Appendix F-J, Solution Manual

Appendix F F.1. Divergence using two systems (a) In Cartesian coordinates we quickly find ∇ · A = 2. In cylindrical coordinates we have ∇ · A = (1/r) ∂(rAr )/∂r = (1/r) ∂(r2 )/∂r = 2, as desired. (b) In Cartesian coordinates we quickly find ∇ · A = 3. In cylindrical coordinates we can write A as A = xx̂ + 2yŷ = r cos θ(r̂ cos θ − θ̂ sin θ) + 2r sin θ(r̂ sin θ + θ̂ cos θ) = r̂r(cos2 θ + 2 sin2 θ) + θ̂r sin θ cos θ. (751) You can check that this agrees with what you would obtain by projecting A onto the r̂ and θ̂ unit vectors. Taking the divergence of A gives ∇·A 1 ∂(rAr ) 1 ∂Aθ + r ∂θ r (∂r ) 2 2 1 ∂(r sin θ cos θ) 1 ∂ r · r(cos θ + 2 sin θ) + = ∂r r r ∂θ = (2 cos2 θ + 4 sin2 θ) + (cos2 θ − sin2 θ) = 3 cos2 θ + 3 sin2 θ = 3, = (752) ∆θ is negative y C ∆r as desired. In Fig. 167, the first-order change in a function f in going from point A to point B is ∆f = (∂f /∂x)∆x. But we can also imagine going from A to B in two steps along the radial and tangential segments shown, via point C. So we can also write ∆f as (∂f /∂r)∆r + (∂f /∂θ)∆θ. Therefore, ∆f = = =⇒ ∆f ∆x r∆θ θ F.2. Cylindrical divergence = 1 ∂f ∂f ∆r + (r∆θ) ∂r r ∂θ ∂f 1 ∂f (∆x cos θ) + (−∆x sin θ) r ∂θ ∂r ∂f 1 ∂f cos θ − sin θ. ∂r r ∂θ (753) But in the limit where A and B are close together, the left-hand side is just ∂f /∂x. Since this equation holds for an arbitrary function f , we can erase the f , which yields the first equation in Eq. (F.28). Similarly, breaking a vertical ∆y segment into the ∆r = ∆y sin θ and r∆θ = ∆y cos θ pieces yields the second equation in Eq. (F.28). A ∆x θ Figure 167 B x Aθ is negative y C Ar Aθ θ A Ax θ Figure 168 B x We can use the same type of reasoning to find the relation between the Cartesian and cylindrical components of a vector. In Fig. 168, Ax is the horizontal component of a vector A. But we can also imagine breaking up Ax x̂ into its radial and tangential components. The horizontal component of the radial vector Arr̂ is Ar cos θ, and the horizontal component of the tangential vector Aθθˆ is −Aθ sin θ (note that Aθ is negative here). So we must have Ax = Ar cos θ − Aθ sin θ, in agreement with the first equation in Eq. (F.29). Likewise, looking at the vertical component Ay gives Ay = Ar sin θ + Aθ cos θ. Alternatively, we can use the expressions for the r̂ and θ̂ unit vectors given in the statement of Exercise F.1. A = Ar r̂ + Aθ θ̂ = Ar (x̂ cos θ + ŷ sin θ) + Aθ (−x̂ sin θ + ŷ cos θ) = x̂(Ar cos θ − Aθ sin θ) + ŷ(Ar sin θ + Aθ cos θ). (754) The x and y components that we read off from this equation agree with those in Eq. (F.29). Now we must calculate ∇ · A = ∂Ax /∂x + ∂Ay /∂y + ∂Az /∂z . Using the above expressions for the partial derivatives and the components, and ignoring the z term, we have (with c ≡ cos θ and s ≡ sin θ) ∇·A ( ) ∂ 1 ∂ cos θ − sin θ (Ar cos θ − Aθ sin θ) r ∂θ ∂r ( ) ∂ 1 ∂ + sin θ + cos θ (Ar sin θ + Aθ cos θ) r ∂θ ∂r ) ) ( ( ✚ ❍1 ∂Ar 1 ∂Aθ ❩ 1 ✚ 1 ∂Aθ ∂Ar + − sc ❍❍ + s2 Ar + s2 + sc❩ − cs ✚ = c2 ✚Aθ r ∂r r ∂θ r ∂θ❍ ✚r ❩ ✚ ∂r ( ) ( ) ❍ 1 ∂Ar ✚ ∂A ❩1 ✚ ✚ θ 21 2 1 ∂Aθ 2 ∂Ar ❍ − cs ❩ + s + c Ar + c + cs ❍ ✚Aθ + sc✚ r ∂θ r ∂θ❍ r ∂r ✚r ❩ ✚ ∂r 1 ∂Aθ ∂Ar 1 + Ar + , (755) = r ∂θ r ∂r = in agreement with the expression in Eq. (F.2) (with the first term expanded out). The z term is the same. F.3. General expression for divergence Consider a small “box” in space, analogous to the one in Fig. F.2. The faces can be grouped in three pairs, with each pair being perpendicular to a particular coordinate axis. Consider one of the faces perpendicular to x̂1 . Let the center of this face be located at (x1 , x2 , x3 ). Its area is [f2 (x1 ) dx2 ][f3 (x1 ) dx3 ], where we have suppressed the x2 and x3 arguments of the f factors. The flux of a vector A inward through this face is therefore A1 (x1 ) · f2 (x1 ) dx2 · f3 (x1 ) dx3 . Similarly, the flux of A outward through the opposite face, whose center is located at the point (x1 + dx1 , x2 , x3 ), is A1 (x1 + dx1 ) · f2 (x1 + dx1 ) dx2 · f3 (x1 + dx1 ) dx3 . (756) This flux is different due to the facts that both the value of A1 and the area of the face at (x1 + dx1 , x2 , x3 ) are different (in general) from what they are at (x1 , x2 , x3 ). The volume of the box is f1 dx1 · f2 dx2 · f3 dx3 , so the net contribution from these two sides to the divergence is = A1 (x1 + dx1 ) · f2 (x1 + dx1 ) dx2 · f3 (x1 + dx1 ) dx3 − A1 (x1 ) · f2 (x1 ) dx2 · f3 (x1 ) dx3 f1 dx1 · f2 dx2 · f3 dx3 1 A1 (x1 + dx1 ) · f2 (x1 + dx1 ) · f3 (x1 + dx1 ) − A1 (x1 ) · f2 (x1 ) · f3 (x1 ) . (757) dx1 f 1 f2 f3 By the definition of the partial derivative, this is simply equal to 1 ∂(A1 f2 f3 ) , f 1 f2 f 3 ∂x1 (758) which agrees with the first term in Eq. (F.31). The other two pairs of faces work out in exactly the same manner. In spherical coordinates, f1 , f2 , f3 are equal to 1, r, r sin θ. So the given expression becomes (with the indices 1, 2, 3 changed to r, θ, ϕ) ) ( ∂(r2 sin θAr ) 1 ∂(rAϕ ) ∂(r sin θAθ ) ∇·A = + + 2 ∂ϕ r sin θ ∂θ ∂r 1 ∂(r2 Ar ) 1 ∂Aϕ 1 ∂(sin θAθ ) = + + , (759) r sin θ ∂ϕ r sin θ ∂θ ∂r r2 which agrees with the expression in Eq. (F.3). F.4. Laplacian using two systems (a) In Cartesian coordinates we quickly find ∇2 f = 2 + 2 = 4. In cylindrical coordinates we have ∇2 f = (1/r) ∂(r ∂f /∂r)/∂r = (1/r) ∂(r · 2r)/∂r = 4, as desired. (b) In Cartesian coordinates we have ∇2 f = 12x2 + 12y 2 = 12r2 . In cylindrical coordinates we can write x = r sin θ and y = r cos θ, so x4 +y 4 = r4 (sin4 θ +cos4 θ). We need to calculate ( ) ∂f 1 ∂2f 1∂ (760) ∇2 f = r + 2 2. r ∂r r ∂θ ∂r The first term is quickly found to be 16r2 (sin4 θ +cos4 θ). The second term equals (letting sin θ → s and cos θ → c) ∂(4s3 c − 4c3 s) = r2 (12s2 c2 − 4s4 + 12c2 s2 − 4c4 ) = 24r2 s2 c2 − 4r2 (s4 + c4 ). ∂θ (761) Putting it all together gives r2 ∇2 f = 12r2 (s4 + c4 ) + 24r2 s2 c2 = 12(s2 + c2 )2 = 12r2 , (762) as desired. F.5. “Sphere” averages in one and two dimensions In one and two dimensions, the procedure leading up to Eq. (F.22) is essentially the same. In 2D, the area in Eq. (F.19) is the area 2πrz of a cylinder, so we have ∫ dfavg,r 1 (763) ∇2 f dV. = 2πrz dr In special case where the cylinder is very small, we can write the volume integral ∫ the ∇2f dV as (πr2 z)(∇2f )center. Therefore, 1 dfavg,r r = (πr 2 z)(∇2f )center = (∇2f )center. dr 2πrz 2 (764) Since (∇2f )center is a constant, we can integrate with respect to r to obtain favg,r = fcenter + r2 2 (∇ f )center . 4 (765) Similarly, in 1D the area in Eq. (F.19) is the area 2yz of the two faces of the slab. So we have ∫ dfavg,±x 1 (766) ∇2f dV. = 2yz dx In special case where the slab is very small, we can write the volume integral ∫ the ∇2f dV as (2xyz)(∇2f )center, assuming that the slab extends from −x to x. Therefore, 1 dfavg,±x (2xyz )(∇2f )center = x(∇2f )center. (767) = 2yz dx Integrating with respect to x gives favg,±x = fcenter + x2 2 (∇ f )center . 