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and linear and angular momenta in simple electromagnetic systems Masud Mansuripur College of Optical Sciences, The University of Arizona, Tucson, Arizona 85721 [Published in Optical Trapping and Optical Micromanipulation XII, Proceedings of SPIE 9548, 95480K~1-24 (2015)]

Abstract. We present examples of simple electromagnetic systems in which energy, linear , and exhibit interesting behavior. The systems are sufficiently simple to allow exact solutions of Maxwell’s equations in conjunction with the electrodynamic laws of , , energy, and momentum. In all the cases examined, conservation of energy and momentum is confirmed.

1. Introduction. This paper presents several examples of simple electromagnetic (EM) systems, whose complete understanding requires careful attention to the detailed behavior of EM force, torque, energy and linear as well as angular momentum. The concept of “hidden momentum” appears in some of the examples, which helps to illustrate the differences between the Einstein- Laub formulation and the standard (i.e., Lorentz) formulation of classical electrodynamics [1-3]. In Sec.2 we analyze the energy and angular momentum content of a finite-duration, finite- diameter -packet (or “ bullet”) propagating in free and carrying both spin and orbital angular momentum. Among other findings, we obtain a simple formula for the total angular momentum per of the light bullet. The spin contribution to the total angular momentum per photon turns out to be , where = = is the normalized component of the -vector along (i.e., the propagation direction), is the angular of 𝑧𝑧 𝑧𝑧 𝑧𝑧 0 𝑧𝑧 the light wave, is the of light in ,ℏ𝝈𝝈 and 𝝈𝝈is the 𝒌𝒌reduced⁄𝑘𝑘 Planck𝑐𝑐𝒌𝒌 ⁄𝜔𝜔 constant. Section 3 is devoted𝑘𝑘 to an analysis𝑧𝑧 of the energy and angular 𝜔𝜔momentum of a quasi-static system containing𝑐𝑐 electrical charge and current. At firstℏ sight, the system appears to violate the law of conservation of angular momentum. We pinpoint the cause of this discrepancy, and point out the whereabouts of the missing angular momentum. The case of a slowly rotating permanent in an external is taken up in Sec.4. This quasi-static problem reveals the perils and pitfalls of ignoring the contribution of the physical source of the -field (i.e., a pair of electrically-charged parallel plates) when balancing the mechanical momentum imparted to the magnet against the EM momentum contained in the field. Additionally, the 𝐸𝐸same problem, when analyzed in accordance with the Lorentz formalism, shows the important contribution of “hidden momentum” to the physical behavior of the magnet. In Sec.5 we examine the case of a slowly rotating electric dipole in an external . Here, once again, the physical source of the external field (i.e., a ) turns out to be of crucial significance when one attempts to verify the conservation of linear momentum. Section 6 describes the connection between the energy of a dipole in an external EM field and the energy content of the field in the region surrounding the dipole. Both cases of an electric dipole in an external -field and a magnetic dipole in an external -field (or -field) will be considered. As before, the physical source of the external field will be seen to play an important role when confirming the𝐸𝐸 conservation of energy. 𝐻𝐻 𝐵𝐵 The final section contains a few concluding remarks and general observations. The integrals needed throughout the paper are listed in Appendix A.

2. A light bullet described by Maxwell’s equations in cylindrical coordinates. Consider a monochromatic wave of frequency , propagating in free space along the z-axis. The wave- number is = = ( / ) , where 0 < < 1. The wave also exhibits an azimuthal phase 𝜔𝜔 𝑘𝑘𝑧𝑧 𝑘𝑘0𝜎𝜎𝑧𝑧 𝜔𝜔 𝑐𝑐 𝜎𝜎𝑧𝑧 𝜎𝜎𝑧𝑧 1

of 2 around the z-axis, where , an integer, may be positive, zero, or negative. In cylindrical coordinates, the electric field ( , ) and the magnetic field ( , ) of the wave are given by 𝜋𝜋𝜋𝜋 𝑚𝑚 ( , , , ) = ( ) 𝑬𝑬+𝒓𝒓 𝑡𝑡 ( ) + ( ) exp [i( 𝑯𝑯 𝒓𝒓+𝑡𝑡 )], (1a)

𝑬𝑬 (𝑟𝑟 ,𝜑𝜑 ,𝑧𝑧 ,𝑡𝑡 ) = �𝐸𝐸𝑟𝑟 (𝑟𝑟 )𝒓𝒓� +𝐸𝐸𝜑𝜑 (𝑟𝑟 )𝝋𝝋� +𝐸𝐸𝑧𝑧 𝑟𝑟( 𝒛𝒛)�� exp [i𝑚𝑚𝑚𝑚( +𝑘𝑘𝑧𝑧𝑧𝑧 − 𝜔𝜔𝜔𝜔 )]. (1b)

Maxwell’s𝑯𝑯 𝑟𝑟 𝜑𝜑 𝑧𝑧equations𝑡𝑡 �𝐻𝐻 𝑟𝑟[4𝑟𝑟-8]𝒓𝒓� can𝐻𝐻 𝜑𝜑now𝑟𝑟 𝝋𝝋be� used𝐻𝐻𝑧𝑧 to𝑟𝑟 𝒛𝒛�relate� the𝑚𝑚𝑚𝑚 various𝑘𝑘𝑧𝑧 𝑧𝑧components− 𝜔𝜔𝜔𝜔 of the E and H fields to each other, that is, + + i + i = 0, (2) −1 ′ −1 𝑟𝑟 𝐸𝐸𝑟𝑟 𝐸𝐸𝑟𝑟 𝑚𝑚=𝑟𝑟 𝐸𝐸𝜑𝜑 , 𝑘𝑘𝑧𝑧𝐸𝐸𝑧𝑧 (3a) −1 𝑚𝑚𝑟𝑟 +𝐸𝐸𝑧𝑧i− 𝑘𝑘=𝑧𝑧𝐸𝐸𝜑𝜑 𝜇𝜇0,𝜔𝜔 𝐻𝐻𝑟𝑟 (3b) ′ 𝑘𝑘i 𝑧𝑧𝐸𝐸𝑟𝑟 +𝐸𝐸𝑧𝑧i 𝜇𝜇+0𝜔𝜔𝐻𝐻𝜑𝜑 = , (3c) −1 ′ −1 𝜑𝜑 𝜑𝜑 𝑟𝑟 0 𝑧𝑧 𝑟𝑟 𝐸𝐸 𝐸𝐸 𝑚𝑚=𝑟𝑟 𝐸𝐸 −, 𝜇𝜇 𝜔𝜔𝜔𝜔 (4a) −1 𝑚𝑚𝑟𝑟 +𝐻𝐻𝑧𝑧i− 𝑘𝑘=𝑧𝑧𝐻𝐻𝜑𝜑 −𝜀𝜀0, 𝜔𝜔𝐸𝐸𝑟𝑟 (4b) ′ 𝑘𝑘i 𝑧𝑧𝐻𝐻𝑟𝑟 +𝐻𝐻i𝑧𝑧 −+𝜀𝜀0𝜔𝜔𝐸𝐸𝜑𝜑 = , (4c) −1 ′ −1 𝑟𝑟 𝐻𝐻𝜑𝜑+ 𝐻𝐻+𝜑𝜑 i 𝑚𝑚𝑟𝑟 𝐻𝐻𝑟𝑟+ i 𝜀𝜀0𝜔𝜔𝐸𝐸=𝑧𝑧 0. (5) −1 ′ −1 With the𝑟𝑟 help𝐻𝐻𝑟𝑟 of Eqs.(3a)𝐻𝐻𝑟𝑟 𝑚𝑚 𝑟𝑟and𝐻𝐻 (3b),𝜑𝜑 Eq𝑘𝑘𝑧𝑧.𝐻𝐻(4𝑧𝑧c) may be written as i ( + i ) + i( + i ) + = ( / ) . −1 ′ ′ ″ −1 −1 2 Rearranging𝑟𝑟 the𝑘𝑘𝑧𝑧𝐸𝐸 terms𝑟𝑟 𝐸𝐸 of𝑧𝑧 the above𝑘𝑘𝑧𝑧𝐸𝐸𝑟𝑟 equation𝐸𝐸𝑧𝑧 yields𝑚𝑚𝑟𝑟 �𝑚𝑚𝑟𝑟 𝐸𝐸𝑧𝑧 − 𝑘𝑘𝑧𝑧𝐸𝐸𝜑𝜑� 𝜔𝜔 𝑐𝑐 𝐸𝐸𝑧𝑧 i + + i + ( / ) = 0. −1 ′ −1 ″ −1 ′ 2 −2 2 Invoking𝑘𝑘 𝑧𝑧Eq.(�𝑟𝑟 2),𝐸𝐸 𝑟𝑟we find𝐸𝐸𝑟𝑟 𝑚𝑚𝑟𝑟 𝐸𝐸𝜑𝜑� − 𝐸𝐸𝑧𝑧 − 𝑟𝑟 𝐸𝐸𝑧𝑧 𝑚𝑚 𝑟𝑟 𝐸𝐸𝑧𝑧 − 𝜔𝜔 𝑐𝑐 𝐸𝐸𝑧𝑧 + + [( / ) ( / ) ] = 0. (6) ″ −1 ′ 2 2 2 A similar𝑧𝑧 operation𝑧𝑧, starting with Eq.(𝑧𝑧 3c) and with𝑧𝑧 the help of Eqs.(4a), (4b) and (5), yields 𝐸𝐸 𝑟𝑟 𝐸𝐸 𝜔𝜔 𝑐𝑐 − 𝑘𝑘 − 𝑚𝑚 𝑟𝑟 𝐸𝐸 + + [( / ) ( / ) ] = 0. (7) ″ −1 ′ 2 2 2 We𝐻𝐻 thus𝑧𝑧 𝑟𝑟find𝐻𝐻 two𝑧𝑧 sets𝜔𝜔 of𝑐𝑐 solutions− 𝑘𝑘𝑧𝑧 − to𝑚𝑚 Maxwell’s𝑟𝑟 𝐻𝐻𝑧𝑧 equations, one with = 0, referred to as transverse electric (TE), and another with = 0, known as transverse magnetic (TM). Defining 𝑧𝑧 = = ( / ) , where 0 < < 1 and + = 1, the solutions to𝐸𝐸 Eqs.(6) and (7) may 𝑧𝑧 be written in terms of Bessel functions of the𝐻𝐻 first2 kind2, order , as 𝑘𝑘𝑟𝑟 𝑘𝑘0𝜎𝜎𝑟𝑟 𝜔𝜔 𝑐𝑐 𝜎𝜎𝑟𝑟 𝜎𝜎𝑟𝑟 𝜎𝜎𝑟𝑟 𝜎𝜎𝑧𝑧 ( ) = ( ); ( = 0; TM) 𝑚𝑚, (8) ( ) = ( ); ( = 0; TE) 𝐸𝐸𝑧𝑧 𝑟𝑟 𝐸𝐸𝑧𝑧0 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 𝐻𝐻𝑧𝑧 . (9) In what follows, we will determine the remaining components of the E and H fields for the 𝐻𝐻𝑧𝑧 𝑟𝑟 𝐻𝐻𝑧𝑧0 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 𝐸𝐸𝑧𝑧 two cases of TE and TM modes. In each case we also find the energy and angular momentum contained in a finite-diameter, finite duration light pulse.

2

2.1. Transverse Electric (TE) modes: Setting = 0 in Eq.(3a), we find = ( / ) . Substitution into Eq.(4b) then yields 𝐸𝐸𝑧𝑧 𝐸𝐸𝜑𝜑 − 𝜇𝜇0𝜔𝜔 𝑘𝑘𝑧𝑧 𝐻𝐻𝑟𝑟 ( ) = i( / ) ( ), ( ) = i [( / ) ] ( ) ′ (8) 𝑟𝑟 ( ) = i(𝑧𝑧 /𝑟𝑟 )𝑧𝑧0 𝑚𝑚 (𝑟𝑟 ). 2 2 −1 ′ 𝐻𝐻 𝑟𝑟 𝜎𝜎 𝜎𝜎 𝐻𝐻 𝐽𝐽 𝑘𝑘 𝑟𝑟 𝐻𝐻𝑟𝑟 𝑟𝑟 𝑘𝑘𝑧𝑧 𝜔𝜔 𝑐𝑐 − 𝑘𝑘𝑧𝑧 𝐻𝐻𝑧𝑧 𝑟𝑟 → � ′ In the above equation, = / is the impedance𝐸𝐸𝜑𝜑 𝑟𝑟 of free− 𝑍𝑍 space.0 𝜎𝜎𝑟𝑟 𝐻𝐻Subsequently,𝑧𝑧0 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 Eq.(3c) yields 𝑍𝑍0 �𝜇𝜇0 𝜀𝜀0 ( ) = [ /( )] ( ). (9) Similarly, from Eq.(4a) we find 𝐸𝐸𝑟𝑟 𝑟𝑟 − 𝑚𝑚𝑍𝑍0 𝜎𝜎𝑟𝑟𝑘𝑘𝑟𝑟𝑟𝑟 𝐻𝐻𝑧𝑧0 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 ( ) = [ /( )] ( ). (10)

