These corrections will bring the 4th and 5th printings to the contents of the 6th printing.

Special thanks to Ralf Kissman for many of the corrections.

It is interesting that although contributions have been made by many, there has been almost zero overlap of their corrections and suggestions. All updates will be posted on a separate page.

There will not be changes in any printing after the 6th. Changes to the 6th and later printings will only appear on the Web.

Additional corrections to the 6th printing are listed in a separate file.

page #
Comment on the cover figure The cover figure shows the motion of a particle around a central force source. There are two central forces acting; an attractive force proportional to 1/r3 and a repulsive force proportional to 1/r4. The figure was probably inspired by the hard core repulsive nuclear force. Another possibility, which is computationly simpler is to make the attractive force 1/r2 and the repulsive force 1/r3.
add 2 paragraphs (first and second new on page) We thank E. Barreto, P. M. Brown, C. Chien, C. Chou, F. Du, R. F. Gans, I. R. Gatland, C. G. Gray, E.J. Guala, Jr., S. Gutti, D. H. Hartman, M. Horbatsch, J. Howard, K. Jagannathan, R. Kissman, L. Kramer, O. Lehtonen, N. A. Lemos, J. Palacios, R. E. Reynolds, D. V. Sathe, G. T, Seidler, J. Suzuki, A. Tenne-Sens, J. Williams. We also thank Martin Tiersten for pointing out the problems with Fig. 3.7 (description) and 3.13 (misleading figure).
10th line above (1.7) ...inertial system (or inertial frame) by...
title of figure 1.2 ...vector rij...
9 integrand of left side of eq. after Eq(1.29) Fi dsi
13 lhs of first of eqs (1.38) r1
14 4th line above second paragraph ...may be invoked to serve
15 equations above (1.39) add "," after first of pair
15 3rd line after (1.39) replace word "perfect" with "exact"
17 Eq. (1.42) last F should be f
20 Eq. (1.52) summation index "j"
20 2nd line after (1.52) ... T with respect to qj (and therefore also with respect to dqj/dt)vanishes. ...
20 last sentence in paragraph after Eq. (1.52) ... with respect to the angle coordinate ...
21 Eqs. (1.54) and (1.55) both qi should be qj
21 Eq. (1.47) after Qj replace "=" with "defined as" symbol. See (1.47)
22 2nd & 3rd lines after Eq. (1.60) 4 places -- change (t,x,y,z) to (x,y,z,t) for normal order
23 below Eq. (1.64) the partial time ...
23 last equation needs a subscript i to the x on both sides.
24 first line of Eq. 1.69, after last equality subscript should be j not i
24 Eq. (1.69) all sums are from i =1,n where n is the number of particles in the system
24 equation on the 2nd line below Eq. (1.69) "a" should not be bold, "v" should be bold, add a minus sign on the right
26 three lines below Eq. (1.75) replace n with theta-hat
26 5th line from the bottom ... cyclindrical coordinates, restricted to the z = 0 plane, where ...
27 line just above 2. Atwood's N(e) =
27 Figure 1.6 add x to horizontal axis and y to vertical axis. replace n with theta-hat, replace r(theta) with r(theta)r-hat, etc.
28 6th line below figure greek delta should be partial derivative symbol
29 2nd display equation first "=" should be "-"
29 5th line above Derivations r = roewt for a bead initally at rest on the wire shows that ...
29 5th & 4th line above Derivations ... bead moves exponentially outwards. Again, the ...
29 2th line above Derivations L = m r2w = m w ro2 e2wt

L, F and N are not bold.

