Course 41: Advanced echanics Scholarship Questions This covers the first sixteen questions from part I up to the end of rigid bodies, and a selection of the questions from part II. I wouldn t be too bothered by it there s bound to be loads of mistakes. Chris Blair Part I Questions I.1 Hamilton s Principle and Lagrange s Equations of otion Hamilton s Principle states that every mechanical system is characterised by a definite function Lq, q, t of the generalised coordinates, velocities and time, such that the integral S = t t 1 Lq, q, t dt is a minimum for the evolution of the system between times t 1 and t, endpoints q 1 and q. The function L is known as the Lagrangian, and S is known as the action. Consider a small deviation q + δq from the true path q of the system, such that δq 1 = δq 1 = 0 that is, the endpoints are the same. We then have δs = t [ ] L L δl dt = δq + t 1 t 1 q q δ q dt t [ L = t 1 q + d L dt q δq d L dt q [ ] t L t [ L = q δq + δq t }{{} 1 t 1 q d L dt q =0 t ] δq dt ] dt So for arbitrary δq to have δs = 0 as it must be if the path of the system minimises the action we must have that the integrand is identically zero, that is L d L = 0 q i dt q i where i = 1... s. These s second-order equations are known as Lagrange s equations of motion. The Lagrangian is additive and unchanged by adding the total time derivative of some function of q and t. Scaling the Lagrangian does not change Lagrange s equations of motion. The Lagrangian of a relativistic free particle is L = m 0 c 1 ẋ i c. 1
I. Noether s Theorem and Conservation Laws Roughly speaking, Noether s Theorem states that for every continuous symmetry there exists a conserved quantity. Consider a transformation of the generalised coordinates such that q i q i = q i + ɛ α q iα and L L = L + d dt Lα ɛ α, plus terms vanishing on-shell i.e. if equations of motion satisfied, where ɛ α is some parameter of the transformation. Then we have the conserved quantity or Proof: Consider δs = t 1 t t 1 δl dt = = = t t 1 t t 1 t t 1 J α = L q i q iα L α J α ɛ α = L q i δq i L α ɛ α [ L L δq + q [ ] d L dt q δq d dt and also δl = d dt Lα ɛ α hence we have t d dt [ ] L t q δq dt = ] q δ q dt q d ] L dt dt q }{{} =0 eom [ L + [ ] L q δq dt d t 1 dt Lα d L dt q qiα ɛ α L α ɛ α = 0 so as ɛ α is arbitrary we have the conserved quantity J α = L q qiα L α. Note that if δl = 0 then J α = L q qiα is conserved. Time/Energy Consider an infinitesimal time translation, t t + ɛ, q i t + ɛ = q i t + ɛ q i, δl = d dt Lɛ. We then have Jɛ hence = L q i qi ɛ Lɛ J = p i q i L E is conserved. Homogeneity of Space/Linear omentum Consider an infinitesimal spatial translation r a r a = r a + ɛ, and L = 1 m r a U r a r b thus δl = 0, giving hence we have the conserved quantity J ɛ = L r a ɛ = a p a ɛ J = P = a p a
