Point charges in damped periodic motion

Problem:

(a) An electron orbits initially, at time t = 0 around a proton at a radius a₀ equal to the Bohr radius.  Using classical mechanics and classical electromagnetism derive an expression for the time it takes for the radius of the orbiting electron to decrease to zero due to radiation.  Here you may assume that the energy loss per revolution is small compared to the total energy of the atom.
(b)  What implication can you draw from this calculation?  Give a qualitative argument on the need to modify the above estimate.

Solution:

• Concepts:
  The radiation field of a point charge moving non-relativistically, the Larmor formula
• Reasoning:
  The electron orbits the "infinitely heavy" proton.  Its acceleration is a = v²/r, and it therefore radiates away energy,
  P = -dE/dt = [q²/(6πε₀c³)]a².  (SI units)
  We assume that the energy loss per revolution is very small so that we can use the equations for a circular orbit.
• Details of the calculation:
  (a)  To solve for the total time it takes the electron to spiral into the nucleus, we can express the acceleration a in terms of E.  We then have
  -dE/dt = f(E), ∫dt = Δt = ∫_E0^-∞dE/f(E).
  Alternatively, we can express E as a function of the radius r and derive an expression -dr/dt = f(r).  We then have
  ∫dt = ΔT = ∫_r0⁰dr/f(r).
  
  (i)  We find a in terms of E for a circular orbit.
  F = q²/(4πε₀r²) = mv²/r.  mv² = q²/(4πε₀r).
  E = ½mv² - q²/(4πε₀r) = -q²/(8πε₀r) = -½mv².
  a = v²/r = q²/(4πε₀r²m) = 2E²8πε₀/(q²m).
  -dE/dt = E⁴16²πε₀/(6c³q²m²).  dt = -(dE/E⁴) (6c³q²m²)/(16²πε₀).
  Δt = (6c³q²m²)/(16²πε₀)∫_E0^-∞(dE/E⁴)  = -(6c³q²m²)/(16²πε₀)(E₀^-3/3),
  where E₀ = -13.6 eV,  Δt = 1.5*10^-11s.
  or
  (ii)  We express a and E as a function of the radius r for a circular orbit.
  a = F/m = q²/(4πε₀r²m),  E = -q²/(8πε₀r),
  dE = [q²/(8πε₀r²)]dr = -Pdt.
  dt = -[q²/(8πε₀r²)]dr/P = -[q²/(8πε₀r²)]dr/[q²a²/(6πε₀c³)].
  dt = -dr[3c³/(4r²a²)] = -r²dr[12c³π²ε₀²m²/q⁴].
  Δt = -[12c³π²ε₀²m²/q⁴]∫_r0⁰r²dr = [12c³π²ε₀²m²/q⁴](r₀³/3),
  where r₀ = Bohr radius = 5.29*10^-11m,  Δt = 1.5*10^-11s.
  (b)  This result implies that the atom is unstable.  Since atoms exist, the classical theory needs to be modified.

Problem:

A non-relativistic particle of mass m and charge q is initially moving with velocity v(0) = v₀i in a uniform magnetic field B = Bk.
(a)   At subsequent times the particle (neglecting radiation) moves in a circle of radius R.  Find R in terms of m, v₀, q, and B.
(b)  Now treat the loss of kinetic energy through radiation as a small perturbation on this circular orbit.  Use the Larmor formula and show that the radius of the circular orbit evolves approximately as R(t) = R(0)exp(-t/τ), where τ = (6πε₀c³m³)/(q⁴B²) in SI units.
(c)  Calculate the total energy radiated by the particle and show that it equals ½mv₀².

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,  mv₀²/R = qv₀B,  R = mv₀/(qB) = p₀/(qB).
  (b)  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 = q²B²R/m², so dE/dt = -(2/3)[q²/(4πε₀c³)]a² = -(2/3)[q²/(4πε₀c³)]q⁴B⁴R²/m⁴.
  We can also write the energy of the particle as a function of the orbital radius
  E = ½mv² = ½m(qBR/m)² = q²B²R²/(2m).
  Therefore
  dE/dt = [q²B²/m]R(dR/dt).
  Combining the two expressions for dE/dt we have
  [q²B²/m]R(dR/dt) = -(2/3)[q²/(4πε₀c³)]q⁴B⁴R²/m⁴.
  dR/dt = -(2/3)[q⁴B²/(4πε₀c³m³)] R.

Problem:

(a)  A particle of mass m and charge q is suspended from a light spring of spring constant k.  If it is set into oscillation, how long will it takes the oscillator to radiate away half its energy?
(b)  A proton is accelerated by a uniform field to an energy of 20 MeV over a distance of 20 m.  How much energy is radiated away?

