Relativististic Mechanics Jan 10, 2019

1. Relativististic Mechanics Jan 10, 2019 • With respect to classical Newtonian mechanics: • Special Theory of Relativity: = v/c • m = m0 / sqrt(1- 2) • Introduction of ElectronVolt as new unit of energy: 1 eV = |q| Joules

2. Efield = V / D • With electron charge q: • F = q . Efield • electron kinetic energy: • Ee- =  F dD = q.V • Ee- independent of: • distance D • particle mass heated filament distance D Potential diffence V Energy E of accelerated particle with unit charge equals qV Joule equal to V eV

3. Klassiek: Newton: • massa in rust, kinetische energie is 0 • eenparige beweging • versnelling: F = m . a • Einstein’s Speciale Relativiteitstheorie:

5.  = v / c • = v / c, and the Lorentz factor γ: relativistic mass mr = γ m0 γ = 1 / sqrt(1- 2), and  = sqrt(γ2 -1) / γ

7. Use Taylor expansion: m c2 = m0 c2 / sqrt(1- 2) = m0c2 + ½ m0 c2 v2/c2+ 3/8 m0 c2 v4/c4+ . . . . Eerste term: E = m0c2 rustmassa energie-equivalent Tweede term: ½ mv2: klassieke kinetische energie

8. Essential in Relativistic Mechanics: Total Energy of system For a moving particle: Total Energy = Kinetic Energy + Rest Mass Energy equivalent

9. The ‘energy’ of a particle is its kinetic energy!!! Total Energy = Kinetic Energy + Rest Mass Energy equivalent Total Energy ET= γ m0c2 = mrc2 ‘gamma factor’ Note: γ = 1: particle at rest. γ = 1 / sqrt(1- 2) and  = sqrt(γ2 -1) / γ

10. Be Aware! This is very confusing: By saying ‘The energy’ of a particle’ we mean its kinetic energy. ! May be smaller than rest mass energy eq.; it may be even zero! Total Energy = Kinetic energy + Rest Mass Energy eq. But: Total Energy Squared: ET2 = mo2 c4 + p2c2 looks like ‘rest mass energy’ looks like ‘kinetic energy’ but this equation concerns ET2!!

11. With van de Graaff accelerator: simple: • E = q V, so E = V eV • From Einstein’s Special Theory on Relativity: • E2 = mo2 c4 + p2c2 • With: • = v / c, and the Lorentz factor γ: relativistic mass mr = γ m0 γ = 1 / sqrt(1- 2), and  = sqrt(γ2 -1) / γ So: total energy E = m0c2 sqrt(1+ 2γ2) [= rest mass + kinetic energy] = γ m0c2 = mrc2

12. Speed of electron after being accelerated with 25 kV? • Total energy = 511 keV + 25 keV = 536 keV • = γ . 511 keV • So γ = 536/511 = 1.048 • = sqrt(γ2 -1) / γ = 0.30 • Dus v = 0.30 c

13. Remember: TOTAL energy E2 = mo2 c4 + p2c2 Note ‘restmass’ term and ‘kinetic’ term (squared!) relativistic mass mr = γ m0 for high energy particles: p = γ m0 c γ = 1 / sqrt(1- 2) For high-energy particles (E >> m0c2): E2 = mo2 c4 + p2c2 = E2 = p2c2  E = pc  p = E/c independent of particle’s mass! Rest Masses photon: 0 (zero) electron: 511 keV/c2 muon: 0.105 GeV/c2 pion+-: 0.139 GeV/c2 pion0: 0.134 GeV/c2 proton: 0.938 GeV/c2 neutron: 0.939 GeV/c2

14. Examples: electron: rest mass m0 = 511 keV/c2 With total energy 1 GeV: kinetic energy = almost 1 GeV Momentum p: 1 GeV/c Other example: electron with [kinetic] energy of 1 MeV (~2 m0 c2) Total energy: 1 MeV + 511 keV = 1511 keV Momentum and speed follows from E2 = mo2 c4 + p2c2 Total Energy = Kinetic Enery + RestMass*c2

15. Photons: energy en momentum Einstein: postulated light quanta with energy E, frequency ν = E /h, (h is Plankc’s Constant) wavelength λ = c/ν Photons travel with speed of light, and their rest mass is zero. With E2 = mo2 c4 + p2c2 we find, for photons: E = pc, or p = E/c With eV as unit: Photon with energy E (eV) has momentum E (eV/c)

16. Positron Annahilation e+ plus e- forms ‘positronium’: unstable, has a certain decay time, thus ‘half life’ time. Decay, if positronium is at rest: 2 photons (γ-quanta) of 511 keV • Conservation of energy and momentum: • angle between photons: 180 deg. • equal energy, dus 2 photons 511 keV each

17. Suppose the electron has a certain energy such that the positronium, formed with a positron, has a (kinetic!) energy of 12 eV. What is the angle between the emitted photons? This will depend on the difference between the direction of the positronium momentum, and the direction(s) of the photons. Let us first assume that the positronium momentum is perpendicular to the axis of the direction of both photons. The photons will have identical energy: Ephoton = (1022 keV + 0.012 keV) / 2 = 511.006 keV What is the momentum p of the positronium? E2 = mo2 c4 + p2c2 p2c2 = E2 - mo2 c4 = (1022.012)2 – (1022)2 (keV)2 = p2 = 24.5 (keV/c)2 p = 4.95 keV/c

18. In the figure the momenta are shown. The momentum of a photon is the vector sum of 511 keV/c in Y and 4.95/2 keV/c in X. This leads to a photon momentum of √(5112+2.482) = 511.006 keV/c This is in agreement met the energy of 12 eV of the positronium. The angle between the two photons: phi = 2 x atg(551/2.48) = 179.44 deg (almost 180 deg). 4.95 keV/c 511 keV/c

19. Now we consider the direction of the positron momentum to be the same as the axis of direction of the photons: E1, p1 E2, p2 ppos = 4.95 keV/c Here: E1 + E2 = 1022.012 keV E2/c – E1/c = 4.95 keV/c, so E2 – E1 = 4.95 keV so E1 = 508.53 keV E2 = 513.48 keV emitted under 180 deg.

20. Summary of the parameters of a moving particle • Charge: q [usually you don’t need, use, or write q] • RestMass: m0 • Relativistic Mass mr • Speed: v [NOTE: Ek ≠ ½ γm0v2] • Relative speedβ • momentum p • Lorentz factor γ • Total Energy ET • Kinetic Energy Ek • Energy of particle: we mean kinetic energy. • Kinetic energy of particle equals Total Energy minus rest mass energy equivalent. • ET2 = mo2 c4 + p2c2 • β = v/c = γ m0v c / γ m0 c2 = p c/ET • = sqrt(γ2 -1) / γ mr = γ m0 γ = 1 / sqrt(1- 2) ET = m0c2 sqrt(1+ 2γ2)= γ m0c2 = mrc2 EK = ET - m0 c2 p2/2m0 = Ekin[for low velocity particles: non-relativistic Check!!!!!!!]

21. Relation between momentum p, radius of curvature and magnetic field: If magnetic field direction perpendicular to the velocity: (p in GeV/c, B in T, r in m, for 1 elementary charge unit = 1.602177x10-19 C, and obtained using 1 eV/c2 = 1.782663x10-36 kg and c = 299792458 m/s ) p = 0.2998 B r Particle trajectory Relation between radius-of-curvature and sagitta: sagitta (small for high momentum particle) r2 = (r-s) 2 +l2/4 → r = l2/8s + s/2