Two equal magnitude, opposite sign charges are located at either end of a molecule of mass M and length l. The molecule rotates end over end (a nonrelativistic tumbling motion) with an initial rotational period T (cT >> l). How long will it take the molecule to lose 1/10 of its rotational energy by electromagnetic radiation?

Solution:

- Concepts:

The radiation field of an accelerating dipole, the Larmor formula - Reasoning:

The energy lost per unit time given by the Larmor formula

P = -dE/dt = <(d^{2}**p**/dt^{2})^{2}>/(6πε_{0}c^{3}). (SI units)

The two point charges move in a circle and produce a rotating dipole. Let the dipole rotate about the x-axis.**p**(t) = p_{0}cos(ωt)**k**- p_{0}sin(ωt)**j**.

d^{2}**p**/dt^{2}= -ω^{2}**p**, (d^{2}p/dt^{2})^{2}= ω^{4}p_{0}^{2}. - Details of the calculation:

P = -dE/dt = ω^{4}p^{2}/(6πε_{0}c^{3}) is the radiated power.

The rotational kinetic energy is E = ½Iω^{2}, with I = Ml^{2}/12.

(I assume that the mass M is distributed uniformly along the length l. you could also read the problem as implying that a mass M/2 is located at the position of each charge and the rod is massless. The I = Ml^{2}/4.)

Approximate solution: -∆E = -(dE/dt)∆t = -E/10.

∆t = Iω^{2}6πε_{0}c^{3}/(20ω^{4}p^{2}) = Ml^{2}πε_{0}c^{3}/(40ω^{2}p^{2}) = MT^{2}ε_{0}c^{3}/(160 πq^{2}).

Exact solution:

-dE/dt = ω^{4}p^{2}/(6πε_{0}c^{3}) = 4E^{2}p^{2}/(6 I^{2}πε_{0}c^{3}).

-dE/E^{2}= ω^{4}p^{2}/(6πε_{0}c^{3}) = 4p^{2}dt/(6 I^{2}πε_{0}c^{3}).

-(6 I^{2}πε_{0}c^{3}/(4p^{2}))∫_{E0}^{0.9E0}dE/E^{2}= ∫dt = ∆t.

∆t = [MT^{2}ε_{0}c^{3}/(16 πq^{2})] [1/0.9 - 1].

Let **E**_{0} = E_{0} **k**. The Abraham-Lorentz
force equation for a damped, charged, oscillator driven by an electric field
**E**_{0}exp(-iωt) in the dipole approximation is

d^{2}**r**'/dt^{2}
+ Γ d**r**'/dt - τ d^{3}**r**'/dt^{3} + ω_{0}^{2}
**r**' = (q/m)**E**_{0}exp(-iωt),

where Γ, τ, and ω_{0}
are constants, q is the charge and m is the mass of the oscillator.

Using
this and the expression for the radiation electric field,

**E**_{rad}(**r**,t)
= -(4πε_{0})^{-1}[(q/(c^{2}r'')]**a**_{⊥}(t -
r''/c),

where **r**'' = **r** - **r**'(t - |**r
**- **r**'|/c),
show that the differential cross section for scattering of radiation of
frequency ω and polarization **n **= (**θ**/θ)**
** is

dσ/dΩ = (e^{2}/(mc^{2}))^{2
}(**k**∙**n**)^{2}[ω^{4}/((ω_{0}^{2}
- ω^{2})^{2} + ω^{2}Γ_{t}^{2})],

where e^{2} = q^{2}/(4πε_{0}) and Γ_{t} = Γ
+ ω^{2}τ.

Solution:

- Concepts:

The scattering cross section - Reasoning:

Power radiated into the solid angle dΩ = <dP> = <incoming intensity> * dσ. - Details of the calculation:

Solve the equation of motion.

d^{2}**r**'/dt^{2}+ Γ d**r**'/dt - τ d^{3}**r**'/dt^{3}+ ω_{0}^{2}**r**' = (q/m)E_{0}exp(-iωt)**k**.