Isotropy of Space/Angular omentum Consider an infinitesimal rotation, with δ r = δ φ r, and impose the condition δl = 0. Here, ɛ = δ φ and so we have the conserved quantity J ɛ = J δ φ = p a δ φ r a = δ φ r a p a J = = a r a p a I.3 One-Dimensional and Central Field otion The general Lagrangian for one-dimensional motion is L = m gq q Uq, and we can always find a transformation such that q q x, q = qx, gq = 1 x This gives us L = m ẋ Ux. The total energy is E = m ẋ Ux, so t = m dx E U The motion can be unbounded and infinite if E > U always, bounded on one side but infinite on the other, or finite and oscillatory not necessarily harmonic. The latter case occurs if the particle if the particle s energy E < U over the range of the motion i.e. if it is in a potential well. In that case the turning points of the motion occur when the E = Ux, i.e. when ẋ = 0. The period of the motion is given by T E = x m x 1 dx E U where the limits of the integration are the turning points of the motion. A central field potential is a potential of the form U r = Ur where r = r. In this case angular momentum = r p is conserved, and as r it follows that r always lies in a plane perpendicular to. If we impose the condition z0 = ż0 = 0 then the Lagrangian can be written L = 1 mṙ + 1 mr φ Ur and we have p φ = mr φ =. The energy is also conserved, from which we obtain E = 1 mṙ + 1 mr φ + Ur = 1 mṙ + t = φ = dr m E U mr r dr me U r mr + Ur 3
The motion in a central field may be infinite, closed always so for U = αr or U = α r, bounded but not closed. For a bounded orbit we have that when the path of the particle moves from r min to r max and back again the angle changes by rmax φ = r min r dr me U r and the orbit is closed if φ is a rational fraction of π. In general the effective potential U eff = Ur + mr known as the centrifugal energy prevents the particle from falling to the centre of the field. This may however occur in certain situations; 1 mṙ = E U mr > 0 r U + m < Er so a particle can reach the centre of the field if [r U] r 0 < m, which can occur if U as α r with α > 1 m, or as r, n >. n I.4 Kepler s Problem This is the problem of a particle moving in a central field potential U = α r with α > 0 this describes gravity and the attractive Coulomb force. The effective potential is U eff = α r + mr : Hence the motion is finite for E < 0 and infinite for E > 0. We have r dr φ = me + α r r Let x = 1 r dx = dr r, so φ = and upon completing the square φ = dx me + mαx x dx me + m α x mα 4
which integrates to give φ + constant = arccos r mα me + m α or 1 r = mα + 1 me + m α cos φ choosing our origins such that the constant of integration vanishes. We rewrite this further as or 1 mα r = 1 + E mα + 1 cos φ p r = 1 + e cos φ where p = mα and e = E mα + 1, known as the eccentricity. For E < 0, e < 1 and the motion is an ellipse. If E = 0, e = 1 and the motion is a parabola, coming from and returning to infinity. If E > 0, e > 1 and the motion is a hyperbola, again coming from and returning to infinity. We can also integrate the expression for time, but the result is fairly complicated and not very instructive. The Runge-Lenz vector is a further integral of the motion D = v + α r r which exists due to the degeneracy of the motion. Differentiating we find D = v + α v 1 r α r r r = v m r v + α v v r r α r r 3 = m r v v m v r v + α v r α r v r r 3 using the cross-product identity a b c = b a c c a b. Now from the equations of motion we have m v = α r 3 r and substituting this in gives D = 0. I.5 Scattering in a Central Field The scattering angle is χ, and φ 0 is the angle made by the line of closest approach, which has length r 0. The impact parameter ρ is the distance from the initial path of the particle to the field centre. The path of the particle is symmetric about the line of closest approach, hence χ = π φ 0. The angle φ 0 is given by φ = r 0 r dr me Ur r 5
