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1 PHYS 110B - HW #8 due to gravity. Fall 2005, Solutions by David Pace Equations referenced as ”Eq. #” are from Griffiths Problem statements are paraphrased 1 x − x0 = vo t + at2 |{z} 2 (2) 0 1 −h = − |g|t2 2 s 2h t = |g| [1.] Problem 11.10 from Griffiths An electron is released from rest and allowed to fall under the influence of gravity. During the first centimeter of its free fall what fraction of the lost potential energy is dissipated through radiation? (4) The radiated power is given by, Solution Recall the kinematic equations in one dimension with constant acceleration. v − vo = at (3) 1 x − xo = vo t + at2 2 P = µo q 2 a2 6πc Eq. 11.61 (5) (1) These equations underlie a physical discrepancy with this problem. The kinematic equations above rely on an ideal system in which all of the potential energy of the falling object is converted into kinematic energy. There are corrections to these equations in the case where radiation is significant compared to the kinetic energy. It turns out that the energy lost due to radiation in this situation pales in comparison to the kinetic term and therefore these equations are valid as a very good approximation. If you considered this initially, then it is still best to assume the radiation term will be very small and use the equations above. If the radiation term turns out to be very large, then you must start over and think about the new physical issues. Let the distance fallen by the electron be equal to h. The actual value of this parameter is given as 1 cm, but it is easier to use the variable throughout the algebra and then plug in the numerical value for the final calculation. The goal of this problem is to determine the fraction Urad /∆Upot where Urad is energy radiated during the 1 cm fall, and ∆Upot is the total amount of gravitational potential energy that is lost during this fall. Treat ∆Upot as a positive value because we are admitting that the energy is decreasing and we only wish to know how this absolute amount relates to the radiated amount. Power is a measure of energy per unit time. The total energy radiated by this falling electron is the power radiated multiplied by the time it takes for it to fall the specified distance. Solving for this time requires the use of (1). Since the electron is dropped from rest we have vo = 0. Setting the ground as my reference requires that the acceleration be a = −g, where g is the acceleration where this is known as the Larmor formula. Griffiths explains in section 11.2.1 that this expression is valid for v c. This condition is met here because the final velocity of any object falling (from rest) through a distance of 1 cm in the Earth’s gravitational field is not going to be appreciable compared to the speed of light. Multiplying (4) and (5) gives the total energy radiated by the electron in this measured fall. Urad (6) = Pt = µo q 2 a2 · 6πc = µo q 2 g 2 6πc s s 2h |g| 2h |g| (7) (8) The total change in gravitational potential energy is (leaving this value positive for the reason given previously), ∆Upot = mgxf − mgxo (9) = mg(xf − xo ) (10) = mgh (11) 2 so the time is now written in terms of parameters given in the problem statement. The desired ratio is, 2 2 Urad ∆Upot = = = µo q g 6πc q 2h |g| (12) mgh µo q 2 6πcm r The total radiated power, Urad , is given from (5) multiplied by this time. 2|g| h (13) Urad (4π × 10−7 )(1.6 × 10−19 )2 6π(3.0 × 108 )(9.11 × 10−31 ) ≈ 2.8 × 10−22 r 2(9.8) 0.01 (14) (15) The energy lost due to radiation is insignificant compared to the general kinetic loss. = µo q 2 a2 vo · 6πc |a| (21) = µo q 2 vo |a| 6πc (22) Now to solve for the desired fraction, Urad K.E.i = µo q 2 vo |a| 6πc 1 2 2 mvo (23) = 2 µo q 2 vo |a| · 6πc mvo2 (24) = µo q 2 |a| 3πcmvo (25) [2.] Problem 11.13 from Griffiths (a) Consider an electron decelerating (a = constant) from initial velocity vo to final velocity v = 0. Assuming vo c so that you may apply the Larmor formula, what fraction of the initial