2 (768) Note that the denominator in the second term in the results in Eqs. (768), (765), and (F.25) equals 3d, where d is the dimension of the space. The result in Eq. (768) makes sense, because in 1D, ∇2f is simply d2 f /dx2 , so the result is consistent with what we obtain by adding together the Taylor series, f (x) = f (0) + xf ′ (0) + x2 ′′ f (0) + · · · , 2 (769) evaluated at x and −x, and dividing by 2. The first-order terms cancel, while the zeroth- and second-order terms add, yielding Eq. (768). F.6. Average over a cube To second order, the Taylor series in three Cartesian coordinates, expanded around the origin, looks like f (x, y, z) = f0 + xfx + yfy + zfz +xyfxy + xzfxz + yzfyz x2 y2 z2 + fxx + fyy + fzz . 2 2 2 (770) The subscripts denote the coordinates that the partial derivatives are taken with respect to. All partial derivatives are evaluated at the origin. To check that this is indeed the correct expansion, we can take various partial derivatives of both sides and then evaluate both sides at the origin. For example, taking ∂ 2 /∂x ∂y of both sides and then setting (x, y, z) = (0, 0, 0) generates an equality; all terms except the fxy term vanish on the right-hand side. (Remember that the partial derivatives are constants.) This procedure shows that at the origin, the right-hand side has the correct value of the function and its first two derivatives. So it must in fact be the correct function (at least to second order). When we take the average of the above expression over the surface of the given cube, the x, y, and z terms vanish, because for every point with a given value of x, there is a point with the value −x. For the same reason, the xy, xz, and yz terms vanish. Our task therefore reduces to finding the average value of, say, x2 over the surface of the cube. Two of the faces have x = ±ℓ, so the average value of x2 over these faces 2 is simply ℓ2 . For the (other ∫ ℓ four) faces, x ranges from −ℓ to ℓ, so the average of x is given by the integral −ℓ x2 dx /2ℓ, which equals ℓ2 /3. The average of x2 over all six faces is therefore ( ) ℓ2 5ℓ2 1 2 · ℓ2 + 4 · x2 = = . (771) 6 3 9 The same result holds for the averages of y 2 and z 2 , so from Eq. (770) the average of f over the entire surface of the cube is favg = fcenter + 5ℓ2 1 5ℓ2 (fxx + fyy + fzz ) = fcenter + (∇2f )center, 18 2 9 (772) as desired. A sphere with radius ℓ lies completely inside the given cube of side 2ℓ. From Eq. (F.25), this sphere would have a 1/6 in place of the 5/18. Consistent with this, we have 5/18 > 1/6; the average value of f over the surface of the cube is larger 2 than the average value over the √ surface of the sphere (assuming (∇ f )center is positive). Also, a sphere with radius 3ℓ lies completely the given cube of side 2ℓ (it √ outside touches its corers). This sphere would have a ( 3)2 /6 in place of the 5/18. Consistent with this, we have 3/6 > 5/18. Appendix H H.1. Ratio of energies From Eq. (H.7) the power radiated is P = e2 a2 /6πϵ0 c3 . The total energy radiated is P t. Therefore, since a = v/t, the desired ratio of energies is e2 (v/t)2 ·t e2 e2 4r 4 1 6πϵ0 c3 = 0. = = 3 ct 3 4πϵ0 mc2 ct 3πϵ0 mc3 t mv2 2 (773) Note that ct is larger than vt, which is twice as large as the stopping distance vt/2. So unless the electron stops within a distance that is of order r0 , the radiated energy will be negligible compared with the initial kinetic energy. H.2. Simple harmonic motion (a) If the position is given by x = A cos ωt, then the acceleration is a(t) = d2 x/dt2 = −Aω2 cos ωt. The average of cos2 ωt over a period is 1/2, so the average of a2 is A2 ω 4 /2. From Eq. (H.7) the average power radiated is then P = e 2 A2 ω 4 e2 (A2 ω 4 /2) = . 