The Fields of Moving Charges, Solution Manual

The fields of moving charges Capacitor plates in two frames The electric field in the lab frame is E0 = V0 /d = (300 V)/(.02 m) = 15, 000 V/m. The charge density is σ0 = ϵ0 E0 , so the charge on the plates is Q0 = σ0 A0 = ϵ0 E0 A0 . The number of excess electrons on the negative plate in therefore N = Q0 /e = ϵ0 E0 A0 /e, which yields ) ( s2 C2 2 8.85 · 10−12 kg m3 (15, 000 V/m)(0.02 m ) (380) = 1.66 · 1010. N= 1.6 · 10−19 C In the frame F1 moving east at v = 0.6c, the plates are moving west at 0.6c. The γ factor associated with 0.6c is 5/4. So the EW dimension is shrunk to (20 cm)/γ = 16 cm. The NS dimension is unchanged, as is the vertical separation. The number of electrons on the negative plate is the same. But the area of the plates is smaller by 1/γ, so σ1 (and hence E1 ) is larger by a factor γ. Therefore, E1 = γE0 = 18, 750 V/m. In the frame F2 moving upward at v = 0.6c, the plates are moving downward at 0.6c. The plate dimensions are still 20 cm and 10 cm, but the vertical separation is shrunk to (2 cm)/γ = 1.6 cm. The number of electrons is the same. And the density is the same also, so E2 = E0 = 15, 000 V/m. Electron beam (a) 0.05 µA equals 5 · 10−8 C/s, so the number of electrons passing a given point per second is n = (5 · 10−8 C/s)/(1.6 · 10−19 C) = 3.1 · 1011 s−1 . They move at essentially speed c, so the average distance between them is d = c/n = (3 · 108 m/s)/(3.1 · 1011 s−1 ) ≈ 0.001 m = 1 mm. The linear charge density is then λ = e/d = (1.6 · 10−19 C)/(0.001 m) = 1.6 · 10−16 C/m. The electric field 1 cm from the beam is therefore E= 1.6 · 10−16 C/m λ ) ( = 2.88 · 10−4 V/m. = s 2 C2 2πϵ0 r 2π 8.85 · 10−12 kg m3 (0.01 m) (381) Note that the distance between the electrons (1 mm) is small compared with the distance from the beam (1 cm), so the beam looks roughly like a uniform distribution of charge. (b) In the electron rest frame, the electrons are “uncontracted” compared with the lab frame, so their average separation is d′ = γd = (20)(0.001 m) = 0.02 m = 2 cm. The linear charge density, and hence electric field, is therefore decreased by a factor 1/γ. So the field 1 cm from the beam is E′ = E 2.88 · 10−4 V/m = 1.44 · 10−5 V/m. = 20 γ (382) This is the average (along a line parallel to the beam) of the radial component of the field. Because the electrons are relatively far apart (2 cm, compared with the 1 cm distance from the beam), there is a large variation in the field as the position varies along the line parallel to the beam. Tilted sheet In√the notation of the example in Section 5.5, the distance between A′ and B ′ is ℓ 1 + (1/γ)2 . So the same charge-invariance argument gives σ′ √ √ √ 2γ σ. 1 + (1/γ)2 = σ 2 =⇒ σ ′ = √ 1 + γ2 (383) The factor here correctly equals 1.1043 if γ = 5/4 (that is, if v = 3c/5). √ √ The same reasoning also gives the magnitude of the field in F ′ as E ′ = (E/ 2) 1 + γ 2 . ′ , we must multiply this by cos(2θ − 90◦ ), where To find the normal component, En tan θ = γ. This trig factor can alternatively be written as sin 2θ, which in turn can be written as 2 sin θ cos θ. So we have E ′n = = E √ E ′ cos(2θ − 90◦ ) = √ 1 + γ 2 · 2 sin θ cos θ 2 √ 1 E √ γ 2γ √ √ 1 + γ2 · 2 √ E. = √ 2 2 1+γ 2 1 + γ2 1+γ (384) This is the same factor that relates σ ′ to σ. So if E = σ/2ϵ0 is true (which it is), then En′ = σ ′ /2ϵ0 is also true. That is, Gauss’s law holds in F ′ . ′ ′ Limits: If γ = 1 (that is, v = 0), we obtain √ σ = σ and En√= E, as expected. And if γ → ∞ (that is, v → c), we obtain σ ′ = 2σ and En′ = 2E; the sheet√is vertical, so a given amount of charge in F ′ is now located within a span that is 1/ 2 times as large as it was in F . z Adding the fields e a a -e 0.8c P Figure 100 x All fields in this problem involve the factor e/4πϵ0 a2 , so let’s ignore that for now. At the point P = (a, 0, 0), √ the field from the proton points down to the√right in Fig. 100 with magnitude (1/ 2)2 . The components of this are Exp = 1/(2 2) and √ Ezp = −1/(2 2). We’ll assume that the negative muon is moving in the positive x direction, although this doesn’t