and as we also have E = 1 mv and = mρv where v is the velocity of the particle at infinity, giving ρ r dr φ = r 0 1 ρ r U E The lower limit r 0 is calculated from the expression for the total energy using the fact that at r = r 0, ṙ = 0. The integral then gives φ 0 and hence χ as a function of ρ. Rutherford scattering is the scattering of a particle in a repulsive Coulomb field, U = α r, α > 0. The end result is α Eρ φ 0 = arccos 1 + α Eρ I.6 Small Oscillations with s Degrees of Freedom The Lagrangian is L = 1 m ijẋ i ẋ j + 1 k ijx i x j, m ij = m ji, k ij = k ji. Writing the total differential, dl = 1 m ijẋ i dẋ j + m ij dẋ i ẋ j + k ij x i dx j + k ij dx i x j = m ij ẋ j dẋ i + k ij x j dx i, hence we have equations of motion m ij ẍ j + k ij x j = 0 We guess solution x j = A j e iωt, giving ω m ij + k ij Aj = 0 For this to have non-trivial solutions we need det k ij ω m ij = 0 which is the characteristic equation of the system, a polynomial of degree s in the eigenfrequencies ωα. Having found the eigenfrequencies we substitute them back into the above expression to find the A j, which are proportional to the minors of the determinant, jα if all roots are distinct. The general solution is then s x j = Re jα θ α where α=1 jα C α e iωαt = α θ α = Re [ C α e iωαt] are the normal modes of the system, satisfying the simple harmonic equation θ α + ω αθ α = 0. I.7 One-Dimensional Harmonic Forced Oscillations We have equation of motion mẍ + kx = f cosγt + β ẍ + ω x = f cosγt + β m and guess solution x = A cosγt + β, hence γ A + ω A = f m A = f 1 m ω γ 6
and our complete solution is x = a cosωt + α + f 1 m ω cosγt + β γ Resonance occurs when γ = ω, and this solution is no longer valid. To analyse the situation we write our solution as x = a cosωt + α + f 1 m ω γ cosγt + β cosωt + β where a now takes a different value, and take the limit γ ω. We have lim γ ω so we use L hopital s rule, hence lim γ ω f 1 m ω γ f 1 m ω γ cosγt + β cosωt + β = 0 0 f t sinγt + β cosγt + β cosωt + β = lim = f t sinωt + β γ ω m γ mω and so x increases linearly with t. x = a cosωt + α + f t sinωt + β mω I.8 One-Dimensional Damped Oscillations The equation of motion is mẍ + mλẋ + kx = 0 or ẍ + λẋ + ω x = 0 This cannot be derived from any Lagrangian. We guess solution x = Ae rt, giving There are 3 cases: r + λr + ω = 0 r = λ ± 4λ 4ω = λ ± λ ω 1. Light damping: λ < ω, then x = C + e λt e iω f t + C + e λt e iω f t where ω f = λ ω.. Heavy damping: λ > ω, then x = C + exp λ + λ ω t + C exp λ λ ω t 3. Critical damping: λ = ω, then x = C 1 e λt + C te λt 7
I.9 One-Dimensional Forced Damped Oscillator We write the driving force as an exponential, giving equation of motion mẍ + mλẋ + kx = f m eiγt+β and guess solution x = Be iγt+β, hence γ B + iγλb + ω B = f m thus B = f 1 m ω γ + iλγ = beiδ where b = f 1 m ω γ + λγ and tan δ = λγ ω γ We take the real part of this for our general solution in the underdamped case. Note that we have x = ae λt cosω f t + α + b cosγt + β + φ bγ = f 1 m ω γ + λγ and b = 0, b0 = f mω. The maximum value of b occurs at resonance, and by differentiating the bottom line of our expression to find its minimum value, this