kinetic energy of this electron is radiated away? (b) Apply the following values to the problem in part (a). vo ≈ 105 m/s d = 30 Å (16) where d is the total distance traveled. What does this result tell us about radiation losses for electrons in ordinary conductors? (b) In this part we want to apply given values to (25). This requires solving for the absolute value of the acceleration because it is not given. Using (1), (16), and (20), xf − xi = Solution 1 vo t + at2 2 The initial kinetic energy of the electron is, vo |a| a − 2 (26) vo2 a2 (27) d = vo where m is the electron mass. d = vo2 1 vo2 − |a| 2 |a| (28) The total power radiated away is found using (6). The time it takes for the electron to come to rest is found using (1). The a term is written negative because this is a deceleration. d = 1 vo2 2 |a| (29) |a| = vo2 2d (30) K.E.i = 1 mv 2 2 o (17) v − vo = −at (18) −vo −a = t (19) t = vo |a| (20) where this derivation also assumes the initial location is the origin and the velocity is v ≥ 0. Solving for the numerical fraction of kinetic energy lost 3 to radiation, 2 vo2 Urad µo q = · K.E.i 3πcmvo 2d (31) = µo q 2 vo 6πcmd = (4π × 10−7 )(1.6 × 10−19 )2 (105 ) (33) 6π(3.0 × 108 )(9.11 × 10−31 )(30 × 10−10 ) (32) ≈ 2.1 × 10−10 where the vector form is written in spherical coordinates. The magnitude of the acceleration is written because the next step is to solve for the electron velocity in its orbits. In the case of centripetal acceleration and circular orbits, the tangential velocity of the particle is found using a = v 2 /r. The distance r changes as the electron moves closer to the proton. As r decreases along this path the initial velocity must be the lowest velocity throughout the trip. This initial velocity is found using, v2 r (34) This is another instance in which the energy lost to radiative processes is negligible compared to other loss mechanisms. = q2 4πo me r2 s v = [3.] Problem 11.14 from Griffiths Consider the Bohr model of the hydrogen atom. In this model the ground state electron travels around the proton in a circular orbit of radius ro = 5 × 10−11 m, due to the Coulomb attraction between the particles. Classical electrodynamics requires that this accelerating electron emit radiation and therefore continuously lose energy. As it loses energy this electron will move inward and eventually impact the proton. Show that the Larmor formula is valid for most of the electron’s inward motion (i.e. v c). Calculate the typical lifetime of such an atom if the electron’s orbits can always be considered circular. Solution = where me is the electron’s mass, q is the electron’s ~ is the electric field due only to the procharge, and E ton. Recall that the force experienced by the electron does not include its own fields in this problem. The electric field of the proton is known, but it’s also given as Eq. 2.10 if you wish to review. The acceleration is, q2 r̂ 4πo me r2 ~a = − a = q2 4πo me r2 (36) (37) r 1 πo me r (39) (40) All of the parameters in (40) are known numerically. The initial (and lowest) velocity of this electron occurs at its initial position where r = ro , vo = 1.6 × 10−19 2 ≈ To show that the Larmor formula is valid I will solve for the initial velocity of the electron in functional form and then determine how this changes as the electron approaches the proton. This is accomplished by looking at the forces involved, ~ F~ = me~a = −q E (35) q 2 q2 4πo me r (38) s π(8.85 × 2.2 × 106 1 × 10−31 )(5 × 10−11 ) (41) 10−12 )(9.11 (42) This is a large velocity compared to everyday experience, but vo /c ≈ 0.007, which satisfies the small velocity condition. As the electron approaches the proton this √ velocity increases by a factor of 1/ r. This factor causes the velocity to slowly increase and we may therefore claim that the electron’s velocity is less than the speed of light for most of the trip. Of course, once the electron gets incredibly close to the proton the distance goes to zero and the velocity goes to infinity. We ignore this situation here. The lifetime of the electron in this configuration is given by the time necessary for the electron to radiate away all of its energy. The total energy of the electron, Ee , is its kinetic energy plus its potential energy. The kinetic energy of the electron is given by (17) replacing the initial velocity there with the instantaneous velocity here. The potential energy is P.E.