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Exercises

Chapter 1: Elementary Newtonian Mechanics

1.1 Under the assumption that the orbital angular momentum 1 = r x P of aparticle is conserved show that its motion takes place in aplane spanned by ro.the initial position, and Po, the initial momentum. Which of the orbits of Fig. Iare possible in this case? (0 denotes the origin of the coordinate system.)

•o

(0) (b) (c)

o•

o(d)

. ~

o(e)

j 4

(f)

o•

Fig.l

1.2 In the plane of motion of Exercise 1.1 introduce polar coordinates {r(t) ,cp(t ) }. Calculate the line element (ds)2 = (dx)2 + (dy)2, as weIl as v 2 = ;j;2 + iland 12 , in polar coordinates. Express the kinetic energy in terms of rand 12.

1.3 For the description of motions in R3 one may use Cartesian coordinatesr(t) = { x(t) , y(t) , z(t ) }, or spherical coordinates {r(t) , O(t) , cp(t )}. Calculate theinfinitesimal line element (ds? = (dx)2 + (dy)2 + (dz)2 in spherical coordinates.Use this result to derive the square of the velocity v2 =1;2 + il + i 2 in thesecoordinates.

1.4 Let ex , ey , ez be Cartesian unit vectors. They then fulfill e; =€; =e; = I ,ex • ey = € x • €z = €y . €z =0, €z =ex x €y (plus cyclic permutations). Introduce three, mutually orthogonal unit vectors er, e'l" eil as indicated in Fig.2.

z

404 Exercises

Fig.2

e/

x

Detennine er and eep from the geometry of this figure. Confinn that er · eep = O.Assurne ee = <lex + ße y + ,ez and detennine the coefficients <l, ß" such thate~ = 1, ee . eep = 0 = ee . er. Calculate v = i: = d(rer)/dt in this basis as weIlas v 2

•

1.5 A particle is assumed to move according to r(t) =vOt with vO = {O, v, O},with respect to the inertial system K. Sketch the same motion as seen fromanother reference frame K', which is rotated about the z-axis of K by an angleiP ,

x ' = x cos iP + y sin iP ,

y' = - x sin iP + Ycos iP , z ' = z ,

for the cases iP =w and iP =wt , where w is a constant.

1.6 A particle of mass m is subject to a central force F = F(r)r [r . Show thatthe angular momentum 1= mr x r is conserved (i.e, its magnitude and direction)and that the orbit lies in a plane perpendicular to I .

1.7 (i) In an N -particle system that is subject to internal forces only, the potentials V;k depend only on the vector differences r ik =r i - r k. but not on theindividual vectors r i o Which quantities are conserved in this system?

(ii) If V;k depends only on the modulus Irik I the force acts along the straight. line joining i to k. There is one more integral of the motion.

1.8 Sketch the one-dimensional potential

U(q) = -5qe-q + q-4 + 2/q for q ~ 0

and the corresponding phase portraits for a particle of mass m = 1 as a functionof energy and initial position qo. In particular, find and discuss the .two points ofequilibrium. Why are the phase portraits symmetrie with respect to the abscissa?

Chapter 1: Elementary Newtonian Mechanics 405

1.9 Study two identieal pendula of length land mass m, coupled by a harmoniespring, the spring being inactive when both pendulums are at rest. For smalldeviations from the vertieal the energy reads

1 (2 2) 1 2(2 2) 1 2 2E = 2m X2 + X4 + Zmwo Xl + X3 + ZmWl (Xl - X3)

with X2 =mXl, X4 =mX3 . Identify the individual tenns of this equation. Derivefrom it the equations of motion in phase space,

d:z: .-=M:z:.dt

The transfonnation

with A- 1 (ß ß)- V2 ß -ß and

u= (~ ~)decouples these equations . Write the equations obtained in this way in dimensionless form and solve them.

1.10 The one-dimensional harmonie oscillator satisfies the differential equation

mx(t) = -AX(t) , (1.1)

with m the inertial mass, A a positive constant, and x(t) the deviation fromequilibrium. Equivalently, (1.1) can be written as

.. 2 0X +w X = 1 (1.2)

Solve the differential equation (1.2) by means of x(t) =acos(f-tt) + bsin(f-tt) forthe initial condition

X(O) = xo and p(O) = mx(O) = PO . (1.3)

Let x (t ) be the abscissa and p(t) the ordinate of a Cartesian coordinate system.Draw the graph of the solution with w =0.8 that goes through the point (xo =1,Po =0).

1.11 Adding a weak friction force to the system of Exercise 1.10 yields theequation of motion

x+ K;X + w2x = 0 .

"Weak" means that K; < 2w. Solve the differential equation by means of

x(t) = e4 t [xocoswt + (PO/mW) sinwtJ .

Draw the graph (x(t) ,P(t» of the solution with w = 0.8 that goes through (xo =I,PO=O).

406 Exercises

y Fig.3

for x< 0for x> O .

---r--::::;~-----'----- X

1.12 A mass point of mass m moves in the piecewise constant potential (seeFig.3)

U= {UlU2

In crossing from the domain x < 0, where its velocity was VI, to the domainx > 0, it changes its velocity (modulus and direction). Express U2 in terms ofthe quantities Ui, VI, 0'1, and 0'2. What is the relation of 0'1 to 0'2 when (i)Ui < U2 and (ii) U, > U2? Work out the relationship to the law of refraction ofgeometrical optics.

Hint: Make use of the principle of energy conservation and show that one component of the momentum remains unchanged in crossing from x < 0 10 x > o.

I

I

-'rs IIIIla Fig.4

1.13 In a system of three mas points ml, m2, m3 let 5 12 be the centerof-mass of 1 and 2 and 5 the center-of-mass of the whole system. Express thecoordinates rl , r2 , r3 in terms of r 8 , sa, and Sb, as defined in Fig.4. Calculatethe total kinetic energy in terms of the new coordinates and interpret the result.Write the total angular momentum in terms of the new coordinates and show thatL:i l, = 18 +Ia +Ib, where 18 is the angular momentum of the center-of-mass and

Chapter 1: Elementary Newtonian Mechanics 407

l « and h are relative angular momenta. By considering a Galilei transfonnationr' ::: r +wt + a, t' ::: t + s show that I s depends on the choiee of the origin, whilet, and lb do not.

1.14 Geometrie similarity, Let the potential U(r) be a homogeneous functionof degree 0' in the coordinates (x , y, z), i.e. U('xr)::: ,XOIU(r).

(i) Show by making the replacements r -. 'xr and t -. p.t, and choosingp.::: ,XI-0I/2, that the energy is modified by a factor ,XOI and that the equation ofmotion remains unchanged.

The consequence is that the equation of motion admits solutions that aregeometrically similar, i.e. the time differences (Llt)a and (Llth of points thatcorrespond to each other on geometrically similar orbits (a) and (b) and thecorresponding linear dimensions La and Lb are related by

(Llth ::: (Lb )1-01/2(Llt)a La

(ii) What are the consequences of this relationship for

- the period of harmonie oscillation?- the relation between time and height of free fall in the neighborhood of the

earth's surface?- the relation between the periods and the semimajor axes of planetary el

lipses?

(iii) What is the relation of the energies of two geometrically similar orbits for

- the harmonie oscillation?- the Kepler problem?

1.15 The Kepler problem. (i) Show that the differential equation for P(r), inthe case of finite orbits, has the following form:

dp 1-:::-

dr r

rprA

(r - rp)(rA - r) ,(1.4)

where rp and rA denote the perihelion and the aphelion, respectively. Calculaterp and rA and integrate (1.4) with the boundary condition p(rp) ::: O.

(ii) Change the potential to U(r) ::: (-A/r) + (B /r 2) with IBI ~ P'f2p. . Determine the new perihelion rp and the new aphelion r~ and write the differentialequation for p(r) in a form analogous to (1.4). Integrate this equation as in (i)and detennine two successive perihelion positions for B > 0 and for B < O.

Hint:

d 0' 1-darccos(O'/(x + ß»)::: ---;:=====.:====::::::::;::

x x Jx2(l - ß2) - 20'ßx - 0'2

(A = Gmtm2) ,

408 Exercises

1.16 The most general solution of the Kepler problem reads, in terms of polarcoordinates r and rp ,

r(~) = p .1 + c: cos ( 4> - 4>0)

The parameters are given by

12

p= Ap'

c:=2Ez2

1+--2 ,pA (

mtm2 )p = mt +m2

What values of the energy arepossible if the angular momentum is given? Calculate the semimajor axis of the earth's orbit under the assumption ms un ~ mEarth;

G =6.672 X 10-11 m3 kg- t S-2 ,

ms = 1.989 x loJ° kg ,

mE = 5.97 x 1024 kg .

Calculate the semimajor axis of the ellipse along which the sun moves about theeenter-of-mass of the sun and the earth and eompare the result to the solar radius(6.96 x 108 m).

1.17 Determine the interaction of two electric dipoles Pt and P2 (example fornoncentral potential force) .

Hints: Calculate the potential of a single dipole Pt, making use of the followingapproximation. The dipole consists of two eharges ±et at a distance d t . Let ettend to infinity and Idtl to zero, in such a way that their product Pt = dt et staysconstant. Then calculate the potential energy of a finite dipole P2 in the field ofthe first and perform the same limit ez - 00, Id21 - 0, with P2 = d2e2 constant,as above. Caleulate the forces that aet on the two dipoles.

Answer:

W(l,2) = (Pt' P2) /r3- 3 (Pt' r) (P2 ' r) /r5

,

F =-V'tW = [3 (Pt ' p2)/r5

- 15 (Pt' r) (P2' r)/r7] r

+ 3 [Pt (P2 . r) + P2 (PI . r)] /r5 =-F 12 •

1.18 Let the motion of a point mass be govemed by the law

v =v x a, a =const . (1.5)

Show that r .a =v(O) . a holds for all t and reduce (1.5) to an inhomogeneousdifferential equation of the form r + w2 r = j(t). Solve this equation by means

Chapter 1: E1ementary Newtonian Mechanics 409

of the substitution rinhom(t) = ct + d. Express the integration constants in termsof the initial values r(O) and v(O) . Describe the curve r(t) == rhom(t)+ rinhom(t).

Hint:

at X (az x a3) = az (at · a3) - a3 (at· az) .

1.19 An iron ball falls vertically onto a horizontal plane from which it is reflected. At every bounce it loses the nth fraction of its kinetic energy. Discussthe orbit x =x (t ) of the bouncing ball and derive the relation between X max andtmax•

Hint: Study the orbit between two successive bounces and sum over previoustimes.

1.20 Consider the foIlowing transformations of the coordinate system:

{t,r}-+{t ,r} , {t,r} -+{t ,-r}, {t,r}-+{-t ,r},E P T

as weIl as the transformation P . T that is generated by performing first T andthen P. Write these transformations in the form of matrices that act on the fourcomponent vector (;) . Show that {E, P, T, PT} form a group.

1.21 Let the potential U(r) of a two-body system be CZ (twice continuouslydifferentiable). For fixed relative angular momentum, under which additionalcondition on U(r) are there circular orbits? Let Eo be the energy of such anorbit. Discuss the motion for E = Eo +e for smaIl positive c. Study the specialcases

U(r) =r" and U( r) = >'I r .

1.22 Following the methods explained in Sect. 1.26 show the following.

(i) In the northern hemisphere a falling object experiences a southward deviationof second order (in addition to the first-order eastward deviation).

(ii) A stone thrown vertically upward falls down west of its point of departure,the deviation being four times the eastward deviation of the falIing stone.

1.23 Let a two-body system be subject to the potential

aU(r) =--

rZ

in the relative coordinate r, with positive o . Calculate the scattering orbits r(4)) .For fixed angular momentum what are the values of a for which the particlemakes one (two) revolutions about the center of force? FoIlow and discuss anorbit that collapses to r = O.

1.24 A pointlike comet of mass m moves in the gravitational field of a sunwith mass M and radius R. What is the total cross section for the comet to crashon the sun?

410 Exercises

1.25 Solve the equations of motion for the example of Sect. 1.21.2 (Lorentzforce with constant fields) for the case

1.26 Consider the Kepler problem. Let Px and py denote the components ofthe momentum in the plane of the orbit. Show: In momentum space, i.e., in thespace spanned by (Px,Py) that all bound orbits are eire/es. Give the radius ofthese circles and the position of their center as functions of angular momentumand energy.

Chapter 2: The Principles of Canonical Mechanics

2.1 The energy E(q,p) is an integral of a finite, one-dimensional, periodicmotion. Why is the portrait symmetrie with respect to the q-axis? The surfaceenclosed by the periodie orbit is

f r:F(E) = pdq = 2 . pdq .qmm

Show that the change of F(E) with E equals the period T of the orbit, T =dF(E)/dE, Calculate Fand T for the example

E(q,p) = pZ12m + mwZqz12.

2.2 A weight glides without friction along a plane inclined by the angle a withrespect to the horizontal. Study this system by means of d' Alembert's principle.

2.3 A ball rolls without friction on the inside of a circular annulus. The annulusis put upright in the earth's gravitational field. Use d' Alembert's principle toderive the equation of motion and discuss its solutions.

2.4 A mass point m that can only move along a straight line is tied to the pointA by means of aspring. The distance of A to the straight Une is I (cf. Fig.5).Calculate (approximately) the frequency of oscillation of the mass point.

2.5 Two equal masses m are connected by means of a (massless) spring withspring constant x. They move without friction along a rail, their distance being1 when the spring is inactive. Calculate the deviations Xl (t) and Xz(t ) from theequilibrium positions, for the following initial conditions:

X l (0) = 0 ,

Xz(O) = 1,

;h (O) = Vo ,

Xz(O) = 0 .

Chapter 2: The Principles of Canonical Mechanics 411

A

m Fig.S

2.6 Given a function F(XI , .. . , x f) that is homogeneous and of degree N inits f variables, show that

f aFL "j):x; =NF.;=1 x,

2.7 If in the integral

lx 2

I[y] = dx f(y , y')Xl

f does not depend explicitly on x, show that

,af ,y ay' - f(y , y) = const .

Apply this result to L(q,q) = T - U and identify the constant. T is assumed tobe a homogeneous quadratic form in q.2.8 Solve the following two problems (whose solutions are weIl known) bymeans of variational calculus :

(i) the shortest connection between two points (XI, y]) and (X2, Y2) in the Euclidean plane;

(ii) the shape of a homogeneous, fine-grained chain suspended at its end points(XI, YI) and (X2, Y2) in the gravitational field.

Hints: Make use of the result of Exercise 2.7. The equilibrium shape of the chainis determined by the lowest position of its center of mass. The line element isgiven by

ds = V(dx)2 + (dy)2 = J1+Y'2dx .

2.9 Two coupled pendula can be described by means of the Lagrangianfunction

(i) Show that the Lagrangian function

412 Exercises

L' =1m (Xl - iwoxd + 1m(X2 - iWOX2r

- tm (w? - w~) (Xl - xd

leads to the same equations of motion. Why is this so?

(ii) Show that transfonning to the eigenmodes of the system leaves the Lagrangeequations form invariant.

2.10 The force acting on a body in three-dimensional space is assumed to beaxially symmetric with respect to the z-axis. Show that

(i) its potential has the form U =U(r,z), where {r,ep, z} are cylindrical coordinates,

X =r cos ip , y =rsin ep, . z =z ;

(ii) the force always lies in a plane containing the z-axis.

2.11 With respect to an inertial system Ko the Lagrangian function of a particleis

The frame of reference K has the same origin as Ko but rotates about the latterwith constant angular velocity w. Show that the Lagrangian with respect to Kreads

m:i:2 mL =-2- + m:i: . (w x e) + 2"(w X z)2 - U(z) .

Derive the equations of motion of Sect. 1.25 from this.

2.12 A planar pendulum is suspended such that its point of suspension glideswithout friction along a horizontal axis. Construct the kinetie and potential energies and a Lagrangian function for this problem.

2.13 A pearl of mass m glides (without friction) along a planar curve s =s(<1')put up vertically. s is the length of arc and <1' the angle between the tangent tothe curve and the horizontal line (see Fig.6.).

(i) Derive the equation for s(t) for harmonie oscillations.

(ii) What is the relation between s(t) and <1'(t)? Discuss this relation and themotion that follows from it. What happens in the limit where s can reach itsmaximal amplitude?

(iii) From the explicit solution calculate the force of constraint and the effectiveforce that acts on the pearl.

2.14 Geometrical interpretation of the Legendre transformation. Given f(x)with f"(x) > o. Construct (.Cf)(x) = xf'(x) - f(x) = xz - f(x) == F(x, z),where z = f'(x). The inverse X = x(z) of the latter exists and so does theLegendre transform of f(x), which is zx(z) - f(x(z» = lf(z) = <1'(z) .

y=zx

Fig.6

Chapter 2: The Principles of Canonica1 Mechanics

"---------,~---x

Fig. 7

413

(i) Comparing the graphs of the functions y = f(x) and y = zx (for fixed z) onesees with

8F(x, z) =08x

that x = x(z) is the point where the vertical distance between the two graphs ismaximal (see Fig.7).

(ii) Take the Legendre transform of ~(z), i.e. (.C~)(z) = z~'(z) - ~(z) = z x ~(z) == G(z, x) with ~'(z) = x . ldentify the straight line y = G(z, x) for fixed zand with x =x(z) show that one has G(z, x) = f(x) . Sketch the picture that oneobtains if one keeps x = Xo fixed and varies z.

2.15 (i) Let

L(Ql, fJ2, 41, 42, t) = T - U with2 2

T = L Cik4i4k + L h4k + a .i,k=l k=1

Under what condition can one construct H(q, P, t) and what are PI, P2, and H?Confirm that the Legendre transform of H is again Land that

( 82L) (82H)

det 84k8lji

det 8pn8pm =1 .

Hint: Take dll = 2Cll, d12 = d21 = C12 + C2l, d22 = 2C22, 7ri = Pi - b io

(ii) Assume now that L = L(XI == 41, X2 == 42, ql, ~, t) == L(Xl , X2, u) with

U ~f(ql' q2, t) to be an arbitrary Lagrangian function. We expect the momentaPi = Pi(Xl, X2, u) derived from L to be independent functions of Xl and X2, i.e.that there is no function F(PI (Xl, X2,u), P2(Xl, X2, u)) that vanishes identically.Show that, if PI and P2 were dependent, the determinant of the second derivativesof L with respect to the Xi would vanish .

414 Exercises

Hint: Consider dF/dX\ and dF/dx2.

2.16 A particle of mass m is described by the Lagrangian function

L I (.2 .2 .2) w 1= 2m x + y + z + "2 3 ,

where h is the z-component of angular momentum and w is a constant frequency.Find the equations of motion, write them in terms of the complex variable x +iy and of z, and solve them. Construct the Hamiltonian function and find thekinematical and canonical momenta. Show that the particle has only kineticenergy and that the latter is conserved.

2.17 Invariance under time translations anti Noether's theorem. The theoremof E. Noether can be applied to the case of translations in time by means of thefollowing trick. Make t a coordinate-like variable by parametrizing both q and tby q =q(r), t = t(r) and by defining the following Lagrangian function:

- ( dq dt ) der ( 1 dq ) dtL q,t, dr ' dr =L q, dt/dr dr ' t dr "

(i) Show that Hamilton's variational principle applied to L yields the same equations of motion as for L.

(ii) Assume L to be invariant under time translations

hS(q, t) = (q, t + s) . (2.1)

Apply Noether's theorem to L and find the constant of the motion correspondingto the invariance (2.1).

2.18 A mass point is scattered elastically on a sphere with center P and radiusR (see Fig. 8). Show that the physically possible orbit has maximallength.

Hints: Show first that the angles a and ß must be equal and construct the actionintegral. Show that any other path AB'n would be shorter than for those points


where the sum of the distanees to A and n is eonstant and equal to the lengthof the physieal orbit.

2.19 (i) Show that eanonieal transformations leave the physieal dimension ofthe product Piqi unehanged, i.e, [PiQi] = [Piqi]. Let ~ be the generating funetionfor a eanonieal transformation. Show that

where H is the Hamiltonian funetion and t the time.

(ii) In the Hamiltonian funetion H = pZ/2m+mwzl /2 of the harmonie oseillatorintroduee the variables

def t= def t= defXt =wymq, xz=p/ym, T=wt,

thus obtaining H = (xf + x~)/2 . What is the generating funetion 4i(Xt, Yt) forthe eanonieal transformation x--;:tY that eorresponds to the funetion ~(q , Q) =

~

(mwlf2)cotQ? Caleulate the matrix M ik = ox;jOYk and eonfirm detM = 1and M JM = J .

2.20 The group SP2/ is partieularly simple for f = 1, i.e. in two dimensions.

(i) Show that every matrix

M = (att atz)aZI an

is symplectie if and only if a n a22 - a12aZt = 1.

(ii) Therefore, the orthogonal matriees

o = ( CO~ 0' sin 0' )

-SIOO' eosO'

and the symmetrie matriees

S = (~ ;) with xz - yZ =1

belong to SP2/. Show that every M E SP2/ ean be written as a produet

M=S·O

of a symmetrie matrix S with determinant 1 and an orthogonal matrix O.

2.21 (i) Evaluate the following Poisson braekets for a single particle:

{li ,rk} , {l j,pk} , {li,r} , {li 'pZ} .

(ii) If the Hamiltonian funetion in its natural form H = T + U is invariant underrotations, what quantities ean U depend on?

416 Exercises

2.22 Making use of the Poisson brackets show that for the system H = T+U(r)with U(r) = "YI r and "Y a constant, the vector

A =P x 1+ zm"YIr

is an integral of the motion (Lenz's vectori.

2.23 The motion of a particle of mass m is described by

o =const .1 (2 2)H =2m PI + P2 + maql ,

Construct the solution of the equations of motion for the initial conditions

making use of Poisson brackets.

2.24 For a three-body system with masses m i, coordinates ri, and momentaPi introduce the following coordinates Uacobian coordinates):

deftt'l =r2 - rl

def ml rl + m2 r2tt'2 =r3 -

ml +m2

def ml rl + m2r2 + m3r3tt'3 =

ml +m2+m3

defmlP2 - m2PI1t't = ,

ml +m2

def(ml + m2) P3 - m3 (Pt + P2)1t'2= ,

mt +m2+m3

(relative coordinate of particles land 2) ,

(relative coordinate of particle 3

and the center of mass of the first two) ,

(center of mass of the three particles),

def1t'3 = PI +P2 +P3 .

(i) What is the physical interpretation of the momenta 1t'1, 1t'2, 1t'3 ?

(ii) How would you define such coordinates for four or more particles?

(iii) Show in at least two (equivalent) ways that the transformation

is canonical.

(2.1)for all o .

2.25 Given a Lagrangian function L for which oLIßt = 0, study onlythose variations of the orbits qk(t, o) which belong to a fixed energy E =L:kqk(oLloqk) - L and whose end points are kept fixed irrespective of thetime (t2 - tl) that the system needs to move from the initial to the end point, i.e.

{qk (tl (a) , o) = q~)

qk (Ma), o) = q?)

Chapter 2: The Principles of Canonica1 Mechanics 417

Thus, initial and final times are also varied, t; = t ;(a) .

(i) Calculate the variation of I(a),

dI(a) I 1t 2(<»öl = -d- da = L (qk(t, o), qk(t, o) dt .

a <>0=() t1(<»

(ii) Show that the variational principle

(2.2)

together with the prescriptions (2.1) is equivalent to the Lagrange equations (thePrinciple 0/Euler anti Maupertuis).

2.26 The kinetic energy

is assumed to be a positive symmetrie quadratic form in the q;. The orbit in thespace spanned by the qk is described by the length of arc s such that T =(ds/dtf.With E == T +U the integral K of Exercise 2.25 can be replaced with an integralover s. Show that the integral principle obtained in this way is equivalent toFermat's principle of geometrie optics,

lX 2

fJ n(z, v)ds = 0Xl

(n : index of refraction, v: frequency).

2.27 Let H = p2/2 + U(q), where the potential is such that it has a localminimum at qo . Thus, in an interval ql < qo < q2 the potential forms a potentialweIl. Sketch a potential with this property and show that there is an intervalU(qo) < E :::; E max where there are periodic orbits. Consider the characteristicequation of Hamilton and Jacobi (2.154). If S(q, E) is a complete integral thent - to =8S/8E. Take the integral

I(E)d;,f_l 1 pdq271" JrE

over the periodie orbit FE with energy E (this is the surface enclosed by FE).Write I(E) as an integral over time and show that

dI T(E)dE =2;"" '

2.28 In Exercise 2.27 replace S(q, E) by S(q, I) with 1= I(E) as defined there.

Izl = 112.

