Goldstein Classical Mechanics Notes Michael Good May 30, 2004 1 Chapter 1: Elementary Principles 1.1 Mechanics of a Single Particle Classical mechanics incorporates special relativity.
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- Classical Mechanics Pdf
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Shed the societal and cultural narratives holding you back and let step-by-step Classical Mechanics textbook solutions reorient your old paradigms. NOW is the time to make today the first day of the rest of your life. Unlock your Classical Mechanics PDF (Profound Dynamic Fulfillment) today. YOU are the protagonist of your own life. Sep 25, 2017 This particular MANUAL SOLUTIONS CLASSICAL MECHANICS GOLDSTEIN 3RD EDITION Document is documented in our data source as -, with file size for. Shed the societal and cultural narratives holding you back and let step-by-step Classical Mechanics textbook solutions reorient your old paradigms. NOW is the time to make today the first day of the rest of your life. Unlock your Classical Mechanics PDF (Profound Dynamic Fulfillment) today. YOU are the protagonist of your own life. Jun 18, 2015 Solutions to Problems in Goldstein, Classical Mechanics, Second Edition Homer Reid December 1, 2001 Chapter 3 Problem 3.1 A particle of mass m is constrained to move under gravity without friction on the inside of a paraboloid of revolution whose axis is vertical. Find the one-dimensional problem equivalent to its motion.
Author | Herbert Goldstein |
---|---|
Country | United States of America |
Language | English |
Subject | Classical mechanics |
Genre | Non-fiction |
Publisher | Addison-Wesley |
Publication date | 1951, 1980, 2002 |
Media type | |
Pages | 638 |
ISBN | 978-0-201-65702-9 |
Classical Mechanics is a textbook about that subject written by Herbert Goldstein, a professor at Columbia University. Intended for advanced undergraduate and beginning graduate students, it has been one of the standard references in its subject around the world since its first publication in 1951.[1][2]
Overview[edit]
In the second edition, Goldstein corrected all the errors that had been pointed out, added a new chapter on perturbation theory, a new section on Bertrand's theorem, and another on Noether's theorem. Other arguments and proofs were simplified and supplemented.[3]
Before the death of its primary author in 2005, a new (third) edition of the book was released, with the collaboration of Charles P. Poole and John L. Safko from the University of South Carolina.[4] In the third edition, the book discusses at length various mathematically sophisticated reformations of Newtonian mechanics, namely analytical mechanics, as applied to particles, rigid bodies and continua. In addition, it covers in some detail classical electromagnetism, special relativity, and field theory, both classical and relativistic. There is an appendix on group theory. New to the third edition include a chapter on nonlinear dynamics and chaos, a section on the exact solutions to the three-body problem obtained by Euler and Lagrange, a discussion of the damped driven pendulum that explains the Josephson junctions. This is counterbalanced by the reduction of several existing chapters motivated by the desire to prevent this edition from exceeding the previous one in length. For example, the discussions of Hermitian and unitary matrices were omitted because they are more relevant to quantum mechanics rather than classical mechanics, while those of Routh's procedure and time-independent perturbation theory were reduced.[5]
Table of Contents (3rd Edition)[edit]
- Preface
- Chapter 1: Survey of Elementary Principles
- Chapter 2: Variational Principles and Lagrange's Equations
- Chapter 3: The Central Force Problem
- Chapter 4: The Kinematics of Rigid Body Motion
- Chapter 5: The Rigid Body Equations of Motion
- Chapter 6: Oscillations
- Chapter 7: The Classical Mechanics of the Special Theory of Relativity
- Chapter 8: The Hamilton Equations of Motion
- Chapter 9: Canonical Transformations
- Chapter 10: Hamilton–Jacobi Theory and Action-Angle Coordinates
- Chapter 11: Classical Chaos
- Chapter 12: Canonical Perturbation Theory
- Chapter 13: Introduction to the Lagrangian and Hamiltonian Formulations for Continuous Systems and Fields
- Appendix A: Euler Angles in Alternate Conventions and Cayley–Klein Parameters
- Appendix B: Groups and Algebras
- Appendix C: Solutions to Select Exercises
- Select Bibliography
- Author Index
- Subject Index
Editions[edit]
- Goldstein, Herbert (1951). Classical Mechanics (1st ed.). Addison-Wesley. ASINB000OL8LOM.
- Goldstein, Herbert (1980). Classical Mechanics (2nd ed.). Addison-Wesley. ISBN978-0-201-02918-5.
- Goldstein, Herbert; Poole, C. P.; Safko, J. L. (2001). Classical Mechanics (3rd ed.). Addison-Wesley. ISBN978-0-201-65702-9.
Reception[edit]
First edition[edit]
S.L. Quimby of Columbia University noted that the first half of the first edition of the book is dedicated to the development of Lagrangian mechanics with the treatment of velocity-dependent potentials, which are important in electromagnetism, and the use of the Cayley-Klein parameters and matrix algebra for rigid-body dynamics. This is followed by a comprehensive and clear discussion of Hamiltonian mechanics. End-of-chapter references improve the value of the book. Quimby pointed out that although this book is suitable for students preparing for quantum mechanics, it is not helpful for those interested in analytical mechanics because its treatment omits too much. Quimby praised the quality of printing and binding which make the book attractive.[6]
In the Journal of the Franklin Institute, Rupen Eskergian noted that the first edition of Classical Mechanics offers a mature take on the subject using vector and tensor notations and with a welcome emphasis on variational methods. This book begins with a review of elementary concepts, then introduces the principle of virtual work, constraints, generalized coordinates, and Lagrangian mechanics. Scattering is treated in the same chapter as central forces and the two-body problem. Unlike most other books on mechanics, this one elaborates upon the virial theorem. The discussion of canonical and contact transformations, the Hamilton-Jacobi theory, and action-angle coordinates is followed by a presentation of geometric optics and wave mechanics. Eskergian believed this book serves as a bridge to modern physics.[7]
Writing for The Mathematical Gazette on the first edition, L. Rosenhead congratulated Goldstein for a lucid account of classical mechanics leading to modern theoretical physics, which he believed would stand the test of time alongside acknowledged classics such as E.T. Whittaker's Analytical Dynamics and Arnold Sommerfeld's Lectures on Theoretical Physics. This book is self-contained and is suitable for students who have completed courses in mathematics and physics of the first two years of university. End-of-chapter references with comments and some example problems enhance the book. Rosenhead also liked the diagrams, index, and printing.[8]
Second edition[edit]
Front cover of the second edition.
Concerning the second printing of the first edition, Vic Twersky of the Mathematical Research Group at New York University considered the book to be of pedagogical merit because it explains things in a clear and simple manner, and its humor is not forced. Published in the 1950s, this book replaced the outdated and fragmented treatises and supplements typically assigned to beginning graduate students as a modern text on classical mechanics with exercises and examples demonstrating the link between this and other branches of physics, including acoustics, electrodynamics, thermodynamics, geometric optics, and quantum mechanics. It also has a chapter on the mechanics of fields and continua. At the end of each chapter, there is a list of references with the author's candid reviews of each. Twersky said that Goldstein's Classical Mechanics is more suitable for physicists compared to the much older treatise Analytical Dynamics by E.T. Whittaker, which he deemed more appropriate for mathematicians.[1]
Herbert Goldstein Classical Mechanics Pdf
E. W. Banhagel, an instructor from Detroit, Michigan, observed that despite requiring no more than multivariable and vector calculus, the first edition of Classical Mechanics successfully introduces some sophisticated new ideas in physics to students. Mathematical tools are introduced as needed. He believed that the annotated references at the end of each chapter are of great value.[9]
Third edition[edit]
Stephen R. Addison from the University of Central Arkansas commented that while the first edition of Classical Mechanics was essentially a treatise with exercises, the third has become less scholarly and more of a textbook. This book is most useful for students who are interested in learning the necessary material in preparation for quantum mechanics. The presentation of most materials in the third edition remain unchanged compared to that of the second, though many of the old references and footnotes were removed. Sections on the relations between the action-angle coordinates and the Hamilton-Jacobi equation with the old quantum theory, wave mechanics, and geometric optics were removed. Chapter 7, which deals with special relativity, has been heavily revised and could prove to be more useful to students who want to study general relativity than its equivalent in previous editions. Chapter 11 provides a clear, if somewhat dated, survey of classical chaos. Appendix B could help advanced students refresh their memories but may be too short to learn from. In all, Addison believed that this book remains a classic text on the eighteenth- and nineteenth-century approaches to theoretical mechanics; those interested in a more modern approach – expressed in the language of differential geometry and Lie groups – should refer to Mathematical Methods of Classical Mechanics by Vladimir Arnold.[4]
Corrected Figure 3.13. Original caption: Orbit for motion in a central force deviating slightly from a circular orbit for .
