MATHEMATICAL PHYSICS

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1 MATHEMATICAL PHYSICS Third Year SEMESTER Classical Mechanics MP350 Prof. S. J. Hands, Prof. D. M. Heffernan, Dr. J.-I. Skullerud and Dr. M. Fremling Time allowed: 1 1 hours Answer two questions All questions carry equal marks

2 Model solutions 1. a The Euler Lagrange equation is d L dt ẋ = d mẋe γt dt = L = m ẍe γt + γẋe γt 1.1 x = mω xe γt = ẍ + γẋ + ω x = If γ > 0, this is the equation for a damped harmonic oscillator. [ 15 marks ] b The canonical momentum is The hamiltonian is p = L ẋ = mẋeγt 1.3 H = pẋ L = 1 m ẋ ẋ + ω x e γt = 1 m ẋ + ω x e γt 1.4 = p m e γt + mω x e γt. 1.5 The hamiltonian is not conserved, since it and the lagrangian depends explicitly on time. Physically, the damping of the oscillator dissipates energy. [ 1 marks ] c With q = xe γt/ we have x = qe γt/ and q = ẋe γt/ + x γ eγt/ = ẋ = qe γt/ xγ = qe γt/ γ qe γt/ 1.6 ẋ = q γ qq + γ 4 q e γt 1.7 L = 1 mẋ ω x e γt = 1 m q γ qq ω γ /4q 1.8 [ 13 marks ] d The canonical momentum is p = L q = m q mγ q. 1.9

3 We can use this to eliminate q: q = 1 m The hamiltonian is, as a function of p and q, p + mγ q H = p q L = p m p + mγ q 1 m [ 1 m p + mγ q γ m = p m + γ pq + m ω q. If we instead write H in terms of q and q we get mγ p + q q ω γ q ] 4 H = p q L = m q mγ q q 1 m q + mγ q q + 1 mω γ /4q = 1 m q + 1 mω γ /4q There is no explicit time dependence in this hamiltonian, so H is conserved. However, H is not the total energy: even if we make the dubious assumption that we can identify the two terms in the original lagrangian with T and V, the kinetic energy is not quadratic in q: it contains the linear term 1 mγ qq. Alternatively, we may note that the coordinate q is introduced in a time-dependent way, ie there is a time-dependent transformation equation between the original coordinate x and the new coordinate q. [ 10 marks ]. a Any operation which can be achieved by a combination of elementary rotations about some axis is a proper rotation. The rotation matrix A for a proper rotation will always have det A = 1. A reflection means taking x x, y y, z z. It is given by the matrix diag 1, 1, 1 which has determinant -1. It is not possible in 3 dimensions to construct a reflection from any number of proper rotations. A combination of a proper rotation and an odd number of reflections is called an improper rotation, and is given by a rotation matrix with determinant -1. [ 5 marks ] b The kinetic energy for a collection of N particles with masses m and positions r forming

4 a rigid body is c E rot = 1 N m r ω = 1 m r ω r ω =1 = 1 m r ωi x i ω i x j ω j i ij = 1 m r δ ij ω i ω j x i ω i x j ω j ij = 1 m r δ ij x i x j ω i ω j 1 I ijω i ω j..1 Equation.1 defines the inertia tensor. ij [ 0 marks ] The principal axes of inertia are body coordinate axes ê 1, ê, ê 3 such that the inertia tensor in the coordinate system defined by these axes is diagonal, I = diagi 1, I, I 3. The corresponding values I 1, I, I 3 are the principal moments of inertia. d The inertial tensors for each of the three particles are computed as I 1 = m 10a ma = ma I = m 5a ma 0 4 = ma I 3 = m 9a ma 1 = ma [ 5 marks ] The total inertia tensor can be written in matrix form as 16 7 I = ma [ 0 marks ]

5 3. a Hamilton s equations of motion are ṙ = H p r = p r m, θ = H p θ = p θ mr, ṗ r = H r = p θ mr 3 k r, 3.1 ṗ θ = H θ = From 3. we see that p θ = l is constant, so the last two terms in H depend only on r, and can be taken to define an effective potential. The first term is the kinetic energy for the motion in the r-direction, T r = p r/m, and the equations of motion arising from the hamiltonian Hr, p r = T r + V eff r are exactly those of 3.1. [ 1 marks ] b The Poisson brackets {H, A} are given by {H, A} = H A p r r H r = p r m p θ r 3 = p rp θ mr 3 A + H A p r p θ + mk r + kp r mr + p rp θ 3 mr 3 θ H θ A p θ p θ mr 3 + k r pr kp r mr 3 = 0. Since A does not depend explicitly on time, its time derivative is given by 3.3 c and A is thus conserved. The parameter e is the eccentricity of the orbit. If e = 0 the orbit is a circle. 0 < e < 1 the orbit is an ellipse. e = 1 the orbit is a parabola. e > 1 the orbit is a hyperbola. da dt = A + {H, A} = {H, A} = 0, 3.4 t [ 10 marks ] The circle and the ellipse are closed orbits, while the parabola and hyperbola are open orbits. [ 8 marks ]

6 d,i The maximum speed is the speed at perihelion the closest approach to the sun. To find the speed at perihelion, we need to find the angular momentum l. Since at perihelion the velocity is orthogonal to the distance vector, we have that l = mv p r min, where v p is the perihelion speed and m = µ is the mass of the comet. The constant k = GmM. From the expressions for and r, we find that r min = 1 + e = l mk = m v pr min m GM = r min 1 + e 3.5 = v p = 1 + egm r min = m /s = v p = m/s. 3.6 d,ii The aphelion distance is given by r max = [ 1 marks ] 1 e = 1 + e 1 e r min = m = m. 3.7 The speed v a at aphelion may be found using conservation of angular momentum, l = mv p r min = mv a r max v a = r min r max v p = 7, m/s. 3.8 [ 8 marks ]