418 Exercises

S generates the canonical transfonnation (q,p,H) --+ (O,I,fI = E(il) . Whatare the canonical equations in the new variables? Can they be integrated?

2.29 Let HO = p2/2 + q2/2. Calculate the integral I(E) defined in Exercise2.27. Solve the characteristic equation of Hamilton and Jacobi (2.154) and writethe solution as S(q,1). Then 0= 8S/81. Show that (q,p) and (0,1) are relatedby the canonical transfonnation (2.95) of Sect. 2.24 (ii).

2.30 We assume that the Langrangian of a mechanical system with one degreeof freedom does not depend explicitly on time. In Harnilton's variational principlewe make a smooth change of the end points qa and qb, as weIl as of the runningtime t = t2 - t), in the sense that the solution <p(t) for the values (qa, qb,t)and the solution 4>(8, t) which belongs to the values (q'a, q'b, t') are related ina smooth manner: <p(t) 1--+ 4>(8 , t) such that 4>(8, t) is differentiable in 8 and4>(8 =0, t) =<p(t).

Show that the corresponding change of the action integral 10 into whichthe physical solution is inserted (this function is called Hamilton's principalfunction) , is given by the foIlowing expression

810 =-E bt + pb 8l- pa Sq" .

Chapter 3: The Mechanics of Rigid Bodies

3.1 Let two systems of reference K and K be fixed in the center of mass of arigid body, the axes of the fonner being fixed in space, those of the lauer fixed inthe body. If J is the inertia tensor with respect to K and ] the one as calculatedin K, show that (i) J and ] have the same eigenvalues . (Use the characteristicpolynomiaI.)

(ii) K is now assumed to be a system of principal axes of inertia. What is theform of ]? Calculate J for the case of rotation of the body about the 3-axis.

3.2 Two particles with masses ml and m2 are held by a rigid but masslessstraight connection with length 1. What are the principal axes and what are themoments of inertia?

3.3 The inertia tensor of a rigid body is found to have the form

112 0)I~ I~3

Detennine the three moments of inertia and consider the foIlowing special cases.

(i) 111 = 122 =A, 112 =B . Can 133 be arbitrary?

(ii) 111 =A, 122 =4A , 112 =2A. What can you say about h3? What is the shapeof the body in this example?

Chapter 3: The Mechanics of Rigid Bodies 419

Fig.9

b

a

3.4 Construct the Lagrangian function for general, force-free motion of a conieal top (heigth h, mass M , radius of base cirele R). What are the equations ofmotion? Are there integrals of the motion and what is their physieal interpretation?

3.5 Calculate the moments of inertia of a torus filled homogeneously with mass.Its main radius is R; the radius of its section is r.

3.6 Calculate the moment of inertia h for two arrangements of four balls, twoheavy (radius R, mass M) and two light (radius r, mass m) with homogeneousmass density, as shown in Fig.9. As a model of a dancer's pirouette comparethe angular velocity for the two arrangements, with L3 fixed and equal in thetwo cases.

3.7 (i) Let the boundary of a homogeneous body be defined by the formula (inspherical coordinates)

R(() = Ro(l + a cos () ,

i.e. e(r, (),p) = eo = const for r ~ R«() and alt () and P, and e(r, (), p) = 0 forr > R«(). If M is the total mass, calculate eo and the moments of inertia.

(ii) Perform the same calculation for a homogeneous body whose shape is givenby

R«() =Ro (l + ßY20«() )

with Y20«() = J5/161r(3cos2 () - 1) being the spherical harmonie with 1 = 2,m =O. In both examples sketch the body.

3.8 Determine the moments of inertia of a rigid body whose inertia tensor withrespect to a system of reference K) (fixed in the body) is given by

420 Exercises

-----Fig.l0

p

9 1 -J38 4 8

J= 1 3 -J34 2 4

-J3 -J3 11

8 4 8

Can one indicate the relative position of the principal inertia system Ko relativeto K t?

3.9 A ball with radius a is filled homogeneously with mass such that the densityis {!O. The total mass is M .

(i) Write the mass density e with respect to a body-fixed system centered inthe center of mass and express {!O in terms of M . Let the ball rotate about a pointP on its surface (see Fig. 10).

(ii) What is the same density function e(r , t) as seen from a space-fixed systemcentered on P?

(iii) Give the inertia tensor in the body-fixed system of (i) . What is the momentof inertia for rotation about a tangent to the ball in P?

Hint: Use the step function B(x) = 1 for x ~ 0, B(x) =0 for x < O.

3.10 A homogeneous circular cylinder with length h, radius r , and mass m

rolls along an inclined plane in theearth 's gravitational field.

(i) Construct the full kinetic energy of the cylinder and find the moment of inertiarelevant to the described motion.

(ii) Construct the Lagrangian function and solve the equation of motion.

3.11 Manifold 0/ motions 0/ the rigid body. A rotation R E SO(3) can bedetermined by a unit vector rp (the direction about which the rotation takesplace) and an angle ep.

(i) Why is the interval 0 ~ ep ~ 1r sufficient for describing every rotation?

(ii) Show that the parameter space (rp ,ep) fills the interior of a sphere with radius1r in R3• This ball is denoted by D2 • Confirm that antipodal points on the ball'ssurface represent the 'same rotation.

Chapter 4: Re1ativistic Mechanies 421

(iii) There are two types of closed orbit in D2, namely those whieh can be contracted to a point and those whieh connect two antipodal points, Show by meansof a sketch that every closed curve can be reduced by continuous deformationto either the former or the latter type.

3.12 Calculate the Poisson brackets (3.92-95).

Chapter 4: Relativistic Mechanics

4.1 (i) A neutral 71" meson (71"0) has constant velocity Vo along the x3-direction.Write its energy-momentum vector. Construct the special Lorentz transformationthat leads to the particle's rest system.

(ii) The particle decays isotropically into two photons, i.e. with respect to itsrest system the two photons are emitted in all directions with equal probability.Study their decay distribution in the laboratory system.

4.2 The decay 71" ---t J1 + v (cf. Example (i) of Sect. 4.9.2) is isotropie in thepion's rest system. Show that above a certain fixed energy of the pion in thelaboratory system there is a maximal angle beyond which no muons are emitted.Calculate that energy and the maximal emission angle as a function of ffi" andffi,. (see Fig. 11). Where do muons go in the laboratory system that in the pion'srest system were emitted forward, backward, or transversely with respect to thepions's velocity in the laboratory?

4.3 Consider a two-body reaction A+B ---t A+B for whieh the relative velocityof A (the projectile) and B (the target) is not small compared to the speed oflight. Examples are

Denoting the four-momenta before and after the scattering by qA , qB andq~ , q~ the following quantities are Lorentz scalars, i.e. they have the samevalues in every system of reference,

dCf 2( ,)2t=c qA-qA .

~------.p:

Fig.ll

422 Exercises

Conservation of energy and momentum requires q~+qß = qA+qB. Furthermore,we have q~ = q1 = (mAc2)2, q1 = qZ = (m BCl)2.

(i) Express sand t in terms of the energies and momenta of the particles in thecenter-of-mass frame. Denoting the modulus of the 3-momentum by q* and thescattering angle by (J*, write s and t in terms of these variables.

(ii) Define u = C2(qA - qß)2 and show that

s+t+u=2(m~+m1)c4 .

4.4 Calculate the variables s and t (as defined in Exercise 4.3) in the laboratorysystem, i.e. in that system where B is at rest before the scattering. What is therelation between the scattering angle (J in the laboratory system and (J* in thecenter-of-mass frame? Compare to the nonrelativistic expression (1.80).

4.5 In its rest system the electron's spin is described by the 4-vector sO< = (0, s) .What is the form of this vector in a frame where the electron has the momentump? Calculate the scalar product (s· p) =sO<po<'

4.6 Show that

(i) every lightlike vector z (z2 = 0) can be brought to the form (1,1,0,0) by meansof Lorentz transformations;

(ii) every spacelike vector can be transformed to the form (0, zl, 0, 0), wherezl = J- z2 .

Indicate the necessary transformations in both cases.

4.7 If J, and Kj denote the generators of rotations and boosts, respectively (cf.Sect, 4.5.2 (iii» define

Making use of the commutation mies (4.59) calculate [Ap, Aq], [B,, Bq], and[Ap , Bq] and compare to (4.59).

4.8 Study the behavior of J, and Kj with respect to space inversion, i.e. determine PJjP- I , PKjP- I

.

4.9 In quantum theory one prefers to use the quantities

JA def.J KA der 'Kj=lj, j=-1 i :

What are the commutators (4.59) for these matrices? Show that the matrices Jjare Hermitian, i.e. that O~)* = Jj •

4.10 A muon decays predominantly into an electron and two massless neutrinos,u: -+ e- + VI + V2. If the muon is at rest, show that the electron assumes itsmaximal momentum whenever the neutrinos are emitted parallel to each other.

Chapter 4: Relativistic Mechanies 423

Calculate the maximal and minimal energies of the electron as functions of m lland me .

Answer:

Draw the corresponding momenta in the two cases.

4.11 A particle of mass M is assumed to decay into three particles (1,2,3) withmasses rni , m2, m3. Detennine the maximal energy of particle 1 in the restsystem of the decaying particle as folIows. Set

PI =- f(x)n, P2 =x f (x )n , P3 =(1 - x )f (x )n ,

where n is a unit vector and x is a number between 0 and 1. Find the maximumof f(x) from the principle of energy conservation.

Examples:

(i) p,- -t e- + VI + 1"2 (cf. Exercise 4.10),

(ii) Neutron decay: n -t p + e + u.

What is the maximal energy of the electron? What is the value of ß= Ivl/c forthe electron? mn - m p = 2.53 me, mp = 1836 me.

4.12 Pions 7r+, 7r- have the mean lifetime r ~ 2.6 x 10-8 sand decay predominantly into a muon and a neutrino. Over what distance can they fly, on average,before decaying if their momentum is P1r = X • m 1rc with x = 1, 10, or l000?(m 1r ~ 140 MeV /c2 = 2.50 X 10-28 kg) .

4.13 The free neutron is unstable. 118 mean lifetime is r ~ 900 s. How far cana neutron fly on average if its energy is E = 10-2 m n c;2 or E =1014 mn c;2 ?

4.14 Show that a free electron cannot radiate a single photon, i.e, the process

e-te+'Y

cannot take place because of energy and momentum conservation.

4.15 The following transfonnation

R2

I : x" H xll =2XllX

implies the relation R..,fi2 =R2• This is an obvious generalization of the wellknown inversion at the circle of radius R, r · f =R2. Show that the sequenceof transfonnations: inversion I of x ll , translation T of the image by the vectorR2cll , and another inversion of the result, i.e.,

x' =(I 0 T 0 I)x

is precisely the special confonnal transformation (4.102).

424 Exercises

4.16 Consider the following Lagrangian

I ( . 2 2("p - 1)2) .L =2m "pq - c0 "p =. L(q , ,,p)

which contains the additional, dimensionless, degree of freedom "p . The parameterCo has the physical dimension of a veloc ity. Show: The extremum of the actionintegral yields a theory obeying special relativity for which CO is the maximalvelocity, in other words, one obtains the Lagrangian (4.97) with the velocity oflight c replaced by Co. Consider the limit Co ~ 00 .

Chapter 5: Geometrie Aspeets of Meehanics

5.1 Let ~ be an exterior k-fonn, ~ an exterior [-fonn. Show that their exteriorproduct is symmetrie if k and/or l are even and antisymmetrie if both are odd,i.e.

k' k.,' kwt\w=(-l) wt\w .

5.2 Let Xl, X2, X3 belocal coordinates in the Euclidean space R3, ds 2 = EI dxf+E7. dx~ + E:3 dx~ the square of the line element, and el , e2, e3 unit vectors alongthe coordinate directions. What is the value of dXi(ej) , i.e. of the action of theone-fonn dz, on the unit vector ej?

5.3 Let a = L:~=I ai(x)ei be a vector field with ai (z) smooth functions on M.

To every such vector field we associate a one-fonn ~a and a two-form ~a suchthat

I 2wa(e) = (a . e), wa(e, 11) = (a . (e x 11» .

Show that

3

~a = La,(X)v!E:dxi,i=1

~a = ai(x)J~E:3dx2 t\ dx3 +cyclic pennutations,

5.4 Making use of the results of Exercise 5.3 detennine the components of V fin the basis {eI, e2, e3}Answer:

3 I 8f ~Vf= L rF: -8i e i·

i=1 V Ei X

5.5 Detennine the functions Ei for the case of Cartesian, cylindrical, and spherical coordinates. In each case give the components of V f.

Chapter 5: Geometrie Aspects of Meehanies 425

5.6 To the force F = (Ft, F2) in the plane we associate the one-form w =Ft dx I + F2dx2 . When we apply w onto a displacement vector, w(e) is the workdone by the force. What is the dual *w of the form w? What is its interpretation?

5.7 The Hodge star operator assigns to every k-form w the (n - k)-form *w.Show that

*(*w) = (_l)k .(n-k)w .

5.8 Let E = (EI ,~, ß) and B = (BI , B2, B3 ) be electric and magneticfields that in general depend on x and t. We assign the following exterior formsto them:

3der,", E d 'ir.p= L.J i x ,

i=1

wd~rBt dx2 1\ dx3 + B2dx3 1\ dx l + B3 dx l 1\ dx2 •

Write the homogeneous Maxwell equation curl E + iJ/ c = 0 as an equationbetween the forms r.p and w.

5.9 If d denotes the exterior derivatives and * the Hodge star operator, thecodifferential S is defined by

örJ;f*d* .

Show that L1dg,fd 0 ö+ Ö0 d, when applied to functions, is the Laplacian operator

5.10 Let

/:.; = L Wi l .. . i t (~) dXi l 1\ .. . 1\ dx i t

i l< ..· <i t

be an exterior k-form over a vector space W. Let F : V -t W be a smooth

mapping of the vector space V onto W. Show that the pullback F* (/:.; 1\ ~) ofthe exterior product of two such forms is equal to the exterior product of the

pullback of the individual forms (F*/:.;) 1\ (F*~) .

5.11 With the same assumptions as in Exercise 5.10 show that the exteriorderivative and the pullback commute,

d(F*w) =F*(dw) .

5.12 Let x and y be Cartesian coordinates in R2 , V = y8x and W = x8y twovector fields on R2. Calculate the Lie bracket [V, W]. Sketch the vector fields V,W, and [V, W] along circles about the origin.

426 Exercises

5.13 Prove the following assertions.

(i) The set of all tangent vectors to the smooth manifold M at the point p E Mform areal vector space, denoted by TpM, whose dimension is n =dirn M .

(ii) If M is Rn, TpM is isomorphie to that space.

5.14 The canonical two-form for a system with two degrees of freedom readsW = L:~=1 dqi /\ dpi. Calculate w /\ wand confinn that this product is proportionalto the oriented volume element in phase space.

5.15 Let u» = p2/2 + Cl - cos q) and H(2) = p2/2 + q(q2 - 3)/6 be theHamiltonian functions for two systems with one degree of freedom. Constructthe corresponding Hamiltonian vector fields and sketch them along some of thesolution curves.

5.16 Let H = HO + H' with HO = (p2 + q2)/2 and H' = cl / 3. Construct theHamiltonian vector fields X HO and X Hand calculate w(XH , X HO).

5.17 Let Land L' be two Lagrangian functions on TQ for whieh iJ! L and iJ! L'

are regular. The corresponding vector fields and canonieal two-forms are XE,XE', W L, and w L' . Show that each of the following assertions implies the other:

(i) L' =L + 0', where 0': TQ ---+ R is a closed one-form, i.e. da =0;

(ii) XE =XE' and wt. =WL'.

Show that in local coordinates this is the result obtained in Sect. 2.10.

Chapter 6: Stability and Chaos

6.1 Study the two-dimensional linear system ~ = A '!f, where A has one of the

Jordan normal fonns

(ii) A = (: b)- b a ' (iii) A = (~ ~).

In all three cases detennine the characteristic exponents and the flow (6.13) withs = o. Suppose the system is obtained by linearizing a dynamieal system inthe neighborhood of an equilibrium position. (i) corresponds to the situationsshown in Figs.6.2a-c. Draw the analogous pietures for (ii) for (a = 0, b > 0)and (a < 0, b > 0), and for (iii) with A < O.

6.2 The variables 0' and ß on the torus T 2 = SI X SI define the dynamiealsystem

ä = a/27r, /3 = b/21r, 0:::; 0', ß :::; 1 ,

where a and bare real constants. Cutting the torus at (0' =1, ß) and at (0' , ß = 1)yields a square of length 1. Draw the solutions with initial condition (0'0, ßo) inthis square for b/ a rational and irrational.

Chapter 6: Stability and Chaos 427

6.3 Show that in an autonomous Hamiltonian system with one degree of freedom (and hence two-dimensional phase space) neighboring trajectories can diverge at most linearly with increasing time as long as one keeps clear from saddlepoints.

Hint : Make use of the characteristics equation (2.154) of Hamilton and Jacobi.

6.4 Study the system

eh = - Jlql - )..q2 + qlq2

q2 = )..ql - Jlq2 + (qf - qi) /2 ,

where 0 ~ fl «:: 1 is a damping term and ).. with 1)..1 «:: 1 is adetuning parameter.Show that if fl =0 the system is Hamiltonian. Find a Hamiltonian function forthis case. Draw the projection of its phase portraits for ).. > 0 onto the (ql, q2)plane and determine the position and the nature of the critical points.

Show that the picture obtained above is structurally unstable when fl is chosento be different from zero and positive, by studying the change of the critical pointsfor flfO.

6.5 Given the Hamiltonian function on R4

show that this system possesses two independent integrals of the motion andsketch the structure of its flow.

6.6 Study the flow of the equations of motion p = q, p = q - q3 - P anddetermine the position and the nature of its critical points. Two of these areattractors. Determine their basin of attraction by means of the Liapunov functionV = p2/2 -l/2 + q4/4.

6.7 Dynamical systems of the type

x= -au/ax == -u_ _ , x

are called gradient fiows. They are quite different from the flows of Hamiltoniansystems. Making use of a Liapunov function show that if U has an isolatedminimum at ~o, then ~o is an asymptotically stable equilibrium position. Studythe example

Xl = -2xI (XI - 1)(2xI -1), X2 = -2X2.

6.8 Consider the equations of motion

of a system with f = 1. Sketch the phase portrait of typical solutions with givenenergy. Study its critical points.

428 Exercises

6.9 By numerieal integration find the solutions of the Van der Pool equation(6.36) for initial eonditions close to (0,0) and for various values of e in theinterval 0 < e ~ 0.4. Draw q(t) as a funetion of time, as in Fig.6.7 . Use theresult to find out empirieally at what rate the orbit approaehes the attraetor.

6.10 Choose the straight line p = q as the transverse seetion for the system(6.36), Fig.6.6. Determine numerieally the points of interseetion of the orbitwith initial eondition (0.01,0) with that line and plot the result as a funetion oftime.

6.11 The system in R2

XI = XI,. 2

X2 = - X2 + XI

has a critical point in X I =0 = X2 . Show that for the linearized system the lineX I = 0 is a stable submanifold and the line X2 = 0 an unstable one. Find theeorresponding manifolds for the exaet system by integrating the latter.

6.12 Study the mapping Xi+1 = f( Xi) with f( x) = I - 2x 2• Substitute u =

(4lrr)aresin J(x + 1)/2 and show that there are no stable fixed points. Calculatenumerieally 50000 iterations of this mapping for various initial values X I f 0 andplot the histogram of the points that land in one of the intervals [n/loo, (n +1)/100] with n = -100, -99, . .. , +99. Follow the development of two closeinitial values X I, x~, and verify that they diverge in the course of the iteration.(For a diseussion see Collet, Eekmann 1990.)

6.13 Study the flow of Roessler's model

X = -y - z, iJ = X + ay, Z = b+ xz - cz

for a = b = 0.2, c = 5.7 by numerieal integration. The graphs of x, y , z asfunetions of time and their projections onto the (x, y)-plane and the (x, x)-planeare partieularly interesting. Consider the Poincare mapping for the transverseseetion y + z = O. As X = 0, X has an extremum on the section. Plot the value ofthe extremum Xi+1 as a funetion of the previous extremum Xi (see also Berge,Pomeau, Vidal 1984 and referenees therein).

6.14 Although this is more than an exereise, the reader is strongly eneouragedto study the system known as Henon's attraetor. It provides a good illustrationof chaotic behavior and extreme sensitivity to initial eonditions (see also, Berge,Pomeau, Vidal 1984, Sect, 3.2 and Devaney 1989, Sect. 2.6, Exereise 10).

6.15 Show that

~exp [i2:am] =n8mo , (m =0, ... , n - 1) .

Use this result to prove (6.63), (6.65), and (6.66).


6.16 Show that by a linear substitution y = Q'X + ß the system (6.67) can betransfonned to Yi+l = 1 - "tyf .Detennine I in tenns of fl and show that Y lies inthe interval (-1 ,1] and I in (0,2] (cf. also Exercise 6.12 above). Making use ofthis transfonned equation derive the values of the first bifurcation points (6.68)and (6.70).

Solution of Exercises

Cross-references to a specific section or equation in the main text of the book aremarked with a capital M preceding the number of that section or equation . Forinstance, Sect. M3.7 refers to Chap.3, Sect. 7, of the main text, while (M4.100)refers to eq. (4.100) in Chap.4. Cross references within this set of solutionsshould be fairly obvious.

Chapter 1: Elementary Newtonian Mechanics

1.1 The time derivative of angular momentum is i = r x p + r X p =mi: X

r x r + r X F. By assumption this is zero which implies that the force F mustbe proportional to r , F = ar, a E R. If we decompose the velocity into acomponent along rand a component perpendicular to it, then F will changeonly the former. Therefore, the motion takes place in a spatially fixed planeperpendicular to the angular momentum I =mr(t) x r(t) =mro X vo, itself aconstant. Motion along (a), (b), (e), and (f) is possible. Motion along (c) is notpossible because I would vanish at the tuming point but would be different fromzero before and after passing through that point. Similarly (d) is not possiblebecause I would vanish in 0 but not before and after.

•oo•

(a) (b) (c)

~ 0.. • .. • • •0 0

(d) (e) (f) Fig.l

432 Solution of Exercises

1.2 We note that x(t) = r(t) cos <p(t), y(t) = r(t) sin <p(t) and, hence dx =dr cos <p - rdip sin <p, dy =dr sin <p + rdip cos <p . In taking (ds)2 =(dx)2 + (dy)2the mixed terms cancel so that (ds)2 = (dr)2 + r 2(d<p)2. Thus, the velocity isv 2 =.;.2+ r2ep2 . As neither r nor v have a z-component, the x- and y-componentsof 1= mr X v vanish. The z-component is

Iz =m(xvz - YVx )

=mrti: sin e cos <p + rep cos2 <p - .;. cos <p sin <p + rep sin2 <p)

=mr2ep .

Thus one finds

If I is constant this means that the product r2 ep = const., thus correlating theangular velocity ep with the radial distance, cf. the examples (a), (b), (e), and (f),of Exercise 1.1. A motion of type (d) could only be possible if, on approaching0, ep were to go to infinity in such a way that the product r 2ep stays finite. Butthen the shape of the orbit would be different, see Exercise 1.23.

1.3 In analogy to the solution of the previous exercise one finds (ds)2 =(dr)2 +r 2(dO)2 + r 2sin2 O(d<pf. Thus, v =.;.2 + r 2(j2 + r 2 sin2 Oep2 .

1.4 Having solved Exercise 1.3 one first reads off er from Fig.2: er =€x sin 0 cos <p+€y sin 0 sin <p+€z cos O. At the point with azimuth <p , €<p is tangentto a great cirlce, see Fig. 3. Hence, e<p = -ex sin e + ey cos e (check the specialcases <p = 0 and 1r /2!) . One verifies that

Starting from the given ansatz for €o the coefficients a, ß" are determined fromthe equations

eo . er = a sin 0 cos <p +ßsin 0 sin <p +, cos 0 = 0 ,

eo . e<p = -a sin <p + ßcos <p = 0 ,

keeping in mind that eo has norm 1, i.e. that a 2 + ß2 +,2 =1. Furthermore, fromFig.2 and for 0 = 0, sp = 0 one has eo = ex, for 0 = 0, <p = 1r /2 one has eo = ey,while for 0 =1r /2 one has always eo =-€ z. The solution of the above equationwhich meets these conditions, reads

a = cos 0 cos ip , ß= cos 0 sin ip ; , = - sin 0 .