Martin Tiersten from the City University of New York pointed out a serious error in the book that persisted in all three editions and even got promoted to the front cover of the book. Such a closed orbit, depicted in a diagram on page 80 (as Figure 3.7) is impossible for an attractive central force because the path cannot be concave away from the center of force. A similarly erroneous diagram appears on page 91 (as Figure 3.13). Tiersten suggested that the reason why this error remained unnoticed for so long is because of the fact that advanced mechanics texts typically do not use vectors in their treatment of central-force problems, in particular the tangential and normal components of the acceleration vector. He wrote, 'Because an attractive force is always directed in toward the center of force, the direction toward the center of curvature at the turning points must be toward the center of force.' In response, Poole and Safko acknowledged the error and stated they were working on a list of errata.[2]
See also[edit]
Solution Manual Classical Mechanics Goldstein Pdf
- Introduction to Electrodynamics (Griffiths)
- Classical Electrodynamics (Jackson)
External links[edit]
- Errata, corrections, and comments on the third edition. John L. Safko and Charles P. Poole. University of South Carolina.
References[edit]
Classical Mechanics Pdf
- ^ abGoldstein, Herbert; Twersky, Vic (September 1952). 'Classical Mechanics'. Physics Today. 5 (9): 19–20. Bibcode:1952PhT.....5i..19G. doi:10.1063/1.3067728.
- ^ abTiersten, Martin (February 2003). 'Errors in Goldstein's Classical Mechanics'. American Journal of Physics. American Association of Physics Teachers. 71 (2): 103. Bibcode:2003AmJPh..71..103T. doi:10.1119/1.1533731. ISSN0002-9505.
- ^Goldstein, Herbert (1980). 'Preface to the Second Edition'. Classical Mechanics. Addison-Wesley. ISBN0-201-02918-9.
- ^ abAddison, Stephen R. (July 2002). 'Classical Mechanics, 3rd ed'. American Journal of Physics. 70 (7): 782–3. Bibcode:2002AmJPh..70..782G. doi:10.1119/1.1484149. ISSN0002-9505.
- ^Goldstein, Herbert; Safko, John; Poole, Charles (2002). 'Preface to the Third Edition'. Classical Mechanics. Addison-Wesley. ISBN978-0-201-65702-9.
- ^Quimby, S.L. (July 21, 1950). 'Classical Mechanics by Herbert Goldstein'. Book Reviews. Science. American Association for the Advancement of Science (AAAS). 112 (2899): 95. JSTOR1678638.
- ^Eskergian, Rupen (September 1950). 'Classical Mechanics, by Herbert Goldstein'. Journal of the Franklin Institute. 250 (3): 273. doi:10.1016/0016-0032(50)90712-5.
- ^Rosenhead, L. (February 1951). 'Classical Mechanics by Herbert Goldstein'. Review. The Mathematical Gazette. The Mathematical Association. 35 (311): 66–7. doi:10.2307/3610571. JSTOR3610571.
- ^Banhagel, E. W. (October 1952). 'Classical Mechanics by Herbert Goldstein'. Review. The Mathematics Teacher. National Council of Teachers of Mathematics. 45 (6): 485. JSTOR27954117.
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Solution Manual Classical Mechanics Goldstein Homework Solutions
Classical Mechanics Goldstein Solution Manual Download Pdf
Solutions to Problems in Goldstein, Classical Mechanics, Second Edition Homer Reid August 22, 2000
Chapter 1 Problem 1.1 A nucleus, originally at rest, decays radioactively by emitting an electron of momentum 1.73 MeV/c, and at right angles to the direction of the electron a neutrino with momentum 1.00 MeV/c. ( The MeV (million electron volt) is a unit of energy, used in modern physics, equal to 1.60 x 10−6 erg. Correspondingly, MeV/c is a unit of linear momentum equal to 5.34 x 10−17 gm-cm/sec.) In what direction does the nucleus recoil? What is its momentum in MeV/c? If the mass of the residual nucleus is 3.90 x 10−22 gm, what is its kinetic energy, in electron volts? Place the nucleus at the origin, and suppose the electron is emitted in the positive y direction, and the neutrino in the positive x direction. Then the resultant of the electron and neutrino momenta has magnitude p |pe+ν | = (1.73)2 + 12 = 2 MeV/c,
and its direction makes an angle
θ = tan−1
1.73 = 60◦ 1
with the x axis. The nucleus must acquire a momentum of equal magnitude and directed in the opposite direction. The kinetic energy of the nucleus is T =
4 MeV2 c−2 1.78 · 10−27 gm p2 = · = 9.1 ev −22 2m 2 · 3.9 · 10 gm 1 MeV c−2
This is much smaller than the nucleus rest energy of several hundred GeV, so the non-relativistic approximation is justified.
1
2
Homer Reid’s Solutions to Goldstein Problems: Chapter 1
Problem 1.2 The escape velocity of a particle on the earth is the minimum velocity required at the surface of the earth in order that the particle can escape from the earth’s gravitational field. Neglecting the resistance of the atmosphere, the system is conservative. From the conservation theorem for potential plus kinetic energy show that the escape velocity for the earth, ignoring the presence of the moon, is 6.95 mi/sec. If the particle starts at the earth’s surface with the escape velocity, it will just manage to break free of the earth’s field and have nothing left. Thus after it has escaped the earth’s field it will have no kinetic energy left, and also no potential energy since it’s out of the earth’s field, so its total energy will be zero. Since the particle’s total energy must be constant, it must also have zero total energy at the surface of the earth. This means that the kinetic energy it has at the surface of the earth must exactly cancel the gravitational potential energy it has there: 1 mMR mve2 − G =0 2 RR so v=
s0012
2GMR RR
= 11.2 km/s ·
0013
=
0012
2 · (6.67 · 1011 m3 kg−3 s−2 ) · (5.98 · 1024 kg) 6.38 · 106 m
1m = 6.95 mi/s. 1.61 km
00131/2
Homer Reid’s Solutions to Goldstein Problems: Chapter 1
3
Problem 1.3 Rockets are propelled by the momentum reaction of the exhaust gases expelled from the tail. Since these gases arise from the reaction of the fuels carried in the rocket the mass of the rocket is not constant, but decreases as the fuel is expended. Show that the equation of motion for a rocket projected vertically upward in a uniform gravitational field, neglecting atmospheric resistance, is m
dv dm = −v 0 − mg, dt dt
where m is the mass of the rocket and v 0 is the velocity of the escaping gases relative to the rocket. Integrate this equation to obtain v as a function of m, assuming a constant time rate of loss of mass. Show, for a rocket starting initially from rest, with v 0 equal to 6800 ft/sec and a mass loss per second equal to 1/60th of the initial mass, that in order to reach the escape velocity the ratio of the weight of the fuel to the weight of the empty rocket must be almost 300! Suppose that, at time t, the rocket has mass m(t) and velocity v(t). The total external force on the rocket is then F = gm(t), with g = 32.1 ft/s2 , pointed downwards, so that the total change in momentum between t and t + dt is F dt = −gm(t)dt.
(1)
At time t, the rocket has momentum p(t) = m(t)v(t).
