Homework 11 # 10.7 b, 10.17, 10.26

Homework 11: # 10.7 b, 10.17, 10.26 10.7 • A single particle moves in space under a conservative potential. Set up the Hamilton-Jacobi equation in ellipsoidal coordinates u, v , φ defined in terms of the usual cylindrical coordinates r, z, φ by the equations. r = a sinh v sin u z= a cosh v cos u For what forms of V (u,v,φ) is the equation separable. • Use the results above to reduce to quadratures the problem of point particle of mass m moving in the gravitational field of two unequal mass points fixed on the z axis a distance 2a apart. Answer: Let’s obtain the Hamilton Jacobi equation. This will be used to reduce the problem to quadratures. This is an old usage of the word quadratures, and means to just get the problem into a form where the only thing left to do is take an integral. Here T = 1 1 1 m ṙ2 + m ż 2 + mr 2 φ̇2 2 2 2 r = a sinh v sin u ṙ = a cosh v sin u v̇ + a sinh v cos uu̇ z = a cosh v cos u ż = a sinh v cos u v̇ − a cosh v sin u u̇ Here ṙ2 + ż 2 = a2 (cosh2 v sin2 u+sinh2 v cos2 u)( v̇ 2 + u̇2 ) = a2 (sin2 u+sinh2 v)( v̇ 2 + u̇2 ) 1 To express in terms of momenta use pv = ∂L = ma2 (sin2 u + sinh2 v) v̇ ∂ v̇ ∂L = ma2 (sin2 u + sinh2 v) u̇ ∂ u̇ because the potential does not depend on v̇ or u̇. The cyclic coordinate φ yields a constant I’ll call αφ pu = pφ = mr 2 ˙ φ = αφ So our Hamiltonian is H= p2v + p2u 2ma2 (sin2 u + sinh2 v) + p2 2ma2 φ sinh2 v sin2 u +V To find our Hamilton Jacobi expression, the principle function applies S = Wu + Wv + αφ φ − Et So our Hamilton Jacobi equation is 1 ∂Wu 2 ∂Wv 2 ∂Wφ 2 1 ) +( ) ]+ [( ( ) +V (u,v,φ) = E ∂v ∂u ∂φ 2ma2 (sin2 u + sinh2 v) 2ma2 sinh2 v sin2 u This is ∂Wu 2 ∂Wv 2 1 1 1 1 [( ) +( ) ]+ ( + )α2 +(sin2 u+sinh2 v)V (u,v,φ) = (sin2 u+sinh2 v)E ∂v 2ma2 ∂u 2ma2 sinh2 v sin2 u φ A little bit more work is necessary. Once we solve for V (u,v,φ) we can then separate this equation into u, v and φ parts, at which point we will have only integrals to take. I suggest drawing a picture, with two point masses on the z axis, with the origin being between them, so they are each a distance a from the origin. The potential is then formed from two pieces V =− GmM1 GmM2 − z| |~r− aˆ |~r+ aˆ z| To solve for the denominators use the Pythagorean theorem, remembering we are in cylindrical coordinates, z|2 = (z ∓ a)2 + r2 |~r∓ aˆ Using the results from part (a) for r and z , z|2 = a2 (cosh v cos u ∓ 1)2 + a2 sinh2 v sin2 u |~r∓ aˆ 2 z|2 = a2 (cosh2 v cos2 u ∓ 2 cosh v cos u + 1 + sinh2 v sin2 u) |~r∓ aˆ Lets rearrange this to make it easy to see the next step, z|2 = a2 (sinh2 v sin2 u + cosh2 v cos2 u + 1 ∓ 2 cosh v cos u) |~r∓ aˆ Now convert the sin2 u = 1 − cos2 u and convert the cosh2 v = 1 + sinh2 v z|2 = a2 (sinh2 v + cos2 u + 1 ∓ 2 cosh v cos u) |~r∓ aˆ Add the 1 and cosh2 v z|2 = a2 (cosh2 v + cos2 u ∓ 2 cosh v cos u) |~r∓ aˆ z|2 = (a(cosh v ∓ cos u))2 |~r∓ aˆ So our potential is now V =− V =− GmM2 GmM1 − a(cosh v − cos u) a(cosh v + cos u) 1 GmM1 (cosh v + cos u) + GmM2 (cosh v − cos u) a cosh2 v − cos2 u Note the very helpful substitution cosh2 v − cos2 u = sin2 u + sinh2 v Allowing us to write V V =− 1 GmM1 (cosh v + cos u) + GmM2 (cosh v − cos u) a sin2 u + sinh2 v Plug this into our Hamilton Jacobi equation, and go ahead and separate out u and v terms, introducing another constant, A: 1 ∂Wu 2 ( ) + 2ma2 ∂u 1 ∂Wv 2 ( ) + 2ma2 ∂v α2φ 1 1 − Gm(M1 − M2 ) cos u − E sin2 u = A 2ma2 sin2 u a αφ2 1 1 − Gm(M1 − M2 ) cosh v − E sinh2 v = −A 2ma2 sinh2 v a The problem has been reduced to quadratures. 