In this basis we find

z Fig.2

/

x

Chapter 1: Elementary Newtonian Mechanies 433

Fig.3

y

= re r + r((O cos 0cos <p - ep sin 0 sin <p )ex

+ (0 cos 0 sin <p + ep sin 0 cos <p )ey - 0sin Oe z)

=re r + r(Oeo + ep sin <pe",) ,

from which follows the result v 2 =r2 + r2(02 + ep2 sin2 0) that we found in theprevious exercise.

1.5 With respect to the frame K, 1'(t) = vt ey , i.e., x(t) = 0 = z(t ) and y(t) = vt .In the rotating frame

j;' =X cos </J +Ysin </J + ~(-x sin </J + Y cos <p)

y' :::: - j; sin </J + Ycos </J - ~(x cos </J + Y sin </J)

Zl =Z =O.

In the first case, </J = w = const., the particle moves uniformly along a straight linewith velocity v' = (v sin w, v cos w, 0). In the second case, </J = wt, x' = v sin wt +wvtcoswt, y' :::: vcoswt - wvtsinwt. Integrating over time, x' (t ) =vtsinwt ,y'(t) =vtcoswt, and Zl(t ) =O. The apparent motion as seen by an observer inthe accelerated frame K /, is sketched in Fig.4.

1.6 The equation of motion of the particle reads

l'mr=F=f(r)- .

r

Take the time derivative of the angular momentum, i =rni: X r + m1' x r . Thefirst term is always zero. The second term vanishes because, by the equation ofmotion, the acceleration is proportional to 1'. Hence, i = 0, which means thatthe magnitude and the direction of the angular momentum are conserved . As Iis perpendicular to r and the velocity r this proves the assertion.


y

. . . .

Fig.4

1.7 (i) By Newton' s third law the forces between two bodies fulfill Fik =-Fkior -ViVik(ri ,rk) =VkVik(ri,rk). Hence, V can only depend on (r , - rk).Constants ofthe motion are: total momentum P , energy E; furthermore, we havefor the center-of-mass motion

r s (t ) - P / Mt = r s(O) = const. .

(ii) When Vij depends only on the modulus Ir i - rkl, we have

Fj i = -ViVij(lr i - rkJ) = -V;j(lri - rkJ)V;jri - rkl

I I I ri - rk=-V;j(ri-rk)1 I '

ri - rk

In this case the total angular momentum is another constant of the motion.

1.8 For q -+ 0 the potential goes to infinity like 1/q4, while for q -+ 00 ittends to zero. Between these points it has two extrema as sketched in Fig.5.As the energy E =p2/2 + U(q) is conserved, the phase portraits are given byp =[2(E - U(q))]I/2 . The figure shows a few examples. The minimum at q =2is a stable equilibrium point, the maximum just beyond q =6 is unstable. Theorbits with E ~ 0.2603 are separatrices . The phase portraits are symmetrie withrespect to reflection in the q-axis because (q,p =+F) and (q,p = -F)belong to the same portrait.

1.9 The term (x~ + x~)/ (2m) is the total kinetic energy while U(XI, X3) =m(wö(xI + x~ ) +wI(xl - x3? )/ 2 is the potential energy. The forces acting onpendula 1 and 2 are, respectively, -au/ aXI , and -au/ aX3. Thus, the equationsof motion are

l/m 0o mWIo 0o -m(w5+wy)

U(q) p


Fig.5

or, for short, :i.: =M z . The transformation as given above

q

A-1=A .

leads to sums and differences of the original coordinates and momenta. We notethat the matrix M has the structure

where Band C are 2 x 2 matrices . Furthermore the transformation A is invertibleand, in fact, the inverse equals A. Thus

du -I.dt = AMA U with

It is useful to note that one can do the calculations in terms of the 2 x 2 submatricesas if these were (possibly noncommuting) numbers. For example,

(B +C I 0)AMA-1=AMA= ---

. 0 B-C

B + C =(0 1/m) and-mw~ 0

B C- ( 0- - -m(w~ + 2wr)

with


This system now separates into two independent oscillators that can be solvedin the usual manner. The first has frequency w(J) = Wo (the two pendula performparallel, in-phase oscillations); the second has frequency w(2) = (w~2) + 2wr)'/2(the pendula swing in antiphase). The general solution is

u, = a, cos(w(J)t + !Pü , U3 = a2 cos(w(2)t + !P2) .

As an example, consider the initial configuration

x , (O) = a , X2(0) = 0 , X3(0) =0 , X4(0) = 0 ,

which means that, initially, pendulum 1 is at maximal elongation with vanishingvelocity while pendulum 2 is at rest. The initial configuration is realized bytaking a2 = a, = a~, !P' = !P2 = O. This gives

w(l) + w(2) w(2) - w(J)x,(t)=acos 2 tcos 2 t=acosiltcoswt ,

w(l) +w(2) w(2) - w(l)x3(t ) = a sin 2 tsin 2 t=asiniltsinwt

Where il := (w(J) + w(2» /2, w := (w(2) - w(J)/2. If il / w = p]q with p , q E Z andp > q, hence rational, the system returns to its initial configuration after timet = 27rp/ il = 27rq/ w . For earlier times one has t = 7rp/ (2il): x , = 0, X3 = a(pendulum 1 at rest, pendulum 2 has maximal elongation); t = np] il: x , = -a,X3 = 0; t = 37rp/(2il) : X I = 0, X3 = -a. The oscillation moves back and forthbetween pendulum 1 and pendulum 2. If il / w is not rational, the system willcome close, at a later time, to the initial configuration but will never assume itexactly (cf. Exercise 6.2). In the example considered here, this wHI happen ifilt ~ 27rn and wt ~ 27rm (with m , nE Z), i.e. if il/w can be approximated bythe ratio of two integers. It may happen that these integers are large so that the"return time" becomes very Iarge.

1.10 As the differential equation is linear, the two terms are solutions preciselywhen J.L = w; a and b are integration constants which are fixed by the initialcondition as follows

x (t) = a cos wt + bsin wt ,

pet) =-amw sinwt + mbwcoswt .

x(O) = Xo gives a = xo, p(O) = Po gives b = po/(mw). The solution with w = 0.8,Xo = 1, Po = 0 reads x(t) = cos(0.8t).

1.11 From the ansatz one has

x(t ) = ax(t) + eO' t(-wxo sin wt +po/m cos wt)

x (t ) = a 2x(t) + 2a eO' t(-wxo sin wt +po/m coswt)

- eO'tw2(xocoswt + po/mw sinwt)

= -a2x + 2ax - w2x .


Inserting and comparing coefficients one finds

Q =- i, w=Jw 2 - Q2 =Jw 2 - ",2/4 .

The special solution x(t) =e-Kt / 2 cos( JO.64 - ",2/4 t), approaches the origin ina spiraling motion as t -t 00.

1.12 Energy conservation formulated for the two domains yields

m 2 m 22"VI +UI =E= 2"V2+U2.

As the potential energy U depends on x only there can be no force perpendicularto the x-axis. Therefore, the component of the momentum along the directionperpendicular to that axis cannot change in going from x < 0 to x > 0: Vll.. =Va . The law of conservation of energy hence reads

m2 m2 m 2 m 22" Vll.. + 2" v IIi+ UI =2" va + 2" v211 + U2 , or

m 2 m 22" VIl I + UI =2" v 211 + U2 .

from which follows

directly yielding2

. 2 V 2J.sm Q2 =-2-'

V 2

v 2

sin2QI = It,VI

sin oj IV21=SinQ2 lv.l '

For UI < U2 we find lvIi> IV21. hence QI < Q2. For UI < U2 all inequalitiesare reversed.

1.13 Let M =ml + m2 + m3 be the total mass and ml2 =ml + m2. From thefigure one sees that r2 + Sa = rJ, Sl2 + Sb = r3, where SI2 is the center-of-masscoordinate of particles land 2. Solving for r 1, r2, r3 we find

m3 m2r r <re> -Sb+--Sa ,

M ml2

m3 mlr2 =r s - -Sb - --Sa ,

M ml2

ml2r3=rS+ MSb.

Inserting these into the kinetic energy all mixed terms cancel. The result containsonly terms quadratic in rs , Sa, Sb

withmlm2

J.La=--,ml2


Fig.6

I

I

II

r ISI

I

II

0 1

Ts is the kinetic energy of the center-of-mass motion, /Ja is the reduced massof the subsystem consisting of partieles 1 and 2. /Jb is the reduced mass of thesubsystem consisting of partic1e 3 and the center-of-mass S!2 of particles 1 and2, n is the kinetic energy of the relative motion of partic1e 3 and S!2.

In an analogous way, the angular momentum is found to be

all mixed terms having cancelled.By a special (and proper) Galilei transformation, r s -t r~ =r S + wt + a,

r s -t r~ =rs+w, Sa -t Sa, Sb -t Sb and, hence,

1~=ls+M(a x (rs+w)+(rs -trs) x w) ,

while l~ =la , l~ =lb remain unchanged .

1.14 (i) With U()..r) = )..<'U(r) and r' = )..r the forces from U(r') := U()..r) andfrom U(r), respectively, differ by the factor )..a-I. Indeed

Integrating F' .dr' over a path in r' space and comparing with the correspondingintegral over F . dr , the work done in the two cases differs by the factor ).. o .

Changing t to t' = )..1-a/2t,

(dr

l

) 2 =)..2)..a-2 (dr)2dt' dt '

which means that the kinetic energy

Chapter I: Elementary Newtonian Mechanies 439

differs from the original one by the same faetor ,\" . Thus, this holds for thetotal energy, too, E' =,\"E. The indieated relation between time differenees andlinear dimensions of geometrieally similar orbits follows.

(ii) For harmonie oseillation the assumption holds with 0' =2. The ratio ofthe periods of two geometrieally similar orbits is Ta/n = 1, independently ofthe linear dimensions.

In the homogeneous gravitational field U(z) =mgz and, henee, 0' =1. Timesof free fall and initial height H are related by T cx HI/2.

In the ease of the Kepler problem U =- A/r and, henee, 0' =-1 . Two geometrieally similar ellipses with semimajor axes aa and bb have eireumferenee Uaand Ui, respeetivley, sueh that Ua/Ub =aa/ab. Therefore the ratio of the periods Ta and n is Ta/n = (Ua/Ub)3/2 from whieh follows (Ta/Tb)2 = (aa/ab)3,Kepler 's third law.

(iii) The general relation is Ea/ Eb =(La/ Lb)". If Ai denotes the amplitudeof harmonie oseillation, Ea/Eb =A~/A~ . In the ease of Kepler motion Ea/ Eb =ab/aa: the energy is inversely proportional to the semimajor axis.

1.15 (i) From the equations of Seet. M1.24

p A 1 - &2 Ar P = 1+ & =- 2E~ =- 2E (l - s) ;

ArA =- 2E (l + s) .

From these we ealculate

Inserting this into the differential equation we obtain

d<p =-r=======dr

This is precisely eq. (MI.67) with Ueff =-A/r + [2/2pr2. Integration of eq. (4)with the boundary eondition as indieated implies

L1 ( ) 1/2rprA<p(r) - <p(rp) = - ( )( ) dr .

rpr r r re rA-r

We make use of the indieated formula with

0' =2_r.:.:A_r

Pe....rA - r»

and obtain

ß=rA +rp

rA - rp


./..( ) 2rAr p - (rA + r p )r'f' r =arccos .

(rA - rp)r

(ii) There are two possibilities for solving this equation: (a) the new equationsare obtained by replacing 12 with P. =12+2J.LB. For the remainder, the solution isexactly the same as for the Kepler problem. If B > O(B < 0), then 1> 1(1 < 1),i.e., in the case of repulsion (attraction) the orbit becomes larger (smaller). (b)With U(r) =Uo(r) + B/r2, Uo(r) = -A/r, the differential equation for eP(r ) iswritten in the same form as above

where r~, r~ denote perihelion and aphelion, respectively, for the perturbedpotential. They are obtained from the formula (r - rp)(rA - r ) + B / E =(r - r~)(r~ - r). Multiplying the differential equation by «r~r~)/(rprA))1/2 andintegrating as before

eP(r) = rprA . 2r~r~ - r(r~ + r~)-,-, arccos " .rprA r(r A - r p)

From this solution follows reeP) =2r~r~/[r~+r~ +(r~ - r~)cos Jr~r~/rprA eP ].The first passage through perhelion is set to ePPI = O. The second is ePP2 =271"«rprA)/(r~r~))1/2 = 271"I/Jl2+2J.LB ~ 271"(1 - J.LB/12). The perhelion precession is (ePP2 - 271"). It is independent of the energy E. For B > 0 (additionalrepulsion) the motion lags behind, and for B < 0 (additional attraction) themotion advances as compared to the Kepler case.

1.16 For fixed 1, the energy must fulfill E ~ -J.LA2/(212). The lower limitis assumed for circular orbits with radius ro =12/ J.LA . The semimajor axis (inrelative motion) follows from Kepler's third law a3 =GN(mE + ms)T2/(471"2).This gives a =1.495 X 1011 m (T =1Y=3.1536 X 107 s). This is approximatelyequal to aE, the semimajor axis of the earth in the center-of-mass system. Thesun moves on an ellipse with semimajor axis

mEas = a ~ 449 km .

mE+mS

This is far within the sun' s radius Rs ~ 7 x 105 km.

1.17 We arrange the two dipoles as sketched in Fig. 7. The potential created bythe first dipole at a point situated at r is

Here, we have expanded


(a)

x

z (b)

x

Jf-------y

Fig.7

up to the term linear in d-: In the limit we obtain PI =r 0 pJ!r30 The potential

energy of the second dipole in the field of the first reads

(PI · (r +dz) PI · r)

W= et(PI(r+dz)-PI(r))= ez Ir+dzl3 -----;:3

Expanding again up to terms linear in dz

Finally, taking the limit ez ---T 00, dz ---T 0, with ezdz = Pz finite, this yields

From this expression one calculates the components of F zi =- \1 1W =- F 12,

making use of relations such as

ar a X I - Xz a= =

aXI ar r ar 'etc..

So, for example

aW(l ,2) 3 X I - Xz 3 ( x--::-'-'---'- =-(PI 0 pz)- - - PI (Pz 0 r)aXI r 4 r r 5

x 15 XI - Xz+(Pt 0 r)pz) + (PI 0 r)(pz 0 r)6" 0

r r

1.18 Take the time derivative of r 0 a,

d . .. . ( ) 0dt r 0 a = r . a = v 0 a = v x a . a = 0

Thus, r 0 a is constant in time and the indicated relation holds for all times. Takingthe time derivative of (5) and inserting 11, we find ii =11 x a =(v x a) x a =


-a2v + (v· a)a. The second term is constant as shown above. Thus, integratingthis equation over t from 0 to t, we have f(t) - f(O) =- w2(r (t ) - r(O» + (v(O) ·a)at, where w2 := a 2. By eq. (5) f(O) = v(O) x a, so that we may write

f(t) +w 2r (t ) =(v (O) . a)at + v(O) x a +w2r (O) .

This is the desired form, the general solution of the homogeneous differentialequations is

rhom(t) =CI sinwt + C2 coswt .

With the given ansatz for a special solution of the inhomogeneous equation theconstants are found to be

I ICI = 3" (a 2v(O) - (v(O)· a )a) = 3"(a x (v(O) x a»

w w

IC2 = -2v(O) x a

w

IC =2 (v (O) . a )a

w

Id= 2 v(O) x a+r(O) .

w

The solution therefore reads

I 0 . 1 0r(t) =3"a x (v( ) x a )sm wt + 2(v( ). a )atw w

I+ 2 v(O) x a (1 - coswt ) + r (O) .

w

It represents a helix winding around the vector a.

1.19 The ball falls from initial height ho. It hits the plane for the first time att l =~, the velocity then being U I =-vrnog =-gtl . Furthermore, with

a := v (n - l )/n

2 ViVi =- a u i , Ui+l =- Vi , ti+1 - i, =- .

g

The first two equations give VI =agt l and Vi =a ig t l • The third equation yields

o Vi 0 0 vi+lt i - t i =- =t i+1 - t i and t i+1 - ti+l =- ,

g g

and, from there, t?+1 - t? =(Vi+1+ Vi )/ g =t l(a + l)a i. With tg =0 we have atonce

i - I

t?=tl(l+ a)L a v.

v=O

From h i =vl!(2g), finally, h i =a 2i ho.


1.20 The answer is contained in the following table giving the products of theelernents

E P T P·T

E E P T P·TP P E P ·T TT T P ·T E P

P·T P ·T T P E

1.21 Let Rand Eo denote the radius and the energy of a circular orbit, respectively. The differential equation for the radial motion reads

:: =~.JEO - Uclf(r), Uor(r) =U(r) + 2~:2 .From this follows Eo =Ueff(R), U~fflr=R =0, U~~lr=R > 0 or, for U(r) ,

, [2 I " 3[2 IU (R)= -- and U (R) > ---.

P, R3 P, R4

IfE=Eo+c,

ddrt -_ f2-;;2 I R 2U" RY-;; e - "2('1' - ) eff( ) .

Setting r: := U~~(R) we obtain, choosing r = R,

t - to = fi.l r

-

Rdc; = fi. arcsin (('I' - R) E)V~ ro-R J2cl r: - <;2 V~ V2(

Solving for 'I' - R yields

'I' - R =~ sin ;; (t - to) .

Thus, the radial distance oscillates around the value R. More specifically, onefinds

(i) Uir) =r", U'(r) =nrn - I , U"(r) =n(n-l)rn - 2• This yields the equation

nRn

-1

= p~3 => R = "+if!.; ,3[2 3[2 (n + 2)[2

~ =n(n - l)Rn-

2 + --4 > 0 {::} n(n - 1)Rn+

2 + - = > 0 .p.H ~ p, p,

t2/<l'n)

(ii) U(r) = >'1'1', U'Ir) = _>'/'1'2, U"(r) =2>'/'1'3. From this R = _[2/(p,>.),r: =->./R3 • This is greater than zero of >. is negative.


1.22 (i) The eastward deviation follows from the formula given in Sect . M1 .26,Ll ~ (2-/2 13)g-I/2H3/2w cose. With w =27r/O day) =7.27 x 10-5 S-I andg =9.81 ms-2 one finds Ll ~ 2.2cm.

(ii) We proceed as in Sect. M1.26 (b) and determine the eastward deviation u

from the linearized ansatz r(t) = r(O)(t) +wu(t), inserting here the unperturbedsolution, rO(t) =gt(T - tt)e v • This gives (d'2Idt2)u(t) ~ 2gcos<p(t - T)e v .

Integrating twice,

u(t) =~g COS <p(t3 - 3Tt2)eo .

The stone returns to the surfaee of the earth after time t =2T. The eastwarddeviation is found to be negative, Ll ~ - ~g eos <pT3, which means , in reality ,that it is a westward deviation. Its magnitude is four times larger than in case (i) .

(iii) Denote the eastward deviation by u as before (direeted from west toeast) , the southward deviation by s (directed from north to south). Aloeal,earth -bound, coordinate system is given by (et, eo, ev ), el defining the directionN-S, eo and ev being defined as in SecLM1.26(b). Thus, u =ueo, s =set. Theequation of motion (M1.74'), together with w = w( - cos <p , 0, sin <p), implies

s=2wsin <pu .

Inserting the approximate solution u ~ ~gt3 cos <p and integrating over timetwice, one obtains

1.23 For E > °all orbits are scattering orbits . If z2 > Luo;

0)

dr'1 l r

4J - 4Jo =-- -r===;;:;=::;;::======:=======,j2jiE ro Jr'2 - ([2 - 2p.cx)/(2p.E)

_ r(O) [ r dr'- P } ro r' Jr'2 - r~ ,

where p. is the reduced mass, rp = J([2 - 2p.cx)/(2p.E) the perihelion and r~O) =1IV2p.E . The particle is assumed to come from infinity, traveling parallel to

the z-axis. Then the solution is 4J(r) = l1J[2 - Zua arcsin(rp/r) . If cx = 0,the corresponding solution is 4J(O)(r) =arcsin(r~O) I r); the particle moves along a

straight line parallel to the z-axis, at the distance r~O). For cx =I°1 'Fr

4J(r =rp) = J[2 _ 2p.cx "2 '

that is, after the scattering and asymptotically, the particle moves in the direction11 J[2 - 2p.cx 'Fr. Before that it travels around the center of force n times if theeondition


[ (. rp . rp ) r~) ( 1r)arcsm - - arcsm - =- 1r - - > n1rJ[2 - 2fLQ 00 rp rp 2

is fulfilled. The number

n = [r~O)]2rp

is independent of energy E.In the case z2 < 2J.LQ eq. (I) can also be integrated. With the same initial

condition one obtains

r(O) b+ Jb2 + r2r/J(r) =1- In r '

where b = «2J.LQ - [2)/(2fLE))1/2. The particle travels around the force centeron a spiral-like orbit, towards the center. As the radial distance tends to zero,the angular velocity increases in such a way as to respect Kepler' s second law(MI.22).

1.24 Let the comet and the sun approach each other with energy E. Longbefore the collision the relative momentum has the magnitude q =J2fLE, withfL the reduced mass, the angular momentum has the magnitude l =qb. The cometcrashes when the perihelion rp of its hyperbola is smaller or equal R, i.e., whenb :::; bmax with bmax following from the condition rp = R, viz.

and A = GmM. One finds bmax = JI + A/(ER) and, hence,

a =t: 21rbdb =1rR2 (I + EAR) .

For A = 0 this is the area of the sun seen by the comet. With increasing gravitational attraction (A > 0) this surface increases by the ratio (potential energy atthe SUI1'S edge)/(energy of relative motion).

1.25 As explained in Sect. M1.21.2 the equation of motion reads

:C =Az +b,

with A as given in eq. (MI.50), and

0 0 0 I/rn 0 0 00 0 0 0 I/rn 0 0

A=0 0 0 0 0 I/rn b=

00 0 0 0 J{ 0 00 0 0 -K 0 0 00 0 0 0 0 0 E


The last of the six equations is integrated immediately, giving X6 =eEt + CI .Inserting this into the third and integrating yields

eE 2X3 =z =- t +Clt + C2 •

2m

The initial conditions z(o) = z(O), i(O) = v~O) give C2 = z(O), CI = v~O) . Theremaining equations are coupled equations. Taking the time derivative of thefourth and replacing Xs by the right-hand side of the fifth gives X4 =-K2x4 +e.KE y which is integrated to X4 =C3 sin K t + C4 COS K t + eEy/ K. Making useof the fifth equation once more yields X3 = C3 COS K t - C4 sin K t + Cs. Alsothe fourth equation yields the condition Cs = -eEx / K . These two expressionsare inserted into the first and second equations so that these can be integratedyielding

C3 1:.' C4 . 1:.' e E CXl = - r;;-- COS r; t + r;;-- sm 1\. t + K y t + 6

l\.m l\.m

C3 . C4 eX2 =+-T- smKt + K cosKt - },/Ext + C7 •

Am m •

Upon insertion of the initial conditions x(O) = x(O), y(O) = y(O), x(O) = v~O),

y(O) =v~O) we finally obtain

(0)_ (0) vy e

C6 - x + K + K2 Ex ,(0)

_ (0) V x eC7 - y - K + K2 E y •

(I)

1.26 Using cartesian coordinates in the plane of the orbit the solution (MI.21)reads

x(t) = P cos(4' - 4'0) ,I + e cos(4' - 4'0)

y(t) = P4' 4' sin(4' - 4'0) .1 +ecos( - 0)

In taking the derivative with respect to time, the factor ~ is replaced by 1/(J-lr2),

by virtue of eq. (M1.19). Furthermore, as p = [2/(AJ-l) we obtain

Px = J-lx = - ~J-l sin(4' - 4'0) ,

AJ-lPy = J-lY = -1- {cos(4' - 4'0) + e ] .

