Haverford College Haverford Scholarship
Faculty Publications Astronomy
2009
The Lorentz force and the radiation pressure of light
Tony Rothman
Stephen P. Boughn Haverford College, [email protected]
Follow this and additional works at: https://scholarship.haverford.edu/astronomy_facpubs
Repository Citation "The Lorentz Force and the Radiation Pressure of Light" (with Tony Rothman), Am. J. Phys, 77, 122 (2009).
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Citation: Am. J. Phys. 77, 122 (2009); doi: 10.1119/1.3027432 View online: http://dx.doi.org/10.1119/1.3027432 View Table of Contents: http://ajp.aapt.org/resource/1/AJPIAS/v77/i2 Published by the American Association of Physics Teachers
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Downloaded 22 Mar 2013 to 165.82.168.47. Redistribution subject to AAPT license or copyright; see http://ajp.aapt.org/authors/copyright_permission The Lorentz force and the radiation pressure of light ͒ Tony Rothmana Department of Physics, Princeton University, Princeton, New Jersey 08544 ͒ Stephen Boughnb Department of Astronomy, Haverford College, Haverford, Pennsylvania 19041 ͑Received 8 July 2008; accepted 27 October 2008͒ To make plausible the idea that light exerts a pressure on matter, some introductory physics texts consider the force exerted by an electromagnetic wave on an electron. The argument as presented is mathematically incorrect and has several serious conceptual difficulties without obvious resolution at the classical, yet alone introductory, level. We discuss these difficulties and propose an alternative argument. © 2009 American Association of Physics Teachers. ͓DOI: 10.1119/1.3027432͔
I. THE FRESHMAN ARGUMENT ated by the E-field in the +x-direction and acquires a velocity Ͼ vx 0. The magnetic field then exerts a force on the charge The interaction of light and matter plays a central role, not equal to qvϫB, which points in the +z-direction, the direc- only in physics itself, but in any introductory electricity and tion of propagation of the wave. The electromagnetic wave magnetism course. To develop this topic, most courses intro- therefore carries a momentum in this direction. Reference 2 duce the Lorentz force law, which gives the electromagnetic states that “…the motion of the charge is mainly due to E. force acting on a charged particle, and later discuss Max- Thus v is along E and reverses direction at the same rate that well’s equations. Students are then persuaded that Maxwell’s E reverses direction. But B reverses whenever E reverses. equations admit wave solutions that travel at the speed of Thus vϫB always has the same sign.”4 light, thus establishing the connection between light and A moment’s reflection shows that the last assertion is electromagnetic waves. At this point instructors generally false. After one-half cycle, both E and B change sign. But state that electromagnetic waves carry momentum in the di- because during this time the E-field has accelerated the rection of propagation via the Poynting flux vector and that charge entirely in the +x-direction, the electron at that point light therefore exerts a radiation pressure on matter. The as- ͑ 1 still has a positive x-velocity. In other words, the velocity sertion is not controversial: Maxwell recognized that light and acceleration are essentially 90° out of phase, as in a should cause a radiation pressure, but his demonstration is harmonic oscillator.͒ A similar argument holds for the not immediately transparent to present-day students. z-velocity. Thus the cross product vϫB reverses sign and At least two texts, the Berkeley Physics Course2 and Tipler 3 now points in the negative z-direction, opposite to the direc- and Mosca’s Physics for Scientists and Engineers, attempt tion of propagation. Furthermore, because there is an to make the suggestion that light carries momentum more x-component to the force, one needs to argue that on average plausible by calculating the Lorentz force exerted by an elec- it is zero. tromagnetic wave on an electron. In their discussion the au- Reference 2 claims that the first two terms in Eq. ͑2͒ av- thors claim—with differing degrees of rigor—to show that a erage to zero, the first because E varies sinusoidally, the light wave exerts an average force on the electron in the second because B varies sinusoidally as well and because we direction of propagation. Tipler and Mosca, for example, “…can assume that the increment of velocity along z during then derive an expression for