The component𝐻𝐻𝜑𝜑 𝑟𝑟 of− the𝑚𝑚𝑚𝑚 Poynting𝑧𝑧 𝜎𝜎𝑟𝑟𝑘𝑘𝑟𝑟𝑟𝑟 vector𝐻𝐻𝑧𝑧0 𝐽𝐽along𝑚𝑚 𝑘𝑘𝑟𝑟 𝑟𝑟the z-axis is readily obtained as follows: ( , ) = ½Re ∗ ∗ 〈𝑆𝑆𝑧𝑧 𝒓𝒓 𝑡𝑡 〉 = ½[( �𝐸𝐸𝑟𝑟𝐻𝐻𝜑𝜑 −)/(𝐸𝐸𝜑𝜑𝐻𝐻𝑟𝑟�) ] ( ) + ½( / ) ( ) 2 2 2 2 2 2 ′ 2 = ½( 𝑚𝑚 𝜎𝜎𝑧𝑧𝑍𝑍0/ 𝑘𝑘) 𝑟𝑟𝜎𝜎(𝑟𝑟𝑟𝑟/ 𝐻𝐻𝑧𝑧)0 𝐽𝐽𝑚𝑚 (𝑘𝑘𝑟𝑟𝑟𝑟) + 𝑍𝑍(0𝜎𝜎𝑧𝑧 )𝜎𝜎.𝑟𝑟 𝐻𝐻𝑧𝑧0 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 (11) 2 2 2 2 ′ 2 Integration over the cross𝑍𝑍0-𝐻𝐻sectional𝑧𝑧0𝜎𝜎𝑧𝑧 𝜎𝜎𝑟𝑟 � 𝑚𝑚 of 𝑘𝑘the𝑟𝑟𝑟𝑟 beam𝐽𝐽𝑚𝑚 from𝑘𝑘𝑟𝑟𝑟𝑟 =𝐽𝐽𝑚𝑚0 to𝑘𝑘 𝑟𝑟𝑟𝑟 yields� 2 ( , ) = ( / ) ( / ) 𝑟𝑟( ) + 𝑅𝑅 ( ) 𝑅𝑅 2 2 𝑅𝑅 2 2 ′ 2 ∫0= (𝜋𝜋〈𝑆𝑆𝑧𝑧 𝒓𝒓 𝑡𝑡 〉𝑟𝑟𝑟𝑟𝑟𝑟/ ) 𝜋𝜋𝑍𝑍0𝐻𝐻 𝑧𝑧0𝜎𝜎(𝑧𝑧 /𝜎𝜎𝑟𝑟) ∫0 (𝑟𝑟�) 𝑚𝑚+ 𝑘𝑘𝑟𝑟𝑟𝑟( )𝐽𝐽𝑚𝑚 𝑘𝑘 𝑟𝑟𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 �𝑑𝑑𝑑𝑑 𝑟𝑟 2 2 2 𝑘𝑘 𝑅𝑅 2 2 ′ 2 = (𝜋𝜋½𝑍𝑍0𝐻𝐻𝑧𝑧0𝜎𝜎𝑧𝑧 𝑘𝑘𝑟𝑟 𝜎𝜎/𝑟𝑟 ∫)0{ 𝑥𝑥�(𝑚𝑚 𝑥𝑥) 𝐽𝐽𝑚𝑚 𝑥𝑥( )𝐽𝐽𝑚𝑚 𝑥𝑥( �𝑑𝑑𝑑𝑑) + 2 [ ( )/( )] }. (12) 2 2 2 2 2 Since𝜋𝜋 𝑅𝑅Bessel𝑍𝑍0𝐻𝐻𝑧𝑧 0beams𝜎𝜎𝑧𝑧 𝜎𝜎𝑟𝑟 in 𝐽𝐽free𝑚𝑚+1 space𝑘𝑘𝑟𝑟𝑅𝑅 travel− 𝐽𝐽𝑚𝑚 𝑘𝑘along𝑟𝑟𝑅𝑅 𝐽𝐽 𝑚𝑚the+2 z𝑘𝑘-axis𝑟𝑟𝑅𝑅 at the𝑚𝑚 𝐽𝐽speed𝑚𝑚 𝑘𝑘𝑟𝑟 𝑅𝑅of 𝑘𝑘𝑟𝑟,𝑅𝑅 the integrated in Eq.(12) should be equal to the EM energy contained in a cylindrical 𝑧𝑧 of radius and length , where = 1/ is the in vacuum.𝑐𝑐𝜎𝜎 Alternatively, one could calculate the energy content of the cylindrical volume directly, using the total - 𝑧𝑧 0 0 averaged ene𝑅𝑅 rgy-density𝑐𝑐 𝜎𝜎of the E and𝑐𝑐 H fields,�𝜇𝜇 𝜀𝜀as follows: ( , ) = ¼ Re + + ¼ Re + + ∗ ∗ ∗ ∗ ∗ 〈ℰ 𝒓𝒓=𝑡𝑡 ¼〉 [ 𝜀𝜀0 /(�𝐸𝐸𝑟𝑟𝐸𝐸𝑟𝑟 )] 𝐸𝐸𝜑𝜑𝐸𝐸 𝜑𝜑�( )𝜇𝜇+0 ¼�𝐻𝐻𝑟𝑟(𝐻𝐻𝑟𝑟/ 𝐻𝐻)𝜑𝜑𝐻𝐻𝜑𝜑 𝐻𝐻(𝑧𝑧𝐻𝐻𝑧𝑧 �) 2 2 2 2 2 ′ 2 +¼𝜀𝜀0 𝑚𝑚(𝑍𝑍0 / 𝜎𝜎𝑟𝑟)𝑘𝑘𝑟𝑟𝑟𝑟 𝐻𝐻𝑧𝑧0( 𝐽𝐽𝑚𝑚 )𝑘𝑘+𝑟𝑟𝑟𝑟¼ [ 𝜀𝜀0 𝑍𝑍/0( 𝜎𝜎𝑟𝑟 𝐻𝐻)]𝑧𝑧0 𝐽𝐽𝑚𝑚 𝑘𝑘(𝑟𝑟𝑟𝑟 ) + ¼ ( ) 2 2 ′ 2 2 2 2 2 2 = ¼ 𝜇𝜇[20 𝜎𝜎𝑧𝑧/𝜎𝜎( 𝑟𝑟 𝐻𝐻𝑧𝑧)0 𝐽𝐽𝑚𝑚 (𝑘𝑘𝑟𝑟/𝑟𝑟 ) +𝜇𝜇01𝑚𝑚𝑚𝑚] 𝑧𝑧 𝜎𝜎(𝑟𝑟𝑘𝑘𝑟𝑟𝑟𝑟) +𝐻𝐻¼𝑧𝑧0 𝐽𝐽𝑚𝑚[(2𝑘𝑘/𝑟𝑟𝑟𝑟 ) 1𝜇𝜇] 0𝐻𝐻𝑧𝑧 0 𝐽𝐽𝑚𝑚(𝑘𝑘𝑟𝑟𝑟𝑟) 2 2 2 2 2 2 2 ′ 2 0 𝑟𝑟 𝑟𝑟 𝑟𝑟 𝑧𝑧0 𝑚𝑚 𝑟𝑟 0 𝑟𝑟 𝑧𝑧0 𝑚𝑚 𝑟𝑟 = ½𝜇𝜇( 𝑚𝑚 / 𝜎𝜎) 𝑘𝑘( 𝑟𝑟/ − )𝑚𝑚 𝑘𝑘( 𝑟𝑟 ) + 𝐻𝐻 ( 𝐽𝐽 )𝑘𝑘 𝑟𝑟 𝜇𝜇 𝜎𝜎 − 𝐻𝐻 𝐽𝐽 𝑘𝑘 𝑟𝑟 2 2 2 2 ′ 2 +¼𝜇𝜇0𝐻𝐻𝑧𝑧0 𝜎𝜎𝑟𝑟 (� 𝑚𝑚) 𝑘𝑘𝑟𝑟𝑟𝑟( /𝐽𝐽𝑚𝑚 𝑘𝑘)𝑟𝑟𝑟𝑟 ( 𝐽𝐽𝑚𝑚) 𝑘𝑘𝑟𝑟 𝑟𝑟 �( ) . (13) 2 2 2 2 ′ 2 In unit time,𝜇𝜇0𝐻𝐻𝑧𝑧 0the�𝐽𝐽𝑚𝑚 Bessel𝑘𝑘𝑟𝑟𝑟𝑟 −beam𝑚𝑚 propagates𝑘𝑘𝑟𝑟𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘 𝑟𝑟a𝑟𝑟 − 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 along� the z-axis. Multiplying into Eq.(13) yields a first term that is identical to Eq.(11). The term, therefore, must 𝑐𝑐𝜎𝜎𝑧𝑧 𝑐𝑐𝜎𝜎𝑧𝑧 3

integrate to zero over the cross-sectional area of the beam. The integral of the second term on the right-hand-side of Eq.(13) over the cylindrical volume of radius and length may be written

½ { ( ) [( / ) ( ) ( )] 2( 𝑅𝑅/ ) ( 𝑐𝑐)𝜎𝜎 𝑧𝑧 ( )} 2 𝑅𝑅 2 ′ 2 ′ 𝜋𝜋𝜎𝜎𝑧𝑧𝑍𝑍0𝐻𝐻𝑧𝑧0 ∫0= ½𝐽𝐽𝑚𝑚 𝑘𝑘(𝑟𝑟𝑟𝑟 /− )𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 𝐽𝐽𝑚𝑚 [𝑘𝑘𝑟𝑟𝑟𝑟 (− )𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 (− ) 𝑚𝑚2𝑘𝑘𝑟𝑟 𝑟𝑟 𝐽𝐽(𝑚𝑚 )𝑘𝑘 𝑟𝑟𝑟𝑟(𝐽𝐽𝑚𝑚)]𝑘𝑘𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟 2 2 𝑘𝑘 𝑅𝑅 2 2 ′ = ½𝜋𝜋(𝜎𝜎𝑧𝑧/𝑘𝑘𝑟𝑟 )𝑍𝑍0𝐻𝐻𝑧𝑧0[∫0 𝑥𝑥𝐽𝐽𝑚𝑚 (𝑥𝑥 −)𝑥𝑥𝐽𝐽𝑚𝑚+1 𝑥𝑥( − )𝑚𝑚] 𝐽𝐽𝑚𝑚( 𝑥𝑥 )𝐽𝐽𝑚𝑚. 𝑥𝑥 𝑑𝑑𝑑𝑑 (14) 2 2 The above expression𝜋𝜋 𝜎𝜎𝑧𝑧 𝑘𝑘 𝑟𝑟vanishes𝑍𝑍0𝐻𝐻𝑧𝑧0 𝑘𝑘 when𝑟𝑟𝑅𝑅 𝐽𝐽𝑚𝑚 +1 is𝑘𝑘𝑟𝑟 𝑅𝑅chosen− 𝑚𝑚 𝐽𝐽to𝑚𝑚 correspond𝑘𝑘𝑟𝑟𝑅𝑅 𝐽𝐽𝑚𝑚 𝑘𝑘 𝑟𝑟to𝑅𝑅 one of the zeroes of ( ). The energy content of a cylindrical volume of radius and length is thus given by Eq.(12) provided that is chosen such that 𝑅𝑅 ( ) = 0. 𝑚𝑚 𝑟𝑟 𝑧𝑧 𝐽𝐽 𝑘𝑘 𝑅𝑅 𝑅𝑅 𝑐𝑐𝜎𝜎 𝑚𝑚 𝑟𝑟 Next we compute the a𝑅𝑅ngular momentum density𝐽𝐽 𝑘𝑘 of𝑅𝑅 the TE mode, as follows: ( , ) = × ½Re( / ) = ½[ /( )] ( ) . (15) ∗ 2 2 2 2 = 0 Integrat〈ing𝐿𝐿𝑧𝑧 the𝒓𝒓 𝑡𝑡 above〉𝒛𝒛� 𝑟𝑟expression𝒓𝒓� −over𝐸𝐸𝑟𝑟𝐻𝐻 the𝑧𝑧 𝝋𝝋� cross𝑐𝑐 -sectional𝑚𝑚𝑍𝑍 0area𝐻𝐻𝑧𝑧0 of𝑐𝑐 the𝑘𝑘𝑟𝑟 beam𝜎𝜎𝑟𝑟 𝐽𝐽 𝑚𝑚from𝑘𝑘𝑟𝑟 𝑟𝑟 𝒛𝒛� to yields 2 ( , ) = [ /( )] ( ) 𝑟𝑟 𝑅𝑅 𝑅𝑅 2 2 𝑅𝑅 2 ∫0 𝜋𝜋𝜋𝜋〈𝐿𝐿𝑧𝑧 𝒓𝒓 𝑡𝑡 〉𝑑𝑑𝑑𝑑 = [𝜋𝜋𝜋𝜋𝑍𝑍0𝐻𝐻𝑧𝑧0 /(𝑐𝑐 𝑘𝑘𝑟𝑟𝜎𝜎𝑟𝑟 )]∫0 𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟(𝑟𝑟 )𝑑𝑑𝑑𝑑 𝑟𝑟 2 2 3 𝑘𝑘 𝑅𝑅 2 = [𝜋𝜋𝜋𝜋½ 𝑍𝑍(0𝐻𝐻𝑧𝑧0 𝑐𝑐 𝑘𝑘𝑟𝑟)/𝜎𝜎(𝑟𝑟 ∫0 )𝑥𝑥][𝐽𝐽𝑚𝑚 (𝑥𝑥 𝑑𝑑𝑑𝑑) ( ) ( )] = [½ ( 2 2 )/( 2 )][ 2( ) ( ) ( )] 𝑚𝑚 𝜋𝜋𝑅𝑅 𝑍𝑍0𝐻𝐻𝑧𝑧0 𝑐𝑐 𝑘𝑘𝑟𝑟𝜎𝜎𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑅𝑅 − 𝐽𝐽𝑚𝑚−1 𝑘𝑘𝑟𝑟𝑅𝑅 𝐽𝐽𝑚𝑚+1 𝑘𝑘𝑟𝑟𝑅𝑅 . (16) 2 2 2 2 In unit time the beam propagates𝑚𝑚 𝜋𝜋𝑅𝑅 𝑍𝑍0𝐻𝐻 a𝑧𝑧0 distance𝑐𝑐𝑐𝑐𝜎𝜎𝑟𝑟 of 𝐽𝐽𝑚𝑚 𝑘𝑘, 𝑟𝑟so𝑅𝑅 the− 𝐽𝐽angular𝑚𝑚−1 𝑘𝑘𝑟𝑟 𝑅𝑅momentum𝐽𝐽𝑚𝑚+1 𝑘𝑘𝑟𝑟𝑅𝑅 in Eq.(16) must be multiplied by in order to correspond to the same amount of energy that is 𝑧𝑧 represented by Eq.(12). The ratio of angular momentum𝑐𝑐𝜎𝜎 to energy content of a cylindrical 𝑧𝑧 volume of radius and𝑐𝑐𝜎𝜎 arbitrary length, assuming that ( ) = 0, is thus given by ( ) ( )/ ( ). For sufficiently large , the asymptotic value of ( ) is 𝑚𝑚 𝑟𝑟 2/( ) cos( 𝑅𝑅½ ¼ ), which leads to ( )/𝐽𝐽 𝑘𝑘 (𝑅𝑅 ) 1. Consequently, ⁄ 𝑚𝑚−1 𝑟𝑟 𝑚𝑚+1 𝑟𝑟 𝑚𝑚 𝑟𝑟 the− 𝑚𝑚 ratio𝜔𝜔 𝐽𝐽of angular𝑘𝑘 𝑅𝑅 momentum𝐽𝐽 𝑘𝑘 𝑅𝑅 to energy for a sufficiently𝑅𝑅 large number of rings of the𝐽𝐽 TE𝑘𝑘 mode𝑅𝑅 𝑟𝑟 𝑟𝑟 𝑚𝑚−1 𝑟𝑟 𝑚𝑚+1 𝑟𝑟 of� a Bessel𝜋𝜋𝑘𝑘 𝑅𝑅 beam𝑘𝑘 𝑅𝑅in− vacuum𝑚𝑚𝑚𝑚 − is 𝜋𝜋given by / . In quantum𝐽𝐽 𝑘𝑘 -𝑅𝑅optical𝐽𝐽 terms,𝑘𝑘 𝑅𝑅 ≅where− the energy of each photon is , the angular momentum per photon will be . 𝑚𝑚 𝜔𝜔 2.2. Transverseℏ 𝜔𝜔Magnetic (TM) modes: Setting = 0 in Eq.(𝑚𝑚ℏ4a), we find = ( / ) . Substitution into Eq.(3b) then yields 𝑧𝑧 𝜑𝜑 0 𝑧𝑧 𝑟𝑟 𝐻𝐻 ( ) = i( / ) 𝐻𝐻( ), 𝜀𝜀 𝜔𝜔 𝑘𝑘 𝐸𝐸 ( ) = i [( / ) ] ( ) ′ (17) 𝑟𝑟 ( ) = i( 𝑧𝑧 /𝑟𝑟 𝑧𝑧0) 𝑚𝑚 ( 𝑟𝑟 ). 2 2 −1 ′ 𝐸𝐸 𝑟𝑟 𝜎𝜎 𝜎𝜎 𝐸𝐸 𝐽𝐽 𝑘𝑘 𝑟𝑟 𝑟𝑟 𝑧𝑧 𝑧𝑧 𝑧𝑧 Subsequently,𝐸𝐸 𝑟𝑟 Eq.(𝑘𝑘4c)𝜔𝜔 yields𝑐𝑐 − 𝑘𝑘 𝐸𝐸 𝑟𝑟 → � ′ 𝐻𝐻𝜑𝜑 𝑟𝑟 𝐸𝐸𝑧𝑧0 𝑍𝑍0𝜎𝜎𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 ( ) = ( / ) ( ). (18) Similarly, from Eq.(3a) we find 𝐻𝐻𝑟𝑟 𝑟𝑟 𝑚𝑚𝐸𝐸𝑧𝑧0 𝑍𝑍0𝑘𝑘𝑟𝑟𝜎𝜎𝑟𝑟𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 ( ) = ( / ) ( ). (19)

Having𝐸𝐸𝜑𝜑 𝑟𝑟 determined− 𝑚𝑚𝑚𝑚𝑧𝑧 the𝜎𝜎𝑟𝑟𝑘𝑘 various𝑟𝑟𝑟𝑟 𝐸𝐸𝑧𝑧0 𝐽𝐽components𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 of the E and H fields of the TM mode of a Bessel beam in vacuum, we proceed to compute the Poynting vector along the z-axis, as follows:

4

( , ) = ½Re ∗ ∗ 〈=𝑆𝑆𝑧𝑧½𝒓𝒓[(𝑡𝑡 〉 )/( �𝐸𝐸𝑟𝑟)𝐻𝐻] 𝜑𝜑 − (𝐸𝐸𝜑𝜑𝐻𝐻)𝑟𝑟�+ ½( / )[ /( )] ( ) 2 2 ′ 2 2 2 2 = ½[(𝜎𝜎𝑧𝑧𝐸𝐸𝑧𝑧0)/(𝑍𝑍0𝜎𝜎𝑟𝑟 )] 𝐽𝐽𝑚𝑚 (𝑘𝑘𝑟𝑟𝑟𝑟 ) + ( 𝜎𝜎/𝑧𝑧𝐸𝐸𝑧𝑧0)𝑍𝑍 0 (𝑚𝑚 )𝜎𝜎𝑟𝑟.𝑘𝑘 𝑟𝑟𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 (20) 2 2 ′ 2 2 2 Integration over𝜎𝜎𝑧𝑧 𝐸𝐸the𝑧𝑧0 cross𝑍𝑍0-𝜎𝜎sectional𝑟𝑟 �𝐽𝐽𝑚𝑚 area𝑘𝑘𝑟𝑟𝑟𝑟 of the𝑚𝑚 beam𝑘𝑘𝑟𝑟𝑟𝑟 from𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟=𝑟𝑟 0� to yields 2 ( , ) = ½( / ) 𝑟𝑟 𝑅𝑅 𝑅𝑅 2 2 2 ∫0 𝜋𝜋𝜋𝜋〈𝑆𝑆𝑧𝑧 𝒓𝒓 𝑡𝑡 〉𝑑𝑑𝑑𝑑 × { 𝜋𝜋𝑅𝑅 𝜎𝜎(𝑧𝑧𝐸𝐸𝑧𝑧0 )𝑍𝑍0𝜎𝜎𝑟𝑟 ( ) ( ) + 2 [ ( )/( )] }. (21) 2 2 Aside from the coefficient𝐽𝐽𝑚𝑚+ 1 𝑘𝑘𝑟𝑟/𝑅𝑅 ,− this𝐽𝐽𝑚𝑚 result𝑘𝑘𝑟𝑟𝑅𝑅 𝐽𝐽is𝑚𝑚 +identical2 𝑘𝑘𝑟𝑟𝑅𝑅 to that𝑚𝑚 𝐽𝐽obtained𝑚𝑚 𝑘𝑘𝑟𝑟𝑅𝑅 in𝑘𝑘 Eq.(12)𝑟𝑟𝑅𝑅 for a TE beam, where the corresponding coefficient2 was . 𝑧𝑧0 0 𝐸𝐸 𝑍𝑍 2 0 In similar fashion, the angular momentum density𝑍𝑍 𝐻𝐻of𝑧𝑧 0a TM Bessel beam is found to be ( , ) = × ½Re( / ) = ½[ /( )] ( ) . (22) ∗ 2 2 = 02 2 Integration〈𝐿𝐿𝑧𝑧 over𝒓𝒓 𝑡𝑡 〉the𝒛𝒛� cross𝑟𝑟𝒓𝒓� -sectional𝐸𝐸𝑧𝑧 𝐻𝐻area𝑟𝑟 𝝋𝝋� of𝑐𝑐 the beam𝑚𝑚 from𝐸𝐸𝑧𝑧0 𝑍𝑍0𝑐𝑐 𝑘𝑘 to𝑟𝑟𝜎𝜎 𝑟𝑟 yields𝐽𝐽𝑚𝑚 𝑘𝑘 𝑟𝑟𝑟𝑟 𝒛𝒛� 2 ( , ) = [ /( )] ( 𝑟𝑟) 𝑅𝑅 𝑅𝑅 2 2 𝑅𝑅 2 ∫0 𝜋𝜋𝜋𝜋〈𝐿𝐿𝑧𝑧 𝒓𝒓 𝑡𝑡 〉𝑑𝑑𝑑𝑑 = [𝜋𝜋𝜋𝜋𝐸𝐸𝑧𝑧0 /(𝑍𝑍0𝑐𝑐 𝑘𝑘𝑟𝑟𝜎𝜎𝑟𝑟 )]∫0 𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟(𝑟𝑟 )𝑑𝑑𝑑𝑑 𝑟𝑟 2 2 3 𝑘𝑘 𝑅𝑅 2 = ½𝜋𝜋𝜋𝜋[𝐸𝐸(𝑧𝑧0 𝑍𝑍0𝑐𝑐)/𝑘𝑘(𝑟𝑟 𝜎𝜎𝑟𝑟 ∫0 )𝑥𝑥][𝐽𝐽 𝑚𝑚 (𝑥𝑥 𝑑𝑑𝑑𝑑) ( ) ( )] 2 2 2 2 = ½𝑚𝑚[(𝜋𝜋𝑅𝑅 𝐸𝐸𝑧𝑧0)/(𝑍𝑍0𝑐𝑐 𝑘𝑘𝑟𝑟𝜎𝜎)𝑟𝑟][ 𝐽𝐽(𝑚𝑚 𝑘𝑘𝑟𝑟)𝑅𝑅 − 𝐽𝐽𝑚𝑚−(1 𝑘𝑘𝑟𝑟)𝑅𝑅 𝐽𝐽𝑚𝑚+(1 𝑘𝑘𝑟𝑟)𝑅𝑅]. (23) 2 2 2 2 In unit time the beam 𝑚𝑚propagates𝜋𝜋𝑅𝑅 𝐸𝐸𝑧𝑧0 a distance𝑍𝑍0𝑐𝑐𝑐𝑐𝜎𝜎𝑟𝑟 of 𝐽𝐽𝑚𝑚 𝑘𝑘, 𝑟𝑟so𝑅𝑅 the− 𝐽𝐽 𝑚𝑚angular−1 𝑘𝑘𝑟𝑟 𝑅𝑅momentum𝐽𝐽𝑚𝑚+1 𝑘𝑘𝑟𝑟𝑅𝑅 in Eq.(23) must be multiplied by in order to correspond to the same amount of energy that is 𝑧𝑧 represented by Eq.(21). Once again, the ratio of angular𝑐𝑐𝜎𝜎 momentum to energy contained in a 𝑧𝑧 cylindrical volume of radius𝑐𝑐𝜎𝜎 and arbitrary length, where ( ) = 0, is seen to approach / in the limit of large . In quantum-optical terms, each photon of energy in the TM 𝑚𝑚 𝑟𝑟 mode of the Bessel beam carries𝑅𝑅 of orbital angular momentum.𝐽𝐽 𝑘𝑘 𝑅𝑅 𝑚𝑚 𝜔𝜔 𝑅𝑅 ℏ𝜔𝜔 2.3. The case of circular 𝑚𝑚ℏ . We define a circularly-polarized Bessel beam as an equal mixture of TE and TM modes, with the phase difference between the two eigen-modes being ±90° for right/left circular states. Thus, in the -space, all plane- that constitute the resulting Bessel beam will be circularly polarized. For a pair of TE and TM beams having the same , , and , and also having = ±i 𝑘𝑘, the superposed and fields are given by ( ) = ±i[ /( )] ( ) + i( / ) ( ) 𝜔𝜔 𝑘𝑘𝑧𝑧 𝑚𝑚 𝐸𝐸𝑧𝑧0 𝑍𝑍0𝐻𝐻𝑧𝑧0 𝑬𝑬. 𝑯𝑯 (24) ( ) = ( / ) ( ) ( / ) (′ ) 𝐸𝐸𝑟𝑟 𝑟𝑟 𝑚𝑚𝐸𝐸𝑧𝑧0 𝜎𝜎𝑟𝑟𝑘𝑘𝑟𝑟𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 𝜎𝜎𝑧𝑧 𝜎𝜎𝑟𝑟 𝐸𝐸𝑧𝑧0 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟.𝑟𝑟 (25) ′ 𝐸𝐸𝜑𝜑( 𝑟𝑟) = ∓ 𝐸𝐸 𝑧𝑧0( 𝜎𝜎𝑟𝑟 )𝐽𝐽.𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 − 𝑚𝑚𝑚𝑚𝑧𝑧 𝜎𝜎𝑟𝑟𝑘𝑘𝑟𝑟𝑟𝑟 𝐸𝐸𝑧𝑧0 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 (26) ( ) = ±( / ) ( ) + ( / ) ( ) 𝐸𝐸𝑧𝑧 𝑟𝑟 𝐸𝐸𝑧𝑧0 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 . (27) ( ) = ±i[ /( ′ )] ( ) + i( / ) ( ) 𝐻𝐻𝑟𝑟 𝑟𝑟 𝐸𝐸𝑧𝑧0𝜎𝜎𝑧𝑧 𝑍𝑍0𝜎𝜎𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 𝑚𝑚𝐸𝐸𝑧𝑧0 𝑍𝑍0𝑘𝑘𝑟𝑟𝜎𝜎𝑟𝑟𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 . (28) ′ 𝐻𝐻𝜑𝜑( 𝑟𝑟) = i(𝑚𝑚𝑚𝑚/𝑧𝑧𝐸𝐸𝑧𝑧0) 𝑍𝑍(0𝜎𝜎𝑟𝑟𝑘𝑘)𝑟𝑟. 𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 𝐸𝐸𝑧𝑧0 𝑍𝑍0𝜎𝜎𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 (29) 𝐻𝐻𝑧𝑧 𝑟𝑟 ∓ 𝐸𝐸𝑧𝑧0 𝑍𝑍0 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 5

The angular momentum density of this circularly-polarized Bessel beam is readily found to be ( , ) = × ½Re[( ) / ] = [ /( ∗)][ (∗ ) ±2 ( ) ( )] 〈𝐿𝐿𝑧𝑧 𝒓𝒓 𝑡𝑡 〉𝒛𝒛� 𝑟𝑟𝒓𝒓� 𝐸𝐸𝑧𝑧𝐻𝐻𝑟𝑟 − 𝐸𝐸𝑟𝑟𝐻𝐻𝑧𝑧 𝝋𝝋� 𝑐𝑐 . (30) 2 2 2 = 0′ ) Upon integration over the𝐸𝐸𝑧𝑧0 cross𝑍𝑍0𝑐𝑐-sectional𝜎𝜎𝑟𝑟𝑘𝑘𝑟𝑟 𝑚𝑚area𝐽𝐽𝑚𝑚 of𝑘𝑘𝑟𝑟 the𝑟𝑟 beam𝜎𝜎𝑧𝑧𝑘𝑘 𝑟𝑟(𝑟𝑟from𝐽𝐽𝑚𝑚 𝑘𝑘 𝑟𝑟𝑟𝑟 𝐽𝐽𝑚𝑚 to𝑘𝑘 𝑟𝑟𝑟𝑟 we𝒛𝒛� find 2 ( , ) = 𝑟𝑟 𝑅𝑅 𝑅𝑅 ∫0 =𝜋𝜋𝜋𝜋[2〈𝐿𝐿𝑧𝑧 𝒓𝒓 /𝑡𝑡(〉𝑑𝑑𝑑𝑑 )] [ ( ) ± ( ) ( )] 2 2 𝑅𝑅 2 ′ = [2𝜋𝜋𝐸𝐸𝑧𝑧0/(𝑍𝑍0𝑐𝑐 𝜎𝜎𝑟𝑟𝑘𝑘𝑟𝑟 )]∫0 𝑚𝑚[𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟(𝑟𝑟 ) ±𝜎𝜎𝑧𝑧𝑘𝑘𝑟𝑟𝑟𝑟𝐽𝐽𝑚𝑚(𝑘𝑘𝑟𝑟)𝑟𝑟 𝐽𝐽(𝑚𝑚 )𝑘𝑘] 𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟 2 2 3 𝑘𝑘 𝑅𝑅 2 2 ′ = [2𝜋𝜋𝐸𝐸𝑧𝑧0/(𝑍𝑍0𝑐𝑐 𝜎𝜎𝑟𝑟𝑘𝑘𝑟𝑟 )] ∫0 𝑚𝑚𝑚𝑚𝐽𝐽𝑚𝑚 𝑥𝑥 𝜎𝜎𝑧𝑧𝑥𝑥 𝐽𝐽𝑚𝑚 𝑥𝑥 𝐽𝐽𝑚𝑚 𝑥𝑥 𝑑𝑑𝑑𝑑 × {½ 2( ) [2 ( 3 ) ( ) ( )] ± ½ ( ) ( ) ( )} 𝜋𝜋𝐸𝐸𝑧𝑧0 𝑍𝑍0𝑐𝑐 𝜎𝜎𝑟𝑟𝑘𝑘𝑟𝑟 2 2 2 = [ 𝑚𝑚 𝑘𝑘𝑟𝑟/𝑅𝑅( 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟)𝑅𝑅][(−±𝐽𝐽𝑚𝑚−1 𝑘𝑘𝑟𝑟)𝑅𝑅 𝐽𝐽𝑚𝑚+(1 𝑘𝑘𝑟𝑟𝑅𝑅) ( 𝜎𝜎𝑧𝑧 𝑘𝑘)𝑟𝑟+𝑅𝑅 𝐽𝐽𝑚𝑚−(1 𝑘𝑘𝑟𝑟𝑅𝑅)]𝐽𝐽.𝑚𝑚 +1 𝑘𝑘𝑟𝑟𝑅𝑅 (31) 2 2 2 ( ) 2 If is𝜋𝜋 chosen𝑅𝑅 𝐸𝐸𝑧𝑧0 to𝑍𝑍 correspond0𝑐𝑐𝑐𝑐𝜎𝜎𝑟𝑟 to𝜎𝜎 𝑧𝑧a −zero𝑚𝑚 of𝐽𝐽𝑚𝑚 −1 𝑘𝑘𝑟𝑟𝑅𝑅 𝐽𝐽, 𝑚𝑚the+1 last𝑘𝑘𝑟𝑟𝑅𝑅 term 𝑚𝑚in𝐽𝐽 𝑚𝑚Eq.(31)𝑘𝑘𝑟𝑟𝑅𝑅 vanishes, and the remaining terms give the angular momentum due to both vorticity and circular polarization. 𝑚𝑚 𝑟𝑟 Finally,𝑅𝑅 we compute the z-component of𝐽𝐽 the𝑘𝑘 𝑅𝑅Poynting vector of the circularly-polarized Bessel beam, as follows: ( , ) = ½Re ∗ ∗ 〈𝑆𝑆𝑧𝑧 𝒓𝒓 𝑡𝑡 〉 = ½{[ �𝐸𝐸𝑟𝑟𝐻𝐻𝜑𝜑 −/𝐸𝐸𝜑𝜑(𝐻𝐻𝑟𝑟� ) ] ( ) ± [ /( )] ( ) ( ) 2 2 2 2 2 2 ′ ±[ 𝑚𝑚 𝜎𝜎𝑧𝑧𝐸𝐸𝑧𝑧/0( 𝑍𝑍0 𝜎𝜎𝑟𝑟𝑘𝑘𝑟𝑟)𝑟𝑟] (𝐽𝐽𝑚𝑚 𝑘𝑘)𝑟𝑟𝑟𝑟 ( 𝑚𝑚)𝐸𝐸+𝑧𝑧0( 𝑍𝑍0𝜎𝜎𝑟𝑟/𝑘𝑘𝑟𝑟𝑟𝑟 )𝐽𝐽𝑚𝑚 𝑘𝑘( 𝑟𝑟𝑟𝑟 )𝐽𝐽 𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 2 2 2 ′ 2 2 ′ 2 + (𝑚𝑚𝑚𝑚𝑧𝑧 𝐸𝐸𝑧𝑧/0 𝑍𝑍0)𝜎𝜎 𝑟𝑟 𝑘𝑘𝑟𝑟𝑟𝑟( 𝐽𝐽𝑚𝑚) ±𝑘𝑘𝑟𝑟(𝑟𝑟 𝐽𝐽𝑚𝑚 /𝑘𝑘𝑟𝑟𝑟𝑟 𝜎𝜎)𝑧𝑧𝐸𝐸𝑧𝑧0( 𝑍𝑍0)𝜎𝜎𝑟𝑟 (𝐽𝐽𝑚𝑚 )𝑘𝑘 𝑟𝑟𝑟𝑟 2 ±( 2 / 2 ′ ) ( ) (2 ) + [ 2 / (′ ) ] ( )} 𝜎𝜎𝑧𝑧𝐸𝐸𝑧𝑧0 𝑍𝑍0𝜎𝜎𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 𝑚𝑚𝐸𝐸𝑧𝑧0 𝑍𝑍0𝑘𝑘𝑟𝑟𝜎𝜎𝑟𝑟 𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 2 2 2 ′ 2 2 2 2 = ( 𝑚𝑚𝜎𝜎𝑧𝑧/𝐸𝐸𝑧𝑧0 𝑍𝑍)0𝜎𝜎(𝑟𝑟 𝑘𝑘/𝑟𝑟𝑟𝑟 𝐽𝐽)𝑚𝑚 𝑘𝑘𝑟𝑟(𝑟𝑟 𝐽𝐽𝑚𝑚) +𝑘𝑘𝑟𝑟𝑟𝑟 ( 𝑚𝑚)𝜎𝜎 𝑧𝑧𝐸𝐸𝑧𝑧0 𝑍𝑍0 𝜎𝜎𝑟𝑟𝑘𝑘𝑟𝑟𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 2 2 2 2 ′ 2 [𝑧𝑧 0 𝑟𝑟 ( 𝑟𝑟 𝑚𝑚 )]𝑟𝑟 ( 𝑚𝑚) (𝑟𝑟 ) ±𝜎𝜎 𝐸𝐸𝑧𝑧(01 +𝑍𝑍 𝜎𝜎 )� 𝑚𝑚/ 𝑘𝑘 𝑟𝑟 𝐽𝐽 𝑘𝑘 𝑟𝑟 𝐽𝐽 𝑘𝑘 𝑟𝑟 � . (32) The first term on the right2 2-hand-side2 of the above′ equation is twice the expression in 𝑚𝑚 𝜎𝜎𝑧𝑧 𝐸𝐸𝑧𝑧0 𝑍𝑍0𝜎𝜎𝑟𝑟 𝑘𝑘𝑟𝑟𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 𝐽𝐽𝑚𝑚 𝑘𝑘𝑟𝑟𝑟𝑟 Eq.(20). Therefore, upon integration, this term yields the total of the beam, as before. The second term integrates to ±[2 (1 + ) /( )] ( ) ( ) = ±[ (1 + ) /( )] ( ). (33) 2 2 2 2 𝑘𝑘𝑟𝑟𝑅𝑅 ′ 2 2 2 2 2 This𝜋𝜋𝜋𝜋 function𝜎𝜎𝑧𝑧 𝐸𝐸𝑧𝑧 0vanishes𝑍𝑍0𝜎𝜎𝑟𝑟 𝑘𝑘 𝑟𝑟if ∫0 is chosen𝐽𝐽𝑚𝑚 𝑥𝑥 𝐽𝐽𝑚𝑚 to𝑥𝑥 coincide𝑑𝑑𝑑𝑑 𝜋𝜋𝜋𝜋with a zero𝜎𝜎𝑧𝑧 𝐸𝐸of𝑧𝑧0 𝑍𝑍(0𝜎𝜎𝑟𝑟 𝑘𝑘)𝑟𝑟. Thus𝐽𝐽𝑚𝑚 𝑘𝑘 𝑟𝑟the𝑅𝑅 energy content of our circularly-polarized Bessel beam is equal to the sum of the of its TE and 𝑚𝑚 𝑟𝑟 TM components. The ratio of angular𝑅𝑅 momentum to energy (in the limit𝐽𝐽 of𝑘𝑘 large𝑅𝑅 ) is now seen to be ( )/ , which includes the contribution ± / of spin to the orbital angular momentum / . The ± sign is related to the sense of circular polarization, which𝑅𝑅 could be the 𝑧𝑧 𝑧𝑧 same as 𝑚𝑚or ∓opposite𝜎𝜎 𝜔𝜔 to the sense of the orbital angular momentum𝜎𝜎 𝜔𝜔 of the Bessel beam of order . When the Bessel𝑚𝑚 𝜔𝜔 beam has a large diameter (compared to a = 2 / ), its will be close to 1.0, in which case the spin angular momentum per photon will be nearly ± . For𝑚𝑚 𝑧𝑧 more compact Bessel beams, however, the spin angular momentum per photon𝜆𝜆 𝜋𝜋will𝜋𝜋 𝜔𝜔 be ± 𝜎𝜎 . ℏ ℏ𝜎𝜎𝑧𝑧 6