affect the answer; the field is the same in front or behind. The field directly in front of the muon is obtained by setting θ = 0 in Eq. (5.15). After stripping off the e/4πϵ0 a2 we have Exµ = −(1 − β 2 ) = −0.36. Adding the fields from the two particles gives a total field with Ex = −0.00645 and Ez = −0.354 (times e/4πϵ0 a2 ). Note that Ex is very close to zero. A β value of about 0.80402 would make Ex exactly equal to zero. By continuity, such a speed must exist, because if β = 0 then the muon’s field dominates and the net Ex is negative, whereas if β → 1 then the x component of the muon’s field is zero, so the net Ex (due only to the proton) is positive. Forgetting relativity Assuming β = 0, the factor (1 − β 2 )/(1 − β 2 sin2 θ)3/2 in Eq. (5.15) is smaller than 1 when θ = 0 and larger than 1 when θ = 90◦ . So there must be an angle in between where the factor equals 1. This occurs when (1 − β 2 sin2 θ) = (1 − β 2 )2/3 =⇒ sin2 θ = 1 − (1 − β 2 )2/3 . β2 (385) If β ≈ 1 then sin2 θ ≈ (1 − 0)/1 = 1. So θ ≈ 90◦ . (The field drops off very quickly from its maximum value at θ = 90◦ .) If β ≈ 0( then we need to ) use the Taylor − ϵ)n ≈ 1 − nϵ. This gives sin2 θ ≈ 1 − (1 − 2β 2 /3) /β 2 = 2/3. So series, (1 √ sin θ = 2/3 =⇒ θ = 54.7◦ . Interestingly, this is the same result as for the β ≈ 0 limit in Problem 5.2. Gauss’s law for a moving charge Consider the sphere in Fig. 5.14, where the charge moves in the x direction (we’ll drop the primes on the coordinates). The angle θ in Eq. (5.15) is measured with respect to the x axis. If we slice the sphere into circular strips with constant θ, the radius of a strip at angle θ is 2πr sin θ. So if the strip subtends an angle dθ, its area is dA = (2πr sin θ)(r dθ). The field points radially, so the total flux through the sphere is ∫ π ∫ 1 − β2 q 2πr2 sin θ dθ E da = 2 (1 − β 2 sin2 θ)3/2 4πϵ r 0 0 ∫ q(1 − β 2 ) π sin θ dθ = . (386) 2 2 3/2 2ϵ0 0 (1 − β sin θ) Using the integral table in Appendix K, this becomes π 2  q(1 − β 2 ) 2 q − cos θ  = q(1 − β ) · √ = , ·  ϵ0 2ϵ0 (1 − β 2 ) 2ϵ0 (1 − β 2 ) 1 − β 2 sin2 θ 0 (387) as desired. Note: having shown that Gauss’s law holds for a sphere, the same reasoning as in Section 1.10 can be used to show that it holds for any shape. The critical property of the field is the 1/r2 dependence on r. Cosmic rays ′ ◦ The maximum value of the √field in Eq. (5.15) is achieved when θ = 90 , in which case the value is (Q/4πϵ0 r′2 )/ 1 − β 2 = γQ/4πϵ0 r′2 . Therefore, Emax = =⇒ r′2 = 1 4πϵ0 1 4πϵ0 γe r′2 ( γe kg m3 ) (1010 )(1.6 · 10−19 C) = 9 · 109 2 2 = 14.4 m2 , (388) E s C 1 V/m so r = 3.8 m. The thickness of the pancake at this distance is r ∆θ ≈ r/γ ≈ (4 m)/1010 = 4 · 10−10 m. Very thin! Reversing the motion c/2 c/2 ct Let t = 0 be the time of the collision at the origin. Outside a sphere of radius ct centered at the origin, the field at time t is that of a uniformly moving proton that would have been located at x = −(c/2)t. See Fig. 101; we have drawn just the top half of the picture. Inside a sphere of radius ct, the field at time t is that of the actual uniformly moving proton that is located at x = (c/2)t. At t = 10−10 s, the radius of the sphere is ct = (3 · 108 m/s)(10−10 s) = 0.03 m, or 3 cm. Note that the interior and exterior field lines are indeed connected in the manner shown, with lines of equal slope being connected by a (nearly) tangential line. This follows from the fact that in Fig. 5.19 the flux through the spherical cap F D equals the flux through the spherical cap EA. Figure 101 A nonuniformly moving electron (a) The electron had been traveling in the positive direction along the negative x axis toward the origin, where it rather suddenly stopped. Since the speed of light is c = 3 · 1010 cm/s, and since the “transition” sphere surrounding the static E field has a radius of about r = 15 cm, the stopping must have taken place at t = −r/c = −5 · 10−10 s. (From the figure, it looks like the stopping commenced at this time, and then lasted for perhaps 0.5 · 10−10 s.) If the electron hadn’t stopped, it would be located at x = 12 cm, because that is where