Solution:

• Concepts:
  The radiation field of a point charge moving non-relativistically, the Larmor formula
• Reasoning:
  Accelerating charges radiate.  At large distance the radiation fields, decreasing inversely with distance, dominate.
• Details of the calculation:
  (a)  For an undamped oscillator suspended from a spring near the surface of the earth we have
  d²y/dt² = -(k/m)y,  where y is the displacement from the equilibrium position.
  [The equilibrium position is not the length L of the unstretched spring but L + mg/k below the support.]
  An oscillating charge is an accelerating charge.  The Larmor formula gives the power radiated by an accelerating charge.
  P = 2e²a²/(3c³).
  To solve for the time it takes the charge to radiate away half its energy, we try to express the acceleration a in terms of the energy E.
  We then have -dE/dt = f(E), ∫dt = Δt = ∫_E0^E0/2dE/f(E).
  We assume that the oscillator does not loose a significant fraction of its energy during one cycle.  We can then assume that the average power radiated is proportional to the average of the square of the acceleration, both averages over one cycle.
  y = A cos(ωt),  a = -ω²A cos(ωt), <a²> = ω⁴A²/2.
  For an oscillator E = ½mω²A², therefore <a²> = ω²E/m.
  P = q²a²/(6πε₀c³) = q²ω²E/(6πε₀c³m) =-dE/dt.
  E = E₀exp(-t/τ), where τ = (6πε₀c³m)/(q²ω²)
  E = E₀/2 if t = t_½ = τ*ln2,  t_½ = (6πε₀c³m)/(q²ω²)ln2.
  This is the time it takes the oscillator to radiate away half its energy.
  (b)  The rest energy of the proton is 938 MeV, its maximum kinetic energy is 20 MeV.
  g_max938 = 958,  g_max = 1.02 ≈ 1.  The problem can be treated nonrelativistically.
  An accelerating charged particle radiates away energy.  If for a time T the magnitude of the acceleration is a, then the energy radiated E_rad = 2e²a²T/(3c³) is given by the Larmor formula.  If the particle is initially at rest, and we neglect radiation, then the kinetic energy of the particle after time T will be E_kin = ½ma²T².
  E_rad << E_kin, if 2e²a²T/(3c³) >> ½ma²T²,  T >> (4/3)e²/(mc³) = (4/3)r₀/c, with r₀ = e²/(mc³).  For a proton r₀ ≈ 1.5*10^-18m.
  If T is much greater than the time it takes light to travel ~2*10^-18m then we can neglect radiation losses when considering the short term motion.
  Here ½ma²T² = 20*1.6*10^-13J. 
  a = qE/m = (20*10⁶V/20m)(1.6*10^-19C/1.67*10^-27kg).
  T = 6.46*10^-7s.
  E_rad = 2e²a²T/(3c³) = 2(q²/(4πε₀))a²T/(3c³) = 3.4*10^-32J = 2.1*10^-13eV = energy radiated away.

Problem:

Consider a fixed spherical charge distribution of radius R and uniform negative charge density -ρ, centered at the origin.  A positive point charge q with mass m interacts with this distribution via the Coulomb Interaction.  Neglect gravity and friction.
(a)  Find the electric field produced by the charge distribution everywhere.
(b)  Find the equilibrium position of the point charge.
(c)  Find the motion of the point charge if it is displaced from its equilibrium position by a distance d < R.
(d)  The point charge is accelerating and therefore radiating.  For an initial displacement d = R, find the frequency of the radiation emitted and the initial rate of energy loss of the of the point charge.
(e)  How long will it takes the charge to radiate away half its energy?

Solution:

• Concepts:
  Gauss' law, radiation produced by accelerating charges
• Reasoning:
  The magnitude electric field inside a uniform charge distribution is proportional to the distance from the center of the charge distribution.  The point charge experiences a linear restoring force and therefore oscillates.  Oscillating charges radiate.
• Details of the calculation:
  (a)  From Gauss' law: E = ρr/(3ε₀) for r < R,  E = ρR³/(3ε₀r²)  for r > R.
  E = -E(r/r),  E points towards the center of the spherical charge distribution.
  (b)  The equilibrium position of the point charge is at the center of the spherical charge distribution.
  (c)  F = -kr = -qρr/(3ε₀),  k = qρ/(3ε₀).
  The point charge oscillates along a line through the center of the distribution with angular frequency ω, where ω² = k/m = qρ/(3ε₀m).
  Place the center of the charge distribution at the origin.  Then r(t) = rcos(ωt + δ) describes the position of the point charge as a function of time.
  (d)  The frequency of the radiation emitted is f = ω/(2π), where ω² = k/m = qρ/(3ε₀m).
  The initial Power radiated is given by the Larmor formula.
  <P> = q²ω⁴R²/(12πc³ε₀)
  (e)  -dE/dt = q²ω⁴A²/(12πc³ε₀), where A is the amplitude of the oscillation.
  To solve for the time it takes the charge to radiate away half its energy, we try to express A in terms of the energy E.
  We then have -dE/dt = f(E), ∫dt = Δt = ∫_E0^E0/2dE/f(E).  We assume that the oscillator does not lose a significant fraction of its energy during one cycle. 
  For an oscillator E = ½mω²A², therefore -dE/dt = q²ω²E/(6πε₀c³m).
  E = E₀exp(-t/τ), where τ = (6πε₀c³m)/(q²ω²)
  E = E₀/2 if t = t_½ = τ *ln2,  t_½ = (6πε₀c³m)/(q²ω²)ln2.
  This is the time it takes the oscillator to radiate away half its energy.