Maxwell’s Equations and E&M, Solution Manual

Maxwell’s equations and E&M waves Displacement-current flux The electric field between the plates equals σ/ϵ0 , so the displacement current is Jd = ϵ0 (∂E/∂t) = dσ/dt. The flux through S ′ is therefore Φ = Jd A = (dσ/dt)A, where A is the area of each plate. Hence, dσ d(σA) dQ A= = = I, (593) dt dt dt as desired. We haven’t paid attention to signs, but if the right plate in Fig. 9.4 is positive, and if the capacitor is discharging, then the displacement current points to the right. (The E field between the plates points to the left, but it is decreasing, so ∂E/∂t points to the right.) The displacement-current flux therefore passes from left to right through S ′ , just as the real-current flux passes from left to right through S . The total flux through the closed volume bounded by S and S ′ is zero, as it should be, because a closed surface has no boundary, so the line integral of B around this (non-existent) boundary is zero. Φ= Sphere with a hole ∫ Very close to the wire, the magnetic field is B = µ0 I/2πr. Therefore C B · ds = (µ0 I/2πr)(2πr) = µ0 I. On the right-hand side of Maxwell’s equation, the term involving J is zero because no current pierces the surface S (the sphere-minus-hole). To calculate the term involving ∂E/∂t, we know that the electric field at points on the surface S is E = Q/4πϵ0 R2 , where Q is the point charge and R is the radius of the sphere. Hence dE/dt = (dQ/dt)/4πϵ0 R2 = I/4πϵ0 R2 . Integrating this over the surface of the sphere brings in a factor of 4πR2 . Remembering the factor of µ0 ϵ0 out front, the right-hand side of Maxwell’s equation equals µ0 ϵ0 (I/4πϵ0 R2 )(4πR2 ) + 0 = µ0 I, in agreement with the left-hand side. Field inside a discharging capacitor Written in terms of the displacement current, the integral law reads ∫ ∫ B · ds = µ0 (Jd + J) · da. C S (594) Since s ≪ b we can neglect the edge fields, in which case the displacement current Jd is uniformly distributed in the gap. The integral of Jd over the area of the plates equals ∫ the conduction current I in the wire (see Exercise 9.13). The fraction of Jd · da = I 2 2 that is enclosed in a circle through P , centered on the axis, is πr /πb . The integral law applied to this circle therefore gives (with the conduction current J = 0 inside the capacitor) ( 2 ) r µ Ir 2πrB = µ0 I 2 + 0 =⇒ B = 0 2 , (595) 2πb b as desired. The similarity of this calculation to the calculation of the E field in Fig. 7.16 is the following. If we solve the problem straight from Maxwell’s equation, without invoking the definition of the displacement current, we can write (with J = 0 inside the capacitor) ∫ ∫ ∂E dΦE B · ds = µ0 ϵ0 , (596) · da =⇒ 2πrB = µ0 ϵ0 dt S ∂t C where ΦE is the flux of the electric field through the given surface. This equation is exactly analogous to Faraday’s law of induction, which we used in the example of Fig. 7.16 (among many other places), ∫ ∫ dΦB ∂B · da =⇒ 2πrE = − . (597) E · ds = − dt C S ∂t The similarity arises because of the symmetry of the two “curl” Maxwell’s equations; and also because there is no current J of real electric charges inside the capacitor in the present problem, and likewise there is no current of real magnetic charges in Fig. 7.16 (or anywhere else) because magnetic monopoles don’t exist (as far as we know). v dt Changing flux from a moving charge r v q θ θ Figure 147 A time dt later, the whole field pattern has moved a distance v dt to the right. So all of the electric flux that initially passed through the circle