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March 4, 2011 Turn in HW 5; Pick up HW 6 Today: Finish R&L Chapter 3 Next: Special Relativity.

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Presentation on theme: "March 4, 2011 Turn in HW 5; Pick up HW 6 Today: Finish R&L Chapter 3 Next: Special Relativity."— Presentation transcript:

1 March 4, 2011 Turn in HW 5; Pick up HW 6 Today: Finish R&L Chapter 3 Next: Special Relativity

2 R&L Problem 3.2: Cyclotron Radiation A particle of mass m, charge e, moves in a circle of radius a at speed V ┴ << c. Define x-y-z coordinate system such that n is in the y-z plane

3 b. Polarization

4 (c) What is the spectrum of the radiation? Only cos ωt and sin ωt terms  spectrum is monochromatic at frequency ω

5 (d) Suppose the particle is moving in a constant magnetic field, B What is ω, and total power P? F B Lorentz force is balanced by centripetal force Lorentz force So gyro frequency of particle in B field

6 from part (a) where

7 (e) What is the differential and total cross-section for Thomson scattering of circularly polarized radiation? Equate the electric part of the Lorentz force = eE with centripetal force = mrω 2 and use our expression for for a circularly moving charge

8 Then Recall So, differential cross section Total cross-section Thomson cross-section

9 Rybicki & Lightman Problem 3.4 Consider an optically thin cloud surrounding a luminous source. The cloud consists of ionized gas. Assume that Thomson scattering is the only important source of optical depth, and that the luminous source emits unpolarized radiation. (a) If the cloud is unresolved, what polarization is observed? If the angular size of the cloud is smaller than the angular resolution of the detector, the polarization of the different parts of the cloud cancel  no net polarization R=1pc “Geometrical dilution”

10 (b) If the cloud is resolved, what is the direction of the polarization as a function of position on the sky? Assume only “single-scattering” – i.e. each photon scatters only once. θ At each θ, the incident, unpolarized wave can be decomposed into 2 linearly polarized waves: one in the plane of the paper, one normal to the plane of the paper. These scatter into new waves in ratio cos 2 θ : 1 Thus, the component normal to the paper dominates the other Integrating over every θ along a given line of sight results in net polarization which is normal to radial line:

11 Net result:

12 (c) If the central object is clearly seen, what is an upper bound for the electron density of the cloud, assuming that the cloud is homogeneous? To see the central object,

13 Radiation Reaction Rybicki & Lightman Sections 3.5 & 3.6

14 When an accelerating charged particle radiates, it is losing energy.  there must be a force acting on the particle This is the “radiation reaction force” Power radiated = energy lost mass cm^2/sec^3  ergs/sec

15 When is this important? When the particle motion is significantly modified on the time-scale of interest Define T=time scale of change in kinetic energy due to F rad in the dipole approximationso Let where So

16 For the electron, So on time scales longer than 10 -23 s, the radiation reaction is a small perturbation

17 we wrote This can’t be right because P(rad) depends on only, and in the above equation, F(rad) depends not only on but also

18 Instead, consider the work done by F(rad) in time interval (t 1,t 2 ) Integrate by parts If the motion is periodic, then so = 0 Collecting terms:

19 so This is the radiation force in a time averaged sense for periodic motion, so one needs to be careful how one uses this expression. Valid for “classical” non-relativistic electron Abraham-Lorentz-Dirac Force: Relativistic Abraham-Lorentz-Dirac-Langevin Force: Relativistic, quantum mechanical scales Abraham-Lorentz Force

20 The equation of motion for the charged particle, with some external force (i.e. the EM wave) applied is then Abraham-Lorentz Eqn. of Motion F=ma radiation force external force The radiation force is a damping term which will damp out the oscillations of the particle induced by the external force F ext

21 Radiation from Harmonically Bound Particles Up to now, we’ve been talking about the motion of free electrons in an oscillation E, B field Now we will consider a particle which is harmonically bound to a center of force, i.e. where so that it oscillates sinusodially around the origin The important applications of this situation include: Rayleigh Scattering Classical treatment of spectral line transitions in atoms

22 First, we consider Undriven, harmonically bound particles damped by radiation force then Driven, harmonically bound particles, where forced oscillations occur due to incident radiation

23 Undriven, harmonically bound particle For oscillations along the x-axis, the equation of motion is Radiation force F=ma restoring force

24 Butis very small, sois small The harmonic solution in first order is (ignoring ) and requiring that at t=0, Further, this implies So that we have we keep the cosine term and

25 so the equation of motion becomes Let the solution Then the equation for α is

26 So we need to solve From the quadratic formula:

27 Letso then the solution for x(t) can be written where x 0 = arbitrary constant = position at t=0

28 The spectrum will be related to the Fourier Transform of x(t) becomes large at and We are only interested in the physically valid ω>0 so

29 Power per frequency or Initial potential energy Line shape (Lorentz profile)

30 The shape of the emitted spectrum is the Lorentz Profile Γ = FWHM FREE OSCILLATIONS OF A HARMONIC OSCILLATOR

31 Driven Harmonically Bound Particles Now consider the forced oscillations of a harmonic oscillator, where the forcing is due to an incident beam of radiation: F=ma Harmonic oscillator restoring force Radiation damping force External driving force: sinusoidally varying E field Note: ω 0 = frequency of harmonic oscillator ω = frequency of incoming EM wave (1)

32 To find the solution, we let (2) To derive y 0, substitute (2) into (1) (1) becomes

33 So where phase shift between driving force and particle oscillation So

34 Note: (1) x(t) is an oscillating dipole, for charge q and frequency ω (2) Phase shift, δ, is not symmetric about ±ω 0 Particle leads F ext Particle lags F ext (3) “Resonance” at ω = ±ω 0

35 Time averaged, total power in dipole approximation: dipole oscillating with frequency ω, amplitude Divide by time-averaged Poynting vector to get the cross-section

36 Note: (1) ω>>ω 0 electron becomes unbound (2) Resonance Looks like free-oscillation Lorentzian

37 (3) ω<<ω 0 Rayleigh Scattering Since ω<<ω 0 the E-field appears static, and the driving force is effectively constant  the electron responds directly to the incident field

38 ω 4 dependance  blue sky, red sun at sunrise and sunset blue light scattered Colorat zenithat sunrise - sunset Red (6500 A)0.960.21 Green (5200 A)0.900.024 Violet (4100 A)0.760.000065 Jackson E&M

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