Solutions Problems for Chapter 2 2.1 We obtain directly dr / dz = f(1 + f2 - rr) / (1 + f2)3/2. The equation of the curve is 1 + f2 - rr = 0, from which the result follows. Therefore r(z) = avl + f(z)2. Setting f(z) = sinh(¢(z)), we obtain r(z) = acosh(¢(z)); i.e., f = a¢(z) sinh(¢(z)), and therefore a¢(z) = 1 and the solution r(z) = acosh((z - zo)/a). This is a particular case of the use of conserved quantities discussed in Chapter 3. 2.2 Lagrange Multipliers We must minimize with the constraints z(O) = zo, z(a) = Zl, and i. B VI + z(x)2dx = L. One can transform the problem into min V = loa (p,gz + >')Vl + z(x)2dx, with z(O) = zo, z(a) = Zl. The conserved quantity (p,gz + >.) = C VI + z(x)2 (7.68) (7.69) (7.70)
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Solutions - CERN · PDF fileSolutions Problems for Chapter 2 2.1 We obtain directly dr / dz = f(1 + f2 - rr) / (1 + f2)3/2. The equation of the curve is 1 + f2 - rr = 0, from which
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Solutions
Problems for Chapter 2
2.1 We obtain directly dr / dz = f(1 + f2 - rr) / (1 + f2)3/2. The equation of the curve is 1 + f2 - rr = 0, from which the result follows.
Therefore r(z) = avl + f(z)2.
Setting f(z) = sinh(¢(z)), we obtain
r(z) = acosh(¢(z)); i.e., f = a¢(z) sinh(¢(z)),
and therefore a¢(z) = 1 and the solution r(z) = acosh((z - zo)/a). This is a particular case of the use of conserved quantities discussed in Chapter 3.
2.2 Lagrange Multipliers We must minimize
with the constraints
z(O) = zo, z(a) = Zl, and i.B VI + z(x)2dx = L.
One can transform the problem into
min V = loa (p,gz + >')Vl + z(x)2dx,
with z(O) = zo, z(a) = Zl.
The conserved quantity
(p,gz + >.) = C VI + z(x)2
(7.68)
(7.69)
(7.70)
168 Solutions
yields z = sinh cf>(x) , i.e., j.£gZ + A = Ccoshcf> with C¢ = j.£g. The solution is
A C (J-Lg ) Z = -- + - cosh -(x - xo) . J-Lg J-Lg C
The constants xo, C, and A are fixed by the conditions z(O) = zo, and Joa JI + z(x)2dx = L.
2.3 Brachistochrone Energy conservation gives
I (dS)2 - - + g(z - a) = O. 2 dt
We want to minimize
T- dx l b ( I + Z2 )
- a 2g(a-z)
(7.71)
z(a) = Zl,
(7.72)
(7.73)
with the constraints z(a) = a, z(b) = {3. The Lagrange function I:- = ylr'cI-+---'-,z2"-j""-2-g-;-( a---z--:-) does not depend on x, and
therefore there is conservation of
(7.74)
where we introduce a positive constant R. Setting z = tan(cf>j2), we obtain the parametric form
Rcoscf> Z-Zo = ---,
2
which is the equation of a cycloid.
2.4 Win a Slalom
R(cf> + sincf» x - Xo = 2 ' (7.75)
1. With this definition of the variable x, we have (z - zo) = (x - xo) sina and the potential energy is V = mg(z - zo) = -mgxsina.
2. The total energy is E = ~m(j;2+1?)-mgxsina. Since energy is conserved, and since it is taken to be zero initially, we have j;2 + iP = 2gx sin a.
3. Therefore dt2 = (dx2 + dy2)j(2gx sin a). 4. The total time to get from 0 to A is therefore
T = fA dt = I fA J I + (y')2 dx o v'2gsina 0 x
5. Using the Lagrange-Euler equation, we obtain
d y' o = - "'t=::;=====;=~ dx ylx(1 + (y')2)
Solutions 169
6. We deduce
where C is a constant. However,
y' dy if =C xyf2g sin 0: ' y'x(l + (y')2) y'x(dx2 + dy2)
(7.76) and therefore if = Kx with K = Cyf2g sin 0:.