Eq. (I) describes a circle with radius Ap/l whose center has the coordinates

(o,eA:)

= (O ,V(AJ-l/l?+2J-lE) .

Chapter 2: The Principles of Canonical Mechanies 447

Chapter 2: The Principles of Canonical Mechanics

2.1 We take the derivative of F(E) with respect to E

dF d r: r: m- =2- . V2m(E - U(q» dq =2 dqdE dE qmin (E ) qm in ( E ) .j2m(E - U(q»

,------- dqmax dqmin+2 2m~ dE -2 2m~ dE .

=0 =0

To find T we must calculate the time integral over one period. In doing so wenote that

dq '--=---=-:0--m dt =P =V2m (E - U(q» , and hence ,

dt = mdq.j2m(E - U(q»

Therefore,

t: mT=2 ~.qm in (E ) .j2m(E - U(q»

This, however, is precisely the expression calculated above. For the exampleof the oscillator with q =qosin wt, p =m wqocos wt , one finds F =m W7rq5 =(27r/w)E and T =27r/w.

2.2 Choose the plane as sketched in Fig. 8. D' Alembert's principle (F -jJ)·ör =0, with F =-mge3, admits virtual displacements along the line of intersectionof the inclined plane and the (l ,3)-plane as weil as long the 2-axis. Denotingthe two indenpendent variables by ql , qz, this means that 8r = 8ql€o + 8q2€2with €o = €I cos a - €3sin a . Inserting this yields the equations of motioniil =-mg sin o , ii2 =0 whose solutions read

t2 •

ql (t) =-(mg sin a)"2 + v lt +al, q2(t) =V2+ a2 .

3

Fig.8


2.3 Choose the (1,3)-plane to coincide with the plane of the annulus and takeits center to be the origin. Choosing the unit vectors t and n as shown in Fig. 9,viz.

we find

the force acting on the system being F =-mge3.0'Alembert's ·principle (F - p) . or = 0 yields the equation of motion ~ +

9 sin ifJ/R =O. This is the equation of motion of the planar pendulum that wasstudied in Sect. M 1.17.

3

i

Fig.9

2.4 Let do be the length of the spring in its rest state and let K be the stringconstant. When the mass point is at the position x the length of the string isd =Vx 2 + [2. The corresponding potential energy is

1 2U(x) = '2K(d - do) .

For do ::; 1 the only stahle equilibrium position is x =O. For do > 1, x =0 is

unstable, while the points x =±Jdij - [2 are stable equilibrium positions.

As an example we study here the case do ::; 1. Expanding U( x) around x =0,

I ( x2

x4) 2 I ( 2 1- do 2 do 4)U( x) ~ -K 1- do+ - - - ~ -K (1 - do) +-- x + - x .

2 21 8[3 2 1 4[3

From this expression we would conclude that the frequency of oscillation isapproximately

A


Fig.l0

m

w=V",l-dO •m 1

However, this does not hold for all values of do. For do=1 the quadratic termvanishes, and x4 is the leading order. In the other extreme, do = 0, we haveU(x) =K(X2 + 12)/2, i.e. a purely harmonic potential (the constant terms in thepotential are irrelevant). Thus, the approximation is acceptable only when do issmall compared to 1.

2.5 A suitable Lagrangian function for this system reads

I .2. 2 1 2L =2:m(x, + x2) + 2:"'(x, - X2) .

"'----v--"~T U

Introduce the following coordinates : u, := x, + X2, U2 := x, - X2. Except fora factor 1/2 these are the center-of-mass and relative coordinates, respectively.The Lagrangian becomes L =m(üi + ü~)/4 + KuV2. The equations of motionthat follow from it are

ü,=O, mÜ2+2Ku2=0 .

The solutions are u, =C,t + C2, U2=C3 sinwt + C4 coswt, with w =J2K/m.It is not difficult to rewrite the initial conditions in the new coordinates, viz.

u,(O) =+1

U2(0) =-1

ü,(O) =Vo

Ü2(0) =Vo .

The constants are deterrnined from these so that the final solution is

Vo ( I. ) 1x,(t) = - t + - smwt - - (1 - coswt)2 w 2

vo( 1. ) 1X2(t) =- t - - smwt + -(1 +coswt) .2 w 2

2.6 By hypothesis F(>.x" .. . , >.xn ) = >.NF(x" .. . , XN) ' We take the firstderivative of this equation with respect to >. and set X=1. The left-hand side is


The same operation on the right-hand side gives N F.

2.7 In the general case the Euler-Lagrange equation reads

Bf d BfBy = dx By' .

Multiply this equation by y' and add the term y" Bf j By' on both sides. Theright-hand side is combined to

y' Bf + y" Bf =.!!:.... (y' Bf )By By' dx By'

If f does not depend explicitly on x then the left-hand side is df(y, y')jdx.The whole equation can be integrated directly and yields the desired relation.Applying this result to L(i], q) =T(i]) - U(q) gives

~ . BT(i])L...Jqi~ - T + U =const. .

i q,

If T is a homogeneous, quadratic form in i] the solution to Exercise 2.6 teIls usthat the first term equals 2T. Therefore, the constant is the energy E =T + U.

2.8 (i) We must minimize the arc length

i.e. we must choose f(y , y') = J1"+Y'2. Applying the result of the precedingexercise we obtain

R r;-:--;:;y' - V 1+ y,2 =const. ,J1"+Y'2

or y' =const. Thus, y =ax + b. Inserting the boundary conditions y(xl) =Yl>y(X2) =Y2 gives

y(x) = Y2 - YI (x - x})+YI .X2 - XI

(ii) The position of the center of mass is determined by the equation

u-, =Jrdm,

M denoting the mass of the chain, dm the mass element. If >. is the mass perunit length, dm =>. ds. As the x-coordinate of the center of mass is irrelevant,


the problem is to find the shape for which its y-coordinate is lowest. Thus wehave to minimize the functional

The result of the preceding exercise leads to

12 JYyy -y~=- y =Co .

~ ~

This equation can be solved for y',

y' = JCy2 - 1 .

This is a separable differential equation whose general solution is

y( X) = Je cosh (VC x + C') .The constants C and C', finally, must be chosen such that the boundary conditionsy(xl) = YI, y(X2) = X2 are fulfilled.

2.9 (i) In either case the equations of motion read

.. 2 I 2 2X I = -mwOXI - 2:m(WI - Wo)(XI - X2)

.. 2 I 2 2X2 = - mwoX2 + 2:m(WI - WO)(XI - X2 ) .

The reason for this result becomes c1ear when we calculate the difference L' - L:

The two Lagrangian functions differ by the total time derivative of a function which depends on the coordinates only. By the general considerations ofSects. M2.9 and M2.10 such an addition does not alter the equations of motion.

(ii) The transformation to proper oscillations reads

This transformation

is one-to-one. Both Fand F- 1 are differentiable. Thus, F, being a diffeomorphism, leaves the Lagrange equations invariant.


2.10 The axial symmetry of the force suggests the use of cylindrieal coordinates.In these coordinates the force must not have a component along the unit vectorec/>. Furthermore, since

t"7U au A I au A au A

v (r ,<p, z)=-a er + - -a

»:»: ez ,r r <p z

U must not depend on ip , The unit vectors er and ez span a plane that containsthe z-axis.

2.11 By a (passive) infinitesimal rotation we have

:c ~:Co - (ep x :Co) e or :Co ~:c + (ep x :c)e .

Here ep is the direction about whieh the rotation takes pIace, e is the angle ofrotation, so that epe =w dt. Thus :i:o =:i: + (w x :C), the dot denoting the timederivative in the system of reference that one considers. Inserting this into thekinetic energy one finds

T =m (:i:2 + 2:i: . (w x :c) + (w x :c?) /2 .

Meanwhile U(:Co) becomes U(:c) =Ü(R-1(t):c). We calculate

aL-a' =mXi +m(w x :C)iXi

aL aü .-a =--a +m(:c x W)i +m«w x e) x W)i.Xi Xi

This leads to the equation of motion

m~ =-V'U - 2m(w x e) - mw x (w x z ) - m(w x e) •

2.12 Let the coordinates of the point of suspension be (x A, 0), <p the anglebetween the pendulum and the vertical, with -'Fr :::; <p :::; 'Fr. The coordinates ofthe mass point are

X =X A + 1sin <p, y =-1 cos <p ,

m denoting the mass, 1 the length of the pendulum. Inserting these into

L =~ (x 2 + 1;2) - mg(y + 1)

gives the answer

L =~ (x~ + 1\?2 + 21cos <pXAep) + mgl(cos <p - 1) .

2.13 (i) If the oscillation is to be harmonie s(t) must obey the following equation


Fig.ll

y

IIIII

IIII

x(z)

y=zx

Fig.12

(ii) The Lagrangian function reads

L =m 82 - U2

where the potential energy is given by (see Fig. 11)

U =mgy =mg 1" sin </> ds .

The Euler-Lagrange equation reads m s+mg sin </> =O. Inserting the above relationfor s(t) we obtain the equation So 11:2 sin II:t =g sin </>.

Since the absolute value of the sine function is always smaller than, or equalto, 1 one has

11:2

A := So - ::; 1 .g

Thus we obtain the equation </>(t) =arcsin(A sin II:t ) whose derivatives are

~ = AII:COSII:t ,

JI - A2 sin2 II:t

~ = -AII:2(l - A2) sin ll:t

(l - A2sin2 II:t )3/2

In the limit A -t I, </> goes to zero and ~ goes to 11:, except if II:t =(2n + 1)/27rwhere they are singular.


(iii) The force of constraint is the one perpendicular to the orbit. It is

(- sin 1J)Z(1J) == mg cos 1J cos 1J .

The effective force is then

(0) . (cos 1J)E == -mg 1 + Z(1J) == -mg sm 1J sin 1J

2.14 (i) The condition 8F(x , z )j 8x == °implies z - 8fj8z == 0, i.e., z == !,(x).Therefore, x == x (z ) is the point where the vertical distance between y == zx (withz fixed) and y == f(x) is largest (see Fig .12).

(ii) The figure shows that (.C4i)(z) == zx - 4i(z) == G(x , z ), z fixed, is tangentto f(x) at the point x == x( z) (the derivative being z ).

Keeping x == Xo fixed and varying z yields the picture shown in Fig. 13. Forfixed z y == G(x, z) is the tangent to f( x) in x(z ). G(xo, z) is the ordinate of theintersection point of that tangent with the straight line x == xo. The maximum isat Xo == x(z), i.e. z(xo) == !'(x)lx=xQ' As f" > 0 all tangents are below the curve .The envelope of this set of straight lines is the curve y == f(x) .

y

Fig.13

x == Xo

2.15 (i) In a first step we determine the canonically conjugate momenta

8LPI == -8' == 2cll(i! + (C12 + C21)Q2 + bl

ql

8L . .P2 == -8' == (C12 + C21)ql + 2c22Q2 + b2 .

Q2

Using the given abbreviations this can be written in the form

7r1 == dllQI + d 12(h , 7r2 == d211iJ + d 22Q2 .


For this to be solvable in terms of qi the determinant

must be different from zero. The qi can then be expressed in terms of the 'lri :

We construct the Hamiltonian function and obtain

IH = Plql +P2q2 - L = D (C22'1rf - (CI2 + C21)'lrI'lr2 + CIl'lrD - a + U .

The above determinant is found to be

(d12 + d2d/21=~ .dll D

The inverse transformation, the construction of L from H , proceeds along thesame lines.

(ii) Assurne that there exists a function

which vanishes identically in the domain of definition of the Xi, u fixed. Takethe derivatives

0= dF =8F 8pI + 8F 8P2dXI 8pI8xI 8P28xt

0= dF" = 8F 8Pl + 8F 8P2

dX2 8PI 8X2 8P2 8 X2

By assumption the partial derivatives of F with respect to Pi do not vanish(otherwise the system of equations would be trivial). Therefore, the determinant

must be different from zero. This proves the assertion.

2.16 We introduce the complex variable w := X + iy. Then

x = (w + w*)/2, y = -i(w - w*)/2, x2 +1/ = iinb" .


13 is calculated to be

m13 =m(xy - yx) =2i (töw* - wtö*) .

Expressed in the new coordinates the Lagrangian function reads

L m ( .. * .2) imw (. * . *)="2 ww + z - 4 ww - ww .

The equation of motion are

m ..* imw .* imw .*-w --- w =--w2 4 4

mz=O.

The first of these is written in tenns of the variable u := ib", It becomes ü = iwu,its solution being u =eiw t

• w* is the time integral of this function, viz.

i .w* = __ e1w t +C,

w

where C is a complex constant. Take the complex conjugate

i .w = - e-Iwt + C* ,

w

from which follow the solutions for x and y

I. Cx =- smwt+ 1 ,w

1y =- cos wt + C2 ,

w

where CI =Re{C}, C2 =-Im{C} .The solution for the z coordinate is simple: z =C3t +C4,. i.e., uniform motion

along a straight line. The canonically conjugate momenta are

. mPx =mx- "2 w y ,

mPy = my + "2 wx, pz = m i ,

while the kinetic momenta are given by Plein =mx. In order to construct theHamiltonian function the velocities are expressed in tenns of the canonical momenta

. 1 wx = m Px +"2 Y '

Then H is found to be

I wy =m Py -"2 x ,

. 1z =- pz .

m

H . L 1 2=P . :c - =- 7l.-.2m rKm·

2.17 (i) Harnilton's variational principle, when applied to L, requires


to be an extremum. Now, since

l T2 jt2Ldr = Ldt with

Tl tlt;=t(r;), i=1,2 ,

the action integral t is extremal if and only if the Lagrange equations that followfrom L are fulfilled.

(ii) We define q = (q1, " " qf), t = qf+1' From Noether 's theorem the quantity

f+1 BL d1= L ~ ds h

S(q1,""

qf+dls=o;=1 q,

is an integral of the motion provided L is invariant under (q1,"" qf+d--+ hS(q1 , .. . , qf +1), i.e., in the case considered here, under (Q1 , .. . ,Qf+d --+

(Q1 , ... , Qf+1 + s). Here

dhS I-d =(0, .. . , 0, 1)

S s=o

and

- - f ( f~ = BL = L + BL _ I dQ; dt = L _ BL dQ; .Bqf+1 B(dt/dr) ~ Bq; (dt/dr)2) dr dr ~ Bq; dt

The integral of the motion is

Except for a sign this is the expression for the energy.

Fig.14

2.18 The points for which the sum of their distances to A and to n is constantHe on the ellipsoid with foci A, n, semi-major axis JR2 + a2 , and serni-minor

.axis R . The reflecting sphere lies inside that ellipsoid and is tangent to it in B.


Thus , any other light path than the one shown in Fig. 14 would be shorter thanthe one through B for which a = ß.

2.19 (i) As usual we set Xa = (q\, · · · , q/; PI , · · . ,Pj), and Yß = (Q\, . . . , Q / ;P t> .. . , p / ), as weIl as M aß = 8xa/8Yß. We have

MTJM = J and J = (ü/x/ n/x / ) . (I )-n/x / ü/x/

The equation always relates 8Pk/8pi to 8q;j8Qk, 8Qj/8pl to 8qL/8Pj etc.From this foIlows that [Pk . Qk] = [Pj . qj]. Let <p(x , y) be generating functionof the canonical transformation. As H = H + 8<p / 8t , the funct ion <P has thedimension of the product H . t . The assertion then foIlows from the canonicalequations.

(ii) With the canonical transformation <P and using T := wt, H goes over intoH = H + 8<p/8T. Hence [<p] = [H] = [XIX2] = [w][Pq]. The new generalizedcoordinate YI = Q has no dimension. As YIY2 has the same dimension as XI X2 ,Y2 must have the dimension of H, or H, that is, Y2 must equal wP. Therefore,

~ I 2<P(XI , YI) = 2" XI cOtYI .

From this one calculates

8$x2 = -8 = XlcOt YI ,

XI

or,

XI = A sin YI , X2 = A COS YI .

Using these formul as one finds

8x a (2y2) 1/ 2 COSYlM aß = 8Yß = -(2Y2)1 /2 sin YI

One easi ly verifies the conditions det M = 1 and MTJM = J.

2.20 (i) For f = 1 the condition det M = I is necessary and sufficient because,quite generaIly ,

(ii) We calculate 5 . 0 , set it equal to M, and obtain the equations

X cos a - Y sin a = al1

X sin a + Ycos a = a\2

Ycos o - z sin a = a21

Y sin a + z cos o = a22

(I )

(2)

(3)

(4)


From the combination «2) - (3))/((1) + (4)) of the equations

This allows us to calculate sin a and cos o, so that the subsystems «(1), (2)) and«3), (4)) can be solved for z , y, and z . One finds x = all cos a + al2 sin o ,z = a22cosa - a21 sin o, y2 = xz - 1.

There is a special case, however, that must be studied separately: This iswhen all + a22 = O. If a12 =I a21 we take the reciprocal of the above relation

«n + a22cota= ---

al2 - a21

If, however, at2 = a21 the matrix M is symmetric and 0 can be taken to be theunit matrix, i.e., a = O.

2.21 (i)

{lj , Tk} ={cjmnTmPn, Tk} =CjmnTm {Pn,Tk} +CjmnPn

{Tm, Tk} =CjmnTmOnk =CjmkTm

and, in a similar fashion, {lj ,pd = CjkmPm. In calculating the third Poissonbracket we note that

{lj ,T} ={CjmnTmPn,T} =CjmnTm{Pn ,T} + CjmnPn{Tm, T}

{JT 1= CjmnTm~ = CjmnTmTn- = 0 .

UTn T

Finally, we have

{lj,p2} = {CjmnTmPn,Pkpd = CjmnTm{Pn,Pkpd +cjmnPn{rm ,Pkpd

= 2CjmnPnPkOmk = CjmnPnPm = 0 .

(ii) U can only depend on T .

2.22 The vector A is a constant of the motion precisely when the Poissonbracket of each of its components with the Hamiltonian function vanishes. Therefore, we calculate

{1 2 , m, }

{H,Ad= -p +-, cklmPllm+-Tk2m T T

=2~ ek/m {p2'Pl lm} +,cklm{1/T,Pl lm}

+ ~ {p2, Tk/T} + m,2{1/T,Tk/T} .

The fourth bracket vanishes. The first three are calculated as follows


{p2 ,PIlm} ={p2,pl} l,« + {p2,lm} PI =0

{I/r, PIlm} = {I/r, Pdlm + {I/r, lm}PI = rl/r31m

{p2 , rk/r} = I/r {p2, rk} + rk {p2 , I/r} = 2pk/r - 2rkP' x/r3 .

Inserting these results we obtain

2.23 Calculation of the Poisson brackets yields differential equations which aresolved taking proper account of the initial conditions as folIows:

PI = {H,pd = -ma :::} PI = <mai + Px ,

P2 = {H,P2} = O:::} P2 = Ps ,

I I 2 Pxli! = {H, qd = - PI :::} ql = --at + - t + Xo ,

m 2 m

{ }I Py

lh = H, q2 = - P2 :::} q2 = - t + Yo .m m

2.24 (i) Let PI =mlm2/(ml + m2) and P2 =(m I + m2)m 3/(m l + m2 + m 3)be the reduced masses of the two two-body systems (1,2) and (Icenter-of-rnassof land 2), 3), respectively . Then 11"1 = PI el and 11"2 = P2e2. This explainsthe meaning of these two momenta. The momentum 71"3 is the center-of-massmomentum.

(ii) We define

j

Mj:=Lmj ,;=1

i.e., M j is the total mass of particles 1,2, .. . , j. We can then write

I j

ej=Tj+l- M .LmjTj , j=I , . . . ,N - IJ ;=1

I Ne» = M Lm;T; ,

N ;=1

1T'j= MI. (MjPj+l-mj+ltPj) , j = l , oo . ,N - IJ+I ;=1

N

1T'n =LP; .;=1


(iii) We choose the following possibilities :

a) As the Poisson bracket of Ti and Pk ist {Pk , Ti} = ll3x30ik we mustalso have {1I"k ,ud = ll3x30ik. We use this suggestive short hand notation for{(Pk)m, (q i)n} = OikOmn with Om denoting the mth cartesian coordinate. Usingthe former brackets one calculates the latter brackets from the defining formulas .For instance, with m\2 := ml + m2, M := ml + m2 + m3 ,

{m 3 m 3){11"2, ud = M - M II =0 , etc.

b) In the 18-dimensional phase space introduce the variables x =(TI,T2 ,T3,PI,P2,P3) and Y = (UI,U2,U3,1I"1,1I"2,1I"3), calculate the matrix Maß := 8Ya/8xßand verify that this matrix is symplectic, i.e. that it satisfies eq. (M2.1l3). Thiscalculation can be simplified by noting that M has the form

such that

Thus, it suffices to verify that ATB=ll9X9 . One finds

[

-m2/mI2ll

B = -m3/Mll

II

the entries being themselves 3 x 3 matrices. In a next step one calculates(ATB)ik =:LI AliBlk . For instance, one finds

(ATB) _ m2 mlm3 ml _ I11---+--+--M\2 m\2M M '

and verifies, eventually, that ATB =ll9x9 '

etc.


2.25 (i) In the situation described in the exercise the variation of I (a ) is

8I = dI(a) I dada 0'=0

dt 2(a) I= L (qk(t2(0), 0), lik(t2(0), 0»~ 0'=0 da

- L (qk(t\(O), 0), lik(t\(O), 0» dtd\ (a) I daa 0'=0

+112(0) (:L 8L 8qk (t , a) I da + :L 8~ 8 Iik (t , a) I da ) dt.

1,(0) k 8qk 8a 0'=0 k 8qk 8a 0'=0

We define, as usual

8qk I 8qk I d8a 0 da =Öqk and 8a 0 da =Öqk =dt Öqk ,

and, in addition, dt j(a )/daloda = ötj , i = 1,2. The time derivative dö qk/dt,by partial integration, is shifted onto 8L/8qk . Here, however, the terms at theboundaries do not vanish because the variations öt j do not vanish. One has

112(0) 8L d [8L d ] 12(0) 112(0) ( d 8L)

-- Öqk dt= -- Öqk - -- Öqkdt .1,(0) 8qk dt 8qk dt 1,(0) 1,(0) dt 8qk

The end points are kept fixed which means that

dqk(t j(a ), o ) I -_ 0 , i = 1,2 .da 0'=0

Taking the derivative with respect to a this implies

dqk(tj(a), a) I = 8qk I dt j(a )I da + 8qk I dada 0 8 t 1=1, da 0'=0 8a 1=1, ,0'=0

== qk(tj)8t j + Öqk1/=1, =0

Inserting this into 8I one obtains the result

8I = [(L - :L 88~ qk) öt j] 12(0) + ( 2(0) dt :L (88L - dd 88~ ) öqk .k qk 1, (0 ) ll,(o) k qk t s»

(ii) One calculates öl< in exactly the way, viz.

112

öl< =K (L + E) dtI,

=[(L - :L :~ qk) öt j ] 12k qk I,

112 (8L d 8L)+ dt:L -8 - -d -8' Öqk + [Eöt j]~~ =0 .I, k qk t qk


Now, by assumption E =L:k qk(8L/8qk) - L is constant. As a consequence thefirst and third tenns of the equation cancel. As the variations Öqk are independentone finds indeed the implication

s« ,), °{::} 8L _ .!!:.- 8L = ° k f8qk dt 8qk ' =1, . .. , .

2.26 We write

and obtain T dt = (ds/dt)ds = JE - Uds. The principle of Euler and Maupertuis , öl{ =0, requires

lq2

Ö JE - Uds =0 .q'

On the other hand, Fermat' s principle states the following: A light pulse traversesthe path ds in the time dt = (n(x , v)/c)ds. The path it chooses is such thatthe integral Jdt is an extremum, i.e . that ÖJn( x , v )ds = O. The analogy isestablished if we associate with the particle an "index of refraction" which isgiven by the dimensionless quantity «E- U)/mc2)1/2; (see also Exercise 1.12).

E~

EI---+.-------f

ql qo qFig.15

2.27 For U(qo) < E ::; Emax the points of intersection ql and q2 of the curvesy =U(q) and y =E are tuming points, q(t) oscillates periodically between ql andq2, cf. Fig. 15. Write the characteristic equation of Hamilton and Jacobi (M2.154)as

(1)


We know that the transfonned momentum obeys the differential equation P =0,i.e., that P = a: = const. We are free to choose this constant to be the energy,P =E. Taking the derivative of eq. (1) with respect to P =E,

en 02Sop oqoP =I .