is found to occur when γ = ω λ. I.10 Inertia Tensor Let P be a point in some rigid body, with r being the radius vector of the point with respect to a fixed frame of reference. Let R be the centre of mass radius vector, and r the radius vector of P with respect to the centre of mass, so that r = R + r. An infinitesimal displacement of P may be written d r = d R + d φ r v = V + Ω r Now the kinetic energy is T = 1 mv, so for the whole body we have T = 1 m V + Ω r = 1 mv + mv Ω r + [ ] m Ω r Ω r = 1 µv + m r V Ω + 1 [ ] m Ω r Ω R }{{} =0C = T trans + T rot 8
where T trans = 1 µv is the kinetic energy due to the translational motion of the body, with µ being the total mass, and T rot is the kinetic energy due to its rotational motion. In tensor form we can write it as T rot = 1 m Ω i x i Ω i x i Ω k x k = 1 m Ωi Ω k δ ik x l Ω i Ω k x i x k = 1 I ikω i Ω k where I ik is the inertia tensor, I ik = m δ ik x l x ix k, or explicitly, my + z mxy mxz I ik = myx mx + z myz mzx mzy mx + y The diagonal entries I xx, I yy, I zz are known as the moments of inertia about the x,y and z axes. We can always diagonalise the inertia tensor by an appropriate choice of axes, known as the principal axes of inertia. In this case the diagonal entries I 1, I, I 3 are known as the principal moments of inertia. A symmetric top is a body with I 1 = I I 3. A spherical top is a body with all principal moments of inertia equal. I.11 Angular omentum of a Rigid Body The angular momentum of a rigid body is given by = m r v = m r V + Ω r If our origin is the centre of mass this becomes = m r Ω r and in component form = m r Ω r Ω r i = m x l Ω i x i Ω k x k = m x l Ω k δ ik x i x k Ω k = Iik Ω k hence = I Ω If x 1, x, x 3 are principal axes then 1 = I 1 Ω 1 = I Ω 3 = I 3 Ω 3 For a symmetric top I 1 = I, and taking to be vertical, making an angle of θ with the x 3 axis, and with the x axis perpendicular to the and x 1 plane we have = 0 1 = I 1 Ω 1 = sin θ Ω 1 = I 1 sin θ 9
3 = I 3 Ω 3 = cos θ Ω 3 = I 3 cos θ We can decompose Ω into Ω pr + Ω 3 where Ω pr lies along the axis and is called the precessional angular velocity, i.e. the angular velocity with which the top rotates about the vertical. We have that Ω 1 = Ω pr sin θ Ω pr = I 1 I.1 Compound Pendulum We have Ω x = φ cos α, Ω y = φ cos β, Ω z = φ cos γ where α, β and γ are the angles between the moving axes and the axis of oscillations which is horizontal. φ is the angle between the vertical and a line through the centre of mass, perpendicular to the axis of rotation. The small angle approximation gives us the Lagrangian L = 1 µl + I 1 cos α + I cos β + I 3 cos γ φ 1 µglφ hence ω = µgl µl + I 1 cos α + I cos β + I 3 cos γ I.13 Equations of otion of a Rigid Body The equations of motion of a rigid body in a fixed system of reference are obtained from ṗ α = f α where f α is the force on the α th particle. We have P = α ṗ α = α f α = F the sum of the external forces acting on the body internal forces cancel out in the sum. Now for the angular momentum we have α = r α p α = m r α v α which is zero in the centre of mass frame as then r = v. Thus we have the torque K = = r α f α The torque depends on choice origin, consider a displacement a so that r = r a and α K = K + a F If the force is perpendicular to the torque it is possible to find some a such that K = a F and so the torque acts along a line through the body. An example of this is a uniform field of force, described by F = α e α E, then K = α e α r α E = r 0 E α e α = r 0 F 10