29 line above Derivations F = 2 m ro w2 ewt
29 Derivation 2 right hand summation add i not equal to j below the i,j
31 first equation in Exercise 9 both A's and grad sign should be bold
31 Exercise 9 Since the text material uses c = 1, remove the c here or add it in the text equations.
32 6th line to 2.1 km/s and a mass...
33 3rd line Eq.(1.56)
33 Exercise 20 -- last term in the equation - V2(x)
36 Add above Figure Eq. (2.3) assumes that x is unidirectional between x1 and x2. Otherwise the integral must be broken into undirectional sub-integrals and the results algebraicly added.
40 5th line in "2. Minimum..." Replace x with y and y with x. It is conceptually important to use a variable which is unidirectional over the range of integration. As the following shows, correct results may (but not will) be obtained in some cases even if this principle is ignored.
41 Figure title -- add Note that this curve is unidirectional in y but not x, so it is best to redo the calculation as described on page 40.
43 first new paragraph The parametric solution ...
43 below last correction (2nd display equation from bottom) x = a(phi-sin(phi)), y = a(1-cos(phi))
43 last equation on page (y/a)3 = (9/2) (x/a)2
43 Figure 2.4b caption Cycloid solution to the ....
45 This section needs to be titled and revised on this page up to the last line as shown 2.4 EXTENSION OF HAMILTON’S PRINCIPLE TO SYSTEMS WITH CONSTRAINTS

We discussed in Section 1.3 how problems with holonomic constraints, those that satisfy Eq. (1.37), may be solved by choosing generalized coordinates such that the constraint equations (1.37) become 0 = 0 in this coordinate system. In this section we show that Hamilton’s principle can be extended to solve not only holonomic but, ay t least in a formal sense, to cover …(pick up the last line on page 45)
46 add at the end of the first full paragraph In particular, should we set the time integral of the variation of the Lagrangian equal to zero or should we set the variation of the time integral of the Lagrangian equal to zero? These are equivalent for holonomic constraints but give different results for nonholonomic constraints. Studies of the results of these two choices shows that the former is the correct choice. (See footnotes on pg 47)
46 modify Eq. (2.20) to include time Replace ") = 0" with "; t) = 0"
46 line below Eq. (2.20) …called semiholonomic. If there is no dependence on the velocities these are just the holonomic constrains of Eq. (1.37)
46 2nd line above Eq. (2.20') to Eq. (20.20') Equation (2.20) is more general than the commonly used restricted form, linear in d(qa)/dt,
fa = (Sum over k)(bak(q's,t) (d(qk)/dt) ) + ba(q's,t) = 0 (2.20')
46 in first line of last paragraph …extra virtual displacements for both holonomic and nonholonomic problems is the method



footnote should read--This is more than the 6th has. J Ray, Amer J. Phys. 34 (1202),1966; E. J. Saletan & A. H. Comer, Amer J. Phys. 38(892-897), 1970. A detailed discussion of nonholonomic constraints is given by M.R. Flannery, "The enigma of nonholomic constraints", Am. J. Phys. 73 (3), March 2005, pp. (265-272) 2005
47 Two lines before Eq. (2.23) If the constraints are holonomic, we can combine (2.21) with (2.2) giving
47 following Eq. (2.23) where the quantity inside the parenthesis can be considered an effective Lagrangian.
47 2nd through the 4th lines following (2.23) variables.
If the constraints are semiholonomic, the variation must be taken before the integral since we cannot consider the term inside to parenthesis to be an effective Lagrangian. Leaving details to the references*, , the resulting …
47 revise Eq. (2.25) Qk = (Sum over b from 1 to m) (lambdab times (partial of fb with respect to the time derivative of qk)
47 3rd line following (2.25) … as an n+m nonholomonic system …forces Qk.
(delete sentence starting on 3rd line and all of fourth line)
47 Eq. (2.28) the 2nd and 3rd terms are replaced by
- (lambda)times (time derivative of y)
47 Eq. (2.29) the 2nd and 3rd terms are replaced by
-(lambda) times (time derivative of x)
47 footnote should read *J. Ray, Amer. J. Phys. 34 (1202), 1966; E. J. Saletan & A. H. Cromer, Amer. J. Phys. 38 (892-897), 1970
50 third equation from the bottom no period
52 5th line, eq. for Ff put "dot" over y
57 equation after (2.49) add i below summation
58 top of page add i below summation
59 last summation on page add i below summation
65 Exercise 10, 2nd line, grammar change ... are not known...
66 Problem 16 s = exp(gamma t/2)q
68 Exercise 23, 2nd line ...Let m2 be confined to move
69 add problem (suggested by C Gray 27. (a) Show that the constraint Eq. (2.27) is truly
nonholonomic by showing that it can not
be integrated to a holonomic form.
(b) Show that the corresponding constraint
forces are virtu;ary workless
(c) Find one or more solutions to
(2-27)-(2.30) for V =0 and show that
they conserve energy.
76 2nd line + minor rewording on lines 10 & 11 to keep page breaks ...quadratures (evaluating integrals), with ...
80 Caption for Figure 3.7 This shows the motion of a particle in a central force which is a combination of a 1/r3 attractive force and a 1/r4 repulsive force. Although there are no stable circular orbits for this set of forces, there are bound, non-closed orbits, such as the example shown. Thanks to Thomas Fischaleck and Ian Gatland for corrections.
80 Figure 3.7 This figure should match the cover drawing