If oH/op =r' 0 (this holds locally if Eis larger than U(qo», then (02S)/(oqoP) =r'O. Thus, the equation Q =oS(q,P)/oP can be solved locally for q =q(Q,P).This yields

(oS )-H q(Q, P), oq (q(Q, P), P) == H(Q, P) =E == P .

From this we conclude

. ob oSP =- oQ =0 :::} Q =t - to =oE .

The integral I(E) becomes

1 i 1 l t o+

T(E )I(E)=- pdq=- p ·qdt

21t' rE 21t' to

so that dI(E)/dE =T(E)/(21t') == w(E), in agreement with Exercise 2.1.

2.28 The function S(q, I) , with I as in Exercise 2.27, generates the transformation from (q,p) to the action and angle variables (0, I),

os(q, I)p= oq

0= oS(q, I) with b =E(I) .oI '

We then have Ö = oE/ 0I = const., t = 0, which are integrated to O(t) =(oE/oI)t + 00 , I = const. Call the circular frequency w(E) := oE/oI so thatO(t) = wt + 00 , I = const.

2.29 We calculate the integral I(E) of Exercise 2.27 for the case H =p2/2 +q2/2 : With p =V 2E _ q2

I(E) =_I 1 pdq =_I 1 pV2E _ q2dq21t' IrE 21t' IrE

=~l+A VA2 _ q2dq, (A=v2E).tt -A

Using

JA lt'A2

VA2- x2dx =--A 2


one finds I(E) =A 2/2 =E, i.e., H = I. The characteristic equation (M2.154)reads in the present example

Its solution can be written as an indefinite integral S = JJ2E - q,2 dq', or

S(q,I) =JJ2I - q,2 dq'. The angle variable follows from this

e es J I , . q=- = dq =arcsm--et J2I - q,2 -..fil '

giving q =-..fil sin (). In a sirnilar fashion one calculates

These are identical with the formulas that follow from the canonical transformation iP(q , Q) = (q2/2)cot Q, cf. eq. (M2.95).

2.30 Following Exercise 2.17 we take the time variable t to be a generalizedcoordinate t =qf+1 and introduce in its place a new variable r such that

- ( dq dt) ( I dq) dtL q, t , dr ' dr := L q, (dt/dr) dr dr '

By assumption f = I, i.e., ql =q and q2 == qf+1 = t . The action integral (theprincipal function) with the boundary conditions modified accordingly, reads

lT~

10= dr L(<PI (s , r ) , <P f+1 (s , r ) , <p~ (s, r ), <Pf +1(s , r)) ,T '

1

the prime denoting the derivative with respect to r. We now take the derivativeof Iawith respect to s , at s =0,

d I l T 2

dIa = drs 8 = 0 Tl

X { aL d<pI + aL d<p~ + aL d<Pf+1 + aL d<Pf+1 } . (1)a<PI ds a<p~ ds af/J f+1 ds af/Jf+1 ds

Replacing aL/af/JI in the first term by (d/dr)(aL/af/JD , through the equationsof motion, the first two terms of the curly brackets in eq. (1) can be combinedto a total derivative with respect to r. The integral over r can be rewritten as anintegral over t , so that the first two terms give the contribution


(Note that here the dot means the derivative with respect to t , as before.) In thethird term we replace 8Lj8ifJf+1 by (djdT)(8Lj8ifJf+I), (again by virtue of theequation of motion) . In this term and in the last term of the curly brackets ineq. (1) we make use of the solution to Exercise 2.17, viz.

8L = 8L = L _ 8~ difJI8(8ifJf+I/8T) 8(8t j 8T) 8ifJ I dt

so that these terms can also be combined to a total derivative

~ ((L _8~ difJ1) difJ f+l ) •dr 8ifJ I dt ds

(2)

As above, in doing the integral one replaces T by the variable t . The inner bracketin eq. (2) is the energy (to within the sign) so that we obtain the difference ofthe term -E(difJf+I/ds) at the two boundary points, that is, (-E) times thederivative of the time t =t2 - t l by s. Summing up one indeed obtains the resultof the assertion. Finally, the generalization to f > 1 is obvious.

Chapter 3: The Mechanics of Rigid Bodies

3.1 (i) As K and K differ by a time-dependent rotation, J is related to J byJ =R(t)JR-1(t), with R(t) the rotation matrix that describes the relative rotationof the two coordinate systems. The characteristic polynomial of J is invariantunder similarity transformations. Indeed, by the multiplication law for determinants,

det IJ - >'lll = det IR(t)JR-1(t) - >'lll

=det IR(t)(J - >'ll)R-1 (t) 1

=det IJ - xII I .

The characteristic polynomials of J and J are the same. Hence, their eigenvaluesare pairwise equal.

(ii) If K is a principal-axes system, J has the form

o 0]h 0 .o h

A rotation about the 3-axis reads

[

cos ifJ(t) sin ifJ(t ) 0]R(t) = - si~ ifJ(t ) cost(t) ~ .

This allows us to compute J with the result

Chapter 3: The Mechanies of Rigid Bodies 467

(1z - IdsinifJcosifJ 0]11 sinz ifJ + IzcosZ ifJ 0

o 13

3.2 The straight line connecting the two atoms is a principal axis. The remainingaxes are chosen perpendicular to the first and perpendicular to each other, assketched in Fig. 16. Using the notation of that figure we have mlal = mZa2,

al + az =I, and therefore

with f.l denoting the reduced mass.

s1 2

m2 Fig.16

3.3 The moments of inertia are determined from the equation

1

111 - xdet(J - >.ll) = I~I

whose solutions are

oo =0 ,

133 - >.

I _ 111 + 122 ±1,2 - 2

(111 - hz)2 + I I4 IZ 21 ,

(i) 11,2 =A±B. Thus it follows that B ::; A and A+B 2 O. Since 11-t, 2 13we also have h ::; 2A, i.e., A 2 O.

(ii) 11 =5A, h =O. From 11+12 2 hand h +h 2 11 follows 13 =5A. Thebody is axially symmetrie with respect to the 2-axis.

3.4 The motion being free we choose a principal-axes system attached to thecenter-of-mass, letting the 3-axis coincide with the symmetry axis. The momentsof inertia are easily calculated, the result being

3 zh= lOMR .

(2)

(I)

(4)


A Lagrangian function is

1 3

L =Trol =2'I:JiWr ,i=t

where Wi are the components of the angular velocity in the body fixed system.They are related to the Eulerian angles and their time derivatives by the formulae(M3.82),

WI =() cos 'l/J + ~ sin 8 sin 'l/J

W2 =-() sin 'l/J + ~ sin 8 cos 'l/J

W3 =~ cos 8 + 'if; .Inserting these into Land noting that II =h we have

. . . I ( ' 2 . 2 2) I ( " ) 2L(1),8,'l/J,1> ,8,'l/J)=2I1 8 +1> sin 8 +213 'l/J+1>cos8 .

The variables 1> and 'l/J are cyclic, hence

8L . 2 (" )Pr/> = - . = II 1> sin 8 +!J 'l/J + 1> cos 8 cos 8 ,81>8L (" )Pt/1 =8'if; =13 'l/J + 1> cos 8 ,

are conserved. Furthermore, the energy E = Tro l = L is conserved . Note thatPr/> =II (WI sin'l/J + W2 cos 'l/J) sin 8 + !JW3 cos 8. This is the scalar product L . e30

of the angular momentum and the unit vector in the 3-direction of the laboratorysystem, i.e. PrP =L3. Regarding Pt/1 we have Pt/1 =!JW3 =L3. The equations ofmotion read

d ddt PrP = dt (II (WI sin'l/J +W2 cos 'l/J) sin 8 + !JW3 cos 8) =0

d d (" )dt Pt/1 =13dt 'l/J + 1> cos 8 =0

d 8L 8L .. .2 (") .dt 8(} - 88 =118 - 111> sin 8 cos 8 + 13 'l/J + 1> cos 8 1> sin 8 ::: 0 . (3)

From the first of these ~ = (L3 - L3cos 8)/(I1 sin2 8). Inserting this into theLagrangian function gives

1'2 I ( - ) 2 I - 2L =- I,8 + . 2 L3 - L3cos 8 + - L3=E =const.2 2I, sm 8 2!J

If, on the other hand, ~ is inserted into the third equation of motion (3), oneobtains

.. cos 8 ( - ) 2 I - ( - )I,8 - . 3 L3 - L3cos 8 +-I. 8 L3 L3 - L3cos 8 =0 ,I, sm 8 I sm

which is nothing but the time derivative of eq. (4).

./

/(\

<,

-3


Fig.17

3.5 We choose the 3-axis to be the symmetry axis of the torus. Let (r' , </» bepolar coordinates in a section of the torus and "p be the azimuth in the planeof the torus as sketched in Fig. 17. The coordinates (r', ,,p,</» are related to thecartesian coordinates by

x ] =(R + r' cos </» cos "p , X 2 =(R + r' cos </» sin e , X 3 =r' sin </> .

The Jacobian is

8(X] , X2 , X3) _ '(R+' )..)8 ' ./.).. - r r cos '/' .(r , '/', ,/,)

The volume of the torus is calculated to be

v =1T

dr'12

'" d"p12

'" d</> r ' (R + r' cos </» = 27r2r 2R ,

so that the mass density is l!o =M / (27r2r 2R). Thus

3.6 In the first position we have Ija) =2(2/5)(MR 2 +mr2 ) and w~a ) = L 3/ I ja).In the second position the contribution of the two smaller balls is caIculatedby means of Steiner's theorem, I:= I, + m(a2 - aD, now with 13 = 2mr2/5,

a =±(b/2)e]:

I jb ) = 2 ((2/5)(MR 2 + mr2) + m b2 /4) .


Fig.18

b

a

It follows that w~b) = L3/1~b), and w~a) /w~b) = 1 + mb2/(2I~a» . One rotatesfaster if the arms are close to one 's body than if they are stretehed out. Makinguse of Steiner's theorem onee more we finally ealculate 1\ = 12• For the twoarrangements one obtains the result

3.7 The relation between density and mass reads

M =Je(r) d3r =12 1r

d4>11r

sin 0 dO100

r2

dr e(r,0, 4» .

In our ease, where e depends on 0 only, this means

M =12 1r

d4> l 1r

sin 0 dO l R

(Q) r 2 dr eo =237r

eo l 1r

sin 0 dO R(O)3 .

The moments of inertia 13, 1\ =12 are ealculated from the formulas

h = Jd3x (Jr

2 (1 - eos2 0), 1\ + t, + t, = 2Jd3x (Jr

2•

(i) Integration gives the results

47r 3 23MM =-3 eoRQ(1 + a) , i.e., eo =-4 3 2 ·

7r Ro(1 +a )

2MR2

{ 9} 2MR2

{ 3}1\ =h =5(1 +a02) 1+4a

2+ "7 a 2 , t, =5(1 +a02) 1+2a

2+ "7 a 4

(ii) Substituting z =eos 0 the integrals are easily evaluated. With the abbreviation "/ := J5/l67r ß we obtain

M 47r p3 (16 3 12 2 1)="3 eO~'il 35 "/ + 5 "/ + ,

that is

( )- \

3 M 16 3 12 2eo= -- -"/ +-"/ +1

47r RÖ 35 5

[

cos ljI sin ljI 0]and R3(ljI)= -sin ljl cos ljl 0

001


I _ I _ 2MRij . 1 + "y + 64"Y2/7 + 8"Y3+ 688"Y4/77 + 2512"Y5 / 1001\- 2- 5 1+ 12"Y2/5+16"Y3/35 '

I _ 2MRij 1-2"Y+40")'2/7-16"Y3/7+208"Y4/77-32"Y5/1oo13 - -5- . 1+ 12"Y2/5 + 16"Y3/35

3.8 The (principal) moments of inertia are the eigenvalues of the given tensorand, hence, are the roots of the characteristic polynomial det(>'TI - J). Calculatingthis determinant we are led to the cubic equation >.3 -4>.2+5>'-2 =O. Its solutionsare >.\ = >'2 = 1, >'3 = 2. The inertia tensor in diagonal form reads

[1 0 0]

J= 0 1 0 .002

We write J = RJRT and decompose the rotation matrix according to R(ljI, (), </»= R3(ljI)R2«()R3(</» . As the factor R3(</» leaves J invariant we can choose </> = o.With

[

cos () 0 - sin () ]R2«() = 0 1 0

sin () 0 cos ()

we calculate

[

cos2 lj1 sin2 ()RJRT = TI + - sin ljI cos ljI sin2 ()

- cos ljI cos ()sin ()

- sin ljI cos ljI sin2()sin2 lj1 sin2 ()

sin ljI cos ~ sin ()

- cos ljI cos ()sin ()1sin ljI cos ()sin ()

cos2()

Fig.19

If this is set equal to J as given, we find cos2() =3/8, and, with the followingchoice of signs for () : cos() = -/3/(2-/2) and sin() = VS/(2-/2), the resultcos ljI = 1/VS, sin ljI = - V4ß.3.9 (i) e(r) =eo()(a - Irl). The total mass is equal to the volume integral ofe(r), viz.

471" 3 3MM =3 a eo =} eo =471"a3 .

(ii) We choose the 3-axis to be the axis of rotation. Let the coordinate in thebody fixed system be (x , y , z), the coordinates in the space fixed system are

x' = x cos wt - (y + a) sin wt, y' = x sin wt + (y + a) cos wt , z' = z .


Inverting these equations we have

x =+x' coswt + y' sin wt , y =- x' sinwt + y' coswt - a ,

whence

x 2+ y2 =X/2+ y /2 + a2 + 2a(x' sin wt - y' cos wt) .

From this we get

{!(r /, t ) ={!o(}( a - J r /2 + a2 + 2a(x' sinwt - y' coswt ) ) .

(iii) In the case of a homogeneous sphere the inertia tensor is diagonal , allthree moments of inerta are equal, 1\ = Iz= h == I. Hence,

J 6Ma231 =1\ + Iz+ h =2 d3r {!(r )r2 =- 5- .

Making use of Steiner's theorem (M3.23) we find

I ( 2 2) 7Ma2

13 = h + M a 833 - a 3 = -5- .

3.10 (i ) The volume of the cylinder is V =11T 2 h, hence the mass density is(!o =m /('lrr2h). The moment of inertia relevant for rotations about the symmetryaxis is best calculated using cylindrical coordinates,

h ={!o r difJ rhdz r (!3 d{! =~mr2 .

Jo Jo Jo 2

Call q(t ) the projection of the center-of-mass' orbit onto the inclined plane.When the center-of-mass moves by an amount dq, the cylinder rotates by anangle difJ = dq]r , Therefore , the total kinetic energy is

1 .2 1 (p 3 .2T = "2 mq + "2 13r 2 = 4 mq .

(ii) A Lagrangian function is

L = T - U = 3mr//4 - mg(qo - q) sin a ,

where qo is the length of the inclined plane and a its angle of inclination. Theequation of motion reads 3mijl2 =mg sin o, the general solution being q(t) =q(O)+ v(O)t + (g sin a)t2/3 .

3.11 (i) The rotation R( 4) . 4» is a right-handed rotation by an angle ifJ aboutthe direction 4>, with 0 ~ ifJ ~ 'Ir . Any desired position is reached by means ofrotations about 4> and -4>.

(ii) With 4> an arbitrary direction in R, and with ifJ between 0 and 'Ir, theparameter space (ifJ, 4» is the ball D 2 (surface and interior of the unit sphere inR2). Every point p E D 2 represents a rotation, the direction 4> being given by thepolar coordinates of p, and the angle ifJ being given by its distance fromcenter.Note, however, that A : (4), ifJ ='Ir) and B : (- 4>, ifJ ='Ir) represent the samerotation.


(iii) There are two types of closed curves in D 2 : curves of the type of Clas shown in Fig. 20, which can be contracted, by a continuous deformation, to apoint, and curves such as C2, which do not have this property . C2 connects theantipodes A and B. As these points represent the same rotation, C2 is a closedcurve. Any continuous deformation of C2 which shifts A to A' also shifts B toB', the antipodal point of A'.

While Cl contains no jumps between antipodes, C2 contains one such jump.One easily convinces oneself, by means of a drawing, that any closed curve withan even number of antipodal jumps can be deformed continuously into Cl or,equivalently, into a point.

.-l

'",//

//

~P/ ..

I/

/C2,'"B

Fig.20

Take the example of a closed curve with two such jumps as shown in Fig. 21.One can let Al move to B2 in such a way that the are B l A2 goes to zero, thesections AIBI and A2B2 become equal and opposite so that the curve A lB2

becomes like Cl in Fig. 20. In a similar fashion one shows that all closed curveswith an odd number of jumps can be continuously deformed into C2 • (One saysthat the two types of curves form homotopy classes.)

3.12 We do the calculation for the example of (M3.93). Equations (M3.89)yield expressions for the components of angular momentum in the body fixedsystem .


Making use of the relation {Pi,!(qj)} =Dijf'(qj) which follows from the definition of the Poisson brackets, we calculate readily

{ - -} { sin 'IjJ .L\,L2 = Pr/>-:--(} - Pt/J sine cots +Pecos 'IjJ ,

sm

cos 'IjJ . }Pr/> ----=----(} - Pt/J cos 'IjJ cot () - pe sm 'IjJ

sm

=Pe c.o~ () (- {sin 'IjJ, Pt/J cos 'IjJ} - {P r/> sin 'IjJ, cos 'IjJ })sm ()

+ Pr/> { - sin2

'IjJ { Si~ () ' ve } + cos2

'IjJ {pe , Si~ () } )

+ cor' (p,p sin 'IjJ ,Pt/J cos 'IjJ } + (sin 'IjJ {p,p cot (),Pesin 'IjJ }

- cos 'IjJ {pe cos 'IjJ, p,p cot (}} )

cos () cos () 2 I=+Pr/>-.- - Pr/>-.- - p,p cot () + P,P-.-sm2 () sm2

() sm2()

=Pr/> =L3 •

3.13 We have to show that d/dt(L . s) =t. . s + L . s =O. The first term onthe right-hand side can only be different from zero if there is a nonvanishing,extemal torque D. The second term is independent of such a torque. It vanishesbecause, for the case of the symmetric top without extemal forces, both L 3 andL3 are conserved. Therefore, we only need to compute the first term. With FRdenoting the frictional force, D = -s x FR. The scalar product of D and s iszero, hence L . s is constant in time.

Chapter 4: Relativistic Mechanies

4.1 (i) Let the neutral pion fly in the 3-direction with velocity v =vOe3.Thefull energy-momentum vector of the pion is

where ßo = vo/c, ,0 = (I - ß~) - 1 /2 . The special Lorentz transformation whichtakes us to the rest system of the pion, is

['0 0 0 - 'OßO]o 1 0 0

t.; = 0 0 1 0

- ,oßo 0 0 ,0Indeed, l -vq = q* = (m".c, 0).


Fig.22

/c*1

- ---r-,,-'--'----+ 3

/c*2

(ii) In the rest system (Fig. 22, left-hand side) the two photons have the fourmomenta ki = (Ei / c, k i ), i = 1,2. Conservation of energy and momentumrequires q* =ki + ki, i.e. Ei + Ei =m 7r c2 and k ~ + k; =O. As photons aremassless, Ei =Iki lc, and, as k~ =- k;, one has E i =Ei . Denote the absolutevalue of the spatial momenta by K,* . Then Ik~ I=Ik;I=K,* =m 7r c/ 2.

In the rest system the deeay is isotropie. In the laboratory system only thedireetion e3 of the pion' s momentum is singled out. Therefore, in this systemthe deeay distribution is symmetrie with respeet to the 3-axis. We first study thesituation in the (1, 3)-plane and then obtain the eomplete answer by rotation aboutthe 3-axis. We have (k~h = K,* cos lJ* = - (k ; h , (k ~) ] = K,* sin lJ* = -(k;) I,while the 2-components vanish. In the laboratory system we have k; =Lv oki ,viz.

~EI =10K,* (1 + ßocos lJ*) ,

(k l)] = (kj) 1 = K,* sin 8* ,

(klh = lOK,* (ßo+cos lJ*) ,

(klh =0 =(k2h .

~E2 =10K,* (I - ßo cos lJ* ) ,c

(k 2) 1 = (ki) 1= - K,* sin 8* ,

(k 2h =10K,* (ßo - cos lJ*) ,

In the laboratory system we then find

tan lJl = (k l)1 = . sin lJ* ;(klh , 0(ßo +cos lJ*)

- sin e:tan lJ2 =---,-------,--

,0(ßo - cos lJ*)

Examples:

a) lJ* =0 (forward emission of one photon, backward emission of the other): Fromthe formulas above one finds E I =m 7rc

2/0(1 + ßo)/ 2, E2 =m 7rc

2/0(1 - ßo)/2,

k l = m 7r C/o(ßo + l)e3/2, k 2 = m 7r C/o(ßo - l )e3/2, and, as ßo ::; 1, lJl = 0,lJ2 =7r.

b) lJ* =1r/2 (transverse emission): In this case EI =E2 =m 7rc2/0 / 2, k l =

m 7r c(e l + l oßoe3)/ 2, k2 =m 7rc(-eI + , 0ßoe e)/2, tan lJ I =tan lJ2 =1/( /o ßo).


e) ()* = 1r/4 and ßo = I/V2, i.e, 10 = V2: In this ease one findsEI = 3m rr c2

/0/4, E 2 = m rrc2/0/4, k 1 = mrrc(el - 2V2e3)/(2V2}, k 2 =

-mrrceI!(2V2), ()t =aretan(l/(2V2)) ~ 0.1081r, ()2 =1r/2.In the rest system the deeay distribution is isotropie, whieh means that the

differential probability dl' for k~ to He in the interval da* =sin ()* d()* drjJ* isindependent of ()* and of rjJ* . (To see this enclose the deeaying pion by a unitsphere. If one eonsiders a large number of deeays then photon I will hit everysurfaee element da* on that sphere with equal probability.) Thus,

dl' = r oda* with r o = eonst.

In the laboratory system the analogous distribution is no longer isotropie. It isdistorted in the direetion of f1ight but is still axially symmetrie about that axis.We have

1 I.ur I .ur sin e: d()*r o d.I' = da da where da = sin () di .

The faetor sin ()* / sin () is ealculated from the above formula for tan ()I, viz.

. () tan () sin ()*sm = =---=----:-J 1+ tan? () 10(ßO+ eos ()*) ,

and the derivative d()/ d()* is obtained from () = aretan(sin e:ho(ßo +eos ()*)) bymaking use of the relation 7Öß6=16- I. One finds

.ur 2 ( *)2da = 10 I + ßo cos () .

The eosine of ()* is expressed in terms of the eorresponding laboratory angle bythe formula

()* eos () - ßoeos = .

1 - ßoeos()

The shift of the angular distribution is well illustrated by the graph of the funetion

._ .ur _ 2 ( eos () - ßo ) 2F«()).- da - 10 I + ßo 1 _ ßo cos ()

for different values of ßo. Quite generally we have dF/ d()19=0 =O. For ßo -+ 1the value F(O) =(1+ßo)/ (I - ßo) tends to infinity. For small argument ()=c; «: I,on the other hand ,

1+ ßo ( c;2 )F(c;)~-- 1---l- ßo l- ßo '

whieh means that for c;2 ~ (I - ßo) F beeomes very smalI . Therefore, whenßo -+ 1 the function F«()) falls off very quickly with inereasing (). Figure 23shows the examples ßo =0, ßo =1/V2, and ßo =11/13.