the radiation pressure produced 3 one cycle is negligible, that is, we can take the slowly in- by a light wave. A cursory look at their argument shows that 4 creasing velocity vz to be constant during one cycle.” With it is incorrect in several obvious ways and in other respects these assumptions Ref. 2 concludes that the average force on leads rapidly into deep waters. ͗ ͘ ͗ ͘ ˆ The purpose of this note is threefold: to demonstrate why the charge is F =q vxBy k. Although the result might seem the “freshman” argument is incorrect, to show that nontrivial plausible, it is also incorrect because the velocity and mag- physics must be introduced to correct it, and to present a netic field are out of phase and consequently the time aver- more plausible derivation of radiation pressure that is acces- age of their product vanishes. That this is so, as well as the sible to first-year students. Consider, then, the situation previous claim, can be seen by an integration of the equation shown in Fig. 1. We assume that a light wave propagates in of motion. the +z-direction, its E-field oscillates in the x-direction, and its B-field oscillates in the y-direction. The wave impinges on a stationary particle with charge q, exerting on it a force according to the Lorentz force law. In units with c=1 the II. EQUATION OF MOTION Lorentz force is To determine the momentum of the charge, which we take F = q͑E + v ϫ B͒, ͑1͒ to be an electron, assume the electric and magnetic fields ͑ ͒ˆ which reduces to of the light wave are given by E=E0 sin t+ i and ͑ ͒ˆ ͑ ͒ˆ ˆ ͑ ͒ B=B0 sin t+ j, where is an arbitrary phase angle. In F = q Ex − vzBy i + qvxByk. 2 / ͑ ͒ our units E0 =B0. We set FLorentz=mdv dt in Eq. 2 to obtain The freshman argument goes like this: Assume that a pair of coupled ordinary linear first-order equations for the Eϳsin͑t͒ and Bϳsin͑t͒. The particle is initially acceler- electron velocity:
122 Am. J. Phys. 77 ͑2͒, February 2009 http://aapt.org/ajp © 2009 American Association of Physics Teachers 122
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E 0.15 z vx,vz q 0.1 y B 0.05 Fig. 1. An electromagnetic wave traveling in the z-direction strikes a point particle with charge q.TheE-field is taken in the x-direction and the B-field is taken in the y-direction. 0 10 20 30 40 ωt
/ Fig. 2. The x- and z-velocities versus t for c =0.1 and =0. Note that / ͑ ͒ ӷ / from the small c approximation see text vx vz; in this case vz vx dv =0.1. z = sin͑t + ͒v , ͑3a͒ dt c x
dv x = sin͑t + ͓͒1−v ͔, ͑3b͒ Additional insight into the solutions can be obtained by c z /Ӷ dt examining the limit c 1. For ordinary light sources at ϳ 16 / ϵ / optical frequencies 10 rad s, consideration of the where we have let c qB0 m, the cyclotron frequency. ͑ ͒ /ϳ −11 ͑ ͒ Poynting flux in the following section gives c 10 , Equations 3 have the somewhat surprising analytic solu- and so the limit is well satisfied. For high-powered lasers, tions such as those at the National Ignition Facility with pulse ϳ Ӷ energy 2 MJ, it is possible that c exceeds . For c , v ͑t͒ = c cosͫ c cos͑t + ͒ͬ ͑ ͒ / z 1 expansion of the solutions 6 to lowest order in c for =0 yields + c sinͫ c cos͑t + ͒ͬ +1, ͑4a͒ 1 2 2 v Х ͩ c ͪ ͓cos͑t͒ −1͔2, ͑7a͒ z 2 ͑ ͒ ͫ c ͑ ͒ͬ v t = c sin cos t + c x 1 Х ͩ ͓ͪ1 − cos͑t͔͒. ͑7b͒ vx ͫ c ͑ ͒ͬ ͑ ͒ Both v and v are positive definite, as shown in Fig. 2. − c2 cos cos t + , 4b z x Therefore their averages must be as well. This in itself con- ͗ ͘ ͗ ͘ where c and c are the integration constants. tradicts the arguments of Ref. 2 that Fx =0 but that Fz 1 2 ͑ /͒2 If we take v ͑0͒=v ͑0͒=0, which is reasonable and of suf- 0. Note also that vz is of order c , and vx is of order z x / ficient generality for our purposes, we find c . The behavior coincides with the plots, but suggests that because v2 ϳ͑ /͒2, a consistent, relativistic calcula- x c c c 5 c = − cosͫ cos ͬ, c = − sinͫ cos ͬ, ͑5͒ tion will significantly change vz. Moreover, the time aver- 1 2 ages of both v˙ x and v˙ z vanish to all orders, and so it is in fact impossible to exert a net force on the particle. and the full solutions are therefore One might object to the arguments of this section on the grounds that we have taken E and B to be simple harmonic ͑t͒ = − cosͩ c cos ͪcosͫ c cos͑t + ͒ͬ vz ϳsin͑t͒ rather than wavelike ϳsin͑kz−t͒. However, it is evident from Eq. ͑7͒ that kzӶt always and that such cor- c c rections are therefore negligible, an assertion borne out by − sinͩ cos ͪsinͫ cos͑t + ͒ͬ +1, ͑6a͒ numerical calculations.
