Some history. F p. 1/??

Transcription

1 Some history F p. 1/??

2 F p. 1/?? Some history 1600: Galileo Galilei cf. section 7.0 Johannes Kepler cf. section : Isaac Newton cf. section : Leonhard Euler cf. section 1.4 Jean Le Rond d Alembert cf. section 1.4 Joseph-Louis Lagrange cf. section 1.4, : Carl Gustav Jacob Jacobi cf. section 10.1 William Rowan Hamilton cf. section 2.1 Joseph Liouville cf. section : Albert Einstein cf. section 7.1 Emmy Amalie Noether cf. section 8.2

3 F p. 1/?? Some history 1600: Galileo Galilei cf. section 7.0 Johannes Kepler cf. section : Isaac Newton cf. section : Leonhard Euler cf. section 1.4 Jean Le Rond d Alembert cf. section 1.4 Joseph-Louis Lagrange cf. section 1.4, : Carl Gustav Jacob Jacobi cf. section 10.1 William Rowan Hamilton cf. section 2.1 Joseph Liouville cf. section : Albert Einstein cf. section 7.1 Emmy Amalie Noether cf. section : Vladimir Igorevich Arnold cf. section 11.2 Alexandre Aleksandrovich Kirillov 1936 Bertram Kostant 1928 Jean-Marie Souriau Jerrold Eldon Marsden Alan David Weinstein 1943

4 1. Survey of elementary principles F p. 2/??

5 F p. 2/?? 1. Survey of elementary principles 1.1 Mechanics of a particle Concepts space, time scalar, scalar field (function), vector, vector field coordinate system, origin reference frame, inertial frame, Galilean frame kinematics, dynamics, statics position, velocity, acceleration mass point, point mass inertial mass, gravitational mass, rest mass, equivalence principle force, force field, work, friction simply connected region, curl-free field, conservative force potential energy, potential, kinetic energy, total energy momentum, angular momentum, torque conservation law, conserved quantity, conserved charge

6 F p. 2/?? 1. Survey of elementary principles 1.1 Mechanics of a particle Results Newton s second law conservation of momentum conservation of angular momentum conservation of total energy Formulas (1.3) F = p (1.12) W 12 (P) = P F d s (1.16) F( r) = V ( r) F( r) = f(r) e r f(r) = dv dr Elementary fact Physics is independent of the choice of coordinate system

7 F p. 2/?? 1. Survey of elementary principles 1.1 Mechanics of a particle Results Newton s second law conservation of momentum conservation of angular momentum conservation of total energy Formulas (1.3) F = p (1.12) W 12 (P) = P F d s (1.16) F( r) = V ( r) F( r) = f(r) e r f(r) = dv dr Warning Do not mix up the notions frame and coordinate system. There are systems with F = V but V time-dependent.

8 F p. 3/?? 1. Survey of elementary principles 1.2 Mechanics of a system of particles Concepts internal and external forces distance vector center of mass configuration strong law of action and reaction mechanical equilibrium

9 F p. 3/?? 1. Survey of elementary principles 1.2 Mechanics of a system of particles Results Newton s third law (weak law of action and reaction) for conservative forces depending only on distance center of mass motion conservation of total momentum conservation of total angular momentum conservation of total energy angular momentum as sum of c.m. term and term for relative motion kinetic energy as sum of c.m. term and term for relative motion

10 F p. 3/?? 1. Survey of elementary principles 1.2 Mechanics of a system of particles Formulas (1.19) pi = F ext i + j F ji ( ) F ij = F ji (1.22) M d2 dt 2 R = F ext d (1.26) L dt = N ext (1.27) r i = r i + R m i r i = 0 i (1.28) L = R M v + i r i p i (1.31) T = 1 2 M v i m i v 2 i i ri = e x x i + e y y i + e z z i

11 F p. 4/?? 1. Survey of elementary principles 1.3 Constraints Concepts dynamical and non-dynamical parts of a system constraint holonomic and non-holonomic constraints, semi-holonomic constraints skleronomic and rheonomic constraints independent dynamical variables, generalized coordinate, degree of freedom constraint force, applied force Results presence of constraints = particle positions no longer independent constraint forces are usually not known explicitly

12 F p. 4/?? 1. Survey of elementary principles 1.3 Constraints Concepts dynamical and non-dynamical parts of a system constraint holonomic and non-holonomic constraints, semi-holonomic constraints skleronomic and rheonomic constraints independent dynamical variables, generalized coordinate, degree of freedom constraint force, applied force Formulas (1.37) f κ = f κ ( r 1, r 2,...;t) = 0 (1.38) r i = r i (q 1,q 2,...,q N ;t) q j = q j ( r 1, r 2,..., r N ;t)

13 F p. 4/?? 1. Survey of elementary principles 1.3 Constraints Concepts dynamical and non-dynamical parts of a system constraint holonomic and non-holonomic constraints, semi-holonomic constraints skleronomic and rheonomic constraints independent dynamical variables, generalized coordinate, degree of freedom constraint force, applied force Examples rigid body bead sliding on a wire disk rolling on a plane Warning A generalized coordinate need not have dimension of length.

14 3. The central force problem F p. 5/??

15 F p. 5/?? 3. The central force problem 3.10 Scattering in a central force field Concepts scattering, scattering angle, trajectory beam, flux density solid angle, cross section, impact parameter, periapsis total cross section, long and short range potentials Rutherford scattering long range potential spiraling, rainbows, glory scattering Results Rutherford cross section

16 F p. 5/?? 3. The central force problem 3.10 Scattering in a central force field Formulas (3.88) σ( Ω) dω = n I (3.89) dω = 2π sinϑdϑ (3.90) l = s 2mE (3.93) σ(ϑ) = s sinϑ (3.94) ϑ = π 2Ψ (3.96) ϑ = π 2 s ϑ r min r ) (3.101) s(ϑ) = k 2E cot( ϑ 2 (3.102) σ(ϑ) = k2 16E 2 sin 4( ϑ 2 sdr r 2 r 2 V (r) E ) s2 with k = ZZ e 2

17 F p. 6/?? 3. The central force problem 3.10 The scattering problem in laboratory coordinates Concepts recoil, laboratory frame scattering angles in laboratory and center of mass frames elastic and inelastic scattering, excitation energy

18 F p. 6/?? 3. The central force problem 3.10 The scattering problem in laboratory coordinates Concepts recoil, laboratory frame scattering angles in laboratory and center of mass frames elastic and inelastic scattering, excitation energy Formulas (3.108) ρ = µ v 0 m 2 v 1 = m 1 v 0 m 2 v cosϑ + ρ (3.110) cosθ = 1 + 2ρ cosϑ + ρ 2 ( 1 + 2ρ cosϑ + ρ 2 ) 3/2 (3.116) σ lab (θ) = 1 + ρ cosϑ σ c.m. (ϑ) θ = ϑ 2 and θ max = π 2 for ρ=1 σ lab (θ) = 4 cosθσ c.m. (2θ) for ρ=1