3 10.17 Solve the problem of the motion of a point projectile in a vertical plane, using the Hamilton-Jacobi method. Find both the equation of the trajectory and the dependence of the coordinates on time, assuming the projectile is fired off at time t = 0 from the origin with the velocity v0 , making an angle θ with the horizontal. Answer: I’m going to assume the angle is θ because there are too many α’s in the problem to begin with. First we find the Hamiltonian, p2y p2 x + + mgy 2m 2m Following the examples in section 10.2, we set up the Hamiltonian-Jacobi equation by setting p = ∂S/∂qand we get H= 1 ∂S ∂S 1 ∂S 2 ( ) + ( )2 + mgy+ = 0 ∂t 2m ∂x 2m ∂y The principle function is S(x,αx ,y,α,t) = Wx (x,αx ) + Wy (y,α) − αt Because x is not in the Hamiltonian, it is cyclic, and a cyclic coordinate has the characteristic component Wqi = qi αi . S(x,αx ,y,α,t) = xαx + Wy (y,α) − αt Expressed in terms of the characteristic function, we get for our HamiltonianJacobi equation 1 ∂Wy 2 α2x + ( ) + mgy= α 2m 2m ∂y This is ∂Wy x p Integrated, we have ∂y = 2mα− Wy (y,α) = − − 2m2 gy α2 1 (2mα− α2 3m2 g x − 2m2 gy)3/2 Thus our principle function is S(x,αx ,y,α,t) = xαx + − 1 3m2 g Solving for the coordinates, 4 (2mα− α2 x − 2m2 gy)3/2 − αt β= βx = 1 ∂S =− (2mα− α2 ∂α mg x − 2m2 gy)1/2 − t ∂S αx (2mα− α2 =x+ ∂αx m2 g x − 2m2 gy)1/2 Solving for both x(t) and y (t) in terms of the constants β, βx , α and αx α2 x g α − y(t) = − (t + β )2 + 2m2 g mg 2 x(t) = βx + 1 αx (− (2mα− α2 m mg x − 2m2 gy)1/2 ) Our x(t) is αx (β + t) m We can solve for our constants in terms of our initial velocity, and angle θ through initial conditions, x(t) = βx + x(0) = 0 → βx = − αx β m α α2 x g = 0 − y(0) = 0 → − β 2 + 2m2 g 2 mg αx m ẏ(0) = v0 sin θ = −gβ ẋ(0) = v0 cos θ = Thus we have for our constants β= βx = α= v0 sin θ −g v 20 cos θ sin θ g mv 20 mg 2 2 (v sin θ + v 20 cos2 θ) = 2 2g 0 αx = mv0 cos θ Now our y(t) is g v sin θ 2 v 20 v 20 cos2 θ ) + − y(t) = − (t + 0 g g 2g 2 g g v02 sin2 θ v 20 v 20 cos2 θ + − y(t) = − t2 + v0 sin θt− g 2 g2 2g 2 5 g y(t) = − t2 + v0 sin θt 2 and for x(t) x(t) = v 20 v0 sin θ cos θ sin θ + v0 cos θ + v0 cos θt −g g x(t) = v0 cos θt Together we have g y(t) = − t2 + v0 sin θt 2 x(t) = v0 cos θt 10.26 Set up the problem of the heavy symmetrical top, with one point fixed, in the Hamilton-Jacobi mehtod, and obtain the formal solution to the motion as given by Eq. (5.63). Answer: This is the form we are looking for. Z du u(t) u(0) t= p 2 Expressing the Hamiltonian(1in− terms we did in the previous u2 )(α −ofβu)momenta − (b − au)like problem, we get H= p2 ψ p2 (pφ − pψ cos θ)2 + θ + + M ghcos θ 2I3 2I1 2I1 sin2 θ Setting up the principle function, noting the cyclic coordinates, we see S(θ,E,ψ,αψ ,φ,αφ ,t) = Wθ (θ,E ) + ψαψ + φαφ − Et Using ∂S =p ∂q we have for our Hamilton-Jacobi equation, solved for the partial S’s α2ψ 1 ∂Wθ 2 (αφ − αψ cos θ)2 ) + + ( 2I3 2I1 ∂θ 2I1 sin2 θ Turning this inside out: 6 + M ghcos θ = E ∂ ψ I1 s Wθ (θ,E) = 