Let**r**' = r'**k**= r_{0}'exp(-iωt)**k**. Then

dr'/dt = -iωr_{0}'exp(-iωt), d^{2}r'/dt^{2}= -ω^{2}r_{0}'exp(-iωt), d^{3}r'/dt^{3}= iω^{3}r_{0}'exp(-iωt).

(-ω^{2}- iωΓ - iω^{3}τ + ω_{0}^{2})r_{0}' = (q/m)E_{0}.

r' = (q/m)E_{0}exp(-iωt/(-ω^{2}- iωΓ_{t}+ ω_{0}^{2}), where Γ_{t}= Γ + ω^{2}τ.**a**= -ω^{2}(q/m)E_{0}**k**exp(-iωt)/(-ω^{2}- iωΓ_{t}+ ω_{0}^{2}).

Calculate**S**.

For convenience, place the oscillator at the origin. Let**r**=**r**(θ, φ), then r'' = r.

**E**_{rad}(**r**,t) = -(4πε_{0})^{-1}[(q/(c^{2})]**a**_{⊥}(t - r/c)

= (4πε_{0})^{-1}[(q/(c^{2}r)]ω^{2}(q/m)E_{0}sinθ (**θ**/θ) exp(-iωt_{r})/(-ω^{2}- iωΓ_{t}+ ω_{0}^{2})

= (e^{2}/(mc^{2})) (sinθ/r) (**θ**/θ) ω^{2}E_{0}exp(-iωt_{r})/(ω_{0}^{2}- ω^{2}- iωΓ_{t}),

with t_{r}= t - r/c.

**S**((θ, φ) = (**r**/r)(Re(E_{rad}))^{2}/(μ_{0}c).

ω^{2}exp(-iωt_{r})/(ω_{0}^{2}- ω^{2}- iωΓ_{t}) = |ω^{2}/(ω_{0}^{2}- ω^{2}- iωΓ_{t})|exp(-iωt_{r})exp(iξ)

= [ω^{4}/((ω_{0}^{2}- ω^{2})^{2}- ω^{2}Γ_{t}^{2})]^{½}exp(-i(ωt_{r}- ξ))**S**(θ, φ) = (**r**/r )cos^{2}(ωt_{r}- ξ) (e^{4}/(m^{2}c^{4})) (sin^{2}θ/r^{2}) (μ_{0}c)^{-1}E_{0}^{2}ω^{4}/((ω_{0}^{2}- ω^{2})^{2}- ω^{2}Γ_{t}^{2}).

<**S**(θ, φ)> = (**r**/r )½(e^{4}/(m^{2}c^{4})) (sin^{2}θ/r^{2}) (μ_{0}c)^{-1}E_{0}^{2}ω^{4}/((ω_{0}^{2}- ω^{2})^{2}- ω^{2}Γ_{t}^{2}).

<dP> = <S(θ, φ)>r^{2}dΩ = power radiated into the solid angle dΩ.

<dP> = <incoming intensity> * dσ.

dσ/dΩ = <S(θ, φ)>r^{2}/( ½(μ_{0}c)^{-1}E_{0}^{2})

= ½(e^{4}/(m^{2}c^{4})) sin^{2}θ ω^{4}/((ω_{0}^{2}- ω^{2})^{2}- ω^{2}Γ_{t}^{2}).

**k**∙**n**=**k**∙(**θ**/θ) = -sinθ, (**k**∙**n**)^{2}= sin^{2}θ.

For a dipole antenna at the origin with complex electric
dipole moment **p**(t) = p_{0}exp(-iωt) **k** the electric and
magnetic fields in the far (radiation) zone are**
E**(

(a) If the real dipole moment is

(b) If there is a second dipole at -d

What is the time average? What are the minimum and maximum values of the time average with respect to variations of the phase α?

