Radiating dipoles

Problem:

An electric dipole p₀ vibrates with frequency ω.
(a)  How will the total radiated power change if the frequency is doubled?
(b)  Find the ratio of the differential power in the direction of θ = 45^o to the dipole's axis to the differential power in the direction perpendicular to the axis. 

Solution:

• Concepts:
  Radiation of an oscillating dipole
• Reasoning:
  We are asked to characterize the power radiated by an oscillating dipole.
• Details of the calculation:
  (a)  The radiation field of an electric dipole oscillating around the origin at with p(t) = p₀cos(ωt) is  E_R(r,t) = -(1/(4πε₀c²r))(d²p_⊥(t-r/c)/dt²).  (SI units).
  Here p₀ = p₀(z/z), so E_R(r,t) = -(1/(4πε₀c²r))ω²p₀ cos(ω(t - r/c))sinθ (θ/θ).
  E proportional to sinθ.  P proportional to E² proportional to sin²θ.  E proportional to ω².  P proportional to ω⁴. 
  If the frequency is doubled, the total power radiated will increase by a factor of 16.
  (b)  P(45^o)/P(90^o) = sin²(45^o)/sin²(90^o) = ½.

Problem:

An electric dipole p(t) = p₀(z/z)cosωt is placed at the origin. 
(a)  Determine the radiation fields on the z-axis.
(b)  Determine the radiation fields for a direction normal to the z-axis.
(c)  How does the intensity of the radiation vary as a function of θ?
(d)  How does the total power radiated vary as a function of ω?

Solution:

• Concepts:
  The radiation field of a point charge moving non-relativistically, the Larmor formula
• Reasoning:
  A dipole can be viewed as a positive and a negative point charge oscillating 180^o out of phase.  Oscillating charges are accelerating charges.
• Details of the calculation:
  (a)  The radiation field of an electric dipole oscillating around the origin at with p(t) = p₀cos(ωt) is
  E_R(r,t) = -(1/(4πε₀c²r))(d²p_⊥(t - r/c)/dt²).  (SI units)
  Here p₀ = p₀(z/z), so E_R(r,t) = -(1/(4πε₀c²r))ω²p₀ cos(ω(t - r/c))sinθ(θ/θ).
  B_R(r,t) = (r/(rc))×E_R(r,t) = -[1/(4πε₀c³r)]ω²p₀ cos(ω(t - r/c))sinθ(φ/φ).
  On the z-axis sinθ = 0, there is no radiation along the z-axis.
  (b)  Perpendicular to the z-axis we have sinθ = 1,
  E_R(r,t) = -(1/(4πε₀c²r))ω²p₀ cos(ω(t - r/c))(θ/θ),
  B_R(r,t) = -[1/(4πε₀c³r)]ω²p₀ cos(ω( t- r/c))(φ/φ).
  (c)  S(r,t) = [1/(4πε₀)²][1/(c⁵r²μ₀)]ω⁴p₀² cos²(ω(t - r/c))sin²θ (r/r).  The intensity varies as sin²θ.
  (d)  <P> = <∫S∙dA> = ω⁴p₀²/(12πε₀c³), proportional to ω⁴.

Problem:

A thin spherical shell of radius a has a surface potential Φ(a,θ) = Φ₀ cosθ, where θ is the usual polar angle relative to the z-axis, and Φ₀ is a constant. 
(a)  Find the electric potential Φ(r,θ) everywhere outside the shell.  Calculate the equivalent electric dipole moment p that produces this potential Φ.
(b)  Assume that previously calculated dipole moment acquires a cos(ωt) time dependence.  Calculate the magnetic field B and the electric field E in the radiation zone.
(c)  Find the time-averaged power d<P>/dΩ radiated per unit solid angle.  Express your result in terms of ω, θ, a, Φ₀, ε₀ and c.
(d)  Find the total power <P> radiated.

Solution:

• Concepts:
  Boundary value problems, the radiation of an oscillating dipole
• Reasoning:
  The electric field outside the shell is a pure dipole field.  A dipole can be viewed as a positive and a negative point charge oscillating 180^o out of phase.  Oscillating charges are accelerating charges.
  Details of the calculation:
• Details of the calculation:
  (a)  Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ) is the most general solution inside and outside of the sphere, since ρ = 0 inside and outside of the sphere.  To find the specific solution we apply boundary conditions.
  Assume Φ₂(r,θ) = (B/r²)P₁(cosθ) = (B/r²)cosθ outside the sphere.  The symmetry of the situation suggests this form of the solution.  If we can satisfy the boundary conditions, the uniqueness theorem guaranties that we have found the only solution. 
  Boundary conditions:
  The potential vanishes as r goes to infinity. 
  Φ is continuous across the boundary.
  Φ₀ cosθ = (B/a²)cosθ,  B = Φ₀a²,  the potential outside the shell is Φ(r,θ) = Φ₀ (a²/r²)cosθ.
  The potential of a dipole p = pk is  Φ(r,θ) = k p cosθ/r², where k = 1/(4πε₀).
  So p = 4πε₀ a²Φ₀ is the magnitude of the equivalent electric dipole moment.

