HW #8 Due to Gravity

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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 2 x − x0 = vot + at (2) |{z} 2 0 1 [1.] Problem 11.10 from Griffiths −h = − |g|t2 (3) 2 s An electron is released from rest and allowed to fall un- 2h t = (4) der the influence of gravity. During the first centimeter |g| of its free fall what fraction of the lost potential energy is dissipated through radiation?

The radiated power is given by, Solution

Recall the kinematic equations in one dimension with 2 2 µoq a constant acceleration. P = Eq. 11.61 (5) 6πc 1 v − v = at x − x = v t + at2 (1) o o o 2 where this is known as the Larmor formula. Griffiths These equations underlie a physical discrepancy with explains in section 11.2.1 that this expression is valid for this problem. The kinematic equations above rely on an v c. This condition is met here because the final ve- ideal system in which all of the potential energy of the locity of any object falling (from rest) through a distance falling object is converted into kinematic energy. There of 1 cm in the Earth’s gravitational field is not going to are corrections to these equations in the case where ra- be appreciable compared to the speed of light. diation 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 Multiplying (4) and (5) gives the total energy radiated therefore these equations are valid as a very good ap- by the electron in this measured fall. proximation. 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 Urad = P t (6) out to be very large, then you must start over and think about the new physical issues. 2 2 s µoq a 2h = · (7) Let the distance fallen by the electron be equal to h. The 6πc |g| actual value of this parameter is given as 1 cm, but it is easier to use the variable throughout the algebra and 2 2 s then plug in the numerical value for the final calcula- µoq g 2h = (8) tion. The goal of this problem is to determine the frac- 6πc |g| tion 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 total change in gravitational potential energy is the energy is decreasing and we only wish to know how (leaving this value positive for the reason given previ- this absolute amount relates to the radiated amount. ously),

Power is a measure of energy per unit time. The total energy radiated by this falling electron is the power ra- ∆Upot = mgxf − mgxo (9) diated multiplied by the time it takes for it to fall the speciﬁed distance. Solving for this time requires the use = mg(xf − xo) (10) 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 = mgh (11) 2

The desired ratio is, so the time is now written in terms of parameters given

2 2 in the problem statement. µoq g q 2h Urad 6πc |g| = (12) ∆Upot mgh The total radiated power, Urad, is given from (5) multiplied by this time. 2 r µoq 2|g| = (13) 2 2 6πcm h µoq a vo U = · (21) rad 6πc |a| r (4π × 10−7)(1.6 × 10−19)2 2(9.8) = 2 6π(3.0 × 108)(9.11 × 10−31) 0.01 µoq vo|a| = (22) (14) 6πc

Now to solve for the desired fraction, ≈ 2.8 × 10−22 (15)

The energy lost due to radiation is insignificant com- µ q2v |a| U o o pared to the general kinetic loss. rad = 6πc 1 2 (23) K.E.i 2 mvo [2.] Problem 11.13 from Griffiths 2 µoq vo|a| 2 = · 2 (24) (a) Consider an electron decelerating (a = constant) 6πc mvo from initial velocity vo to final velocity v = 0. Assum- ing vo c so that you may apply the Larmor formula, 2 µoq |a| what fraction of the initial kinetic energy of this electron = (25) is radiated away? 3πcmvo

(b) Apply the following values to the problem in part (b) In this part we want to apply given values to (25). (a). This requires solving for the absolute value of the acceleration because it is not given. 5 vo ≈ 10 m/s d = 30 A˚ (16) where d is the total distance traveled. What does this Using (1), (16), and (20), result tell us about radiation losses for electrons in ordi- nary conductors? 1 x − x = v t + at2 (26) f i o 2 Solution

2 vo a v The initial kinetic energy of the electron is, d = v − o (27) o |a| 2 a2 1 K.E. = mv2 (17) i 2 o v2 1 v2 where m is the electron mass. d = o − o (28) |a| 2 |a|

The total power radiated away is found using (6). The 1 v2 time it takes for the electron to come to rest is found us- d = o (29) ing (1). The a term is written negative because this is a 2 |a| deceleration. 2 v − vo = −at (18) v |a| = o (30) 2d

−vo = t (19) −a where this derivation also assumes the initial location is the origin and the velocity is v ≥ 0.

vo t = (20) |a| Solving for the numerical fraction of kinetic energy lost 3 to radiation, where the vector form is written in spherical coordi- 2 2 nates. The magnitude of the acceleration is written be- Urad µoq v = · o (31) cause the next step is to solve for the electron velocity in 3πcmv 2d K.E.i o its orbits.

