Physics 506 Winte 2008 Homewok Assignment #10 Solutions Textbook poblems: Ch. 12: 12.10, 12.13, 12.16, 12.19 12.10 A chaged paticle finds itself instantaneously in the equatoial plane of the eath s magnetic field assumed to be a dipole field) at a distance R fom the cente of the eath. Its velocity vecto at that instant makes an angle α with the equatoial plane v /v = tan α). Assuming that the paticle spials along the lines of foce with a gyation adius a R, and that the flux linked by the obit is a constant of the motion, find an equation fo the maximum magnetic latitude λ eached by the paticle as a function of the angle α. Plot a gaph not a sketch) of λ vesus α. ak paametically along the cuve the values of α fo which a paticle at adius R in the equatoial plane will hit the eath adius R 0 ) fo R/R 0 = 1.2, 1.5, 2.0, 2.5, 3, 4, 5. Since the paticle spials along the lines of foce ie magnetic field lines), we fist set out to calculate what these lines ae. Fo a dipole field with a magnetic dipole moment m = ẑ, the magnetic field is B = 3ˆˆ m) m 3 = ẑ 3 cos θˆ) 3 whee θ is the standad pola angle in spheical coodinates. This expession may be tansfomed entiely into spheical coodinates by witing ẑ = ˆ cos θ ˆθ sin θ. The esult is B = 3 2 cos θˆ + sin θˆθ) 1) Because of azimuthal symmety, we can think of this as a vecto field in the and θ diections. What we want to do now is to come up with a paametic equation = λ), θ = θλ) descibing the field lines. Hee λ is a paamete along the cuve. The key to elating this paametic equation to the magnetic field is to ealize that the tangent to the cuve should be identified with the magnetic field vecto B. Since the tangent to the cuve is given by λ = d dθ ˆ + dλ dλ ˆθ 2) 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) = R sin 2 θ 3)
Note that we have chosen the initial condition that π/2) = R, since θ = π/2 coesponds to the equatoial plane. In addition to the equation fo a magnetic field line, we also need the magnitude of the magnetic field. This may be computed fom 1) B = 1 + 3 cos 2 θ 3 Along the line = R sin 2 θ, this becomes Bθ) = R 3 1 + 3 cos2 θ sin 6 θ 4) Since the flux linked by the obit is a constant of motion an adiabatic invaiant), we end up with the velocity elation v θ) 2 = v0 2 v,0 2 Bθ) = v,0 2 B + v2,0 1 Bθ) ) 0 B 0 whee we have used v 2 0 = v 2,0 + v2,0. The paticle stats at an angle θ 0 = π/2. Fom 4), the initial magnetic field is B 0 = /R 3. hence v θ) 2 = v 2,0 + v2,0 1 ) 1 + 3 cos2 θ sin 6 θ The minimum value of θ is eached at the tuning point when v θ) = 0. This coesponds to ) 1 + 3 v,0 2 + cos2 θ v2,0 1 sin 6 = 0 θ 1 + 3 cos2 θ sin 6 θ = 1 + v2,0 v 2,0 Setting θ = π/2 λ whee λ is the magnetic latitude, and using v,0 /v,0 = tan α then gives 1 + 3 sin 1 + tan 2 2 λ α = cos 6 λ o We may plot λ vesus α as ) 1/2 α = tan 1 1 + 3 sin 2 λ cos 6 1 λ
