Physics 14 Solution Set 4 Winter 017 1. [Jackson, problem 1.3] A particle with mass m and charge e moves in a uniform, static, electric field E 0. (a) Solve for the velocity and position of the particle as explicit functions of time, assuming that the initial velocity v 0 was perpendicular to the electric field. Using eqs. (1.1) and (1.) of Jackson and setting B = 0, we have: d p dt = e E, dw dt = e v E, where W is the total mechanical energy (usually called E, but we have renamed this W in order to better distinguish it from the electric field) and v is the particle velocity (which is denoted as u by Jackson). Clearly, the motion takes place in a plane containing the E-field. Without loss of generality, we assume that E = Eˆx, and assume that the motion takes place in the x y plane. By assumption, v E = 0 at t = 0, in which case p x = 0 at t = 0. Solving the equations, in follows that dp x dt = ee, dp y dt = 0, p x = eet, p y = p 0, where p 0 is a constant. Using p = γm v and E = γmc, it follows that Hence, v x = v = c p W = c p p c +m c 4. c eet (p 0 +e E t )c +m c 4, v y = c p 0 (p 0 +e E t )c +m c 4. Since v = d x/dt, it follows that x = c tdt ee W 0 +(ceet), y = c p 0 dt W 0 +(ceet), (1) where W 0 = p 0 c +m c 4. 1
We shall define the origin of the coordinate system to coincide with t = 0. Then computing the integrals in eq. (1) yields x(t) = 1 ] [W 0 ee +(ceet) W 0, y(t) = p ( ) 0c ceet ee sinh 1. () W 0 (b) Eliminate the time to obtain the trajectory of the particle in space. Discuss the shape of the path for short and long times (define short and long times). We can eliminate t from eq. (), t = W 0 cee sinh ( ) eey. p 0 c Inserting this into the equation for x(t) and using the identity cosh z sinh z = 1, it follows that x = W [ ( ) ] 0 eey cosh 1, ee p 0 c which is the equation for a catenary curve. To describe the shape of the path for short and long times, we note that W 0 /(cee) has units of time. This we can define short and long times relative to this quantity. For t W 0 /(cee), we have W 0 +(ceet) W 0 + (ceet) W 0, Hence the approximate form of eq. () is x(t) c eet W 0, ( ) ceet sinh 1 y(t) p 0c t W 0. Solving for t and inserting the result back into the above equations yields x eew 0y p 0 c. W 0 ceet W 0. Since v 0 = c p 0 /W 0, we can eliminate W 0 from the above expression to obtain, That is, as short times, the motion is parabolic. 1 x eey p 0 v 0. (3) 1 The result of eq. (3) also coincides with the non-relativistic limit (in which case p 0 = mv 0 ). To verify this assertion, we can perform a formal expansion in powers of 1/c. In this limit, W 0 mc and t W 0 cee mc ee, which is always true in the limit of c (which is equivalent to taking the non-relativistic limit).
For t W 0 /(cee), eq. () yields: In the latter case, we used: x(t) ct, sinh 1 z = ln Hence, to a good approximation, or equivalently, y(t) p 0c ee ln ( ) ceet. W 0 ( z + ) z +1 lnz, for z 1. y p 0c ee ln x W 0 ee exp That is, at long times the motion is exponential. ( ) eex, W 0 ( ) eey. p 0 c. [Jackson, problem 1.9] The magnetic field of the earth can be represented approximately by a magnetic dipole of magnetic moment M = 8.1 10 5 gauss-cm 3. Consider the motion of energetic electrons in the neighborhood of the earth under the action of this dipole field (Van Allen electron belts). [Note that M points south.] (a) Show that the equation for a line of magnetic force is r = r 0 sin θ, where θ is the usual polar angle (colatitude) measured from the axis of the dipole, and find an expression for the magnitude of B along any line of force as a function of θ. Let the z-axis point from the origin in the direction of the north pole. Then, the magnetic dipole moment (which points south) is given by M = Mẑ, where M M. The vector potential is given in gaussian units by: where r x. Then, A( x) = M x x 3 = M sinθ r B = A = = 1 ˆx ŷ ẑ r det 0 0 M 3 rsinθcosφ rsinθsinφ rcosθ (ˆx sinφ ŷ cosφ) = M sinθ r ˆφ, ˆr rˆθ rsinθˆφ 1 r sinθ det r θ φ, 3 0 0 rsinθa φ
