1 (a) The kinetic energy of the rolling cylinder is. a(θ φ)

Transcript

1 NATURAL SCIENCES TRIPOS Part II Wednesday 3 January am to 2.30pm THEORETICAL PHYSICS I Answers (a) The kinetic energy of the rolling cylinder is T c = 2 ma2 θ2 + 2 I c θ 2 where I c = ma 2 /2 is its moment of inertia about its axis. The kinetic energy of the plank is T p = 2 m2 (ẋ 2 + ẏ 2 ) + 2 I p φ 2 where I p = ml 2 /2 is its moment of inertia about its centre. The potential energy of the Plank is V p = mgy. Therefore L = T c + T p V p = 3 4 ma2 θ ml2 φ2 + 2 m(ẋ2 + ẏ 2 ) mgy φ a(θ φ) a θ φ aθ (b) The top of the cylinder is at (aθ, 0). Therefore the point of contact of plank and cylinder is at a(θ + sin φ, cos φ ). If there is no slipping, the vector from there to the centre of the plank is a(θ φ)(cos φ, sin φ). Hence x/a = θ + sin φ + (θ φ) cos φ, y/a = + cos φ (θ φ) sin φ. for continuation of question

2 2 (c) We have ẋ/a = θ + φ cos φ + ( θ φ) cos φ + (φ θ) φ sin φ, ẏ/a = φ sin φ ( θ φ) sin φ + (φ θ) φ cos φ. Hence (ẋ 2 + ẏ 2 )/a 2 = [ θ( + cos φ) + (φ θ) φ sin φ] 2 + [ θ sin φ (φ θ) φ cos φ] 2 = 2 θ 2 ( + cos φ) + (φ θ) 2 φ2 + 2(φ θ) θ φ sin φ The canonical momentum p θ is p θ = L θ = 3 2 ma2 θ + 2 ma2 [4 θ( + cos φ) + 2(φ θ) φ sin φ] = 2 ma2 [ θ(7 + 4 cos φ) + 2(φ θ) φ sin φ] The canonical momentum p φ is p φ = L φ = 2 ml2 φ + ma 2 (φ θ)[(φ θ) φ + θ sin φ] (d) To second order in θ and φ, so that y/a 2 φ2 θφ, (ẋ 2 + ẏ 2 )/a 2 4 θ 2 L 4 ma2 θ ml2 φ2 mgaφ(φ 2θ) 2 To find the Hamiltonian to second order, we only need the canonical momenta to first order: Hence p θ 2 ma2 θ, pφ 2 ml2 φ p 2 θ H = p θ θ + pφ φ L ma + 6 p2 θ 2 ml + mgaφ(φ 2θ). 2 2 Hamilton s equations are p θ θ = H = 2 p θ ma, 2 φ = H = 2 p φ p φ ml, 2 ṗ θ = H θ = mgaφ ṗ φ = H φ = mga(φ θ)

3 (e) Hence θ = 2 g a φ, Writing θ = Ae iωt, φ = Be iωt, so and therefore ω 2 A = 2 g a B, 3 ga φ = 2 (θ φ). l2 ω2 B = 2 ga (A B) l2 g 2 ω 2 B = 2 ga l B 24 2 l 2 ω B, 2 ω 4 2 ga l 2 ω2 24 g 2 l = 0, 2 ω 2 = 6 ga l 2 ± l 2 a 2 The negative root corresponds to a runaway solution the plank falls off. The positive root ω + is an oscillation with θ and φ in antiphase and B A = a 2 g ω2 + = 33 a l 2. l 2 33 a 2 2 (a) We have A µ = = ( Q rc, 2 Br sin θ, 2 Br cos θ, 0 ) (φ(r)/c, 2 By, 2 Bx, 0 ) where φ(r) is the electrostatic potential due to the charge at the origin. Now A µ = (φ/c, A) where E = φ A/ t and B = A. Thus in this case E = φ and B = (0, 0, B) as required. (b) The Lagrangian is L = T V = 2 mv2 e(φ v A) In plane polar coordinates v = ṙˆr + r θˆθ and here A = Brˆθ/2, so L = 2 m(ṙ2 + r 2 θ2 ) eq r + 2 ebr2 θ (c) Lagrange s equations of motion are d dt ( ) L = L ṙ r, d dt ( ) L θ = L θ