where and is known as the total torque. r 0 = α e α r α α e α K = r 0 F I.14 Euler Angles Consider a moving frame of reference x 1 x x 3 with its origin coinciding with a fixed frame, XY Z. We can then represent a general rotation of the moving frame as the product of three simple rotations as follows: first, rotate by φ in the XY plane. Second, rotate by θ about the line ON line of nodes, which is the intersection of the x 1 x and XY planes. Third, rotate by ψ about x 3. x 3 Z x θ O Y φ ψ X x 1 N This general rotation can be represented by the following matrix: cos φ sin φ 0 1 0 0 cos ψ sin ψ 0 sin φ cos φ 0 0 cos θ sin θ sin ψ cos ψ 0 0 0 1 0 sin θ cos θ 0 0 1 The angles φ, ψ and θ are known as the Euler angles. We decompose their rates of change onto the moving axes to obtain the components of the angular velocity about these axes: Ω 1 = φ sin θ sin ψ + θ cos θ Ω = φ sin θ cos ψ θ sin θ Ω 3 = φ cos θ + ψ To analyse the free motion of a symmetric top, let the angular momentum correspond to the Z axis, and let ψ = 0 at the moment considered. We then have 1 = 0 as x 1 so therefore θ = 0 θ = constant. Also, = sin θ = I 1 Ω = I 1 φ sin θ, hence φ = I 1, the precessional angular velocity. Lastly we have Ω 3 = I 3 cos θ. 11
I.15 Lagrange s Top x 3 Z θ mg l φ ψ X x x 1 Y N As I 1 = I we have that the rotational energy about the centre of mass is Our Lagrangian is then T rot = 1 I φ 1 sin θ + θ + 1 I 3 ψ + φ cos θ L = 1 I 1 φ sin θ + θ + 1 I 3 ψ + φ cos θ mgl cos θ where I 1 = I 1 + ml, where l is the distance from the origin to the centre of the mass, and the ml term describes the velocity of the centre of the mass. The angles φ and ψ are cyclic so we have the conserved quantities hence The energy is also conserved, p φ = L φ = I φ 1 sin θ + I 3 ψ + φ cos θ cos θ = Z p ψ = L ψ = I 3 ψ + φ cos θ = 3 ψ = 3 I 3 φ cos θ φ = Z 3 cos θ I 1 sin θ ψ = 3 Z 3 cos θ cos θ I 3 I 1 sin θ E = 1 I 1 φ sin θ + θ + 1 I 3 ψ + φ cos θ + mgl cos θ 1
Hence or where E = 1 I 1 t = [ Z 3 cos θ I 1 I 1 sin θ ] + θ + 1 3 I 3 + mgl cos θ dθ [ ] E 1 3 I 3 1 Z 3 cos θ mgl cos θ I 1 sin θ t = dθ I [E U 1 eff ] U eff = 1 Z 3 cos θ mgl1 cos θ I 1 sin θ E = E 1 3 mgl I 3 The effective potential goes to infinity at θ = 0 and θ = π, hence the motion takes place in a range bounded by θ min and θ max which are determined from the equation E = U eff. Therefore the motion consists of a variation between θ min and θ max while precessing around the vertical. The nature of this precession is determined by whether φ changes sign in the course of the motion from θ min to θ max. We have φ = Z 3 cos θ I 1 sin θ and if Z 3 cos θ 0 over the course of the motion then φ is always positive, hence we have nutation; the axis of the top oscillates up and down while rotating monotonically around the vertical. If Z 3 cos θ 0 = 0 for some θ 0 then the axis describes loops a special case of this occurs if Z 3 cos θ 0 = 0 for θ 0 = θ min or θ max. I.16 Euler s Equations We can write the total time derivative of some vector A as da dt = d A dt + Ω A where the prime denotes differentiation with respect to the moving axes. Applying this to the total linear and angular momenta we get d P dt + Ω P = F d dt + Ω = K or in component form and using P = µ v, dvi µ dt + ɛ ijkω j v k = F i and using i = I i Ω i, I i dω i dt + ɛ ijkω j I k Ω k = K i 13
