Physics 506 Winter 2006 Homework Assignment #9 Solutions

Transcription

1 Physics 506 Winte 2006 Homewok Assignment #9 Solutions Textbook poblems: Ch. 12: 12.2, 12.9, 12.13, a) Show fom Hamilton s pinciple that Lagangians that diffe only by a total time deivative of some function of the coodinates and time ae equivalent in the sense that they yield the same Eule-Lagange equations of motion. Suppose Lagangians L 1 and L 2 diffe by a total time deivative of the fom L 2 = L 1 + d dt f(q i(t), t) Witing out the time deivative explicitly gives L 2 = L 1 + f q i + f q i t The Eule-Lagange equations fo L 2 ae deived fom Then L f q j + 2 f q i q i q i q j q i t L 2 + f q i q i q i L 2 q i d dt L 2 q i q i q i q i d L f q j + 2 f dt q i q i q j q i t d f dt q i d L 1 + df dt q i q i dt d f dt q i d L 1 dt q i As a esult, both L 1 and L 2 yield the same Eule-Lagange equations. Note that it is pehaps moe staightfowad to conside the change in the action S 2 = t2 t 1 L 2 dt = t2 t 1 (L 1 + df t2 dt ) dt = L 1 dt + f(q i (t 2 ), t 2 ) f(q i (t 1 ), t 1 ) t 1 In othe wods, the additional of a total time deivative only changes the action by a suface tem. So long as we do not vay the path at its endpoints (δq i (t 1 ) = δq i (t 2 ) = 0) we end up with the same equations of motion.

2 b) Show explicitly that the gauge tansfomation A α A α + α Λ of the potentials in the chaged-paticle Lagangian (12.12) meely geneates anothe equivalent Lagangian. We stat with the Lagangian L = mc 2 1 u 2 /c 2 + e c u A eφ In components, the gauge tansfomation A µ A µ + µ Λ eads Φ Φ + 1 c t Λ, A A Λ In this case, the Lagangian changes by the tem δl = e ( ) + u Λ c t Howeve, fo Λ = Λ( x(t), t), the above is just the total time deivative δl = e c As a esult the Lagangian changes by a total time deivative. Thus the gauge tansfomed Lagangian is equivalent to the oiginal one in the sense of pat a) The magnetic field of the eath can be epesented appoximately by a magnetic dipole of magnetic moment M = gauss-cm 3. Conside the motion of enegetic electons in the neighbohood of the eath unde the action of this dipole field (Van Allen electon belts). [Note that M points south.] a) Show that the equation fo a line of magnetic foce is = 0 sin 2 θ, whee θ is the usual pola angle (colatitude) measued fom the axis of the dipole, and find an expession fo the magnitude of B along any line of foce as a function of θ. Taking a spheical coodinate system with the ẑ axis pointing noth, the magnetic dipole moment of the eath can be epesented by m = Mẑ. This gives ise to a magnetic field 3ˆ(ˆ m) m B = 3 = M (ẑ 3 cos θˆ) 3 To obtain the equation fo a line of magnetic foce, we fist esolve the above into spheical coodinate components using ẑ = ˆ cos θ ˆθ sin θ. This gives dλ dt B = M 3 (2 cos θˆ + sin θˆθ) (1) We now note that the equation fo a line of magnetic foce can be witten paametically as = (λ), θ = θ(λ). In this case, the tangent to the cuve is given by λ = d dθ ˆ + dλ dλ ˆθ (2)

3 Since this tangent vecto must point in the same diection as B, we may take atios of ˆ and ˆθ components of (1) and (2) to obtain 2 cos θ sin θ = d/dλ dθ/dλ = 1 d dθ This gives ise to the sepaable equation d/ = 2 cot θ dθ which may be integated to yield (θ) = 0 sin 2 θ Fom (1), the magnitude of the magnetic field is B = M cos 2 θ 3 Along the line = 0 sin 2 θ, this becomes B(θ) = M cos 2 θ 3 0 sin6 θ (3) b) A positively chaged paticle cicles aound a line of foce in the equatoial plane with a gyation adius a and a mean adius (a ). Show that the paticle s azimuthal position (east longitude) changes appoximately linealy in time accoding to φ(t) = φ 0 3 ( a ωb (t t 0 ) 2 whee ω B is the fequency of gyation at adius. Since the magnetic field is non-unifom, the gyating paticle will pick up a dift velocity. We may use the geneal expession fo the dift velocity v D = 1 ω B c (v v2 )( ˆ c ˆB) whee c is the adius of cuvatue vecto. So long as the paticle mainly cicles aound a line of foce, we assume the paticle s velocity is almost entiely pependicula to the lines of foce. Thus v D 1 2ω B c v 2 (ˆ ( ˆθ)) = ω B 2 c a 2 ˆφ (4) whee we have used the fact that the magnetic field in the equitoial plane points in the ˆθ (ẑ) diection and that the adius of cuvatue vecto points along ˆ. The adius of cuvatue can now be obtained fom the equation fo the magnetic

