Problem 1. Periodic pulses

Consider a periodic motion that repeats itself with period To. Show that the continuousfrequency spectrum becomes a discrete spectrum containing frequencies that are integralmultiples of the fundamental, ωo = 2π/To.

Let the electric field from a single pulse (or period) be E1(t), i.e. where E1(t) is non-zero between 0 and To and vanishes elsewhere, t < 0 and t > To. Let E1(ω) be its fouriertransform.

(a) Suppose that the wave form repeats once so that two pulses are received. E2(t) consistsof the first pulse E1(t), plus a second pulse, E2(t) = E1(t) +E1(t−To). Show that theFourier transform and the power spectrum is

E2(ω) = E1(ω) (1 + eiωTo) |E2(ω)|2 = |E1(ω)|2 (2 + 2 cos(ωTo)) (1)

(b) Now suppose that we have n (with n odd) arranged almost symmetrically around t = 0,i.e.

En(t) = E1(t+(n−1)To/2)+. . .+E1(t+To)+E1(t)+E1(t−To)+. . . E1(t−(n−1)To/2) ,(2)

so that for n = 3E3(t) = E1(t+ To) + E1(t) + E1(t− To) . (3)

Show that

En(ω) = E1(ω)sin(nωTo/2)

sin(ωTo/2)(4)

and

|En(ω)|2 = |E1(ω)|2(

sin(nωTo/2)

sin(ωTo/2)

)2

(5)

(c) By taking limits of your expressions in the previous part show that after n pulses, withn→∞, we find

En(ω) =∑

m

E1(ωm)2π

Toδ(ω − ωm) (6)

and

|En(ω)|2 = nTo︸︷︷︸total time

×∑

m

|E1(ωm)|2 2π

T 2o

δ(ω − ωm) (7)

where ωm = 2πm/To.Remark We have in effect shown that if we define

∆(t) ≡∞∑

n=−∞

δ(t− nTo) . (8)

Then the Fourier transform of ∆(t) is

∆(ω) =∑

n

e−iωnTo =∑

m

2π

Toδ(ω − ωm) . (9)

1

0 5 10 15 20 250

1

2

3

4

0 5 10 15 20 250

5

10

15

20

25

0 5 10 15 20 250

20

40

60

80

100

n = 2

(sin(nωTo/2)

ωTo/2

)2

n = 5

n = 10

2

(d) Show that a general expression for the time averaged power radiated per unit solidangle into each multipole ωm ≡ mωo is:

dPmdΩ

=|rE(ωm)|2T 2o

(10)

OrdPmdΩ

=e2ω4

om2

32π4c3

∣∣∣∣∫ To

0

v(T )× n exp

[iωm(T − n · r∗(T )

c)

]∣∣∣∣2

dT , (11)

Here dPm/dΩ is defined so that over along time period ∆T , the energy per solid angleis

dW

dΩ= ∆T

∞∑

m=1

dPmdΩ

(12)

Also note that we are summing only over the positive values of m which is differentfrom how we had it in class:

dPmdΩ≡ dPm

dΩ+dP−mdΩ

(13)

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Problem 2. Radiation spectrum of a SHO

(a) Show that for the simple harmonic motion of a charge discussed in Problem 2 theaverage power radiated per unit solid angle in the m-th harmonic is

dPmdΩ

=e2cβ2

8π2H2m2 tan2 θ [Jm(mβ cos θ)]2 (14)

(b) Show that in the non-relativistic limit the total power radiated is all in the fundamentaland has the value

P =e2

4π

2

3ω4oH

2 (15)

where H2 is the mean squared amplitude of the oscillation.

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Problem 3. Radiation spectrum from a damped SHO

The non-relativistic motion of a charged particle of charge e is described by a dampedharmonic oscillator

md2z

dt2+mη

dz

dt+mω2

oz = 0 (16)

where η is small, η ωo. Also assume that ∆ω ≡ ω − ωo ωo. Be sure to use theseapproximations at all points of the clculation.

The charge is released from rest with initial amplitude z(t = 0) = H.

(a) On the x axis, far from the charge, how is the light polarized ?

(b) Estimate (i.e. don’t calculate) the energy lost per time to radiation. We will requirethat the energy lost to radation is small compared to energy lost to friction. How doesthis requirement constrain the dimensionful parameters of this problem: m,H, ωo, η, e, c

(c) Determine the spectrum of photons which are emitted

ωdN

dω=

1

~dI

dω=

2

~dW

dω

∣∣∣∣ω>0

(17)

(The factor of two incorporates the contributions with ω < 0, which give an equalcontribution. Why?) Express your final result in terms of the fine structure constantα instead of the charge (squared).