where A φ = M sinθ/r. Evaluating the above determinant yields: B = M r 3 cosθ ˆr M sinθ r 3 ˆθ. (4) Given the magnetic field at every point in space, B( x), one can consider a related vector field, F( x) = q m B( x), which gives the force on a magnetic test charge q m due to the magnetic field at the point x. If we choose our test charge to have q m = 1, then there is no distinction between the lines of magnetic force and the magnetic field lines. We choose to follow this convention in what follows. The lines of force follow a curve x(x), where the arclength s parameterizes the location along the curve. By definition B( x) is tangent to the lines of force. That is, d x ds = B ( ) x(s), (5) B where B B. To understand the normalization on the right hand side above, we note that eq. (5) is equivalent to the three equations, dx ds = B x B, dy ds = B y B, dz ds = B z B. Squaring each equation and summing the three resulting equations yields (ds) = (dx) +(dy) +(dz), which is the well-known formula for the differential arclength. It is convenience to work in spherical coordinates. Consider an infinitesimal displacement d x, where x = rsinθcosφˆx+rsinθsinφŷ +rcosθẑ. By the chain rule, d x = x dr dr + x dθ dθ+ x dφ dφ = (cosφsinθ ˆx+sinθsinφŷ +cosθẑ)dr +r(cosθcosφ ˆx+cosθsinφŷ sinθẑ)dθ +r( sinθsinφ ˆx+sinθcosφŷ)dφ = ˆrdr+ ˆθrdθ+ ˆφrsinθdφ. (6) Of course, magnetic charges do not exist in classical electromagnetism. But the concept of lines of force were developed before this fact was understood. In the case of the electric field, we do have F = q E, so the terminology lines of force makes sense. In the case of magnetic fields, it would be better to refer to the lines of force as the magnetic field lines. Nevertheless, following Jackson, we retain the old terminology in this problem. 4
The tangent to the curve x(s) then takes the form d x ds dr dθ dφ = ˆr + ˆθr + ˆφrsinθ ds ds ds. (7) Using eq. (4), it follows that the line of magnetic force is determined by the equation, d x ds = B( x(s)) B = M M sinθ cosθ ˆr ˆθ, Br3 (8) Br 3 where r, θ and φ are functions of s. Equating eqs. (7) and (8) yields three differential equations, dr cosθ = M, r dθ sinθ dφ = M, = 0. (9) ds Br 3 ds Br 3 ds Dividing the first two equations above yields, which is easily integrated, dr dθ = rcosθ sinθ, dr cosθ r = sinθ dθ, Evaluating the integrals and imposing the condition r = r 0 at θ = 1 θ, we obtain ( ) r ln = lnsinθ, or equivalently r 0 r = r 0 sin θ, (10) which we identify as the equation for the line of magnetic force. Note that the third equation in eq. (9) implies that φ is a constant along the line of magnetic force. Finally, we evaluate the magnitude of B along the line of force. Since B B = B r +Bθ +Bφ = M 4cos r θ+sin θ, 3 We simply plug in eq. (10) to obtain B as a function of θ along the line of magnetic force, B(θ) = M 1+3cos θ r0 3 sin 6, (11) θ after using sin θ = 1 cos θ in the numerator above. (b) A positively charged particle circles around a line of force in the equatorial plane with a gyration radius a and a mean radius R (where a R). Show that the particle s azimuthal position (east longitude) changes approximately linearly in time according to: φ(t) = φ 0 3 ( a ) ωb (t t 0 ). R where ω B is the frequency of gyration at radius R. 5
Assuming that a R, we can use eq. (1.55) of Jackson to obtain an approximate formula for the gradient drift velocity, v G ω B a = a ( ) B B B, (1) where a is the gyration radius and B is the field at the equator (θ = 1 π). Using eq. (4), this means that B = ˆθ M r 3 r=r = M R 3 ˆθ, B = ˆr B r = 3M ˆr, (13) r=r R4 where R is the mean radius. In computing B, we used the fact that B B = M/r 3 and = ˆn = ˆr r + ˆφ 1 rsinθ φ, where ˆn B = 0. Inserting the results of eq. (13) into eq. (1), we end up with v G = ω ( )( )( ) B a R 6 M 3M ˆθ ˆr = 3ω B a M R 3 R 4 R ˆφ. (14) Finally, we can express v G in terms of the angular velocity dφ/dt by v G = R dφ dt ˆφ. Comparing this equation with eq. (14), we conclude that dφ dt = 3a R ω B. Solving this differential equation, and imposing the initial condition φ(t 0 ) = φ 0, we end up with φ(t) = φ 0 3a R ω B (t t 0). (15) (c) If, in addition to its circular motion of part (b), the particle has a small component of velocity parallel to the lines of force, show that it undergoes small oscillations in θ around θ = 1 π with frequency Ω = (3/ )(a/r)ω B. Find the change in longitude per cycle of oscillation in latitude. As discussed in Chapter 1, section 4 of Jackson, the transverse velocity of gyration is v = ω B a [cf. discussion below eq. (1.61) of Jackson]. If in addition, we now include the small component of the velocity parallel to the lines of magnetic force, we may use eq. (1.7) of Jackson to write: v = v 0 v 0 6 B(z) B 0.