(2)
On the other hand, during the time interval dt the rocket releases a mass ∆m of gas at a velocity v 0 with respect to the rocket. In so doing, the rocket’s velocity increases by an amount dv. The total momentum at time t + dt is the sum of the momenta of the rocket and gas: p(t + dt) = pr + pg = [m(t) − ∆m][v(t) + dv] + ∆m[v(t) + v 0 ]
(3)
Subtracting (2) from (3) and equating the difference with (1), we have (to first order in differential quantities) −gm(t)dt = m(t)dv + v 0 ∆m or dv v 0 ∆m = −g − dt m(t) dt
Homer Reid’s Solutions to Goldstein Problems: Chapter 1
4
which we may write as dv v0 = −g − γ dt m(t)
(4)
where ∆m 1 = m0 s−1 . dt 60
γ=
This is a differential equation for the function v(t) giving the velocity of the rocket as a function of time. We would now like to recast this as a differential equation for the function v(m) giving the rocket’s velocity as a function of its mass. To do this, we first observe that since the rocket is releasing the mass ∆m every dt seconds, the time derivative of the rocket’s mass is dm ∆m =− = −γ. dt dt We then have
dv dv dm dv = = −γ . dt dm dt dm Substituting into (4), we obtain −γ
dv v0 = −g − γ dm m
or dv =
dm g dm + v 0 . γ m
Integrating, with the condition that v(m0 ) = 0, v(m) =
g (m − m0 ) + v 0 ln γ
0012
m m0
0013
.
Now, γ=(1/60)m0 s−1 , while v 0 =-6800 ft/s. Then 0012 0013 0010m 0011 m 0 v(m) = 1930 ft/s · − 1 + 6800 ft/s · ln m0 m
For m0 001d m we can neglect the first term in the parentheses of the first term, giving 0010m 0011 0 . v(m) = −1930 ft/s + 6800 ft/s · ln m The escape velocity is v = 6.95 mi/s = 36.7 · 103 ft/s. Plugging this into the equation above and working backwards, we find that escape velocity is achieved when m0 /m=293. Thanks to Brian Hart for pointing out an inconsistency in my original choice of notation for this problem.
Homer Reid’s Solutions to Goldstein Problems: Chapter 1
5
Problem 1.4 Show that for a single particle with constant mass the equation of motion implies the following differential equation for the kinetic energy: dT = F · v, dt while if the mass varies with time the corresponding equation is d(mT ) = F · p. dt
We have F = p˙
(5)
If m is constant, F = mv˙ Dotting v into both sides, F · v = mv · v˙ = =
1 d 2 m |v| 2 dt
dT dt
On the other hand, if m is not constant, instead of v we dot p into (5): F · p = p · p˙ d(mv) 0012 dt 0013 dm dv = mv · v +m dt dt 1 d 1 d = v 2 m2 + m2 (v 2 ) 2 dt 2 dt 1 d d(mT ) = (m2 v 2 ) = . 2 dt dt
= mv ·
(6)
6
Homer Reid’s Solutions to Goldstein Problems: Chapter 1
Problem 1.5 Prove that the magnitude R of the position vector for the center of mass from an arbitrary origin is given by the equation M 2 R2 = M
X
mi ri2 −
i
We have Rx = so
1X 2 mi mj rij . 2 ij
1 X mi xi M i
X X 1 Rx2 = 2 mi mj xi xj m2i x2i + M i i6=j
and similarly
X X 1 Ry2 = 2 mi mj y i y j m2i yi2 + M i i6=j
X X 1 m2i zi2 + mi mj z i z j . Rz2 = 2 M i i6=j
Adding,
X X 1 R2 = 2 m2i ri2 + mi mj (ri · rj ) . M i
(7)
i6=j
On the other hand,
2 rij = ri2 + rj2 − 2ri · rj 2 and, in particular, rii = 0, so X X 2 = mi mj rij [mi mj ri2 + mi mj rj2 − 2mi mj (ri · rj )] i,j
i6=j
=2
X
mi mj ri2 − 2
X
mi mj (ri · rj ).
(8)
i6=j
i6=j
Next, M
X i
mi ri2 =
X j
mj
X i
mi ri2
!
=
X i
m2i ri2 +
X i6=j
mi mj ri2 .
(9)
Homer Reid’s Solutions to Goldstein Problems: Chapter 1
7
r0 , φ0 r, φ
(x, y) θ
Figure 1: My conception of the situation of Problem 1.8
Subtracting half of (8) from (9), we have M
X
mi ri2 −
X X 1X 2 ijmi mj rij = m2i ri2 + mi mj (ri · rj ) 2 i i6=j
and comparing this with (7) we see that we are done.
Problem 1.8 Two wheels of radius a are mounted on the ends of a common axle of length b such that the wheels rotate independently. The whole combination rolls without slipping on a plane. Show that there are two nonholonomic equations of constraint, cos θ dx + sin θ dy = 0 sin θ dx − cos θ dy = a(dφ + dφ0 ) (where θ, φ, and φ0 have meanings similar to the problem of a single vertical disc, and (x, y) are the coordinates of a point on the axle midway between the two wheels) and one holonomic equation of constraint, a θ = C − (φ − φ0 ) b where C is a constant. My conception of the situation is illustrated in Figure 1. θ is the angle between the x axis and the axis of the two wheels. φ and φ0 are the rotation angles of the two wheels, and r and r0 are the locations of their centers. The center of the wheel axis is the point just between r and r0 : (x, y) =
1 (rx + rx0 , ry + ry0 ). 2
Homer Reid’s Solutions to Goldstein Problems: Chapter 1
8
If the φ wheel rotates through an angle dφ, the vector displacement of its center will have magnitude adφ and direction determined by θ. For example, if θ = 0 then the wheel axis is parallel to the x axis, in which case rolling the φ wheel clockwise will cause it to move in the negative y direction. In general, referring to the Figure, we have dr = a dφ[sin θ ˆi − cos θ ˆj] dr0 = a dφ0 [sin θ ˆi − cos θ ˆj]
(10) (11)
Adding these componentwise we have1 a [dφ + dφ0 ] sin θ 2 a dy = − [dφ + dφ0 ] cos θ 2
dx =
Multiplying these by sin θ or − cos θ and adding or subtracting, we obtain sin θ dx − cos θ dy = a[dφ + dφ0 ] cos θ dx + sin θ dy = 0. Next, consider the vector r12 = r − r0 connecting the centers of the two wheels. The definition of θ is such that its tangent must just be the ratio of the y and x components of this vector: y12 x12 1 y12 2 dy12 . → sec θ dθ = − 2 dx12 + x12 x12 tan θ =
Subtracting (11) from (10), 0013 0012 1 y12 cos θ sec2 θdθ = a[dφ − dφ0 ] − 2 sin θ − x12 x12 Again substituting for y12 /x12 in the first term in parentheses, sec2 θdθ = −a[dφ − dφ0 ]
1 (tan θ sin θ + cos θ) x12
or 1 (sin2 θ cos θ + cos3 θ) x12 1 = −a[dφ − dφ0 ] cos θ. x12
dθ = −a[dφ − dφ0 ]
(12)
1 Thanks to Javier Garcia for pointing out a factor-of-two error in the original version of these equations.
Homer Reid’s Solutions to Goldstein Problems: Chapter 1
9
However, considering the definition of θ, we clearly have x12 x12 = cos θ = 2 2 1/2 b (x12 + y12 ) because the magnitude of the distance between r1 and r2 is constrained to be b by the rigid axis. Then (12) becomes a dθ = − [dφ − dφ0 ] b with immediate solution a θ = C − [φ − φ0 ]. b with C a constant of integration.
Problem 1.9 A particle moves in the x − y plane under the constraint that its velocity vector is always directed towards a point on the x axis whose abscissa is some given function of time f (t). Show that for f (t) differentiable, but otherwise arbitrary, the constraint is nonholonomic. The particle’s position is (x(t), y(t)), while the position of the moving point is (f (t), 0). Then the vector d from the particle to the point has components dx = x(t) − f (t)
dy = y(t).
