Problem 1: 

Find the total cross section for a free cosmic object (e.g. meteorite) to fall into the Sun.  The object of mass m₁ and at infinite distance from the Sun has velocity v_∞ relative to the Sun.   The mass of the Sun is m₂ >> m₁ and the Sun's radius is R.  Compare your result with the geometrical cross section of the Sun.  When does your result approach the geometrical cross section, and when is it very different?

Solution:

• Concepts:
  Motion in a central potential.
• Reasoning:
  The object has a hyperbolic orbit.  Energy E and angular momentum L are conserved, the motion is in a plane.
  L = m₁bv_∞ = m₁r_cv_c, where r_c is the distance of closest approach, and v_c is the speed at r_c.
  The potential energy of the object is U(r) = -Gm₂m₁/r.  If r_c < R, the object will crash onto the sun.
  We find the relationship between r_c and b, and find the largest impact parameter b_max for which r_c < R.
  The cross section for crashing onto the sun is σ = πb_max².
• Details of the calculation:
  For motion in a central field energy E and angular momentum L are conserved.
  E = ½m₁v_∞² = ½m₁v_c² - Gm₂m₁/r_c.  v_c² = v_∞² + 2Gm₂/r_c.
  L = m₁bv_∞ = m₁r_cv_c,  v_∞ = r_cv_c/b.
  Solve for b in terms of r_c.
  b = r_cv_c/v_∞ = r_c(1 + 2Gm₂/(r_cv_∞²))^½.  b_max = R(1 + 2Gm₂/(Rv_∞²))^½.
  σ = πR²(1 + 2Gm₂/(Rv_∞²))^½.
  The geometrical cross section is πR².
  When v_∞ --> ∞, the crash cross section approaches the geometrical cross-section.  However, when v_∞ is small, for example when Sun is passing through a galactic gas cloud of cold atoms that is co-moving with the Solar system in the galaxy, the crash cross section can be substantially larger than the geometrical cross section.

Problem 2:

A bead is constrained to move without friction on a helix whose equation in cylindrical polar coordinates is ρ = b, z = aΦ under the influence of the potential V = ½k(ρ² + z²).  
(a)  Use the Lagrange multiplier method and find the appropriate Lagrangian including terms expressing the constraints. 
(b)  Apply the Euler-Lagrange equations to obtain the equations of motion. 
(c)  Next, repeat parts (a) and (b) without using the Lagrange multiplier method.  Instead, build the constraints into the general coordinate(s).

Solution:

• Concepts:
  Lagrangian Mechanics, the Lagrange multiplier method
• Reasoning:
  We are instructed to use the Lagrange multiplier method to solve the problem.
• Details of the calculation:
  (a)  Lagrangian not incorporating the constraints:
  L = ½m[(dρ/dt)² + ρ²(dΦ/dt)² +(dz/dt)²] - ½k(ρ² + z²).  
  Put the equations of constraint into the form
  Σ_k a_lk dq_k + a_lt dt =  0,  l = 1, ..., m.
  We have two equations of constraint.
  (i)  dρ = 0;  a_1ρ = 1, a_1Φ = a_1z = 0.
  (ii)  dz - adΦ = 0 ;  a_2ρ = 0, a_2Φ = -a, a_2z = 1.
  
  (b)  Lagrange's equation:
  d/dt(∂L/∂(dq_k/dt)) -  ∂L/∂q_k = ∑_lλ_la_lk,   Σ_k a_lk dq_k + a_lt dt =  0.
  We have:
  md²ρ/dt² - mρ(dΦ/dt)² + kρ = λ₁,
  mρ²d²Φ/dt² + 2mρ(dρ/dt)(dΦ/dt) = -aλ₂,
  md²z/dt² + kz = λ₂,
  and
  ρ = b,  dρ/dt = d²ρ/dt² = 0,
  z = aΦ,  dz/dt = adΦ/dt, d²z/dt² = ad²Φ/dt².
  
  We therefore have:
  -mb(dΦ/dt)² + kb = λ₁.
  mb² d²Φ/dt² = -aλ₂.
  ma d²Φ/dt² + kaΦ = λ₂.
  Using the last two equations to solve for λ₂ we have λ₂ = Φ kab²/(a² + b²).
  Therefore the equation of motion is
  d²Φ/dt² + (ka²/(m(b² + a²))Φ = 0.
  λ₁ = -mb(dΦ/dt)² + kb.
  λ₁ is the force of constraint in the radial direction, λ₂ is the force of constraint in the z- direction.
  
  (c)  L = ½m[(adΦ/dt)² + (bdΦ/dt)²] - ½k(b² + (aΦ)²).  
  d/dt(∂L/∂(dq_i/dt)) -  ∂L/∂q_i = 0.
  m(b² + a²)d²Φ/dt² + ka²Φ = 0.
  d²Φ/dt² + (ka²/(m(b² + a²))Φ = 0.
  
