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Relativistic Mechanics 1 SR As a Meta­Theory: Laws of Physics Must Be Lorentz­Invariant

Relativistic Mechanics 1 SR As a Meta­Theory: Laws of Physics Must Be Lorentz­Invariant

Relativistic 1 SR as a meta­theory: laws of must be Lorentz­invariant. Laws of 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 (or rest mass) m 0 Define 4­ = 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 ?

When v/c ≪ 1 :

Einstein proposed the equivalence of mass and energy: E = mc2 => P = (p, E / c) Relativistic 3 rest energy: m c2 0 : T = mc2 – m c2 = ( ­ 1) m c2 0 0 Note: Potential energy associated with 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 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 : 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 (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 ?

Define 4­ = 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 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 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 /d2 = (u) ( , c­1 dE/dt) F 0A 0 U 0 R f

=> (u) = m d2 /d2 f 0 r