v ͑t͒ = − cosͩ c cos ͪsinͫ c cos͑t + ͒ͬ x 0.1 vx,vz c c + sinͩ cos ͪcosͫ cos͑t + ͒ͬ. ͑6b͒ 0.05
The behavior of these solutions is not particularly trans- ωt 0 10 20 30 40 parent, but can easily be plotted. In Figs. 2–4 we show sev- / eral graphs for various values of c and phase angle . –0.05 Note that regardless of , vz is always positive, but that there is also a nonzero vx whose average can be positive, negative, or zero depending on . The =0 case is shown in –0.1 Figs. 2 and 4 and the =/2 case in Fig. 3. Also, for Fig. 3. The same plot as in Fig. 2 except that =/2. In this case the time Ӷ ӷ vx 1,vx vz. average of vx =0.
123 Am. J. Phys., Vol. 77, No. 2, February 2009 T. Rothman and S. Boughn 123
Downloaded 22 Mar 2013 to 165.82.168.47. Redistribution subject to AAPT license or copyright; see http://ajp.aapt.org/authors/copyright_permission 0.8 Gauss’s law becomes the source charge distribution and they cannot be equated. In the present situation there is not only a single test charge but no source charges whatsoever. Thus the 0.6 standard derivation cannot be applied. The only volume we vx,vz have at our disposal is the volume of the electron itself, 0.4 which leads quickly into quantum territory. A second difficulty is that the assumption of plane waves 0.2 with constant amplitude is an assumption of constant energy and momentum. If the light wave has constant momentum, how can any be transferred to the electron? There are many 0 instances in physics where we ignore the backreaction of a 10 20ω 30 40 t recoiling particle on the system. For instance, according to / conservation of momentum, a ball should not bounce off a Fig. 4. The same as Fig. 2 except that c =0.5. wall, until it is realized that the ball’s change in momentum is absorbed by Earth. Holding the amplitude constant in the current calculation III. INTERPRETATION might seem a reasonable approximation, but to be totally The question is whether the behavior just discussed can be consistent we should take into consideration the fact that the reconciled with the classical picture of the Poynting flux. The electron is accelerating and consequently emits radiation, Poynting vector in our units is and with that radiation momentum. The customary way to do this calculation in the nonrelativistic limit is via Thomson E ϫ B S = , ͑8͒ scattering. The differential Thomson scattering cross section 4 for a wave polarized in the x-direction is ء and the time average is ͗S͘=Re͑EϫB ͒/8. S points in the d 1 e2 2 = ͩ ͪ ͑cos2 cos2 + sin2 ͒, ͑11͒ direction of propagation of the electromagnetic wave and in d⍀ 2 m units with c=1 can be regarded interchangeably as power per unit area, energy per unit volume ͑or pressure͒, or momen- where is the angle between the incident and scattered tum flux. If the freshman argument is correct, then the par- wave. The differential scattering cross section is defined as ticle should be accelerated in the direction of the Ponyting the ratio of the radiated power per unit solid angle to the vector. But our previous results show that, on the contrary, incident power per unit area. The Thomson cross section, the particle drifts off in some other direction at a constant though, is symmetric with respect to reflection through the average velocity. origin and consequently as much momentum is emitted in Unfortunately, there seem to be several deep inconsisten- the forward as in the backward direction and hence no net cies in the entire approach. One is that the freshman argu- force is exerted on electron. It is therefore not obvious how ment is an invalid attempt to apply the standard classical to remedy this situation in the classical limit. Only when we derivation