4 foce lines = 0 sin 2 θ. Howeve, we take a shotcut given by the elation ight afte (12.60) in Jackson B B = ˆ c c fo a cul-fee field B. In the equitoial plane, the magnitude of the magnetic field is B = M/c 3 and the pependicula gadient diection is the ˆ diection. The above elation then gives c = 3 whee is the distance fom the cente of the eath (taken as the mean adius). Substituting this into (4) gives v D = 3a2 2 ω B ˆφ This dift velocity (in the equitoial plane) is given by v = φ ˆφ. As a esult, a simple integation gives φ φ 0 = 3 ( a ωb (t t 0 ) (5) 2 c) If, in addition to its cicula motion of pat b), the paticle has a small component of velocity paallel to the lines of foce, show that it undegoes small oscillations in θ aound θ = π/2 with a fequency Ω = (3/ 2)(a/)ω B. Find the change in longitude pe cycle of oscillation in lattitude. So long as v v, we can ignoe its effect on the dift velocity. On the othe hand, by consevation of enegy and of flux though obits of the paticle, we have v0 2 = v 2 + B(z) v2,0 v 2 B + ω2 Ba 2 B(z) 0 B 0 whee z is the coodinate paallel to the field line. Using the geometical elation θ π 2 (z/) valid nea the equitoial plane, and substituting it into (3), we obtain B(z) M sin 2 (z/) 3 cos 6 (z/) M 3 (1 + 3 M ( ( z ) 2 Hence B(z)/B (z/)2, and ( z ) 1/2 ( v 2 0 = v 2 + (ω Ba ( ωb a z 2 ( z + ) 6

5 This can be witten in tems of an effective consevation of enegy equation E = 1 2 mv2 + V (z) whee V (z) = 1 2 m ( 3ωB a 2 z 2 This is an effective hamonic oscillato potential V = 1 2 mω2 z 2 with Ω = 3 ( a ) ω B 2 A complete peiod takes a time of T Ω = 2π Ω = 2 2π 3ω B ( ) a (6) Inseting this into (5) gives a change of longitude of φ = 3 2 ( a ove one complete oscillation in lattitude. ωb T Ω = ( a ) 2π d) Fo an electon of 10 MeV kinetic enegy at a mean adius = m, find ω B and a, and so detemine how long it takes to dift once aound the eath and how long it takes to execute one cycle of oscillation in latitude. Calculate the same quantities fo an electon of 10 kev at the same adius. Note that ω B = eb/γmc, while to a good appoximation the stength of the magnetic field is B 0 = M/ 3. As a esult ω B = em γmc 3 = ecm E 3 whee E is the elativistic enegy of the electon, E = γmc 2 = mc 2 + KE. Hee KE = (γ 1)mc 2 is the kinetic enegy of the electon. Solving this fo velocity gives 2KE/mc2 + (KE/mc v = c KE/mc 2 which may be substituted into the elation v v = ω B a to get a = mc2 3 2KE/mc 2 + (KE/mc 2 em The time fo an electon to dift once aound the eath ( φ = 2π) can be obtained fom (5). The esult is T 2π = 4π ( 1 3 a ω B

6 In addition, the peiod fo a lattitude oscillation is given by (6). Using M = gauss-cm 3, = cm, e = statcoul, c = cm/s and mc 2 = 511 kev, 1 ev = eg, as well as KE = 10 MeV gives 10 MeV : ω B = s 1, a = cm T 2π = 110 s, T Ω = 0.30 s Although the adius of gyation is athe lage (120 km), it is less than half a pecent of the distance fom the cente of the eath. As a esult, the gadient of the magnetic field is still quite small ove the obit of the electon, and hence the appoximations we have used ae still valid. On the othe hand, fo KE = 10 kev, we find instead 10 MeV : ω B = s 1, a = cm T 2π = s, T Ω = 1.50 s a) Specialize the Dawin Lagangian (12.82) to the inteaction of two chaged paticles (m 1, q 1 ) and (m 2, q 2 ). Intoduce educed paticle coodinates, = x 1 x 2, v = v 1 v 2 and also cente of mass coodinates. Wite out the Lagangian in the efeence fame in which the velocity of the cente of mass vanishes and evaluate the canonical momentum components, p x = L/ v x, etc. The two paticle Dawin Lagangian eads L = 1 2 m 1v m 2v c 2 (m 1v 4 1+m 2 v 4 2) q 1 q q 1q c 2 [ v 1 v 2 +( v 1 ˆ)( v 2 ˆ)] (7) We take a standad (non-elativistic) tansfomation to cente of mass coodinates = x 1 x 2, = m 1 x 1 + m 2 x 2 M whee M = m 1 + m 2. Inveting this gives x 1 = + m 2 M, x 2 = m 1 M As a esult, the individual tems in the Lagangian (7) become 1 2 m 1v m 2v2 2 = 1 2 MV µv2 (m 1 v1 4 + m 2 v2) 4 8c 2 = 1 ( 8c 2 MV 4 + 6µV 2 v 2 + 4µ m 2 m 1 M ( V v)v 2 + µ m3 1 + m 3 ) 2 M 3 v 4 v 1 v 2 = V 2 + m 2 m 1 M V v µ M v2 ( v 1 ˆ)( v 2 ˆ) = ( V ˆ + m 2 m 1 M ( V ˆ)( v ˆ) µ ( v ˆ)2 M

