Assignment 8

Problem 1: 

A pulsar is a neutron star with mass M ≈ 1.4*M_sun ≈ 2.8 × 10³⁰ kg and radius R ≈ 10 km.  The star rotates with angular velocity ω and has a magnetic moment m, which is, in general, not parallel to the rotation axis.
(a)  Describe the radiation emitted by the pulsar, and find the total radiated power, assuming that the angle between the magnetic moment and the rotation axis is α, as in the figure.
(b)  Find the "spin-down rate" (i.e. find ω(t)) of the pulsar, assuming that energy loss is due to radiation only.

Solution:

• Concepts:
  Magnetic dipole radiation
• Reasoning:
  Treat the neutron star as a rotating magnetic dipole with dipole moment m.
  The rotating magnetic dipole will radiate energy at a rate P_rad =  <(d²m/dt²)²>/(6πε₀c⁵).
• Details of the calculation:
  (a)  Let the rotation axis be the z-axis.
  m(t) = m₀ cosθ k + m₀ sinθ (cos(ωt) i + sin(ωt) j),  where θ is tilt angle.
  d²m/dt² = -ω²m₀ sinθ [cos(ωt) i + sin(ωt) j].
  <(d²m/dt²)²> = ω⁴m₀²sin²θ,  P_rad = ω⁴m₀²sin²θ/(6πε₀c⁵).
  
  (b)  The rotational kinetic energy of the neutron star is ½Iω² = (1/5)MR²ω².
  P_rad(t) = -dE/dt = -(d/dt)((1/5)MR²ω²) = -(2/5)MR²ω dω/dt.
  ω⁴m₀²sin²θ/(6πε₀c⁵) = -(2/5)MR²ω dω/dt.
  -5 m₀²sin²θ dt/(12 MR²πε₀c⁵) =   dω/ω³.
  [-5 m₀²sin²θ /(12 MR²πε₀c⁵)] ∫₀^t dt =  ∫_ω(0) ^ω(t)dω/ω³.
  [5 m₀²sin²θ /(12 MR²πε₀c⁵)] t = ½(ω(t)^-2 - ω(0)^-2).
  5 m₀²sin²θ t/(6 MR²πε₀c⁵) = 1/ω(t)² - 1/ω(0)².
  1/ω(t)² = 1/ω(0)² + 5 m₀²sin²θ t/(6 MR²πε₀c⁵).
  ω(t)² = ω(0)²/(1 + 5 m₀²sin²θ ω(0)²t/(6 MR²πε₀c⁵)).
  ω(t) = ω(0)/(1 + t/τ)^½, with τ = 6 MR²πε₀c⁵/(5 m₀²sin²θ ω(0)²).

Problem 2:

A spherical shell of radius R carrying a uniform surface charge σ is set spinning at angular velocity ω.

(a)  Find the vector potential A that it produces at point P located at a distance s, s > R from the center of the sphere.
(b)  Find the magnetic field B outside the sphere.
(c)  How does the result for A change if r < R?
(d)  Find B inside the sphere.

Solution:

• Concepts:
  Boundary conditions, the uniqueness theorem
• Reasoning:
  One form of the uniqueness theorem for magnetostatics:
  If the current density j is specified throughout a volume V and either A or B are specified at the boundaries of a volume V, then a unique solution exists for B inside V.
  In magnetostatics we choose ∇·A = 0. 
  Then  ∇²A = -μ₀j,  A(r) = (μ₀/(4π)) ∫_V'dV' j(r')/|r - r'|.  (SI units).
  Boundary conditions in magnetostatics:
  (B₂ - B₁)·n₂ = 0,  (B₂ - B₁)·t = μ₀k·n.
  A is continuous across the boundary.
  There are different approaches.  One is outlined below.
  We can solve for the magnetic field of a rotating spherical shell with uniform surface charge density σ, by treating the problem as a boundary value problem. 
  Let the radius of the sphere be R, its center be located at the origin, and let the sphere rotate with angular velocity ω = ωk.
  The rotating surface charge density produces a surface current density, k_m = σω x R, 
  k_m = σωRsinθ e_φ.
  For r >> R we will have a dipole potential,
  A(r) = (μ₀/(4π))m x r/r³, or A_φ = (μ₀/4π)msinθ/r².
  To find the dipole moment m, we consider the rotating sphere to be a stack of current loops.
  For a current loop at θ we have dm = IA = Rσω sinθ Rdθ πR²sin²θ = R⁴σωπ sin³θdθ.
  Integrating dm from θ = 0 to θ = π we find m = (4/3)R⁴σωπ, m = mk.
  [∫sin³xdx = -(1/3)cosx (sin²x + 2)]
  For r >> R we have A_φ = (μ₀/3)R⁴σωsinθ/r².
  For any r > R assume that A(r) only has φ component.
  Assume that the 1/r² term is the only term in the expansion of A in powers of 1/rⁿ outside of the sphere, i.e. assume that outside the sphere the potential is a pure dipole potential.
  Since A must be continuous at r = R and A must be finite at the origin,
  assume that A_φ = Csinθr = (μ₀/3)Rσωsinθr  inside the sphere and A = A_φe_φ.
  If this solution fulfills the boundary conditions,
  (B₂ - B₁)·n₂ = 0,  (B₂ - B₁)·t = μ₀k·n,
  we have the only solution.
  
  B = ∇×A,  
  B_{in_r} = (1/)rsinθ))∂(sinθA_{φ_in})/∂θ = (2μ₀/3)Rσω cosθ.
  B_{in_θ} = -(1/r)∂(rA_{φ_in})/∂r = -(2μ₀/3)Rσω sinθ.
  B_{out_r} = (1/(rsinθ))∂(sinθA_{φ_out})/∂θ = (2μ₀/3)R⁴σω cosθ/r³.
  B_{out_θ} = -(1/r)∂(rA_{φ_out})/∂r = (μ₀/3)R⁴σωsinθ/r³.
  
  Now check that the boundary conditions are satisfied.
  At r = R the radial component of B is continuous.
  B_{out_θ} - B_{in_θ} = μ₀R⁴σω sinθ/R³ = μ₀k_m.
  The boundary conditions are satisfied and we have the only solution.
  The field outside the sphere is a pure dipole field,
  B(r,θ,φ) = (μ₀/3)R⁴σω (2cosθ/r³ e_r + sinθ/r³ e_θ),
  and the field inside the sphere is constant,  B(r,θ,φ) = k(μ₀/4π)2m/R³.
• Details of the calculation:   
  (a)  A_out (r,θ,φ) = (μ₀/3)R⁴σωsinθ/r² e_φ.
  (b)  B_out (r,θ,φ) = (μ₀/3)R⁴σω/r³ (2cosθ e_r + sinθ e_θ).
  (c)  A_in(r,θ,φ) = (μ₀/3)Rσωsinθ r e_φ.
  (d)  B_in(r) = (μ₀/4π)2m/R³ k.

Problem 3:

Assume you have two very small spheres with radius a, made of lih material with magnetic susceptibility Χ_m > 0.
The spheres are separated by a distance  R >> a .
Assume that the magnetic dipole moment m of each sphere is induced by the magnetic field due to the magnetic dipole moment of the other sphere.
Choose your coordinate system so that sphere 1 is located at the origin and its magnetic dipole moment points in the z-direction.
(a)  Where do you have to place sphere 2, so that the dipole moments of the two spheres are aligned?
(b)  Where do you have to place sphere 2, so that the dipole moments of the two spheres are anti-aligned?