or explicitly I 1 dω 1 dt I dω dt I 3 dω 3 dt + I 3 I Ω Ω 3 = K 1 + I 1 I 3 Ω 3 Ω 1 = K + I I 1 Ω 1 Ω = K 3 The above six equations are known as Euler s equations. For a free symmetric top K = 0 and I 1 = I, hence Ω 3 = constant Ω 1 = ωω Ω = ωω 1 where ω = I3 I1 I 1 Ω 3, hence Ω 1 = A cos ωt Ω = A sin ωt Part II Problems II.1 There are two degrees of freedom and the Lagrangian is L = 1 m [ ẋ + ẏ + 4axẋ + yẏ ] + xf x + yf y + ax + y F z II. A general rotation is described by x i x i = A ijx j ; rotational symmetry means we must have x i = x i = A ij x j t A ij x j A t A = I. Infinitesimally this means I + ɛ t I + ɛ = I and to order ɛ this means that ɛ t = ɛ, or ɛ ij = ɛ ji. Our rotations are then parametrised as x i x i = x i + ɛ ij x j. We use Noether s theorem; δl = 0, hence the conserved quantity is 1 J ijɛ ij = L δx i ẋ i where the factor of the half is to prevent us double-counting terms in the sum. So, hence we find conserved charges 1 J ijɛ ij = p i x j ɛ ij = 1 p ix j p j x i ɛ ij J ij = p i x j p j x i 14
For the Lagrangian L = 1 mv i κ r this is For two functions f and g the Poisson bracket is J ij = mẋ i x j mẋ j x i {f, g} = f g q k f q k g and in particular Thus, and {f, p i } = f q i {f, q i } = f p i {J ij, p k } = x k p i x j p j x i = p i δ jk + p j δ ik {J ij, x k } = p i x j p j x i = x j δ ik x i δ jk {J ij, J kl } = {J ij, p k x l p l x k } = {J ij, p k } x l + {J ij, x l } p k {J ij, p l } x k {J ij, x k } p l = p i δ jk + p j δ ik x l + x j δ il x i δ jl p k p i δ jl + p j δ il x k x j δ ik x i δ jk p l = δ ik p j x l p l x j δ jk p i x l p l x i + δ il p k x j p j x k δ jl p k x i p i x k = δ ik J jl δ jk J il + δ il J kj δ jl J ki For D = 3 we have 1 = J 3, = J 31 and 3 = J 1, then { 1, x} = { J 3, x 1 } = 0, { 1, y} = { J 3, x } = x 3 = z, { 1, z} = y in general Similarly, { i, x j } = ɛ ijk x k { 1, p x } = { J 3, p 1 } = 0, { 1, p y } = { J 3, p } = p 3 = p z, { 1, p z } = y and in general Finally, { i, p j } = ɛ ijk p k { 1, } = { J 3, J 31 } = {J 3, J 13 } = J 1 = 3 {, 3 } = { J 31, J 1 } = {J 13, J 1 } = J 3 = 1 or in general { 3, 1 } = { J 1, J 3 } = {J 1, J 3 } = J 31 = { i, j } = ɛ ijk k 15
II.3 We have Ux = V tan ax, and the motion of a particle in this potential is given by m dx t = E V tan ax m cosaxdx = E cos ax V sin ax Let u = sinax du a t = 1 m a = 1 a = cosaxdx, so du E u E + V m E + V du E E+V u = 1 m a E + V sin 1 u E E+V + C so we get [ x = 1 ] E E + V a sin 1 E + V sin a t + C m II.4 We have z = ax +y = ar in polar coordinates. The potential energy is given by U = F dz = zf z. Our Lagrangian is then L = 1 mṙ + r φ + 4a r ṙ + ar F z and as φ is cyclic we have that φ = mr φ is conserved. The total energy is whence E = 1 m1 + 4a r ṙ + φ mr ar F z t = φ = 1 m 1+4a r dr E φ mr + ar F z dr r m 1+4a r E φ mr + ar F z 16