the external field lines point. So its speed was (12/15)c = (0.8)c. (b) At t = −7.5 · 10−10 s, which was ∆t = 2.5 · 10−10 s before the electron stopped, it was located at x = −v ∆t = −(0.8)(3 · 1010 m/s)(2.5 · 10−10 s) = −6 cm, with respect to the origin. (c) To find the field strength at the origin when the electron was at x = −6 cm, we can use Eq. (5.15) with θ′ = 0. This gives E= ( kg m3 ) (1.6 · 10−19 C)(1 − 0.82 ) 1 e(1 − β 2 ) = 9·109 2 2 = 1.44·10−7 V/m. 2 r (0.06 m)2 s C 4πϵ0 (389) Colliding particles ct (−e) (+e) Figure 102 No charge remains after the collision, so the field is zero within a sphere of radius ct centered at the origin. Outside this sphere, the field is what the field would be if the two charges had kept moving. The positive charge would be on the right, at position x = vt, and the negative charge would be on the left, at position x = −vt. The field lines can’t end, so the incoming lines on the left side must connect (via lines along the surface of the sphere with radius ct) with the outgoing lines on the right side; see Fig. 102. The thickness of the shell containing the connecting lines is determined by the duration of the deceleration period. This thickness remains constant. This radiation is called Bremsstrahlung. As the spherical shell expands, it turns out that the E field in the shell decreases like 1/r instead of the usual 1/r2 . This is demonstrated in Appendix H, but we can also understand it in another more qualitative way. We know that the density of radial field lines falls off like 1/r2 . Equivalently, if we paint a collection of dots on a balloon (which represent the intersection of the field lines with the balloon) and then expand the balloon, the density of dots falls off like 1/r2 . However, if we paint a collection of lines on the balloon (which represent the tangential field lines in the present setup), then the density of these lines falls off only like 1/r. This is true because there is now effectively only one dimension that is expanding (at least with regard to the spacing between the lines), namely the dimension lying on the surface of the sphere and perpendicular to the lines. (We have used the fact that the thickness of the shell remains constant.) Since the density of field lines is proportional to the field strength, we arrive at the desired result that the tangential field falls off like 1/r. Relating the angles The flux through the inner cap is (ignoring the 1/4πϵ0 ) ∫ θ0 0 θ0 Q  2πr2 sin θ dθ = −2πQ cos θ = 2πQ(1 − cos θ0 ). 2 0 r (390) The flux through the outer cap is (ignoring the 1/4πϵ0 ) ϕ0 Q 1 − β2 2πr2 sin ϕ dϕ = 2πQ(1 − β 2 ) r2 (1 − β 2 sin2 ϕ)3/2 ϕ0 sin ϕ dϕ . (1 − β 2 + β 2 cos2 ϕ)3/2 (391) With x ≡ cos ϕ =⇒ dx = − sin ϕ dϕ, this becomes (using the integral given in the exercise) ∫ − 0 2πQ(1 − β 2 ) β3 ∫ cos ϕ0 1 dx ( 1−β 2 + x2 β2 )3/2 ∫ 0  cos ϕ0  x 2πQ(1 − β 2 )  ) ( 2 3 1/2  1−β 1−β 2 β 2 1 β2 + x β2 ( ) β cos ϕ0 2πQ β− β (1 − β 2 + β 2 cos2 ϕ0 )1/2 ) ( cos ϕ0 . (392) 2πQ 1 − 2 2 1/2 (1 − β sin ϕ0 ) = − = = Alternatively, we could have obtained this result in a quicker manner by applying Eq. (K.15) in Appendix K to the integral on the left-hand side of Eq. (391). Equating this flux with the flux in Eq. (390) gives cos θ0 = cos ϕ0 (1 − β 2 sin2 ϕ0 )1/2 . (393) Now let’s show that this is equivalent to tan ϕ0 = γ tan θ0 . If tan θ0 = tan ϕ0 /γ, then cos θ0 = = 1 )1/2 = ( ( 1 + tan2 θ0 ( ϕ0 1 + (1 − β 2 ) sin cos2 ϕ0 1 2 1 1+ ) tan2 ϕ0 1/2 γ2 )1/2 = ( (394) cos ϕ0 (cos2 ϕ0 + sin2 ϕ0 ) − β 2 sin2 ϕ0 which equals the righthand side of Eq. (393), as desired. )1/2 , Half of the flux From Eq. (392) in the solution to Exercise 5.20, the flux through a cap that subtends an angle θ is (bringing the 1/4πϵ0 back in) ( ) cos θ Q 1− . (395) 2ϵ0 (1 − β 2 sin2 θ)1/2 With θ ≡ π/2 − δ this becomes ) ( Q sin δ . 1− (1 − β 2 cos2 δ)1/2 2ϵ0 (396) If this flux equals Q/4ϵ0 , then the mirror-image cap will also contain a flux of Q/4ϵ0 , which will leave Q/2ϵ0 for the region between the cones, which is half of the total flux Q/ϵ0 . So we want sin δ 1 = (1 − β 2 cos2 δ)1/2 2 =⇒ 4 sin2 δ = 1 − β 2 (1 − sin2 δ ) =⇒ sin2 δ = 1 − β2 . 