2 µoq vo = (32) In the case of centripetal acceleration and circular or- 6πcmd bits, the tangential velocity of the particle is found using a = v2/r. The distance r changes as the electron moves (4π × 10−7)(1.6 × 10−19)2(105) closer to the proton. As r decreases along this path the = (33) 6π(3.0 × 108)(9.11 × 10−31)(30 × 10−10) initial velocity must be the lowest velocity throughout the trip. This initial velocity is found using,

≈ 2.1 × 10−10 (34) v2 q2 = 2 (38) This is another instance in which the energy lost to ra- r 4πomer diative processes is negligible compared to other loss mechanisms. s q2 v = (39) 4π m r [3.] Problem 11.14 from Grifﬁths o e

r Consider the Bohr model of the hydrogen atom. In this q 1 = (40) model the ground state electron travels around the pro- 2 πomer −11 ton in a circular orbit of radius ro = 5 × 10 m, due to the Coulomb attraction between the particles. Classical All of the parameters in (40) are known numerically. The electrodynamics requires that this accelerating electron initial (and lowest) velocity of this electron occurs at its emit radiation and therefore continuously lose energy. initial position where r = ro, As it loses energy this electron will move inward and eventually impact the proton. Show that the Larmor vo = formula is valid for most of the electron’s inward mo- s tion (i.e. v c). Calculate the typical lifetime of such 1.6 × 10−19 1 an atom if the electron’s orbits can always be considered 2 π(8.85 × 10−12)(9.11 × 10−31)(5 × 10−11) circular. (41)

Solution ≈ 2.2 × 106 (42)

To show that the Larmor formula is valid I will solve This is a large velocity compared to everyday experi- for the initial velocity of the electron in functional form ence, but vo/c ≈ 0.007, which satisfies the small velocity and then determine how this changes as the electron ap- condition. As the electron approaches√ the proton this proaches the proton. This is accomplished by looking at velocity increases by a factor of 1/ r. This factor causes the forces involved, the velocity to slowly increase and we may therefore F~ = me~a = −qE~ (35) claim that the electron’s velocity is less than the speed of light for most of the trip. Of course, once the elec- where me is the electron’s mass, q is the electron’s ~ tron gets incredibly close to the proton the distance goes charge, and E is the electric field due only to the pro- to zero and the velocity goes to infinity. We ignore this ton. Recall that the force experienced by the electron situation here. does not include its own fields in this problem.

The lifetime of the electron in this conﬁguration is given The electric ﬁeld of the proton is known, but it’s also by the time necessary for the electron to radiate away given as Eq. 2.10 if you wish to review. The acceleration all of its energy. The total energy of the electron, E , is is, e its kinetic energy plus its potential energy. The kinetic q2 energy of the electron is given by (17) replacing the ini- ~a = − 2 rˆ (36) 4πomer tial velocity there with the instantaneous velocity here. The potential energy is P.E.= −qV , where V is the scalar

2 potential ﬁeld due to the proton and the negative sign is q included because of the electron charge. The scalar po- a = 2 (37) 4πomer tential is known for a single charged particle, giving the 4 following expression for the energy, Now we equate (48) and (53) so that we can solve for the time it takes for the electron to impact the proton. 1 2 q Ee = mev − q (43) In (48) we have an expression for the amount of energy 2 4πor radiated away as a function of the acceleration (with the Simplify using (39), radial dependence coming from the acceleration term). In (53) we have the same quantity, but have derived it 2 2 me q q from energy principles directly since we know the initial Ee = − (44) 2 4πomer 4πor energy and can quantify its rate of change as a function of time and radial position.