Thanks to Thomas Fischaleck and Ian Gatland

87 Eq. (3.33) "1" before and inside parentheses should be "l"
90 above Eq. (3.45) (from Eq, (3.34)
90 below Eq. (3.49) or

r = ro + a cos (beta times theta).

91 Figure 3.13 --As first pointed out by Martin Tiersten Figure 3.13 has been incorrect in the last edition and previous printings of the text.

The caption should read: Orbit for motion deviating slightly from a circular orbit if the central force is given by (beta) = 5.
The figure should look as shown (In the printed version the circle may be dotted.):

95 line below Eq. (3.60) The coefficient of the linear term in this particular quadratic ...
98 line above Eq. (3.66) (3.55)
99 2nd line equation reference should be (3.56)
100 Eq. (3.69) remove minus sign after =
100 Eq. (3.72) put "dot" over theta
105 fifth line from bottom subscript of omega is theta
109 fast form in Eq. (3.99) and in (3.100) replace e with e2
109 line above Eq. (3.98) ... change than writing ...
110 equation at top of page replace e with e2
118 line below Eq. (3.113) where E = (1/2)m1vo2 is the ...
120 equation below (3.117') square the cosine term on the right hand side
121 Fig. 3.37 Replace the xi with ri or modify the caption to si = xj - xk
126 Derivation 2, equation replace sin wt with sin nwt
( w for omega)
126 last equation in derivation 3 replace sin (...) with sin3(...) and divide rhs by 6
126 derivation 4 (1-...) in deominator is squared
126 derivation 8 last term on rhs should be "-"
129 Exercise 17, line 2 ...on the orbit 180° out of phase
129 equation given is for the potential--lhs V(r) =
132 exercise 35 reverse the two inequalities
135 2nd line from bottom Note that the configuration
165 display equation above (4.68) capital omega should be bold face
166 secondline below (4.71) capital omega should be bold face
168 line beloe Eq. (4.77) ...hence those of V* do not,
169 Eq. (4.77') V*i = ...
174 2nd line above Eqs. (4.87) axis (omegapsi - time derivative of psi). Adding...
174 Eqs. (4.87) need "," after each equation
175 last equation, ratio 366.5/365.5 (366.25/365.25)
181 Derivation 10, line 3 eB
181 Derivation 12, 2nd line replace italic lower case theta with italic lower case phi
190 Equation above (5.15), after third = axi
191 Section 5.3, first line L is boldface L
194 Eq. (5.23) rho(r)...
203 footnote ...which suggests "snakelike."
206 2nd equationin (5.47) ... (I3 - I1) ...
209 2nd paragraph, line one The rates of change...
209 first of three equations giving the time rates of change ...= rotation (or spinning) of the top...
217 above Eq. (5.73) ...Eqs. (5.71) and ...
224 last equation on page factor of 2 Pi missing -- but the result is still zero
224 2nd line Eq. (5.84)
231 line above Eq. (5.104) ... angular velocity for each particle (denoted by i) is
231 above Eq. (5.105) (L is a system quantity so summation is taken over repeated i subscripts for the remainder of this chapter)
231-232 Eqs. (5.105), (5.107), (5.110) replace "+" in front of V with "–"
240 Eq. (6.8) first term -- drop subscript "i" from "eta"
241 between Eqs. (6.11) and (6.12) Change Eq. (6.9) to Eq (6.11) -- 2 places
245 6th line below Eq. (6.24) delta subscripts should be ik
245 equation above (6.25) middle quanity on right hand side should be ajl
246 4th line above (6.27) The (no sum on i) clearly applies to the first equation on this line.
254 3rd of Eqs. (6.55) omega3
256 Eq. (6.58a) the three matrix elements are a11, a21, and a31
257 Eq. (6.59) remember the zeta's and eta's are magnitudes, not unit vectors
257 Eq. (6.59) the corrections on p. 256 lead to:

zeta1 = (m eta1+M eta2 +m eta3)/root(2m+M),

zeta2 = (eta1-eta3)root(m/2)

zeta3 = (eta1 - 2eta2 + eta3) x root(mM/(4m+M)).