Fig.23

4.2 Let the energy-momentum four-vectors of n , u, and v be q, p, and k,respectively. We have always q =p + k. In the pion's rest system

If ",* := Ip* I denotes the magnitude of the momentum of the muon and of theneutrino, then

In the laboratory system the situation is as folIows: The pion has velocity Vo =VO€3 and, therefore,

q =(Eq / c, q) =(,om7rc, ,0m7rvo) =,0m7rc(l, ßO€3)

with ßo =Vo / c, ,0 = 1/JI - ß1i. It is sufficient to study the kinematics in the(I,3)-plane. The transformation from the pion's rest system to the laboratorysystem yields


~Ep = /'0 (~E; + ßop*3) = /'0 (~E; + ßOK* COS(}*) ,pi =p*1 ,

p2 = p*2 = 0 ,

p3 =/'0 (~ßoEo + p*3) =/'0 (~ßoE; + K* COS(}*)

and, therefore, the relation between the angles of emission (}* and (} is (cf. Fig. 24)

or

pi K* sin (}*tan(}=-= ,

p3 /'o(ßoE;/C+K*COS(}*)

1

(1)

(2)

p

2

Fig.24

Making use of ß* := CK* / E; = (m;' - m;)/(m;' + m;) (this is the beta factorof the muon in the rest system of the pion), eq. (1) is rewritten

ß* sin e:tan (} = -----

/,o(ßo + ß* cos (}*)(3)

There exists a maximal angle (} if the muons which are emitted backwards in thepion's rest system «(}* = 7r), have momenta

p3 = /,oE;/ C (ßo + ß" cos (}*) = /'oE;/c (ßo - ß*) > 0 ,

i.e., if ßo > ß*. The magnitude of the maximal angle is obtained from thecondition dtan(}/d(}* =0 which gives cos ß* = - ß*/ ßo and, finally,

(4)

Chapter 4: Relativistic Mechanics 479

4.3 The variables s and t are the squared norms of four-vectors and are thusinvariant under Lorentz transformations, The same is true for u =C

2(qA - q'nf.For OUf calculations it is convenient to choose units such that c = I. It is notdifficult to re-insert the constant c in the final results. (This is important if wewish to expand in terms of v / c.) To reconstruct those factors one must keep inmind that terms Iike (mass times c2

) and (momentum times c) have the physicaldimension of energy.

Conservation of energy and momentum means that the four equations

(1)

(2)

(3)

(4)

must be satisfied. This means that the variables s , t, u can each be expressed intwo different ways (now setting c =I):

2 (' I )2S = (qA +qB) = qA +qB

t = (qA - q~)2 = (q'n _ qB)2

u = (qA - q'n)2 = (q~ - qB)2

(i) In the center-of-mass system we have

I (E'* '*)qA = A , q ,

qe =(EH ' - s') ,q'n=(E~ ,_q'*) ,

(5)

where E'A =)m~ + (q*)2 etc., with q* = Iq* I. Like in the nonrelativistic case

energy conservation requires the magnitudes of the three-momenta in the centerof-mass system to be equaI. However, the simple nonrelativistic formula

( *) mB I lablq n.r. = q Am A+ mB

that foIIows from eq. (M1.79a) no longer holds. This is so because neither thenonrelativistic energy

T =m A + m B ( *)2r 2 q n.r.

mAma

nor the quantity

(qA +qB)2

2(mA +mB)

are conserved. We have

s = (E'A + E B)2 = m~ + m1 + 2(q*)2

+2)(q*)2 +m~) (q*)2 + m1) . (6)

Thus, s is the square of the total energy in the center-of-mass system. Reintroducing the velocity of light,

(8)


S = m~c4 +m~c4 + 2(q*)2c2 +2)(cq*)2c2 +m~ c4) ((q*)2c2 +m~c4) .

In a first step we check that s , when expanded in terms of 1/c, gives the correctnonrelativistic kinetic energy Tr of relative motion (except for the rest masses ,of course)

2(I ( (q*)4))s ~ (mAc2 + mBc2) 1+ (q*)2/ c2+0 ~ ,m AmB m c

and, thus,

2 2 mA+mB 2 ((q*c)4)vfs ~ mAc + mBc + 2 (q*) + 0 --2-4mAmB (mc)

The magnitude of the center-of-mass momentum is obtained from (6)

q*(s) =2~ )(s - (mA +mB)2) (s - (mA - mB)2) . (7)

Clearly, the reaction can take place only if s is at least equal to the square of thesum of the rest energies ,

s ~ So := (mA + mB)2 ~ (mA c2 + mBc2)2 .

So is called the threshold of the reaction . For s =So the momentum vanishes,which means that at threshold the kinetic energy of relative motion vanishes .

The variable t is expressed in terms of q* and the scattering angle 0* asfolIows:

t - ( , ) 2 2 ,2 2 ' 2 2 2E* E '* 2 * ,*= qA - qA =qA + qA - qA ' qA = m A - A A + q . q .

As the magnitudes of q * and of s" are equal, E'A = EH' Therefore,

t =-2(q*?(l - cos 0*) .

Except for the sign, t is the square of the momentum transfer (q * - q' * ) in thecenter-of-mass system . For fixed s ~ So, t varies as follows

_4(q*)2 ~ t ~ 0 .

Examples:

a) e" +e- -t e" +e

s ~so = 4 (meC2) 2 , -(s-so)~t~O .

b) lI+e- -te-+lI

2 I 2S ~ 80 =(mec2

) , - - (8 - 80) ~ t ~ O.8

(ii) Calculating 8 +t +u from the formulas (2)-(4) and making use of (l), wefind s + t + u =2(m~ +m~)c4 . More generally, for the reaction A+B -t C+D,one finds


4.4 In the laboratory system

qA = (EA, qA) ,

q~ =(E~ , q~) ,

qB = (mBc2,0) ,

q~ =(E~, q~) ,(1)

The scattering angle () is the angle between the three-vectors qA and q~ . Fromeq. (3) of the solution to Exercise 4.3 above (using c =1),

(2)

Equation (8) of Exercise 4.3 above gives an alternative expression for t. Theaim is now to express the laboratory .quantities E A, E~, IqAI, Iq~1 in termsof the invariants sand t. Using (1) s is found to be, in the laboratory system,s =m~ +m~ + 2EAmB, that is,

1 ( 2 2 )EA=-- s - m A - m B2mB

From this, using q~ =Ei - m~,

(3)

with q* as given by eq. (7) ofExercise 4.3 above. We now calculate t =(qB _q~)2

in the laboratory system and find E~ =(2m~ - t)/(2m B) and from this E~ =E A + m» - E~ to be equal to

(5)

and, eventually, from q~ =E1- m~

(6)

From (2)

This is used to calculate sin () and tan (), replacing all quantities which are notinvariants by the expressions (3)-(6). With the abbreviations E := (mA + mB)2and .,1 := (mA - mBP we find

() 2mBv-t (st + (s - E)(s - 2\))tan = (2 2)(s - E)(s - .,1) + t s - m A - m B

Finally, cos e: and sin ()* can also be expressed in terms of sand t, starting fromeqs. (8) and (7) of Exercise 4.3 above,


()* _ 2st + (s - 17)(s - .<1)cos - (s - 17)(s - .<1) ,

. ()* _ r: V-t (st + (s - 17)(s - .<1))sm - 2y s (s _ 17)(s _ .<1) .

Replace the square root in the numerator of tan () by sin ()* and insert t in thedenominator, as a function of cos ()*, to obtain the final result

(7)

For s ~ (mA + mB)2 one recovers the nonrelativistic result (M1.80). The caseof two equal masses is particularly interesting. With mA =me == m

2m ()*tan() =-tan- .

JS 2

As JS ~ 2m the scattering angle () is always smaller than in the nonrelativisticsituation.

4.5 If we wish to go from the rest system of a partic1e to another systemwhere its four-momentum is p =(E!e,p), we have to apply a special Lorentztransformation L(v) with v related to p by p =mq» , Solving for v ,

pe pe 2v = ----;:::::::=====:===:c=

vp2+m2e2 E'Insertion into eq. (M4.41) and application to the vector (0, s) gives

s = l(v)(O, s) = (1s· v, s + 2 "'(2 V . sv) .e e (l +"'()

As sapa is a Lorentz scalar, and hence is independent of the frame of referencethat one uses, this quantity may be evaluated most simply in the rest system . Itis found to vanish there and, hence, in any frame of reference.

4.6 In either case the coordinate system can be chosen such that the y- andz-components of the four-vector vanish and the x-component is positive, i.e.,z = (z'', z l , 0, 0) with z l > 0, If zO is smaller than zero we apply the timereversal operation (M4.30) so that, from here on, we assurne zO > 0, withoutloss of generality.

(i) A light-like vector has z2 =0 and, hence, zO=z l . We apply aboost withparameter >. along the x-direction, cf. (M4.39) . In order to obtain the desiredform of the four-vector we must have

zOcosh >. - zOsinh >. = 1 or zOe- '\ = I ,

from which follows >. = In zO.


(ii) For a space-like vector z2 =(zO)2 - (z~)2 < 0, i.e., °< zO < zl. Applyingaboost with parameter >. it is transformed to

(zOcosh >. - zl sinh >. , z' cosh >. - zOsinh >. ,0,0) .

For the time component to vanish, one must have tanh X = zO/ zl. Calculatingsinh X and cosh X from this yields the assertion z l =J- z2.

4.7 The commutation relations (M4.59) can be summarized as folIows, markinguse of the Levi-Civitä symbol:

[Jp , J q] =Cp qr Jr ,

[K p , Kq ] =-cpqrJr ,

[Jp , Kq] =c pqr K r •

From this one obtains

[Ap , Aq] =cpqrAr ,

[Bp, Bq] =CpqrBr ,

4.8 The explicit calculation gives

PJjP- 1 =Jj , PKjP- 1 =-Kj .

This corresponds to the fact that space inversion does not alter the sense ofrotation but reverses the direction of motion.

4.9 The commutators (M4.59) read in this basis

The matrix J, and Kj being real and skew-symmetric, we have for instance

(AT)* . * . AJ j =- (i.L) =d j =J .

4.10 This exercise is a special case of Exercise 4.11. The result is obtainedfrom there by taking m2 = 0= m3 .

4.11 Energy conservation implies (again taking c =I)

M =EI +E2 +E3 =Jmf+ j2 + Jm~ +x2j2 + Jm~ +(1- x )2 j2

=M(xf(x» .


The maximum of f(x) is found from the equation

04: df =_aM/ax =-fEI

xE3-(l-x)E2dx aM/af E2E3+ x2EIE3+(l-x)2EIE2 '

or XE3 = (l - x )E2. Squaring this equation gives x2(m~ + (l - x )2f 2) = (l x)2(m~ + x2f 2), and from this the condition

! m 2x=---

m2+ m3

Taking into account the condition

1- x m3E 3=--E2= -E2 ,

x m2

one obtains

M - EI =m2 + m3 E2 .m 2

The square of this yields

M 2 _ 2MEI + mi + f 2 = (m2 + m 3)2 ( m 2 + m~ f 2)m~ 2 (m2 + m3)2

and from this

Examples:

(i) J1- ---T e + VI + V2 : m2 = m3 = 0, M = m IJ, ml = me. Thus

With mIJ/m e ~ 206.8 one finds (Ee) max ~ 104.4mec2.

(ii) n ---T p + e + V; M = m n , m l = me, m2 = m p, m3 = O. Therefore oneobtains

1 (2 2 2) 2 1 ( 2) 2(Ee)max =-- m n + me - m p c =-- (2m n - Ll)Ll + me c ,2m n 2m n

where Ll := m n - m p ' Inserting numerical values yields (E e)max ~ 2.528m em c2.

Thus , max = 2.528 and ßmax = J,~ax - l/rmax = 0.918. The electron is highlyrelativ istic at the maximal energy.

4.12 The apparent lifetime 7 (v ) in the laboratory system is related to the reallifetime 7 (0) by 7(v ) =,7(0). During this time the partic1e, on average , travels adistance

L =V 7(v) =ß,7(0)C .


Now, the product ß, equals Iplcj(mc2) , cf. eq. (M4.83), so that for Ipl ::;: xm r

there follows the relation

L::;: XT(O)C •

For pions, for instance, one has T~O)c ~ 780 cm.

4.13 From the results of the preceding exercise we obtain T~O)C ~ 2.7 X 1013

cm. For E ::;: 1O-2mnc

2 one has x ::;:~ ::;: 0.142, while for E ::;: 1014mnc

2

one has x ~ 1014 .

4.14 Let PI, P2 be the energy-momentum four-vectors of the incoming andoutgoing electron, respectively, and k that of the photon. Energy and momentumconservation in the reaction e -T e+, requires PI ::;: P2 +k. Squaring this relationand making use of

pi ::;: m ec2

::;:P~, k 2::;: 0 ,

one deduces P2 . k ::;: O. As k is a light-like four-vector this relation can only holdif P2 is light-like, too, i.e., if P~ ::;: O. This is in contradiction with the outgoingelectron being on its mass shell, P~ ::;: m~ . Hence, the reaction cannot take place.

4.15 The first inversion leads from x!' to (R2 j x2)x ll , the translation that followsleads to R2(x ll jx 2 + c"), and the second inversion, finally, to

I R4 (x ll + X2cll ) x4 x!' + x2cll

xll

::;: x2R4(X+x2c)2 ::;: 1+2(c'x)+c2x2 '

The inversion .:J leaves invariant the two halves of the time-like hyperboloidx2 ::;: R2, but interchanges those of the space-like hyperboloid x 2 ::;: _R2. Theimage of the light-cone by the inversion is at infinity. The light-cone as a wholestays invariant under the combined transformation .:J 0 T 0 .:J.

4.16 As L does not depend on q, the equation of motion for this variable reads

d 8L d .dt 8q ::;: m dt ('ljJq) ::;: 0 . (1)

In turn, L is independent of~. The condition for the action integral to be extremalleads to the following equation for 'ljJ,

8L 1 (.2 2 'ljJ2 - 1)8'ljJ ::;: 2m q - »:»: ::;: O.

The solutions of this equations are


Insertion of 'l/J I into the Lagrangian function yields

L(q,'l/J ='l/J» =~m( - 2coJC5 - q2 +2cÖ) =-mcöJl - q2/cÖ+mcÖ'

This is nothing but eq. (M4.97), with c replaced by Co, towhich the constantenergy mcÖ is added.

If we let Co go to infinity, 'l/J I trends to 1 and the Lagrangian function becomesLnr =mq2/2, well-known from nonrelativistic motion. One verifies easily that(I) is the correct equation of motion in either case.

The second solution 'l/J2must be excluded. Obviously, the additional term

I (2 2'l/J- I )-m('l/J - 1) q - co- -2 'l/J .

which is added to the Lagrangian function Lnr takes care of the requirement thatthe velocity q should not exceed the value Co.

Chapter 5: Geometrie Aspeets of Meehanies

5.1 We make use of the decomposition (M5.52) for /:; and ~

2: wil ...i,dx i11\ .. . l\dx ik I\dxi11\ . . . l\dx i ,.

i l <···<i ,

The analogous decomposition of ~ 1\ /:; (k and 1 interchanged) is obtained fromthis by shifting first dxi>, then dx h, and so forth up to dx i" across the productdXi1 1\ dXi2 1\ ... 1\ dXik from the right to the left. Each one of these operationsgives rise to a factor (_)k, so that one obtains the total factor (_ )kl .

5.2 We calculate

3 3

ds2(€i'€i) = 2: Ekdxk(€ i)dxk(€i) = 2:Eka~aj ,i=1 i=1

where we have set a~ := dxk(€i)' As ds2(€i , €k) = 8ik. we must have a~ =b~ / y'JJJ;, where {b~} is an orthogonal matrix . This matrix must be orthogonal because the coordinate axes were chosen orthogonal, Therefore dx k (€i) =8~ / y'JJJ;.

5.3 Consider

31 ~ .

W a =.L..t w i(x)dx' ,i=1

(;j a =bl (x) dx 2 1\ dx 3 + cyclic permutations

whose coefficients, Wi(X) and b,(x), are to be determined.

Chapter 5: Geometrie Aspeets of Meehanies 487

Since, on the other hand,

I "' .wa(O =a- e =LJai(X)e'i

we deduce

Wi(X) =ai(x)j"jjj; .

b) We calculate

2Wa (e , 1]) =b, (x )(dx 2(e) dx3(1]) - dx 2(1]) dx\e)) + cycI. perms.

= bl (x)(e71 3 - 7l2C)/ JE2E3+ cycI. perms. .

Comparing this with the scalar product of a and ex 1] yields

b1(x) =JE2E3 a](x) (cyclic permutations) .

5.4 Denote by (V f)i the components of VI with respect to the orthogonalbasis that we consider. We then have, according to the solution of Exercise 5.3,

~V'f =L(Vf)ij"jjj;dx i .i

With e= L: i ( iei a unit vector, the function ~V' f(~) = L: ;CV f)ie is the directional derivative of 1 along the direction e. This quantity can be calculatedaltematively from the total differential

d1= L g~ dxi

i

to be

A L 01 k i A L 1 01 id1(e) = ~ ( dx (ek) = n;;-~ ~ .ux' yEi ux'

i ,k i t

Comparing the two expressions yields the result

1 01(Vf)i = re sx

yEiux'

5.5 For cartesian coordinates we have EI =E2 =E 3 =1.For cylindrical coordinates (ee' ec/J ' ez ) we have ds2 = de 2+{pd4>2+dz2, i.e.,

EI =E 3 =1, E 2 =e2 and, therefore,

VI = ( 01 ~ 01 ( 1)oe ' e 0 4> ' o z .


For spherical coordinates (er, eo, e,p) we have ds2 =dr2+ r2d(J2 + r sin2 Bd4J2,which means that EI =I, E2 =r2 , E3 =r 2 sin2 Band, hence,

\7f = (~~, ~ ~; , r s:n B~~) .

5.6 The defining equation (M5.58) can be written alternatively as follows

Here ci 1 ... in is the totally antisymmetric Levi-Civitä symbol. It equals +1(-1)if (i l .•. in) is an even (odd) permutation of (1, ... , n) , and vanishes whenevertwo of its indices are equal. Thus, for n =2, *dx l =dx 2, *dx 2 = -dx l , and*w =Fl dx2 - F2dx l . Therefore, w(e) =F· e,while *w(e) =F x e. If eis adisplacement vector e=rA - r B, F a constant force, w(e) is the work of theforce along that displacement. In turn, *w(e) describes the change of the externaltorque.

5.7 For any base k-form dXi1 /\ . . . /\ dXik with i l < .. . < ik

(d il /\ /\ d ik) - . . . . d ik+l /\ /\ d in* X ' " X - CI 1 ... lk 1k+l .. . l n X . .. X .

Here we have assumed the indices on the right-hand side to be ordered, too, viz.i k+1 < . < in. The dual of this form is again a k-form and is given by

** (dXi1 /\ . . . /\ dXik) =cil ...ikik+l...incik+l ...inil ...ikdxit /\ . . . /\ dXik .

All indices i I .. . in must be different. Therefore, the set (h .. .j k) must bea permutation of Cil . . . i k ) . If we choose the ordering i, < ' " < j b theni, = i h · · · ,jk = ik . In the second s-symbol interchange the group of indicesCil, .. . , ik) with the group (ik+h"" in). For i l this requires exactly (n - k) exchanges of neighbors . The same holds true for i2 up to ik. This gives k times asign factor (- )n-k. As (Ci 1 .. . i n ? =1 for all indices different, we conclude

* * (dXi1 /\ . . . /\ dXik) =(_ )k(n-k)dxi1 /\ . . . /\ dXik .

5.8 The exterior derivative of 4J is calculated following the rule (CD3) ofSect. M5.4.4, viz.

d4J = (- ~:; + ~:~ ) dx I /\ dx2+ cyclic permutations

=(curlE), dx l /\ dx 2+ ... .

This yields the result d4J +W/ c =O.

5.9 With f a smooth function df ='L(8 f / 8xi) dx", thus

* df =(8f / 8xI) dx2 /\ dx 3+ cyclic permutations,

d(*df) =(82 f / (8x i)2) dx I /\ dx2 /\ dx3+ cyclic permutations ,

Chapter 5: Geometrie Aspeets of Meehanies 489

and

3

*d(*df) =L82f/(8xi)2 .i

Furthennore, *f = f dx I /\ dx 2 /\ dx 3 and d(*f) = O.

5.10 If (:; is a k-fonn which is applied to k vectors (e" ... ek), then, by thedefinition of the pull-back (special case, for vector spaces, of (M5.41), Sect. 5.4.1)

F* (:;(e" ... €k) =(:;(F(el), ... , F(ek))' Then

F * k 1 A A k 1 F A F A(w /\ w)(el , . . . , ek+l) =(w /\ w)( (eI), ... , (ehl)),

which, in turn, equals (F* (:;) /\ (F* J.,).

5.11 This exercise is solved in close analogy to the solution of Exercise 5.10above.

5.12 With V := y8x and W := x8y we find readily Z := [V, W] = (y8x)(x8y )

(x8y)(y8x ) =y8x - x8x .

5.13 Let VI and V2 be elements of TpM. Addition of vectors and multiplicationby real numbers being defined as in (M5.20), it is clear that both VI +V2 and aVi

with a E R belong to TpM, too. The dimension of TpM is n =dimM. TpM isa vector space. In the case M =Rn, TpM isomorphie to M .

5.14 We have

2 2

W /\ w = L L dq' /\ dp , /\ dqj /\ dpj = -2dql /\ dq2 /\ dp, /\ dP2 .i= 1 j=1

This is so because we interchanged dp, and dq", and because the tenns (i =1,] = 2) and (i = 2,] = 1) are equal.

5.15 H(I) =p2/2+ 1-cos q is the Hamiltonian function that describes the planarmathematical pendulum. The corresponding Hamiltonian vector field reads

(I) _ 8H 8H _ .X H - 8p 8q - 8q 8p - p8q - sm q8p •

A sketch of this vector field will yield the vectors tangent to the curves ofFig. M1.1O. The neighborhood of the point (p =0, q =7r) is particularly interesting as this represents an unstable equilibrium. For

1 1H(2) =_p2 + _ q(q2 _ 3)2 6


the Hamiltonian vector field is

X (2) a 1 2 aH =P q - 2" (q - I) p •

This vector field has two equilibrium points, (p = 0, q = +1) and (p = 0, q = - I) .A sketch of xW will show that the fonner is a stable equilibrium (center),while the latter is unstable (saddle point). Linearization in the neighborhood ofq = +1 means that we set u := q - I and keep up to linear tenns in u only.Then xW ~ pau - uap. This is the vector field of the hannonic oscillator or,

equivalently, the vector field xW above, for small values of q.Linearization of the system in the neighborhood of (p = 0, q = - I), in turn,

means setting u := q + I so that X~) ~ pau + uap. Here the system behaves like

the mathematical pendulum (described by xW above) in the neighborhood ofits unstable equilibrium (p = 0, q = 11") where sin q = - sin(q - 11") ~ - (q - 11") .(See also Exercise 6.8.)

5.16 One finds XHo = paq - qap, XH = paq - (q + cq2)ap, and, finally

W (XH , X HO) = dH (XHO) = cpl = {HO , H} .

5.17 For the proof consult for example Sect. 3.5.18 of Abraham and Marsden(1981).

Chapter 6: Stability and Chaos

6.1 (i) A is already diagonal. The flux is

exp(tA) = (e~l et~2) '

(ii) The characteristic exponents (i.e., the eigenvalues of A) are .A( = a + ib,.A2 = a - ib so that in the diagonalized form the flux reads as folIows: With

~--+'I1:=U~, . Ä=UAU-1=( a+ib 0 .)- - - 0 a - Ib '

we have

o ( et{a+ib) 0)'I1:(t) = exp(rA)'I1:(O) = 0 et(a-ib) '11:(0).

For a = 0, b > 0 we find a (stable) center. For a < 0, b > 0 we find an(asymptotically stable) node.

(iii) The characteristic exponents are equal, .A ( = .A2 == .A . For .A < 0 we againfind anode.

6.2 The flux of this system is

( a (T) = ~ T + ao(mod 1), ß(T) = -.!!..- T + ßo(mod I))211" 211"


If the ratio b/ a is rational, i.e., b/ a = m /n with m, n E Z, the system returnsto its initial position after the time T =T where T folIows from 0:0 + aT / (21r ) =0:0 (mod 1) and ßo+ bT/(27r ) =ßo (mod 1), i.e., T =21rn/a =21rm/b . We studythe example (a =2/3 , b=1) with initial condition (0:0 =1/2, ßo =0). This yieldsthe results shown in Table I and in Fig. 25. In the figure the sections of the orbitare numbered in the order in which they appear.