3. Electromagnetic energy and angular momentum of a rotating cylinder. As a second example of simple EM systems, consider an infinitely long, uniformly-charged, non-conducting, hollow cylinder of radius , having a surface-charge-density , as shown in (a) z (b) z Fig.1(a). The E-field inside the𝑅𝑅 cylinder is 𝑠𝑠 zero, while that outside 𝜎𝜎may be evaluated R R using Gauss’s law [4,5], as follows: R − 1 − + + + − − ( ) = ( ) ; > + + . (34) Ω0 − + − − − + + − 0 𝑠𝑠 ⁄ 0 + + − + Fig.1𝑬𝑬 . (a)𝒓𝒓 Infinitely𝑅𝑅𝜎𝜎 𝒓𝒓�-long𝜀𝜀 hollow𝑟𝑟 cylinder𝑟𝑟 𝑅𝑅 of radius − + − , having a uniform surface-charge-density , is − − + + + − − + brought from rest to a constant angular + 𝑠𝑠 σ − + − around𝑅𝑅 the -axis. (b) A second infinitely-long𝜎𝜎 s − − Ω0 + + + + cylinder of radius and uniform surface-charge- + density 𝑧𝑧resides inside the cylinder of radius . + 1 The inner cylinder,𝑅𝑅 which does not rotate, is 𝑠𝑠1 concentric𝜎𝜎 with the outer (rotating) cylinder. 𝑅𝑅

Let the cylinder start to spin slowly around its axis, so that its is brought from zero to some finite value after a long time, . During the time interval (0, ), when the angular velocity is ( ), the surface-current-density will be ( ) = ( ) . Ignoring 0 max max the retardation effects (justified byΩ the gradual rise of the𝑡𝑡 spinning velocity), the vector potential𝑡𝑡 𝑠𝑠 𝑠𝑠 ( , ) may be calculated as Ωfollows𝑡𝑡 [4-8]: 𝑱𝑱 𝑡𝑡 𝜎𝜎 𝑅𝑅Ω 𝑡𝑡 𝝋𝝋�

𝑨𝑨 𝒓𝒓 𝑡𝑡 ( , ) ( , ) = = | ′ | 𝜋𝜋 ( ) 𝜇𝜇0 𝑱𝑱𝑠𝑠 𝒓𝒓 𝑡𝑡 ′ 𝜇𝜇0 ∞ 2𝜎𝜎𝑠𝑠𝑅𝑅Ω cos 𝜑𝜑 ′ 2 2 2 4𝜋𝜋 𝒓𝒓 − 𝒓𝒓 4𝜋𝜋 𝑧𝑧=−∞ � 𝑟𝑟 +𝑅𝑅 −2𝑟𝑟𝑟𝑟 cos 𝜑𝜑 +𝑧𝑧 𝑨𝑨 𝒓𝒓 𝑡𝑡 � 𝑑𝑑𝑠𝑠 �𝜑𝜑=0 ∫ ( ) 𝝋𝝋�𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 = cos lim ln See Appendix A 2 𝜋𝜋 ( 2 2 ) 2 𝜇𝜇0𝜎𝜎𝑠𝑠𝑅𝑅 Ω 𝝋𝝋� � 𝑟𝑟 +𝑅𝑅 −2𝑟𝑟𝑟𝑟 cos 𝜑𝜑 +𝑧𝑧 +𝑧𝑧 2 2 2 2𝜋𝜋 � 𝜑𝜑 𝑧𝑧→∞ � � 𝑟𝑟 +𝑅𝑅 −2𝑟𝑟𝑟𝑟 cos 𝜑𝜑 +𝑧𝑧 −𝑧𝑧� 𝑑𝑑𝑑𝑑 0 ( ) = 𝜋𝜋 cos lim ln 2 2 2 2 2 𝜇𝜇0𝜎𝜎𝑠𝑠𝑅𝑅 Ω𝝋𝝋� �� 𝑟𝑟 +𝑅𝑅 −2𝑟𝑟𝑟𝑟 cos 𝜑𝜑 +𝑧𝑧 +𝑧𝑧� 2 2 2𝜋𝜋 � 𝜑𝜑 𝑧𝑧→∞ � 𝑟𝑟 +𝑅𝑅 −2𝑟𝑟𝑟𝑟 cos 𝜑𝜑 � 𝑑𝑑𝑑𝑑 = 0 cos lim 2 ln ( + 2 cos ) + + 2 𝜋𝜋 0 𝑠𝑠 𝜇𝜇 𝜎𝜎 𝑅𝑅 Ω𝝋𝝋� 2 2 2 2𝜋𝜋 � 𝜑𝜑 𝑧𝑧→∞ � �� 𝑟𝑟 𝑅𝑅 − 𝑟𝑟𝑟𝑟 𝜑𝜑 𝑧𝑧 𝑧𝑧�� 𝑑𝑑𝑑𝑑 0 cos ln( + 2 cos ) 2 𝜇𝜇0𝜎𝜎𝑠𝑠𝑅𝑅 Ω𝝋𝝋� 𝜋𝜋 2 2 0 =− 2𝜋𝜋 ∫ cos 𝜑𝜑lim 𝑟𝑟ln( 𝑅𝑅) +−ln 𝑟𝑟1𝑟𝑟+ 1𝜑𝜑+𝑑𝑑𝑑𝑑( + 2 cos )/ 2 𝜋𝜋 0 𝑠𝑠 𝜇𝜇 𝜎𝜎 𝑅𝑅 Ω𝝋𝝋� 2 2 2 𝜋𝜋 � 𝜑𝜑 𝑧𝑧→∞ � 𝑧𝑧 � � 𝑟𝑟 𝑅𝑅 − 𝑟𝑟𝑟𝑟 𝜑𝜑 𝑧𝑧 �� 𝑑𝑑𝑑𝑑 0 cos {ln( ) + ln[1 2( / ) cos + ( / ) ]} 2 𝜇𝜇0𝜎𝜎𝑠𝑠𝑅𝑅 Ω𝝋𝝋� 𝜋𝜋 2 2 =− 2𝜋𝜋 [lim∫0 ln𝜑𝜑( ) +𝑅𝑅ln(2) ln( −)] 𝑟𝑟cos𝑅𝑅 𝜑𝜑 𝑟𝑟 𝑅𝑅 𝑑𝑑𝑑𝑑 2 𝜇𝜇0𝜎𝜎𝑠𝑠𝑅𝑅 Ω𝝋𝝋� 𝜋𝜋 𝜋𝜋 𝑧𝑧cos→∞ ln𝑧𝑧[1 2( /−) cos𝑅𝑅 +∫0( / 𝜑𝜑) 𝑑𝑑𝑑𝑑] See Appendix A 2 𝜇𝜇0𝜎𝜎𝑠𝑠𝑅𝑅 Ω𝝋𝝋� 𝜋𝜋 2 − 2𝜋𝜋 ∫0 𝜑𝜑 − 𝑟𝑟 𝑅𝑅 𝜑𝜑 𝑟𝑟 𝑅𝑅 𝑑𝑑𝑑𝑑

7

½( ) , < ; = (35) ½(𝜇𝜇0𝜎𝜎𝑠𝑠𝑅𝑅Ω𝑟𝑟 𝝋𝝋�) , 𝑟𝑟 >𝑅𝑅 . � 3 The magnetic field produced𝜇𝜇0𝜎𝜎𝑠𝑠𝑅𝑅 Ω⁄ by𝑟𝑟 𝝋𝝋� the spinning𝑟𝑟 𝑅𝑅 cylinder is thus found to be ( ) < , ( , ) = ( , ) = × ( , ) = = (36) −1 00 𝑠𝑠 > . 0 𝑟𝑟 𝜑𝜑 𝜇𝜇 𝜎𝜎 𝑅𝑅Ω 𝑡𝑡 𝒛𝒛� 𝑟𝑟 𝑅𝑅 While𝑩𝑩 𝒓𝒓 𝑡𝑡the cylinder𝜇𝜇 𝑯𝑯 𝒓𝒓 spins𝑡𝑡 𝜵𝜵up, an𝑨𝑨 electric𝒓𝒓 𝑡𝑡 𝑟𝑟field𝜕𝜕 is�𝑟𝑟 induced𝐴𝐴 �𝒛𝒛� by� the time-varying vector potential. The magnitude of this -field at the cylinder surface is 𝑟𝑟 𝑅𝑅 ( = , ) = ( = , ) = ½( ) 𝐸𝐸 . (37) 2 ′ The (per unit length𝑡𝑡 along the z-axis)0 needed𝑠𝑠 to bring the cylinder from 𝑬𝑬 𝑟𝑟 𝑅𝑅 𝑡𝑡 −𝜕𝜕 𝑨𝑨 𝑟𝑟 𝑅𝑅 𝑡𝑡 − 𝜇𝜇 𝜎𝜎 𝑅𝑅 Ω 𝝋𝝋� rest to its final spinning speed = ( ) is obtained by integrating over the cylindrical surface and over time from = 0 to . We will have Ω0 Ω 𝑡𝑡max 𝑬𝑬 ∙ 𝑱𝑱𝑠𝑠 = 2𝑡𝑡 𝑡𝑡max( = , ) ( ) 𝑡𝑡max ℰ = 𝜋𝜋𝜋𝜋 ∫0 𝐸𝐸 𝑟𝑟 𝑅𝑅( 𝑡𝑡) 𝜎𝜎(𝑠𝑠𝑅𝑅)Ω 𝑡𝑡=𝑑𝑑𝑑𝑑½ . (38) max 4 2 𝑡𝑡 ′ 4 2 2 This energy is stored in𝜇𝜇 0the𝜋𝜋𝑅𝑅 internal𝜎𝜎𝑠𝑠 ∫0 BΩ-field𝑡𝑡 Ω (or𝑡𝑡 𝑑𝑑𝑑𝑑-field)𝜇𝜇 0of𝜋𝜋 𝑅𝑅the𝜎𝜎 𝑠𝑠cylinder,Ω0 whose density is ½ = ½ ( ) . 2 2 𝐻𝐻 3.1.𝜇𝜇0 Angular𝐻𝐻 𝜇𝜇 momentum0 𝜎𝜎𝑠𝑠𝑅𝑅Ω0 conservation. In the system of Fig.1(a), the torque (per unit length along the z-axis) exerted on the cylinder is obtained by integrating × ( = , ) over the cylindrical surface, that is, 𝒓𝒓 𝜎𝜎𝑠𝑠𝑬𝑬 𝑟𝑟 𝑅𝑅 𝑡𝑡 ( ) = × ( ½ ) = ( ) . (39) 2𝜋𝜋 2 2 ′ 2 4 ′ The total angular𝑻𝑻 𝑡𝑡 momentum∫𝜑𝜑=0 𝑅𝑅𝒓𝒓� picked− 𝜇𝜇 up0𝜎𝜎𝑠𝑠 by𝑅𝑅 theΩ 𝝋𝝋�agen𝑅𝑅𝑅𝑅𝑅𝑅cy in− charge𝜋𝜋𝜇𝜇0𝜎𝜎𝑠𝑠 of𝑅𝑅 spinΩ 𝑡𝑡ning𝒛𝒛� the cylinder is thus given by = ( ) = . (40) max 𝑡𝑡 2 4 At this point, we are faced𝑳𝑳 with∫0 a dilemma.𝑻𝑻 𝑡𝑡 𝑑𝑑𝑑𝑑 In− 𝜋𝜋its𝜇𝜇 0final𝜎𝜎𝑠𝑠 𝑅𝑅 state,Ω0𝒛𝒛� the charged cylinder is spinning at the constant angular velocity , producing a constant, uniform magnetic field = inside the cylinder, and a constant, radially-decaying electric field ( ) given by Eq.(34) 0 𝑠𝑠 0 outside the cylinder. Therefore, theΩ EM field produced by the spinning charged cylinder𝑯𝑯 𝜎𝜎appears𝑅𝑅Ω 𝒛𝒛� 0 to not have any angular momentum. In other words, the existence of𝑬𝑬 the𝑟𝑟 mechanical angular momentum of Eq.(40), which is acquired by the spinning agent, appears to contradict the conservation of angular momentum. This discrepancy, however, is easily resolved if one recognizes the𝑳𝑳 assumed infinite length of the cylinder as the culprit. A finite-length cylinder will always produce a magnetic field in faraway regions ( ), and the EM angular momentum density ( ) = × ( × / ), when integrated over the entire space, turns out to be exactly equal in magnitude and opposite in2 direction to the mechanical𝑟𝑟 ≫ 𝑅𝑅 angular momentum of Eq.(40). 𝑒𝑒𝑒𝑒 A more𝓛𝓛 tractable𝒓𝒓 𝒓𝒓 version𝑬𝑬 𝑯𝑯 of𝑐𝑐 the above problem involves a second charged cylinder of radius < placed inside (and concentric with) the spinning cylinder, as shown in Fig.1(b)𝑳𝑳 . If the surface-charge-density of the inner cylinder is taken to be = / , the static -field will 1 𝑅𝑅be confined𝑅𝑅 to the region between the two cylinders, as follows: 𝜎𝜎𝑠𝑠1 −𝑅𝑅𝜎𝜎𝑠𝑠 𝑅𝑅1 𝐸𝐸