that is invoked to identify the Poynting flux with do a quantum mechanical derivation ͑Compton scattering͒ do electromagnetic momentum, a derivation which breaks down we see an asymmetry in the scattering cross section. In the ប/ ϳ −5 in the limit we have been considering. That is, advanced present circumstance, however, me 10 , so it would texts such as Jackson,6 typically begin by considering the appear that quantum corrections should be unnecessary. Lorentz force on a volume of charges: What we do in practice to obtain the radiation pressure of dp light in, say, astrophysical calculations is to multiply the = ͵ ͑E + J ϫ B͒d3x. ͑9͒ time-averaged Poynting flux ͗S͘ by the total Thomson cross dt vol section T. One can see why this works as follows. A photon scattered off an electron will have a z-momentum The first step is to eliminate the charge density in favor of -E. We also eliminate J in pz =p0 cos for initial momentum p0, and it therefore re· E via Gauss’s law, =͑1/4͒١ moves ͑1−cos ͒p from the original momentum compo- favor of ٌÃB via Ampère’s law to find 0 nent; the electron must gain the same amount. Multiplying dp d 1 1 the differential Thomson scattering cross section in Eq. ͑11͒ E͒ · mech + ͵ ͑E ϫ B͒d3x = ͵ ͓E͑١ by ͑1−cos ͒ and integrating over the sphere gives the total dt dt vol 4 4 vol Thomson scattering cross section .ϫ B͔͒d3x ١͑ B͒ − B ϫ · ϫ E͒ + B͑١ ١͑ E ϫ + 8 e2 2 ͑10͒ = ͩ ͪ . ͑12͒ T 3 m This result is purely formal, which after the elimination of and J relies only on vector identities. Because the second Multiplication by the momentum flux of photons will give term on the left is the only electromagnetic term with a time the total force on the electron. Because the Poynting flux is derivative, we tentatively identify it with the momentum of the momentum flux of photons, the same numerical result is the field. obtained by multiplying the Thomson cross section by the The crucial difference between the “graduate” approach time-averaged Poynting flux. This argument, however, relies and the freshman method, however, is that in the graduate on the quantum nature of photons. The Thomson cross sec- approach we consider a continuous charge distribution. In tion is the nonrelativistic limit of a cross section that must the limit of a single charge, the in the Lorentz force law ultimately be derived from QED, and so we see that the becomes the test charge distribution, whereas the in freshman argument leads quickly to a situation that might
124 Am. J. Phys., Vol. 77, No. 2, February 2009 T. Rothman and S. Boughn 124
Downloaded 22 Mar 2013 to 165.82.168.47. Redistribution subject to AAPT license or copyright; see http://ajp.aapt.org/authors/copyright_permission /Ӷ Ӷ have no resolution in the realm of classical physics! With the assumption that c 1 and 1 the v¨ z term The failing of Thomson scattering is due to the fact that no in Eq. ͑15b͒ can be ignored. Then energy is removed from the original beam. A possible clas- sical “out” to this situation is to assert that the energy radi- ie2 v˙ Х E B ͑1−i͒. ͑18͒ ated by the electron must be that lost by the incoming beam. z m2 x y Therefore because E=p for a classical wave, momentum conservation implies that the electron must acquire a For simplicity, take Ex and By to be real. We want the time z-momentum as for the Compton scattering case just average of the real part of this expression, or discussed.7 Although this argument is valid in terms of con- 4 servation laws, it gives no mechanism for transferring the e E0B0 ͗F ͘ = ͗mv˙ ͘ = = ͗S͘ . ͑19͒ energy from the incident wave to the electron. Unfortunately, z z m2 3 T modeling the process as interference between the incident plane