(a) An electron orbits initially, at time t = 0 around a
proton at a radius a_{0} 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 = v^{2}/r, and it therefore radiates away energy,

P = -dE/dt = [q^{2}/(6πε_{0}c^{3})]a^{2}. (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 = ∫_{r0}^{0}dr/f(r).

(i) We find a in terms of E for a circular orbit.

F = q^{2}/(4πε_{0}r^{2}) = mv^{2}/r. mv^{2}= q^{2}/(4πε_{0}r).

E = ½mv^{2}- q^{2}/(4πε_{0}r) = -q^{2}/(8πε_{0}r) = -½mv^{2}.

a = v^{2}/r = q^{2}/(4πε_{0}r^{2}m) = 2E^{2}8πε_{0}/(q^{2}m).

-dE/dt = E^{4}16^{2}πε_{0}/(6c^{3}q^{2}m^{2}). dt = -(dE/E^{4}) (6c^{3}q^{2}m^{2})/(16^{2}πε_{0}).

Δt = (6c^{3}q^{2}m^{2})/(16^{2}πε_{0})∫_{E0}^{-∞}(dE/E^{4}) = -(6c^{3}q^{2}m^{2})/(16^{2}πε_{0})(E_{0}^{-3}/3),

where E_{0}= -13.6 eV, Δt = 1.5*10^{-11}s.

or

(ii) We express a and E as a function of the radius r for a circular orbit.

a = F/m = q^{2}/(4πε_{0}r^{2}m), E = -q^{2}/(8πε_{0}r),

dE = [q^{2}/(8πε_{0}r^{2})]dr = -Pdt._{ }dt = -[q^{2}/(8πε_{0}r^{2})]dr/P = -[q^{2}/(8πε_{0}r^{2})]dr/[q^{2}a^{2}/(6πε_{0}c^{3})].

dt = -dr[3c^{3}/(4r^{2}a^{2})] = -r^{2}dr[12c^{3}π^{2}ε_{0}^{2}m^{2}/q^{4}].

Δt = -[12c^{3}π^{2}ε_{0}^{2}m^{2}/q^{4}]∫_{r0}^{0}r^{2}dr = [12c^{3}π^{2}ε_{0}^{2}m^{2}/q^{4}](r_{0}^{3}/3),

where r_{0}= Bohr radius = 5.29*10^{-11}m, Δt = 1.5*10^{-11}s.

(b) This result implies that the atom is unstable. Since atoms exist, the classical theory needs to be modified.

A non-relativistic particle of mass m and charge q is initially moving with
velocity v(0) = v_{0}**i** in a uniform magnetic field
**B** = B**k**.

(a) At subsequent times the particle (neglecting radiation) moves in a circle
of radius R. Find R in terms of m, v_{0}, 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 = (6πε_{0}c^{3}m^{3})/(q^{4}B^{2})
in SI units.

(c) Calculate the total energy radiated by the particle and show that it equals
½mv_{0}^{2}.

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**= q**v**×**B**, mv_{0}^{2}/R = qv_{0}B, R = mv_{0}/(qB) = p_{0}/(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 v^{2}/r.

From the Larmor formula we have

P = (2/3)[q^{2}/(4πε_{0}c^{3})]a^{2}= -dE/dt.

a = v^{2}/R = q^{2}B^{2}R/m^{2}, so dE/dt = -(2/3)[q^{2}/(4πε_{0}c^{3})]a^{2}= -(2/3)[q^{2}/(4πε_{0}c^{3})]q^{4}B^{4}R^{2}/m^{4}.

We can also write the energy of the particle as a function of the orbital radius

E = ½mv^{2}= ½m(qBR/m)^{2}= q^{2}B^{2}R^{2}/(2m).

Therefore

dE/dt = [q^{2}B^{2}/m]R(dR/dt).

Combining the two expressions for dE/dt we have

[q^{2}B^{2}/m]R(dR/dt) = -(2/3)[q^{2}/(4πε_{0}c^{3})]q^{4}B^{4}R^{2}/m^{4}.

dR/dt = -(2/3)[q^{4}B^{2}/(4πε_{0}c^{3}m^{3})] R.

(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

d^{2}y/dt^{2}= -(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 = 2e^{2}a^{2}/(3c^{3}).

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}^{E0/2}dE/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 = -ω^{2}A cos(ωt), <a^{2}> = ω^{4}A^{2}/2.

For an oscillator E = ½mω^{2}A^{2}, therefore <a^{2}> = ω^{2}E/m.

P = q^{2}a^{2}/(6πε_{0}c^{3}) = q^{2}ω^{2}E/(6πε_{0}c^{3}m) =-dE/dt.

E = E_{0}exp(-t/τ), where τ = (6πε_{0}c^{3}m)/(q^{2}ω^{2})

E = E_{0}/2 if t = t_{½}= τ*ln2, t_{½}= (6πε_{0}c^{3}m)/(q^{2}ω^{2})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_{max}938 = 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}= 2e^{2}a^{2}T/(3c^{3}) 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}= ½ma^{2}T^{2}.

E_{rad}<< E_{kin}, if 2e^{2}a^{2}T/(3c^{3}) >> ½ma^{2}T^{2}, T >> (4/3)e^{2}/(mc^{3}) = (4/3)r_{0}/c, with r_{0}= e^{2}/(mc^{3}). For a proton r_{0}≈ 1.5*10^{-18}m.

If T is much greater than the time it takes light to travel ~2*10^{-18}m then we can neglect radiation losses when considering the short term motion.

Here ½ma^{2}T^{2}= 20*1.6*10^{-13}J.

a = qE/m = (20*10^{6}V/20m)(1.6*10^{-19}C/1.67*10^{-27}kg).

T = 6.46*10^{-7}s.

E_{rad}= 2e^{2}a^{2}T/(3c^{3}) = 2(q^{2}/(4πε_{0}))a^{2}T/(3c^{3}) = 3.4*10^{-32}J = 2.1*10^{-13}eV = energy radiated away.

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/(3ε_{0}) for r < R, E = ρR^{3}/(3ε_{0}r^{2}) 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/(3ε_{0}), k = qρ/(3ε_{0}).

The point charge oscillates along a line through the center of the distribution with angular frequency ω, where ω^{2}= k/m = qρ/(3ε_{0}m).

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 ω^{2}= k/m = qρ/(3ε_{0}m).

The initial Power radiated is given by the Larmor formula.

<P> = q^{2}ω^{4}R^{2}/(12πc^{3}ε_{0})

(e) -dE/dt = q^{2}ω^{4}A^{2}/(12πc^{3}ε_{0}), 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}^{E0/2}dE/f(E). We assume that the oscillator does not lose a significant fraction of its energy during one cycle.

For an oscillator E = ½mω^{2}A^{2}, therefore -dE/dt = q^{2}ω^{2}E/(6πε_{0}c^{3}m).

E = E_{0}exp(-t/τ), where τ = (6πε_{0}c^{3}m)/(q^{2}ω^{2})

E = E_{0}/2 if t = t_{½}= τ *ln2, t_{½}= (6πε_{0}c^{3}m)/(q^{2}ω^{2})ln2.

This is the time it takes the oscillator to radiate away half its energy.