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)
• 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).

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
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 = ⅔[q2/(4πε0c3)]a2 = -dE/dt.
a = v2/R = q2B2R/m2, so dE/dt = -⅔[q2/(4πε0c3)]a2 = -⅔[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) = -⅔[q2/(4πε0c3)]q4B4R2/m4.
dR/dt = -⅔[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 = ∫E0E0/2dE/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 = t½ = τ*ln2,  t½ = (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.
gmax938 = 958,  gmax = 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 Erad = 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 Ekin = ½ma2T2.
Erad << Ekin, 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.
Erad = 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 = ∫E0E0/2dE/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 = t½ = τ *ln2,  t½ = (6πε0c3m)/(q2ω2)ln2.
This is the time it takes the oscillator to radiate away half its energy.