7. The parametric form x(B) = (1- cos2B)/2C2 = sin2 B/C2, y(B) = (2Bsin2B)/2C2 satisfies the equation (y')2 = C2x/(1- C2x); i.e., (dy/dB)2 = (dx/dB)2tan2B. From if/x = K, we obtain (dy/dB)(dB/dt)/x = K; i.e., dB/dt = K/2 and B = Kt/2 since, for t = 0, B = O.
8. The curve is a portion of a cycloid. We have dy / dx = tan B and therefore y' » 1 for B rv 7r /2. The trajectory starts vertically (dy / dx = 0 for B = 0) and becomes horizontal if y(A) » x(A), as shown in Figure 7.1.
o y
A x
Fig. 7.1. Optimal trajectory from 0 to A.
9. Since point A is fixed, the velocity VA at A is fixed by energy conservation. It is the maximum velocity of the skier. Therefore, the time to get horizontally from y(A) to y(O) is larger than the time (y(A) - y(O))/VA it would take to cover this distance at the maximum velocity. On the other hand, one must start vertically in order to acquire the maximum velocity as quickly as possible. The ideal trajectory comes from an optimization between these two effects.
2.5 Strategy of a Regatta
1. We have by definition x = Vx = V cos B, i = Vz = v sin B, and therefore z' = dz/dx = tanB.
2. We have Vx = vcosB = w/h. This velocity is maximum when h(z') is minimum; i.e., for z' = 1, namely B = 7r/4. We then have Vx = w/2. In fact, it is sufficient to multiply h by a constant to be in the appropriate situation for a given sailboat for which vx,max = )..w.
3. We have dt = dx/vx = h'(z') dx/w(z), and therefore
-lL h'(z') T- dx (). o w z
(7.77)
170 Solutions
4. Setting <fJ = h'(z')/w(z), the Lagrange-Euler equation that optimizes the total time T is
~: = :X (~:,) . 5. The function <fJ does not depend explicitly on x. Therefore, we have
~<fJ _ , 8<fJ ,,8<fJ dx - z 8z + z 8z'·
Consequently,
~ (<fJ - z' 8<fJ) = 0 dx 8z' ,
which gives (h'(z')z' - h(z'))/w(z) = constant. 6. We have z'h' - h = -2/z'. We therefore obtain the first-order differential
equation for the function x(z), (-2/A)dx/dz = w(z), and hence the result
x = L WoZ - wlzoln(l + (z/zo)) WOZI - WIZO In(l + (zI/zo)) '
(7.78)
where we have incorporated the conditions (x = 0, z = 0) and (x = L, z =
zd· 7. We obtain
, dz WOZI - wlzoln(l + (zI/zo)) z - -- - ~~~--~~~~~~~ - dx - woL - wILzo/(z + zo)
If Zl « Land Zl « Zo, the velocity of the wind does not vary appreciably over the whole path, and one has z' '" zI/ L « l.
In the second question, we have seen that the optimal velocity for a constant wind velocity is attained for z' = 1. The present configuration certainly does not correspond to the best strategy. One must tack at some point (Xl, Z) with 0 < Xl < Land Z » Zl, as represented in Figure 7.2 in order to benefit fully from the power of the wind (this possibility was excluded in the text).
z
x:::L Z :::z,
shore x
Fig. 7.2. Path of the boat with a tacking at x = L/2.
Solutions 171
The trajectory drawn with an angle of fJ = 45 degrees (lz'l = 1) and a tacking fJ -+ -fJ at x = L/2 has a total length LV2 and a velocity greater than (wO - wl)/2. The time along this path, Tv = 2LV2/(wO - wI), is obviously shorter than the time along the path with no tacking, T rv
2L(zl/L)/(wO - wI) = 2zl/(wO - wI) . In realistic cases, for instance the America's Cup, one can see how
subtle the regatta problem is. Skippers must make quick decisive choices between very different options.