wave and the spherical wave outgoing from the elec- The earliest paper we have found that proposes this calcula- tron fails to result in any transfer of z-momentum from the tion is by Page in 1920,8 although one suspects that Edding- wave to the charge. To recover the Compton result eventu- ton carried it out earlier. Clearly there are a few things to be ally requires including the radiation-reaction force on the desired in the derivation, but it does show that the radiation- electron, which we now consider, but because this derivation reaction force is necessary to obtain the claimed result. involves the classical radius of the electron, it has already With slightly more work the conclusion can be put on a gone beyond the realm of classical electromagnetism. firmer footing via a perturbation calculation9 as follows. The most straightforward way to deal with the failure of Note that Eq. ͑7͒ is the zeroth-order solution of Eq. ͑15͒, that Ӷ ϵ the classical approaches is via the Abraham-Lorentz model, is, when =0 and vz 1 is neglected. Assume vx vx0 +vx1 ϵ͑ ͒Ӷ which accounts for the energy radiated by the electron, if in and vz vz0 +vz1 vx, where the subscript 0 refers to the a somewhat ad hoc manner. From the Larmor formula the zeroth-order solution and the subscript 1 refers to the pertur- energy radiated by an accelerated electron over a time T is bation. It is then not too difficult to show that the surviving ϳ 2 2 / 2e a T 3. Equating this energy to the kinetic energy lost v˙ z is the perturbative part: by the particle ϳma2T2 gives a characteristic time to lose all Х ͑ ͒Х2 2͑ ͒ ͑ ͒ the energy to radiation: v˙ z1 cvx1 sin t c sin t . 20 2e2 Taking the time average of this expression vindicates the = . ͑13͒ 3m previous result. We emphasize that the Abraham-Lorentz model includes an explicit statement about the structure of This timescale is 2/3 the time for light to cross the clas- the electron and hence cannot be regarded as entirely classi- 2 / sical radius of the electron, rc =e m, and has a value cal; the model is a transition to quantum mechanics and Ϸ10−23 s. The total force acting on a particle is now quantum field theory. mv=Fext+Frad, where Frad is the radiation-reaction force. Conservation of energy considerations led Abraham and Lor- ͑ ͒ entz to propose that Frad=m v¨ see Ref. 7 for more details and consequently obtain the famous formula IV. ALTERNATIVE APPROACH ͑ ͒ ͑ ͒ m v˙ − v¨ = Fext. 14 Despite the many pitfalls revealed by the above methods, ͑ ͒ there is a superior and convincing demonstration that light With sufficient massaging, Eq. 14 can be applied to the exerts a pressure on matter, one that should be accessible to present circumstance to obtain the desired answer, that is, the students who have had a basic exposure to Maxwell’s equa- force imparted to the electron by an electromagnetic wave is ͗ ͘ ͑ ͒ tions. The advantage of this demonstration is that it avoids F= S T. Equation 2 now becomes consideration of the force acting on a point charge and can e therefore be carried out at the purely classical level. For this ͑ ͒ ͑ ͒ v˙ x − v¨ x = Ex − vzBy , 15a reason it should be adopted by introductory textbook authors. m What follows is a simplified version of a calculation de- scribed by Planck.10 e As before, consider a light wave propagating in the v˙ − v¨ = v B . ͑15b͒ z z m x y +z-direction that bounces off a mirror at z=0 ͑see Fig. 5͒.We take the mirror to be a near perfect conductor of height dx, Ӷ In the nonrelativistic regime vz 1 and we ignore the second width dy, and thickness z. The electric field of the light is a ͑ ͒ term on the right in Eq. 15a . We also take both vx and vz to superposition of