(c) A large number, N, of dipoles, as above, is located near the origin with Nkd

Solution:

- Concepts:

Radiation of oscillating dipoles - Reasoning:

We are asked to characterize the power radiated by an oscillating dipoles. - Details of the calculation:

(a) The radiation field of an electric dipole oscillating around the origin at with**p**(t) = p_{0}cos(ωt)**k**is

**E**(**r**,t) = -(1/(4πε_{0}c^{2}r))ω^{2}p_{0}cos(ω(t - r/c))sinθ (**θ**/θ).**B**(**r**,t) = (E/c)(**φ**/φ).

**S**(**r**,t) = (1/μ_{0})(**E**×**B**) = [1/(4αε_{0})^{2}][1/(c^{5}r^{2}μ_{0})]ω^{4}p_{0}^{2}cos^{2}(ω(t - r/c))sin^{2}θ (**r**/r).

dP =**S**∙(**r**/r) r^{2}dΩ = [1/(4πε_{0})] [ω^{4}/(4πc^{3})]p_{0}^{2}cos^{2}(ω(t - r/c))sin^{2}θ dΩ

= instantaneous power radiated into dΩ.

<dP> = <**S**>∙(**r**/r) r^{2}dΩ = [1/(4πε_{0})] [ω^{4}/(8πc^{3})]p_{0}^{2}sin^{2}θ dΩ

= average power radiated into dΩ.

(b)**E**=**E**_{1}+**E**_{2}.

**E**_{1}(**r**,t) = -(1/(4πε_{0}c^{2}r))ω^{2}p_{0}cos(ωt - kr) sinθ (**θ**/θ).

**E**_{2}(**r**,t) = -(1/(4πε_{0}c^{2}r'))ω^{2}p_{0}cos(ωt - kr' + α) sinθ' (**θ**'/θ').

**r**' =**r**+ d_{0}**i**. r' = |**r**+ d_{0}**i**| ≈ r(1 + d_{0}**i**∙**r**/r^{2}) = r + d_{0}cos(π/2 - θ) = r + d_{0}sin(θ).

kr' ≈ kr + kd_{0}sin(θ).

Since kd_{0}<< 1 and r >> d_{0}(radiation zone), we can write

**E**_{2}(**r**,t) ≈ -(1/(4πε_{0}c^{2}r))ω^{2}p_{0}cos(ωt - kr + α) sinθ (**θ**/θ).

**B**_{1}(**r**,t) = (E_{1}/c)(**φ**/φ),**B**_{2}(**r**,t) = (E_{2}/c)(**φ**/φ).

**S**(**r**,t) = (1/μ_{0})(**E**_{1}+**E**_{2})**B**_{1}+**B**_{2}.)

= [1/(4πε_{0})^{2}][1/(c^{5}r^{2}μ_{0})]ω^{4}p_{0}^{2}sin^{2}θ (**r**/r) *

[cos^{2}(ωt - kr) + cos^{2}(ωt - kr + α) + 2 cos(ωt - kr) cos(ωt - kr + α)].

dP =**S**∙(**r**/r) r^{2}dΩ.

<dP> = <**S**>∙(**r**/r) r^{2}dΩ = [1/(4πε_{0})] [ω^{4}/(8πc^{3})]p_{0}^{2}sin^{2}θ [½ + ½ + 2cos(α)]dΩ

= [1/(4πε_{0})] [ω^{4}/(8πc^{3})]p_{0}^{2}sin^{2}θ [2 cos^{2}(α/2)]dΩ

= average power radiated into dΩ.

Minimum: α = nπ, n = odd, cos^{2}(α/2) = 0, <dP>/dΩ = 0.

Maximum: α = nπ, n = even, cos^{2}(α/2) = 1, <dP>/dΩ = [1/(4πε_{0})] [ω^{4}/(2πc^{3})]p_{0}^{2}sin^{2}θ.

(c) For a large number N of dipoles with randomly distributed phases we get

<dP>/dΩ = N * <dP>/dΩ from part (a). Interference effects average out.

A thin linear antenna of length d is excited in
such a way that the sinusoidal current makes a full wavelength of oscillation as shown in
the figure.