4 Hence 4 m r = mr θ 2 + eq r + ebr θ, ( 2 ) d mr 2 θ + dt 2 ebr2 = 0. (d) We see from the equation of motion for θ that the angular momentum p θ = mr 2 θ + 2 ebr2 = J is a constant of the motion. Furthermore the Lagrangian does not depend explicitly on time, so the Hamiltonian is also constant: (e) Writing H = p r ṙ + p θ θ L = 2 m(ṙ2 + r 2 θ2 ) + eq r = E θ = φ ω L t = φ eb 2m t the equations of motion become m r = ( ) eb 2 mr φ + eq 2m r 2 = mr φ 2 + eq d ( mr 2 φ) = 0. dt r e2 B 2 2 4m r, ( ) + ebr eb φ 2m Therefore, to first order in B, the effect of the field is cancelled in a frame rotating with the Larmor frequency. 3 (a) We are given V = S 0 γ + ( ) γ+ ρ ρ 0 where ρ = ρ 0 ( ξ), i.e. ρ/ρ 0 = δ, where δ = ξ is small. Expanding S [ 0 V = ( δ) γ+ ] γ + [ S 0 = (γ + )δ + ] γ + 2 γ(γ + )δ = S 0 [ δ + γ ] 2 δ2 + O(δ 3 ) [ = S 0 ξ + γ ] 2 ( ξ)2

5 5 (b) L = T V = 2 ρ 0 ξ ξ + S 0 ( ξ γ 2 ( ξ)2 ) Lagrange s equation of motion for ξ i is L ξ i j L x j ( ξ i / x j ) L t ξ = 0 i Hence in this case i.e. 0 x i S 0 ( γ ξ) ρ 0 ξi = 0 ρ 0 ξ γs0 ( ξ) = 0. (c) The canonical momentum density π i = L/ ξ i = ρ 0 ξi. Hence H = π i ξi L i = ρ 0 ξ ξ 2 ρ ξ 0 ξ S 0 ( ξ γ ) 2 ( ξ)2 = π2 S 0 ( ξ γ ) 2ρ 0 2 ( ξ)2 The term involving the total derivative ξ gives a contribution to the total Hamiltonian equal to an integral of the field over a surface at infinity, S 0 d 3 r ξ = S 0 since the field vanishes at infinity. Hence H = d 3 r H(r, t) d 2 S ξ = 0 where (d) We have where H(r, t) = 2 ρ 0π 2 + γ 2 S 0( ξ) 2. ρ 0 ξt,l γs0 ( ξ T,L ) = 0 ξ T = 0, ξ L = 0. Thus ξ T obeys the free-particle equation of motion ξ T = 0, while for ξ L we have ρ 0 ξl γs0 ( ξ L ) = 0.

6 6 Now we need the identity which since ξ L = 0 gives ( A) = ( A) 2 A ( ξ L ) = 2 ξ L so that ξ L obeys ρ 0 ξl γs0 2 ξ L = 0 which is the wave equation with wave velocity γs 0 /ρ 0. 4 (a) Hamilton s principle of least action states that δs = 0 for variations of the motion around the classical path, where the action S is S = L dt = L dx dt for a field in one dimension. In this case we have L = L (ϕ t, ϕ xx ) where ϕ t ϕ/ t and ϕ xx 2 ϕ/ x 2. Therefore Now δs = [ L δϕ t + L ] δϕ xx dx dt ϕ t ϕ xx δϕ t = t δϕ, δϕ xx = 2 x δϕ. 2 Therefore, integrating the first term by parts once w.r.t t and the second by parts twice w.r.t x, and dropping boundary terms since the field must vanish at ± : [ δs = L + 2 t ϕ t x 2 L ϕ xx The variation in ϕ is arbitrary and therefore L 2 L = 0 t ϕ t x 2 ϕ xx ] δϕ dx dt = 0 or in this case ρa 2 ϕ t 2 ( ) + 2 EI 2 ϕ = 0. x 2 x 2