still passes through, but some additional flux now passes through. This additional flux is the flux that passes through the left circle in Fig. 147 but doesn’t pass through the right circle. (If you imagine the left circle riding along with the field pattern, it becomes the right (fixed) circle after a time dt.) Equivalently, the additional flux is the flux that passes through the cylindrical ring between the two circles, with radius r and width v dt. The area of this cylinder is 2πrv dt, so the change in electric flux through the fixed right circle is dΦE = (2πrv dt)E sin θ, where the sin θ factor comes from the fact that in finding the flux through the cylinder, we are concerned only with the component perpendicular to the x axis. Maxwell’s equation is therefore satisfied if the magnitudes E and B satisfy ∫ 1 dΦE 1 v B · ds = 2 =⇒ 2πrB = 2 2πrvE sin θ =⇒ B = 2 E sin θ. (598) c c c dt In view of the given relation B = (v/c2 ) × E, this is indeed true. The magnitude is correct because the cross product between v and E generates the sin θ. And the sign is correct because the flux increases to the right, which by the right-hand rule corresponds to B pointing in the counterclockwise direction when viewed from the right. This is consistent with the direction of B obtained from B = (v/c2) × E. Gaussian conditions In free space, the two “curl” Maxwell’s equations in Gaussian units are ∇×E=− 1 ∂B c ∂t and ∇×B= 1 ∂E . c ∂t (599) The given wave is E = ẑE0 sin(y − vt) and B = x̂B0 sin(y − vt). (600) The relevant derivatives are: ∇ × E = x̂E0 cos(y − vt), ∇ × B = −ẑB0 cos(y − vt), ∂E = −ẑvE0 cos(y − vt), ∂t ∂B = −x̂vB0 cos(y − vt). ∂t (601) The first of the equations in Eq. (599) yields E0 = (v/c)B0 , and the second yields B0 = (v/c)E0. Combining these gives v = ±c, and B0 = ±E0. Associated B field The wave is traveling in the −ẑ direction, as shown by the sign in (z +ct); if t increases, then z must decrease to keep the same value of (z + ct). B is perpendicular to both this direction and to E. So B must point in the ±(x̂ − ŷ) direction. But since we know that E × B points in the direction of the wave’s velocity, which is −ẑ, we must pick the “+” sign, as you can quickly verify with the right-hand rule. The magnitude of B is 1/c times the magnitude of E, so the desired B field is B = (E0 /c)(x̂ − ŷ) sin[(2π/λ)(z + ct)]. (602) · 108 m/s) = 6.67 · 10−8 T. The With E0 = 20 V/m, we have B0 = E0 /c = (20 V/m)(3 √ amplitudes of the E and B waves are actually √2 times E0 and B0 /c, respectively, because the magnitude of the (x̂ ± ŷ) vectors is 2. Find the wave It is given that E ⊥ ẑ. And we know that E ⊥ v, where v ∝ −x̂ here. So E must point in the ±ŷ direction. Let’s pick +ŷ. The other direction would simply change the sign of E0 ; the sign is arbitrary, since the trig function switches signs anyway. So we have (a sine would work just as well) E = yE ˆ 0 cos(kx + ωt), (603) where ω = 2πν = 6.28 · 108 s−1 and k = ω/c = 2.09 m−1 . The sign inside the cosine is a “+” because the wave is traveling in the negative x direction. Since E × B points in the direction of v, which is −x̂, and since B0 = E0 /c, the B field must take the form, B = −ẑ(E0 /c) cos(kx + ωt). (604) Kicked by a wave Equation (9.28) gives the electric field as E= 1+ E0 ŷ , (x + ct)2 ℓ2 (605) where E0 = 100 kV/m and∫ ℓ = 1 foot. We are concerned with the field at x = 0. Since F = dp/dt =⇒ p = F dt, the momentum acquired by the proton during the passage of the pulse is ∫ ∞ ∫ ∞ dt py = eEy dt = eE0 2 −∞ −∞ 1 + (ct/ℓ) ] πeE0 ℓ eE0 