Physics 214 Final Exam Solutions Winter 2017

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1 Physics 14 Final Exam Solutions Winter An electron of charge e and mass m moves in a plane perpendicular to a uniform magnetic field B If the energy loss by radiation is neglected, the orbit is a circle of some radius R Let E be the total relativistic energy of the electron, and assume that E mc (corresponding to ultra-relativistic motion) (a) Express B analytically in terms of the parameters given above Compute numerically the required magnetic field B, in gauss, for the case of R = 30 meters and E = 5 GeV For circular motion, a = d v dt = v R ˆr (1) Since the circular motion is in a plane that is perpendicular to the magnetic field B, it follows that B, v and ˆr are mutually orthogonal vectors Moreover, eqs (138) and (139) on p 585 of Jackson yield d v dt = e γmc v B () Thus, if B points in the z-direction, then v = vˆθ and the circular motion is clockwise in the x y plane Combining eqs (1) and (), it follows that B = γmvc er In the ultra-relativistic limit, we have v c so that B γmc er = E er, where E = γmc is the total relativistic energy of the electron Plugging in the numbers, B = (5 109 ev)( ergs/ev) ( esu)( cm) = gauss (b) In fact, the electron radiates electromagnetic energy Suppose that the energy loss per revolution, E, is small compared to E Express the ratio E/E analytically in terms of the parameters given above Then, evaluate this ratio numerically using the values of R and E given in part (a) Using eq (1446) on p 671 of Jackson, P(t ) = e a γ 4, c 3 1

2 where a d v/dt For circular motion, the magnitude of the acceleration is a = v /R One orbit covers a distance of πr in a time δt = πr/v Thus, the energy lost per orbit is E = P(t ) t = e c 3 ( ) v ( πr In the ultra-relativistic limit, v c, so we end up with R v E 4πe 3R γ4 )γ 4 = 4πe 3R ( v c ) 3γ 4 Dividing by the total relativistic energy E = γmc, and substituting γ E/(mc ), it follows that ( ) 3 E E 4πe E 3mc R mc Plugging in the numbers, ( ) E E 4π( esu) ev = ( ev)( ergs/ev)( cm) ev Consider an oversimplified model of an antenna consisting of a thin wire of length l and negligible cross section, carrying a harmonically varying current density flowing in the z direction The (complex) current in the wire is given by Ie iωt, where I is a constant (independent of position) (a) Show that the (complex) current density takes the form: J( x,t) = ẑie iωt δ(x)δ(y)[θ(z + 1 l) Θ(z 1 l)], where the step function Θ(x) 1 if x > 0 and Θ(x) 0 if x < 0 Here, we assume that the point z = 0 corresponds to the midpoint of the antenna First we note that 0, z > l/, Θ(z + 1 l) Θ(z 1 l) = 1, z < l/, 0, z < l/ Thus, J = 0 if z > l/ or z < l/ For z < l/, J z = Ie iωt δ(x)δ(y), J x = J y = 0 The current is obtained by computing J d a = J ẑdxdy = J z dxdy = Ie iωt

3 (b) Prove that there is an oscillating charge density at z = ± 1 l (ie, at both ends of the antenna), but the charge density vanishes at any interior point on the antenna The continuity equation is J + ρ t = 0 For J( x,t) = J( x)e iωt and ρ( x,t) = ρ( x)e iωt, the continuity equation then reads: Using J given in part (a), J = iωρ( x) J = J z z = Iδ(x)δ(y) [ Θ(z + 1 z l) Θ(z 1l)] Setting this result to iωρ( x), we conclude that: = Iδ(x)δ(y) [ δ(z + 1 l) δ(z 1 l)] ρ( x) = ii ω δ(x)δ(y)[ δ(z + 1 l) δ(z 1 l)], which corresponds to two point charges located at the two ends of the antenna Moreover, ρ( x,t) = ρ( x)e iωt indicates that the point charges have magnitudes that oscillate in time (As usual, we take the real part to find the corresponding physical quantity) (c) Calculate the angular distribution of the radiated power, dp/dω, assuming that l, where is the wavelength of the emitted radiation Express your answer in terms of the current I, the antenna length l and the wavelength Integrate over angles to obtain the total radiated power For l, the electric dipole approximation is very accurate We therefore compute: p = xρ( x)d 3 x ( ii = (xˆx+yŷ +zẑ) ω = ii ω ẑ ) δ(x)δ(y) [ δ(z + 1 l) δ(z 1 l)] dxdydz zdz [ δ(z + 1 l) δ(z 1 l)] = iil ω ẑ Then, using eq (93) of Jackson (in MKS units), dp dω = c Z 0 3π k4 p sin θ = c Z 0 3π I l k 4 ω sin θ, 3