Here, the subscript 0 refers to the equator z = 0 (or equivalently to θ = 1 π). In particular, we can write v0 = v 0 +v 0 so that ( v = v 0 +v 0 1 B(z) ). (16) B 0 In part (a), we found that along the lines of magnetic force, B(θ) = M 1+3cos θ r0 3 sin 6, (17) θ where r 0 r(θ = 1 π). In this problem, we are interested in the behavior of the particle at the mean radius R, so we take r 0 = R. To compute B(z), we expand about z = 0. Since z = Rcosθ, we expand about z = 0 by writing θ = 1 π +ǫ. Then, z = Rcosθ = Rcos ( 1 π +ǫ) = Rsinǫ Rǫ. Hence, ǫ z/r and θ 1 π z/r. It follows that ( π cosθ cos R) z = sin z ( π R, sinθ sin z ) = cos z R R. Using eq. (17), B(z) M r 3 0 1+3sin (z/r) cos 6 (z/r) Plugging this result into eq. (16) yields M r 3 0 v (z) = v 0 9 1+3z /R [1 z /(R )] 6 M r 3 0 ] [1+ 9z. R ( ωb a ) z. (18) R As discussed below eq. (1.7) of Jackson, this equation is equivalent to the conservation of energy of a one-dimensional non-relativistic mechanics problem with total mechanical energy, E(z) = 1 mv +V(z), where ( 9ω ) V(z) = 1 m B a z, (19) R is the potential energy of a one-dimensional harmonic oscillator. Indeed, eq. (18) is equivalent to the statement that E(z) = E(0), i.e. conservation of energy. If we write the harmonic oscillator potential in the standard form, V(z) = 1 mω z, the eq. (19) implies that the effective oscillator frequency Ω is given by Ω = 3 ω B a R. 7
That is, the charged particle undergoes small oscillations in θ around θ = 1 π with frequency Ω. One period T of oscillation is given by T = π Ω = πr 3ω B a. (0) Using the results of part (b) [cf. eq. (15)], the change of longitude is φ = 3a R ω B t. (1) Choosing t = T then yields the change of longitude per cycle of oscillation in latitude, πa φ = R. (d) For an electron of 10 MeV kinetic energy at a mean radius of R = 3 10 7 m, find ω and a, and so determine how long it takes to drift once around the earth and how long it takes to execute one cycle of oscillation in latitude. Calculate the same quantities for an electron of 10 kev at the same radius. Given M = 8.1 10 5 gauss-cm 3 and R = 3 10 9 cm, the magnetic field at the equator is Using eq. (1.39) of Jackson, B = M R 3 = 3 10 3 gauss. ω B = eb γmc = ecb γmc. () Although the last step above is rather trivial, it is convenient to write ω B in this form. The numerical value of the quantity ec is given by ec = (4.8 10 10 statcoulombs)(3 10 10 cm s 1 ) = 14.4 statcoulombs cm s 1. (3) It is convenient to eliminate statcoulombs in favor of gauss. That is, 1 gauss = 1 dyne statcoulomb 1 = 1 erg cm 1 statcoulomb 1. Using 1 ev = 1.6 10 1 ergs, we can write: 1 gauss = (1.6 10 1 ) 1 ev cm 1 statcoulomb 1 = 6.5 10 11 ev cm 1 statcoulomb 1. Hence, it follows that 1 statcoulomb = 6.5 10 11 ev cm 1 gauss 1. 8
Inserting this result into eq. (3) yields ec = 9 10 1 ev gauss 1 s 1. Therefore, the gyration frequency can be written as ω B = 9 10 1 1 B (gauss) s γmc (ev). (4) For the electron, we have mc = 511 kev. If the electron has a kinetic energy of K = 10 MeV, then E = γmc = mc +K, which yields K = (γ 1)mc. Hence, It follows from eq. (4) that γ = 1+ K 10 MeV = 1+ mc 0.511 MeV = 0.57. ω B = 9 10 1 s 1 3 10 3 (0.57)(5.11 10 5 ) =.57 103 s 1. Next we use v v = ω B a to determine a. Since γ 1, it follows that v c, so that a = c ω B = 3 1010 cm s 1.57 10 3 s 1 = 117 km. To drift once around the earth requires the longitude (or azimuthal angle φ) to change by π. Inserting φ = π in eq. (1) [the overall sign is not significant here], we obtain t = 4πR 3a ω B = 4π(3 10 9 cm) 3(1.17 10 7 cm) (.57 10 3 s 1 ) = 107 s. Finally, the time it takes to execute one cycle of oscillation in latitude was obtained in part (c) [cf. eq. (0)]: T = πr 3ω B a = π(3 10 9 cm) = 0.3 s. 3(1.17 10 7 cm) (.57 10 3 s 1 ) For an electron with kinetic energy of 10 kev, It follows from eq. (4) that γ = 1+ K 10 kev = 1+ mc 511 kev = 1.0. (5) ω B = (9 101 s 1 )(3 10 3 ) (1.0)(5.11 10 5 ) = 5.18 10 4 s 1. To determine a, we first compute v using eq. (5): 1 1 v /c = 1.0 = v c = 0.195. 9
