Relativistic Mechanics 1 SR as a metatheory: laws of physics must be Lorentzinvariant. Laws of classical mechanics are not: they are invariant under Galilean transformations, not Lorentz transformations. => we need a revised, relativistic, mechanics!

Starting point: Assume each particle has an invariant proper mass (or rest mass) m 0 Define 4momentum = m P 0 U Basic axiom: conservation of in particle collisions: * = 0 P Pn (in the sum, P of each particle going into the collision is counted positively while P of each particle coming out is counted negatively) Since a sum of 4vectors is a 4vector, this equation is Lorentzinvariant. 2 = m = m (u) ( , c) ≡ ( , mc) P 0 U 0 u p relativistic mass: m = (u) m 0 relativistic momentum: p = m u

Conservation of 4momentum yields both: * p = 0 * m = 0 These are Newtonian conservation laws when c ∞ What about conservation of energy?

When v/c ≪ 1 :

Einstein proposed the equivalence of mass and energy: E = mc2 => P = (p, E / c) Relativistic energies 3 rest energy: m c2 0 kinetic energy: T = mc2 – m c2 = ( 1) m c2 0 0 Note: Potential energy associated with position of particle in an external electromagnetic or gravitational field does not contribute to the relativistic mass of a particle. When particle's PE changes, it's actually the energy of the field that's changing.

H atom: energy of electron = 13.6 eV (binding energy) => m c2 = m c2 + m c2 – 13.6 eV atom p e => atom has less mass than its constituents!

1 eV = 1.602 × 1012 ergs = 1.602 × 1019 J m = 1.67 × 1024 g ; m = 9.11 × 1028 g p e => m / m = 13.6 eV / (m + m ) c2 ≈ 1.8 × 108 p e Nuclear physicists use the atomic mass unit u, defined so that the mass 4 of the 12C atom is exactly 12 u.

1 u = 931.502 MeV / c2 m = 1.00727647 u ; m = 1.00866501 u ; m = 5.485803 × 104 u p n e

=> m / m ≈ 0.8% for the 12C nucleus

Binding energy ≈ 90 MeV (≈ 7.5 MeV per nucleon) 5 6 = m (u) ( , c) = ( , mc) = ( , E / c) P 0 u p p

=> P2 = E2/c2 – p2

In particle's rest frame: = ( , m c) => 2 = m 2 c2 P 0 0 P 0

=> E2/c2 – 2 = m 2 c2 => E2 = m 2 c4 + 2c2 p 0 0 p

For a photon: m = 0 => E = pc , = E/c ( , 1) 0 P n E = h = hc/ , n = a unit vector in dir of propagation

Since P is a 4vector, E and p transform as follows under a standard LT:

p ' = (p E/c) E' = (E p c) x x x p ' = p y y p ' = p z z Applications 7

1. Compton scattering: photon strikes stationary electron E'

E e Before: . After: e 8

h / m c is called the e “Compton wavelength” of the electron. SP 4.1 9 2. Inverse Compton scattering: photon scatters off a highly energetic charged particle E Before After 2 E m 0 E 1

As before:

=>

Assume E ≫ m c2 ≫ E 0 10 (E ≫ m c2) 0

(from previous slide)

Maximum energy is transferred to photon when cos = 1:

=> 11 Cosmic Microwave Background (CMB): 3 K blackbody spectrum

=> typical photon energy ~ kT = 3 × 104 eV

Very high energy cosmic ray proton: 1020 eV ; m c2 = 938 MeV p => CMB photon can be upscattered to ~ 1019 eV (source of gamma rays)

3. Particle accelerators: “available energy” for creating new particles is the energy in the CM frame

Consider 2 arrangments: a) a 30 GeV proton collides with a proton at rest

b) two 15 GeV protons collide headon (lab and CM frames are identical => 30 GeV of available energy) 12 a) lab frame: = [ m c, (+1) m c ] CM frame: = , E/c) Pi p p P (0

To reach an available energy of 30 GeV in arrangement (a), we would need a 480 GeV proton! 4. Cosmic ray cutoff: highenergy nucleons can undergo the 13 reaction + N N +

Reactions with CMB photons (E ~ 3 × 104 eV) probably produces a cutoff in the cosmic ray spectrum: no CRs with energy above the threshold energy for the reaction. Assuming a headon collision, what's the threshold energy? We need E = (m + m ) c2 in CM frame => = ( , m c + m c) N Pf 0 N

Lab frame, before collision: 14

(m c2 ≈ 940 MeV , m c2 ≈ 140 MeV) N

=> E ≫ m c2 => LHS 2 E => E ≈ 3 × 1014 MeV N N N N

SP 4.28 15 What about forces?

Define 4force = d /d = d(m ) / d = m + (dm /d) F P 0U 0A 0 U

Limited version of Newton's 3rd Law for contact collisions between 2 particles: during contact, is the same for both particles.

+ = d( + ) / d = 0 => = F1 F2 P1 P2 F2 F1

Relativistic 3force f = dp/dt = d(mu)/dt

f becomes Newtonian force when 0

4th component of F is proportional to the power absorbed by the particle. 16

=> F⋅U = proper rate at which particle's internal energy increases

Also:

Equating above 2 expressions for F⋅U => rest masspreserving forces are characterized by: F⋅U = 0 f ⋅u = dE/dt => f ⋅dr = dE (as in Newtonian theory; the added energy ⋅ F = (u) (f, f u /c) is entirely kinetic) 17 4force transformation: (Q ≡ dE/dt)

Note: for rest masspreserving force, f is invariant among IFs with relative velocities parallel to force. SP 4.9 For a rest masspreserving force: 18

a is not generally parallel to f

A moving particle offers different inertial resistances to the same force, depending on whether it is applied longitudinally or transverse. Also for a rest masspreserving force: 19

= m + dm /d = m d2 /d2 = (u) ( , c1 dE/dt) F 0A 0 U 0 R f

=> (u) = m d2 /d2 f 0 r