4 − β2 (397) For β = 0 we have sin δ = 1/2 =⇒ δ = 30◦ , which is the correct result for a spherically symmetric field, as you can verify. (A cap with a half-angle of 60◦ has half the area of a hemisphere.) For β → 1 (and γ → ∞) we have sin2 δ ≈ (1√− β 2 )/3, so √ δ ≈ sin δ ≈ 1/( 3 γ). The angle between the two cones is then 2δ ≈ 2/( 3 γ ). Electron in an oscilloscope (a) • The kinetic energy of the electron is 250 keV. The total energy, including the rest energy, mc2 = 500 keV, is therefore 750 keV. But the total energy 2 is also given by √ γmc = γ(500 √ keV). So γ = 750/500 = 1.5. The β factor is given by β = 1 − 1/γ2 = 5/3 = 0.745. • The momentum is px = γm(βc) = (1.5)(0.745)mc = (1.12)mc. • The time spent between the plates is t = (0.04 m)/(βc) = 1.79 · 10−10 s. • The transverse force, which has the constant value of Ee, equals the rate of change of the transverse momentum. So py = (Ee)t. But the electric field is E = V /s, where s is the separation between the plates. So py = V et/s. We could plug in the numbers, but since we want to write the result in units of mc anyway, it is easier to take the following route. We have py /mc = V et/smc. Multiply by c/c to obtain eV tc 6 keV (1.79 · 10−10 s)(3 · 108 m/s) py · = = = 0.0805. 500 keV 0.008 m mc mc2 s (398) (Remember that an eV is the change in energy of an electron moving through a potential difference of 1 volt. So to be precise, “eV” should actually be written as “eV.” That is, it is the charge of one electron, e, multiplied by 1 volt. And the V in Eq. (398) is 6 kV.) • The transverse velocity at exit is given by py = γmvy =⇒ vy = py (0.0805)mc = 1.61 · 107 m/s. = γm γm (399) • Since the transverse force (and hence transverse acceleration) is constant, the average transverse velocity is half of the vy we just found, that is, v y = 8.05 · 106 m/s. (vy is small enough so that nonrelativistic kinematics works fine here.) The transverse distance traveled is then y = v y t = (8.05 · 106 m/s)(1.79 · 10−10 s) = 1.44 · 10−3 m, which is 1.44 mm. • The angle of the trajectory at exit is (using tan θ ≈ θ for small θ) θ= 0.0805 vy py = 0.072 rad = 4.1◦ . = = 1.12 vx px (400) (b) In the frame in which the electron is initially at rest, the plates are moving to the left with speed βc = (0.745)c, and their length is (0.04 m)/γ = 0.0267 m. (The other two dimensions are unchanged.) The plates are above and below the electron for a time t′ = (0.0267 m)/(0.745c) = 1.19 · 10−10 s (which can also be obtained from time dilation), during which time the electron is accelerated upward in the field E ′ = γE = γV /s. The upward momentum acquired is E ′ et′ = (γE )e(t/γ ) = Eet = (0.0805)mc. This is the same as in the lab frame, which is consistent with the fact that transverse momenta are unaffected by a Lorentz transformation. In this frame the electron is non-relativistic, to a good approximation, so the final vy′ is v ′y ≈ (0.0805)c = 2.42 · 107 m/s. (If you want, you can show that taking relativity into account would decrease the speed by only about 0.3%.) The average transverse velocity is then vy′ = v ′y /2 = 1.21 · 107 m/s. The total transverse distance is therefore y ′ = vy′ t′ = (1.21 · 107 m/s)(1.19 · 10−10 s) = 1.44 · 10−3 m, in agreement with the result in the lab frame, which is consistent with the fact that transverse distances are unaffected by a Lorentz transformation. In short, the transverse distances are the same in the two frames because in the electron’s frame v ′y is larger by a factor γ, but t′ is smaller by a factor γ . Two views of an oscilloscope The various answers are: [ ] 125,000 eV 125 keV by construction, just e times ∆ϕ [ ] 5/4 γ = (500 + 125)/500 √ [ ] (3/5)c β = 1 − [1/γ 2 ] 2.0 · 10−22 kg m/s px = γm(βc) [ ] 4 3 · 10 V/m E = V /d [ ] 4.8 · 10−15 newtons F = Ee [ ] −10 s t = ℓ/(βc) 2.78 · 10 [ ] py = F 1.33 · 10−24 kg m/s ]t [ 0.0066 radians θ = py /px [ ] 1.8 · 108 m/s same β as above [ ] 0.04 m ℓ′ = ℓ/γ [ ] 2.22 · 10−10 s t′ = ℓ′ /(βc); equivalently, t′ = t/γ [ ′ ] 4 3.75 · 10 V/m E = γE, since the charge density is larger