(a) Calculate the power radiated per unit solid angle and plot the angular
distribution of the radiation.

(b) Determine the total power radiated and find a numerical value for the
radiation resistance R by setting <P> = <I^{2}>R. You can leave your
answer in integral form.

Solution:

- Concepts:

Radiation from antennas, dipole radiation - Reasoning:

We can treat the antenna as made up of small sections, each section emitting dipole radiation. - Details of the calculation:

(a) Let the antenna lie on the z-axis, from z = -λ/2 to z = λ/2.

Consider a small section of the antenna. Treat it like a dipole.

Let**p**= q dz (**z**/z), q = q_{0}(z)_{ }cosωt.

Then I(z) = dq(z)/dt = -ωq_{0}(z) sinωt = I_{0}(z) sin(ωt), I_{0}(z) = -ωq_{0}(z).

p_{0}(z) = q_{0}(z)_{ }dz = -(I_{0}(z)/ω)dz.

For this antenna I_{0}(z) = I_{0}sin(kz).

A small section of the antenna of length dz' located at z' a distance**R**=**r**- (**z**/z)z' from the observation point therefore produces

d**E**_{R}(**r**,t) = (1/(4πε_{0}c^{2}R))ω^{2}(I_{0}(z)/ω) cos(ωt - kR) sinθ' dz' (**θ**'/θ')

= (1/(4πε_{0}c^{2}R))ω^{2}(I_{0}/ω) cos(ωt - kR) sinθ' sin(kz') dz' (**θ**'/θ').

We have r >> λ, r >> z', Therefore θ' ≈ θ, 1/(kR) ≈ 1/(kr), since kR ≈ kr - kz' cosθ.

For the phase we use cos(ωt - kR) ≈ cos(ωt - kr + kz' cosθ)

d**E**_{R}(**r**,t) = [1/(4πε_{0}c^{3}kr)]ω^{2}I_{0 }sinθ cos(ωt - kr + kz' cosθ) sin(kz') dz' (**θ**/θ)

We now can find**E**_{R}(**r**,t) due to the antenna.

**E**_{R}(**r**,t) = (**θ**/θ)[1/(4πε_{0}c^{3}kr)]ω^{2}I_{0 }sinθ ∫_{-λ/2}^{λ/2}cos(ωt - kr + kz' cosθ) sin(kz') dz'.

∫_{-λ/2}^{λ/2}cos(ωt - kr + kz' cosθ) sin(kz') dz'

= (1/k)cos(ωt - kr)∫_{-π}^{π}cos(kz' cosθ) sin(kz') dkz'

- (1/k)sins(ωt - kr)∫_{-π}^{π}sin(kz' cosθ) sin(kz') dkz'.

The first term is an odd function of kz' and therefore zero.

∫_{-π}^{π}sin(ax) sin(x') dx = [½sin (a-1)x/(a-1) - ½sin (a+1)x/(a+1)]|_{-π}^{π}= 2sin(aπ)/(1 - a^{2})

**E**_{R}(**r**,t) = (**θ**/θ) [1/(4πε_{0}c^{3}k^{2}r)]ω^{2}I_{0 }sinθ sin(ωt - kr) 2sin(πcosθ)/sin^{2}θ.

**S**= (**r**/r) (1/(μ_{0}c))**|E**_{R}(**r**,t)|^{2}= (**r**/r)[1/(4π^{2}ε_{0}r^{2}c sin^{2}θ)]I_{0}^{2}sin^{2}(ωt - kr) sin^{2}(πcosθ).

<dP> = <**S**>∙(**r**/r) r^{2}dΩ = [1/(8π^{2}ε_{0}c sin^{2}θ)]I_{0}^{2}sin^{2}(πcosθ)dΩ.

The plot below compares the angular radiation pattern of a dipole (sin^{2}θ) and a full-wave antenna (sin^{2}(πcosθ)/sin^{2}θ).