(13)
The particle’s velocity v has components dx dy vy = (14) dt dt and for the vectors in (13) and (?? to be in the same direction, we require vx =
vy dy = vx dx or
dy y(t) dy/dt = = dx/dt dx x(t) − f (t)
so
dy dx = y x − f (t) For example, if f (t) = αt, then we may integrate to find
(15)
ln y(t) = ln[x(t) − α(t)] + C or y(t) = C · [x(t) − αt] which is a holonomic constraint. But for general f (t) the right side of (15) is not integrable, so the constraint is nonholonomic.
Homer Reid’s Solutions to Goldstein Problems: Chapter 1
10
φ
l θ a
Figure 2: My conception of the situation of Problem 1.10
Problem 1.10 Two points of mass m are joined by a rigid weightless rod of length l, the center of which is constrained to move on a circle of radius a. Set up the kinetic energy in generalized coordinates. My conception of this one is shown in Figure 2. θ is the angle representing how far around the circle the center of the rod has moved. φ is the angle the rod makes with the x axis. The position of the center of the rod is (x, y) = (a cos θ, a sin θ). The positions of the masses relative to the center of the rod are (xrel , yrel ) = ±(1/2)(l cos φ, l sin φ). Then the absolute positions of the masses are (x, y) = (a cos θ ±
l l cos φ, a sin θ ± sin φ) 2 2
and their velocities are (vx , vy ) = (−a sin θ θ˙ ∓
l ˙ a cos θ θ˙ ± l cos φ φ). ˙ sin φ φ, 2 2
The magnitudes of these are l2 ˙ 2 ˙ φ ± al θ˙ φ(sin θ sin φ + cos θ cos φ) 4 l2 = a2 θ˙2 + φ˙ 2 ± al θ˙ φ˙ cos(θ − φ) 4
|v| = a2 θ˙2 +
When we add the kinetic energies of the two masses, the third term cancels, and we have l2 1X mv 2 = m(a2 θ˙2 + φ˙ 2 ). T = 2 4
11
Homer Reid’s Solutions to Goldstein Problems: Chapter 1
Problem 1.13 A particle moves in a plane under the influence of a force, acting toward a center of force, whose magnitude is 0012 0013 1 r˙ 2 − 2¨ rr F = 2 1− , r c2 where r is the distance of the particle to the center of force. Find the generalized potential that will result in such a force, and from that the Lagrangian for the motion in a plane. (The expression for F represents the force between two charges in Weber’s electrodynamics). If we take
1 U (r) = r
then
0012 0013 v2 1 (r) ˙ 2 1+ 2 = + 2 c r c r
∂U 1 r˙ 2 =− 2 − 2 2 ∂r r c r
and
2¨ r 2(r) ˙ 2 − 2 2 2 c r c r 0012 0013 2r¨ r − (r) ˙ 2 d ∂U 1 ∂U + = 2 1+ Qr = − ∂r dt ∂ r˙ r c2 d d ∂U = dt ∂ r˙ dt
so
0012
2r˙ c2 r
0013
=
The Lagrangian for motion in a plane is L=T −V =
1 2 1 2 ˙2 1 mr˙ + mr˙ θ − 2 2 2 r
0012
1+
2r¨ r − (r) ˙ 2 2 c
0013
.
Problem 1.14 If L is a Lagrangian for a system of n degrees of freedom satisfying Lagrange’s equations, show by direct substitution that L0 = L +
dF (q1 , . . . , qn , t) dt
also satisfies Lagrange’s equations, where F is any arbitrary, but differentiable, function of its arguments. We have
∂L0 ∂L ∂ dF = + ∂qi ∂qi ∂qi dt
(16)
Homer Reid’s Solutions to Goldstein Problems: Chapter 1
12
and
∂L ∂ dF ∂L0 = + . ∂ q˙i ∂ q˙i ∂ q˙i dt For the function F we may write
(17)
X ∂F ∂F dF = q˙i + dt ∂qi ∂t i and from this we may read off ∂F ∂ dF = . ∂ q˙i dt ∂qi Then taking the time derivative of (17) gives d ∂L d ∂F d ∂L0 = + dt ∂ q˙i dt ∂ q˙i dt ∂qi so we have ∂L0 d ∂L0 ∂L d ∂L ∂ dF d ∂F − = − + − . ∂qi dt ∂ q˙i ∂qi dt ∂ q˙i ∂qi dt dt ∂qi The first two terms on the RHS cancel because L satisfies the Euler-Lagrange equations, while the second two terms cancel because F is differentiable. Hence L0 satisfies the Euler-Lagrange equations.
Problem 1.16 A Lagrangian for a particular physical system can be written as L0 =
m K (ax˙ 2 + 2bx˙ y˙ + cy˙ 2 ) − (ax2 + 2bxy + cy 2 ), 2 2
where a, b, and c are arbitrary constants but subject to the condition that b2 − ac 6= 0. What are the equations of motion? Examine particularly the two cases a = 0 = c and b = 0, c = −a. What is the physical system described by the above Lagrangian? Show that the usual Lagrangian for this system as defined by Eq. (1-56) is related to L0 by a point transformation (cf. Exercise 15 above). What is the significance of the condition on the value of b2 − ac? Clearly we have ∂L = −Kax − Kby ∂x
∂L = max˙ + mby˙ ∂ x˙
so the Euler-Lagrange equation for x is ∂L d ∂L = ∂x dt ∂ x˙
→
m(a¨ x + b¨ y) = −K(ax + by).
Homer Reid’s Solutions to Goldstein Problems: Chapter 1
13
Similarly, for y we obtain m(b¨ y + c¨ y ) = −K(bx + cy). These are the equations of motion for a particle of mass m undergoing simple harmonic motion in two dimensions, as if bound by two springs of spring constant K. Normally we would express the Lagrangian in unravelled form, by transforming to new coordinates u1 and u2 with u1 = ax + by
u2 = bx + cy.
The condition b2 − ac 6= 0 is the condition that the coordinate transformation not be degenerate, i.e. that there are actually two distinct dimensions in which the particle experiences a restoring force. If b2 = ac then we have just a onedimensional problem.
Problem 1.17 Obtain the Lagrangian equations of motion for a spherical pendulum, i.e. a mass point suspended by a rigid weightless rod. Let m and L be the mass of the particle and the length of the rod. Since the particle is constrained to move on the surface of a sphere of radius L, we may parameterize its position by the angles θ and ϕ, in terms of which the particle’s position and velocity are x = L (sin θ cos ϕi + sin θ sin ϕj + cos θk) 0014 0015 ˙ v = L (cos θ cos ϕθ˙ − sin θ sin ϕϕ)i ˙ + (cos θ sin ϕθ˙ + sin θ cos ϕϕ)j ˙ − (sin θθk) .
so the kinetic energy is
T =
1 mL2 ˙2 mL2 mv2 = θ + sin2 θϕ˙ 2 . 2 2 2
On the other hand, the gravitational potential energy depends only on θ : V = −mgL cos θ where we take the potential at the height of the fulcrum as the zero of potential. Then the Lagrangian is L=T −V =
1 1 mL2 θ˙2 + mL2 sin2 θϕ˙ 2 + mgL cos θ 2 2
and the equations of motion are ϕ¨ = 0 0011 0010g − ϕ˙ 2 cos θ sin θ. θ¨ = − L
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If there is no motion in the azimuthal direction, ϕ˙ = 0 and the systempis the elementary one-dimensional mechanical pendulum with frequency ω0 = g/L. But any finite velocity of azimuthal spinning gives rise to an additional effect which we may think of as mitigating the downward force of gravity, yielding an effective gravitational acceleration g 0 = g − Lϕ˙ 2 cos θ. The mitigating effect is largest near the trough of the pendulum, vanishes as the particle passes through the vertical height of the fulcrum, and becomes an enhancing effect in the upper hemisphere. For small oscillations p about the trough, cos θ ≈ 1 and the pendulum frequency is reduced to ω = ω02 − ϕ˙ 2 . Thanks to Tomasz Szymanski for pointing out an error in an earlier version of this solution.