λ 80 60 40 R /R 0 1.5 2.0 1.2 2.5 4 5 3 20 20 40 60 80 Since the magnetic field line is given by 3), the paticle will hit the eath when R 0 = R sin 2 θ = R cos 2 λ, o λ = cos 1 R 0 /R. These values ae indicated on the plot. 12.13 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 2 1+ 1 2 m 2v 2 2+ 1 8c 2 m 1v 4 1+m 2 v 4 2) q 1 q 2 12 + q 1q 2 2 12 c 2 v 1 v 2 + v 1 ˆ) v 2 ˆ)] 5) We take a standad non-elativistic) tansfomation to cente of mass coodinates = x 1 x 2, R = m 1 x 1 + m 2 x 2 whee = m 1 + m 2. Inveting this gives x 1 = R + m 2, x 2 = R m 1 As a esult, the individual tems in the Lagangian 5) become 1 2 m 1v1 2 + 1 2 m 2v2 2 = 1 2 V 2 + 1 2 µv2 m 1 v1 4 + m 2 v2) 4 8c 2 = 1 8c 2 V 4 + 6µV 2 v 2 + 4µ m 2 m 1 V v)v 2 + µ m3 1 + m 3 ) 2 3 v 4 v 1 v 2 = V 2 + m 2 m 1 V v µ v2 v 1 ˆ) v 2 ˆ) = V ˆ) 2 + m 2 m 1 V ˆ) v ˆ) µ v ˆ)2 α
whee µ = m 1 m 2 / is the educed mass. Fo vanishing cente of mass velocity V = 0) the Lagangian becomes L = 1 2 µv2 + 1 8c 2 µm3 1 + m 3 2 3 v 4 q 1q 2 The canonical momentum is p i = L/ v i, which gives µq 1q 2 2c 2 v2 + v ˆ) 2 ] 6) p = µ v + 1 2c 2 µm3 1 + m 3 2 3 v 2 v µq 1q 2 v + v ˆ)ˆ] 7) 2c2 b) Calculate the Hamiltonian to fist ode in 1/c 2 and show that it is H = p2 2 1 m 1 + 1 m 2 ) + q 1q 2 p4 1 8c 2 m 3 1 + 1 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) 2 ) The Hamiltonian is obtained fom the Lagangian 6) by the tansfomation H = p v L. Note, howeve, that we must invet the elation 7) 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 3 v 4 + q 1q 2 = p2 2µ 1 2µ p µ v)2 1 8c 2 µm3 1 + m 3 2 3 v 4 + q 1q 2 + µq 1q 2 2c 2 v2 + v ˆ) 2 ] + µq 1q 2 2c 2 v2 + v ˆ) 2 ] 8) Since we only wok to fist ode in 1/c 2, we do not need to completely solve 7) fo v in tems of p. Instead, it is sufficient to note that v = 1 µ p + O 1 c 2 ) Inseting this into 8) gives to ode 1/c 2 ) H = p2 2µ 1 m 3 1 + m 3 2 8c 2 3 µ 3 p 4 + q 1q 2 = p2 1 + 1 ) p4 1 2 m 1 m 2 8c 2 m 3 1 + q 1q 2 2µc 2 p2 + p ˆ) 2 ] + 1 ) m 3 + q 1q 2 q 1 q 2 + 2 2m 1 m 2 c 2 p2 + p ˆ) 2 ] 12.16 a) Stating with the Poca Lagangian density 12.91) and following the same pocedue as fo the electomagnetic fields, show that the symmetic stess-enegymomentum tenso fo the Poca fields is Θ αβ = 1 g αγ F γλ F λβ + 14 gαβ F λν F λν + µ 2 A α A β 12 gαβ A λ A λ )]
The Poca Lagangian density is Since we find L = 1 16π F µνf µν + 1 8π µ2 A µ A µ T µν = L µ A λ ν A λ η µν L T µν = 1 F µλ ν A λ + 1 16π ηµν F 2 1 8π µ2 η µν A 2 whee we have used a shothand notation F 2 F µν F µν and A 2 A µ A µ. In ode to convet this canonical stess tenso to the symmetic stess tenso, we wite ν A λ = F ν λ + λ A ν. Then T µν = 1 F µλ F ν λ 1 4 ηµν F 2 + 1 2 µ2 η µν A 2 ] 1 F µλ λ A ν = 1 F µλ F ν λ 1 4 ηµν F 2 + 1 2 µ2 η µν A 2 λ F µλ )A ν ] 1 λf µλ A ν ) Using the Poca equation of motion λ F λµ + µ 2 A µ = 0 then gives whee T µν = Θ µν + λ S λµν Θ µν = 1 F µλ F ν λ 1 4 ηµν F 2 µ 2 A µ A ν 1 2 ηµν A 2 ) ] 9) is the symmetic stess tenso and S λµν = 1/)F λµ A ν is antisymmetic on the fist two indices. b) Fo these fields in inteaction with the extenal souce