= −qV , where V is the scalar potential field due to the proton and the negative sign is included because of the electron charge. The scalar potential is known for a single charged particle, giving the 4 following expression for the energy, 1 q Ee = me v 2 − q 2 4πo r (43) Simplify using (39), Ee Now we equate (48) and (53) so that we can solve for the time it takes for the electron to impact the proton. In (48) we have an expression for the amount of energy radiated away as a function of the acceleration (with the radial dependence coming from the acceleration term). In (53) we have the same quantity, but have derived it from energy principles directly since we know the initial energy and can quantify its rate of change as a function of time and radial position. = me q2 q2 − 2 4πo me r 4πo r (44) = q2 4πo r (45) µo q 6 3 96π 2o cm2e r4 = − q 2 dr 8πo r2 dt (54) = − (46) dt = − 96π 3 2o cm2e r4 q2 · dr 2 8πo r µo q 6 (55) = − 12π 2 o cm2e r2 dr µo q 4 (56) Z 0 me −1 2me q2 8πo r This is the total amount of energy that must be lost to radiation before the electron impacts the proton. The energy per unit time radiated away is given by (5), and we have solved for the acceleration in (37). 2 q2 µo q 2 (47) P = 6πc 4πo me r2 µo q 6 = 3 96π 2o cm2e r4 (48) Z dt ro ∆E ∆t = (49) where E represents energy and not electric field. t = − dE dt t (50) where the negative sign is placed because we know the energy is decreasing in the system but the radiated power must be positive. Using our expression for the total energy that must be lost as given in (46), and noting that r = r(t), we have, dEe d q2 = − − (51) − dt dt 8πo r = q2 8πo 1 − 2 r dr dt q dr 2 8πo r dt Z 0 r2 dr (58) ro = − 0 12π 2 o cm2e r3 µo q 4 3 ro (59) = − 4π 2 o cm2e −ro3 4 µo q (60) = 4π 2 o cm2e ro3 µo q 4 (61) t (62) = 4π 2 (8.85 × 10−12 )(3 × 108 )(9.11 × 10−31 )2 (5 × 10−11 )3 (4π × 10−7 )(1.6 × 10−19 )4 (63) (52) 2 − 12π 2 o cm2e µo q 4 All of these values are known, ≈ = (57) To be more mathematically rigorous we may require that the left side of (58) be t − to , but since we can always adjust the initial time to be to = 0 it will not matter. As we reduce the ∆t to an infinitesimal time period we have, P =− 12π 2 o cm2e r2 dr µo q 4 In (57) the limits of the r integral reflect the fact that the electron begins at the ground state radius and moves into the proton. To be able to compare these energy terms we take advantage of what power represents. Since power is energy per unit time it can be written as, P − = (53) 1.3 × 10−11 s (64) If this truly explained hydrogen, then it would be an incredibly short lived atom. 5 [4.] Problem 11.15 from Griffiths Example 11.3 provides an expression for the power radiated as a function of angle θ, given as Eq. 11.74, Regarding example 11.3 (page 463), find the angle, θmax , at which the maximum radiation emission occurs. Show that, r 1−β ∼ θmax = for v ≈ c (65) 2 Calculate the intensity of the radiation at θmax for the ultrarelativistic case (i.e. where v ≈ c), but write this value in proportion to the same quantity for an instantaneous velocity v = 0. State the ratio in terms of γ. Reference figure 11.14. Solution The parameters in the problem statement are from page 463 of Griffiths, β≡ v c γ≡p 1 1 − v 2 /c2 (66) dP dΩ = µo q 2 a2 sin2 θ 16π 2 c (1 − β cos θ)5 (67) To find the angle at which the maximum radiation occurs we take the relevant derivative of (67) and set it equal to zero. We already know that the minimum in radiation (equal to zero) occurs at θ = 0. So although this method can only show us where extrema occur, unless the solution is θ = 0 we can safely assume we have