2 q me µ q6 q2 dr = − 1 (45) o = − 4π r 2m 3 2 2 4 2 (54) o e 96π ocmer 8πor dt

q2 q2 96π32cm2r4 = − (46) o e dt = − 2 · 6 dr (55) 8πor 8πor µoq This is the total amount of energy that must be lost to radiation before the electron impacts the proton. 2 2 2 12π ocmer = − 4 dr (56) µoq The energy per unit time radiated away is given by (5), and we have solved for the acceleration in (37). Z Z 0 2 2 2 12π ocmer 2 dt = − dr (57) µ q2 q2 4 o ro µoq P = 2 (47) 6πc 4πomer In (57) the limits of the r integral reﬂect the fact that the 6 µoq electron begins at the ground state radius and moves = 3 2 2 4 (48) 96π ocmer into the proton.

To be able to compare these energy terms we take ad- 2 2 Z 0 12π ocm vantage of what power represents. Since power is en- t = − e r2 dr (58) µ q4 ergy per unit time it can be written as, o ro ∆E To be more mathematically rigorous we may require P = (49) ∆t 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. where E represents energy and not electric ﬁeld. 12π2 cm2 r3 0 t = − o e As we reduce the ∆t to an inﬁnitesimal time period we 4 (59) µoq 3 have, ro dE P = − (50) 4π2 cm2 dt o e 3 = − 4 −ro (60) µoq where the negative sign is placed because we know the energy is decreasing in the system but the radiated 2 2 3 power must be positive. 4π ocmero = 4 (61) µoq Using our expression for the total energy that must be All of these values are known, lost as given in (46), and noting that r = r(t), we have, 2 t = dEe d q (62) − = − − (51) 2 −12 8 −31 2 −11 3 dt dt 8πor 4π (8.85 × 10 )(3 × 10 )(9.11 × 10 ) (5 × 10 ) (4π × 10−7)(1.6 × 10−19)4 (63) q2 1 dr = − 2 (52) 8πo r dt ≈ 1.3 × 10−11 s (64) 2 q dr If this truly explained hydrogen, then it would be an in- = − 2 (53) 8πor dt credibly 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, 2 2 2 r dP µoq a sin θ 1 − β = (67) θ =∼ for v ≈ c (65) dΩ 16π2c (1 − β cos θ)5 max 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 To find the angle at which the maximum radiation oc- velocity v = 0. State the ratio in terms of γ. Reference curs we take the relevant derivative of (67) and set it figure 11.14. equal to zero. We already know that the minimum in radiation (equal to zero) occurs at θ = 0. So although Solution this method can only show us where extrema occur, un- less the solution is θ = 0 we can safely assume we have found a maximum. If you are still unsure, then you can The parameters in the problem statement are from page take the second derivative of (67) at the value of θ found 463 of Griffiths, from setting the first derivative to zero and if this result v 1 is negative (positive) then the point represents a maxi- β ≡ γ ≡ (66) mum (minimum). c p1 − v2/c2

Finding θmax,

2 2 2 d µoq a sin θ 0 = (68) dθ 16π2c (1 − β cos θ)5

2 2 2 µoq a d sin θ = (69) 16π2c dθ (1 − β cos θ)5

2 2 5 2 4 µoq a (1 − β cos θ) (2 sin θ cos θ) − sin θ(5(1 − β cos θ) (−β)(− sin θ)) = (70) 16π2c (1 − β cos θ)10

(1 − β cos θ)5(2 sin θ cos θ) − sin2 θ(5(1 − β cos θ)4(−β)(− sin θ)) = (71) (1 − β cos θ)10