Problems for Chapter 3
3.1 Moving Pendulum
3.2 Properties of the Action
1. Free particle s = m (X2 - xd2
2 t2 - h 2. Harmonic oscillator
3. Constant force
with Va = (X2 - Xd/(t2 - td - (1/2)(F/m)(t2 - h). 4. One varies the endpoint of arrival in the integration by parts of
5. One varies t2, taking into account that the variation of the time of arrival yields a variation of the trajectory.
172 Solutions
3.3 Conjugate Momenta in Spherical Coordinates
1. The Lagrangian is C = ~m(f2 + r2 iJ2 + r2 sin2 0 ¢2) - V(r). 2. The conjugate momenta are
ac . Pr = af = mr,
ac 2· ac 2 2 . P9 = aiJ = mr 0, P</> = a¢ = mr sin O¢.
3. Taking the derivative of (3.73) with respect to time, and taking into account that in Cartesian coordinates p = mv, one obtains directly the result Lz = mr2 sin2 O¢ = P</>o
4. The conservation of P</>' or L z , corresponds to the invariance under translation in ¢; i.e., rotation invariance around the z axis.
5. If a charged particle is in a magnetic field B parallel to Oz, there is rotational invariance around the z axis and the component L z is conserved.
Problems for Chapter 4
4.1 Coupled Oscillators
1. One obtains directly
{X,P} = 1 {X,Q} = 0 {Y,P} = 0 {Y,Q} = 1
p 2 mw2 X 2 Q2 m(w2 + ,n2)y2 H = 2m + 2 + 2m + 2 .
2. The eigenfrequencies of the system are therefore WI = wand W2
Jw2 +,n2 . 3. The general form of the motion follows from
4.2 Three Coupled Oscillators We obtain with no difficulty
2. In these variables, which are the same as those used by Dirac in the quantum harmonic oscillator,
H = w(a*a).
3. We obtain {a, a*} = -i. 4. The evolution equation in time of a is
a = {a,H} = -iwa,
which is a first-order differential equation. The general solution is
a(t) = ao exp (-iwt),
where ao is a complex constant. The energy of the oscillator is E = wlaol 2 .
5. For t ::::; 0, we have ao = O. In the presence of Hpoh the Hamiltonian becomes
H = w(a*a) + b(a + a*) sin fit.
Therefore, we have
a = {a, H} = -iwa - ib sin fit.
This is solved by standard techniques. With the condition E(t < 0) = 0, one obtains
e-i(D-w)T _ 1 e-i(D+w)T - 1
E(t > T) = wb21 2i(D _ w) + 2i(D + w) 12.
6. This is a resonance phenomenon at D = w (or at D = -w, which is equivalent). In the vicinity of D = w, the energy acquired by the oscillator is of the form
E( T) = b2sin2(D - w)T/2 t> w (D-W)2 '
which has a peak of height wb2T2/4 at D = w.
4.4 Closed Chain of Coupled Oscillators.
1. a) In the definition, we see that
Yk = y'N-k,
b) We have
The summation over k gives onn' and the result
174 Solutions
N N
L,qkqk = L,p;. (7.80) k=l n=l
Similarly
t qkqk = t ( ~ t e-2ikmr/N pn) ( ~ t e2ikn'7r/N p~) . k=l k=l VN n=l VN n'=l
(7.81) The summation over k gives bnn" and hence the result.
fl,2 ( + * ) .* _ { * H} - m k Yk YN-k _ fl,2 qk - qkl - - 2 - m kYk·
d) We therefore have {Yk(t)} = ak cos(fl\t + ¢k), and hence {xn(t)}.
Solutions 175
3. If, at time t = 0, we have YN(O) = 1, YN(O) = 0 and {Yn(O) = 0, Yn(O) = O}, 'Vn =1= N, then YN(t) = cos(wt) and Yn(t) = 0, 'Vn =1= N. Therefore xn(t) = (l/VN) cos(wt). Oscillators of the same amplitude at a given time are always in phase, and only the global motion with respect to the plane x = 0 with frequency w appears.