found a maximum. If you are still unsure, then you can take the second derivative of (67) at the value of θ found from setting the first derivative to zero and if this result is negative (positive) then the point represents a maximum (minimum). Finding θmax , 0 = d µo q 2 a2 sin2 θ dθ 16π 2 c (1 − β cos θ)5 = µo q 2 a2 d 16π 2 c dθ = µo q 2 a2 (1 − β cos θ)5 (2 sin θ cos θ) − sin2 θ(5(1 − β cos θ)4 (−β)(− sin θ)) 16π 2 c (1 − β cos θ)10 (70) = (1 − β cos θ)5 (2 sin θ cos θ) − sin2 θ(5(1 − β cos θ)4 (−β)(− sin θ)) (1 − β cos θ)10 (71) = (1 − β cos θ)5 (2 sin θ cos θ) − sin2 θ(5(1 − β cos θ)4 (−β)(− sin θ)) (72) = (1 − β cos θ)4 (1 − β cos θ)(2 sin θ cos θ) − 5β sin3 θ (73) = sin θ (1 − β cos θ)(2 cos θ) − 5β sin2 θ (74) = 2 cos θ − 2β cos2 θ − 5β(1 − cos2 θ) (75) = 2 cos θ − 2β cos2 θ − 5β + 5β cos2 θ (76) = 3β cos2 θ + 2 cos θ − 5β (77) sin2 θ (1 − β cos θ)5 (68) (69) 6 This is the form of a quadratic equation and may be solved as such. I have not labeled this angle as θmax , but by the method taken we are finding that value. dure in asymptotic analysis. cos θmax = cos θ = −2 ± p 4 − 4(3β)(−5β) 6β = = −1 ± p 1+ 3β 15β 2 p 1 + 15β 2 3β (84) −1 + p 1 + 15(1 − )2 3(1 − ) (85) (78) = p −2 ± 2 1 + 15β 2 6β −1 + (79) = hp i 1 1 + 15(1 − 2 + 2 ) − 1 3(1 − ) (86) = √ 1 1 + 15 − 30 − 1 3(1 − ) (87) (80) We choose the sign in (80) based on information about θmax provided in example 11.3. From figure 11.14 and a statement on the bottom of page 463 we know that the radiation is concentrated through a narrow cone near the forward direction. Figure 11.14 is turned sideways, but if we look at the location of θmax it is evident that concentrating the radiation in a narrow region near the z axis means that θmax < 45◦ . This requires that cos θ > 1. Meeting this condition requires that we take the plus sign in (80). " r # 1 30 4 1+ −1 = 3(1 − ) 16 (88) " # 1/2 15 1 4 1+ −1 = 3(1 − ) 8 (89) Recall the binomial expansion, which will be applied multiple times in the steps to follow, Finally, (1 + x)n p 1 + 15β 2 cos θ = 3β " # p 2 −1 + 1 + 15β θmax = cos−1 3β −1 + 1 + nx (90) (81) for small x. (82) Applying a binomial expansion to the square root term in (89) provides, For the next part of the problem it is more direct to use the cosine form of θmax as written in (81). Since we are in the ultrarelativistic case β ≡ v/c ≈ 1, but slightly less than one since the velocity is still less than the speed of light. When the velocity is very close to the speed of light we know that β is so close to unity that it may be written as, β ≈1− ≈ (83) where is an expansion parameter and is very small. The rest of this problem concerns rewriting terms and solving for this expansion parameter. To begin, replace β with its expansion in (81). When writing out the terms we should neglect anything that is of order 2 or higher. Since 1, any term containing an 2 is so small it should be ignored. This process is known as expanding to order one in , a common proce- cos θmax 1 15 1 = 4 1+ −1 3(1 − ) 2 8 (91) = 1 15 4+ −1 3(1 − ) 4 (92) = 1 5 1− 1− 4 (93) We can solve further by using the binomial expansion on the 1/(1 − ) term. 1 1− = (1 − )−1 (94) ≈ 1+ (95) 7 5 cos θmax ≈ (1 + ) 1 − 4 (96) 5 5 ≈ 1 − + − 2 4 4 (97) 1 ≈1− 4 (98) We invoke another expansion of the cosine function for comparison with (98). cos x = cos θmax ∞ X (−1)n x2n (2n)! n=0 The direction at which the maximum power is radiated occurs at θ = 90◦ . This is known because the maximum must occur when the sine term is equal to unity. The maximum power radiated in this zero velocity instant is, max dP µo q 2 a2 = (108) dΩ v=0 16π 2 c The other radiated power term is (67) evaluated at the already determined θmax . (99) = 1− x4 x2 + − ... 2! 