ℓ [ arctan(∞) − arctan(−∞) = . (606) = c c The proton’s final speed is then (using 1 ft = 0.305 m) vy = π(1.6 · 10−19 C)(105 V/m)(0.305 m) py πeE0 ℓ = = 3.1 · 104 m/s. = mc (1.67 · 10−27 kg)(3 · 108 m/s) m (607) The displacement during the few nanoseconds of acceleration is negligible. So one microsecond later the proton will be located at y = vy t = (3.1 · 104 m/s)(10−6 s) = 0.031 m, or 3.1 cm. Since B0 = E0 /c, the magnitude of the magnetic force qv × B is smaller than the magnitude of the electric force qE by the factor v/c ≈ 10−4 . So as mentioned in the statement of the exercise, the magnetic force is indeed negligible. Effect of the magnetic field Before the pulse has completely passed, the proton has acquired some velocity in the ŷ direction. It will therefore experience a force ev × B due to the magnetic field of the wave. Since B points in the −ẑ direction, this force points in the −x̂ direction, which is the direction in which the wave is traveling. The force would also be in that direction for a negative particle, because both the q and the v in the qv × B force switch sign. The wave tends to knock the particle along. ∫ Since F = dp/dt =⇒ p = F dt, we can say that in order of magnitude, if τ is the duration of the pulse with amplitude E, then py ∼ (eE)τ , so vy ∼ eEτ /m. The momentum in the x direction due to the magnetic force is px ∼ −(evy B)τ = −evy (E/c)τ . Therefore, px /py ∼ −vy /c ∼ −eEτ /mc. But in order of magnitude, τ equals ℓ/c, so we obtain px /py ∼ −eE0 ℓ/mc2 . For small angles, this ratio is essentially equal to the angle by which the final velocity has changed. We see that the effect is second order in 1/c. Note that px /py equals the ratio of two energies; the numerator is the work the electric field would do on the proton over a distance ℓ, and the denominator is the rest energy of the proton. Plane-wave pulse v E l B d Figure 148 ∫ (a) We want to apply the “displacement current” Maxwell equation, B · ds = µ0 ϵ0 dΦE /dt, to the loop. We’ll trace out the line integral in the counterclockwise direction, in which case the right-hand rule defines positive electric-field flux to point out of the page. Let the transverse length of the loop be ℓ, as shown in Fig. 148. Then since B = 0 outside the slab, the left-hand side of the above Maxwell equation is simply Bℓ. On the right-hand side, the flux increases because there is increasing overlap of the moving slab and the stationary loop. In time dt, the overlap area increases by ℓ(v dt). So the rate of area increase is ℓv, which means that dΦE /dt = Eℓv. Therefore, ∫ dΦE B · ds = µ0 ϵ0 =⇒ Bℓ = µ0 ϵ0 Eℓv =⇒ B = µ0 ϵ0 Ev. (608) dt (b) Using the “Faraday” Maxwell equation, the same argument with a loop perpendicular to the page (lying in the horizontal plane) gives ∫ dΦB =⇒ Eℓ = −(−Bℓv) =⇒ E = Bv. (609) E · ds = − dt The minus sign in the flux comes from the fact that if the loop is traced out in the counterclockwise direction when viewed from above, the right-hand rule defines positive magnetic-field flux to point upward, which is opposite to the direction of B. Plugging E = Bv into the result in part (a) gives B = µ0 ϵ0 (Bv)v, so v = √ 1/ µ0 ϵ0 . This equals c, as we know it should. Field in a box We immediately see that ∇ · E = 0, because Ez has no z dependence. And also ∇ · B = 0, because the ∂Bx /∂x and ∂By /∂y terms cancel. So two of Maxwell’s equations are satisfied. For the other two, we can calculate the curls via the usual determinant method,    x̂ ẑ  ŷ   (610) ∇ × E =  ∂/∂x ∂/∂y ∂/∂y  .  