(b) <P> = 2π∫_{0}^{π}sinθdθ <dP>/dΩ = (1/(4π^{2}ε_{0}c)]I_{0}^{2}∫_{0}^{π}dθ sin^{2}(πcosθ)/sinθ.

<P> = <I^{2}>R. R = [1/(2πε_{0}c)] ∫_{0}^{π}dθ sin^{2}(πcosθ)/sinθ.

Numerical integration: R = 93 ohm.

A non-relativistic positron of charge q_{e} and velocity **v**_{1}
(v_{1 }<< c) impinges head-on on a fixed nucleus of charge Zq_{e}.
The positron which is coming from far away (∞), is decelerated until it comes to
rest and then accelerated again in the opposite direction until it reaches a
terminal velocity **v**_{2}. Taking radiation loss into account
(but assuming it is small), find v_{2} as a function of v_{1}
and Z.

Solution:

- Concepts:

Motion in a central field, energy conservation, the Larmor formula - Reasoning:

The positron moves in a central field. It has no angular momentum about the position of the nucleus (at the origin). If we neglect radiation losses, we can find its velocity and acceleration as a function of distance from energy conservation. Using the Larmor formula, we then can find the energy lost in a small time interval dt or a small distance dr. Integrating these losses over the path, we find E_{2}and v_{2}. This works if the energy radiated is much less than the characteristic energy of the system. - Details of the calculation:

E = ½mv_{1}^{2}= ½mv^{2}+ Ze^{2}/r. v(r) = (v_{1}^{2}- 2Ze^{2}/(mr))^{1/2}, with e^{2}= q_{e}^{2}/(4π^{2}ε_{0}).

dv/dt = (dv/dr)v = Ze^{2}/(mr^{2}).

The Larmor formula gives the power radiated by an accelerating charge.

P = 2e^{2}a^{2}/(3c^{3}) = [2e^{2}/(3c^{3})](Ze^{2}/(mr^{2}))^{2}= -dE/dt = -(dE/dr)v.

dt = dr/v.

ΔE = -∫Pdt = -∫[2e^{2}/(3c^{3})](Ze^{2}/(mr^{2}))^{2}dt = -2∫_{rmin}^{∞}[2e^{2}/(3c^{3})](Ze^{2}/(mr^{2}))^{2}dr/v

= -[4e^{6}Z^{2}/(3c^{3}m^{2})]∫_{rmin}^{∞}r^{-4}dr/v(r).

v(r) = v_{1}(1 - 2Ze^{2}/(mrv_{1}^{2}))^{1/2}. v(r_{min}) = 0.

r_{min}= 2Ze^{2}/(mv_{1}^{2}).

ΔE = -[4e^{6}Z^{2}/(3c^{3}m^{2}v_{1})]∫_{rmin}^{∞}r^{-4}dr/(1 - r_{min}/r)^{1/2}.

Let x = r_{min}/r, dx = -r_{min}dr/r^{2}.

ΔE = -[4e^{6}Z^{2}/(3c^{3}m^{2}v_{1}r_{min}^{3})]∫_{0}^{1}x^{2}dx/(1 - x)^{1/2}.

From an integral table:

∫_{0}^{1}x^{2}dx/(1 - x)^{1/2}= [2(8 + 4x + 3x^{2})/15](1 - x)^{1/2}|_{0}^{1}= -16/15.

ΔE = [4e^{6}Z^{2}/(3c^{3}m^{2}v_{1}r_{min}^{3})]16/15 = (8/45)mv_{1}^{5}/(Zc^{3}).

For a nonrelativistic v_{1}we have ΔE << E.

ΔE/2 is the radiation loss during the approach. The radiation loss on the way out is also ΔE/2.

½mv_{2}^{2}= ½mv_{1}^{2}- ΔE, v_{2}^{2}= v_{1}^{2}- (16/45)v_{1}^{5}/(Zc^{3}).

v_{2}≈ v_{1}(1 - (8/45)v_{1}^{3}/(Zc^{3})).