8

0, < , ( ) = /( ), < < , (41) 1 0, 𝑟𝑟 >𝑅𝑅 . 𝑬𝑬0 𝒓𝒓 �−𝑅𝑅𝜎𝜎𝑠𝑠𝒓𝒓� 𝜀𝜀0𝑟𝑟 𝑅𝑅1 𝑟𝑟 𝑅𝑅 The EM angular momentum (per unit length along z), which is now confined to the region 𝑟𝑟 𝑅𝑅 between the two cylinders, is given by = × ( × / ) = ( ) . (42) 2𝜋𝜋 𝑅𝑅 2 2 2 2 2 The difference𝑳𝑳𝑒𝑒𝑒𝑒 ∫ 𝜑𝜑between=0 ∫𝑟𝑟=𝑅𝑅 1the𝒓𝒓 mechanical𝑬𝑬0 𝑯𝑯 𝑐𝑐 angular𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟momentum𝜋𝜋𝜇𝜇0𝜎𝜎𝑠𝑠 𝑅𝑅 in𝑅𝑅 Eq.(40)− 𝑅𝑅1 Ωand0𝒛𝒛� the EM angular momentum in Eq.(42) is taken up by the external agency responsible for holding the inner cylinder stationary while the outer cylinder is spun from rest to𝑳𝑳 its final angular velocity . Needless to say, no energy is spent in keeping the inner cylinder stationary. However, the 0 induced -field ( = , ) = ( = , ) = ½( ) exerts a torque on Ωthe inner cylinder, which, upon integration over time, yields ′ 𝐸𝐸 𝑬𝑬 𝑟𝑟 𝑅𝑅1 𝑡𝑡 −𝜕𝜕𝑡𝑡𝑨𝑨 𝑟𝑟 𝑅𝑅1 𝑡𝑡 − 𝜇𝜇0𝜎𝜎𝑠𝑠𝑅𝑅𝑅𝑅1Ω 𝝋𝝋� = × ( = , ) = . (43) max 𝑡𝑡 2𝜋𝜋 2 2 2 The total𝑳𝑳1 mechanical∫𝑡𝑡=0 ∫𝜑𝜑=0 𝑅𝑅angular1𝒓𝒓� 𝜎𝜎 𝑠𝑠1momentum𝑬𝑬 𝑟𝑟 𝑅𝑅1 𝑡𝑡 𝑅𝑅+1𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 is thus𝜋𝜋𝜇𝜇0 𝜎𝜎seen𝑠𝑠 𝑅𝑅 𝑅𝑅to1 Ω be0𝒛𝒛� exactly equal in magnitude and opposite in direction to the EM angular momentum stored in the and 1 fields in the region between the two cylinders. 𝑳𝑳 𝑳𝑳 𝑒𝑒𝑒𝑒 A similar argument can be made if the oppositely-charged stationary𝑳𝑳 cylinder has𝑬𝑬 a radius𝑯𝑯 > . The EM momentum now vanishes, and the angular momenta of the cylinders cancel out.

𝑅𝑅3.12. S𝑅𝑅pinning cylinder in an external electric field. Next, we assume that the rotating cylinder of Fig.1(a) is placed inside a uniform electric field = , as shown in Fig.2. The charges on the cylinder and those on the plates attract/repel each other along the x-axis. However, no other 𝑥𝑥 net force is exerted on the cylinder as it spins up from𝑬𝑬 𝐸𝐸=𝒙𝒙�0 to . In contrast, the induced - field during the time-interval (0, ), namely, Ω Ω0 𝐸𝐸 ( , ) = ( , ) = ½( / ) 𝑡𝑡max , (44) exerts a force in the y-direction on both plates. With reference3 ′ to Fig.2 and considering that 𝑬𝑬 𝑟𝑟 𝑡𝑡 −𝜕𝜕𝑡𝑡𝑨𝑨 𝑟𝑟 𝑡𝑡 − 𝜇𝜇0𝜎𝜎𝑠𝑠𝑅𝑅 Ω 𝑟𝑟 𝝋𝝋� = ( 2) tan , the force (per unit-length along z) exerted on each plate is readily seen to be ½ ( ) = = , cos ( 2)(1 + tan ) = ½ 𝑦𝑦 𝑑𝑑⁄ ½𝜑𝜑 . (45) 𝜋𝜋 𝑑𝑑 2 3 ′ 𝑦𝑦 𝜑𝜑=− 𝜋𝜋 0 𝜑𝜑 2 cos 𝜑𝜑 dy 0 0 𝑠𝑠 𝐹𝐹 𝑡𝑡 ∫ −𝜎𝜎 𝐸𝐸 �𝑟𝑟 𝑡𝑡� 𝜑𝜑 𝑑𝑑⁄ 𝜑𝜑 𝑑𝑑𝑑𝑑 𝜋𝜋𝜇𝜇y 𝜎𝜎 𝜎𝜎 𝑅𝑅 Ω + − + σ Ω s − + r − Fig.2. An infinitely-long, uniformly-charged cylinder of radius + and surface-charge-density rotates at an angular velocity + ϕ − around the z-axis. A pair of uniformly-charged parallel plates · x 𝑠𝑠 σ z −σ − pro𝑅𝑅 duces a constant, uniform E𝜎𝜎-field in the cross-sectional planeΩ + 0 0 of the cylinder. The surface-charge-density of the parallel plates + − is ± , the E-field produced in the space between the plates is + − = ( / ) , the H-field inside the cylinder is = , E 0 + and 𝜎𝜎the EM momentum-density within the cylinder is given by − 0 0 𝑠𝑠 + 𝑬𝑬 =𝜎𝜎 𝜀𝜀× 𝒙𝒙�/ = ( ) . 𝑯𝑯 𝜎𝜎 𝑅𝑅Ω 𝒛𝒛� d 2 𝓹𝓹𝑒𝑒𝑒𝑒 𝑬𝑬 𝑯𝑯 𝑐𝑐 − 𝜇𝜇0𝜎𝜎0𝜎𝜎𝑠𝑠𝑅𝑅Ω 𝒚𝒚�

9

When the above force is integrated over time from = 0 to and multiplied by two (to account for both plates), we find the mechanical momentum acquired by the plates at = to max be = . Considering that the E-field produced𝑡𝑡 by𝑡𝑡 the plates is = ( / ) , max and the H-field inside3 the cylinder is = , the EM momentum-density 𝑡𝑡within𝑡𝑡 the 0 0 𝑠𝑠 0 0 0 cylinder𝒑𝒑 𝜋𝜋 will𝜇𝜇 𝜎𝜎 be𝜎𝜎 𝑅𝑅 Ω =𝒚𝒚� × / = ( ) . Multiplication by the cross𝑬𝑬-sectional𝜎𝜎 𝜀𝜀 area𝒙𝒙� 𝑠𝑠 0 of the cylinder then shows that the2 EM momentum𝑯𝑯 𝜎𝜎 𝑅𝑅Ω stored𝒛𝒛� within the cylinder is precisely equal 𝑒𝑒𝑒𝑒 0 0 𝑠𝑠 0 in magnitude and𝓹𝓹 opposite𝑬𝑬 in𝑯𝑯 sign𝑐𝑐 to the− 𝜇𝜇mechanical𝜎𝜎 𝜎𝜎 𝑅𝑅Ω momentum𝒚𝒚� acquired by the plates.

3.3. Hidden momentum of the spinning cylinder. It has been argued that a stationary system such as that of the spinning cylinder residing between the parallel plates of Fig.2 must have an internal “hidden momentum” equal and opposite to its EM momentum, so that the net linear momentum of the stationary system would be zero [9-32]. The acquisition of hidden momentum by the cylinder thus dictates the transfer of a compensatory mechanical momentum to the cylinder as it spins up from rest ( = 0) to its final state where = . The linear mechanical momentum per unit-length of the cylinder at = is thus argued to be = 0 . In this way, theΩ system acquires no net mechanicalΩ Ω momentum, as the max mech cylinder and3 the plates end up moving in opposite 𝑡𝑡directions𝑡𝑡 with equal momentum, while𝒑𝒑 the 0 0 𝑠𝑠 0 EM−𝜋𝜋𝜇𝜇 momentum𝜎𝜎 𝜎𝜎 𝑅𝑅 Ω 𝒚𝒚�is compensated by an equal and opposite amount of hidden momentum. If the spinning cylinder happens to be used as a model for a uniformly-magnetized solid cylinder, then two alternative formulations of classical electrodynamics may be used to describe the situation discussed above. In the Lorentz formalism [4,5], the of the cylinder is equivalent to a current-density = × , which turns out to be the surface 0 current-density = . The H-field inside −the1 cylinder is now zero, but the 𝑀𝑀internal𝒛𝒛� B- 0 field continues to be given−1 by = , which𝑱𝑱 𝜇𝜇is equa𝜵𝜵 l𝑴𝑴 to . In the Lorentz formalism, the 𝑠𝑠 0 0 EM momentum-𝑱𝑱density𝜇𝜇 is𝑀𝑀 𝝋𝝋� = × (known as Livens momentum), whereas the hidden 0 𝑠𝑠 0 momentum-density is =𝑩𝑩 𝜇𝜇× 𝐽𝐽 𝒛𝒛� . Therefore, all the 𝑀𝑀preceding𝒛𝒛� arguments remain intact. 𝑒𝑒𝑒𝑒 0 In the Einstein-Laub formalism𝓹𝓹 𝜀𝜀 [33]𝑬𝑬 , 𝑩𝑩where the EM momentum-density is = × / , hidden 0 there is no need for hidden𝓹𝓹 momentum.𝑴𝑴 𝜀𝜀 𝑬𝑬In this case, the EM momentum of the system is zero.2 𝑒𝑒𝑒𝑒 However, the force-density exerted by an external E-field on the magnetization𝓹𝓹 ( , 𝑬𝑬) is 𝑯𝑯given𝑐𝑐 by ( , ) = × . If rises slowly from zero to , the mechanical momentum- density acquired by the magnetized cylinder will be , which is the same𝑴𝑴 𝒓𝒓 mechanical𝑡𝑡 𝑡𝑡 0 0 momentum𝒇𝒇 𝒓𝒓 𝑡𝑡 as− obtained𝜕𝜕 𝑴𝑴 𝜀𝜀 in𝑬𝑬 the 𝑴𝑴Lorentz formalism — albeit after𝑀𝑀 𝒛𝒛�invoking the notion of hidden 0 0 𝑥𝑥 momentum, an endemic feature of the Lorentz formula−tion𝜀𝜀 𝑀𝑀 in𝐸𝐸 its𝒚𝒚� treatment of magnetic materials.

4. Rotating magnetic dipole in a constant and uniform electric field. As our third example, we consider a small magnetic dipole residing at the origin of the -plane and rotating slowly around the -axis at the constant angular velocity , as shown in Fig. 3. We have 𝒎𝒎0 𝑥𝑥𝑥𝑥 ( ) = [cos( ) + sin( ) ] 𝑧𝑧 𝜔𝜔 . (46) The dipole in Fig.3 sits between a pair of non-conducting, uniformly-charged, infinitely- 𝒎𝒎 𝑡𝑡 𝑚𝑚0 𝜔𝜔𝜔𝜔 𝒙𝒙� 𝜔𝜔𝜔𝜔 𝒚𝒚� large parallel plates, which produce a constant and uniform electric field = ( ) in the region of space between the two plates. The distance between the plates is , and the surface 0 0 0 electrical charge-densities of the two plates are assumed to be ± . The𝑬𝑬 slow 𝜎𝜎⁄𝜀𝜀 𝒙𝒙� of the dipole ensures that the EM radiation is negligible, and that quasi-static treatment𝑑𝑑 of the EM fields 0 is permitted. (Of course, for a magnetic dipole to rotate at a constant𝜎𝜎 angular velocity, a torque must be applied to the dipole. We may assume that a constant, uniform magnetic field is

𝐻𝐻0𝒛𝒛� 10

present at and around the origin of coordinates in the system of Fig.3. The presence of this magnetic field will in no way affect the ensuing analysis.) In the Einstein-Laub formulation [33], the -field exerts no torque on the dipole. However, the dipole experiences a net force ( ) in the presence of the external -field. Denoting the magnetization distribution within the dipole by 𝐸𝐸 ( , ), we will have 𝑭𝑭 𝑡𝑡 𝐸𝐸 ( ) = ( , ) × 𝑴𝑴 𝒓𝒓 𝑡𝑡 = cos( ) . (47) ∞ Next, we compute𝑭𝑭 𝑡𝑡 −the∭ EM−∞ 𝜕𝜕(Abraham)𝑡𝑡𝑴𝑴 𝒓𝒓 𝑡𝑡 momentum𝜀𝜀0𝑬𝑬0 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ( 𝜎𝜎)0 of𝑚𝑚 0the𝜔𝜔 field𝜔𝜔 [31𝜔𝜔 𝒛𝒛�-33], and show that the force ( ) is not rooted in an exchange of momentum between the EM field and the dipole. 𝒑𝒑𝑒𝑒𝑒𝑒 𝑡𝑡 z y 𝑭𝑭 𝑡𝑡

σ0 E0 + − + − + ω + t − − x − + + m0 − − + − −σ 0

d

Fig. 3. A small magnetic dipole rotates slowly in a constant, uniform -field produced by a pair of uniformly-charged parallel plates. The surface electric charge densities of the plates are denoted by ± . 𝒎𝒎0 𝐸𝐸 0 Since the magnetic particle is assumed to rotate slowly, we may take the instantaneous𝜎𝜎 magnetization profile ( , ) and compute the corresponding magnetic charge-density distribution ( , ) = ( , ). The -field distribution will then be obtained using Maxwell’s magnetostatic𝑴𝑴 equations𝒓𝒓 𝑡𝑡 [4-8], as follows: 𝜌𝜌𝑚𝑚 𝒓𝒓 𝑡𝑡 −𝜵𝜵 ∙ 𝑴𝑴 𝒓𝒓 𝑡𝑡 𝐻𝐻 ( , )( ) ( , ) = . (48) ∞ ′ | | ′ 𝜌𝜌𝑚𝑚 𝒓𝒓 𝑡𝑡 𝒓𝒓 − 𝒓𝒓 ′ ′ ′ ′ 3 4𝜋𝜋𝜇𝜇0 𝒓𝒓 − 𝒓𝒓 In Eq.(48) is the vacuum𝑯𝑯 𝒓𝒓 𝑡𝑡 permeability.�−∞ In the region𝑑𝑑𝑥𝑥 𝑑𝑑 𝑦𝑦of 𝑑𝑑space𝑧𝑧 between parallel plates of Fig.3, the Abraham momentum of the EM field may be computed as follows: 𝜇𝜇0 ( ) = ( , ) × ( , ) +𝑑𝑑⁄2 1 ∞ ∞ 2 𝑒𝑒𝑒𝑒 𝑐𝑐 𝑦𝑦=−∞ 𝑧𝑧=−∞ 𝒑𝒑 𝑡𝑡 �𝑥𝑥=−𝑑𝑑⁄2 � ∫ 𝑬𝑬 𝒓𝒓 𝑡𝑡 𝑯𝑯 𝒓𝒓 𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 × ( ) = ∞ ( , ) +𝑑𝑑⁄2 | | ′ 𝜎𝜎0 ′ ∞ ∞ 𝒙𝒙� 𝒓𝒓 − 𝒓𝒓 ′ ′ ′ ′ 3 4𝜋𝜋 𝑚𝑚 𝑦𝑦=−∞ 𝑧𝑧=−∞ 𝒓𝒓 − 𝒓𝒓 � 𝜌𝜌 𝒓𝒓 𝑡𝑡 ��𝑥𝑥=−𝑑𝑑⁄2 ∫ ∫ 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑� 𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧 −∞

11

( ) ( ) = ∞ ( , ) +𝑑𝑑⁄2 [( ) ( ′ ) ( ′ ) ] 𝜎𝜎0 ′ ∞ ∞ 𝑦𝑦 − 𝑦𝑦 𝒛𝒛� − 𝑧𝑧 − 𝑧𝑧 𝒚𝒚� ′ ′ ′ ′ 2 ′ 2 ′ 2 3⁄2 4𝜋𝜋 𝑚𝑚 𝑦𝑦=−∞ 𝑧𝑧=−∞ 𝑥𝑥−𝑥𝑥 + 𝑦𝑦−𝑦𝑦 + 𝑧𝑧−𝑧𝑧 � 𝜌𝜌 𝒓𝒓 𝑡𝑡 ��𝑥𝑥=−𝑑𝑑⁄2 ∫ ∫ 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑� 𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧 −∞ 0 +𝑑𝑑⁄2 = ∞ ( , ) ∞ ( ) ( ) ( ) ∞ 𝜎𝜎0 ′ 𝒚𝒚� ′ 2 ′ 2 ′ 2 4𝜋𝜋 𝑚𝑚 � 𝜌𝜌 𝒓𝒓 𝑡𝑡 � �� �� 𝑥𝑥−𝑥𝑥 + 𝑦𝑦−𝑦𝑦 + 𝑧𝑧−𝑧𝑧 �𝑧𝑧=−∞� 𝑑𝑑𝑑𝑑 −∞ 𝑦𝑦=−∞ 𝑥𝑥=−𝑑𝑑⁄2 0

∞ = 0. (49) ( ) ( ) ( ) ∞ 𝒛𝒛� ′ ′ ′ ′ 2 ′ 2 ′ 2 − � �� 𝑥𝑥−𝑥𝑥 + 𝑦𝑦−𝑦𝑦 + 𝑧𝑧−𝑧𝑧 �𝑦𝑦=−∞� 𝑑𝑑𝑑𝑑� 𝑑𝑑𝑑𝑑𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧 In the absence of a net𝑧𝑧=− EM∞ momentum in the region between the plates, the force ( ) of Eq.(47) cannot be balanced by a corresponding change in the field momentum ( ). We must, therefore, look elsewhere for confirmation of momentum conservation. Specifically, we note𝑭𝑭 𝑡𝑡 that 𝑒𝑒𝑒𝑒 the rotating dipole’s magnetic field, being time-dependent, must give rise to 𝒑𝒑an 𝑡𝑡-field which would then exert a force on the electrically-charged parallel plates. Since, in the absence of retardation effects, the vector potential ( , ) of the slowly-rotating magnetic dipole𝐸𝐸 is given by

( , ) = [cos(𝑨𝑨 )𝒓𝒓 𝑡𝑡+ sin( ) ] × 𝑚𝑚0 3 𝑨𝑨 𝒓𝒓 𝑡𝑡 = 4𝜋𝜋𝑟𝑟 [cos(𝜔𝜔𝜔𝜔) (𝒙𝒙� 𝜔𝜔)𝜔𝜔+𝒚𝒚�sin( 𝒓𝒓 ) ( )], (50) 𝑚𝑚0 3 the corresponding -field is readily4𝜋𝜋𝑟𝑟 evaluated𝜔𝜔𝜔𝜔 𝑦𝑦𝒛𝒛 �as− follows:𝑧𝑧𝒚𝒚� 𝜔𝜔𝜔𝜔 𝑧𝑧𝒙𝒙� − 𝑥𝑥𝒛𝒛� ( , ) =𝐸𝐸 ( , ) = [sin( ) ( ) cos( ) ( )]. (51) 𝑚𝑚0𝜔𝜔 3 To determine𝑬𝑬 𝒓𝒓 𝑡𝑡 the−𝜕𝜕 force𝑡𝑡𝑨𝑨 𝒓𝒓 exerted𝑡𝑡 4𝜋𝜋 by𝑟𝑟 the above𝜔𝜔𝜔𝜔 𝑦𝑦𝒛𝒛�-field− 𝑧𝑧𝒚𝒚� on− each of𝜔𝜔 𝜔𝜔the 𝑧𝑧charged𝒙𝒙� − 𝑥𝑥𝒛𝒛� plates, we fix the coordinate and integrate the field over the -plane to obtain 𝐸𝐸 (𝑥𝑥 , ) 𝑦𝑦𝑦𝑦 ∞ ∬−∞ 𝑬𝑬 =𝒓𝒓 𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 [cos( ) + sin( ) ] cos( ) + sin( ) ∞ 𝑚𝑚0𝜔𝜔 𝑧𝑧 𝑥𝑥 𝑦𝑦 3 3 3 4𝜋𝜋 𝑟𝑟 𝑟𝑟 0 𝑟𝑟 − �−∞� 𝜔𝜔𝜔𝜔 𝒙𝒙� 𝜔𝜔𝜔𝜔 𝒚𝒚� − � 𝜔𝜔𝜔𝜔 𝜔𝜔𝜔𝜔 � 𝒛𝒛�� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = [cos( ) + sin( ) ] ∞ ∞ 𝑚𝑚0𝜔𝜔 1 2 2 2 4𝜋𝜋 𝜔𝜔𝜔𝜔 𝒙𝒙� 𝜔𝜔𝜔𝜔 𝒚𝒚� � ��𝑥𝑥 +𝑦𝑦 +𝑧𝑧 � � 𝑑𝑑𝑑𝑑 𝑦𝑦=−∞ 0 𝑧𝑧=−∞ sin( ) ∞ + cos( ) ∞ ∞ 𝑚𝑚0𝜔𝜔 1 𝑚𝑚0𝜔𝜔 1 2 2 2 3 4𝜋𝜋 �𝑥𝑥 +𝑦𝑦 +𝑧𝑧 4𝜋𝜋 𝑟𝑟 − 𝜔𝜔𝜔𝜔 𝒛𝒛� � � �𝑦𝑦=−∞� 𝑑𝑑𝑑𝑑 𝑥𝑥 𝜔𝜔𝜔𝜔 𝒛𝒛� �−∞ 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑧𝑧=−∞ = cos( ) = ½ cos( ) ∞ ( ) ∞ 𝑚𝑚0𝜔𝜔 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 1 2 2 3⁄2 2 2 4𝜋𝜋 𝑥𝑥 +𝜌𝜌 0 �𝑥𝑥 +𝜌𝜌 = ½ 𝑥𝑥 cos𝜔𝜔( 𝜔𝜔 )𝒛𝒛�(�0 | |) . − 𝑚𝑚 𝜔𝜔𝜔𝜔 𝜔𝜔𝜔𝜔 𝒛𝒛� � ��𝜌𝜌=0 (52)

For the plate𝑚𝑚 on0𝜔𝜔 the left𝜔𝜔𝜔𝜔-hand𝑥𝑥⁄-𝑥𝑥side𝒛𝒛� in Fig.3, is negative and is positive, whereas for the plate on the right-hand-side, is positive while is negative. The total force on both plates is, 𝑥𝑥 𝜎𝜎0 𝑥𝑥 𝜎𝜎0 12

therefore, ( ) = cos( ) . This is exactly equal in magnitude and opposite in sign to the force exerted on the magnetic dipole; see Eq.(47). Note that the force exerted by the rotating 0 0 magnetic di𝑭𝑭pole𝑡𝑡 on− 𝜎𝜎the𝑚𝑚 parallel𝜔𝜔 𝜔𝜔plates𝜔𝜔 𝒛𝒛� is independent of the separation between the plates. Consequently, the effect of the dipole on the plates cannot be ignored, even if the plates happen to be infinitely far away. We have thus confirmed the conservation of mome𝑑𝑑 ntum by showing explicitly that the on the dipole is equal and opposite to the reaction of the parallel plates.

4.1. Hidden momentum of the rotating magnetic dipole. Let us now consider the same problem from the perspective of the Lorentz formulation [1]. In this case, since the dipole has no electric charge, it does not experience any force from the -field, nor does the bound-current of the dipole experience any force in the absence of an external -field. So, it might be thought that, in the system of Fig.3, the Lorentz formulation predicts𝐸𝐸 no acting on the dipole. Now, in the Lorentz formulation, the EM momentum is the Livens momentum𝐵𝐵 [34], whose density is ( , ) × ( , ). Since = + , and, in accordance with Eq.(49), the part of does not contribute to the EM momentum, the Livens momentum in the system of Fig.3 must be 𝜀𝜀0𝑬𝑬 𝒓𝒓 𝑡𝑡 𝑩𝑩 𝒓𝒓 𝑡𝑡 𝑩𝑩 𝜇𝜇0𝑯𝑯 𝑴𝑴 𝜇𝜇0𝑯𝑯 𝑩𝑩 ( ) = × ( ) = sin( ) . (53) Livens [The same result as𝒑𝒑 in𝑒𝑒𝑒𝑒 Eq.(53) 𝜀𝜀may0𝑬𝑬0 be 𝒎𝒎obtained𝑡𝑡 𝜎𝜎 0using𝑚𝑚0 an 𝜔𝜔al𝜔𝜔ternative𝒛𝒛� method of calculation outlined in Appendix B.] It is thus tempting to associate the time--of-change of the Livens momentum of Eq.(53) with the mechanical force exerted on the parallel plates, and to declare that momentum conservation is upheld. However, in the Lorentz formulation there is known to be a certain amount of hidden momentum = ( ) × = sin( ) associated with the magnetic dipole residing in the external -field [29-32]. The time-rate-of-change of the hidden momentum being hidden 0 0 0 0 given by 𝒑𝒑 = 𝒎𝒎 𝑡𝑡 𝜀𝜀cos𝑬𝑬 ( −) 𝜎𝜎, it𝑚𝑚 is clear𝜔𝜔𝜔𝜔 that𝒛𝒛� an equal and opposite force must be experienced by the rotating𝐸𝐸 dipole. The EM force experienced by the dipole in the Lorentz 𝑡𝑡 hidden 0 0 formulation,𝜕𝜕 𝒑𝒑 assuming −the𝜎𝜎 𝑚𝑚contribution𝜔𝜔 𝜔𝜔𝜔𝜔 of𝒛𝒛� hidden momentum is properly taken into account, is thus seen to be in agreement with the result obtained in Eq.(47) via the Einstein-Laub formulation. Once again the total momentum is seen to be conserved, albeit with the inclusion of the hidden momentum and the Livens momentum in addition to the mechanical momenta imparted to the magnetic dipole and to the pair of parallel plates.

5. Rotating electric dipole in a constant and uniform magnetic field. In this section we examine the dual of the system of Fig.3, namely, a rotating electric dipole in a uniform magnetic field. With reference to Fig.4, consider a constant electric dipole , located at the origin of the coordinate system and rotating slowly at a fixed angular velocity in the -plane around the - 0 axis. The dipole may thus be written 𝒑𝒑 𝜔𝜔 𝑥𝑥𝑥𝑥 𝑦𝑦 ( ) = [cos( ) + sin( ) ]. (54) In the presence of a constant, uniform magnetic field ( , ) = , produced by an 𝒑𝒑 𝑡𝑡 𝑝𝑝0 𝜔𝜔𝜔𝜔 𝒙𝒙� 𝜔𝜔𝜔𝜔 𝒛𝒛� infinitely-long cylinder of radius carrying a constant surface-current-density = = , 0 the law yields the EM force exerted on the dipole as𝑯𝑯 follows:𝒓𝒓 𝑡𝑡 𝐻𝐻 𝒛𝒛� 𝑅𝑅𝑐𝑐 𝑱𝑱𝑠𝑠 𝐽𝐽0𝝋𝝋� 𝐻𝐻0𝝋𝝋� ( ) = ( ) × = sin( ) . (55)

𝑭𝑭 𝑡𝑡 𝜕𝜕𝑡𝑡𝒑𝒑 𝑡𝑡 𝜇𝜇0𝑯𝑯 𝜇𝜇0𝐻𝐻0𝑝𝑝0𝜔𝜔 𝜔𝜔𝜔𝜔 𝒚𝒚�

13

To confirm the conservation of momentum, we must first evaluate the EM counter-force exerted by the magnetic field of the rotating dipole on the current-carrying cylinder. Considering that the polarization density of the point-dipole is ( , ) = ( ) ( ) ( ) ( ), its bound current-density [4,5] is given by 𝑷𝑷 𝒓𝒓 𝑡𝑡 𝒑𝒑 𝒕𝒕 𝛿𝛿 𝑥𝑥 𝛿𝛿 𝑦𝑦 𝛿𝛿 𝑧𝑧 ( ) = ( , ) = ( ) ( ) ( )[sin( ) cos( ) ]. (56)