Ex Ey Ez  You can verify that the various derivatives are ) ( ∇ × E = kE0 − x̂ cos kx sin ky + ŷ sin kx cos ky cos ωt, ∂E = −ωẑE0 cos kx cos ky sin ωt, ∂t ∇ × B = −2kẑB0 cos kx cos ky sin ωt, ) ( ∂B = ωB0 x̂ cos kx sin ky − ŷ sin kx cos ky cos ωt. ∂t (611) B = (1/c2 )∂E/∂t Therefore, ∇ × E = −∂B/∂t gives kE0 = ωB0 . And ∇ × √ √ gives 2kB0 = ωE0 /c2 . These two requirements ω = 2ck and E0 = 2cB0 , quickly yield √ √ as desired. (Technically, ω = − 2ck and E0 = − 2cB0 also work, but these relations yield the same wave, as you can verify.) The fields don’t depend on z, so to determine what they look like, let’s consider the square cross section of the box in the xy plane. At all times, E is zero on the boundary of the box where (x, y) = (±π/2k, ±π/2k). At a given instant in time, cos ωt takes on a specific value, so E is proportional to ẑ cos kx cos ky. This function is maximum at the origin. The plot of Ez ∝ cos kx cos ky is basically a bump above the xy plane (or a valley below the xy plane at times when cos ωt is negative). The bump oscillates up and down according to cos ωt. The level curves of constant Ez are given by cos kx cos ky = C. You can show with a Taylor series that these level curves are circles near the origin. So the curves start off as circles and end up as squares. They are shown roughly in Fig. 149. Since E has only a z component, it points perpendicular to the page. (π/2k, π/2k) y x B isn’t quite as clean, but it’s easy to get a handle on its values along the x and y axes, and along the 45◦ lines, and also along the boundary of the box. Some sample vectors at times when sin ωt = 1 are shown in Fig. 150. All the vectors oscillate in phase according to sin ωt. Figure 149 Remark: That’s all that was required, but we can say a little more about the fields. For small x and y, we can use cos θ ≈ 1 and sin θ ≈ θ to obtain B ≈ kB0 (x̂y − ŷx) sin ωt. The field lines associated with this B field are circles, because the vector B ∝ (y, −x) is always perpendicular to the radial vector (x, y). Alternatively, since the tangent to the field line is in the direction of B, we can separate variables and integrate dy/dx = By /Bx = −x/y to obtain x2 + y 2 = C, where C is a constant. The B field goes to zero at the origin. y x Figure 150 What do the B field lines look like for general x and y values? Again, since the tangent to the field line is in the direction of B, we have the general relation, By dy sin kx cos ky . = =− cos kx sin ky dx Bx (612) Separating variables and integrating gives ln(cos ky) = − ln(cos kx) + D, where D is a constant. Exponentiating gives cos kx cos ky = C, where C = eD is another constant. Small values of C yield near-squares close to the boundary of the box, and values close to 1 yield the small near-circles close to the origin we found above. Note that the cos kx cos ky = C curves of the B field lines are also the curves of constant Ez , which we found above and plotted in Fig. 149. This can be traced to the fact that if E has only a z component, then ∇ × E is perpendicular to ∇Ez , as you can verify. Satellite signal The area covered is A = π(5 · 105 m)2 ≈ 8 · 1011 m2 . The power density is therefore S = P/A = (104 W)/(8 · 1011 m2 ) ≈ 10−8 W/m2 . So from Eq. (9.37) we have S= E2 377 Ω =⇒ E 2 = (377 Ω)(10−8 W/m2 ) =⇒ Erms ≈ 0.002 V/m, (613) or 2 millivolts/meter. Microwave background radiation As shown in Section 9.6, the average energy density U of a sinusoidal electromagnetic 2 . So we have wave is U = ϵ0 E02 /2 = ϵ0 Erms E 2rms = 4 · 10−14 J/m3 U = = 4.5 · 10−3 V2 /m2 =⇒ Erms = 0.067 V/m. (614) s2 C 2 ϵ0 8.85 · 10−12 kg m3 If the 1 kilowatt radiated by the transmitter is spread out over a sphere of radius R, then the power density at radius R equals S = (103 W)/4πR2 . The energy density is then U = S/c. We therefore want 1 103 W · = 4 · 10−14 J/m3 =⇒ R = 2600 m, c 4πR2 (615) or 2.6 km. However, the power is undoubtedly emitted in at least a somewhat directed manner, so the distance from an actual radio transmitter would be larger than this. An electromagnetic wave (a) The fields are E = ŷE0 sin(kx + ωt), and B = −ẑ(E0 /c) sin(kx + ωt). (616) We immediately see that ∇ · E = 0 (because the lone y component of E has no y dependence) and ∇ · B = 0 (because the lone z component of B has no z dependence). So two of Maxwell’s equations are satisfied. For the other two, you can verify that ∇ × E = ẑkE0 cos(kx + ωt), ∇ × B = ŷk(E0 /c) cos(kx + ωt), ∂E = ŷωE0 cos(kx + ωt), (617) ∂t ∂B = −ẑω(E0 /c) cos(kx + ωt). ∂t Therefore, ∇ ×E = −∂B/∂t requires k = ω/c. And (using µ0 ϵ0 = 1/c2 ) ∇× B = (1/c2 )∂E/∂t requires k/c = (1/c2 )ω, which again says that k = ω/c. (b) The wavelength is λ= 2π 2π(3 · 108 m/s) 2πc = = = 0.19 m. ω k 1010 s−1 (618) As shown in Section 9.6, the average energy density of a sinusoidal electromagnetic wave is ϵ0 E02/2, which equals 1 s 2 C2 ) 3 1( 8.85 · 10−12 (10 V/m)2 = 4.4 · 10−6 J/m3 . ϵ0 E02 = kg m3 2 2 (619) 1 S = ϵ0 E02 c = (4.4 · 10−6 J/m3 )(3 · 108 m/s) = 1300 J/(m2 s). 2 (620) The power density equals the energy density times the speed, so Reflected wave Let Ei be the amplitude of the oscillating electric field of the incident wave, and√Er that of the reflected wave. If half of the incident energy is reflected, then Er = Ei / 2. As in Section 9.5, the incident electric wave is described by Eq. (9.30) with E0 → Ei , and the reflected wave is described by Eq. (9.29) with E0 → Er . Using the trig sum formulas, you can check that the sum of these two waves is E = ẑ(Ei + Er ) sin 2πy 2πy 2πct 2πct + ẑ(Ei − Er ) cos sin cos . λ λ λ λ (621) At points where y equals λ/4, 3λ/4, 5λ/4, etc., the second term is zero, and the sin(2πy/λ) factor in the first term equals ±1. So E oscillates with amplitude Ei + Er . At these points the two oscillating electric fields are, and remain at all times, in phase. (In the setup in Fig. 9.10 the mirror was a perfect conductor, so Er = Ei .) Similarly, at points where y equals 0, λ/2, λ, 3λ/2, etc., the first term is zero, and the cos(2πy/λ) factor in the second term equals ±1. So E oscillates with amplitude Ei − Er . At these points the two oscillating electric fields are, and remain at all times, exactly 180◦ out of phase. (In Fig. 9.10 these points have E = 0 at all times, because Er = Ei .) Note that at these points E reaches its maximum value a quarter cycle before or after the maximum at the points in the previous paragraph, due to the sin(2πct/λ) versus cos(2πct/λ) dependence. √ In our case with Er = Ei / 2, the ratio of maximum amplitude observed to minimum amplitude observed is √ Ei + E r 1 + 1/ 2 √ = 5.83. = (622) Ei − E r 1 − 1/ 2 Poynting vector and resistance heating The electric field inside the wire is given by E = J/σ. Since the curl of E is zero, we can draw a thin rectangular loop along the surface to show that the electric field right outside the wire is also E = J/σ (and it points in the direction of the current, of course). The magnetic field right outside the wire points tangentially with the usual magnitude of B = µ0 I/2πR, where R is the radius of the wire. E and B are