7 (b) The canonical momentum per unit length is 7 π L = ρa ϕ ϕ t t and the Hamiltonian per unit length is H = π ϕ t L = 2 ρa (c) We have H t = ρa ϕ t Applying the equation of motion, H t = ϕ 2 ( t x 2 2 ϕ t 2 ( ϕ t EI 2 ϕ x 2 ) EI + EI 2 ϕ x 2 Writing EI 2 ϕ/ x 2 ψ, the r.h.s. has the form ϕ 2 ψ t x + ψ 3 ϕ 2 t x = 2 x ) ( 2 ) 2 ϕ. x 2 3 ϕ t x. 2 + EI 2 ϕ x 2 3 ϕ t x. 2 ( ϕ ψ t x ψ 2 ϕ t x and so J = ϕ ψ t x ψ 2 ϕ t x. (d) For EI = constant, substituting a wave solution ϕ = C cos φ with φ = kx ωt in the equation of motion we have and hence the dispersion relation is ρacω 2 + EICk 4 = 0 ω = The wave and group velocities are EI ρa k2. ) c w ω k = EI ρa k, The energy per unit length is c g dω EI dk = 2 ρa k = 2c w. H = 2 ρac2 ω 2 sin 2 φ + 2 EIC2 k 4 cos 2 φ = 2 EIC2 k 4 and ψ = EICk 2 cos φ, so the energy current is J = Cω sin φ EICk 3 sin φ + EICk 2 cos φ Cωk cos φ = EIC 2 ωk 3. Therefore the velocity of energy transfer is J /H = 2ω/k = 2c w = c g.

8 8 5 (a) We have L = ( µ φ )( µ φ) V (φ) The Euler-Lagrange equation for φ is L φ j x j L ( φ/ x j ) L t φ = 0 which gives (in units where c = ) i.e. V φ + 2 φ φ = 0 µ µ φ + V φ = 0. Similarly the Euler-Lagrange equation for φ gives µ µ φ + V φ = 0. (b) Setting c =, the canonical momentum densities are π = L/ φ = φ / t and π = L/ φ = φ/ t. The corresponding Hamiltonian density is H = π φ + π φ L = π π + φ φ + V (φ) (c) A global phase change φ φe iɛ, φ φ e iɛ leaves φ φ and ( µ φ )( µ φ) unchanged. Therefore if the potential V is a function of φ φ, the Lagrangian density is invariant under this change. (d) We are given the Coleman-Weinberg potential, ( φ V (φ) = (φ φ) [ln 2 ) ] φ κ, where Λ and κ are real, positive constants. Let X = φ φ and consider V as a function of X. Then V is continuous for X > 0, V = 0 at X = 0 and V + as X. We have dv dx = 2X [ ln(x/λ 2 ) κ ] [ + X = 2X ln(x/λ 2 ) κ + ] 2 This is zero for X = 0 and X = X 0 where ln(x 0 /Λ 2 ) = κ /2, i.e. Λ 2 X 0 = Λ 2 e κ /2

9 At X = X 0 we have d 2 V [ln(x dx = /Λ 2 ) κ + ] + 2 = Hence X 0 is a minimum of V and the Hamiltonian is bounded from below. (e) The states of minimum energy correspond to the circle in the complex φ plane where φ φ = X 0, i.e. φ = φ 0 where φ 0 = r 0 e iθ, r 0 = Λe (2κ )/4. (f) Considering small field variations around φ = r 0, i.e. φ = r 0 + (χ + iχ 2 )/ 2, we have Then X = φ φ = r r 0 χ + 2 (χ2 + χ 2 2) i.e. V = V (X 0 ) + 2 (X X 0) 2 d2 V dx = V (X 0) + 2r 2 0χ 2 + +O(χ 3 ) V (φ) = V (φ 0 ) + 2 m2 χ 2 + O(χ 3 ). where m = 2r 0. Thus the field χ satisfies a Klein-Gordon equation with mass (in natural units) 2r 0, while the field χ 2 satisfies a massless Klein-Gordon equation. This is an example of Goldstone s theorem: the global phase symmetry is spontaneously broken when the field chooses a particular ground state on the circle, and there is an associated massless Goldstone boson χ 2, while the other degree of freedom of the field χ is massive.