Problem 1.18 A particle of mass m moves in one dimension such that it has the Lagrangian m2 x˙ 4 + mx˙ 2 V (x) − V 2 (x), 12
L=
where V is some differentiable function of x. Find the equation of motion for x(t) and describe the physical nature of the system on the basis of this equation. We have dV ∂L dV = mx˙ 2 − 2V (x) ∂x dx dx ∂L m2 x˙ 3 = + 2mxV ˙ (x) ∂ x˙ 3 d ∂L d = m2 (x) ˙ 2x ¨ + 2m¨ xV (x) + 2mx˙ V (x) dt ∂ x˙ dt In the last equation we can use d dV V (x) = x˙ . dt dx Then the Euler-Lagrange equation is d ∂L ∂L − =0 dt ∂ x˙ ∂x or
→ 0012
m2 (x) ˙ 2 x¨ + 2m¨ xV (x) + mx˙ 2
m¨ x+
dV dx
0013
0001 mx˙ 2 + 2V (x) = 0.
dV dV + 2V (x) dx dx
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If we identify F = −dV /dx and T = mx˙ 2 /2, we may write this as (F − m¨ x)(T + V ) = 0 So, this is saying that, at all times, either the difference between F and ma is zero, or the sum of kinetic and potential energy is zero.
Problem 1.19 Two mass points of mass m1 and m2 are connected by a string passing through a hole in a smooth table so that m1 rests on the table and m2 hangs suspended. Assuming m2 moves only in a vertical line, what are the generalized coordinates for the system? Write down the Lagrange equations for the system and, if possible, discuss the physical significance any of them might have. Reduce the problem to a single second-order differential equation and obtain a first integral of the equation. What is its physical significance? (Consider the motion only so long as neither m1 nor m2 passes through the hole). Let d be the height of m2 above its lowest possible position, so that d = 0 when the string is fully extended beneath the table and m1 is just about to fall through the hole. Also, let θ be the angular coordinate of m1 on the table. Then the kinetic energy of m2 is just m2 d˙2 /2, while the kinetic energy of m1 is m1 d˙2 /2 + m1 d2 θ˙2 /2, and the potential energy of the system is just the gravitational potential energy of m2 , U = m2 gd. Then the Lagrangian is L=
1 1 (m1 + m2 )d˙2 + m1 d2 θ˙2 − m2 gd 2 2
and the Euler-Lagrange equations are d ˙ =0 (m1 d2 θ) dt (m1 + m2 )d¨ = −m2 g + m1 dθ˙2 From the first equation we can identify a first integral, m1 d2 θ˙ = l where l is a constant. With this we can substitute for θ˙ in the second equation: (m1 + m2 )d¨ = −m2 g +
l2 m1 d3
Because the sign of the two terms on the RHS is different, this is saying that, if l is big enough (if m1 is spinning fast enough), the centrifugal force of m1 can balance the downward pull of m2 , and the system can be in equilibrium.
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Problem 1.20 Obtain the Lagrangian and equations of motion for the double pendulum illustrated in Fig. 1-4, where the lengths of the pendula are l1 and l2 with corresponding masses m1 and m2 . Taking the origin at the fulcrum of the first pendulum, we can write down the coordinates of the first mass point: x1 = l1 sin θ1 y1 = −l1 cos θ1 The coordinates of the second mass point are defined relative to the coordinates of the first mass point by exactly analogous expressions, so relative to the coordinate origin we have x2 = x1 + l2 sin θ2 y2 = y1 − l2 cos θ2 Differentiating and doing a little algebra we find x˙ 21 + y˙ 12 = l12 θ˙12 x˙ 2 + y˙ 2 = l2 θ˙2 + l2 θ˙2 − 2l1 l2 θ˙1 θ˙2 cos(θ1 − θ2 ) 2
2
1 1
2 2
The Lagrangian is L=
1 1 (m1 +m2 )l12 θ˙12 + m2 l22 θ˙22 −m2 l1 l2 θ˙1 θ˙2 cos(θ1 −θ2 )+(m1 +m2 )gl2 cos θ1 +m2 gl2 cos θ2 2 2
with equations of motion
and
i d h (m1 + m2 )l12 θ˙1 − m2 l1 l2 θ˙2 cos(θ1 − θ2 ) = −(m1 + m2 )gl2 sin θ1 dt i d h ˙ l2 θ2 − l1 θ˙1 cos(θ1 − θ2 ) = −g sin θ2 . dt
If θ˙1 = 0, so that the fulcrum for the second pendulum is stationary, then the second of these equations reduces to the equation we derived in Problem 1.17.
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Problem 1.21 The electromagnetic field is invariant under a gauge transformation of the scalar and vector potential given by A → A + ∇Ψ(r, t), 1 ∂Ψ Φ→ Φ− , c ∂t where Ψ is arbitrary (but differentiable). What effect does this gauge transformation have on the Lagrangian of a particle moving in the electromagnetic field? Is the motion affected? The Lagrangian for a particle in an electromagnetic field is q L = T − qΦ(x(t)) + A(x(t)) · v(t) c If we make the suggested gauge transformation, this becomes ' # 1 ∂Ψ q → T − q Φ(x(t)) − + [A(x(t)) · v(t) + v · ∇Ψ(x(t))] c ∂t x=x(t) c 0014 0015 q q ∂Ψ = T − qΦ(x(t)) + A(x(t)) · v(t) + + v · ∇Ψ(x(t)) c c ∂t q q d = T − qΦ(x(t)) + A(x(t)) · v(t) + Ψ(x(t)) c c dt q d Ψ(x(t)). =L+ c dt So the transformed Lagrangian equals the original Lagrangian plus a total time derivative. But we proved in Problem 1.15 that adding the total time derivative of any function to the Lagrangian does not affect the equations of motion, so the motion of the particle is unaffected by the gauge transformation.
Problem 1.22 Obtain the equation of motion for a particle falling vertically under the influence of gravity when frictional forces obtainable from a dissipation function 21 kv 2 are present. Integrate the equation to obtain the velocity as a function of time and show that the maximum possible velocity for fall from rest is v = mg/k. The Lagrangian for the particle is L=
1 mz˙ 2 − mgz 2
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and the dissipation function is k z˙ 2 /2, so the equation of motion is 0012 0013 d ∂L ∂L ∂F + → m¨ z = mg − k z. ˙ − dt ∂ z˙ ∂z ∂ z˙ This says that the acceleration goes to zero when mg = k z, ˙ or z˙ = mg/k, so the velocity can never rise above this terminal value (unless the initial value of the velocity is greater than the terminal velocity, in which case the particle will slow down to the terminal velocity and then stay there).
Chapter 1 Problem 1.1 A nucleus, originally at rest, decays radioactively by emitting an electron of momentum 1.73 MeV/c, and at right angles to the direction of the electron a neutrino with momentum 1.00 MeV/c. ( The MeV (million electron volt) is a unit of energy, used in modern physics, equal to 1.60 x 10−6 erg. Correspondingly, MeV/c is a unit of linear momentum equal to 5.34 x 10−17 gm-cm/sec.) In what direction does the nucleus recoil? What is its momentum in MeV/c? If the mass of the residual nucleus is 3.90 x 10−22 gm, what is its kinetic energy, in electron volts? Place the nucleus at the origin, and suppose the electron is emitted in the positive y direction, and the neutrino in the positive x direction. Then the resultant of the electron and neutrino momenta has magnitude p |pe+ν | = (1.73)2 + 12 = 2 MeV/c,
and its direction makes an angle
θ = tan−1
1.73 = 60◦ 1
with the x axis. The nucleus must acquire a momentum of equal magnitude and directed in the opposite direction. The kinetic energy of the nucleus is T =
4 MeV2 c−2 1.78 · 10−27 gm p2 = · = 9.1 ev −22 2m 2 · 3.9 · 10 gm 1 MeV c−2
This is much smaller than the nucleus rest energy of several hundred GeV, so the non-relativistic approximation is justified.
