Relativististic Mechanics Dec 15, 2014 With respect to classical Newtonian mechanics: - Special Theory of Relativity:  = v/c m = m 0 / sqrt(1-  2 ) - Introduction of ElectronVolt as new unit of energy: 1 eV = |q| Joules 1

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

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

4

 = v / c, and the Lorentz factor γ: relativistic mass m r = γ m 0 γ = 1 / sqrt(1-  2 ), and  = sqrt(γ 2 -1) / γ  = v / c 5

6

Essential in Relativistic Mechanics: Total Energy of system For a moving particle: Total Energy = Kinetic Energy + Rest Mass Energy equivalent 7

8

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: E T 2 = m o 2 c 4 + p 2 c 2 looks like ‘rest mass energy’looks like ‘kinetic energy’ but this equation concerns E T 2 !! 9

With van de Graaff accelerator: simple: E = q V, so E = V eV From Einstein’s Special Theory on Relativity: E 2 = m o 2 c 4 + p 2 c 2 With:  = v / c, and the Lorentz factor γ: relativistic mass m r = γ m 0 γ = 1 / sqrt(1-  2 ), and  = sqrt(γ 2 -1) / γ So: total energy E = m 0 c 2 sqrt(1+  2 γ 2 ) [= rest mass + kinetic energy] = γ m 0 c 2 = m r c 2 10

Speed of electron after being accelerated with 25 kV? Total energy = 511 keV + 25 keV = 536 keV = γ. 511 keV So γ = 536/511 =  = sqrt(γ 2 -1) / γ = 0.30 Dus v = 0.30 c 11

Remember: TOTAL energy E 2 = m o 2 c 4 + p 2 c 2 Note ‘restmass’ term and ‘kinetic’ term (squared!) relativistic mass m r = γ m 0 for high energy particles: p = γ m 0 c γ = 1 / sqrt(1-  2 ) For high-energy particles (E >> m 0 c 2 ): E 2 = m o 2 c 4 + p 2 c 2 = E 2 = p 2 c 2  E = pc  p = E/c independent of particle’s mass! Rest Masses photon: 0 (zero) electron: 511 keV/c 2 muon: GeV/c 2 pion +- : GeV/c 2 pion 0 : GeV/c 2 proton: GeV/c 2 neutron: GeV/c 2 12

Examples: electron: rest mass m 0 = 511 keV 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 m 0 c 2 ) Total energy: 1 MeV keV = 1511 keV Momentum and speed follows from E 2 = m o 2 c 4 + p 2 c 2 Total Energy = Kinetic Enery + RestMass*c 2 13

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 E 2 = m o 2 c 4 + p 2 c 2 we find, for photons: E = pc, or p = E/c With eV as unit: Photon with energy E (eV) has momentum E (eV/c) 14

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 15

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: E photon = (1022 keV keV) / 2 = keV What is the momentum p of the positronium? E 2 = m o 2 c 4 + p 2 c 2 p 2 c 2 = E 2 - m o 2 c 4 = ( ) 2 – (1022) 2 (keV) 2 = p 2 = 24.5 (keV/c) 2 p = 4.95 keV/c 16

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 √( ) = 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) = deg (almost 180 deg). 511 keV/c 4.95 keV/c 17

Now we consider the direction of the positron momentum to be the same as the axis of direction of the photons: E 1, p 1 E 2, p 2 p pos = 4.95 keV/c Here: E 1 + E 2 = keV E 2 /c – E 1 /c = 4.95 keV/c, so E 2 – E 1 = 4.95 keV so E 1 = keV E 2 = keV emitted under 180 deg. 18