Solution:

• Concepts:
  The dipole field
• Reasoning:
  The magnetic field outside the tiny sphere is a pure dipole field.  Since the spheres are tiny, we can assume that the external magnetic field inside the spheres is equal to that at the center of the spheres.
  For linear, isotropic, homogeneous (lih) magnetic materials we have
  M = Χ_mH,  B = μ₀(H + M) = μ₀(1 + Χ_m)H =  μ₀κ_mH =  μH.
  Χ_m > 0 for paramagnetic materials.
• Details of the calculation:
  (a)  Assume both spheres have a dipole moment m = mk.  At a distance r, the magnetic field produced by sphere 1 is
  B(r) = (μ₀/4π)(3(m·r)r/r⁵ - m/r³) = (μ₀m/(4πr³))[2 cosθ (r/r) + sinθ (θ/θ)],
  H(r) = (1/4π)(3(m·r)r/r⁵ - m/r³) = (m/(4πr³))[2 cosθ (r/r) + sinθ (θ/θ)].
  On the z - axis H(z) = 2m/(4π|z|³).
  If sphere 2 is located on the z-axis and the distance between the spheres is d, then the magnetic field at the location of each sphere is due to the dipole moment of the other sphere is H(d) = 2m/(4πd³).
  We want m = V(3χ_m/(3 + χ_m))H(d), with V = 4πa³/3.
  Then m = V(3χ_m/(3 + χ_m))2m/(4πd³).
  Solve for d:  d³ = V(6χ_m/(3 + χ_m))/(4π).
  
  (b)  In the xy plane, a distance ρ from the magnetic field produced by sphere 1 points in the opposite direction of its magnetic moment.
  H(ρ) = -m/(4πρ³) k.
  If sphere 2 is located in the xy plane and the distance between the spheres is d, then the magnetic field at the location of each sphere is due to the dipole moment of the other sphere is H(d) = -m/(4πd³).
  We want m = V(3χ_m/(3 + χ_m))H(d), with V = 4πa³/3.
  Then m = V(3χ_m/(3 + χ_m))m/(4πd³). (magnitude)
  Solve for d:  d³ = V(3χ_m/(3 + χ_m))/(4π).

Problem 4:

At t = 0 a particle of mass m and charge q moves in the xy-plane in a circular orbit of radius R in a uniform magnetic field B = Bk.  The particle will loose kinetic energy through radiation.  Assume that the energy loss per revolution is small compared to the total energy of the particle. 
(a)  Neglecting radiation, what is the magnitude of the particles momentum at t = 0?  Does your answer depend on the speed of the particle (relativistic or non-relativistic)?
(b)  Now assume the particle is moving non-relativistically.  Derive an expression for the time it takes the particle to loose half of its kinetic energy.

Solution:

• Concepts:
  Motion of a charged particle in a magnetic field, the radiation field of a point charge moving non-relativistically, the Larmor formula
• Reasoning:
  In a magnetic field a moving charged particle is acted on by a force perpendicular to the direction of its velocity.  Therefore this force changes the direction of the velocity, but not the energy of the particle.  This acceleration results in power loss via radiation.  The power radiated is given by the Larmor formula.
• Details of the calculation:
  (a)  F = qv×B,  dp/dt = γmdv/dt = γma.
  For circular motion a = v²/R pointing towards the center of the circle.
  γmv²/R = qvB,  R = γmv/(qB) = p/(qB).  p = RqB.
  (b)  For a non-relativistic particle mv = rqB, r = mv/qB.
  We assume that the orbital radius r changes very little while the particle completes one revolution, so that the acceleration is always given by v²/r.
  From the Larmor formula we have
  P = (2/3)[q²/(4πε₀c³)]a² = -dE/dt.
  a = v²/r = 2E/(mr).  Since r = mv/qB,we have
  a² = 2Eq²B²/m³.
  dE/dt = -[q⁴B²/(3π m³ε₀c³)]E.
  E = E₀exp(-t/τ), where τ = (3πε₀c³m³)/(q⁴B²)
  ½ = exp(-t/τ), t = τ ln(2) is the time it takes the particle to loose half of its kinetic energy.