Consequently,𝑱𝑱bound the vector𝑡𝑡 𝜕𝜕 𝑡𝑡potential𝑷𝑷 𝒓𝒓 𝑡𝑡 (−𝑝𝑝, 0)𝜔𝜔 of𝛿𝛿 the𝑥𝑥 𝛿𝛿 dipole𝑦𝑦 𝛿𝛿 𝑧𝑧in the quasi𝜔𝜔𝜔𝜔 𝒙𝒙�-static− limit𝜔𝜔𝜔𝜔 may𝒛𝒛� be written as ( , ) =𝑨𝑨 𝒓𝒓 𝑡𝑡 [sin( ) cos( ) ]. (57) 𝜇𝜇0𝑝𝑝0𝜔𝜔 The magnetic field produced𝑨𝑨 𝒓𝒓 𝑡𝑡 in− the4 𝜋𝜋surrounding𝜋𝜋 𝜔𝜔𝜔𝜔 region𝒙𝒙� − by𝜔𝜔 the𝜔𝜔 𝒛𝒛�(slowly) rotating dipole is thus given by ( , ) = ( , ) = × ( , ) = [( ) sin( ) ( ) cos( )]. (58) 𝜇𝜇0𝑝𝑝0𝜔𝜔 0 3 Fig𝑩𝑩. 4𝒓𝒓. A𝑡𝑡 small𝜇𝜇 spherical𝑯𝑯 𝒓𝒓 𝑡𝑡 particle𝜵𝜵 of 𝑨𝑨radius𝒓𝒓 𝑡𝑡 centered4𝜋𝜋𝑟𝑟 at 𝑧𝑧the𝒚𝒚� −origin𝑦𝑦𝒛𝒛� of 𝜔𝜔𝜔𝜔 − 𝑦𝑦𝒙𝒙� − 𝑥𝑥𝒚𝒚� z 𝜔𝜔𝜔𝜔 coordinates has a permanent electric dipole moment , which rotates 𝑠𝑠 slowly at a constant angular velocity in 𝑅𝑅the -plane around the - 0 H0 H0 axis. The dipole is acted upon by a constant, uniform𝒑𝒑 magnetic field , produced by an infinitely long, thin𝜔𝜔 cylinder𝑥𝑥𝑥𝑥 of radius , which𝑦𝑦 carries the constant surface-current-density = = . The 0 𝑐𝑐 dipole𝐻𝐻 𝒛𝒛� experiences the oscillatory Lorentz force along the𝑅𝑅 -axis 𝑠𝑠 0 0 p0 given by Eq.(55). 𝑱𝑱 𝐽𝐽 𝝋𝝋� 𝐻𝐻 𝝋𝝋� 𝑦𝑦 ωt We are now in a to calculate the Lorentz y force exerted by the rotating dipole on the cylinder of = = radius carrying the surface-current-density Js , which generates the magnetic field inside the 𝑐𝑐 𝑠𝑠 0 cylinder.𝑅𝑅 We have 𝑱𝑱 𝐽𝐽 𝝋𝝋� 0 R 𝐻𝐻 𝝋𝝋� 𝐻𝐻0𝒛𝒛� x c ( ) = × ( , , , ) ∞ 2𝜋𝜋 𝑭𝑭𝑐𝑐 𝑡𝑡 = ∫𝑧𝑧=−∞ ∫𝜑𝜑=0 𝐽𝐽0(𝝋𝝋�cos 𝜇𝜇0𝑯𝑯 sin𝑅𝑅𝑐𝑐 𝜑𝜑 )𝑧𝑧×𝑡𝑡 𝑅𝑅𝑐𝑐𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑[( ) sin( ) ( ) cos( )] ∞ 2𝜋𝜋 𝜇𝜇0𝑝𝑝0𝜔𝜔 0 3 = ∫𝑧𝑧=−∞ ∫𝜑𝜑=0 𝐻𝐻0 [sin(𝜑𝜑 𝒚𝒚� )− 𝜑𝜑cos𝒙𝒙� ( 4)𝜋𝜋𝑟𝑟] 𝑧𝑧𝒚𝒚� − 𝑦𝑦𝒛𝒛� 𝜔𝜔𝜔𝜔 − 𝑦𝑦𝒙𝒙� − 𝑥𝑥𝒚𝒚� 𝜔𝜔𝜔𝜔 𝑅𝑅𝑐𝑐𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ( ) 𝜇𝜇0𝐻𝐻0𝑝𝑝0𝜔𝜔𝑅𝑅𝑐𝑐 ∞ 2𝜋𝜋 𝑅𝑅𝑐𝑐 sin 𝜑𝜑 cos 𝜑𝜑 0 2 2 3⁄2 4𝜋𝜋 [ ( ) ( )] 𝑧𝑧=−∞ 𝜑𝜑=0 𝑅𝑅𝑐𝑐 + 𝑧𝑧 +− � 𝜔𝜔𝜔𝜔 𝒙𝒙� − 𝜔𝜔𝜔𝜔 𝒛𝒛� ∫ +∫ sin( ) 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ( ) ( ) ∞ 2𝜋𝜋 𝑥𝑥cos 𝜔𝜔𝜔𝜔 + 𝑧𝑧 sin 𝜔𝜔𝜔𝜔 sin 𝜑𝜑 ∞ 2𝜋𝜋 𝑦𝑦 sin 𝜑𝜑 2 2 3⁄2 2 2 3⁄2 𝒛𝒛� ∫𝑧𝑧=−∞ ∫𝜑𝜑=0 𝑅𝑅𝑐𝑐 + 𝑧𝑧 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝜔𝜔𝜔𝜔 𝒚𝒚� ∫𝑧𝑧=−∞ ∫𝜑𝜑=0 𝑅𝑅𝑐𝑐 + 𝑧𝑧 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑� = sin( ) ( 2) 𝜇𝜇0𝐻𝐻0𝑝𝑝0𝜔𝜔𝑅𝑅𝑐𝑐 ∞ 2𝜋𝜋 𝑅𝑅𝑐𝑐 sin 𝜑𝜑 2 2 3⁄2 = −¼ 4𝜋𝜋 �∫𝑧𝑧=−∞ ∫𝜑𝜑=0 𝑅𝑅𝑐𝑐 +sin𝑧𝑧 ( 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑) =� ¼𝜔𝜔𝜔𝜔 𝒚𝒚� 1 + sin( ) ( ) ∞ 𝑑𝑑𝑑𝑑 ∞ 2 3⁄2 2 = −½𝜇𝜇0𝐻𝐻0𝑝𝑝0𝜔𝜔 �sin∫−∞( 1)+ 𝜁𝜁. � 𝜔𝜔𝜔𝜔 𝒚𝒚� − 𝜇𝜇0𝐻𝐻0𝑝𝑝0𝜔𝜔�𝜁𝜁⁄� 𝜁𝜁 ��−∞ 𝜔𝜔𝜔𝜔 (59𝒚𝒚� )

The cylinder− 𝜇𝜇0𝐻𝐻 thus0𝑝𝑝0𝜔𝜔 experiences𝜔𝜔𝜔𝜔 𝒚𝒚� a force only half as large as that experienced by the dipole, and in the opposite direction; see Eq.(55). The missing momentum must, therefore, come from the EM momentum inside the volume of the cylinder. To verify this, we take the dipole to be a

14

small spherical particle of radius and uniform polarization ( ). In fact, it is best to imagine two overlapping spherical dipoles, one oscillating along , the other along the -axis. Inside each 𝑠𝑠 spherical dipole, the -field is uniform𝑅𝑅 and given by ( )/(𝑷𝑷3 𝑡𝑡 ). The EM momentum inside the spherical dipole oscillating along the -axis is thus equal𝑥𝑥 to ( ) = 𝑧𝑧 cos( ) , 0 whereas that inside the𝐸𝐸 dipole oscillating along the -−axis𝑷𝑷 𝑡𝑡is zero.𝜀𝜀 The EM momentum outside 𝑒𝑒𝑒𝑒 0 0 0 the spherical dipole oscillating along also𝑥𝑥 turns out to be zero. 𝓹𝓹To see𝑡𝑡 this,⅓ note𝜇𝜇 𝐻𝐻 that𝑝𝑝 the 𝜔𝜔quasi𝑡𝑡 𝒚𝒚�- static potential ( , ) associated with ( )𝑧𝑧= sin( ) in the region outside the sphere of radius is given by [4-8] 𝑧𝑧 0 𝜓𝜓 𝒓𝒓 𝑡𝑡 ( ) 𝒑𝒑 𝑡𝑡 (𝑝𝑝 ) 𝜔𝜔𝜔𝜔 𝒛𝒛� 𝑠𝑠 ( , ) = = 𝑅𝑅 ( ) . (60) 𝑝𝑝0 sin 𝜔𝜔𝜔𝜔 𝑧𝑧 𝑝𝑝0 sin 𝜔𝜔𝜔𝜔 𝑧𝑧 3 2 2 3⁄2 In the above equation,𝜓𝜓 𝒓𝒓 is𝑡𝑡 the distance4𝜋𝜋𝜀𝜀0𝑟𝑟 from4𝜋𝜋 the𝜀𝜀0 𝜌𝜌 origin+ 𝑧𝑧 to the observation point (spherical coordinates), whereas is the radial distance from the -axis (cylindrical coordinates). The corresponding -field and EM𝑟𝑟 momentum-density inside the cylinder are thus found to be 𝜌𝜌 𝑧𝑧 𝐸𝐸 ( , ) = ( , ) = ( ) ( ) . (61) ( ) 𝑬𝑬( 𝒓𝒓, 𝑡𝑡) × −(𝜵𝜵,𝜓𝜓) 𝒓𝒓 𝑡𝑡 = − 𝜕𝜕𝜌𝜌𝜓𝜓 𝝆𝝆� − 𝜕𝜕𝑧𝑧𝜓𝜓 𝒛𝒛� ( ) . (62) 2 3𝜇𝜇0𝐻𝐻0𝑝𝑝0 sin 𝜔𝜔𝜔𝜔 𝑧𝑧𝑧𝑧 2 2 5⁄2 Clearly, the integral𝑬𝑬 𝒓𝒓 𝑡𝑡of the𝑯𝑯 𝒓𝒓momentum𝑡𝑡 ⁄𝑐𝑐 −-density� 4𝜋𝜋 𝜌𝜌 in+ 𝑧𝑧 Eq.(62�)𝝋𝝋� over the cylinder’s volume (excluding the particle’s spherical volume) vanishes. We conclude that the dipole oscillating along the -axis does not contribute to the EM momentum of the system of Fig.4. As for the dipole oscillating along the -axis, we have (in the region inside the cylinder and outside the spherical particle)𝑧𝑧 𝑥𝑥 ( ) ( ) ( , ) = = ( ) . (63) 𝑝𝑝0 cos 𝜔𝜔𝜔𝜔 𝑥𝑥 𝑝𝑝0 cos 𝜔𝜔𝜔𝜔 𝜌𝜌 cos 𝜑𝜑 3 2 2 3⁄2 ( , ) =𝜓𝜓 𝒓𝒓 𝑡𝑡( , ) =4𝜋𝜋𝜀𝜀0𝑟𝑟( ) 4𝜋𝜋𝜀𝜀0 𝜌𝜌 +(𝑧𝑧 ) ( ) . (64) −1 ( , ) × ( ,𝑬𝑬) 𝒓𝒓 𝑡𝑡 = −𝜵𝜵𝜓𝜓 𝒓𝒓 𝑡𝑡 − 𝜕𝜕𝜌𝜌𝜓𝜓 𝝆𝝆� − 𝜌𝜌 𝜕𝜕 𝜑𝜑𝜓𝜓 𝝋𝝋� − 𝜕𝜕𝑧𝑧𝜓𝜓 𝒛𝒛� 2 −1 ( ) 𝑬𝑬 𝒓𝒓 𝑡𝑡= 𝑯𝑯 𝒓𝒓 𝑡𝑡 ⁄𝑐𝑐 𝜇𝜇0𝜀𝜀0𝐻𝐻0��𝜕𝜕𝜌𝜌𝜓𝜓�𝝋𝝋� − 𝜌𝜌 cos�𝜕𝜕𝜑𝜑𝜓𝜓�𝝆𝝆�+� ( ) ( 2) ( ) 𝜇𝜇0𝐻𝐻0𝑝𝑝0 cos 𝜔𝜔𝜔𝜔 1 3𝜌𝜌 sin 𝜑𝜑 2 2 3⁄2 2 2 5⁄2 2 2 3⁄2 4𝜋𝜋( ) 𝜌𝜌 + 𝑧𝑧 𝜌𝜌 + 𝑧𝑧 𝜌𝜌 + 𝑧𝑧 = �� − cos (�cos 𝜑𝜑 𝝋𝝋�sin ) + 𝝆𝝆�� (cos + sin ) ( ) ( 2) ( ) 𝜇𝜇0𝐻𝐻0𝑝𝑝0 cos 𝜔𝜔𝜔𝜔 1 3𝜌𝜌 sin 𝜑𝜑 2 2 3⁄2 2 2 5⁄2 2 2 3⁄2 4𝜋𝜋 ( ��)𝜌𝜌 + 𝑧𝑧 − 𝜌𝜌 + 𝑧𝑧 � 𝜑𝜑 𝜑𝜑 𝒚𝒚� − 𝜑𝜑 𝒙𝒙� 𝜌𝜌 + 𝑧𝑧 𝜑𝜑 𝒙𝒙� 𝜑𝜑 𝒚𝒚� � = + . (65) ( ) 2 ( ) 2 𝜇𝜇0𝐻𝐻0𝑝𝑝0 cos 𝜔𝜔𝜔𝜔 𝒚𝒚� 3𝜌𝜌 �sin 𝜑𝜑 cos 𝜑𝜑 𝒙𝒙� − cos 𝜑𝜑 𝒚𝒚�� 2 2 3⁄2 2 2 5⁄2 It is now straightforward4𝜋𝜋 � 𝜌𝜌 + 𝑧𝑧 to integrate the 𝜌𝜌EM+ 𝑧𝑧 momentum�-density of Eq.(65) over the volume of the cylinder (excluding the interior of the spherical particle). The integral must be evaluated in two steps, first when ranges from to , and then when goes from 0 to . We will have

( ) 𝑠𝑠 𝑐𝑐 𝑠𝑠 a) 𝜌𝜌 𝑅𝑅 𝑅𝑅 + 𝜌𝜌 𝑅𝑅 ( ) 2 ( ) 2 𝜇𝜇0𝐻𝐻0𝑝𝑝0 cos 𝜔𝜔𝜔𝜔 𝑅𝑅𝑐𝑐 2𝜋𝜋 ∞ 𝒚𝒚� 3𝜌𝜌 �sin 𝜑𝜑 cos 𝜑𝜑 𝒙𝒙� − cos 𝜑𝜑 𝒚𝒚�� 2 2 3⁄2 2 2 5⁄2 4𝜋𝜋 ∫𝜌𝜌=𝑅𝑅𝑠𝑠 ∫𝜑𝜑=0 ∫𝑧𝑧=−∞ � 𝜌𝜌 + 𝑧𝑧 𝜌𝜌 + 𝑧𝑧 � 𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌 = cos( ) See Appendix A ( ) ( 3 ) 𝑅𝑅𝑐𝑐 ∞ 𝜌𝜌 3𝜌𝜌 2 2 3⁄2 2 2 5⁄2 𝜇𝜇0𝐻𝐻0𝑝𝑝0 𝜔𝜔𝜔𝜔 𝒚𝒚� ∫𝜌𝜌=𝑅𝑅𝑠𝑠 ∫𝑧𝑧=0 � 𝜌𝜌 + 𝑧𝑧 − 2 𝜌𝜌 + 𝑧𝑧 � 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

15

= cos( ) = 0. (66a) 𝑅𝑅𝑐𝑐 1 1 0 0 0 𝜇𝜇 𝐻𝐻 𝑝𝑝 𝜔𝜔𝜔𝜔 𝒚𝒚� ∫𝜌𝜌=𝑅𝑅𝑠𝑠 �𝜌𝜌 − 𝜌𝜌� 𝑑𝑑𝑑𝑑 b) cos( ) ½ See Appendix A ( ) ( 3 ) 𝑅𝑅𝑠𝑠 ∞ 𝜌𝜌 3𝜌𝜌 2 2 2 2 3⁄2 2 2 5⁄2 𝜇𝜇0𝐻𝐻0𝑝𝑝0 𝜔𝜔𝜔𝜔 𝒚𝒚� ∫𝜌𝜌=0 ∫𝑧𝑧=�𝑅𝑅𝑠𝑠 −𝜌𝜌 � � 𝜌𝜌 + 𝑧𝑧 − 2 𝜌𝜌 + 𝑧𝑧 � 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= cos( ) 𝑅𝑅𝑠𝑠 1 1 + 2 2 1⁄2 2 2 1⁄2 2 2 3⁄2 1 �𝑅𝑅𝑠𝑠 −𝜌𝜌 � 3 �𝑅𝑅𝑠𝑠 −𝜌𝜌 � 1 �𝑅𝑅𝑠𝑠 −𝜌𝜌 � 3 𝜇𝜇0𝐻𝐻0𝑝𝑝0 𝜔𝜔𝜔𝜔 𝒚𝒚� � �𝜌𝜌 � − 𝑅𝑅𝑠𝑠 � − 2𝜌𝜌 � − 𝑅𝑅𝑠𝑠 − 3 3𝑅𝑅𝑠𝑠 �� 𝑑𝑑𝑑𝑑 = ½ cos( ) 0 𝑅𝑅𝑠𝑠 2 2 3 0 0 0 𝑠𝑠 𝑠𝑠 = ½𝜇𝜇 𝐻𝐻 𝑝𝑝 cos(𝜔𝜔𝜔𝜔) 𝒚𝒚� �0 �𝜌𝜌�1𝑅𝑅 − 𝜌𝜌 �=𝑅𝑅 �𝑑𝑑𝑑𝑑 cos( ) . See Appendix A (66b) 1 2 1 0 0 0 0 0 0 The total𝜇𝜇 EM𝐻𝐻 𝑝𝑝 momentum𝜔𝜔𝜔𝜔 𝒚𝒚� (including�0 𝜂𝜂� − the𝜂𝜂 𝑑𝑑𝑑𝑑 contribution6 𝜇𝜇 𝐻𝐻 𝑝𝑝 from the𝜔𝜔𝜔𝜔 interior𝒚𝒚� region of the spherical dipole) is thus seen to be ( ) = ½ cos( ) , confirming the conservation of overall momentum (i.e., electromagnetic plus mechanical) in the system of Fig.4. 𝓹𝓹𝑒𝑒𝑒𝑒 𝑡𝑡 𝜇𝜇0𝐻𝐻0𝑝𝑝0 𝜔𝜔𝜔𝜔 𝒚𝒚� 6. Energy of a dipole in an external EM field. Our next example concerns the EM energy of an electric dipole in a constant and uniform external electric field. The E-field of a small, uniformly-polarized spherical particle (i.e., spherical dipole) located at the origin of a spherical coordinate system and oriented along the z-axis is given by [4-8] ( ) , ( ) = 𝑝𝑝0 2 cos 𝜃𝜃𝒓𝒓� + sin 𝜃𝜃𝜽𝜽� 3 (67) 4𝜋𝜋𝜀𝜀0, 𝑟𝑟 𝑟𝑟 ≥< 𝑅𝑅. 𝑬𝑬 𝒓𝒓 � 𝑝𝑝0𝒛𝒛� 3 Here is the radius of the spherical− 4𝜋𝜋𝜀𝜀0𝑅𝑅 particle and 𝑟𝑟 is its𝑅𝑅 (electric) dipole moment. In a static external electric field ( )( ), the energy of the dipole is said to be 𝑅𝑅 𝑝𝑝0𝒛𝒛� ext ( ) ( ) 𝑬𝑬= 𝒓𝒓 (0) = (0). (68) ext ( ) ext This is because the torque exerted by (0 )𝑧𝑧 on the dipole oriented in an arbitrary ℰ −(𝒑𝒑 ∙ )𝑬𝑬 −𝑝𝑝 𝐸𝐸 ( ) direction is given by = × (0). Denotingext the between and (0) by , the ( ) magnitude of the torque is readilyext seen to be𝑬𝑬 𝒓𝒓 (0) sin . If now𝒑𝒑 the angleext changes by , the done by 𝝉𝝉the E𝒑𝒑-field𝑬𝑬 on the dipole will extbe 𝒑𝒑 𝑬𝑬 𝜑𝜑 𝑝𝑝0𝐸𝐸 𝜑𝜑 𝜑𝜑 ( )( ) ( )( ) ( )( ) 𝛿𝛿 𝛿𝛿 = 0 sin = 0 cos = 0 . (69) ext ext ext ( ) The𝛿𝛿𝛿𝛿 energy𝑝𝑝0 of𝐸𝐸 the dipole 𝜑𝜑in𝛿𝛿 the𝛿𝛿 external−𝛿𝛿�𝑝𝑝0𝐸𝐸 E-field is thus𝜑𝜑� said− 𝛿𝛿to� 𝒑𝒑be∙ 𝑬𝑬 � (0), with the zero of energy assigned (arbitrarily) to that orientation of the dipole which is extperpendicular to ( )(0). This energy is exchanged between the dipole’s rotational kinetic−𝒑𝒑 energy∙ 𝑬𝑬 and the stored energyext in the E-field in the region of space surrounding the dipole. To see this, note that the energy𝑬𝑬 -density of the E-field is generally given by [4,5]