by γ [ ] 1.46 · 106 m/s vy′ = ay′ t = (E ′ e/m)t′ , non-relativistic is fine here [ ′ ] −24 ′ kg m/s py = mv y for non-relativistic speeds 1.33 · 10 The y components of the momenta in the two frames are equal, as dictated by the Lorentz transformations. In short, this is due to the fact that in the electron-neutron frame the transverse field E ′ (and hence transverse force Fy′ ) is larger by a factor γ , but the time t′ is smaller by a factor γ. So the product Fy′t′ is the same as in the lab frame. Acquiring transverse momentum (a) Let E1y be the y component of the electric field due to q1 , at the location of q2 ; see Fig. 103. Let P2y be the y component of the momentum of particle q2 . Then ∫ dP2y /dt = q2 E1y =⇒ ∆P2y = q2 E1y dt. We can exchange the dt here for dx E1y b q2 E1 v y x q1 Figure 103 y q1 E2 v q2 x E2y Figure 104 b via dt = dx/v. Therefore, since P2y = 0 at x = −∞, the value of P2y at x = +∞ ∫∞ is P2y = (q2 /v) −∞ E1y dx. By Gauss’s law, the surface integral of E1 over the infinitely long cylinder of radius b with ∫q1 lying on the axis (represented ∫ ∞ by the dotted lines in Fig. 103) ∞ is q1 /ϵ0 . So −∞ E1y 2πb dx = q1 /ϵ0 =⇒ −∞ E1y dx = q1 /2πϵ0 b. Therefore, P2y = q1 q2 /2πϵ0 vb, as desired. Also, P2x = 0 by symmetry (E1 points the right just as much as it points to the left along the cylinder). So P2 points in the y direction, that is, it is perpendicular to v. By Newton’s third law, the momentum gained by q1 is equal and opposite to the momentum gained by q2 . But we can also calculate this in the same manner as above, by considering a cylinder with radius b, drawn with the path of q2 as its axis; see Fig. 104. At any given instant, the flux through this ∫ ∞ cylinder is q2 /ϵ0 (Gauss’s law still holds for relativistic speeds). So as above, −∞ E2y dx = q2 /2πϵ0 b. (The E2y here is the magnitude of the y component.) In this integral, we are considering a snapshot in time, with x running from −∞ to ∞. But we are free to write the integral in terms of the time t, for which dt = dx/v . If we multiply both ∫ ∞ sides of the previous equation by q1 and substitute v dt for dx, we obtain q1 −∞ E2y dt = q1 q2 /2πϵ0 vb. But the left side here is simply the expression for the magnitude of the final P1y . The direction is downward, so the final P1y equals −q1 q2 /2πϵ0 vb. In short, the surface integral over the cylinder is the same whether you calculate it for a stationary cylinder, or stand at a fixed position (at q1 ) and have the cylinder (which you can imagine as attached to q2 ) slide past you, adding up the integrals over the circular strips as they pass by. (b) We want the transverse distances that the particles move to be much smaller than b. Consider q2 . Its final y speed is Py /γm, so in order of magnitude, its y speed throughout the main part of the interaction (when the particles are a distance of order b apart) is also Py /γm. The time of the main part of the interaction is on the order of b/v, since the force is negligible when the particles are far apart. The rough transverse distance traveled by q2 is the product of the rate and time, so we want Py b · ≪ b =⇒ γm v q1 q 2 b · ≪ b =⇒ 2πϵ0 vb · γm v q1 q2 ≪ m. 2πϵ0 γv 2 b (401) This is the desired condition on m. Note, interestingly, that the transverse distance (P/γm)(b/v ) is independent of b. This is due to the facts that the force decreases like 1/b2 , but the distance (and hence time) of the interaction increases like b. And the latter matters twice in the nonrelativistic expression for distance, y = ay t2 /2. To do the same calculation for q1 , we must remember that since the x speed of q2 is relativistic, we need to take into account the fact that q2 ’s field lines are squashed into a “pancake.” The result from Problem 5.6 or Exercise 5.21 shows that the angular width of the pancake is on the order of 1/γ. So the time of the main part of the interaction (as far as the force on q1 is concerned) is only on the order of b/γv. But since q1 is moving slowly, we now have vy ≈ Py /m, with no need for a γ factor in the denominator. We therefore end up with the