right- and left-traveling waves: −it be of the form v=v0e , which is of course manifestly un- ͑ ͒ ͑ ͒ ͑ ͒ true according to the results of Sec. II. Then v˙ x =−i vx and Ex = E0 cos kz − t − E0 cos kz + t , 21 v¨ =−2v . Equation ͑15a͒ becomes x x where k=2/ł is the wave number, and we have included a e phase change on reflection. ͑This solution ensures that E=0 − iv ͑1+i͒Х E , ͑16͒ x m x at the surface of the conductor. Recall that the tangential component of an E-field must be continuous across a bound- or with Ӷ1 ary, and because the interior field essentially vanishes for a good conductor, the exterior field at the boundary must also.͒ ie 11 v Х E ͑1−i͒. ͑17͒ From the differential form of Faraday’s law, t, we haveץ/Bץ−=١ϫE x m x
125 Am. J. Phys., Vol. 77, No. 2, February 2009 T. Rothman and S. Boughn 125
Downloaded 22 Mar 2013 to 165.82.168.47. Redistribution subject to AAPT license or copyright; see http://ajp.aapt.org/authors/copyright_permission tor where the falloff is exponential, the total pressure on the mirror should be ϱ 1 1 1 P =− ͵ BdB=+ B͑0͒2 = B2 cos2͑t͒, ͑26͒ J 0 4 0 8 2 dy where the last equality follows from Eq. ͑23͒ and the conti- By nuity of the tangential component of B across the boundary. The time average of Eq. ͑26͒ gives dx dz E B E P = 0 0 =2͗S͘ ͑27͒ 4 incident as desired. Note that the factor of 2 is expected due to the B z recoil of the wave off the mirror. There are a few tacit assumptions in this derivation that should be made explicit. One might wonder, for example, why we used Ampère’s law ͑24͒ to calculate the conduction current, rather than Faraday’s law, d/dt=−͛E·ds=E, for Fig. 5. A light wave traveling in the z-direction strikes an almost perfectly the magnetic flux =Bdxdz and the induced EMF E. Nor- conducting mirror of thickness z, width dy, and height dx. An Ampèrian loop in the yz plane is also shown, with the direction of B given by the mally, we would have students use this law to calculate the right-hand rule. induced current I=E/R in, for example, a wire loop of resis- tance R. However, in a good conductor EӶB and hence ͉d/dt͉= ͛E·dsӶ ͛B·ds=4I, the last equality represent- ing Ampère’s law. -E Furthermore, the B-field in Eq. ͑24͒ includes both the inץ ϫ E = xˆj =−E k͓sin͑kz − t͒ − sin͑kz + t͔͒ ˆj ١ z 0 cident field and that generated by the induced currents. Itץ seems unreasonable that the portion of the B-field generated -B by the induced currents can result in a net force on the curץ =− . ͑22͒ 12 -t rents themselves ͑no “Munchausen effect” ͒. A detailed calץ culation demonstrates that the integrated force exerted on the Integrating with respect to t and remembering that k= in induced currents by the induced B-field vanishes. With these units where c=1 gives assumptions the simpler derivation we have presented is sound and shows that light waves do exert a pressure on B = E ͓cos͑kz − t͒ + cos͑kz + t͔͒ ˆj 0 matter in the direction of propagation. ͑ ͒ ͑ ͒ ˆ ͑ ͒ In conclusion, although one does not, and cannot, expect =2B0 cos kz cos t j. 23 derivations at the introductory level to be uniformly rigorous, ͑ ͒ Notice that at the boundary, B=2B0 cos t 0 and that this case is of particular interest because the interaction of therefore by Ampère’s law, ͛B·ds=4I, oscillating currents light with matter is of fundamental importance. Moreover, must be induced near the surface of the mirror. Because B is the explanation presented in some textbooks is so seriously in the Ϯy-direction, the right-hand-rule tells us that these flawed that even students sometimes notice the difficulties. currents will be in the Ϯx-direction, and that IϫB will al- Rather than try to paper over these problems with what must ways