perpendicular, and you can show with the right-hand rule that the Poynting vector S = E × B/µ0 points radially into the wire. So the direction is correct; the energy in the wire increases, consistent with the fact that it heats up. The magnitude of S equals JI 1 1 J µ0 I = . (623) S= EB = 2πRσ µ0 µ0 σ 2πR To obtain the power flux into the wire through the surface, we must multiply by 2πRℓ, where ℓ is the length of a given section of the wire. So the total energy flow per time into a length ℓ of the wire is Pℓ = S · 2πRℓ = JI (I/A)I ℓ ρℓ JI 2πRℓ = ℓ= ℓ = I2 = I2 = I 2 R, 2πRσ σ σ σA A (624) where R is the resistance of the length ℓ of the wire. We have used the fact that the resistivity ρ is given by ρ = 1/σ. As desired, Pℓ equals the rate of resistance heating in the length ℓ of the wire. Pℓ can also be written as I(IR) = IV , of course, where V is the voltage drop along the length ℓ of the wire. Alternatively, we never actually had to use the J/σ form of E. A quicker method is: Pℓ = S · 2πRℓ = (625) because V = Eℓ. b E 1 µ0 I E · 2πRℓ = IEℓ = IV, µ0 2πR I r B Figure 151 S Energy flow in a capacitor We can find the magnetic field inside the capacitor by integrating the ∇ × B = µ0 J + ϵ0 µ0 ∂E/∂t Maxwell equation over a disk of radius r. Since J points upward in Fig. 151, the upper plate is positive, so ∂E/∂t points downward. We therefore have (using Stokes’ theorem) ∫ ∂E ∂E B · ds = µ0 I − ϵ0 µ0 (area) =⇒ B(2πr) = µ0 I − ϵ0 µ0 (πr2 ) ∂t ∂t µ0 I ϵ0 µ0 r ∂E =⇒ B = − . (626) 2 ∂t 2πr It will be helpful to write I in terms of E (or rather ∂E/∂t). We have I= dQ d(σA) d(ϵ0 EA) dE = = = ϵ0 (πb2 ) . dt dt dt dt Therefore, B= ϵ0 µ0 dE 2 dt ( ) b2 −r . r (627) (628) This is positive since we are concerned only with r values smaller than b. This magnetic field points tangentially around the circle of radius r, counterclockwise when viewed from above, as you can check with the right-hand rule. So it points into the page on the right. Since E points downward,1 you can quickly verify with the right-hand rule that the Poynting vector S = (E × B)/µ0 points away from the wire. Energy flows out from the wire into the surrounding space inside the capacitor. Since E is perpendicular to B, the magnitude of S is ) ( EB ϵ0 dE b2 S= = E −r . (629) 2 r dt µ0 The total energy per time (that is, the power) flowing out of a cylinder of radius r is S times the area 2πrh of the cylinder (where h is the separation between the plates). 1 Very close to the wire, the electric field actually points upward due to surface charges on the wire, but a little farther away the downward field from the capacitor dominates. See the remark at the end of the solution to Problem 9.10. The power is then ( )) ) ( ( ϵ0 dE b2 d ϵ0 E 2 dE P = (πb2 − πr2 )h = E (πb2 − πr2 )h . − r 2πrh = ϵ0 E 2 dt 2 r dt dt (630) This is correctly the rate of change of the energy stored in the electric field, because ϵ0 E02/2 is energy density and (πb2 − πr2 )h is the volume contained between radius r and radius b. If r ≈ 0, we obtain the statement that the power flowing away from the wire equals the rate of change of the total energy stored in the πb2 h volume between the plates. If r = b, we obtain zero power flow, which is correct. (As usual, we are ignoring edge effects in the capacitor and assuming that the electric field drops abruptly to zero at r = b.) As mentioned at the end of the example in Section 9.6.2, we don’t need to worry about the energy stored in the magnetic field. Comparing the energy densities If E(t) = E0 cos ωt, then ∂E/∂t = −ωE0 