4! (100) ≈ 1− 2 θmax 2 (101) dP dΩ max µo q 2 a2 sin2 θmax 2 16π c (1 − β cos θmax )5 = θ=θmax The desired ratio is, dP max dΩ θ=θmax dP max dΩ v=0 = sin2 θmax µo q 2 a2 16π 2 c (1−β cos θmax )5 µo q 2 a2 16π 2 c (110) = sin2 θmax (1 − β cos θmax )5 (111) 4 since keeping the θmax term seems unnecessary. Equate (98) and (101), 2 θmax 2 1 1− 4 = 1− 2 = 2 θmax (102) (103) (109) Since this is evaluated at θmax we know that β can be expanded as in (83). The plan is to write (111) in terms of and then relate this result to γ as requested. The numerator of (111) can be rewritten using yet another trigonometric property; sin x ≈ x for small x. We know that θmax is small in this case because we solved in terms of β ≈ 1. As such, we know from (105) that the angle is small. The denominator is rewritten using (98). 2 r 2 = (104) θmax Using (83) to rewrite the above, r 1−β θmax ≈ 2 = (105) In the next part of the problem we want to write the radiated power in the direction of θmax as a fraction of the radiated power in the equivalent maximum direction for a particle of instantaneous velocity equal to zero. Notice that the particle is still accelerating, as it must in order to radiate at all, but that we are concerned with the instant at which its velocity reaches zero. Begin by writing (67) for v = 0, which is the same as setting β = 0. dP dΩ 2 2 = = (112) 2 θmax (1 − (1 − )(1 − 4 ))5 (113) p 2 = 2 (1 − (1 − 4 −+ 2 5 4 )) (114) Once again, drop the 2 term. p 2 2 (1 − (1 − 4 −+ 2 4 ))5 = 2 5 5 ( 4 ) = 1 2 = 0.16384−4 2 µo q a sin θ 16π 2 c (1 − (0) cos θ)5 µo q 2 a2 sin2 θ 16π 2 c sin2 θmax (sin θmax ) ≈ 5 (1 − β cos θmax ) (1 − β(1 − 4 ))5 5 4 −4 5 (115) (116) (106) (107) (117) where inverse powers of small numbers are large and therefore kept. 8 It is necessary to express in terms of γ. γ = arrive at a solution of, 1 S floor (118) p 1 − β2 = µo q 2 d2 ω 4 R2 h 5/2 32π 2 c (R2 + h2 ) (126) = 1 p 1 − (1 − )2 (119) (b) Using your result from (a), calculate the energy per unit time across the entire (infinite) floor. Comment on this value and describe whether it makes sense. = 1 p 1 − (1 − 2 + 2 ) (120) ≈ 1 √ 2 (121) (c) The spring/charge system is losing energy through this radiation. Solve for the time τ at which the oscillation amplitude has been reduced to d/ exp, assuming that the energy loss per cycle is very small. Solution ≈ 1 2γ 2 (122) Our final answer is given in (117), and we can now write this in terms of γ, dP max dΩ θ=θmax dP max dΩ v=0 = 0.16384 1 2γ 2 −4 = 0.16384(24 )γ 8 ≈ 2.6γ 8 (123) (a) From Eq. 9.63 we know that the intensity of EM waves (which compose radiation) is equivalent to the time averaged value of their Poynting flux. The hint reminds us to make sure we only solve for the intensity that actually strikes the floor. This is the amount of energy flux that passes through the floor. Defining a geometry in which the floor is at a constant z = 0 value means that the area vector of the floor is given by −ẑ. The intensity striking the floor is, (124) (125) There are probably more equations in this solution where it is appropriate to have an “approximately equal” sign instead of an “equal to” sign, but in all the steps above the approximations are so good that there is little error. You can use the equal signs in this case, ultrarelativistic, without large error. hSi floor Griffiths’ Hints: Assume d λ h. The calculation only concerns the average energy per unit time that strikes the floor. Neglect radiative damping and (127) The oscillating charge is essentially an electric dipole. The intensity of the radiation from such a system is given by Eq. 11.21, [5.] Problem 11.21 from Griffiths (a) Solve for the radiation intensity that strikes the floor as a function or R, the distance away from the floor point located directly beneath the charge. Solve for the value of R at which the radiation intensity is a maximum. ~ · ẑ hSi where I am taking the absolute value of this expression because we already know the radiated energy is leaving the oscillating charge. ~ hSi A particle (mass = m and charge = q) is attached to a spring and suspended from a ceiling. Reference figure 11.19 in Griffiths. The equilibrium position of this spring-mass system is a distance h above the floor. At time t = 0 the mass is pulled downward a distance d and then released to begin