6πϵ0 c3 12πϵ0 c3 (774) (b) The velocity of the electron is v(t) = −Aω sin ωt, so the initial speed is v = Aω. The initial energy of the oscillator is therefore U = mv 2 /2 = mA2 ω 2 /2. As time goes on, the amplitude will decrease. But in terms of the amplitude at any instant, U and P are given by the above expressions. They are therefore always related by P = (e2 ω 2 /6πϵ0 mc3 )U . So with 6πϵ0 mc3 /e2 ω 2 ≡ T , we have dU = −P dt =⇒ =⇒ ∫ U ∫ t ′ dU U dU ′ dt =− =⇒ =− ′ dt T U U0 0 T ( ) U t =⇒ U (t) = U0 e−t/T . ln =− T U0 (775) So U falls to 1/e of its initial value after a time T = 6πϵ0 mc3 /e2 ω 2 . There was actually no need to separate variables and integrate as we did, because we know that the solution to the differential equation, dU/dt ∝ −U , is simply an exponential. Note that ωT , which is the number of radians of oscillation during the time T , can be written as ωT = 3 λ 6πϵ0 mc3 λ 3 c 6πϵ0 mc2 = , = = 2π 2r0 4π r0 ω e2 e2 ω (776) where λ is the wavelength of the emitted light (which satisfied λν = c), and r0 is the classical electron radius, r0 = e2 /4πϵ0 mc2 . So if λ is much larger than r0 , then many oscillations will occur before the amplitude decays significantly. H.3. Thompson scattering If the maximum acceleration is E0 e/m, the acceleration as a function of time looks like a(t) = (E0 e/m) cos ωt. Since the average value of cos2 ωt is 1/2, the average value of a2 is E02 e2 /2m2 . From Eq. (H.7) the average power radiated is therefore P = e2 (E 02 e2 /2m2 ) E 20 e4 . = 12πϵ0 m2 c3 6πϵ0 c3 (777) Dividing this result by the power density (power per unit area), ϵ0 E02 c/2, we find that the area that receives an amount of power P is σ= e4 . 6πϵ20 m2 c4 (778) This is the scattering cross section. In terms of the classical electron radius, r0 = e2 /4πϵ0 mc2 , we can write σ as σ = (8π/3)r02. Since r0 = 2.8 · 10−15 m, we have σ = 6.6 · 10−29 m2 . As far as energy absorption goes, the electron looks like it takes up this much area, from the wave’s point of view. H.4. Synchrotron radiation Let the electron’s velocity in the lab frame be v. Consider the inertial frame F ′ moving along with the electron at a given instant. Using the Lorentz transformations, ′ the electric field in F ′ is E⊥ = γv × B⊥ . (B⊥ here is simply the B field in the lab frame.) This electric field causes the electron to accelerate in frame F ′ (where its initial velocity was zero) with an acceleration of a = eE⊥ /m. From Eq. (H.7) this acceleration causes the electron to radiate energy in F ′ at a rate P′ = γ 2 e4 v 2 B 2 γ 2 e4 B 2 e2 (eE⊥ /m)2 = ≈ , 6πϵ0 c3 6πϵ0 m2 c3 6πϵ0 m2 c (779) where we have used the fact that β ≈ c, since we are told that the electron is highly relativistic. We now claim that the power is the same in both the frame F ′ and the lab frame F , in which case transforming back to the lab frame doesn’t change the answer. This claim is true because power is energy per time, and both energy U and time t transform the same way under a Lorentz transformation; they are both the fourth (or first, depending on the convention) component of a 4-vector. More precisely, in this particular case the x-t Lorentz transformation gives ∆t = γ∆t′ , because ∆x′ = 0 in F ′ . And the p-E Lorentz transformation (E here is energy, not electric field) gives E = γE ′ , because p′ = 0 in F ′ . This then implies ∆E = γ∆E ′ . Therefore, ∆E/∆t = ∆E ′ /∆t′ =⇒ P = P ′. As time goes on, we will need to continually pick new inertial frames F ′ that move along with the electron. In any one of these frames the power equals the P ′ we found above, so the power in the lab frame takes on the constant value P = P ′ . Appendix J J.1. Emf from a proton At an instant when the magnetic moment is perpendicular to the plane of the coil, the B field at points in the plane of the coil is perpendicular to the