Hence, a = v ω B = (0.195)(3 1010 cm s 1 ) 5.18 10 4 s 1 = 1.13 km. Finally, following the previous computation, t = 4πR 3a ω B = 4π(3 10 9 cm) 3(1.13 10 5 cm) (5.18 10 4 s 1 ) = 5.7 104 s, and T = πr 3ω B a = π(3 10 9 cm) = 1.5 s. 3(1.13 10 5 cm) (5.18 10 4 s 1 ) Note that in both computations above, we have a R, which implies that the gradient of the magnetic field is small over the orbit of the electrons. Hence, the approximations introduced in Chapter 1, sections 4 and 5 of Jackson are valid for the charged particle motions examined in this problem. 3. [Jackson, problem 1.11] Consider the precession of the spin of a muon, initially longitudinally polarized, as the muon moves in a circular orbit in a plane perpendicular to a uniform magnetic field B. (a) Show that the difference Ω of the spin precession frequency and the orbital gyration frequency is Ω = eba m µ c, independent of the muon s energy, where a = 1 (g ) is the magnetic moment anomaly. Find the equations of motion for the components of the spin along the mutually perpendicular directions defined by the particle s velocity, the radius vector from the center of the circle to the particle, and the magnetic field. Our starting point is the Thomas equation, which Jackson writes in the following form [cf. eq. (11.170) of Jackson]: d s dt = e {( g mc s 1+ 1 ) ( g ) ( γ B γ 1 γ +1 ( β B) β g γ ) } β E, γ +1 (6) where the time derivative of the velocity vector is given by [cf. eq. (11.168) of Jackson]: dβ dt = e [ E + β B β( β E)]. (7) γmc For a particle moving in a circular orbit in a plane perpendicular to a uniform magnetic field B, we have β B = 0, where v cβ is the particle velocity. Hence, eqs. (6) and (7) reduce to d s dt = e ( g mc 1+ 1 ) s B, γ 10 d v dt = e γmc v B, (8)
since by assumption there is no electric field present ( E = 0). That is, eq. (8) can be written in the form of precession equations, d s dt = s ω, d v dt = v ω B, where the spin precession frequency ω and the orbital gyration frequency ω B are given by: ω e [ ( ) ] g 1+ γ B, ω γmc B e B. γmc The difference of these two frequencies is Ω ω ω B = e ( ) g mc B, and the magnitude of this frequency difference is given by Ω = eba mc, where a = 1 (g ). Tofindtheequationsofmotionforthecomponentsofthespinvector, wefirstdecompose this vector into longitudinal and transverse components with respect to the direction of the velocity, ˆβ β/β. That is, s = s + s, where s = (ˆβ s)ˆβ, s = s s. By construction, We first work out d s /dt. s ˆβ = 0. (9) d s dt = d ( ) (ˆβ s)ˆβ = dt ˆβ d ) (ˆβ s + s dt Jackson gives the following result in his eq. (11.171), d (ˆβ s) = e [ (g ) dt mc s 1 ˆβ B + Setting E = 0, we obtain d ) (ˆβ s = eb dt mc ( g ˆβ dˆβ dt. (30) ( gβ 1 ) ] E. β ) s (ˆβ ˆB). (31) We also need to work out dˆβ/dt. dˆβ dt = d dt ( ) β = 1 β β d β dt β β dβ dt. (3) 11