= (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) 6

This is the form of a quadratic equation and may be dure in asymptotic analysis. solved as such. I have not labeled this angle as θmax, but by the method taken we are ﬁnding that value. −1 + p1 + 15β2 cos θ = (84) max 3β −2 ± p4 − 4(3β)(−5β) cos θ = (78) 6β −1 + p1 + 15(1 − )2 = (85) 3(1 − ) −2 ± 2p1 + 15β2 = (79) 6β 1 hp i = 1 + 15(1 − 2 + 2) − 1 (86) 3(1 − ) −1 ± p1 + 15β2 = (80) 3β 1 √ = 1 + 15 − 30 − 1 (87) 3(1 − ) We choose the sign in (80) based on information about θ max provided in example 11.3. From ﬁgure 11.14 and a " r # statement on the bottom of page 463 we know that the 1 30 = 4 1 + − 1 (88) radiation is concentrated through a narrow cone near 3(1 − ) 16 the forward direction. Figure 11.14 is turned sideways, but if we look at the location of θ it is evident that max " # concentrating the radiation in a narrow region near the z 1 15 1/2 ◦ = 4 1 + − 1 (89) axis means that θmax < 45 . This requires that cos θ > 1. 3(1 − ) 8 Meeting this condition requires that we take the plus sign in (80). Recall the binomial expansion, which will be applied multiple times in the steps to follow, Finally,

(1 + x)n ≈ 1 + nx (90) −1 + p1 + 15β2 cos θ = (81) 3β for small x. " # −1 + p1 + 15β2 θ = cos−1 (82) max 3β 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 1 1 15 cos θ = 4 1 + − 1 (91) in the ultrarelativistic case β ≡ v/c ≈ 1, but slightly less max 3(1 − ) 2 8 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 1 15 = 4 + − 1 (92) written as, 3(1 − ) 4

β ≈ 1 − (83) 1 5 = 1 − (93) 1 − 4 where is an expansion parameter and is very small. The rest of this problem concerns rewriting terms and We can solve further by using the binomial expansion solving for this expansion parameter. on the 1/(1 − ) term.

To begin, replace β with its expansion in (81). When 1 = (1 − )−1 (94) writing out the terms we should neglect anything that 1 − is of order 2 or higher. Since 1, any term contain- ing an 2 is so small it should be ignored. This process is known as expanding to order one in , a common proce- ≈ 1 + (95) 7 5 The direction at which the maximum power is radiated cos θmax ≈ (1 + ) 1 − (96) 4 occurs at θ = 90◦. This is known because the maximum must occur when the sine term is equal to unity. The 5 5 maximum power radiated in this zero velocity instant ≈ 1 − + − 2 (97) is, 4 4 max 2 2 dP µoq a = (108) 1 dΩ 16π2c ≈ 1 − (98) v=0 4 The other radiated power term is (67) evaluated at the We invoke another expansion of the cosine function for already determined θmax. comparison with (98). max 2 2 2 ∞ dP µoq a sin θmax X (−1)nx2n = (109) cos x = (99) dΩ 16π2c (1 − β cos θ )5 (2n)! θ=θmax max n=0 The desired ratio is,

2 4 max 2 2 2 x x dP µoq a sin θmax = 1 − + − ... (100) dΩ θ=θmax 16π2c (1−β cos θmax)5 2! 4! = 2 2 (110) dP max µoq a dΩ v=0 16π2c 2 θmax cos θmax ≈ 1 − (101) 2 2 sin θmax = 5 (111) 4 (1 − β cos θmax) since keeping the θmax term seems unnecessary. Since this is evaluated at θmax we know that β can be Equate (98) and (101), expanded as in (83). The plan is to write (111) in terms of and then relate this result to γ as requested. The 1 θ2 1 − = 1 − max (102) numerator of (111) can be rewritten using yet another 4 2 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 = θ2 (103) 2 max small. The denominator is rewritten using (98).

sin2 θ (sin θ )2 r max ≈ max (112) = θ (104) (1 − β cos θ )5 (1 − β(1 − ))5 2 max max 4