1
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Homer Reid’s Solutions to Goldstein Problems: Chapter 1
Problem 1.2 The escape velocity of a particle on the earth is the minimum velocity required at the surface of the earth in order that the particle can escape from the earth’s gravitational field. Neglecting the resistance of the atmosphere, the system is conservative. From the conservation theorem for potential plus kinetic energy show that the escape velocity for the earth, ignoring the presence of the moon, is 6.95 mi/sec. If the particle starts at the earth’s surface with the escape velocity, it will just manage to break free of the earth’s field and have nothing left. Thus after it has escaped the earth’s field it will have no kinetic energy left, and also no potential energy since it’s out of the earth’s field, so its total energy will be zero. Since the particle’s total energy must be constant, it must also have zero total energy at the surface of the earth. This means that the kinetic energy it has at the surface of the earth must exactly cancel the gravitational potential energy it has there: 1 mMR mve2 − G =0 2 RR so v=
s0012
2GMR RR
= 11.2 km/s ·
0013
=
0012
2 · (6.67 · 1011 m3 kg−3 s−2 ) · (5.98 · 1024 kg) 6.38 · 106 m
1m = 6.95 mi/s. 1.61 km
00131/2
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Problem 1.3 Rockets are propelled by the momentum reaction of the exhaust gases expelled from the tail. Since these gases arise from the reaction of the fuels carried in the rocket the mass of the rocket is not constant, but decreases as the fuel is expended. Show that the equation of motion for a rocket projected vertically upward in a uniform gravitational field, neglecting atmospheric resistance, is m
dv dm = −v 0 − mg, dt dt
where m is the mass of the rocket and v 0 is the velocity of the escaping gases relative to the rocket. Integrate this equation to obtain v as a function of m, assuming a constant time rate of loss of mass. Show, for a rocket starting initially from rest, with v 0 equal to 6800 ft/sec and a mass loss per second equal to 1/60th of the initial mass, that in order to reach the escape velocity the ratio of the weight of the fuel to the weight of the empty rocket must be almost 300! Suppose that, at time t, the rocket has mass m(t) and velocity v(t). The total external force on the rocket is then F = gm(t), with g = 32.1 ft/s2 , pointed downwards, so that the total change in momentum between t and t + dt is F dt = −gm(t)dt.
(1)
At time t, the rocket has momentum p(t) = m(t)v(t).
(2)
On the other hand, during the time interval dt the rocket releases a mass ∆m of gas at a velocity v 0 with respect to the rocket. In so doing, the rocket’s velocity increases by an amount dv. The total momentum at time t + dt is the sum of the momenta of the rocket and gas: p(t + dt) = pr + pg = [m(t) − ∆m][v(t) + dv] + ∆m[v(t) + v 0 ]
(3)
Subtracting (2) from (3) and equating the difference with (1), we have (to first order in differential quantities) −gm(t)dt = m(t)dv + v 0 ∆m or dv v 0 ∆m = −g − dt m(t) dt
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which we may write as dv v0 = −g − γ dt m(t)
(4)
where ∆m 1 = m0 s−1 . dt 60
γ=
This is a differential equation for the function v(t) giving the velocity of the rocket as a function of time. We would now like to recast this as a differential equation for the function v(m) giving the rocket’s velocity as a function of its mass. To do this, we first observe that since the rocket is releasing the mass ∆m every dt seconds, the time derivative of the rocket’s mass is dm ∆m =− = −γ. dt dt We then have
dv dv dm dv = = −γ . dt dm dt dm Substituting into (4), we obtain −γ
dv v0 = −g − γ dm m
or dv =
dm g dm + v 0 . γ m
Integrating, with the condition that v(m0 ) = 0, v(m) =
g (m − m0 ) + v 0 ln γ
0012
m m0
0013
.
Now, γ=(1/60)m0 s−1 , while v 0 =-6800 ft/s. Then 0012 0013 0010m 0011 m 0 v(m) = 1930 ft/s · − 1 + 6800 ft/s · ln m0 m
For m0 001d m we can neglect the first term in the parentheses of the first term, giving 0010m 0011 0 . v(m) = −1930 ft/s + 6800 ft/s · ln m The escape velocity is v = 6.95 mi/s = 36.7 · 103 ft/s. Plugging this into the equation above and working backwards, we find that escape velocity is achieved when m0 /m=293. Thanks to Brian Hart for pointing out an inconsistency in my original choice of notation for this problem.
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Problem 1.4 Show that for a single particle with constant mass the equation of motion implies the following differential equation for the kinetic energy: dT = F · v, dt while if the mass varies with time the corresponding equation is d(mT ) = F · p. dt
We have F = p˙
(5)
If m is constant, F = mv˙ Dotting v into both sides, F · v = mv · v˙ = =
1 d 2 m |v| 2 dt
dT dt
On the other hand, if m is not constant, instead of v we dot p into (5): F · p = p · p˙ d(mv) 0012 dt 0013 dm dv = mv · v +m dt dt 1 d 1 d = v 2 m2 + m2 (v 2 ) 2 dt 2 dt 1 d d(mT ) = (m2 v 2 ) = . 2 dt dt
= mv ·
(6)
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Homer Reid’s Solutions to Goldstein Problems: Chapter 1
Problem 1.5 Prove that the magnitude R of the position vector for the center of mass from an arbitrary origin is given by the equation M 2 R2 = M
X
mi ri2 −
i
We have Rx = so
1X 2 mi mj rij . 2 ij
1 X mi xi M i
X X 1 Rx2 = 2 mi mj xi xj m2i x2i + M i i6=j
and similarly
X X 1 Ry2 = 2 mi mj y i y j m2i yi2 + M i i6=j
X X 1 m2i zi2 + mi mj z i z j . Rz2 = 2 M i i6=j
Adding,
X X 1 R2 = 2 m2i ri2 + mi mj (ri · rj ) . M i
(7)
i6=j
On the other hand,
2 rij = ri2 + rj2 − 2ri · rj 2 and, in particular, rii = 0, so X X 2 = mi mj rij [mi mj ri2 + mi mj rj2 − 2mi mj (ri · rj )] i,j
i6=j
=2
X
mi mj ri2 − 2
X
mi mj (ri · rj ).
(8)
i6=j
i6=j
Next, M
X i
mi ri2 =
X j
mj
X i
mi ri2
!
=
X i
m2i ri2 +
X i6=j
mi mj ri2 .
(9)
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r0 , φ0 r, φ
(x, y) θ
Figure 1: My conception of the situation of Problem 1.8
Subtracting half of (8) from (9), we have M
X
mi ri2 −
X X 1X 2 ijmi mj rij = m2i ri2 + mi mj (ri · rj ) 2 i i6=j
and comparing this with (7) we see that we are done.
Problem 1.8 Two wheels of radius a are mounted on the ends of a common axle of length b such that the wheels rotate independently. The whole combination rolls without slipping on a plane. Show that there are two nonholonomic equations of constraint, cos θ dx + sin θ dy = 0 sin θ dx − cos θ dy = a(dφ + dφ0 ) (where θ, φ, and φ0 have meanings similar to the problem of a single vertical disc, and (x, y) are the coordinates of a point on the axle midway between the two wheels) and one holonomic equation of constraint, a θ = C − (φ − φ0 ) b where C is a constant. My conception of the situation is illustrated in Figure 1. θ is the angle between the x axis and the axis of the two wheels. φ and φ0 are the rotation angles of the two wheels, and r and r0 are the locations of their centers. The center of the wheel axis is the point just between r and r0 : (x, y) =
1 (rx + rx0 , ry + ry0 ). 2
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If the φ wheel rotates through an angle dφ, the vector displacement of its center will have magnitude adφ and direction determined by θ. For example, if θ = 0 then the wheel axis is parallel to the x axis, in which case rolling the φ wheel clockwise will cause it to move in the negative y direction. In general, referring to the Figure, we have dr = a dφ[sin θ ˆi − cos θ ˆj] dr0 = a dφ0 [sin θ ˆi − cos θ ˆj]
(10) (11)
Adding these componentwise we have1 a [dφ + dφ0 ] sin θ 2 a dy = − [dφ + dφ0 ] cos θ 2
dx =
Multiplying these by sin θ or − cos θ and adding or subtracting, we obtain sin θ dx − cos θ dy = a[dφ + dφ0 ] cos θ dx + sin θ dy = 0. Next, consider the vector r12 = r − r0 connecting the centers of the two wheels. The definition of θ is such that its tangent must just be the ratio of the y and x components of this vector: y12 x12 1 y12 2 dy12 . → sec θ dθ = − 2 dx12 + x12 x12 tan θ =
Subtracting (11) from (10), 0013 0012 1 y12 cos θ sec2 θdθ = a[dφ − dφ0 ] − 2 sin θ − x12 x12 Again substituting for y12 /x12 in the first term in parentheses, sec2 θdθ = −a[dφ − dφ0 ]
1 (tan θ sin θ + cos θ) x12
or 1 (sin2 θ cos θ + cos3 θ) x12 1 = −a[dφ − dφ0 ] cos θ. x12
dθ = −a[dφ − dφ0 ]
(12)