Using dβ dt = d ( ) β 1/ 1 ( ) β = 1/ d ( β β β β) = 1 dt dt β β d β dt = ˆβ d β dt, in eq. (3), we conclude that [ ( )] dˆβ dt = 1 d β β dt ˆβ ˆβ d β. dt From eq. (8), we obtain Hence ˆβ d β/dt = 0, and we end up with d β dt = e γmc β B. dˆβ dt = eb γmc ˆβ ˆB. (33) Inserting eqs. (31) and (33) into eq. (30), we obtain ( ) ds g dt = eb [ s (ˆβ mc ˆB)]ˆβ + eb γmc s ˆβ(ˆβ ˆB). Since s ( s ˆβ)ˆβ, it immediately follows that s ˆβ(ˆβ ˆB) = s B. We can further simplify the quantity [ s (ˆβ ˆB)]ˆβ by using s ˆβ = 0 [cf. eq. (9)] and ˆβ ˆB = 0. First, consider the triple cross product ] s [ˆβ (ˆβ ˆB) = [ s (ˆβ ˆB)]ˆβ (ˆβ ˆB) s ˆβ = [ s (ˆβ ˆB)]ˆβ. However, ˆβ (ˆβ ˆB) = ˆβ(ˆβ ˆB) ˆB = ˆB. Hence, [ s (ˆβ ˆB)]ˆβ = s ˆB. Inserting eqs. (35) and (36) into eq. (34) then yields d s dt = eb mc [( g ) s + 1 ] γ s ˆB Using this result, we can evaluate d s /dt. d s = d ) ( s s dt dt = eb ( g mc 1+ 1 ) ( s γ + s ) B = eb mc 1 [( g ) s + 1 ] γ s ˆB,
which simplifies to d s dt = eb mc [( g ) s + 1 ] γ s ˆB Finally, we need to further decompose s into components along the direction of the magneticfieldandalongtheradiusvector r whichpointstothecenterofthecircularpathof the moving spin. In light of eq. (7) [with E = 0], d v/dt ˆβ ˆB. But for circular motion, ˆr ˆβ = 0 and the acceleration d v/dt points radially into the origin, i.e. d v/dt ˆr. It follows that ˆr = ˆB ˆβ, and we conclude that the unit vectors { ˆB, ˆβ, ˆr} form a mutually orthogonal right-handed triad of vectors. Thus, we can write: Note that s s B + s r, where s B ( s ˆB) ˆB and s r ( s ˆr)ˆr. (34) d s B dt ( = ˆB d s ) dt since B is time-independent by assumption and B d s dt B ( s B) = 0, ˆB = 0, (35) in light of eq. (8). Thus, s B is a constant in time, from which it follows that d s r dt = d dt ( s + s B ) = d s dt. (36) Hence, the equations of motion for the components of the spin vector are: d s B dt = 0, d s r dt = eb mc d s dt = eb mc [( g [( g after using s B ˆB = ( s ˆB) ˆB ˆB = 0. ) s + 1 ] γ s r ˆB, ) s r + 1 ] γ s ˆB, (b) For the CERN Muon Storage Ring, the orbit radius is R =.5 meters and B = 17 10 3 gauss. What is the momentum of the muon? What is the time dilation factor γ? How many periods of precession T = π/ω occur per observed laboratory mean lifetime of the muons? [Relevant data: m µ = 105.66 MeV, τ 0 =. 10 6 s, a α/(π) where α 1/137.] For circular motion, a = d v dt = v R ˆr. (37) 13
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. (1.38) and(1.39) of Jackson yield d v dt = e γmc v B. (38) Thus, if B points in the z-direction, then v = vˆθ and the circular motion is clockwise in the x y plane. Combining eqs. (37) and (38), it follows that γmv = ebr c, (39) which we recognize as the relativistic momentum of the muon, p µ. Using eq. (1.4) of Jackson, we can rewrite eq. (39) as 3 Hence, The γ-factor is p µ (MeV/c) = 3 10 4 BR (gauss-cm). p µ = (3 10 4 )(1.7 10 4 )(50) MeV/c = 1.75 10 3 MeV/c. γ = E ( ) mc = (p c +m c 4 ) 1/ p 1/ = mc m c +1. The muon rest energy is mc = 105.66 MeV. Hence, ] γ = [1+ (1.75 103 ) 1/ = 1.11. (105.66) The number of periods of precession, T = π/ω, occurring per observed mean muon lifetime, γτ 0 = γ(. 10 6 s), is given by 4 γτ 0 T = γτ 0Ω π = γτ 0eBa πmc = γ τ 0 va πr, where eq. (39)was usedtoarriveatthefinalresult above. Sinceγ 1, wecanapproximate v c. In addition, we take a = 1 (g ) α π, where α 1 137, as predicted at lowest non-trivial order in quantum electrodynamics. Hence, γτ 0 T γ τ 0 cα 4π R = (1.11) (. 10 6 s)(3 10 10 cm s 1 ) 4π (50 cm)(137) = 7.156. 3 The factor of 3 10 4 arises as follows. In gaussian units, e = 4.8 10 10 esu and 1 MeV= 1.6 10 6 ergs. Hence, the conversion factor between ergs and MeV is 4.8 10 10 /1.6 10 6 = 3 10 4. 4 Note that in the laboratory frame, the observed muon lifetime is given by γτ 0, where τ 0 is the muon lifetime in the muon rest frame. 14