4. Wave propagation. If w = 0, the eigenfrequencies are !?~ = 2!?sin(k1r/N) rv 2!?(k1r/N) for k « N. The boundary conditions give Y1 = cos 2!?7rt / N, Y N -1 = cos2!?7rt/N, and Yn = 0 otherwise. a) Therefore, we obtain
2 (2!?7rt) lXn = XN-n = VN cos ~ cos2n7rN (7.84)
_ 1 [ (2!?7rt + 2n7r) (2!?7rt - 2n7r)] ( ) - VN cos N + cos N . 7.85
b) We observe a propagation phenomenon in both directions since
in the notation above. The point xn+m has the same amplitude at time t + m/!? as the point Xn at time t.
c) If we write xn(t) = f(t, Y = na), the function f is
f( ) __ 1_ [ (2!?7rt + 2Y7r/a) (2!?7rt - 2Y7r/a)] t,y - VN cos N + cos N
and satisfies the wave equation
1 82 f 82 f ------=0. !?2a2 8t2 8x2
In this chain of coupled oscillators, a progressive wave of velocity !?a propagates.
4.5 Virial Theorem
1. One obtains p2
{A, H} = - - r . V'V. m
The time evolution of A is simply
dA p2 dt = {A, H} = m - r . V'v.
2. We have (.,4) (A(T) - A(O))/T = O. Therefore, inserting this in the result above, we obtain
2 (::) = (r· V'V).
176 Solutions
3. If V = grn , we have 8V
r· V'V = ra;: = nV.
We therefore obtain 2(Ec) = n(V). 4. The total energy is E = Ec + V. We therefore obtain
a) For a harmonic oscillator, E = 2(Ec) = 2(V). b) For a Newtonian potential, E = -(Ec) = (1/2)(V), which is obvious on a circular trajectory, but holds for any elliptic trajectory.
5. In general, for an arbitrary potential, the orbits of bound states are not closed. However, they remain confined in a given region of space at any time. The generalization of the averaging (4.107) is
1 rT (I) = lim (T---+oo) T Jo f(t) dt.
With this definition, we have
(A) = lim (T---+oo) (A(T) - A(O))/T = 0
since A(t) is bounded for any t. With this definition, the result remains true.
4.6 {Lx,Ly} = Lz
4.7 We obtain
and cyclic permutations.
Problems for Chapter 5
5.1 Telegraph Equation The Lagrangian density is
(7.86)
where 'lj;* is the "mirror" density which concentrates instead of diffusing. This leads to the propagation equation
3 82'lj; 28'lj; 2~ -i1'lj;+a ~ =0. v ut ut
(7.87)
This equation can be solved by Fourier transformation if the coefficients v and a are constants. (This is not the case if the medium is inhomogeneous or discontinuous. )
Solutions 177
Problems of Chapter 6
6.2 Geodesics Solutions exist only for p 2: R (which is explained by equation (6.136)).
The energy is
(7.88)
The calculation is similar to previous cases such as (2). We define the parameters wand , as before:
2 2E w = mR2' (7.89)
We obtain
(7.90)
and tanh(¢(t) - ¢o) = ,tanhw(t - to). (7.91)
Problems for Chapter 7
7.1 Propagator of a Harmonic Oscillator The classical action for a harmonic one-dimensional oscillator is
The calculation of the propagator involves only Gaussian integrals, and the result follows directly. One recovers (7.61).
References
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2. Arthur Koestler, The Act of Creation, Hutchinson & Co., London (1964).
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5. Izrail Moiseevich Gelfand and Sergei Vasilevich Fomin, Calculus of Variations, Rev. English ed. Prentice-Hall, Englewood Cliffs, NJ, (1963). Andrew Russell Forsyth, Calculus of Variations, Dover, New York (1960). Jean-Pierre Bourguignon, Calcul Variationnel, Ecole Poly technique, Palaiseau (1990).
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University Press, Cambridge (1982). 12. Max Born and Emil Wolf, Principles of Optics, Pergamon Press, Oxford
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bridge (1982).
180 References
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20. R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals, McGraw-Hill, New York (1965).
21. Lawrence S. Schulman, Techniques and Applications of Path Integration, John Wiley & Sons, New York (1981).
22. Julian Schwinger, Selected Papers on Quantum Electrodynamics, Dover, New York, (1958).