Using (83) to rewrite the above, θ2 = max r 5 (113) 1 − β (1 − (1 − )(1 − 4 )) θ ≈ (105) max 2 p 2 In the next part of the problem we want to write the radi- 2 = 2 (114) ated power in the direction of θmax as a fraction of the ra- (1 − (1 − − + ))5 diated power in the equivalent maximum direction for 4 4 a particle of instantaneous velocity equal to zero. Notice Once again, drop the 2 term. that the particle is still accelerating, as it must in order to radiate at all, but that we are concerned with the instant p 2 2 2 at which its velocity reaches zero. 2 = (115) 5 ( 5 )5 (1 − (1 − 4 − + 4 )) 4 Begin by writing (67) for v = 0, which is the same as setting β = 0. 1 45 = −4 (116) 2 2 2 2 5 dP µoq a sin θ = (106) dΩ 16π2c (1 − (0) cos θ)5 = 0.16384−4 (117) 2 2 µoq a = sin2 θ (107) where inverse powers of small numbers are large and 16π2c therefore kept. 8

It is necessary to express in terms of γ. arrive at a solution of, 2 2 4 2 µoq d ω R h 1 S = (126) γ = (118) ﬂoor 2 2 2 5/2 p1 − β2 32π c (R + h )

1 = (119) (b) Using your result from (a), calculate the energy per p1 − (1 − )2 unit time across the entire (infinite) floor. Comment on this value and describe whether it makes sense. 1 = (120) (c) The spring/charge system is losing energy through p1 − (1 − 2 + 2) this radiation. Solve for the time τ at which the oscillation amplitude has been reduced to d/ exp, assuming 1 that the energy loss per cycle is very small. ≈ √ (121) 2 Solution 1 ≈ (122) 2γ2 (a) From Eq. 9.63 we know that the intensity of EM waves (which compose radiation) is equivalent to the Our final answer is given in (117), and we can now write time averaged value of their Poynting flux. The hint re- this in terms of γ, minds us to make sure we only solve for the intensity that actually strikes the floor. This is the amount of en- dP max −4 ergy flux that passes through the floor. Defining a ge- dΩ θ=θ 1 max = 0.16384 (123) ometry in which the floor is at a constant z = 0 value dP max 2γ2 means that the area vector of the floor is given by −zˆ. dΩ v=0 The intensity striking the floor is, = 0.16384(24)γ8 (124) ~ hSi floor = hSi · zˆ (127) ≈ 2.6γ8 (125) where I am taking the absolute value of this expression because we already know the radiated energy is leaving There are probably more equations in this solution the oscillating charge. where it is appropriate to have an “approximately equal” sign instead of an “equal to” sign, but in all the The oscillating charge is essentially an electric dipole. steps above the approximations are so good that there The intensity of the radiation from such a system is is little error. You can use the equal signs in this case, given by Eq. 11.21, ultrarelativistic, without large error. 2 4 2 ~ µopoω sin θ hSi = 2 2 rˆ (128) [5.] Problem 11.21 from Griffiths 32π cr The dipole moment amplitude is po = qd because d is the maximum displacement from the origin. The origin A particle (mass = m and charge = q) is attached to is set as the equilibrium position of the charge. The an- a spring and suspended from a ceiling. Reference fig- gular frequency of a spring/mass system is known to be ure 11.19 in Griffiths. The equilibrium position of this p spring-mass system is a distance h above the floor. At ω = k/m. Inserting this information into the expres- time t = 0 the mass is pulled downward a distance d sion for intensity gives, and then released to begin oscillatory motion. 4 2 r ! µo sin θ k hS~i = (qd)2 rˆ (129) 32π2cr2 m (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 2 2 2 2 µoq d k sin θ for the value of R at which the radiation intensity is a = rˆ (130) 32π2cm2r2 maximum. The R dependence is contained in the r and sin θ terms. Griffiths’ Hints: Assume d λ h. The calcu- The following image shows the charge (solid black cir- lation only concerns the average energy per unit time cle) suspended from the spring (coil). The charge is that strikes the floor. Neglect radiative damping and placed at the center of a spherical coordinate system. 9

which an extrema occurs and this will be the location of maximum intensity.