(c) Express the difference frequency Ω in units of orbital rotation frequency and compute how many precessional periods (at the difference frequency) occur per rotation for a 300 MeV muon, a 300 MeV electron, a 5 GeV electron (this last typical of the e + e storage ring at Cornell). NOTE: The energy values above correspond to the total relativistic energies. For a 300 MeV muon, γ = E mc = 300 105.66 =.839, and Ω = eba mc = γω B a γω B α π = 3.3 10 3 ω B. One revolution occurs in time t = πr/v. In this time, the number of periods of precession, T = π/ω, is given by ( )( ) t πr Ω T = = ΩR v π v. We can rewrite the above result using eq. (39), which yields Hence, for a 300 MeV muon, we have Hence, R v = γmc eb = 1 ω B. t T = Ω γα ω B π = 3.3 10 3. For a 300 MeV electron, we use m e c = 511 kev to obtain Finally, for a 5 GeV electron, we have γ = 300 0.511 = 587. t T = Ω ω B γα π = 0.68. It follows that γ = 5000 0.511 = 9.785 103. t T = Ω γα ω B π = 11.37. 15
4. [Jackson, problem 14.4] Using the Liénard-Wiechert fields, discuss the time-averaged power radiated per unit solid angle in nonrelativistic motion of a particle with charge e, moving: (a) along the z axis with instantaneous position z(t) = acosω 0 (t), (b) in a circle of radius R in the x y plane with constant angular frequency ω 0. Sketch the angular distribution of the radiation of the radiation and determine the total power radiated in each case. (a) Case 1: Non-relativistic motion of a particle with charge e moving along the z-axis with instantaneous position z(t) = acosω 0 (t). where We make use of eq. (14.0) of Jackson, which is relevant for non-relativistic motion, ( ) dp dω = e 4πc ˆn ˆn d β, (40) dt In this case, we have β = v c = 1 c d x dt. x(t) = ẑacosω 0 t, which yields dβ dt = ẑ aω 0 cosω 0 t. c Working out the absolute square of the triple product in eq. (40), ( ) ( ) ˆn ˆn d β = dt ˆn ˆn d β d β d = ( ) β ˆn d β (41) dt dt dt dt = a ω 4 0 c cos ω 0 t [ 1 (ˆn ẑ) ] = a ω 4 0 c cos ω 0 t sin θ. In obtaining the final result above, we chose to work in a coordinate system in which the origin corresponds to the instantaneous position of the charged particle, and the unit vector ˆn has polar angle θ and azimuthal angle φ with respect to the z-axis, ˆn = ˆx sinθ cosφ+ŷ sinθ sinφ+ẑ cosθ. (4) The time-averaged power is easily obtained by noting that 5 cos ω 0 t = 1. 5 To compute the time-average of cos ω 0 t, note that the time averages satisfy cos ω 0 t = sin ω 0 t, and cos ω 0 t+sin ω 0 t = 1. 16
Hence, it follows that dp = e a 0 ω4 0 sin θ. (43) dω 8πc 3 In Figure 1, the angular distribution of the radiated power is exhibited as a polar plot. Figure 1: A polar plot of the angular distribution of the power radiated by a charged particle moving non-relativistically along the z axis with instantaneous position z(t) = acosω 0 (t). The angular distribution is given by eq. (43) and is proportional to sin θ. This plot was created with Maple 15 software. Integrating over the solid angle yields the total radiated power, P = e a ω 4 0 3c 3. (b) Case : Non-relativistic motionof a particle with charge e moving in a circle of radius R in the x y plane with constant angular frequency ω 0. For circular motion in the x y plane, the trajectory of the particle is given by x(t) = R(ˆx cosω 0 t+ŷ sinω 0 t). Then, we easily compute dβ dt = 1 d x c dt = ω 0 c x(t). We again choose to work in a coordinate system in which the origin corresponds to the instantaneous position of the charged particle, and the unit vector ˆn given by eq. (4) has polar angle θ and azimuthal angle φ with respect to the z-axis. Consequently, ˆn d β dt = ω 0 R (cosω 0 t sinθ cosφ+sinω 0 t sinθ sinφ). c 17