oscillatory motion. = = µo p2o ω 4 sin2 θ r̂ 32π 2 cr2 (128) The dipole moment amplitude is po = qd because d is the maximum displacement from the origin. The origin is set as the equilibrium position of the charge. The angularp frequency of a spring/mass system is known to be ω = k/m. Inserting this information into the expression for intensity gives, r !4 2 µ sin θ k o 2 ~ hSi = r̂ (129) (qd) 32π 2 cr2 m = µo q 2 d2 k 2 sin2 θ r̂ 32π 2 cm2 r2 (130) The R dependence is contained in the r and sin θ terms. The following image shows the charge (solid black circle) suspended from the spring (coil). The charge is placed at the center of a spherical coordinate system. 9 which an extrema occurs and this will be the location of maximum intensity. 0= = d hSi floor dR d dR = (139) µo q 2 d2 k 2 R2 h 2 32π cm2 (h2 + R2 )5/2 µo q 2 d2 k 2 h 32π 2 cm2 (140) d R2 2 dR (h + R2 )5/2 (141) From this drawing we see that, sin θ = R r r = p (131) h2 + R 2 (132) = (h2 + R2 )5/2 (2R) − R2 (5/2)(h2 + R2 )3/2 (2R) (h2 + R2 )5 (142) = 2R(h2 + R2 )5/2 − 5R3 (h2 + R2 )3/2 (143) = (h2 + R2 )3/2 2R(h2 + R2 ) − 5R3 (144) Returning to (130) with this information, 2 2 2 ~ = µo q d k hSi 32π 2 cm2 = R r µo q 2 d2 k 2 R2 32π 2 cm2 2 √ 1 √ 2 h + R2 1 h2 + R 2 2 2 (133) r̂ = R 2h2 + 2R2 − 5R2 1 h2 + R 2 r̂ (134) = µo q 2 d2 k 2 R2 2 32π cm2 (h2 + R2 )2 (135) r̂ · ẑ µo q 2 d2 k 2 R2 cos θ 32π 2 cm2 (h2 + R2 )2 R=h = (146) 2 3 (147) The radial distance to the position of maximum intensity is slightly shorter than the distance of the particle from the floor. (136) (137) (b) The total energy per unit time (the power) across the floor is found by integrating the expression in (138) over the floor. This is written as, Z µo q 2 d2 k 2 R2 h 32π 2 cm2 (h2 + R2 )5/2 (145) = 2h2 − 3R2 r The intensity striking the floor is the z component of this value, = µo q 2 d2 k 2 R2 = r̂ 2 32π cm2 (h2 + R2 )2 hSi floor Pfloor = ~ floor · dA ~ hSi (148) (138) Z where I have used the drawing to determine that cos θ = h/r and included the value of r as given in (132). This answer agrees with the solution provided by Griffiths if we replaced the k and m factors with their equivalent representation in terms of ω. It is good form to always solve problems in terms of the specific variables given. To solve for the value of R at which this intensity is a maximum, set the R derivative of (138) equal to zero. One of the solutions should be R = 0, which we already know to be a minimum. Find the other value of R for ∞ Z 2π hSiẑ · R dR dφẑ = 0 Z ∞ Z = 0 (149) 0 0 2π µo q 2 d2 k 2 R2 h (R 2 32π cm2 (h2 + R2 )5/2 dR dφ) (150) = µo hq 2 d2 k 2 (2π) 32cπ 2 m2 Z 0 ∞ (h2 R3 dR + R2 )5/2 (151) 10 The remaining integral is found in a table of integrals, Z a 1 x3 dx = 2 − √ 2 5/2 2 3/2 2 (a + cx ) 3c (a + cx ) c a + cx2 (152) Considering our specific integral, Z ∞ R3 dR (153) 2 2 5/2 0 (h + R ) = h2 1 −√ 2 3(h + R2 )3/2 h2 + R 2 = 0−0− 1 h2 −√ 3(h2 )3/2 h2 ∞ (154) determine when it reaches the specified value. One expression that includes the oscillation amplitude is the potential energy of the spring/charge system, U = 1 2 kz 2 (162) where z = d in the first parts of this problem and z = z(t) now. The time derivative of this energy is, 0 dU dt = k d 2 z 2 dt (163) (155) h2 1 + 3h3 h = − = 1 1 − h 3h (157) = 2 3h (158) (156) The power radiated by the dipole is also an expression for the energy change per unit time. Note, however, that the radiation emitted by the dipole carries energy away to both the ceiling and the floor. Using (161) with the k and m dependence, P dU dt = = µo q 2 k 2 z 2 12πcm2 (164) and the d term has been replaced with the time dependent z. This result is now applied to (151), Pfloor = = µo hq 2 d2 k 2 16cπm2 µo q 2 d2 k 2 24cπm2 2 3h Since the spring system is losing energy we can write, (159) k d 2 z 2 dt (160) Consider Eq. 11.22, the total power radiated by an electric dipole, hP i = µo p2o ω 4 12πc = − µo q 2 k 2 z 2 12πcm2 (165) which leads to a familiar differential equation and solution, d 2 z dt = − µo