d 0 = hSi (139) dR ﬂoor

2 2 2 2 d µoq d k R h = (140) dR 32π2cm2(h2 + R2)5/2

2 2 2 2 µoq d k h d R = (141) 32π2cm2 dR (h2 + R2)5/2 From this drawing we see that, (h2 + R2)5/2(2R) − R2(5/2)(h2 + R2)3/2(2R) R = sin θ = (131) (h2 + R2)5 r (142) p r = h2 + R2 (132) = 2R(h2 + R2)5/2 − 5R3(h2 + R2)3/2 (143) Returning to (130) with this information, 2 2 3/2 2 2 3 2 2 2 2 2 = (h + R ) 2R(h + R ) − 5R (144) µoq d k R 1 hS~i = √ rˆ (133) 32π2cm2 r h2 + R2 = R 2h2 + 2R2 − 5R2 (145)

2 2 2 2 2 µoq d k R 1 1 = √ rˆ (134) 2 2 32π2cm2 h2 + R2 h2 + R2 = 2h − 3R (146)

2 2 2 2 r µoq d k R 2 = rˆ (135) R = h (147) 32π2cm2(h2 + R2)2 3

The intensity striking the ﬂoor is the z component of this The radial distance to the position of maximum intensity value, is slightly shorter than the distance of the particle from

2 2 2 2 the floor. µoq d k R hSi = rˆ · zˆ (136) floor 32π2cm2(h2 + R2)2 (b) The total energy per unit time (the power) across the floor is found by integrating the expression in (138) over 2 2 2 2 µoq d k R the floor. This is written as, = cos θ (137) 32π2cm2(h2 + R2)2 Z ~ ~ Pfloor = hSifloor · dA (148) 2 2 2 2 µoq d k R h = (138) 32π2cm2(h2 + R2)5/2 Z ∞ Z 2π where I have used the drawing to determine that cos θ = = hSizˆ · R dR dφzˆ (149) 0 0 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 Z ∞ Z 2π µ q2d2k2R2h = o (R dR dφ) ω 2 2 2 2 5/2 representation in terms of . It is good form to always 0 0 32π cm (h + R ) solve problems in terms of the specific variables given. (150)

R To solve for the value of at which this intensity is a µ hq2d2k2 Z ∞ R3 maximum, set the R derivative of (138) equal to zero. = o (2π) dR 2 2 2 2 5/2 One of the solutions should be R = 0, which we already 32cπ m 0 (h + R ) know to be a minimum. Find the other value of R for (151) 10

The remaining integral is found in a table of integrals, determine when it reaches the specified value. One expression that includes the oscillation amplitude is the Z x3 a 1 dx = − √ potential energy of the spring/charge system, (a + cx2)5/2 3c2(a + cx2)3/2 c2 a + cx2 (152) 1 U = kz2 (162) Considering our specific integral, 2 Z ∞ R3 dR where z = d in the first parts of this problem and z = 2 2 5/2 (153) 0 (h + R ) z(t) now.

h2 1 ∞ = − √ (154) The time derivative of this energy is, 2 2 3/2 2 2 3(h + R ) h + R 0 dU k d = z2 (163) h2 1 dt 2 dt = 0 − 0 − − √ (155) 3(h2)3/2 2 h The power radiated by the dipole is also an expression for the energy change per unit time. Note, however, that h2 1 the radiation emitted by the dipole carries energy away = − + (156) 3h3 h to both the ceiling and the ﬂoor. Using (161) with the k and m dependence,

1 1 2 2 2 = − (157) dU µoq k z h 3h P = = (164) dt 12πcm2 2 = (158) and the d term has been replaced with the time depen- 3h dent z. This result is now applied to (151),