Evaluating the absolute square of the triple cross product as in part (a) [cf. eq. (41)], we obtain: ( ) ˆn ˆn d β = ω4 0 R [ 1 sin θ(cosφ cosω dt c 0 t+sinφ sinω 0 t) ] Using eq. (40), it follows that = ω4 0 R c [ 1 sin θ cos (ω 0 t φ) ]. dp dω = e ω0r 4 [ 1 sin θ cos (ω 4πc 3 0 t φ) ]. The time-averaged power is easily obtained by noting that cos (ω 0 t φ) = 1. Employing the trigonometric identity, 1 1 sin θ = 1 (1+cos θ), it follows that dp dω = e ω 4 0 R 8πc 3 ( 1+cos θ ). (44) In Figure, the angular distribution of the radiated power is exhibited as a polar plot. Figure : A polar plot of the angular distribution of the power radiated by a charged particle moving non-relativistically in a circle of radius R in the x y plane with constant angular frequency ω 0. The angular distribution is given by eq. (44) and is proportional to 1+cos θ. This plot was created with Maple 15 software. Integrating over solid angles yields the total radiated power, P = e ω 4 0R 3c 3. 5. [Jackson, problem 14.5] A nonrelativistic particle of charge ze, mass m and kinetic energy E makes a head-on collision with a fixed central force field of finite range. The interaction is repulsive and described by a potential V(r), which becomes greater than E at close distances. 18
(a) Show that the total energy radiated is given by W = 4 z e m dv dr 3 m c 3 dr V(rmin ) V(r), where is the closest distance of approach in the collision. Consider a particle with kinetic energy E at t = that is initially an infinite distance away and is headed in a radial direction toward the origin. Because the potential V(r) is repulsive and becomes greater than E at close distances, there is a distance of closest approach,, where the particle s radial velocity drops to zero. At this point, the particle collides head on with the central force field and is turned around. It now travels back along its original radial path until it reaches its original starting point (an infinite distance away) at t =. If the particle does not radiate, then we can use energy conservation to compute the instantaneous velocity of the particle at all points along its trajectory. In particular, the conservation of the sum of the kinetic and potential energy yields E = 1 m[v(r)] +V(r), for r <, (45) since the potential is assumed to be of finite range which means that lim r V(r) = 0. At the point of closes approach to the origin, v( ) = 0. Hence, it follows from eq. (45) that E = V( ). (46) Writing v(r) = dr/dt, we can solve for the velocity using eq. (45), v(t) = dr dt = ± E V(r), (47) m where we employ theminus signas theparticle moves towardthe origin andtheplus sign as the particle moves away from the origin. The acceleration can be obtained by differentiating eq. (47) with respect to t. However, a more direct computation uses Newton s second law, F = m a = V(r) = ˆr dv dr, which yields a = ˆr 1 dv m dr. (48) During the period of acceleration, the particle radiates and hence loses energy, W. Thus, it is not justified to ignore this energy loss in the energy balance equation given in eq. (45). Nevertheless, if W E, then it is justified in first approximation to ignore the radiated energy loss in deriving the acceleration given in eq. (48). 6 Thus, we shall assume 6 To properly take the radiation loss into account, we must address the question of radiation reaction, which is treated in Chapter 16 of Jackson. It turns out that this topic involves numerous subtleties, not all of which are completely understood. For further details, check out the first few sections of Chapter 16. 19