2I1 E − ∂θ integrated, When I3 α2 − (αφ − αψ cos θ)2 − 2I1 M ghcos θ sin2 θ (αφ − αψ cos θ)2 ψ I1 − α 2 I3 sin2 θ Z − 2I1 M ghcos θ)1/2 dθ Wθ = (2I1 E − Now we are in a position to solve ∂S ∂Wθ = −t ∂E ∂E βθ = ∂Wθ Z α2 I 1 (α −α cosθ)2 dθ2 θ 2(2I1 E − ψI3 − φ 2Iψ1sin = βθ + t = ∂E Using the same constants Goldstein uses 2E − α= α2 − 2I1 M ghcos θ)1/2 α2 ψ 2E − I1 I3 I1 2M gl β= I1 ψ I3 I1 = where αφ = I1 b αψ = I1 a and making these substitutions Z βθ + t = (I1 (2E − α2 ψ I3 cosθ ) dθ2 θ ) − I 2 1 (b−aI1sin 2 − I1 2M ghcos θ)1/2 Z cosθ)2 1/2 (α − (b−a sin 2 θdθ − β cos θ) βθ + t = For time t, the value of θ is θ(t) Z θ(t) θ(0) (α − (b−a cosθ)2 sin2 θ dθ − β cos θ)1/2 t= The integrand is the exact expression as Goldstein’s (5.62). Making the substitution u = cos θ we arrive home Z t= du u(t) u(0) p (1 − u2 )(α − βu) − (b − au)2 7 Homework 12: # 10.13, 10.27, Cylinder Michael Good Nov 28, 2004 10.13 A particle moves in periodic motion in one dimension under the influence of a potential V (x) = F |x|, where F is a constant. Using action-angle variables, find the period of the motion as a function of the particle’s energy. Solution: Define the Hamiltonian of the particle p2 H≡E= + F |q| 2m Using the action variable definition, which is Goldstein’s (10.82): I J= we have p dq I p (E − F q) dq J = we2m For F is greater than zero, have only the first quadrant, integrated from q = 0 to q = E/F (where p = 0). Multiply this by 4 for all of phase space and our action variable J becomes Z E/F √ p 0 2m E − F q dq J =4 A lovely u-substitution helps out nicely here. u=E−F q → √ du = −F dq 1 2mu1/2 −F du J =4 √ √ 4 2m Z 8 2m E 3/2 1/2 J= du = E u 3F 0 Z 0 E F 1 Goldstein’s (10.95) may help us remember that ∂H =ν ∂J and because E = H and τ = 1/ν , τ= ∂J ∂E This is τ= √ 8 2m 3/2 [ 3F E ] ∂E ∂ And our period is τ= √ 4 2mE F 10.27 Describe the phenomenon of small radial oscillations about steady circular motion in a central force potential as a one-dimensional problem in the action-angle formalism. With a suitable Taylor series expansion of the potential, find the period of the small oscillations. Express the motion in terms of J and its conjugate angle variable. Solution: As a reminder, Taylor series go like f (x) = f (a) + (x − a)f ′ (a) + 1 ′′ (x − a)2 f (a) + ... 2! Lets expand around some r0 for our potential, U (r) = U (r0 ) + (r − r0 )U ′ (r0 ) + 1 ′′ (r − r0 )2 U (r0 ) + ... 2 Using the form of the Hamiltonian, involving two degrees of freedom, in polar coordinates,(eq’n 10.65) we have l2 1 (p2 + 2 ) + V (r) 2m r r Defining a new equivalent potential, U (r) the Hamiltonian becomes H= H= 1 2 p + U (r) 2m r The r0 from above will be some minimum of U (r), U ′ (r0 ) = − l2 + V ′ (r0 ) = 0 mr 30 2 The second derivative is the only contribution 3l2 + V ′′ (r0 ) = k mr 40 U ′′ = where k > 0 because we are at a minimum that is concave up. If there is a small oscillation about circular motion we may let r = r0 + λ where λ will be very small compared to r0 . Thus our Hamiltonian becomes 1 2 p + U (r0 + λ) 2m r H= This is H= 1 1 2 p + U (r0 ) + (r − r0 )2 U ′′ (r0 ) 2m r 2 H= 1 2 1 2 ′′ p + U (r0 ) + λ U (r0 ) = E 2m r 2 If we use the small energy ǫ defined as ǫ = E − U (r0 ) We see 1 2 1 2 p + λ k 2m r 2 This energy is