zeta1 is center of mass motion

258 general comments on 3rd paragraph Angular momentum conservation is not the relevent argument. The motion must be such that all three molecules lie in a plane and the end molecules are always on the same side of the original symmetry axis while the central atom has a smaller displacement to the opposite side. The submitted figure for the 3rd edition was correct but we missed the displacement of the central molecule in proof. This figure may have been corrected in the 4th printing of the 3rd edition.
266 line above Eq. (6.88) value ...
275 Exercise 22 ...V11 > V22 >0 and...
276 5th line implied a preferred inertial frame. ...
276 7th line electromagnetic theory without this implication. After...
276 Eqs. (7.1) add commas between equations
276 replace line F' = (d/dt)P' (7.2)
279 9th line ...time interval measured by a clock at rest with respect to that body the proper time of that clock,
279 4th below Eq. (7.6) time into three regions,...
279 right hand side of equation above (7.6) (1 - v2/c2)
280 first line of Sec. 7.2 ...of transformations between inertial frames that preserve...
280 second line of Sec 7.2 ...are linear in Minkowski coordinates...
281 2nd line above Eqs. (7.9) ...=(ct, r) allows...
287 2nd line above Eq. (7.30) two slots (both of which are linear) into which ...
297 2nd line above (7.66) the Lorenz condition ...
298 line below Eq. (7.67b) holdover from old units. Replace +1/c2 with "-" and 1/c with 1. Also replace e with q for generality
298 Eqs. (7.71), (7.71') and (7.71") multiply lhs by c or divide each element on rhs by c.
314 Eq. (7.140) starting with the 2nd "=+ sign =T + mc2 + V = E
320 Eq. (7.159) leading term on rhs x'o/c ...
322 Eq. (7.165) 2nd term on rhs, (q/c)
323 Eq. (7.166) after first inequality change u superscript to (nu) and multiply expression by g(mu)(nu)
324 9thline of Section 7.11 ...(paths of extremal distance)...
328 sign of lambda There are several notational approaches to GR. Our choice makes k a negative number, hence a negative lambda causes expansion.
329 Derivations 12 ...are said to form a group (see Appendix B) if ...
331 Exercise 21 symbol associated with h should be an italic greek nu. Font makes it hard to tell.
332 Exercises 29 & 30 insert TM where appropriate
334 3rd line below Eq. (8.1) ... or the n qi's ...
337 Eq. (8.16) add dt to the last term on the rhs
340 add above Eq. (8.25) gives us, since T is symmetric
347 8th line below the figure title ... where cm denotes ...
351 display equation below (8.56) superscript should be italic greek nu. Font makes it hard to tell.
353 Eq. (8.61), right-hand-side of both equalities replace superscript in the demoninator by (sigma) and multiply each right-hand-side by g(nu)(sigma)
353 Equation above (8.62), after "=" sign replace superscript (0) with (mu) and multiply expression by g0(mu)
362 Derivation 3, last line, first p dot should be a (q dot)i
364 #14, 5th term on rhs (y-dot)2
365 last display equation needs a 2 in the denominator
367 1st display for Exercise 34, 2nd term both lamba and nu are equal subscripts
379 Eq. (9.39b) the p in the square root should be P
381 1st line ... use of canonical transformations ...
382 Eq. (9.52) zeta is bold face
384 2nd line below last display equation J is bold face and Sans Serif
386 last line 1 is bold Sans Serif
388 Eq. (9.70) subscript eta is bold face
388 rhs of Eq. (9.69) [pj,pk]q,p
389 11th line from bottom eta is bold
392 Eq. (9.78a) A and B are bold face
392 Eq. (9.78b) A and B are bold face and Sans Serif
394 footnote Carathedory has ' over the e
416 line above Eq, (9.142) A is bold face and Sans Serif
423 Derivations 13., first line ...transformations forms a group (Appendix B). ...
425 last eqn on page ...=qB
428 line above the equation in 36. (b) ...find a general expression for
