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 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 gravita- tional 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) multi- plied 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 accel- eration 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.