(d) Optional – but extremely good practice for exam Integrate the results of theprevious part over frequency to determine the total energy that is emitted. Calculatethe same result by integrating the Larmour formula

P (te) =q2

4π

2

3

a2(te)

c3(18)

over time.

(e) Optional In part (c) you determine the frequency spectrum for ∆ω ωo. In part(d) you integrated over ∆ω (from −∞ . . .∞) to determine the total power. Estimatethe error made by extending this integral over the full frequency range instead of justa narrow range around ωo. Similarly estimate the error in your approximate formulafor the acceleration.

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Problem 4. Soft bremsstrahlung during a decay

In a collision or decay that happens at location ro over an infinitessimally short time scale,τaccel, the charged particles moving with velocity, v1,v2, . . . before the collsions and thecharged particles moving with v1′ ,v2′ , . . ., after the collision each contribute to the radiationfield. (The total radiation field is just a sum of the radiation fields from each particle.)

(a) Show that for frequencies low ω 1/τaccel the total radiation field is

Erad(ω, r) = eiω(r−n·ro)/c

( ∑

j′ ∈ final

qj′

4πrc2

n× n× vj′

1− n · βj′−

∑

j ∈ initial

qj4πrc2

n× n× vj1− n · βj

)

(19)This generalizes the result of Lecture 46.

Hint. You may encounter an integral like∫ ∞

0

n× n× v eiωT (1−n·v/c) . (20)

To give this integral definite meaning insert a convergence factor e−ε|T | and then takethe limit ε → 0 after integration. In any real experiment the velocity v(T ) would becut off in time, and provide this convergence factor naturally.

(b) A neutral ωo meson of mass Mωc2 = 784 MeV has a relatively rare decay mode ωo →

π+π−, with branching fraction of 1.53%. (98.5% of the time it decays to somethingelse.) It has another rare decay mode ωo → e+e− with branching ratio 7.28× 10−3%.(These are pretty rare decays for the ωo meson – most of the time it decays to π+π−π0

with a branching fraction of 89.2%). The mass of a pion is mc2 = 140 MeV, while theelectron mass is . . .

(i) Compute the frequency spectrum of the soft electromagnetic radiation per solidangle that accompanies both of these decay modes

dI

dωdΩ= 2

dW

dωdΩ

∣∣∣∣ω>0

, (21)

Describe your result qualitatively.

(ii) Show that for both of these decay modes the frequency spectrum of radiatedenergy at low frequencies is

dI

dω=

e2

4π2c

[(1 + β2

β

)ln

1 + β

1− β − 2

]' e2

π2c

[ln

(Mω

m

)− 1

2

](22)

where Mω is the mass of the ωo meson, m is the mass of one of the decay products,and β is the velocity/c of the decay products.

(iii) Roughly evaluate the total energy radiated in each decay by integrating the spec-trum up to a point where the photon’s momentum is half of the momentum ofthe decay products. (Beyond this point the recoil of the charged decay products

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would need to be considered. This lies outside of classical electrodynamics. Inclassical electrodynamics we specify the currents and solve for the fields.). Youshould find in a leading log(Mω/m) approximation

Irough

Mωc2' α

πlog

(Mω

m

)(23)

Using this rough evaluation, what fraction of the rest energy of the ωo is carriedaway by soft radiation in the two decay modes

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Problem 5. Thomson Scattering

We will do this in class. It is very important, especially for astrophysics.

(i) Polarized light with linear polarization vector εo, is propagating in the z-directionwith electric field amplitude Eo and is incident upon an electron at rest. Assumethat ~ω is much less than the electron mass mec

2. Show that the time averagepower radiated into light with polarization ε is

⟨dPpol

dΩ

⟩= 1

2cE2

o

(e2

4πmec2

)2

|ε∗ · εo|2 (24)

where ε is the polarization of the outgoing radiation, i.e. n · ε = z · εo = 0.

(ii) Show that the time averaged power radiated into light of any polarization by anincident beam with polarization εo is

⟨dPunpol

dΩ

⟩= 1

2cE2

o

(e2

4πmec2

)2

|n× εo|2 (25)

(iii) Show that the polarized and unpolarized cross sections for incident light withpolarization εo are

dσpol

dΩ= r2

e |ε∗ · εo|2 (26)

anddσunpol

dΩ= r2

e |n× εo|2 , (27)

respectively. Here the classical electromagnetic radius is

re =e2

(4π)mec2(28)

(iv) By sticking in appropriate powers of ~, show that re is 137 times smaller than thecompton wavelength, λC = ~/mec. Show that re is (137)2 times smaller than theBohr radius.