4 where Z 0 µ 0 /ε 0 is the impedance of free space Recalling that ω = kc and k = π/, the above result can be written as: dp dω = Z ( ) 0I l sin θ 8 Integrating over angles yields: P = πz 0I 3 ( ) l (3) Note that we can also derive eq (3) by using eq (94) of Jackson, ( ) P = c Z 0 k 4 4 ( ) ( ) 1π p = c Z 0 π I l = πz ( ) 0I l 1π c π 3 3 A charged particle of mass m with relativistic velocity v 0 enters a medium where it is slowed down by a force that is proportional to its velocity That is, F = d p/dt = η v, where η is a positive dimensionful constant The time t refers to the moving charge and t = 0 when the particle enters the medium (a) Using relativistic mechanics, determine the acceleration of the charged particle as a function of its velocity, mass and η This is a one dimensional problem, since the particle moves in a straight line The relativistic equation of motion for the particle is given by: F = d p dt = d dt (γm v), where p = γm v is the relativistic momentum Writing γ (1 v /c ) 1/, where v v, and remembering that the velocity v and hence γ depends on time, [ ( d dγ d v (γm v) = m v dt dt +γmd v dt = γm dt + γ c v v d v )], (4) dt where we have used dγ dt = d dt ) 1/ (1 v = γ3 For linear motion, v and d v/dt are parallel vectors Thus, ( v v d v ) = v d v dt dt c c ( v d v ) dt Inserting this result back into eq (4) yields, F = d ) (γm v) = γmd v (1+ v /c dt dt 1 v /c 4 = γ 3 m d v dt (5)

5 Using F = η v, we can solve eq (5) for the acceleration a d v/dt, a = η γ 3 m v If we denote a = aˆv and v = vˆv, where ˆv is the direction of the motion, then which indicates that the particle is decelerating a = ηv γ 3 m, (6) (b) Determine the angular distribution of the instantaneous power radiated once the particle has entered the medium and slowed down to a velocity v The polar and azimuthal angles of the emitted radiation are defined relative a z-axis, which can be chosen to lie along the direction of the particle velocity In your calculation, you may neglect the effect of the medium on the emitted radiation (ie, you should treat the radiation as if it is emitted in the vacuum) Using the relativistic Larmour formula for an accelerating charge in linear motion given in eq (1439) on p 669 of Jackson, 1 dp dω = e a sin θ 4πc 3 (1 βcosθ), (7) 5 where θ is the direction of emitted radiation and β v/c Inserting the result for the acceleration a obtained in eq (6) into eq (7), one obtains dp dω = e η v 1 cos θ 4πγ 6 m c 3 (1 βcosθ) (8) 5 (c) How much energy is emitted in the form of electromagnetic radiation from the time the particle enters the medium until it slows down and reaches zero velocity? To obtain the total instantaneous power emitted by the particle after slowing down to a velocity v, we simply integrate eq (8) over the polar and azimuthal angles, Setting x = cosθ and using P = e η v 4πγ 6 m c 3 π cos θ (1 βcosθ) 5 dcosθ 1 x (1 βx) 5 dx = 4 3 γ6, 1 In this problem, the time t refers to the moving charge This is what Jackson calls t in section 3 of Chapter 14 5

6 we end up with P = e η v 3m c 3 (9) Note that the instantaneous radiated power integrated over all solid angles can be directly obtained from eq (146) on p 666 of Jackson Since the velocity and acceleration are parallel, it follows that P = e a 3c 3 γ6 Using eq (6) for the acceleration a in the above formula, we immediately recover eq (9) The power is defined by P = de/dt Starting from eq (6), it follows that: or equivalently, dv = ηv γ 3 m dt, v dt = mγ3 vdv = mγ3 η η dv Hence, E = Pdt = e η v dt = e η 0 dv 3m c 3 3mc 3 v0 (1 v /c ), 3/ where v 0 v(t = 0) The last integral is elementary, and the final result is, where γ 0 (1 v 0/c ) 1/ E = e η 3mc (γ 0 1), 4 Consider the scattering of an electromagnetic wave of (angular) frequency ω and polarization ˆε 0 off of an electron bound in an atom The wavelength of the incoming wave is assumed to be significantly larger that the size of the atom You may also assume that the non-relativistic limit is a good approximation Onecanmodel theelectron byassuming thatitisboundbyadampedharmonicoscillator force with oscillation frequency ω 0 and damping coefficient η The electron, with mass m and charge e, also experiences a force due to the electric field of the incoming wave The response of the electron to the initial wave is the generation of a time-dependent electric dipole moment, p(t) = Re( pe iωt ), where the complex vector p is given by p = e m ( E 0ˆǫ 0 ω 0 ω iηω ) (10) (a) Assuming that the incoming wave is left circularly polarized and propagates in the positive z-direction, compute the angular distribution of the scattering cross section, under the assumption that the final state polarization ˆε is not observed Express the coefficient of the angular factor as a function of the frequency of the incoming wave 6