that one can neglect the energy loss due to radiation and check for consistency at the end of the computation. In this case, we can use the Larmor formula [cf. eq. (14.) of Jackson] for the instantaneous power emitted by a nonrelativistic, accelerated charge, P = 3 z e c 3 a = 3 z e m c 3 dv dr. (49) where eq. (48) has been used for the acceleration. To compute W, we first consider the energy emitted by the radiation from the initial position of the particle at r = until the point of closest approach to the origin,. Then, 7 W = P ( r(t) ) dt = rmin P(r) dt m dr dr = dr P(r) E V(r). (50) Since we are neglecting the energy loss in computing the instantaneous acceleration of the particle, theenergylossoftheparticleasitmovesfromthedistanceofclosestapproachback out to infinity again yields eq. (50). Hence the total energy radiated during < t < is just twice that of eq. (50), m dr W = P(r) V( ) V(r), after employing eq. (46). Finally, we substitute for P using eq. (49), which yields W = 4 3 z e m m c 3 dv dr To justify this computation, we would have to show that dr V(rmin ) V(r), (51) W 1 mv 0, (5) where E 1 mv 0 and v 0 is the initial velocity at time t =. Since the motion is nonrelativistic, we have v 0 c, and one can check that for for reasonable potentials, eq. (5) is satisfied [cf. eq. (56)]. (b) If the interaction is a Coulomb potential V(r) = zze /r, show that the total energy radiated is W = 8 zmv0 5 45 Zc, 3 where v 0 is the velocity of the charge at infinity. 7 Note that we use the minus sign in eq. (47) when the particle moves toward the origin. This minus sign is then used to reverse the limits of integration in eq. (50). 0
Substituting V(r) = zze /r into eq. (51) yields W = 4 z e 5 m 3 m c 3 (zz)3/ It is more convenient to rewrite this as W = 4 3 z e 5 m c 3 (zz)3/ We now change variables by defining m dr 1 r 4 1 1 r u = r 1. Then dr = du and r = (u+1). Hence, dr 1 r 7/ r 1 = 1 r 5/ min dr 1 r 7/ r 1 0 du u(u+1) 7/.. (53) We make one more change of variables by defining u = x. Then du = xdx = udu so that dx = du/ u. Hence, dr 1 = dx. (54) r 7/ r 1 r 5/ min 0 (x +1) 7/ To evaluate this integral, we first consider the well-known result, dx (x +a ) 3/ = x a x +a. The desired integral can be obtained by differentiating twice with respect to a. Alternatively, one can use an integral table to obtain dx (x +a ) = 1 { x 7/ a 6 x +a x 3 3 (x +a ) + 1 } x 5. 3/ 5 (x +a ) 5/ Hence, 0 It follows that dx (x +1) 7/ = x x +1 3 x 3 (x +1) 3/ + 1 5 dr 1 r 7/ r 1 1 x 5 (x +1) 5/ = 16. 15r 5/ min 0. = 1 3 + 1 5 = 8 15.
Plugging this result back into eq. (53), we end up with W = 4 3 z e 5 m m c 3 (zz)3/ 16 15r 5/ min. (55) We can simplify this expression by writing E = 1 mv 0, where v 0 is the initial velocity of the particle at t =. Using eq. (46), 1 mv 0 = V() = zze. Solving for and inserting this result back into eq. (55) yields our final result, W = 64 45 z e 5 m m c 3 (zz)3/ The condition of eq. (5) then implies that 16z 45Z ( ) mv 5/ 0 = 8 zmv0 5 zze 45 Zc. 3 ( v0 ) 3 1, (56) c which is always satisfied for non-relativistic motion [assuming that z/z O(1)].