plane and has magnitude µ0 m/4πr3 . As mentioned in the caption of Fig. J.2, the flux through the coil is determined by the field outside the coil. Demonstrating this fact was the task of Exercise 7.37. In short, every field line of the dipole passes through the tiny region inside the dipole’s current loop (or whatever is going on inside a quantum mechanical spin). The lines that then loop around and pass back through the inside of the coil form a closed loop inside the coil and hence produce no net flux. The lines that loop around outside the coil have an uncanceled flux through the inside of the coil. See Figure 137. Since there are four turns in the given coil, the maximum flux through the coil has magnitude ∫ ∞ 2µ0 m µ0 m Φmax = 4 2πr dr = . (780) 4πr3 a a If m(t) precesses sinusoidally, then Φ(t) is likewise a sinusoidal function. Therefore Φ(t) = (2µ0 m/a) cos ωp t, and the induced emf is E(t) = −dΦ/dt = (2µ0 mωp /a) sin ωp t. The amplitude of the emf is then ) ( (1.411 · 10−26 J/T)ωp 2µ0 mωp 2 4π · 10−7 kgCm 2 E0 = = a a ωp = (3.55 · 10−32 V m s) . (781) a Alternatively, you can use the result from Exercise 11.19. Since the magnetic field at the center of a ring is µ0 I/2a, the result E1 = (ω/I1 )B1 · m0 from that exercise becomes E = ωp (µ0 /2a)m. Multiplying by 4 due to the four turns yields the above result. J.2. Emf from a bottle (a) There are essentially 6 · 1023 nucleons in a gram of anything. This number (Avogadro’s number) is the inverse of the proton mass 1.67 · 10−24 g (the neutron mass is nearly the same). Since 2/18 of the nucleons in water are the protons in the hydrogen atoms, the number of protons in 200 cm3 of water (which is the same as 200 g) is 2 N = (200 g)(6 · 1023 g−1 ) = 1.33 · 1025 . (782) 18 In order of magnitude, the fractional excess of magnetic moments pointing along the B0 = 0.1 T magnetic field is f= mB0 (1.41 · 10−26 J/T)(0.1 T) = 3.5 · 10−7 . = 4 · 10−21 J kT (783) So we expect the net magnetic moment in the sample to be roughly mnet = f Nm = (3.5 · 10−7 )(1.33 · 1025 )(1.41 · 10−26 J/T) = 6.6 · 10−8 J/T. (784) If you want to be a little more precise, you can use the Boltzmann distribution. Let N/2 + n protons point in the direction of B0 , and N/2 − n point in the opposite direction. The difference in energy of these two states is 2mB0 . So in thermal equilibrium we have N/2 − n = e−2mB0 /kT N/2 + n =⇒ =⇒ 2mB0 1 − 2n/N ≈1− kT 1 + 2n/N 4n 2mB0 mB0 1− ≈1− =⇒ n ≈ N(785) . N kT 2kT The net magnetic moment is then mnet = nm − n(−m) = 2nm = mB0 N m. kT (786) This agrees with the mnet = f N m result we obtained above, even though that was just an order-of-magnitude calculation. (b) From page 822 in the text, the proton spin precesses at a frequency of 4258 revolutions per second in a field of 1 gauss. Since this frequency is proportional to B (it equals mB/J), the angular frequency in a field of 0.4 gauss is ωp = 2π · 0.4 · 4258 s−1 = 1.07 · 104 s−1 . For a rough estimate of the signal voltage, assume that the protons are all near the center of a ring coil. Then the only modification to Exercise J.1 we need to make is to multiply the result by 500/4, since we now have 500 turns instead of 4. So we find ( ) kg m 500 2µ0 mnet ωp 500 2 4π · 10−7 C2 (6.6 · 10−8 J/T)(1.07 · 104 s−1 ) E0 = = 0.04 m 4 4 a (787) = 5.5 · 10−6 V. Actually, the coil in Fig. J.3 is more like a solenoid than a ring, and the water almost fills it. But the result from Exercise 11.19, combined with the fact that the field inside the solenoid is somewhat uniform, implies that the actual E0 won’t be too much different from the one we calculated.