2 2 2 Since the spring system is losing energy we can write, µohq d k 2 Pﬂoor = 2 (159) 16cπm 3h 2 2 2 k d µoq k z z2 = − (165) 2 dt 12πcm2 2 2 2 µoq d k = (160) 24cπm2 which leads to a familiar differential equation and solution, Consider Eq. 11.22, the total power radiated by an electric dipole, 2 d µokq z2 = − z2 (166) 2 4 dt 6πcm2 µop ω hP i = o (161) 12πc 2 2 d(z ) µokq where Grifﬁths correctly applies the angle brackets to = − dt (167) z2 6πcm2 indicate that this is a time average. All of the work in this solution should have included these angle brackets 2 as well. We knew from the beginning that these were µokq ln z2 = − t + z (168) time averages, however, so I left off the brackets. 6πcm2 0

2 If we rewrite (160) in terms of the variables given in µokq z2 = d2 exp − t (169) (161), replacing the ω and dipole moment amplitude 6πcm2 dependencies, then we see that the power striking the floor is equal to half of the total energy radiated by this 2 dipole. We integrated out to infinity so any radiation µokq z = d exp − t (170) that does not reach the floor must hit the ceiling. 12πcm2

(c) The oscillation amplitude is given by d, the distance where the initial amplitude, zo = d, is given in the prob- by which the charge is initially displaced. We’ll need to lem statement. The amplitude is equal to d/e when the solve for this amplitude as a function of time and then exponential argument above is equal to −1. This occurs 11 when, moment of this particle/image pair is, ~ d ~p = qd (176) z = (171) e = q(2~z) (177) 2 µokq − t = −1 (172) 12πcm2 p(t) = 2qz(t) (178) where in the ﬁnal step we keep the magnitude of the 2 12πcm dipole moment and I indicated the time dependence t = 2 (173) µokq contained therein. The factor of 2 comes from the separation between the particle and its image. where my t is the τ that we have been asked to ﬁnd. Notice that this time decreases with increasing value of For the term in (175), the charge q. The radiated power increase with charge so this is a sensible result. p¨ = 2qz¨ (179)

[6.] Problem 11.25 from Griffiths It remains to solve for z¨. This is done by examining the force on the particle due to its image. Considering mag- A charged particle (mass m and charge q) is a distance nitudes only, z from a conducting surface. As this particle moves to- d2 ward (or away) from the conducting surface it emits ra- F = ma = m z = mz¨ (180) dt2 diation. 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 mo- This force is due to the electric influence of the image. ment of which changes in time. Find the total radiated The primed field is that due to the image particle, which power as a function of the particle’s height z above the must have charge −q (review ch. 3 in the text if you do conducting surface. not believe this). F = qE0 (181) Griffiths Hint: The solution is, −q 2 3 1 µocq = q 2 (182) P (z) = (174) 4πor 6m2z4 4π q2 = − 2 (183) 4πo(2z) Solution q2 = − 2 (184) 16πoz 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 Equating (180) and (184), expression for the power radiated by an electric dipole q2 is, mz¨ = − 2 (185) 16πoz

2 2 µop¨ q P =∼ 11.60 (175) z¨ = − 2 (186) 6πc 16mπoz where p is the dipole moment and the double dot represents a double time derivative. Solving for the second time derivative of the dipole moment provides, This may be treated as a one dimensional problem along q2 p¨ = 2q − the line that connects the charged particle to its image. 2 (187) 16mπoz This particle will accelerate as it approaches the conducting surface due to the greater amount of induced q3 = − 2 (188) surface charge that will exert a force on it. The dipole 8mπoz 12

2 Back to (175), tionship µoo = 1/c .

µ q3 2 1 P = o − 4 2 2 (189) 2 = c µo (192) 6πc 8mπoz o µ q6 µ q6 = o o 4 2 2 2 2 4 (190) P = 2 3 4 (c µo) (193) 6πc 64m π oz 384cm π z µ q6 µ3q6c3 = o o 2 3 2 4 (191) = 2 3 4 (194) 384cm π oz 384m π z

2 3 1 µocq = (195) This does not look like the solution Grifﬁths provided. 6m2z4 4π His solution, (174), has no o dependence so it is probably best to remove this in our solution noting the rela- Our solution is equivalent to that provided by Grifﬁths.