1 Thanks to Javier Garcia for pointing out a factor-of-two error in the original version of these equations.
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However, considering the definition of θ, we clearly have x12 x12 = cos θ = 2 2 1/2 b (x12 + y12 ) because the magnitude of the distance between r1 and r2 is constrained to be b by the rigid axis. Then (12) becomes a dθ = − [dφ − dφ0 ] b with immediate solution a θ = C − [φ − φ0 ]. b with C a constant of integration.
Problem 1.9 A particle moves in the x − y plane under the constraint that its velocity vector is always directed towards a point on the x axis whose abscissa is some given function of time f (t). Show that for f (t) differentiable, but otherwise arbitrary, the constraint is nonholonomic. The particle’s position is (x(t), y(t)), while the position of the moving point is (f (t), 0). Then the vector d from the particle to the point has components dx = x(t) − f (t)
dy = y(t).
(13)
The particle’s velocity v has components dx dy vy = (14) dt dt and for the vectors in (13) and (?? to be in the same direction, we require vx =
vy dy = vx dx or
dy y(t) dy/dt = = dx/dt dx x(t) − f (t)
so
dy dx = y x − f (t) For example, if f (t) = αt, then we may integrate to find
(15)
ln y(t) = ln[x(t) − α(t)] + C or y(t) = C · [x(t) − αt] which is a holonomic constraint. But for general f (t) the right side of (15) is not integrable, so the constraint is nonholonomic.
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φ
l θ a
Figure 2: My conception of the situation of Problem 1.10
Problem 1.10 Two points of mass m are joined by a rigid weightless rod of length l, the center of which is constrained to move on a circle of radius a. Set up the kinetic energy in generalized coordinates. My conception of this one is shown in Figure 2. θ is the angle representing how far around the circle the center of the rod has moved. φ is the angle the rod makes with the x axis. The position of the center of the rod is (x, y) = (a cos θ, a sin θ). The positions of the masses relative to the center of the rod are (xrel , yrel ) = ±(1/2)(l cos φ, l sin φ). Then the absolute positions of the masses are (x, y) = (a cos θ ±
l l cos φ, a sin θ ± sin φ) 2 2
and their velocities are (vx , vy ) = (−a sin θ θ˙ ∓
l ˙ a cos θ θ˙ ± l cos φ φ). ˙ sin φ φ, 2 2
The magnitudes of these are l2 ˙ 2 ˙ φ ± al θ˙ φ(sin θ sin φ + cos θ cos φ) 4 l2 = a2 θ˙2 + φ˙ 2 ± al θ˙ φ˙ cos(θ − φ) 4
|v| = a2 θ˙2 +
When we add the kinetic energies of the two masses, the third term cancels, and we have l2 1X mv 2 = m(a2 θ˙2 + φ˙ 2 ). T = 2 4
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Homer Reid’s Solutions to Goldstein Problems: Chapter 1
Problem 1.13 A particle moves in a plane under the influence of a force, acting toward a center of force, whose magnitude is 0012 0013 1 r˙ 2 − 2¨ rr F = 2 1− , r c2 where r is the distance of the particle to the center of force. Find the generalized potential that will result in such a force, and from that the Lagrangian for the motion in a plane. (The expression for F represents the force between two charges in Weber’s electrodynamics). If we take
1 U (r) = r
then
0012 0013 v2 1 (r) ˙ 2 1+ 2 = + 2 c r c r
∂U 1 r˙ 2 =− 2 − 2 2 ∂r r c r
and
2¨ r 2(r) ˙ 2 − 2 2 2 c r c r 0012 0013 2r¨ r − (r) ˙ 2 d ∂U 1 ∂U + = 2 1+ Qr = − ∂r dt ∂ r˙ r c2 d d ∂U = dt ∂ r˙ dt
so
0012
2r˙ c2 r
0013
=
The Lagrangian for motion in a plane is L=T −V =
1 2 1 2 ˙2 1 mr˙ + mr˙ θ − 2 2 2 r
0012
1+
2r¨ r − (r) ˙ 2 2 c
0013
.
Problem 1.14 If L is a Lagrangian for a system of n degrees of freedom satisfying Lagrange’s equations, show by direct substitution that L0 = L +
dF (q1 , . . . , qn , t) dt
also satisfies Lagrange’s equations, where F is any arbitrary, but differentiable, function of its arguments. We have
∂L0 ∂L ∂ dF = + ∂qi ∂qi ∂qi dt
(16)
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and
∂L ∂ dF ∂L0 = + . ∂ q˙i ∂ q˙i ∂ q˙i dt For the function F we may write
(17)
X ∂F ∂F dF = q˙i + dt ∂qi ∂t i and from this we may read off ∂F ∂ dF = . ∂ q˙i dt ∂qi Then taking the time derivative of (17) gives d ∂L d ∂F d ∂L0 = + dt ∂ q˙i dt ∂ q˙i dt ∂qi so we have ∂L0 d ∂L0 ∂L d ∂L ∂ dF d ∂F − = − + − . ∂qi dt ∂ q˙i ∂qi dt ∂ q˙i ∂qi dt dt ∂qi The first two terms on the RHS cancel because L satisfies the Euler-Lagrange equations, while the second two terms cancel because F is differentiable. Hence L0 satisfies the Euler-Lagrange equations.
Problem 1.16 A Lagrangian for a particular physical system can be written as L0 =
m K (ax˙ 2 + 2bx˙ y˙ + cy˙ 2 ) − (ax2 + 2bxy + cy 2 ), 2 2
where a, b, and c are arbitrary constants but subject to the condition that b2 − ac 6= 0. What are the equations of motion? Examine particularly the two cases a = 0 = c and b = 0, c = −a. What is the physical system described by the above Lagrangian? Show that the usual Lagrangian for this system as defined by Eq. (1-56) is related to L0 by a point transformation (cf. Exercise 15 above). What is the significance of the condition on the value of b2 − ac? Clearly we have ∂L = −Kax − Kby ∂x
∂L = max˙ + mby˙ ∂ x˙
so the Euler-Lagrange equation for x is ∂L d ∂L = ∂x dt ∂ x˙
→
m(a¨ x + b¨ y) = −K(ax + by).
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Similarly, for y we obtain m(b¨ y + c¨ y ) = −K(bx + cy). These are the equations of motion for a particle of mass m undergoing simple harmonic motion in two dimensions, as if bound by two springs of spring constant K. Normally we would express the Lagrangian in unravelled form, by transforming to new coordinates u1 and u2 with u1 = ax + by
u2 = bx + cy.
The condition b2 − ac 6= 0 is the condition that the coordinate transformation not be degenerate, i.e. that there are actually two distinct dimensions in which the particle experiences a restoring force. If b2 = ac then we have just a onedimensional problem.