  (b)  The radiation field of an electric dipole oscillating around the origin at with p(t) = p₀cos(ωt) is
  E_R(r,t) = -(1/(4πε₀c²r))(d²p_⊥(t - r/c)/dt²).  (SI units)
  Here p₀ = p₀(z/z), so E_R(r,t) = -(1/(4πε₀c²r))ω²p₀ cos(ω(t - r/c))sinθ (θ/θ).
  B_R(r,t) = (r/(rc))×E_R(r,t) = -[1/(4πε₀c³r)]ω²p₀ cos(ω(t - r/c))sinθ (φ/φ).
  
  (c)  S(r,t) = [1/(4πε₀)²][1/(c⁵r²μ₀)]ω⁴p₀² cos²(ω(t - r/c))sin²θ (r/r).  The intensity varies as sin²θ.
  d<P>/dΩ = r²<S(r,t)> = 2πε₀²ω⁴a⁴Φ₀²sin²θ/c³.

  (d)  <P> = <∫S·dA> = ω⁴p₀²/(12πε₀c³) =  4πε₀ω⁴a⁴Φ₀²/(3c³), proportional to ω⁴.

Problem:

An electric dipole p₀ makes an angle of θ degrees with respect to the z-axis as it rotates about the z-axis with angular speed ω.  Find its initial rate of energy loss.

Solution:

• Concepts:
  Electric dipole radiation
• Reasoning:
  The x and y components of the dipole moment are oscillating with angular frequency ω.  The rotating electric dipole will radiate energy at a rate P_rad = <(d²p/dt²)²>/(6πε₀c³).
• Details of the calculation:
  p(t) = p₀ cosθ k + p₀ sinθ (cos(ωt) i + sin(ωt) j), where θ is the tilt angle.
  d²p/dt² = -ω²p₀ sinθ [cos(ωt) i + sin(ωt) j].
  <(d²p/dt²)²> = ω⁴p₀²sin²θ,   P_rad = ω⁴p₀²sin²θ/(6πε₀c³).

Problem:

A small current loop of area A and N turns carries a sinusoidally varying current I(t) = I₀sin(ωt).
(a)  Find the magnetic moment of the loop.
(b)  Find the average power radiated by the loop.

Solution:

• Concepts:
  Magnetic dipole radiation
• Reasoning:
  We are asked to find the average power radiated by a sinusoidally varying magnetic dipole.
• Details of the calculation:
  (a)  m(t) = NAI₀sin(ωt).  The direction of m is given by the right-hand rule.
  (b)  Larmor formula:  The magnetic dipole will radiate energy at a rate P_rad = <(d²m/dt²)²>/(6πε₀c⁵).
  d²m(t)/dt² = -ω²NAI₀sin(ωt). 
  <P_rad> = ω⁴N²A²I₀²/(12πε₀c⁵).

Problem:

A rotating neutron star slows down due to dipole radiation.   Assume that a neutron star has a surface magnetic field strength of 100 Mega Tesla at the poles, a radius of 10 km, and nuclear density.  The neutron star's magnetic axis is tilted by 45 degrees relative to the axis of rotation of the neutron star.  How long will it take for the neutron star's rotation rate to decrease by a factor of 3 if the initial rotation rate is 10⁵ revolutions per second?

Solution:

• Concepts:
  Magnetic dipole radiation
• Reasoning:
  Treat the neutron star as a uniform magnetized sphere.  The magnetic field inside the sphere is uniform and the magnetic field outside the sphere is a pure dipole field.
  B_out = (μ₀/4π)[3(m₀∙r)r/r⁵ - m₀/r³].  Her m₀ is the dipole moment of the sphere.
  At the poles the sphere B_out = B_in = (μ₀/4π)(2m₀/R³). 
  The dipole moment of the sphere therefore has magnitude m₀ = (4π/μ₀)(B_inR³/2).
  We therefore have m₀ = 5*10²⁶ Am².
  The rotating magnetic dipole will radiate energy at a rate P_rad =  <(d²m/dt²)²>/(6πε₀c⁵).
• Details of the calculation:
  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).
  Initially P_rad(0) = 2.41*10⁴³ J/s.
  P_rad(t) = -dE/dt = -(d/dt)(½(2/5)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 =  ∫_ω0 ^ω0/3 dω/ω³.
  [5 m₀²sin²θ/(12 MR²πε₀c⁵)] T = ½((ω₀/3)^-2 - ω₀^-2).
  T = ((ω₀/3)^-2 - ω₀^-2)(16 ρ R⁵π²ε₀c⁵)/(10 m₀²sin²θ ).
  The average radius of a nucleus with A nucleons is R = R₀A^(1/3), where R₀ = 1.2*10^-15 m.
  The mass of a nucleus is A times the mass of a nucleon, m_nucleon ~ 1.6*10^-27 kg.
  The nuclear density is therefore ρ = m_nucleon/((4/3)πR₀³) ~ 2*10¹⁷ kg/m³.
  Plugging in numbers:  T ~ 1.1*10⁶ s  ~ 13 days.