the effect on the frequency, so following section 10.6 ǫ= ǫ= Jω 2π We have for the action variable r J = 2πǫ and a period τ= with motion given by ∂J ∂ǫ r r = r0 + r pr = mk = 2π J πmω mJωπ 3 r k m sin 2πω cos 2πω A particle is constrained to the edge of a cylinder. It is released and bounces around the perimeter. Find the two frequencies of its motion using the action angle variable formulation. Solution: Trivially, we know the frequency around the cylinder to be its angular speed divided by 2π because it goes 2π radians in one revolution. ˙θ 2π And also simply, we may find the frequency of its up and down bouncing through Newtonian’s equation of motion. νθ = h= 1 2 gt 2 s 2h g t= Multiply this by 2 because the period will be measured from a point on the bottom of the cylinder to when it next hits the bottom of the cylinder again. The time it takes to fall is the same time it takes to bounce up, by symmetry. s → νz = 1 r 2h g T = 2 2 formulation we first start by To derive these frequencies via the action-angle writing down the Hamiltonian for the particle. 2hg p2θ p2 z + 2m 2mR2 Noting that pθ is constant because there is no external forces on the system, and because θ does not appear in the Hamiltonian, therefore it is cyclic and its conjugate momentum is constant. H ≡ E = mgz + pθ = m ˙θR2 we may write Jθ = 2πpθ based on Goldstein’s (10.101), and his very fine explanation. Breaking the energy into two parts, one for θ and one for z, we may express the Eθ part as a function of Jθ . Eθ = p2θ = 2mR2 4 J2 θ 4π 2 2mR2 The frequency is νθ = νθ = ∂Eθ Jθ = ∂Jθ 4π 2 mR2 Jθ m ˙ θR2 2πpθ pθ = = = 2πmR2 4π 2 mR2 4π 2 mR2 2πmR2 Thus we have ˙θ 2π The second part is a bit more involved algebraically. Expressing the energy for z : νθ = Ez = mgz + p2 z 2m Solving for pz and plugging into I J= we get p dq √ I Jz = 2m (Ez − mgz )1/2 dz we can do this closed integral by integrating from 0 to h and multiplying by 2. h √ 2 −1  Jz = 2 2m (Ez − mgz )3/2 ( 0 3 )  mg will be mgh, its potential The original energy given to it in the z direction energy when released from rest. Thus the first part of this evaluated integral is zero. Only the second part remains: Jz = 4 √ 3 3/2 1 2m mg Ez Solved in terms of Ez Ez = ( The frequency is 3 4 νz = g 2/3 r 2 Jz ) m ∂Ez 2 3 1 = ( ( r )2/3 ) 1/3 ∂Jz 3 J m2 g 4 5 All we have to do now is plug what Jz is into this expression and simplify the algebra. As you may already see there are many different steps to take to simplify, I’ll show one. 2 3 1 νz = ( ( r )2/3 ) 3 (mgh)3/2 ]1/3 m2 [4 m q g 2 4 Now we have a wonderful mess. Lets3g gather the numbers, and the constants to one side νz = 2 3 2/3 1 ( 3 4) 2 1/ 3 ( 34 )1/3 21/6 g 2/3 m1/3 g 1/3 m1/6 m1/2 g 1/2 h1/2 You may see, with some careful observation, that the m’s cancel, and the constant part becomes g 1/2 h1/2 The number part simplifies down to 1 √ 2 2 Thus we have 1 1 = r g2h g h √ as we were looking for from Newton’s trivial method. Yay! Our two fre2 2 2 quencies together νz = r νθ = νz = 1 ˙θ 2π r 2h g The condition for the same path to2 be retraced is that the ratio of the frequencies to be a rational number. This is explained via closed Lissajous figures and two commensurate expressions at the bottom of page 462 in Goldstein. 6