Problem 1.17 Obtain the Lagrangian equations of motion for a spherical pendulum, i.e. a mass point suspended by a rigid weightless rod. Let m and L be the mass of the particle and the length of the rod. Since the particle is constrained to move on the surface of a sphere of radius L, we may parameterize its position by the angles θ and ϕ, in terms of which the particle’s position and velocity are x = L (sin θ cos ϕi + sin θ sin ϕj + cos θk) 0014 0015 ˙ v = L (cos θ cos ϕθ˙ − sin θ sin ϕϕ)i ˙ + (cos θ sin ϕθ˙ + sin θ cos ϕϕ)j ˙ − (sin θθk) .
so the kinetic energy is
T =
1 mL2 ˙2 mL2 mv2 = θ + sin2 θϕ˙ 2 . 2 2 2
On the other hand, the gravitational potential energy depends only on θ : V = −mgL cos θ where we take the potential at the height of the fulcrum as the zero of potential. Then the Lagrangian is L=T −V =
1 1 mL2 θ˙2 + mL2 sin2 θϕ˙ 2 + mgL cos θ 2 2
and the equations of motion are ϕ¨ = 0 0011 0010g − ϕ˙ 2 cos θ sin θ. θ¨ = − L
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If there is no motion in the azimuthal direction, ϕ˙ = 0 and the systempis the elementary one-dimensional mechanical pendulum with frequency ω0 = g/L. But any finite velocity of azimuthal spinning gives rise to an additional effect which we may think of as mitigating the downward force of gravity, yielding an effective gravitational acceleration g 0 = g − Lϕ˙ 2 cos θ. The mitigating effect is largest near the trough of the pendulum, vanishes as the particle passes through the vertical height of the fulcrum, and becomes an enhancing effect in the upper hemisphere. For small oscillations p about the trough, cos θ ≈ 1 and the pendulum frequency is reduced to ω = ω02 − ϕ˙ 2 . Thanks to Tomasz Szymanski for pointing out an error in an earlier version of this solution.
Problem 1.18 A particle of mass m moves in one dimension such that it has the Lagrangian m2 x˙ 4 + mx˙ 2 V (x) − V 2 (x), 12
L=
where V is some differentiable function of x. Find the equation of motion for x(t) and describe the physical nature of the system on the basis of this equation. We have dV ∂L dV = mx˙ 2 − 2V (x) ∂x dx dx ∂L m2 x˙ 3 = + 2mxV ˙ (x) ∂ x˙ 3 d ∂L d = m2 (x) ˙ 2x ¨ + 2m¨ xV (x) + 2mx˙ V (x) dt ∂ x˙ dt In the last equation we can use d dV V (x) = x˙ . dt dx Then the Euler-Lagrange equation is d ∂L ∂L − =0 dt ∂ x˙ ∂x or
→ 0012
m2 (x) ˙ 2 x¨ + 2m¨ xV (x) + mx˙ 2
m¨ x+
dV dx
0013
0001 mx˙ 2 + 2V (x) = 0.
dV dV + 2V (x) dx dx
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If we identify F = −dV /dx and T = mx˙ 2 /2, we may write this as (F − m¨ x)(T + V ) = 0 So, this is saying that, at all times, either the difference between F and ma is zero, or the sum of kinetic and potential energy is zero.
Problem 1.19 Two mass points of mass m1 and m2 are connected by a string passing through a hole in a smooth table so that m1 rests on the table and m2 hangs suspended. Assuming m2 moves only in a vertical line, what are the generalized coordinates for the system? Write down the Lagrange equations for the system and, if possible, discuss the physical significance any of them might have. Reduce the problem to a single second-order differential equation and obtain a first integral of the equation. What is its physical significance? (Consider the motion only so long as neither m1 nor m2 passes through the hole). Let d be the height of m2 above its lowest possible position, so that d = 0 when the string is fully extended beneath the table and m1 is just about to fall through the hole. Also, let θ be the angular coordinate of m1 on the table. Then the kinetic energy of m2 is just m2 d˙2 /2, while the kinetic energy of m1 is m1 d˙2 /2 + m1 d2 θ˙2 /2, and the potential energy of the system is just the gravitational potential energy of m2 , U = m2 gd. Then the Lagrangian is L=
1 1 (m1 + m2 )d˙2 + m1 d2 θ˙2 − m2 gd 2 2
and the Euler-Lagrange equations are d ˙ =0 (m1 d2 θ) dt (m1 + m2 )d¨ = −m2 g + m1 dθ˙2 From the first equation we can identify a first integral, m1 d2 θ˙ = l where l is a constant. With this we can substitute for θ˙ in the second equation: (m1 + m2 )d¨ = −m2 g +
l2 m1 d3
Because the sign of the two terms on the RHS is different, this is saying that, if l is big enough (if m1 is spinning fast enough), the centrifugal force of m1 can balance the downward pull of m2 , and the system can be in equilibrium.
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Problem 1.20 Obtain the Lagrangian and equations of motion for the double pendulum illustrated in Fig. 1-4, where the lengths of the pendula are l1 and l2 with corresponding masses m1 and m2 . Taking the origin at the fulcrum of the first pendulum, we can write down the coordinates of the first mass point: x1 = l1 sin θ1 y1 = −l1 cos θ1 The coordinates of the second mass point are defined relative to the coordinates of the first mass point by exactly analogous expressions, so relative to the coordinate origin we have x2 = x1 + l2 sin θ2 y2 = y1 − l2 cos θ2 Differentiating and doing a little algebra we find x˙ 21 + y˙ 12 = l12 θ˙12 x˙ 2 + y˙ 2 = l2 θ˙2 + l2 θ˙2 − 2l1 l2 θ˙1 θ˙2 cos(θ1 − θ2 ) 2
2
1 1
2 2
The Lagrangian is L=
1 1 (m1 +m2 )l12 θ˙12 + m2 l22 θ˙22 −m2 l1 l2 θ˙1 θ˙2 cos(θ1 −θ2 )+(m1 +m2 )gl2 cos θ1 +m2 gl2 cos θ2 2 2
with equations of motion
and
i d h (m1 + m2 )l12 θ˙1 − m2 l1 l2 θ˙2 cos(θ1 − θ2 ) = −(m1 + m2 )gl2 sin θ1 dt i d h ˙ l2 θ2 − l1 θ˙1 cos(θ1 − θ2 ) = −g sin θ2 . dt
If θ˙1 = 0, so that the fulcrum for the second pendulum is stationary, then the second of these equations reduces to the equation we derived in Problem 1.17.
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Problem 1.21 The electromagnetic field is invariant under a gauge transformation of the scalar and vector potential given by A → A + ∇Ψ(r, t), 1 ∂Ψ Φ→ Φ− , c ∂t where Ψ is arbitrary (but differentiable). What effect does this gauge transformation have on the Lagrangian of a particle moving in the electromagnetic field? Is the motion affected? The Lagrangian for a particle in an electromagnetic field is q L = T − qΦ(x(t)) + A(x(t)) · v(t) c If we make the suggested gauge transformation, this becomes ' # 1 ∂Ψ q → T − q Φ(x(t)) − + [A(x(t)) · v(t) + v · ∇Ψ(x(t))] c ∂t x=x(t) c 0014 0015 q q ∂Ψ = T − qΦ(x(t)) + A(x(t)) · v(t) + + v · ∇Ψ(x(t)) c c ∂t q q d = T − qΦ(x(t)) + A(x(t)) · v(t) + Ψ(x(t)) c c dt q d Ψ(x(t)). =L+ c dt So the transformed Lagrangian equals the original Lagrangian plus a total time derivative. But we proved in Problem 1.15 that adding the total time derivative of any function to the Lagrangian does not affect the equations of motion, so the motion of the particle is unaffected by the gauge transformation.
Problem 1.22 Obtain the equation of motion for a particle falling vertically under the influence of gravity when frictional forces obtainable from a dissipation function 21 kv 2 are present. Integrate the equation to obtain the velocity as a function of time and show that the maximum possible velocity for fall from rest is v = mg/k. The Lagrangian for the particle is L=
1 mz˙ 2 − mgz 2
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and the dissipation function is k z˙ 2 /2, so the equation of motion is 0012 0013 d ∂L ∂L ∂F + → m¨ z = mg − k z. ˙ − dt ∂ z˙ ∂z ∂ z˙ This says that the acceleration goes to zero when mg = k z, ˙ or z˙ = mg/k, so the velocity can never rise above this terminal value (unless the initial value of the velocity is greater than the terminal velocity, in which case the particle will slow down to the terminal velocity and then stay there).