II.5 We have central field potential U = α r, so t = We let x = 1 r dx = dr r There are four cases to consider: φ = giving φ = 1. If E > 0 and > mα we get m dr E + α r mr dr r me + 1 r mα dx me + x mα φ + constant = mα arccos x me mα 1 me r = mα cos mα φ choosing origins such that the constant vanishes.. If E > 0 and < mα we get φ + constant = mα sinh 1 x me mα 1 me mα r = mα sinh φ choosing origins such that the constant vanishes. 3. If E < 0 and > mα we get φ + constant = mα cosh 1 x m E mα 1 m E r = mα cosh mα φ choosing origins such that the constant vanishes, and as cosh is even. 4. If E < 0 and < mα, there is no real solution. 17
For the t integral we let x = r dr = dx, which gives x II.6 Particle oscillating on parabola. t = 1 m dx Ex + α m t = m Ex + α E m m t = m Er E m + α m m II.7 Forced oscillations. II.8 Three masses on a circle. II.9 Forced oscillations in the presence of friction. II.10 oments of inertia. II.11 The cavity has radius r and we denote its mass by m = 4 3 πρr3. The large sphere has radius R and mass = 4 3 πρr3. The distance from the centre of the large sphere to the centre of the small sphere blue line is a, and the distance from the centre of the large sphere to the new centre of mass is z red line. This is given by r 3 z + ma z = 0 z = a R 3 r 3 18
The moments of inertia of the cavity if it were a sphere with respect to the new centre of mass are I 1 = I = 5 mr + ma z I 3 = 5 mr and so the moments of inertia of the large sphere taking the cavity into account are I 1 = I = 5 R + z 5 mr ma z Now a z = ar3 R 3 r 3 so these are I 3 = 5 R 5 mr I 1 = I = 8 15 πρr5 r 5 4 3 πρa R3 r 3 R 3 r 3 I 3 = 8 15 πρr5 r 5 For an axis through the centre of the sphere we have Lagrangian using small angle approximation for cos φ and for an axis through the centre of the cavity L = 1 mz φ + 1 I φ 1 1 m z φ the frequencies following from these. L = 1 ma z φ + 1 I φ 1 1 m a z φ II.1 We let the torque K lie along the Z axis. Now = K = Kt as Ω0 = 0 0 = 0. By our choice of axes we have X = Y = 0 Z = Kt We decompose the angular momentum onto the moving axes and use the Euler angles to obtain 1 = Z sin θ sin ψ = I 1 Ω 1 = I 1 φ sin θ sin ψ + θ cos ψ = Z sin θ cos ψ = I 1 Ω = I 1 φ sin θ cos ψ θ sin ψ 3 = Z cos θ = I 3 Ω 3 = I 3 φ cos θ + ψ 19
x 3 Z, K, θ φ ψ X x x 1 Y hence N θ = Z sin θ sin ψ I 1 cos ψ φ sin θ sin ψ cos ψ Z sin θ cos ψ = I 1 φ sin θ cos ψ Z sin θ sin ψ I 1 cos ψ Z I 1 so the precessional angular velocity is and we then have cos ψ sin ψ = cos ψ φ φ = Z I 1 = Kt I 1 θ = 0 θ = constant cos ψ sin ψ cos ψ φ sin θ sin ψ cos ψ and finally the angular velocity with which the top rotates about its own axis is II.13 Lagrange s top. II.14 Find Hamiltonian. II.15 Find Hamiltonian of relativistic free particle. Ω 3 = Kt cos θ I 3 0
II.16 We have old coordinates p, q and new coordinates P, Q. We have our action principle δs = P dq H dt = δs = pdq Hdt which implies that the two integrands can differ up to the total derivative of some function, that is, so we have S 1 = S 1 q, Q, t and We also have so and also so and so In all cases we have II.17 pdq Hdt = P dq H dt ds 1 ds 1 = pdq + P dq H Hdt S 1 q = p S 1 Q = P d S 1 P Q = pdq QdP + H Hdt }{{} S q,p,t S q = p S 1 P = Q d S 1 + pq = qdp + P dq H Hdt }{{} S 3p,Q,t S 3 p = q S 3 Q = P d S 1 + pq P Q = qdp QdP H Hdt }{{} S 4p,P,t { Q i, Q j} = Qi Q j p,q q k Qi Q j q k S 4 p = q S 4 P = Q S i t = H H = Qi S q k P j + Qi q k S 4 P j as S P j = Q j, = Qi P j + Qi q k q k P j as S q k = p k, = Qi P j = 0 S 4 P j = Q j S 4 = q k 1
