_Model.pdf. This expression is simplified again on the cup. To get an impression of what it means, the individual sub-expressions are labeled in Fig. 17. Fig. 16 F22 F23 F24 F25 F26 F27 F28 F29 F30 F31 F32 F33 F34 For testing purposes only - property of the publisher öbv

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