Remark: A heuristic way to understand why re is smaller than the “the sizeof an electron” , ~/mec, is that the cross section is the cross-sectional area ∝(~/mec)

2 of the electron times the probability that the light will actually interactwith the electron, wich is α2.

(v) Now consider unpolarized incident light (light which is equally likely to be polar-ized in the x or y directions). Let the radiation be scattered at an angle θ in thexz plane, where n ·no = cos θ. Depending on the scattering angle θ, the outgoinglight will be partially polarized in the xz plane, or out of the xz plane (i.e. in they direction).

Show that the cross-section for unpolarized light to produce in-plane polarizedlight is

dσ‖dΩ

= 12r2e cos2 θ (29)

8

while the cross-section to produce out-of-plane polarized light is

dσ⊥dΩ

= 12r2e (30)

And conclude that the cross-section for unpolarized light to produce light of anypolarization is

dσ

dΩ= r2

e

1 + cos2 θ

2(31)

(vi) By using the results of this problem and integrating over angles, or appealingdirectly to the Larmour formula, determine the total electromagnetic cross sectionfor light electron scattering. This is known as the Thomson cross section:

σT =8π

3r2e (32)

Evaluate the Thomson cross section numerically, without looking up any numbers.

(vii) Plot the polarization asymmetry

dσ‖dΩ− dσ⊥

dΩdσ‖dΩ

+ dσ⊥dΩ

(33)

as a function of scattering angle θ.

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Problem 6. Scattering from a perfectly conducting sphere

Consider light of wavenumber k scattering off a perfectly conducting sphere of radiusa. Assume that ka 1 and that the skin depth is much less than the size of the sphereThe incident light propagates along the z-direction.

(i) Optional Show that the external field E = Eoe−iωtεo and H = Hoe

−iωtn × εoinduces a time dependent electric and magnetic dipole moment :

p = 4πa3Eoe−iωt m = −2πa3Hoe

−iωt (34)

For the magnetic case you can look at the solutions to homework 5 (pages 2-6).For the electric case you can look at lecture 3.

(ii) By computing the radiated power from the time dependent magnetic and electricdipole, show that for arbitrary initial polarization εo of the incoming light, thescattering cross section off the sphere, summed over outgoing polarizations is givenby:

dσ

dΩ(εo,no,n) = k4a6

[5

4− |εo · n|2 −

1

4|n · (no × εo)|2 − no · n

](35)

where no and n are the directions of the incident and scattered radiations, whileεo is the (perhaps complex) unit polarization vector of the incident radiation(ε∗o · εo = 1; no · εo = 0).

Hint: as an intermediate step in the calculation show that

Erad =−ω2

4πc2

e−iωt+kr

rDo

[−εo + n(n · εo)− 1

2n× (no × εo)

](36)

where Do = 4πa3Eo. Then square this result (repeating to yourself like the thelittle engine . . . “I think I can, I think I can, think I can”) using the front coverof Jackson.

(iii) If the incident radiation is linearly polarized, show that the cross section is

dσ

dΩ(εo,no,n) = k4a6

[5

8(1 + cos2 θ)− cos θ − 3

8sin2 θ cos 2φ

](37)

where n ·no = cos θ and the azimuthal angle φ is measured from the direction ofthe linear polarization.

(iv) What is the ratio of the scattered intensities at θ = π/2, φ = 0 and θ = π/2, φ =π/2? Explain physically in terms of the induced multipoles and their radiationpatterns.

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Problem 7. (Optional) Estimates

Without looking up numbers make the following estimates1. Explain qualitatively howyou arrived at your estimate from the Lienard-Wiechert potentials.

(i) The light source NSLS II at BNL circulates electrons at 3 GeV with a circumfer-ence of 792 m. (i) Estimate the energy lost per turn to radiation. (ii) Estimatethe energy of the typical photon which is emitted, and compare this energy withthe energy of the electron. (iii) Estimate the angular width of the radiation cone.

(ii) The LHC at CERN circulates protons at 7 TeV with a circumference of 27 km.(i) Estimate the energy lost per turn for a proton at the LHC. (ii) Estimate theenergy a typical photon that is emitted at the LHC due to synchrotron radiation,and compare this to the proton energy. (iii) Estimate the angular width of theradiation cone.

1 You really need to know these numbers to get through life:

α =e2

4π~c' 1

137~c = 197 eV nm (38)

mec2 = 0.511 MeV (half an MeV) mpc

2 = 0.938 MeV (2000 times the electron mass ) (39)

Seriously. . . they wont be given on the final and you may need them, togethewith the Bohr model estimates.

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