sin ωt, so the amplitude of the B field given in Eq. (9.46) is B0 = (ϵ0 µ0 r/2)(ωE0 ). The ratio of the magnetic energy density to the electric energy density is therefore )2 1 ( ϵ0 µ 0 r B 20 ωE0 µ0 ϵ0 r2 ω 2 ( πr )2 2µ0 2µ0 2 = = , 2 = ϵ 0 4 cT ϵ0 E 0 E 20 2 2 (631) where we have used ω = 2π/T and 1/µ0 ϵ0 = c2 . As desired, this result is small if the period T much larger than r/c, which is (half) the time it takes light to travel across the capacitor disks. As in Problem 9.6, we have ignored the high-order feedback effects between E and B. These effects are negligible if the current doesn’t change too quickly. E Field momentum of a moving charge We know that the electric field points radially with a magnitude essentially equal to the Coulomb value of E = q/4πϵ0 r2 . And Eq. (6.81) gives the magnetic field as B = (v/c2 )×E, so B points out of the page at the point shown in Fig. 152. (We’re using v here as the velocity of the charged particle, not the velocity of the primed frame in Eq. (6.81).) With θ defined as shown, the magnitude of B is (v/c2 )(q/4πϵ0 r2 ) sin θ = µ0 qv sin θ/4πr 2 , where we have used 1/ϵ0 c2 = µ0 . Problem 9.11 tells us that the momentum density is p̃ = S/c2 = (E × B)/µ0 c2 . At the point shown in Fig. 152, E×B points down and to the right. When we integrate p̃ over all space, only the rightward component survives; you can check that the transverse components associated with the angles θ and π − θ cancel (and likewise for the angles θ and −θ). So this brings in another factor of sin θ. The magnitude of the total momentum is therefore ∫ ∫ 1 1 S sin θ dv = p = EB sin θ dv 2 c c2 µ 0 ∫ ∞∫ π 1 1 q µ0 qv sin θ = sin θ(2πr sin θ)(r dθ) dr 2 2 4πr2 a 0 c µ0 4πϵ0 r ∫ ∫ π ∞ q2 v dr = sin3 θ dθ 8πϵ0 c2 a r2 0 2 14 q = 2 v, (632) c 3 8πϵ0 a r q θ v Figure 152 B as desired. We evaluated the trig integral here by writing sin3 θ as sin θ(1 − cos2 θ). Alternatively, you can use the integral table in Appendix K. A Lorentz invariant In terms of the parallel and perpendicular components, we have E ′2 − c2 B ′2 = = where we have used E∥′ ′2 ) − c2 (B ′2 ′2 (E ′2 ∥ + E⊥ ∥ + B⊥ ) ′ ′ ), (E∥2 − c2 B∥2) + (E′⊥ · E⊥ − c2 B′⊥ · B⊥ = E∥ and B∥′ (633) = B∥ . We must now deal with the “⊥” terms. Our goal is to show that they combine to form 2 − c2 B 2 . Using the expression for E′ in Eq. (6.76), the E′ · E′ E⊥ ⊥ ⊥ ⊥ ⊥ term becomes [ ] ( ) ( ) 2 2 2 γ E⊥ + v × B⊥ · E⊥ + v × B⊥ = γ E ⊥ + 2E⊥ · (v × B⊥ ) + (v × B⊥ )2 . (634) Since B⊥ is perpendicular to v by definition, we have (v × B⊥ )2 = v 2 B⊥2 . Hence ] [ ′ 2 + 2E · (v × B ) + v 2 B 2 E′⊥ · E⊥ (635) = γ 2 E⊥ ⊥ ⊥ ⊥ . In a similar manner we find [ ] ′ −c2 B′⊥ · B⊥ = c2 γ 2 − B 2⊥ + 2(1/c2 )B⊥ · (v × E⊥ ) − (v 2 /c4 )E⊥2 . (636) When we add the previous two equations, the two middle terms will cancel if E⊥ · (v × B⊥ ) = −B⊥ · (v × E⊥ ). This is indeed true, because each of these “dot-cross” products can be transformed into the other by cyclicly permuting the vectors (which doesn’t change anything) and then reversing the order of the cross product (which brings in a minus sign). The sum of Eqs. (635) and (636) is therefore ′ ′ E⊥ · E′⊥ − c2 B′⊥ · B⊥ 2 2 = γ 2 (1 − v 2 /c2 )E⊥ − c2 γ 2 (1 − v 2 /c2 )B⊥ 2 2 2 = E⊥ − c B ⊥ . (637) Substituting this into Eq. (633) gives E ′2 − c2 B ′2 = = = (E∥2 − c2 B∥2) + (E 2⊥ − c2 B⊥2 ) 2 2 ) (E∥2 + E⊥ ) − c2 (B 2∥ + B⊥ E 2 − c2 B 2 , (638) as desired. Alternatively, it doesn’t take too long to solve this exercise by explicitly writing out the squares of all the components of E and B in Eq. (6.74).