errata for Panofsky and Phillips

Errata for Panofsky and Phillips Classical Electricity and Magnetism p. 33 Eq. 2–19 The constant is not arbitrary but is proportion al to the magnitude of any surface dipole layer (Eq. 1–61). pp. 39–40 Properly, an isolated molecule is in a state of defin ite parity, such as one of its ro- tational eigenstates, and therefore has no permanent elect ric dipole moment (the expectation value of an operator of odd parity, such as the el ectric dipole moment, in a state of definite parity is zero). However, a proper quantum mechanical calculation gives the same result at the classical calculation here. The details are in P. Debye’s 1929 book Polar Molecules , but not in most modern textbooks. p. 40 Eq. 2–45 fails for many polar liquids, like H 2O, because of short-range correlations resulting from strong angle-dependent (covalent) intermo lecular forces. p. 48 It is clearer to say that the triangles with sides rldandl′r′d′are similar. p. 85 Eq. 5–16 the last = should be a −. p. 92 Eq. 5–42 e0should be ǫ0. p. 93 Exc. 3 Pνshould be Pn. p. 99 The derivation of the symmetry of καβ(last paragraph, Eqs. 6–18 and 6–19) is incorrect. The symmetry of καβfollows from defining dielectric susceptibility in a thermodynamically consistent manner, as the second deriva tive tensor of the free energy with respect to the electric field; see Landau and Lifs chitzElectrodynamics of Continuous Media §11. p. 113 Eq. 6–76 the upper limit should be B. p. 117 Exc. 7 (6–44) should be (6–45). p. 128 Eq. 7–42/contintegraltext should be/integraltext . p. 168 Exc. 9–4 can be solved if the orbital radius is taken to b e constant, which is inconsis- tent with the conditions given. It is possible to solve this p roblem if the central force field is ignored, the initial /vectorBis nonzero, and its variation is slow compared to the gyroperiod. The second part of Exc. 9–7 shows that for a speci al spatial distribution of/vectorBand no central force field the orbital radius remains constan t. p. 168 Exc. 9–7 (first part) requires the assumption that˙/vectorBis independent of ϕforanyaxis z. Do only the second part. p. 174 Eq. 10–14∂ ∂tshould bed dt. p. 321 Eq. 17–73 −2c2p1·¯ p1should be +2 c2p1·¯ p1. Eq. 17–74 should have a −sign on one side. p. 364 Eq. 20–35 ( u2/c)2should be ( u2/c2). p. 365 Eq. 20–38 should have a factor m0c2on the right. p. 370 The reason real steady currents flowing through wires d on’t radiate significantly is their low speed and the Pauli exclusion principle; the curre nt distribution is not mi- croscopically continuous, contrary to the assertion here. p. 378 One line above 21–9, (17–32) should be (18–32). Eq. 21–9 The −sign should be +. p. 389 Eq. 21-56 The first −sign on the left hand side should be +. p. 401 Eq. 22–1 In the middle and right hand side (two places) xshould be...x. p. 404 One line after 22–20, (19–20) should be (20–18). p. 412 After the unnumbered equation (22–9) should be (22–19 ). p. 419 Second line from bottom, ǫ|H|2should be ǫ|E|2. p. 420 Eq. 22–76 The first exponential is e−ikr(1−cosθ). p. 422 Eq. 22–89 This equation appears to give the erroneous r esult that for κ2= 0 (as is true to high accuracy for optical glass) but κ1/negationslash= 1 the scattering cross-section must be zero. This paradox arises from the use of the infinite-medi um relation between p andE. In the case of a small particle (or interface) the proper rel ation must include radiation damping, implying κ2/negationslash= 0 (with its actual value determined by 22–89). p. 465 1 farad = 9 ×1011cm (cgs capacitance). 1 ohm =1 9×10−11sec/cm (cgs resistance; = esu).