7 We may use eq (104) of Jackson Since there is no magnetic moment, we shall set m = 0 (The higher l moments can be neglected in the long wavelength approximation) In this problem I prefer to employ gaussian units, so I will remove the factor of 4πǫ 0 from eq (104) of Jackson Thus, dσ dω = k4 E 0 ˆǫ p (11) Inserting the expression for p given in eq (10), and noting that k = ω/c, we find 3 ( ) dσ e dω = ˆǫ mc ˆǫ 0 ω 4 (ω0 ω ) +η ω (1) Since the final state polarization is not observed, we must sum over the two possible polarization states We shall employ the polarization sum formula derived in class, ǫ () i ǫ () j = δ ij n i n j, (13) where the n i are the components of the unit vector ˆn k/k Thus, ˆǫ () ˆǫ 0 = ǫ () i ǫ () j (ǫ 0 ) i (ǫ 0) j = (ǫ 0 ) i (ǫ 0) j [δ ij n i n j ] = 1 ˆn ˆǫ 0, (14) after using ˆǫ 0 ˆǫ 0 = 1 in the final step ) For a left circularly polarized incident wave moving in the ẑ direction, ˆǫ 0 = 1 (ˆx+iŷ, following the optics conventions introduced by Jackson on the top of p 300 We shall define the angles of the scattered wave using spherical coordinates with respect to the fixed z-axis, ˆn = ˆxsinθcosφ+ŷsinθsinφ+ẑcosθ Consequently, ˆn ˆǫ 0 = 1 sinθe iφ Inserting this result into eq (14) yields: 4 ˆǫ () ˆǫ 0 = 1 1 sin θ = 1 1 (1 cos θ) = 1 (1+cos θ) (15) Using the above result in eq (1), we end up with dσ dω = 1 ( ) e ω 4 mc (ω0 ω ) +η ω (1+cos θ) (16) Since E 0 may be complex, one should write E 0 in eq (11) as I did in class rather than E0 as Jackson does in eq (104) 3 To convert the cross section formulae of this problem to SI units, simply replace e e /(4πǫ 0 ) 4 Note that ˆǫ() ˆǫ 0 = 1 ˆn ˆǫ 0 ) = ˆn ˆǫ 0 Thus, for a left circularly [ polarized incoming wave moving in the ẑ direction, ˆǫ 0 = 1 (ˆx + iŷ, and we have ˆn ˆǫ0 = ] i (ˆx+iŷ)cosθ ẑsinθe iφ It [ follows that ˆn ˆǫ 0 = 1 cos θ +sin θ ] = 1 (1+cos θ), and we recover the result of eq (15) 7

8 (b) Integrate the differential cross section over angles to obtain the total scattering cross section Compare your result to the Thomson cross section in the following three limiting cases: (i) ω ω 0 η; (ii) ω = ω 0 η; and (iii) ω ω 0 η Integrating the differential cross section obtained in eq (16) over angles over solid angles using, 1 dω(1+cos θ) = π dcosθ(1+cos θ) = 16π 1 3, we obtain the total cross section, σ = 8π ( ) e ω 4 3 mc (ω0 ω ) +η ω (17) We recognize the Thomson cross section [cf eq (1416) of Jackson]: ( e ) Hence, one can rewrite eq (17) as: σ = σ T 8π 3 Three limiting cases will now be considered mc ω 4 (ω 0 ω ) +η ω σ T Case 1: ω ω 0 η In this limit, the factor ω 4 (ω 0 ω ) +η ω 1 σ σ T Case : ω = ω 0 η In this limit, σ ω 0 η σ T σ T This is a case of resonant enhancement due to the fact that the frequency of the incoming wave matches the natural frequency of the harmonic potential that binds the electron to the atom Case 3: ω ω 0 η In this limit, σ ω4 σ ω0 4 T, which corresponds to Rayleigh scattering in the long wavelength approximation 8

9 REMARK: A more proper analysis would include the effects of radiation reaction For a discussion of these effects, see section 168 of Jackson BONUS MATERIAL: A derivation of eq (10) The electron is assumed to be bound by a damped harmonic oscillator force, F = mω 0 x(t) ηmd x(t) dt, (18) where x(t) is the position (at time t) of the electron, ω 0 is the natural frequency of the oscillator, and η is the damping coefficient Note that the minus signs above indicate that this is a restoring force Using Newton s second law, it is straightforward to show that the response of the electron to the initial wave can be modeled by a time-dependent electric dipole moment, p(t) = Re( pe iωt ), where the complex vector p is given by eq (10) Consider a charged particle of mass m and charge e subject to a damped harmonic oscillator force given in eq (18), which also experiences the force due to the electric field of the incoming wave Newton s second law then takes the following form, m d x(t) dt = e E(t) mω 0 x(t) ηm d x(t) dt where the time-dependent electric field of the incident wave is given by, E(t) = Re( Ee iωt ), with E = E 0ˆǫ 0, (19) Assuming a solution to eq (19) of the form x(t) = Re( xe iωt ), we insert this ansatz into eq (19) and obtain, mω x = ee 0ˆǫ 0 mω 0 x+iηωm x Solving for x, we get x = e ( ) E 0ˆǫ 0 m ω0 ω iηω Thus, the time-dependent electric dipole moment is given by p(t) = Re( pe iωt ), where ( ) p = e x = e E 0ˆǫ 0, m ω0 ω iηω which establishes eq (10) 9