condition (Py /m)(b/γm) ≪ b =⇒ q1 q2 /2πϵ0 γv2 b ≪ m, in agreement with the above result. Note that this condition can be written as q1 q2 /4πϵ0 b ≪ γmv2 /2. For nonrelativistic motion, this says that the kinetic energy of q2 must be much larger than the electrical potential energy of the particles at closest approach. It is reasonable that the condition takes this form, because these are the only two energy scales in the problem. Decreasing velocity Force equals the rate of change of momentum. Therefore, since there is no force in the x direction in the lab frame, px must be constant. Now, px = γmvx , so if vx were to remain constant, then the increase in γ (due to the increase in vy due to the Ey field) would cause px to increase, in contradiction with the fact that px is constant. Therefore, vx must actually decrease. Remember that the γ factor in px = γmvx involves the square of the entire velocity, not just the vx component. So px is indeed affected by the y motion. We can also demonstrate this result by looking directly at the forces, without mentioning momentum. Let F be the lab frame, and let F ′ be the particle frame. The key point is that although the electric force on the particle points in the y direction in F , it does not point in the y ′ direction in F ′ . This can be seen as follows. Except right at the start, the particle’s velocity v will be angled upward in F , as shown in Fig. 105. The E∥ and E⊥ components (with respect to the velocity v) of the given vertical E field in F are shown. If we transform these components to the particle frame F ′ , the ⊥ component is larger by a factor γ (because it is smallest in the frame of the source, which is the lab frame; imagine two large capacitor plates). So we end up with the E′ vector shown. Since F′ = qE′ , the force on the particle in the particle’s frame is therefore tilted leftward. Physically, the slightly leftward-pointing field in F ′ arises from the tilted pancakes that we encountered in Fig. 5.26. The particle therefore has an acceleration in the negative x′ direction in its frame, so it picks up a negative x′ velocity relative to the inertial frame it was just in. The x speed in the lab frame therefore decreases (via the velocity addition formula). Charges in a wire √ The β value associated with γ = 1.2 is β = 1 − 1/γ 2 = 0.553. So the test charge is moving at speed v = (0.553)c with respect to the lab frame. We also know that the electrons are moving at speed v0 = β0 c = (0.8)c with respect to the lab frame. β0′ , which gives the speed of the electrons with respect to the test charge, is therefore given by the velocity addition (or subtraction) formula, β0′ = β0 − β 0.8 − 0.553 = 0.443. = 1 − (0.8)(0.553) 1 − β0 β (402) From Eq. (5.24) we have λ′ = γββ0 λ0 = (1.2)(0.553)(0.8)λ0 = (0.531)λ0 . (403) Equal velocities If β = β0 , then γ = γ0 . So the positive charge density γλ0 in the test-charge frame becomes γ0 λ0 . And the negative charge density −γλ0 (1 − ββ0 ) in Eq. (5.24) (the second term in that equation) becomes −γ0 λ0 (1 − β02) = −λ0 /γ0 . These results make sense, because in the test charge’s frame the positive charges are moving backward at speed v0 , so their separation is contracted by γ0 (because they were at rest in the lab frame). And the negative charges are at rest in the test charge’s frame, so their separation is uncontracted by a factor γ0 (because they were moving at speed v0 in the lab frame). E' E E' E E = E' v Figure 105 Stationary rod and moving charge (a) The force is always largest in the rest frame of the particle. It is smaller in any other frame by the γ factor associated with the speed v of the particle. So the force in the new frame (the charge’s frame) is larger; it is γqλ/2πrϵ0 . This force is repulsive, assuming q and λ have the same sign. (b) In the new frame the charge q isn’t moving, so even though the magnetic field is nonzero, the magnetic force is zero, due to the v in FB = qvB . We therefore need only worry about the electric force. In the new frame the linear charge density on the rod is increased to γλ, due to length