point in the +z-direction. Consequently, the Lorentz be regarded as nonsensical arguments, the occasion would be force due to the light, F=IdxϫB for a mirror of height dx better exploited to point out that physics is composed of a and total current I, will produce a force in the direction of collection of models that are brought to bear in explaining propagation. physical phenomena, but that these models have limited do- We can calculate the magnitude of the force simply and mains of applicability and as often as not are inconsistent. plausibly. The magnitude of the Lorentz force is dF=IdxB, or dF=JdxdydzB for current density J. The differential form of Ampère’s law tells us that ACKNOWLEDGMENT Bץ -ϫ B =− yˆi =4J, ͑24͒ The authors would like to thank Jim Peebles for suggest ١ .z ing the mirror argumentץ ץ/ ץ͒ / ͑ or J=− 1 4 By z. The Lorentz force therefore becomes ͒ a Electronic mail: [email protected] B b͒ץ dF 1 =− y B dz. ͑25͒ Electronic mail: [email protected] z y 1 James C. Maxwell, Treatise on Electricity and Magnetism, reprint of 3rdץ dxdy 4 edition, 1891 ͑Dover, New York, 1954͒. The quantity on the left is dP, where P is the pressure. Be- 2 Frank S. Crawford, The Berkeley Course in Physics, Waves ͑McGraw- cause the only spatial dependence of B is on z, we can ignore Hill, New York, 1965͒, Vol. 3, pp. 362–363. 3 the distinction between the partial and full differentials. Evi- Paul A. Tipler and Gene Mosca, Physics for Scientists and Engineers,5th ed. ͑Freeman, New York, 2004͒, Vol. 2, pp. 982–983. z is connected to J, we must interpret B 4ץ/ Bץ dently, because y Reference 2, p. 363. as being the field exerting a force on a given slice within the 5 Hanno Essén, “The pushing force of a propagating electromagnetic conductor. If we assume that the magnetic field drops off to wave,” arXiv:physics/0308007v1. zero at infinity, which is certainly true inside a good conduc- 6 John D. Jackson, Classical Electrodynamics, 2nd ed. ͑Wiley, New York,
126 Am. J. Phys., Vol. 77, No. 2, February 2009 T. Rothman and S. Boughn 126
Downloaded 22 Mar 2013 to 165.82.168.47. Redistribution subject to AAPT license or copyright; see http://ajp.aapt.org/authors/copyright_permission 1975͒. The derivation can easily be carried out by considering infinitesimal 7 David Bohm, Quantum Theory ͑Prentice Hall, New York, 1951͒,p.34. loops in the xz and yz planes as follows: The integral form of 8 Leigh Page, “Radiation pressure on electrons and atoms,” Astrophys. J. Faraday’s law is ͛E·ds=−d/dt for magnetic flux . For the 52, 65–72 ͑1920͒. case of our mirror the right-hand rule gives 9 ͛ ͑ ͒ ͑ ͒ ͑ / ͒ ͑ / ͒ ͑ / ͒ Reference 7,p.34. E·ds=Ex z+dz dx−Ex z dx= dEx dz dzdx=− dBy dt dxdz=− d dt 10 Max Planck, The Theory of Heat Radiation, reprint of 1912 edition ͑Do- which leads immediately to Eq. ͑22͒. Similarly, the integral form of ver, New York, 1991͒, Part II, Chap. 1. Ampère’s law ͛B·ds=4I leads to Eq. ͑24͒. 11 Most introductory texts use the integral form of Maxwell’s equations. 12 Anonymous, 1781.
Vacuum Tube Testing Outfit. The 1937 catalogue of the Central Scientific Company of Chicago shows this appa- ratus at $14.00 to “aid the elementary student in the usually confusing study of radio vacuum tube characteristics.” The student was able to obtain curves of the plate current as a function of the plate voltage for various values of the negative grid voltage. This one, in the Greenslade Collection, has a type 201A triode tube installed. The patent date is from 1935. ͑Photograph and Notes by Thomas B. Greenslade, Jr., Kenyon College͒
127 Am. J. Phys., Vol. 77, No. 2, February 2009 T. Rothman and S. Boughn 127
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