Table 1

T 3 9e=- 0 - 1 2 - 3

271" 4 4

1 1 5 1o - 1 - - 1 -

2 6 6 2

ß3 1

0 - 1 1 - 14 4

1 .----r----r-----r---,

ß

Fig.25

If the ratio b/ a is irrational then the flux wilI cover the torus, or the squareof Fig. 25, densely. As an example one may choose a "out of tune" at the valuea = I/Vi ~ 0.7071, keeping b= 1 fixed, and plot the flux in the square. Aspecific example is provided by two coupled oscillators, cf. Exercises 1.9 and2.9. The modes of the system obey the differential equation Üi + w (i)2u i = 0,i = 1,2. These are rewritten in terms of action and angle variables I, and Bi,respectively, by means of the canonical transformation (M2.95). They becomet, = 0, 8i =w(i), i = 1,2. Embedded in the phase space which is now fourdimensional, we find two two-dimensional tori which are determined by the givenvalues of I, =I?=const. Each of these tori carries the flux described above.

6.3 The Hamiltonian function has the form H =p2/(2m) + U(q). The characteristic equation

_1_ (8So (q, 0:» )2 + U(q) =Eo2m 8q


is integrated by quadrature:

So(q,a) = r J2m (Eo - U(q')) dq' .lqo

We have p =m4 =aso/aq =V2m(Eo - U(q)) and hence

lq m , es« ,

t(q) - t(qo)=. dq = - .qo V2m (Eo - U(q')) aEo

(This holds away from equilibrium positions, in case the system possesses anyequilibria.) Choose P == a =E o. Then Q=aso/ aEo =t - to, or, alternatively,(F =0, Q= I). Thus, we have achieved rectification of the Hamiltonian vectorfield: In the coordinates (P, Q) the particle moves on the straight line P =Eowith velocity Q=1.

Consider now an energy in the vicinity of E o, E = ßEo, with ß not farfrom I. Let the particle travel from qo to a point q' in such a way that thetime t(q') - t(qo) is the same as for Eo. Of course, p~ =v2m(E - U(qo)), andp' =v2m(E - U(q')), and

r mdxt - to =lqo v2m(E - U(x)) .

The new momentum is chosen to be P =Eo (the energy of the first orbit). Thenone has

Q = aS(q, E) =ßaS(q, E) =ß(t _ t )aEo oE 0 ,

i.e., (F =0, Q=ß) . In the new coordinates the particle again moves along thestraight line P =Eo, this time, however, with velocity ß. The particles on theorbits to be compared move apart linearly in time. If U(q) is such that in someregion of phase space all orbits are periodic, one should transform to action andangle variables, I(E) =const., 8 =w(E)t + 8 0• Also in this situation one seesthat the orbits separate at most Iinearly in time. In the case of the oscillator,where w is independent of E or I, their distance remains constant.

The integration described above is possible only if Eis [arger than the maximum of U(q). For E =Umax(q) the running time goes to infinity logarithmically(cf. Sect. M1.23) . The statement of this exercise does not apply when one of thetrajectories is a separatrix .

6.4 For Il =°the system becomes 4\ = aH/ aq2 and 42 = -aH/ aq\, withH =-)..(qr+q~)/2+(q\q~-qU3)/2. The critical points (where the Hamiltonianvector field vanishes) are obtained from the system of equations - )..q2 +q\q2 =0,)..q\ + (qr - q~)/2 = 0. One finds the following solutions: Po = (q\ = 0, q2 =0), P\/2 = (q\ = ).., q2 = ±v'3 )..), P3 = (q\ = -2)", q2 = 0). Linearization inthe neighborhood of Po leads to 4\ ~ -)..q2, 42 ~ )..q\. Thus Po is a center.


Linearization in the neighborhood of PI is achieved by the transfonnation U I :=ql - >., U2 := q2 - v'3 >., whereby the differential equations become.

UI =v'3 >'UI + UIU2 ~ v'3 >'UI ;

Uz = 2>'ul - v'3 >'uz + (uT - uD /2 ~ 2>'ul - v'3 >'~z .

The flux tends to PI along UI =0 but tends away from it along Uz =O. Thus PIis a saddle point. The same is true for P2 and P3. One easily verifies that thesethree points belong to the same energy E =H(P;) =-2>.3/3 and that they arepairwise connected by separatrices. Indeed, the straight lines q2 =±(ql+2>')/v'3and ql = >. are curves with constant energy E = -2>.3/3 and build up the triangle(PI, Pz, P3 ) .

If one switches on the damping tenns by means of 1~ 1-" .> 0, Po is still anequilibrium point because in the neighborhood of (ql =0, q2 =0) we have

(:~) ~ (-: =~) (:~) == A ( :~) .

From the equation det(x TI - A) =0 one finds the characteristic exponents to beXI/2 =-I-" ± i>' . Thus Po becomes anode (a sink). The points PI, P2, and P3,however, are no longer equilibrium positions. The lines which connect them arebroken up.

6.5 We write H in two equivalent fonns

(i) H =11 + Iz with 11 = (PT + ql) /2 + qi!3 ,

Iz =(p~ + qD /2 - qV3 ,

(ii) H=(PT-t:pD/2+U(ql ,q2) with U=(qT+qD/2+(qi-qD/3

= (17z+<:1Z) /4+17z<:1/4+<:13/12 ,

where 17:= ql + q2 , <:1 := ql - qz .

Then the equations of motion are

(it =PI, q2 =P2 ,. 2 · ZPI = -ql - ql, P2 = -qz + q2 .

The critical points of this system are Po : (ql = 0, q2 = 0, PI = 0, P2 = 0),PI : (0, 1,0,0), Pz : (-1,0,0,0), P3 : (-1, 1,0,0). One easily verifies thatd1;jdt =0, i =I, 2, i.e., 11 and Iz are independent integrals of the motion. Thepoints PI and Pz Heon two equipotential lines, viz. the straight line ql - q2 =-I,and the ellipse 3(ql + q2? + (ql - q2)2 + 2(ql - q2)- 2 =O. In either case U =1/6.As an example and using these results, one may sketch the projection of the fluxonto the plane (ql , q2).

6.6 The critical points of the system q = P, P= q - q3- P are Po : (q = 0, P = 0),PI : Cl, 0), and P2 =(-1 ,0). Linearization around Po gives


The eigenvalue of Aare AI/2=(-1 ± 0)/2, hence AI > °and A2 < °whichmeans that Po is a saddle point. Linearizing in the neighborhood of PI andintroducing the variables u := q - 1, v := p, the system becomes

The characteristic exponents are J-LI/2 =(-1 ±h!7)/2. The same values are foundfor the system linearized in the neighborhood of P2• This means that both PIand P2 are sinks.

The Liapunov function V (q, p) has the value °in Po, and the value -1/4 inPI and P2• One easily verifies that PI and P2 are minima and that V increasesmonotonically in a neighborhood of these points. For instance, close to PI takeu := q-l , v := p. Then PI(U, v) := V(q = u+l ,p = v)+1/4 = v2/2+u2+u3+u4/4.Indeed, at the point PI we find PI (0,0) =°while PI is positive in a neighborhoodof PI.

Along solutions the function V(q,p), or, equivalently, PI(U,V), decreasesmonotonically. Let us check this for V :

dV dV . sv . sv 3 sv 2- =-p+-q= -(q-q -p)+-p= -p .dt ßp ßq ßp ßq

In order to find out towards which of the two sinks a given initial configurationwill tend, one has to calculate the two separatrices that end in Po. They form theboundaries of the basins of PI and P2 as indicated in Fig. 26 by the blank anddotted areas, respectively .

6.7 As xo, by assumption, is an isolated minimum, a Liapunov function ischosen as fOllOWS: V(x) := U(x) - U(xo). In a certain neighborhood M of xo,V(x) is positive semi-definite and we have

d n sv n (ßU)2- V (X) = '"' -Xi = - '"' - .dt - L B», L B»,

i=1 I i=1 I

Ifwe follow a solution in the domain M\{xo}, V(x) decreases, i.e., all solutioncurves tend "inwards", towards xo. Thus, this point is asymptotically stable.

In the example U(Xt,X2) = xI(xl - 1)2 + x~. The points Xo = (0,0) andx~ =(l, 0) are isolated minima and, hence, are asymptotically stable equilibria.

6.8 This system is Hamiltonian. A Hamiltonian function is H =p2/2 + q(q2 3)/6. The phase portraits are obtained by drawing the curves H(q,p) =E =const.The Hamiltonian vector field VH =(p, (l - q2)/2) has two critical points whosenature is easily identified by linearizing in their neighborhoods . One finds


~ ~ ~ ~ ~> ~ ~ ~ ~ ~ ~ ~ ~ Fig.26......... .. ............... .. .......... .. ..... ..... .. .......... .. .......... .. .......... .. .......... .. ....... ... ., ... ...... . .. .......... , ......... . ., .......... .. .......... , .......... .. .......... , .......... .. .......... .. .......... .. .......... .. .

....... .. ........ .. ....... , ....... .. .:: ::: : ".:q::

...................................

..............................................

PI : (q == -I ,p == 0) and, with u :== q + I , v :== p : (ü ~ v , iJ ~ u). Thus, PI is asaddle point.

P2 : (q == I ,p == 0) and, with 'Ü :== q - I , v :== p : (ü ~ v, iJ ~ - u). Thus, P2 is acenter. In the neighborhood of P2 there will be hannonic oscillations withperiod 211" (see also Exercise 5.15).

6.9 The differential equation ij == f(q , q), with f(q, q) == -q+(c - q2)q, is solvednumerically by means of a Runge-Kutta procedure as follows

qn+1 == qn + h (qn+ ~ (kl + k2 + k3)) + O(h5)

1qn+1 == qn + 6(kl + 2k2 + 2k3 + k4 ) ,

h being the integration step in the time variable, and the auxiliary quantities k,being defined by

k l == hf(qn, qn) ,

k2 == hf (qn + %qn + ~k"qn + ~kl) ,

k3 == hf (qn + %qn + ~kl ,qn + ~k2) ,k4 == hf (qn + hqn + %k3, qn+ k3) .

One lets the dimensionless time variable T == wt run from 0 to, say, 611", insteps of 0.1, or 0.05, or 0.01. This will produce pictures of the type shown


in Figs. (M6.6)-(M6.8). Alternatively, one may follow the generation of thesefigures on the screen of a pe. One will notice that all of them tend quickly tothe attractor.

6.10 The program developed in Exercise 6.9 may be used to print out, for agiven initial configuration, the time T and the distance from the origin d, eachtime the orbit crosses the line p =q. One finds the following result:

T 5.46 11.87 18.26 24.54 30.80 37.13 43.47p=q>O:

d 0.034 0.121 0.414 1.018 1.334 1.375 1.378

T 2.25 8.86 15.07 21.42 27.66 33.96 40.30p=q<O:

d 0.018 0.064 0.227 0.701 1.238 1.366 1.378

Plotting In d versus T shows that this function increases approximately linearly(with slope ~ 0.1) until it has reached the attractor. Thus, the point of intersection of the orbit and the straight line p =q wanders towards the attractor atan approximately exponential rate. One finds a similar result for orbits whichapproach the attractor from the outside.

6.11 The motion manifold of this system is R2• For the linearized system

(XI = XI, X2 = -X2) the straight line Ustab = (XI = 0, X2 ) is a stable submanifold.Indeed , the velocity field points towards the equilibrium point (0, 0) and thecharacteristic exponent is -1. The straight line Uunst = (Xl , X2 = 0), in turn,is an unstable submanifold: The velocity field points away from (0, 0) and thecharacteristic exponent is +1. The full system can be transformed to

2 •X I =X2 + X2 ,

whose general solution is

X2(t ) =aexp(2t) + bexp(-t) , XI (t ) =~expt ,

or, equivalently,

1 2 ;;;-:-11 2 CX2 =-XI + bv 3a- == - x I + - .

3 XI 3 XI

Among this set of solutions the orbit with c =0 goes through the point (0,0) andis tangent to Uunst in that point. On the submanifold Vunst = (x I , X2 = xV 3) thevelocity field moves away from (0,0).

The corresponding stable submanifold of the full system coincides with Ustab

because, with a = 0, X I (t) = 0, X2(t ) = bexp(-t) which means that "Vstab = (XI =0, X2).

6.12 We have Xn+1 = 1 - 2x ;, and Yi = 4/1r arcsin V(Xi + 1)/2 - 1. With-1 S; Xi S; 0 also -1 S; Yi S; 0, and with 0 S; X i S; 1 also 0 S; Yi S; 1. We

-l~x~O,

for 0 ~ x ~ 1 .


wish to know the relation between Yn+1 and Yn' First, the relation Xn -+ Yn+1 isYn+1 =4/1r arcsin(l- x;Y/2 - 1. Using the addition theorem arcsin u +arcsin v =arcsin(uv'f=V2 + vVf=U2) and setting u =v =(l + x)/2 one shows

arcsin~ =2 arcsin Jx ; 1 for

• r;--::;12 • r;+1arcsm VI - z- =1r - 2arcsm V~-2-

In the first case Yn ~ 0 and Yn+1 = 1 + 2Yn, in the second case Yn 2: 0 andYn+1 = 1 - 2Yn' These can be combined to Yn+1 =1 - 21Ynl . The derivative ofthis iterative mapping is ±2; its magnitude is 1arger than 1. There are no stab1efixed points.

6.15 If, for instance, m = 1, then Zrr := exp(i21ro"/n) are the roots of theequation z" - 1 = (z - zd . . . (z - zn) = O. They 1ie on the unit circ1e in thecomp1ex plane and neighboring roots are separated bythe angle 21rIn. Expandingthe product (z - Zl)'" (z - zn) =z" - Z 2:;=1 Zrr + . . ., one sees that, indeed,2:;=1 Zrr = O. For the va1ues m = 2, .. . , n - 1 we renumber the roots andobtain the same resu1t. For m := 0 and for m =n, however, the sum is equa1 toO. Mu1tip1ying xrr =2:;=1 x rexp(-2i1r17T/n)/vn by 1/vnexp(2i1r17A/n) andsumming over 17 one obtains

This is used to calcu1ate

1 n 1 .9 - - ~ x x - - ~ x x* ~ e217r / n (rr(p +v)+AV)

A - n W a rr+A - n2 W P n-v W .rr=1 P ,v a

The orthogonality relation imp1ies J.l + v =0 (mod n), and we have used x~_v =xv. Furthermore, Xpmodn =xp and, fina11y, xp and xn - p have the same modu1us.It follows that 9A = l/n 2::=llxpI2cos(21rAJ.l/n). The inverse of this formu1a

Ix rrl2 =2:~=1 9A cOS(21r17A/n) is obtained in the same way.

6.16 If we set Y =ax + ß, i.e., x =Y/ a - ß/o, the relation

Xi+1 =J.lXi(l - Xi) =J.l (± Yi - ~) (1 + ~ - ±Yi)

will take the desired form provided a and ß are chosen such that they fulfillthe equations o + 2ß =0, ß(l - J.l(a + ß)/a) =1. These give a =4/(J.l - 2),ß = -a/2, and, therefore, , = J.l(J.l - 2)/4. From 0 ~ J.l < 4 follows 0 ~

, < 2. One then sees easily that Yi E [-1,+1] is mapped onto Yi+1 in the sameinterval. Let h(y,,) := 1_,y2• The first bifurcation occurs when h(y,,) = Y and

(2)

(I)


oh(y,,)/oy = 1, i.e., when ,0 = 3/4, Yo = 2/3, or, correspondingly, J1.0 = 3/4,Xo = 2/3. Take then k := ho h, i.e., k(y,,) = 1 -,(I _,y2)2. The secondbifurcation occurs at ,I =5/4. The corresponding value YI of y is calculatedfrom the system

k (y , ~) =_~ + 2: y2 _ ~ y4 =Y ,

ok (y ~) =25 y (1 _~ y2) =-1oy ' 4 4 4 .

Combining these equations according to y . (2)-4 . (I), one finds the quadraticequation

2 4 4Y -Sy-

25= 0 ,

whose solutions are YI/2 = 2(1 ± V2)/5. From ,I = 5/4 we have J1.1 = 1 + J6,from this and from YI/2, finally

IXI/2= 1O(4+V6± (2V3-h)) =0.8499 and 0.4400.

Appendix

A. Some Mathematical Notions

"Order" and "Modulo". The symbol O(c:n) stands for terms of the order s"

and higher powers thereof that are being neglected.

Example. If in a Taylor series one wishes to take account only of the first threeterms, one writes

f(x) = f(O) + f'(O)x + i f"(0)x2 + O(x3) • (A.l)

This means that the right-hand side is valid up to terms of order x3 and higher.The notation y =x (mod a) means that x and x + na should be identified, n

being any integer. Equivalently, this means that one should add to x or subtractfrom it the number a as many times as are necessary to have y fall in a giveninterval.

Example. Suppose two angles (}' and ß are defined in the interval [0, 27r]. Theequation (}' = f(ß) (mod 27r) means that one must add to the value of the functionf(ß), or subtract from it, an integer number of terms 27r such that (}' does notfall outside its interval of definition.

Mappings. A mapping f that maps a set A onto a set B is denoted as folIows:

f:A--+B: al-+b. (A.2)

It assigns to a given element a E A the element bEB. The element b is said tobe the image and a its preimage.

Examples . (i) The real function sin x maps the real x-axis onto the interval[-1, +1],

sin: R --+ [-1 , +1] : x f-4 y = sin x .

(ii) The curve 'Y: x = cos wt, Y = sinwt in R2 is a mapping from the real t-axisonto the unit circle Si in R2,

'Y : Rt --+ Si : t f-4 (x = coswt, Y = sinwt) .

A mapping is called surjective if f(A) = B, i.e. if Bis covered completely. Themapping is called injective if two distinct elements in A also have distinct imagesin B, in other words, if every bEB has one and only one original a E A. Ifit has both properties it is said to be bijective . In this case every element of Ahas exactly one image in B, and for every element of B there is exactly onepreimage in A. In other words the mapping is then one-to-one.

500 Appendix

The composition fog of two mappings fand 9 means that 9 should beapplied first and then f should be applied to the result of the first mapping, viz.

If g: A -+ B and f: B -+ C, then fog : A -+ C . (A.3)

(AA)

Example. Suppose f and gare functions over the reals. Then, with y = g(x) andz = f(y) we have z =(f 0 g)(x) = f(g(x)) .

The identicalmapping is often denoted by "id", i.e.

id: A -+ A: a I-t a .

Special Properties of Mappings. A mapping f: A -+ B is said 10 be continuousat the point u E A if for every neighborhood V of its image v = f(u) E B thereexists a neighborhood U of u such that f(U) c V. The mapping is continuousif this property holds at every point of A.

Homeomorphisms are bijective mappings f: A -+ B that are such that bothfand its inverse f-1 are continuous.

Diffeomorphisms are differentiable, bijective mappings f that are such thatboth fand f-1 are smooth (i.e. differentiable, COO).

Derivatives. Let f(x 1, x2, ... , z ") be a function overthe space Rn, {€1 , ... , €n},a set of orthogonal unit vectors. The partialderivative with respect to the variablex i is defined as follows:

o f. = lim f(x + h€ i) - f( x)o x' h -.O h

Thus one takes the derivative with respect to xi while keeping all other arguments1 i - I i+l n fi edx, .. . ,x , x , ... ,x x .

Collecting all partial derivatives yields a vector field called the gradient,

v f = (:~ , .. . , :~) . (A.5)

Since any direction n in Rn can be decomposed in terms of the basis vectors€1, ... , €n, one can take the directional derivative of f along that direction, viz.

(A.6)

(A.7)

The total differential of the function f(x 1, ••• , z ") is defined as follows:

df = of dx1 + of dx2 + . . . + of dz" .ox1 OX2 Ox n

Examples. (i) Let f(x , y) = ~(x2 + y2) and let (z = r cos 4> , y = r sin 4» withfixed r and 0 ~ 4> ~ 271" be a circle in R2• The normalized vector tangent tothe circle at the point (x , y) is given by V t = (- sin 4>, cos 4» . Similarly, thenormalized normal vector at the same point is given by V n = (cos 4>, sin 4» . The

A. Some Mathematical Notions 501

total differential of f is df = xdx + ydy and its directional derivative along Vt

is Vt • V f = -x sin rP + Y cos rP = 0; the directional derivative along V n is givenby V n • V f =x cos rP + y sin rP =r. For an arbitrary unit vector v =(cos a, sin o)we find v . V f = r(cos rP cos o + sin rP sin o), For fixed rP the absolute value ofthis real number becomes maximal if a = rP (mod 1l'). Thus, the gradient definesthe direction along which the function f grows or falls fastest.

(ii) Let U(x, y) = xy be a potential in the plane R2. The curves along whichthe potential U is constant (they are called equipoteniial /ines) are obtained bytaking U(x, y) = c, with c a constant real number. They are given by y = cf»,i.e. by hyperbolas whose center of symmetry is the origin. Along these curvesdU(x, y) = ydx + xdy =0 because dy = -(c/x2)dx = -(y/x)dx. The gradientis given by VU = (aUlax, au/ay) = (y , x) and is perpendicular to the curvesU(x, y) = c at any point (z, y). It is the tangent vector field of another set ofcurves that obey the differential equation

dy x-=-dx y

These latter curves are given by y2 - x2 = a.

DitTerentiability of Functions. The function f(x 1, • • • , xi, ... , z ") is said to beer with respect to xi if it is r-times continuously differentiable in the argumentxi. A function is said to be e= in some or all of its arguments if it is differentiablean infinite number of times. It is then also said to be smooth.

Variables and Parameters. Physical quantities often depend on two kinds ofarguments, the variables and the parameters. This distinction is usually madeon the basis of a physical picture and, therefore, is not canonical. Generally,variables are dynamical quantities whose time evolution one wishes to study.Parameters, on the other hand, are given numbers that define the system underconsideration. In the example of a forced and damped oscillator, the deviationx(t) from the equilibrium position is taken to be the dynamical variable whilethe spring constant, the damping factor, and the frequency of the driving sourceare parameters.

Lie Groups. The definition assurnes that the reader is familiar with the notion ofdifferentiable manifold, cf. Sect. 5.2. A Lie group is a finite dimensional, smooth,manifold Gwhich in addition is a group and for which the product " . ",

G x G --t G : (g,g') t-t 9 . g' ,

as well as the transition to the inverse,

G --t G : 9 t-t g-I

are smooth operations.G being a group means that it fulfiIls the group axioms, cf. e.g., Sect . 1.13:

There exists an associative product; G contains the unit element e; for every

502 Appendix

g E G there exists an inverse g-I E G. Loosely speaking, smoothness means thatthe group elements depend differentiably on parameters, which may thought ofas angles for instance, and that group elements can be deformed in a continuousand differentiable manner.

A simple example is the unitary group

U(l) ={eiala E [0,27rl} .

This is an Abelian group (i.e ., a commutative group). Further examples areprovided by the rotation group SO(3) in three real dimensions, and the unitarygroup SU(2) which are dealt with in Sects. 2.21 and 5.2.3 (iv). The Galilei groupis defined in Sect. 1.13, the Lorentz group is discussed in Sects.4.4 and 4.5 .

B. Historical Notes

There follow some biographical notes on scientists who made important contributions to mechanics. Without striving for completeness, the founders of mechanicsproper are marked by an asterisk and are treated in somewhat more detail.

*d'Alembert, Jean-Baptlste: born 17 November 1717 in Paris, died 29 October1783 in Paris. Writer, philosopher, and mathematician. Co-founder of the Encyclopedie. Important contributions to mathematics, mathematical physics and astronomy. Hisprincipal work "Traite de dynamique" contains the principle whichbears his name.

Arnol'd, Vladimir Igorevich: 1936-, Russian mathematician.

Cartan, EHe: 1869-1951, French mathematician.

Corlolis, Gustave-Gaspard: 1792-1843, French mathematician.

Coulomb, Charles Augustin: 1736-1806, French physicist.

*Descartes, Rene (Cartesius): born 31 March 1596 in La Haye (Touraine), diedII February 1650 in Stockholm. French philosopher, mathematician and naturalphilosopher. In spite of the fact that Descartes' contributions to mechanics werenot too successful (for instance, he proposed incorrect laws of collision), hecontributed decisively to the development of analytic thinking without whichmodem natural science would not be possible. In this regard his book Discoursde La Methode pour bien conduire sa Raison, published in 1637, was particularlyimportant. Also his imaginative conceptions - ether whirls carrying the planetsaround the sun; God having given eternal motion to the atoms relative to theatoms of the ether which span our space; the state of motion of atoms beingable to change only by collisions - inspired the amateur researchers of the 17thcentury considerably. It was in this community where the real scholars of sciencefound the resonance and support which they did not obtain from the scholasticand rigid attitude of the universities of their time .