contraction. So the electric field is γλ/2πrϵ0 . This field produces a repulsive electric force of FE = γqλ/2πrϵ0 , which agrees with the result in part (a). Protons moving in opposite directions In the rest frame of one of the protons, the other proton moves with a speed given by β2 = 2β/(1 + β 2 ), from the velocity addition formula. In this frame, Eq. (5.15) gives the electric field of the moving proton as γ2 e/4πϵ0 r2 . That is, the field is larger by a factor γ2 in the transverse direction, compared with the naive Coulomb result. The γ2 factor equals γ2 = √ 1 = √ 1 − β 22 1− 1 ( 2β 1+β 2 1 + β2 1 + β2 )2 = √(1 + β 2 )2 − 4β 2 = 1 − β 2 . (404) The force on each proton in its rest frame is therefore Frest = where γ ≡ 1/ √ 1 + β2 e2 e2 , · = γ 2 (1 + β 2 ) 4πϵ0 r2 1 − β 2 4πϵ0 r2 (405) 1 − β2. The relative speed of the proton rest frame and the√ lab frame is βc, so the force in the lab frame is smaller than Frest by a factor 1/γ = 1 − β 2 . (Remember, the force is always largest in the rest frame of the particle on which it acts.) The repulsive force on each of the protons in the lab frame is therefore Flab = e2 Frest = γ(1 + β 2 ) . γ 4πϵ0 r2 (406) This is the correct total force in the lab frame. As stated in the problem, it is not equal to γe2 /4πϵ0 r2 . The task now is to explain the force as the sum of the electric and magnetic forces in the lab frame. As mentioned in the problem, Eq. (5.15) tells us that the repulsive electric force in the lab frame is Flab,E = γe2 /4πϵ0 r2 . This must not be the whole force, because it doesn’t equal the correct total force in Eq. (406). Apparently there must be an additional repulsive force in the lab frame that equals the difference between these two repulsive forces. This difference is Flab − Flab,E = γ(1 + β 2 ) e2 e2 e2 . −γ = γβ 2 4πϵ0 r2 4πϵ0 r2 4πϵ0 r2 (407) This is exactly the force produced by a magnetic field with strength B = (β/c)γe/4πϵ0 r2 . (This is consistent with the Lorentz transformations in Chapter 6.) This field does indeed produce the desired force, because the proton is moving with speed βc through this B field, which yields a magnetic force with magnitude Flab,B = qvB = e(βc)B = γβ 2 e2 . 4πϵ0 r2 (408) This is the quantity that appears on the right-hand side of Eq. (407), so that equation is correctly the statement that Flab − Flab,E = Flab,B =⇒ Flab = Flab,E + Flab,B . (409) The direction of the magnetic force will be correct (repulsive) provided that the B field at the location of each proton, due to the other proton, points out of the page. Note that the electric and magnetic forces in the lab frame have exactly the same magnitudes that they had in the second example in Section 5.9, because the speeds are the same in the two setups. But the difference in direction of the top charge’s motion in the two setups causes its magnetic field to point in the opposite direction. The qv × B forces on both charges switch sign, so in the present setup the magnetic force is added to the electric force instead of subtracted from it. The total force in the lab frame is therefore different in the two setups. Transformations of λ and I The speed of the charges in the new frame is given by βk′ = (βk + β)/(1 + βk β). The γ factor associated with this speed is γk′ = γk γ(1 + βk β) (see below). The density in the rest frame of the charges is λk /γk , so the density in frame F ′ is ( ) ) ( β λk ′ λk βI γk γ(1 + βk β) = γ λk + (λk βk c) = γ λk + k , λ′k = γk = (410) c c γk γk as desired. The current in F ′ is Ik′ = = ( ) βk + β c 1 + βk β λk γ(βk + β)c = γ(λk βk c + βcλk ) = γ (Ik + βcλk ), λk′ βk′ c = λk γ(1 + βk β) (411) as desired. Since these two transformations are linear in Ik and λk , they hold for any corresponding linear combinations of the Ik and λk (for example, 2I1 − 7I5 + 3I8 and 2λ1 − 7λ5 + 3λ8 ). In particular, they hold for the sums of the Ik and the λk . But these sums are just the total current I and the total density λ. So the Lorentz transformations hold for the total I and λ, as we wanted to show. Here is the algebra that produces γk′ : γ ′k = √ 1 1− ( βk +β 1+βk β )2 Moving perpendicular to a wire = = √ √ 1 + βk β ✟ ✟ 2 2 2 2 2β✟ β✟ (1 + ✟ kβ + β ) k β + βk β ) − (β k + 2 ✟ 1 + βk β = γk γ(1 + βk β). (1 − β k2 )(1 − β 2 ) (412) In the notation of Fig. 5.25, the velocity components of the electrons in the charge’s frame are vx = v0 /γ and vy = v. So βe2c2 = v02 /γ 2 + v 2 . And the angle α in Fig. 5.26 is given by tan α = v/(v0 /γ) = γv/v0 . The angle θ′ in Eq. (5.15) is measured with respect to the direction of motion of the electrons. So for the left electron in Fig. 5.26 we have θ′ = ϕ − α. This holds for the right electron too, provided that we consistently measure ϕ with respect to the positive x axis (unlike how it is defined for the right electron in Fig. 5.26). The angle α is fixed by the parameters in the setup; the variable we will integrate over is ϕ. The range 0 ≤ ϕ ≤ π corresponds to the range −∞ ≤ x ≤ ∞ on the wire. If the charge q is a distance ℓ from the wire in the lab frame, then it is ℓ′ = ℓ/γ from the wire in its own frame, due to length contraction. (Imagine a transverse stick of length ℓ attached to the wire as it moves toward the charge in the charge’s frame, and consider the moment when the end of the stick coincides with the charge.) In the charge’s frame, the distance r′ from the charge q to an electron is r′ = ℓ′ / sin ϕ = ℓ/γ sin ϕ. The position of an electron along the wire is given by x = −ℓ′ / tan ϕ = −ℓ/γ tan ϕ. Taking the differential of this gives dx = ℓ dϕ/γ sin2 ϕ. We now have a handle on all the necessary quantities, so we can use Eq. (5.15) to find the force on the charge q (as measured in its own frame) due to a small interval of the wire with length dx. The (negative) electron charge contained in this length is (−λ0 )dx = −λ0 ℓ dϕ/γ sin2 ϕ. (There is no length contraction along the wire in the charge’s frame.) We are concerned with the x component of the force, which brings in a factor of cos ϕ, so from Eq. (5.15) we obtain dFx q(−λ0 ℓ dϕ/γ sin2 ϕ) 1 − βe2 cos ϕ 4πϵ0 (ℓ/γ sin ϕ)2 (1 − βe2 sin2 (ϕ − α))3/2 2 γqλ0 (1 − βe ) cos ϕ dϕ − . 4πϵ0 ℓ (1 − βe2 sin2 (ϕ − α))3/2 = = (413) Note that the γ here is associated with the speed v of the charge q in the lab frame, whereas the βe is the total speed of the electrons in the charge q frame, which is given by β 2e c2 = v02/γ 2 + v 2 . To find the total force on the charge q in its own frame, we need to integrate Eq. (413) from ϕ = 0 to ϕ = π. The integral is given in Appendix K, so the total horizontal force from all of the electrons in the wire is F =− γqλ0 (1 − β 2e ) (2 − βe2) sin ϕ + β 2e sin(2α − ϕ) √ 4πϵ0 ℓ 2(1 − β e2 ) 1 − β e2 sin2 (α − ϕ) π    .  (414) 0 In the numerator, the sin ϕ term is zero at both limits, and sin(2α − π) = − sin 2α. In the denominator, sin(α − π) = − sin α, but we’re squaring this, so the minus sign doesn’t matter. The denominators are therefore equal at both limits. Hence we obtain twice the value at the upper limit: F = γqλ0 γqλ0 βe2 sin α cos α β 2 sin 2α = . √ e √ 4πϵ0 ℓ 2πϵ 2 0ℓ 1 − βe2 sin α 1 − β 2e sin2 α From tan α = γv/v0 we have sin α = γv/ √ γ 2 v 2 + v 02 and cos α = v0 / (415) √ γ 2 v 2 + v 20 . Since βe2 can be written as (γ 2 v 2 + v02 )/γ 2 c2 , we therefore have F = = γqλ0 √ 2πϵ0 ℓ γ 2 v 2 + v 20 γv · v0 γ 2 c2 γ 2 v 2 + v 20 1− γ 2 v 2 + v02 γ 2 v 2 γ 2 c2 γ 2 v 2 + v02 1 qλ0 vv0 γqλ0 vv0 √ = . 2πϵ0 ℓc2 2πϵ0 ℓc2 1 − v 2 /c2 (416) This is the total force on the charge q in its own frame. The force is positive, so it points rightward, consistent with the field lines in Fig. 5.26. Now, back in the lab frame, λ0 v0 equals the current I in the wire (which is directed to the left, since the electrons are moving to the right). Transforming the above force to the lab frame involves dividing by the γ factor associated with the speed v, which is the γ factor that appears in Eq. (416). So we end up with a rightward directed force with magnitude qvI/2πϵ0 ℓc2 , as desired. As mentioned in the text, if the magnetic field B points out of the page, then v × B points to the right. So the force can be interpreted as the qv × B force from a magnetic field with strength I/2πϵ0 ℓc2 .