  Let C² = ka²/(m(b² + a²)).
  Φ = A exp(-iCt) + exp(iCt) = Φ₀sin(Ct + δ).
  The particle oscillates about z = Φ = 0.  The amplitude and phase of the oscillation are determined by the initial conditions.  The angular frequency of the oscillation is C.

Problem 3:

A particle is constrained to move on the surface of a sphere of radius R₀ centered about the origin (0, 0, 0) in the usual Cartesian coordinates x, y, z. 
A two-dimensional harmonic oscillator potential is applied in the y-z plane.
U(x, y, z) = ½k(y² + z²).
The equation of constraint is x² + y² + z² = R₀².

(a)  Use the Lagrange multiplier method and find the appropriate Lagrangian including terms expressing the constraint. 
(b)  Apply the Euler-Lagrange equations to obtain the equations of motion which must be solved together with the equation of constraint.
(c)  For the initial conditions r(0) = (0, 0, R₀), v(0) = (0, ωR₀, 0), find the complete solution for r(t) and the Cartesian components of the force of constraint for all t.

Solution:

• Concepts:
  Lagrangian Mechanics, the Lagrange multiplier method
  d/dt(∂L/∂(dq_k/dt)) -  ∂L/∂q_k = ∑_lλ_la_lk,    Σ_ka_lk dq_k + a_lt dt =  0.
  Visualizing:
  
• Reasoning:
  The Lagrange multiplier method yields the forces of constraint.
• Details of the calculation:
  (a)  Lagrangian not incorporating the constraints:
  L = ½m[(dx/dt)² + (dy/dt)² +(dz/dt)²] - ½k(y² + z²).  
  Put the equations of constraint into the form
  Σ_ka_lk dq_k + a_lt dt =  0,  l = 1, ..., m.
  We have one equations of constraint.  We may write it as
  xdx + ydy + zdz = 0, λ(a_xdx + a_ydy +a_z dz) = 0.
  
  (b)  The equations of motion are
  md²x/dt² = λx,  md²y/dt² = (λ - k)y,  md²z/dt² = (λ - k)z.
  These must be solved simultaneously with the equation of constraint
  x² + y² + z² = R₀².

  (c) The given initial conditions
  x(0) = 0, v_x(0) = 0, and the equation of motion
  md²x/dt² = -λx
  imply that x(t) = 0.
  The motion is in the yz plane.
  
  The initial conditions
  y(0) = v_z(0) = 0, z(0) = R₀, v_y(0) = ωR₀, and the equations of motion
  md²y/dt² = -(k - λ)y,  md²z/dt² = -(k - λ)z,
  together with the equation of constraint
  y² + z² = R₀²
  imply y = R₀ sinωt, z = R₀ cosωt, ω² = (k - λ)/m, λ = k - mω²,
  
  Let F_c be the force of constraint.  For the given initial conditions we have
  F_cx = 0, F_cy = (k - mω²)y, F_cz = (k - mω²)z.  F_c = (k - mω²)R₀ e_r.
  If k > mω² the force of constraint pushes on the particle, If k < mω² the force of constraint pulls on the particle.
   

Problem 4:

In a scattering experiment, a beam of particles of mass m and energy E is sent towards a target.  The number of particles scattered per unit area per unit time into a specific direction is measured.  The potential energy function U(r) is that of a central attractive potential with a repulsive core.

(a)  Give the definition of the differential scattering cross section dσ/dΩ in terms of the impact parameter b and the scattering angle θ.
(b)  Relate the impact parameter b to the angular momentum, and thus find the dependence of θ(b) on U(r) and b in the form of an integral.
(c)  What is the value of the scattering angle θ for the special cases b = 0 and b = ∞?
(d)  Show that θ should become negative for suitable E and b.
(e)  Given the result of (d)  does the differential cross section ever diverge for any value of b?  Explain!

This problem just asks for definitions and reasoning, not for calculations.

Solution:

• Concepts:
  The scattering cross section, scattering from a spherically symmetric potential
• Reasoning:
  We are asked to think qualitatively about properties of the scattering cross section.
• Details of the calculation:
  (a)  σ(θ) =|(b/sinθ)(db/dθ)|.
  (b)  M = mv₀b = b(2mE)^½,  θ = π - 2φ₀ for a repulsive potential, 
  φ₀ = b∫₀^umaxdu/[1 - b²u² - U(u)/E]^½, with u = 1/r.
  E =  b²Eu²_max + U(u_max) yields u_max.
  (c)  b = 0 yields  φ₀ = 0, θ = π.  (The particle scatters backward, M = 0.)
  b = ∞ yields φ₀ = π/2, θ = 0.  (The particle is out of range of the scattering potential.)
  (d)  If we choose the impact parameter and the energy such that r_min = 1/u_max is greater than the value of r where the potential energy U(r) has a minimum, then the particle never sees the repulsive core and is scattered by an attractive potential, θ = π - 2φ₀ is negative.
  (e)  If we decrease the impact parameter, r_min decreases, the particle will see the repulsive core and θ will become positive.  For some finite impact parameter we will therefore have θ = 0, and σ(θ) diverges.