{P i, P j } p,q = P i P j q k P i q k P j = P i S 1 q k Q j P i q k S 1 Q j as S 1 Q j = P j, = P i Q j P i q k q k Q j as S 1 q k = p k, = P i Q j = 0 S 3 Q j = P j S 3 = q k II.18 Find canonical transformation. II.19 Find canonical transformations. II.0 odify canonical transformations. II.1 We take spiral motion here to mean a particle performing circular motion in the xy plane while moving in the upwards z direction. We have the equations zt = z0 + vt xt = R cosωt + ϕ = R cos ϕ cos ωt R sin ϕ sin ωt = x0 cos ωt y0 sin ωt yt = R sinωt + ϕ = R cos ϕ sin ωt + R sin ϕ cos ωt = x0 sin ωt + y0 cos ωt and so we are looking for a generating function corresponding to transforming from x, y, z x0, y0, z0 to X, Y, Z xt, yt, zt, and similarly for the momenta. We may write our transformation explicitly as X = x cos ωt y sin ωt Y = x sin ωt + y cos ωt Z = z + vt P x = p x cos ωt p y sin ωt P y = p x sin ωt + p y cos ωt P z = p z To check this is canonical, we compute {P x, X} = P x X p x x P x X + P x X x p x p y y P x X + P x X y p y p z z P x X z p z
{P x, X} = cos ωt + sin ωt = 1 and similarly for the other brackets. We now seek a generating function of the form Sx, y, z, P x, P y, z, t = S 0 x, y, P x, P y, t + S z z, P z, t so we have Similarly, S z z = p z S z P z = Z S z = P z z + vt S 0 x = p x S 0 P x = X so we get and we can check that as we want. S 0 y = p y S 0 P y = Y S 0 = P x x cos ωt y sin ωt P y x sin ωt + y cos ωt S 0 x = P x cos ωt P y sin ωt = p x cos ωt p y sin ωt cos ωt p x sin ωt + p y cos ωt sin ωt = p x II. We have U r = F x giving Hamiltonian The Hamilton-Jacobi equation is H = 1 p m x + p y + p z F x S t x, y, z, P x, P y, P z, t + H x, y, z, S x, S y, S = 0 z and as there is no time-dependence in the Hamiltonian it follows that we seek a solution of the form Now y and z are cyclic and so we have where Sx, y, z, P x, P y, P z, t = S 0 x, y, z, P x, P y, P z Et S 0 x, y, z, P x, P y, P z = S x x, P x + S y y, P y + S z z, P z S y y, P y = yp y S z z, P z = zp z 3
We also have 1 m Sx F x = P x x and so S = 3 S x x = mp x + F x S x = 3 m P x + F x 3 Our new constant coordinates are found using F m P x + F x 3 F E = P x + 1 m P y + P z + yp y + zp z P x + 1 m P y + Pz t X = S = 1 mpx + F x t P x F and the old momenta are given by Y = S P z = y P y m t Z = S P z = z P z m t giving p x = S x = mp x + F x p y = S y = P y p z = S z = P z X + t x = m F P x F y = Y + v y t z = z + v y t p x = F X + t p y = P y p z = P z II.3 For a harmonic oscillator we have Hamiltonian H = p m + kx = E Now we can write the expression for the energy as p me + x E k 4 = 1
the equation of an ellipse in phase space. We know that I = 1 π pdq = A π, where A is the area of the region in phase space bounded by the path of the particle, hence A = π me E k = π E ω I = E ω We now effect a canonical transformation such that I is the new momentum, with generating function Sx, I. Hence 1 S = Iω kx x S = miω m ω x dx The angle variable α is then found from α = S I = mωdx miω m ω x = α = arcsin x I mω dx I mω x and then and also x = I mω sin α p = miω cos α I = mωx + p mω 1 mωx α = tan p II.4 Relativistic particle in electric field. II.5 Relativistic particle in magnetic field. 5