434 line above Eq. (10.14) principal function ...
436 Eq. (10.27) add "." at end
438 Eqs. (10.35), (10.36), (10.38) add "," between equations where appropriate
438 line above Eq. (10.38) ... The principal function ...
439 Eqs. (10.35') and (10.42) add spearating "," as needed
441 1st of Eqs. (10.46) Q1 = t + (beta)1 ...
446 last paragraph, 1st line ... a coordinate qj is separable if ...
446 third line above Eq. (10.56') "S" should be "s"
448 line below Eq. (10.63) Staeckel
457 3rd line below Eq. (10.99) interchange p and q
458 2nd line below Eq. (10.100) p... in the (qi, pi) plane ...
460 Line below (10.108) The differential operator ...
461 Eq. (10.111) add an "i" in exponent
464 9th line a solution of the Hamilton-Jacobi equation.
465 line above Eq. (10.122) written (where jki are positive or negative integers)
471 1st of Eqs. (10.144) add "," a end
481 Exercise 25 Show, by the method...
488 1st of Eqs. (11.9) add "," after equation
506 Eqs. (11.29) add "," after each equation
509 line below Eq. (11.31) ... with the variable x restricted to ...
511 4th line in table caption replace "=" with "-"
511 3rd line from bottom produce seemingly random ...
514 3rd line from bottom ... and that are considerably ...
523 Exercises 6 (b) and display for 8 add "." at end of lines
525 Exercise 16, 2nd line ... , such as those
538 line below Eq. (12.51) ...gravitational radius of the Sun
543 first of Eqs. (12.69) and Eq.(12.75a) add "," at end
543 Eqs. (12.71b) and (12.71c) symbol following = is a bold face, greek nu
544 Eq. (12.71d) symbol following = is a bold face, greek nu
548 14th line of (Jo' ...
550 Eq. (12.96) add a bar above the lhs to show this is an average and replace the bar over the a on the left hand side with two dots to show a second derivative
553 Eq. (12.107) lower case greek omega is not bold face
555 Exercise 2, 1st line ... vertically hung Hooke's-law ...
556 Exercise 6. (b) Use first-order perturbation ...
556 Exercise 7, below display equation ... constant. Use first-order time-independent ...
558 2nd line ... afinite, or at ... infinite, number ...
564 Eq. (13.18) place a "," before the x on the rhs. Note that the comma in the middle tewrm is a subscript as it should be.
567 Eq. (13.32), denominator of 2nd term on right dxj
583 line below Eq. (13.94) move and from in front of two equations to between the two equations
593 last eqn on page ...+O(a(dot), a(double dot))
594 2nd below Eq. (13.142) ... helpful, however, to ...
596 6th line below Eq. (13.154) ... for the (spinless) Klein-Gordon ...
601 2nd line above Eqs. (A.1y) ... Temporarily using ...
606 5th and 6th lines below Table B.1 DELETE "We shall use h for the group order."
608 2nd line above Eq. (B.10) and 2nd line below Eq. (B.11) add a "," in each
611 following 6th line below Table B.4 Add caption TABLE B.5 Characteristics of the Dihedral Group D3
611 3rd line before LIE GROUPS AND ALGEBRAS ... = 0. In quantum...
612 line following Eq. (B.16) ... cijk = - cjik) ...
612 Eq. (B.18) first k is a subscript to the summation -- i.e. sum over k
616 first line TEXTBOOKS
624 entry for Lagrange, J. L. 14, 123, 198, ...
625 Apsidal >> vector page 87 not 86
626 Boost Boost, 282, ...
626 Coriolis >> effect 125 not 126
629 Force .. centrifugal 175, 176
631 Inertial >> force, 5 Inertial >> frame, 2
632 Lorentz >> boost 282 not 284
635 Quadrature 75, 76, 211
636 Scattering >> laboratory coordinates, 114-120
Comment on the cover figure The cover figure shows the motion of a particle around a central force source. There are two central forces acting; an attractive force proportional to 1/r3 and a repulsive force proportional to 1/r4. The figure was probably inspired by the hard core repulsive nuclear force.
add paragraph additional akowledgements and a reference to this site

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