kq 2 2 z 6πcm2 (166) d(z 2 ) z2 = − µo kq 2 dt 6πcm2 (167) = − µo kq 2 t + z0 6πcm2 (168) z2 = µo kq 2 d2 exp − t 6πcm2 (169) z = µo kq 2 t d exp − 12πcm2 (170) (161) where Griffiths correctly applies the angle brackets to indicate that this is a time average. All of the work in this solution should have included these angle brackets as well. We knew from the beginning that these were time averages, however, so I left off the brackets. If we rewrite (160) in terms of the variables given in (161), replacing the ω and dipole moment amplitude dependencies, then we see that the power striking the floor is equal to half of the total energy radiated by this dipole. We integrated out to infinity so any radiation that does not reach the floor must hit the ceiling. (c) The oscillation amplitude is given by d, the distance by which the charge is initially displaced. We’ll need to solve for this amplitude as a function of time and then ln z 2 where the initial amplitude, zo = d, is given in the problem statement. The amplitude is equal to d/e when the exponential argument above is equal to −1. This occurs 11 when, moment of this particle/image pair is, z p~ = q d~ d e = (176) (171) (177) = q(2~z) 2 − µo kq t 12πcm2 t = −1 = 12πcm2 µo kq 2 (172) (173) where my t is the τ that we have been asked to find. Notice that this time decreases with increasing value of the charge q. The radiated power increase with charge so this is a sensible result. [6.] Problem 11.25 from Griffiths A charged particle (mass m and charge q) is a distance z from a conducting surface. As this particle moves toward (or away) from the conducting surface it emits radiation. This process may be treated as an interaction between the particle and its image due to the conducting surface. This pair forms an effective dipole, the moment of which changes in time. Find the total radiated power as a function of the particle’s height z above the conducting surface. = 1 6m2 z 4 µo cq 4π 2 (174) The problem statement gives the method by which to solve it. Treat this situation as an electric dipole where the separation distance is changing in time. The general expression for the power radiated by an electric dipole is, ∼ = µo p̈2 6πc For the term in (175), (179) p̈ = 2qz̈ It remains to solve for z̈. This is done by examining the force on the particle due to its image. Considering magnitudes only, F = ma = m d2 z dt2 = 11.60 (175) where p is the dipole moment and the double dot represents a double time derivative. This may be treated as a one dimensional problem along the line that connects the charged particle to its image. This particle will accelerate as it approaches the conducting surface due to the greater amount of induced surface charge that will exert a force on it. The dipole mz̈ (180) This force is due to the electric influence of the image. The primed field is that due to the image particle, which must have charge −q (review ch. 3 in the text if you do not believe this). 3 Solution P where in the final step we keep the magnitude of the dipole moment and I indicated the time dependence contained therein. The factor of 2 comes from the separation between the particle and its image. F = qE 0 Griffiths Hint: The solution is, P (z) (178) p(t) = 2qz(t) (181) = q −q 4πo r2 = − q2 4πo (2z)2 (183) = − q2 16πo z 2 (184) (182) Equating (180) and (184), mz̈ = − z̈ = − q2 16πo z 2 (185) q2 16mπo z 2 (186) Solving for the second time derivative of the dipole moment provides, q2 p̈ = 2q − (187) 16mπo z 2 = − q3 8mπo z 2 (188) 12 tionship µo o = 1/c2 . Back to (175), µo P = 6πc µo = 6πc = q3 − 8mπo z 2 2 q6 64m2 π 2 2o z 4 µo q 6 384cm2 π 3 2o z 4 (189) 1 = c4 µ2o 2o (190) P = (191) = This does not look like the solution Griffiths provided. His solution, (174), has no o dependence so it is probably best to remove this in our solution noting the rela- µo q 6 (c4 µ2o ) 384cm2 π 3 z 4 µ3o q 6 c3 384m2 π 3 z 4 3 µo cq 2 1 = 6m2 z 4 4π (192) (193) (194) (195) Our solution is equivalent to that provided by Griffiths.