10 6 The propagator G(t) must vanish for t < 0 and so we can write 0 G(t) = Θ(t)g(t) where g(t) can be chosen to be either an odd or even function of t. By the convolution theorem, it follows that dω G(ω) = Θ(ω ω ) g(ω ) 2π dω [ ] = πδ(ω ω ) + ip g(ω ) 2π ω ω where g(ω) = + dt g(t)(cos ωt + i sin ωt) Hence if we choose g(t) to be an odd function, g(ω) will be purely imaginary, say g(ω) = i h(ω). Then equating real parts in the above equation Re G(ω) dω h(ω ) = P 2π ω ω and equating imaginary parts Im G(ω) dω = 2π πδ(ω ω ) h(ω ) = h(ω). 2 Substituting this in the above then gives the Kramers-Kronig relation. (a) We have Re G(ω) ω ω 0 = (ω ω 0 ) 2 + γ 2 /4 Im G(ω) = γ 2 (ω ω 0 ) 2 + γ 2 /4. The r.h.s. of the Kramers-Kronig relation is thus dω γ P 2π (ω ω) (ω ω 0 ) 2 + γ 2 /4. The integrand has poles at ω = ω and at ω = ω 0 ± iγ/2. However, it vanishes rapidly at, so we can complete the contour with a large semicircle in either the upper or the lower half-plane. Using P = lim ɛ 0 2 ( + +iɛ +iɛ + + iɛ ) iɛ and choosing the upper half plane, the first integral encloses only the pole at ω = ω 0 + iγ/2, which gives iπ 2π γ (ω 0 + iγ/2 ω) (iγ) = 2 (ω 0 + iγ/2 ω).,

11 Choosing the lower half plane for the second integral encloses only the pole at ω = ω 0 iγ/2 (in a negative sense), giving iπ 2π γ (ω 0 iγ/2 ω) ( iγ) = 2 (ω 0 iγ/2 ω). The sum of these is 2 (ω 0 iγ/2 ω) 2 (ω 0 + iγ/2 ω) = ω ω 0 (ω ω 0 ) 2 + γ 2 /4 = Re G(ω) as required. (b) The propagator is G(r r, t t ) where G(r, t) = d 3 k dω (2π) 4 exp(ik r iωt) ω hk 2 /2m + iγ/2. For t < 0 we can complete the contour with a large semicircle in the upper half-plane, which encloses no singularities and so gives G(r, t) = 0 for t < 0. For t > 0 we must instead choose the lower half-plane, which encloses the pole at ω = hk 2 /2m iγ/2 (in a negative sense), giving G(r, t) = i for t > 0. Now we can write ik r ihk2 t 2m Hence, changing the variable of integration to d 3 ( k (2π) exp ik r ihk2 t 3 2m γ ) 2 t = iht ( k m ) 2 2m ht r im + 2ht r2 k = k m ht r we have But so d 3 k ( G(r, t) = i (2π) exp iht 3 2m k 2 + im 2ht r2 γ ) 2 t d 3 k exp ( ak 2) ( ) π 3/2 = a G(r, t) = i ( 2πm (2π) 3 iht ( ) im 3/2 = exp 2πht ) 3/2 ( im exp ( im 2ht r2 γ 2 t ) 2ht r2 γ 2 t )

12 2 for t > 0 and G(r, t) = 0 for t < 0. Correspondingly [ ] 3/2 ( G(r, r, t, t im im(r r ) 2 ) = exp 2πh(t t ) 2h(t t ) γ 2 (t t ) ) for t > t and G(r, r, t, t ) = 0 for t < t. END OF PAPER