7 whee µ = m 1 m 2 /M is the educed mass. Fo vanishing cente of mass velocity ( V = 0) the Lagangian becomes L = 1 2 µv c 2 µm3 1 + m 3 2 M 3 v 4 q 1q 2 The canonical momentum is µq 1q 2 2Mc 2 [v2 + ( v ˆ ] (8) p = v L = µ v + 1 2c 2 µm3 1 + m 3 2 M 3 v 2 v µq 1q 2 [ v + ( v ˆ)ˆ] (9) 2Mc2 b) Calculate the Hamiltonian to fist ode in 1/c 2 and show that it is H = p2 2 ( 1 m m 2 ) + q 1q 2 ( p4 1 8c 2 m m 3 2 ) + q 1q 2 2m 1 m 2 c 2 [You may disegad the compaison with Bethe and Salpete.] ( p 2 + ( p ˆ ) The Hamiltonian is obtained fom the Lagangian (8) by the tansfomation H = p v L. Note, howeve, that we must invet the elation (9) to wite the esulting H as a function of p and. We stat with H = p v 1 2 µv2 1 8c 2 µm3 1 + m 3 2 M 3 v 4 + q 1q 2 = p2 2µ 1 2µ ( p µ v)2 1 8c 2 µm3 1 + m 3 2 M 3 v 4 + q 1q 2 + µq 1q 2 2Mc 2 [v2 + ( v ˆ ] + µq 1q 2 2Mc 2 [v2 + ( v ˆ ] (10) Since we only wok to fist ode in 1/c 2, we do not need to completely solve (9) fo v in tems of p. Instead, it is sufficient to note that v = 1 µ p + O ( 1 c 2 ) Inseting this into (10) gives (to ode 1/c 2 ) H = p2 2µ 1 m m 3 2 8c 2 M 3 µ 3 p 4 + q 1q 2 ( = p ) ( p4 1 2 m 1 m 2 8c 2 m q 1q 2 2Mµc 2 [p2 + ( p ˆ ] + 1 ) m 3 + q 1q 2 q 1 q m 1 m 2 c 2 [p2 + ( p ˆ ] An altenative Lagangian density fo the electomagnetic field is L = 1 8π αa β α A β 1 c J αa α

8 a) Deive the Eule-Lagange equations of motion. Ae they the Maxwell equations? Unde what assumptions? The vaiations of the Lagangian density ae L A µ = 1 c J ν, This leads to the Eule-Lagange equations o L ( ν A µ ) = 1 4π ν A µ 1 c J µ + 1 4π ν ν A µ = 0 ν ν A µ = 4π c J ν (11) This can be ecognized as the equation of motion fo the vecto potential A µ in the Loenz gauge µ A µ = 0. To see this, ecall that if we define F µν = µ A ν ν A µ then the Maxwell equation µ F µν = (4π/c)A ν becomes which yields (11) povided µ A µ = 0. µ µ A ν ν ( µ A µ ) = 4π c J ν b) Show explicitly, and with what assumptions, that this Lagangian density diffes fom (12.85) by a 4-divegence. Does this added 4-divegence affect the action o the equations of motion? The diffeence between this altenative Lagangian density and the standad Maxwell Lagangian density can be expessed as L 0 = 1 16π F µνf µν 1 c J µ A m u L = 1 16π F µνf µν 1 8π µa ν µ A ν = 1 8π µa ν ν A µ = 1 8π µ(a ν ν A µ A µ ν A ν ) 1 8π ( µ A µ whee the last line follows by simple manipulation of deivatives. This demonstates that the altenative Lagangian density diffes fom the coect one by a 4-divegence so long as we ae esticted to Loenz gauge, µ A µ = 0. Finally, we ecall that the action is the fou-dimensional integal of the Lagangian density. In this case, since the integal of a 4-divegence gives a suface tem, the action is only affected by a possible suface tem. Assuming the vecto potential A µ falls off sufficiently at infinity, this suface tem will in fact vanish, so the action is actually unchanged. This then demonstates that the equations of motion ae unaffected by the addition of a 4-divegence to the Lagangian density.