J β, as in 12.91), show that the diffeential consevation laws take the same fom as fo the electomagnetic fields, namely α Θ αβ = J λf λβ c Taking a 4-divegence of the symmetic stess tenso 9) gives µ Θ µν = 1 µ F µλ F ν λ + F µλ µ F ν λ 1 2 F ρλ ν F ρλ µ 2 µ A µ A ν + A µ µ A ν A λ ν A λ ) ] = 1 µ F µλ F ν λ + 1 2 F ρλ2 ρ F νλ ν F ρλ ) + µ 2 A λ ν A λ λ A ν ) ] = 1 µ F µλ + µ 2 A λ )F ν λ + 1 2 F ρλ ρ F νλ + λ F ρν + ν F λρ ) ] = 1 c J λ F ν λ = 1 c J λf λν
Note that in the second line we have used the fact that µ A µ = 0, which is automatic fo the Poca equation. To obtain the last line, we used the Bianchi identity 3 ρ F νλ] = 0 as well as the Poca equation of motion. c) Show explicitly that the time-time and space-time components of Θ αβ ae Θ 00 = 1 8π E2 + B 2 + µ 2 A 0 A 0 + A A)] Θ i0 = 1 E B) i + µ 2 A i A 0 ] Given the explicit fom of the axwell tenso, it is staightfowad to show that Thus Θµν = 1 F 2 F µν F µν = 2E 2 B 2 ), A 2 A µ A µ = A 0 ) 2 A 2 F µλ F ν λ + 1 2 ηµν E 2 B 2 ) µ 2 A µ A ν 1 2 ηµν A 0 ) 2 A ] 2 )) The time-time component of this is Θ 00 = 1 F 0i F 0 i + 1 2 E2 B 2 ) µ 2 A 0 ) 2 1 2 A0 ) 2 A ] 2 )) = 1 1 2 E2 + B 2 ) 1 2 µ2 A 0 ) 2 + A ] 2 ) = 1 8π E 2 + B 2 + µ 2 A 0 ) 2 + A 2 ) Similaly, the time-space components ae Θ 0i = 1 F 0 j F ij µ 2 A 0 A i] = 1 = 1 ɛijk E j B k µ 2 A 0 A i] = 1 ] E j ɛ ijk B k ) µ 2 A 0 A i] E B) i + µ 2 A 0 A i] 12.19 Souce-fee electomagnetic fields exist in a localized egion of space. Conside the vaious consevation laws that ae contained in the integal of α αβγ = 0 ove all space, whee αβγ is defined by 12.117). a) Show that when β and γ ae both space indices consevation of the total field angula momentum follows. Note that Hence αβγ = Θ αβ x γ Θ αγ x β 0ij = Θ 0i x j Θ 0j x i = cg i x j g j x i ) = cɛ ijk g x ) k = cɛ ijk x g ) k
whee g is the linea momentum density of the electomagnetic field. Since x g is the angula momentum density, integating 0ij ove 3-space gives the field angula momentum ij 0ij d 3 x = cɛ ijk x g ) k d 3 x = cɛ ijk L k The consevation law µ µij = 0 then coesponds to the consevation of angula momentum in the electomagnetic field. b) Show that when β = 0 the consevation law is d X dt = c2 Pem E em whee X is the coodinate of the cente of mass of the electomagnetic fields, defined by X u d 3 x = xu d 3 x whee u is the electomagnetic enegy density and E em and P em ae the total enegy and momentum of the fields. In this case, we have 0i 00i d 3 x = = Θ 00 x i Θ 0i x 0 ) d 3 x ux i cg i x 0 ) d 3 x = ux i c 2 tg i ) d 3 x aking use of the definition ux i d 3 x = EX i whee E = u d 3 x is the total field enegy, we have simply 0i = EX i c 2 tp i whee P = g d 3 x is the linea) field momentum. chage, its time deivative must vanish. This gives Since 0i is a conseved 0 = d dt E X) c 2 d dt t P ) = E d X dt c2 P whee we used the fact that enegy and momentum ae conseved, namely de/dt = 0 and d P /dt = 0). The esult d X/dt = c 2 P /E then follows.