B. Historical Notes 503

*Einstein, Albert: born 14 March 1879 in Ulm (Germany), died 18 April 1955in Princeton, N.J. (U.S.A.). German-Swiss physicist, 1940 naturalized in theU.S.A. His most important contribution to mechanics is Special Relativity whichhe published between 1905 and 1907. In his General Relativity, published between 1914 and 1916, he succeeded in establishing a (classical) description ofgravitation as one of the fundamental interactions. While Special Relativity isbased on the assumption that space-time is the flat space R4, General Relativity is a dynamical and geometrie field theory which allows one to determine themetric on space-time from the sourees, i.e., from the given distribution of massesin the universe.

*Euler, Leonhard: born 15 April 1707 in Basel (Switzerland), died 18 September 1783 in St. Petersburg (Russia). Swiss mathematician. Professor initially ofphysics, then of mathematics at the Academy of Seiences in St. Petersburg (from1730 until 1741, and again from 1766 onwards), and, at the invitation of Frederick the Great, member of the Berlin Academy (1741-1766). Among his giganticscientific work particularly relevant for mechanics: development of variationalcalculus; law of conservation of angular momentum as an independent principle; equations of motion for the top. He also made numerous contributions tocontinuum mechanics.

*Galilei, Galileo: born 15 February 1564 in Pisa (Ita1y), died 8 January 1642in Arcetri near Florence (Italy). Italian:mathematician, natural philosopher, andphilosopher, who belongs to the founding-fathers of natural sciences in the modem sense. Professor in Pisa (1589-1592), in Padua (1592-1610), mathematicianand physicist at the court of the duke of Florence (1610-1633). From 1633 onunder confinement to his house in Arcetri, as a consequence of the conflict withPope Urban VIIland the Inquisition, which was caused by his defence of theCopernican, heliocentric, p1anetary system. Galilei made important contributionsto the mechanics of simple machines and to observationa1 astronomy. He deve1oped the laws of free fall.

*Hamilton, William Rowan: born 4 August 1805 in Dublin (Ireland), died 2September 1865 in Dunsink near Dublin. Irish mathematician, physicist, and astronomer. At barely 22 years of age he became professor at the university ofDub1in. Important contributions to optics and to dynamics. Developed the variational principle which was cast in its 1aterand more elegant form by c.G.J. Jacobi.

*Huygens, Christiaaru born 14 April 1629 in The Hague (Netherlands), died 8July 1695 in The Hague. Dutch mathematician, physicist and astronomer. From1666 unti1 1681 member of the Academy of Seiences in Paris. Although Huygens is not mentioned explicit1y in this book he made essential contributionsto mechanics: among others the correct 1aws for elastic, centra1 collisions and,building on Galilei's discoveries, the classical principle of relativity.

504 Appendix

Jacobi, Carl Gustav Jakob: 1804-1851, German mathematician.

*Kepler, Johannes: born 27 December 1571 in Weil der Stadt (Germany), died15 November 1630 in Regensburg (Gennany). German astronomer and mathematician. Led a restless live, in part due to numerous misfortunes during theturbulent times before and during the Thirty Years War, but also for reasons to befound in his character. Of greatest importance for hirn was his acquaintance withthe Danish astronomer Tyge (Tycho) Brahe, in the year 1600 in Prague, whoseastronomical data were the basis for Kepler' s most important works. SucceedingT. Brahe, Kepler became mathematician and astrologer at the imperial court, firstunder emperor Rudolf II, later under Mathias. Finally, from 1628 until his deathin 1630, he was astrologer of the duke of Friedland and Sagan, A. von Wallenstein. Kepler's first two laws are contained in his Astronomia nova (1609), thethird is contained in his main work Harmonices Mundi (1619). They were notgenerally accepted, however, until Newton's work who gave them a new, purelymechanical foundation . Kepler' s essential achievement was to overcome the ancient opposition between celestial mechanics (where the circle was believed tobe the natural inertial type of movement) on one hand and terrestial mechanicson the other.

Kolmogorov, Andrei Nikolaevic: 1903-, Russian mathematician.

*Lagrange, Joseph Louis: born 25 January 1736 in Torino, died 10 April 1813in Paris. Italian-French mathematician . At the early age of 19 professor for mathematics at the Royal School of Artillery in Torino, from 1766 member of theBerlin Academy, succeeding d' Alembert, then from 1786 member of the FrenchAcademy of Sciences, professor at Ecole Normale, Paris, in 1795 and at EcolePolytechnique in 1797. His major work Mecanique Analytique which appearedin 1788, after Newton's Principia (1688) and Euler 's Mechanica (1736), is thethird of the historically important treatises of mechanics . Of special relevancefor mechanics were his completion of variational calculus which he used to derive the Euler-Lagrange equations of motion, and his contributions to celestialperturbation theory.

*Laplace, Pierre Simon de: born 28 March 1749 in Beaumont-en-Auge(France), died 5 March 1827 in Paris. French mathematician and physicist. From1785 on member of the Academy, he became professor for mathematics at EcoleNormale (Paris) in 1794. He must have been rather flexible because he survivedfour political systems without hann. Under Napoleon he was for a short timeSecretary of the Interior. Besides important contribut ions to celestial mechanicson the basis of which the stability of the planetary system was rendered plausible, he developed potential theory, along with Gauß and Poisson. Other importantpublications of his deal with the physics of vibrations and with thennodynamics.

Legendre, Adrien Marie: 1752-1833, French mathematician.

B. Historical Notes 505

Leibniz, Gottfried Wilhelm: 1646-1716, German natural philosopher and philosopher.

Lie, Marius Sophus: 1842-1899, Norwegian mathematician.

Lyapunov, Aleksandr Mikhailovich: 1857-1918, Russian mathematician.

Liouville, Joseph: 1809-1882, French mathematician.

Lorentz, Hendrik Antoon: 1853-1928, Dutch physicist.

*Maupertuis, Pierre Louis Moreau de: born 28 September 1698 in St. Malo(Britanny, France) , died 27 July 1759 in Basel (Switzerland). French mathematician and natural philosopher. From 1731 paid member of the Academy of Seiences of France, in 1746 he became the first president of the Prussian AcademieRoyale des Seiences et BeIles Lettres, newly founded by Frederick the Great inBerlin who called many important scientists to the academy, notably L. Euler. In1756, seriously iII, Maupertuis returned first to France but then joined his friendJoh.II Bernoulli in Basel where he died in 1759. Along with Voltaire, Maupertiuswas a supporter of Newton's theory of gravitation which he had come to knowwhile visiting London in 1728, and fought against Descartes' ether whirls. Ofdecisive importance for the development of mechanics was his principle of leastaction, fonnulated in 1747, although his own fonnulation was still somewhatvague (the principle was fonnulated in precise form by Euler and Lagrange). Awidely noticed dispute of priority started by the Swiss mathematician SamuelKönig who attributed the principle to Leibniz, was eventually decided in favorof Maupertuis. This dispute alienated Mauptertuis from the Prussian Academyand contributed much to his bad state of health.

Minkowski, Hermann: 1864-1909, German mathematician.

Moser, Jürgen: 1928-, German mathematician.

*Newton,Isaac: born 25 December 1643 in Whoolsthorpe (Lincolnshire, England), died 20 March 1726 in Kensington (London), (both dates according tothe Julian calender which was used in England until 1752). Newton, who hadstudied theology at Trinity College Cambridge, learnt mathematics and naturalsciences essentially by hirnself. His first great discoveries, differential calculus,dispersion of light, and the law of gravity which he communicated to a smallcircle of experts in 1699, so impressed Isaac Barrow, then holding the "LucasianChair" of mathematics at Trinity College, that he renounced his chair in favor ofNewton. In 1696 Newton was called at the Royal Mint and became its directorin 1699. The Royal Society of London elected hirn president in 1703. Veneratedand admired as the greatest English natural philosopher, Newton was buried inWestminster Abbey.

506 Appendix

His principal work, with regard to mechanics, is the three-volume PhilosophiaNaturalis Principia Mathematica (1687) which he wrote at the instigation ofhis pupil Halley and which was edited by Halley. Until this day the Principiahave been studied and completely understood only by very few people. Thereason for this is that Newton 's presentation uses a highly geometricallanguage,divides matters into "definitions" and "axioms" which mutually complete andexplain one another, following examples from antiquity, in a manner difficult tounderstand for us, and because he presupposes notions of scholastic and Cartesianphilosophy we are normally not familiar with. Even during Newton 's lifetime ittook a long time before his contemporaries learned to appreciate this difficult andcomprehensive work which, in addition to Newton 's laws, deals with a wealthof problems in mechanics and celestial mechanics, and which contains Newton'soriginal ideas about space and time that have stimulated our thinking ever since.Newton completed a long development that began during antiquity and wasinitiated by astronomy, by showing that celestial mechanics is determined onlyby the principle of inertia and the gravitational force and, hence, that it followsthe same laws as the mechanics of our everyday world. At the same time he laidthe foundation for a development which to this day is not concluded.

Noether, Emmy Amalie: 1882-1935, German mathematician. Belongs to thegreat scientific personalities in the mathematics of the 20th century.

"Poincare, Jules Henri: born 29 April 1854 in Nantes (France) , died 17 July1912 in Paris . French mathematician , professor at Sorbonne university in Paris .Poincare, very broad and extraordinarily productive, made important contributions to the many-body problem in celestial mechanics for which he received aprize donated by King Oscar 11 of Sweden. (Originally the prize was announcedfor the problem of convergence of the celestial perturbation series, the famousproblem of the "small denominators" that was solved much later by Kolmogorov,Arnol'd, and Moser.) Poincare may be considered the founder of qualitative dynamics and of the modem theory of dynamical systems.

Poisson, Sirneon Denis: 1781-1840, French mathematician .

Bibliography

Mechanics Textbooks and Exercise Books

Fetter, A.L., Walecka , J.D.: Theoret ical Mechan ics 0/ Particles and Continua (McGraw Hill, NewYork 1980)

Goldstein, H.: ClassicalMechanics (Addison-Wesley, Reading 1980)

Honerkamp, J., Römer, H.: Theoretical Physics (Springer, Berlin , Heidelberg 1993)

Landau , L.D., Lifshitz, E.M.: Mechanics (Pergamon , Oxford and Addison-Wesley, Reading 1971)

Spiegel , M.R.: General Mechanics, Theory and Applications (McGraw Hill, New York 1976)

Some General Mathematical Texts

Amol'd, V.I.: Ordinary Different ial Equations (Springer, New York, Berlin , Heidelberg 1992)Amol'd, V.I. : Geometrical Methods in the Theory ofOrdinary Differential Equations (Springer, New

York, Berlin, Heidelberg 2nd ed. 1988)

Flanders, H.: Differential Forms (Academi c, New York 1963)

Hamermesh, M.: Group Theory and lts Applications to Physical Problems (Addison-Wesley, Reading1962)

Special Functions and Mathematical Formulae

Abramowitz, M., Stegun , LA.: Handb ook 0/ Math emat ical Funct ions (Dover, New York 1965)Gradshteyn, I.S., Ryzhik, I.M.: Tables 0/Integrals. Series and Products (Academic, New York 1965)

Chapter 2

Rüßmann, H.: Konvergente Reihenentwicklungen in der Störungstheorie der Himmelsmechanik, Selecta Mathematica V (Springer, Berlin, Heide1berg 1979)

Rüßrnann, H.: Non-degeneracy in the Perturbation Theory 0/fntegrable Dynamical Systems, in Dodson, M.M., Vickers , J.A.G. (eds.) Number Theory and Dynam ical Systems, London MathematicalSociety Lecture Note Series 134 (Cambridge University Press 1989)

Chapter 4

Hagedorn. R.: Relativistic Kinematics (Benjamin, New York 1963)

Jackson, J.D.: Classical Electrodynamics (Wiley, New York 1975)

Sexl, R.U., Urbantke, H.K.: Relativität. Gruppen. Teilchen (Springer, Berlin, Heide1berg 1976)

Weinberg , S.: Gravitation and Cosmology, Principles and Applications 0/ the General Theory 0/Relativity (Wiley, New York 1972)

508 Bibliography

Chapter 5

Abraham, R., Marsden , J.E.: Foundations 0/Mechanics (Benjamin Cummings, Reading 1981)Amol 'd , V.I.: Mathematical Methods ofClassical Mechanics (Springer, New York, Berlin, Heidelberg

2nd ed. 1989)Guillemin , V., Stemberg, S. : Symplectic Techniques in Physics (Cambridge University Press, New

York 1990)O'Neill, B.: Semi-Riemannian Geometry , With Appli cations to Relativity (Academic, New York 1983)Thirring , W.: A Course in Mathematical Physics , Vol. 1: Classical Dynamical Systems (Springer,

Berlin, Heidelberg 2nd ed. 1992)

Chapter 6

Amol'd, V.I.: Catastrophe Theory (Springer, Berlin, Heidelberg 1986)Berge, P., Pomeau, Y., Vidal, C.: Order within Chaos; Towards a Deterministic Approach to Turbu

lence (Wiley, New York 1986) French original (Hermann, Paris 1984)Chirikov, B.V.: A Universallnstability 0/Many-Dimensional Oscillator Systems, Physics Reports 52,

263 (1979)Collet, P., Eckmann, J.P.: Iterated Maps on the Interval as Dynamical Systems (Birkhäuser, Boston

1990)Devaney, R.L.: An lntroduction to Chaotic Dynamical Systems (Benjamin Cummings, Reading 1989)Feigenbaum, M.: J. Stat, Phys. 19,25 (1978) and 21, 669 (1979)Guckenheimer, J., Holmes, P.: Nonlinear Oscillations , Dynamical Systems, and Bifurcations ofYector

Fields (Springer, Berlin, Heidelberg 1990)Henon, M., Heiles, C.: Astron. Joum. 69, 73 (1964)Hirsch, M.W., Smale, S.: Differential Equations , Dynamical Systems and Linear Algebra (Academic,

New York 1974)Palis, J., de Melo, W.: Geometrie Theory 0/Dynamical Systems (Springer, Berlin, Heidelberg 1982)Peitgen, H.O., Richter, P.H.: The Beauty 0/ Fractals, Images ofComplex Dynamical Systems (Springer,

Berlin, Heidelberg 1986)Ruelle, D.: D.: Elements 0/ Differential Dynamics and Bifurcation Theory (Academic, New York

1989)Schuster, H.G.: Deterministic Chaos, An lntroduc tion (Physik-Verlag, Weinheim 1987)Wisdom, J.: Chaotic Behaviour in the Solar System, Nucl. Phys. B (Proc. Suppl.) 2, 391 (1987)

Subject Index

Acceleration 4Action 88,312Action function 312Action integral 88Action-angle variables 114,151 ,349Addition of velocities 234Adiabatic invariants 159Angular momentum- of rigid body 181- of single particle 13Angular-momentum conservation,

principle of 20Angular velocity 170Aphelion 46Apocenter 46Asymptotic stability 329,340Atlas 262-, complete 263Attractor 337,342-, strange 337,361,373Atwood's machine 84Autonomous systems 37Averaging principle 158Axial vector 289

Base differential forms 278Base integral curve 3.15Base space 272Base vector fields 271,275Basin of attractor 343Bifurcations 350--.:; Hopf 353-, pitchfork 353- , saddle-node 352-, transcritical 352Body cone 196Boost, see special Lorentz transformationBundle chart 273

Canonical equations 98,115,301Canonical one-form 294Canonical systems 98

Canonical transformations 110, 117-, infinitesimal 133Canonical two-form 296Cantor's middle third set 374Cartan derivative 286Center 332Center-of-mass

coordinates 11momentum 18motion, principle of 19system 55

Central force 13Centrifugal force 51Chaos 356, 358Characteristic exponents 328Characteristic multipliers 348Chart 261Children's top 178,203Circular orbit 16,405Closed form 296Commutators

of generators of Lorentz group 232- of generators of rotation group 109- of vector fields 276Composition of mappings 426Cone- , body 196- of nutation 196-, space 196Conformal group 253Constant of the motion, see integral

of the motionConstraints 77Contravariant index 217Coordinate functions, natural 260Coordinate manifold 258, 290Coordinate system (for rigid bodies)- , body-fixed 169-, intrinsic 169- , space-fixed 169Coordinates-, canonical 300- , cyclic 95, 110

510 Subject Index

-, generalized 79-, local 258, 261Coriolis force 51, 409Cotangent bundle 258, 277Cotangent space 257,277Coulomb force 8, 62Coulomb scattering 61, 62Covariant derivative 94Covariant equation 242Covariant index 217Critical elements (of vector fields) 340Critical points 326,333Cross section-, differential 59- , total 60Curl 28Curves on manifolds 268Cyclic coordinates 95, 110

D' Alembert ' s principle 80Darboux 's theorem 300Decomposition theorem 228Degree of freedom 35Derivative

directional 270,500- , exterior 286- , exterior in R3 287-, Lie 276,303-, partial 500-, total 500-, variational 87Deterministic motion 36Diffeomorphism 93,326,500Differential equation with separable

variables 40Differential form 277Differential mapping 280Dilatations 253Dissipative systems 338Dynamical systems 322,326

Eigenfrequencies 162Energy conservation, principle of 20Energy function 312Energy-momentum vector 244Equilibrium position 144,326Eulerian angles 185,186Euler-Lagrange equations 89,391Euler's differential equation 87Euler's equations 190, 192Event, see world pointExterior forms 285Exterior product 284

Fermat's principle 86,417Fiber 238,272Fiber bundle 238,272Fibonacci 's numbers 370Field strengths, tensor of 247Finite orbit 16,41Flow- , global 326-, local 325-, maximal 325

in phase space 37of a vector field 324

Flow fronts 325Force- , centrifugal 51

conservative 28of constraint 80

- , external 7-, fictitious 5,51- , generalized 81-, internal 7-, Lorentz 90Force field 7Force law, relativistic 242Form invariance 5,242Forms-, canonical, of mechanics 294,296-, exterior 285Fractal 373Frequency doubling 367Functions on manifolds 268

Galilei group 24- proper, orthochronous 23Galilean space-tirne 221,237Gauge transformations 91,92Generating function

for canonical transformations 111- for infinitesimal canonical

transformations 133Generators

for infinitesimal Lorentz transformations 231for infinitesimal rotations 104,108for infinitesimal symplectictransformations 333

Golden Mean 370Gradient 500Gradient flow 348,427Gravitational constant 8,10Gravitational force 8Group axioms 24, 222Group-, conformal 253

-, Galilei 24-, K1ein's 223,409-, Lorentz 222-, Poincare 222

special orthogonal 106-, special unitary 265-, symplectic 119

Hamiltonian density 393Hamilton-Jacobi differential equation 137Hamiltonian function- for relativistic motion 252- for rigid body 20IHamiltonian systems 98,302,338-, time-dependent 307Hamiltonian vector field 269Hamilton's variational principle 88,391Hausdorff dimension 374Hausdorff space 259Hodge's star operator 288,425Holonomic constraints 77Homeomorphism 500Homogeneity of space 3,25Hopf bifurcation 353Hyperion 376

Impact parameter 58Impact vector 58Inertia, moment of 174Inertia tensor 172Inertial frame 5Infinitesimal transformations, see generatorsInner product 304Integrable systems 148Integral curve- , complete 283-, maximal 283,325- of vector fields 282,324Integral of the motion 102, 134Intermittency 367Invariant plane 197Involution 147Isotropy of space 3,25Iterative mappings 356

Jacobi identity 132,306Jacobian coordinates 416Jacobian matrix 275

KAM theorem 155KAM tori 155,381Kep1er ellipse 16Kepler's laws 14,16

Subject Index 511

Kirkwood gaps 375,382Klein-Gordon equation 393Klein's group 223,409

Laboratory system 55Lagrange equations 82, 89,315Lagrangian density 390Lagrangian function-, natural form of 83,94- , relativistic 251- for rigid body 202Lagrangian mechanics in geometric

language 309Lagrangian two-form 311Lagrangian vector field 313Legendre transformation 95,97,316,412Leibniz's rule 270Length contraction 249Lenz's vector 416Liapunov characteristic exponent 367,372,380Liapunov function 335Liapunov stability 340Liapunov stable point 329Lie algebra

of Lorentz group 232- of rotation group 109- of smooth functions 306Lie derivative 276,303Lie group 109Lie product 306Light cone 220Lightlike vectors 220Limit curve 344Linear systems 38,116Linearization 327,347Line element 241Liouville' s theorem 123,303Logistic equation 362Lorentz contraction 249Lorentz force 90, 246Lorentz group

homogeneous 222-, inhomogeneous 222-, proper orthochronous 224Lorentz transformations-, proper 224- , orthochronous 222-, special 224,226Lorentzian space-time manifold 239

Manifoldof coordinates 256

-, differentiable 262

512 Subject Index

- , space-tirne 221,237symplectic 300

Mapping- , iterative 357- of manifolds 280-, Poincare 347Mass

gravitational 8-, inertial 4

moving 243reduced 12

Mass density 66, 167Mass point 2Mass shell 246Massless particles 213Metric 291Metric tensor 217Minkowskian space-time 221,237Momentum 6-, canonically conjugate 95, 100,393-, kinematic 6, 100Momentum conservation, principle of 21Motion reversal 25Muon 214,244,422

Neutrino 214Newton's laws 2Noether's theorem- for space transformations 102- for time translations 414Normal coordinates 163n-particle system, closed 21,89Nutation 205

One-form 277- , canonical, of mechanics 294Orbital stability 340Orthochronous transformations 222Oscillations, small 161Oscillator, harmonic 30

Parameter 11, 50IParity, see space rellectionParticle 2Pendulum- , planar 33,41-, spherical 149Pendulum chain 394Pericenter 46Perihelion 46-, precession of 72Period doubling 354Perturbation theory 152

Phase portrait 30,32Phase space 30, 290-, extended 38Photon 212Pion decay 213,244,423Poincare group 222Poincare mapping 347Poinsot' s construction 197Poisson bracket 128,304Poisson's theorem 132Potential 13,28Potential force 28Precession 183, 195Pressure 9Principal axes 173Principal function, Hamilton's 418Principle-, Hamilton's 88- of Euler and Maupertuis 416Projectile 55Proper time 3,26,240Puli-back 280

Quasiperiodic motion 149,369

Rapidity 226Rectification theorem 144Relative angular momentum 18Relative momentum 18Rest energy 244Rest mass 7,242Rest system 228Rheonomic constraints 78Rigid rod 182Rosette orbits 49Rotation group, see SO(3)Rutherford scattering 61,75

Saddle point 331Scattering angle 56Scattering orbits 46,54Scleronomic constraints 78Section 274Separatrix 34Sine-Gordon equation 395Snowllake set 374SO(3) 106,109,167,265Soliton 397SP2j(R) II 9, 300Space cone 196Space rellection 25Spacelike vectors 220Special conformal transformations 253

Special Lorentz transformations 224Special orthogonal group, see SO(3)Special relativity, postulate of 220Special unitary group, see SU(2)Speed of light 212Spherieal hacmonies 68Stability- of equilibrium positions 329- of orbits 340Steiner's theorem 176SU(2) 109,265Suberitical bifureation 353Summation eonvention 218Supereritieal bifureation 353Sympleetie charts 300Symplectie form 299Sympleetic manifold 300Sympleetic transformations 299,333Sympleetie veetor spaee 299

Tangent bundle 258, 272Tangent spaee 257,272Tangent veetor 270Target 55Tensor fields 291Tensors-, contravariant 291-, eovariant 291Thomas preeession 233Three-body problem, restrieted 152Time dilatation 248Time reversal 25Timelike veetors 220Tippe top 207Top-, asymmetrie 175, 192- , sleeping 204

Subjeet Index 513

-, spherieal 175,182- , symmetrie 175,182,193 ,206Tori- , nonresonant 155,342-, resonant 155,342Torque 20,377Torus 265Translations- in spaee 22, 137- in time 23Transverse seetion 346Tuming points 41Two-body system 11,44Two-form, eanonieal, of meehanics 296

Uniform reetilinear motion 4Unitary unimodular group, see SU(2)Units 9

Van der Pool's equation 343Variable 11, 501Variational ealculus 86Variational derivative 87Vector field 257,274,324- , complete 325Velocity 4Virial 69Virtual displacement 79Virtual work 80Volume form 298

Wave equation 390Wedge produet 284Winding number 370World line 215World point 10,215

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