( ) = ½ | ( )| = ½ ( ) + ( )( ) 2 ext 2 ℰ 𝒓𝒓 = ½𝜀𝜀0|𝑬𝑬(total)| 𝒓𝒓+ ½ 𝜀𝜀(0�𝑬𝑬)(𝒓𝒓) +𝑬𝑬 (𝒓𝒓)� ( )( ). (70) 2 ext 2 ext 𝜀𝜀0 𝑬𝑬 𝒓𝒓 𝜀𝜀0�𝑬𝑬 𝒓𝒓 � 𝜀𝜀0𝑬𝑬 𝒓𝒓 ∙ 𝑬𝑬 𝒓𝒓

16

The first two terms on the right-hand-side of the above equation are independent of the orientation of the dipole. Thus the stored energy in the E-field varies as the integral of the third term, ( ) ( )( ), over the entire space. When the dipoleext is aligned with the z-axis, its E-field given by Eq.(67) maybe equivalently 0 expressed𝜀𝜀 𝑬𝑬 as𝒓𝒓 follows:∙ 𝑬𝑬 𝒓𝒓

( ) = . (71) 𝑝𝑝0 𝜕𝜕 𝒓𝒓 3 This should be obvious, considering𝑬𝑬 𝒓𝒓 that− 4the𝜋𝜋𝜀𝜀0 dipole𝜕𝜕𝜕𝜕 �𝑟𝑟 � consists of a pair of point-charges, ± , separated by a short distance along the z-axis. The E-field produced by a charge located at the origin of the coordinate system is given by /(4 ). Separating the ± charges by𝑞𝑞 a short distance along the z-axis𝑑𝑑 and recalling that = yields3 the dipolar E-field𝑞𝑞 of Eq.(71). 0 Note, however, that Eq.(71) does not exactly reproduce𝑞𝑞𝒓𝒓 𝜋𝜋𝜀𝜀 the𝑟𝑟 internal field of the𝑞𝑞 dipole in the 0 region < 𝑑𝑑as specified in Eq.(67). Nevertheless,𝑝𝑝 in 𝑞𝑞the𝑞𝑞 vicinity of the origin, the general behavior of the E-field of Eq.(71) is consistent with that of the spherical dipole described by Eq.(67).𝑟𝑟 𝑅𝑅 The relevant part of the E-field energy in Eq.(70), the third term on the right-hand-side (i.e., the cross-term), may now be written as follows:

= ( ) ( )( ) = ( )( ) ∞ ext 𝑝𝑝0 ∞ 𝜕𝜕 𝒓𝒓 ext 0 ( )( ) 3 ∆ℰ = ∭−∞ 𝜀𝜀 𝑬𝑬 𝒓𝒓 ∙ 𝑬𝑬 𝒓𝒓 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 − 4𝜋𝜋 ∭−∞ 𝜕𝜕𝜕𝜕 �𝑟𝑟 � ∙ 𝑬𝑬 𝒓𝒓 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ext . (72) 𝑝𝑝0 ∞ 𝒓𝒓 𝜕𝜕𝑬𝑬 𝒓𝒓 3 The final step4𝜋𝜋 ∭ in−∞ the𝑟𝑟 ∙above𝜕𝜕𝜕𝜕 derivation𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 involves an integration by parts followed by the assumption that the external E-field at ± is sufficiently weak to be negligible. We now invoke ( ) ( ) the static nature of ( )( ) and the fact that × ( )( ) = 0 to replace by ( ) ( ext) ∞ ext ext ext and by . We will have 𝑧𝑧 𝑦𝑦 𝑦𝑦 𝑧𝑧 ext 𝑬𝑬ext 𝒓𝒓 𝜵𝜵 𝑬𝑬 𝒓𝒓 𝜕𝜕 𝐸𝐸 𝜕𝜕 𝐸𝐸 ( ) ( ) ( ) 𝜕𝜕𝑧𝑧𝐸𝐸𝑥𝑥 𝜕𝜕𝑥𝑥𝐸𝐸=𝑧𝑧 + + 𝑝𝑝0 ∞ 𝒓𝒓 ext ext ext 3 𝑥𝑥 ( ) 𝑦𝑦 𝑧𝑧 ∆ℰ = 4𝜋𝜋 ∭−∞ 𝑟𝑟 ∙ �𝜕𝜕 𝐸𝐸𝑧𝑧 𝒙𝒙� 𝜕𝜕( 𝐸𝐸)𝑧𝑧 𝒚𝒚� 𝜕𝜕 𝐸𝐸𝑧𝑧 𝒛𝒛��𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑝𝑝0 ∞ 𝒓𝒓 ext 3 ( ) ( ) = − 4𝜋𝜋 ∭−∞ �4𝜵𝜵 ∙ (�𝑟𝑟)�� 𝐸𝐸𝑧𝑧 ( )𝒓𝒓 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑= (0). (73) 𝑝𝑝0 ∞ ext ext − 4𝜋𝜋 ∭−∞ 𝜋𝜋𝜋𝜋 𝒓𝒓 𝐸𝐸𝑧𝑧 𝒓𝒓 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 −𝑝𝑝0𝐸𝐸𝑧𝑧 z − − − − − − − − − − − − − − − − −

Fig.5. A small spherical dipole = is placed in the uniform E-field between two uniformly- p 0 E0 charged parallel plates. The EM𝒑𝒑 energy𝑝𝑝 𝒛𝒛� stored in the E-field contains a contribution from the cross- ( ) ( ) y + + + + + + + + + + + + + + + + + + term ( ) ( ), where ( ) = . x ext ext 0 0 𝜀𝜀Thus,𝑬𝑬 𝒓𝒓 ∙when𝑬𝑬 the𝒓𝒓 dipole𝑬𝑬 is anti𝒓𝒓 -parallel𝐸𝐸 𝒛𝒛� to ( )(0), the energy stored in the E-field is at its peak value. Allowing the dipole to rotate will enableext the torque to bring it into alignment with the local external field. In the process the dipole𝑬𝑬 acquires rotational while the energy stored in the E-field is reduced by an equal amount. A simple scenario involving a dipole placed between two uniformly-charged, infinitely large, parallel plates, as shown in Fig.5,

𝒑𝒑 17

would be instructive. It is easy to verify by direct integration that, when = and the dipolar E-field ( ) is given by Eq.(67) while ( )( ) = in the region between the parallel plates, ( ) 0 the cross-term ( ) ( ) appearingext in Eq.(70) would integrate to𝒑𝒑 𝑝𝑝 𝒛𝒛�. 0 𝑬𝑬 𝒓𝒓 ext 𝑬𝑬 𝒓𝒓 𝐸𝐸 𝒛𝒛� 6.1. The case𝜀𝜀 0of𝑬𝑬 a𝒓𝒓 relativistically∙ 𝑬𝑬 𝒓𝒓 -induced electric dipole. Next, with−𝑝𝑝 0reference𝐸𝐸0 to Fig.6, consider a small, uniformly-magnetized, spherical particle (i.e., a magnetic dipole) located at the origin of the coordinate system, with its magnetization aligned with the -axis. The magnetic field of the dipole is given by ′ 𝑥𝑥′𝑦𝑦′𝑧𝑧′ ( ) 𝑧𝑧 , ′ ′ 0 � ( ) = 𝑚𝑚 2 cos 𝜃𝜃 𝒓𝒓� + sin 𝜃𝜃 𝜽𝜽 ′ (74) ′3 ′ 4𝜋𝜋𝜇𝜇, 0 𝑟𝑟 𝑟𝑟 ≥< 𝑅𝑅. 𝑯𝑯 𝒓𝒓 � 𝑚𝑚0𝒛𝒛� ′ 3 − 4𝜋𝜋𝜇𝜇0𝑅𝑅 𝑟𝑟 𝑅𝑅 z z′

Fig.6. A small spherical magnetic dipole = m V moves with constant velocity along the y-axis of a y′ 0 y Cartesian coordinate system. The rest-frame𝒎𝒎 of 𝑚𝑚the𝒛𝒛� particle is identified as 𝑉𝑉. x′ ′ ′ ′ x Note that, in order𝑥𝑥 𝑦𝑦 𝑧𝑧to maintain complete symmetry with the case of an electric dipole, we are defining the magnetic dipole moment and its corresponding magnetization = 3 /(4 ) such that, in the SI system of units, = + . In this way, the magnetic 0 dipole moment3 of a small loop of area carrying𝑚𝑚 𝒛𝒛� a constant in the -plane𝑴𝑴 0 0 will𝑚𝑚 be𝒛𝒛� 𝜋𝜋𝑅𝑅= . 𝑩𝑩 𝜇𝜇 𝑯𝑯 𝑴𝑴 Suppose now that the frame 𝐴𝐴moves with constant velocity along𝐼𝐼 the y-axis.𝑥𝑥𝑥𝑥 From 0 0 the perspective𝒎𝒎 𝜇𝜇 of𝐼𝐼𝐼𝐼 an𝒛𝒛� observer in the stationary frame (i.e., the lab frame), the magnetic dipole will be accompanied𝑥𝑥′𝑦𝑦′𝑧𝑧′ by a relativistically-induced electric dipole𝑉𝑉 = . The E-field in the frame is readily obtained from a Lorentz𝑥𝑥𝑥𝑥𝑥𝑥 transformation of the B-field in the 0 0 0 frame,𝑚𝑚 𝒛𝒛� as follows: 𝒑𝒑 𝜀𝜀 𝑉𝑉𝑚𝑚 𝒙𝒙� 𝑥𝑥𝑥𝑥𝑥𝑥 + = [( + 2 ) + 3 ], 𝑥𝑥′𝑦𝑦′𝑧𝑧′ ( ) ( , = 0) = 𝛾𝛾𝛾𝛾𝑚𝑚0 2 2 2 2 2 2 2 2 5⁄2 (75) −𝛾𝛾𝛾𝛾𝐵𝐵𝑧𝑧𝒙𝒙� = 𝛾𝛾𝛾𝛾𝐵𝐵𝑥𝑥𝒛𝒛� , 4 𝜋𝜋 𝑥𝑥 + 𝛾𝛾 𝑦𝑦 + 𝑧𝑧 𝑥𝑥 𝛾𝛾 𝑦𝑦 − 𝑧𝑧 𝒙𝒙� 𝑥𝑥 𝑥𝑥 𝒛𝒛 � 𝑟𝑟′ <≥ 𝑅𝑅. 𝑬𝑬 𝒓𝒓 𝑡𝑡 � 𝛾𝛾𝛾𝛾𝑚𝑚0 ′ 3 Clearly, this−𝛾𝛾 field𝛾𝛾𝐵𝐵𝑧𝑧𝒙𝒙� is very− 2𝜋𝜋 𝑅𝑅different𝒙𝒙� from the E-field produced by an ordinary𝑟𝑟 electric𝑅𝑅 dipole = placed at the origin of the coordinate system. In the presence of a static external E-field ( )( ), the E-field energy associated with the third term on the right-hand- 0 0 side𝒑𝒑 of𝜀𝜀 𝑉𝑉Eq.(𝑚𝑚 70𝒙𝒙� ) willext be 𝑥𝑥𝑥𝑥𝑥𝑥 𝑬𝑬 𝒓𝒓 ( = 0) = ( , , , = 0) ( )( , , ) ∞ ext ( ) ∆ℰ 𝑡𝑡 = ∭−∞ 𝜀𝜀0𝑬𝑬 𝑥𝑥 𝑦𝑦( 𝑧𝑧, 𝑡𝑡 , ,∙ 𝑬𝑬= 0) 𝑥𝑥 𝑦𝑦 𝑧𝑧 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑, , ) . (76) −1 ∞ −1 ext −1 It is readily seen from Eq.(𝜀𝜀075𝛾𝛾 ) that,∭− ∞in𝑬𝑬 the𝑥𝑥 region𝛾𝛾 𝑦𝑦 𝑧𝑧 𝑡𝑡 , the∙ 𝑬𝑬 dipolar𝑥𝑥 E𝛾𝛾-field𝑦𝑦 𝑧𝑧 may𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 be expressed as ′ ( , , , = 0) =𝑟𝑟 ≥ 𝑅𝑅 . (77) −1 𝛾𝛾𝛾𝛾𝑚𝑚0 𝜕𝜕 𝑧𝑧𝒙𝒙�−𝑥𝑥𝒛𝒛� 3 𝑬𝑬 𝑥𝑥 𝛾𝛾 𝑦𝑦 𝑧𝑧 𝑡𝑡 4𝜋𝜋 𝜕𝜕𝜕𝜕 � 𝑟𝑟 �

18

The above expression, however, does not reproduce the correct behavior for the internal E- field of the dipole (i.e., in the region < ), as given by Eq.(75). The reason being that ( ) is an appropriate representation ′for the H-field of the magnetic dipole, whereas the E- field of3 the induced electric dipole given𝑟𝑟 by𝑅𝑅 Eq.(75) is intimately tied to the B-field of the 𝑧𝑧 magnetic𝜕𝜕 𝒓𝒓⁄𝑟𝑟 dipole. We thus proceed to evaluate ( = 0) of Eq.(76) by ignoring the difference between the E-field of Eq.(77) and the true internal field of the dipole. The result will subsequently be adjusted to accommodate the∆ℰ correction𝑡𝑡 to the internal field. Substituting Eq.(77) into Eq.(76), we find

( = 0) = ( )( , , ) 𝜀𝜀0𝑉𝑉𝑚𝑚0 ∞ 𝜕𝜕 𝑧𝑧𝒙𝒙�−𝑥𝑥𝒛𝒛� ext −1 3 ( ) ( ) ∆ℰ 𝑡𝑡 4𝜋𝜋 ∭−∞ 𝜕𝜕𝜕𝜕 � 𝑟𝑟 � ∙ 𝑬𝑬 𝑥𝑥 𝛾𝛾 𝑦𝑦 𝑧𝑧 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = ext + ext 𝜀𝜀0𝑉𝑉𝑚𝑚0 ∞ 𝑧𝑧𝒙𝒙�−𝑥𝑥𝒛𝒛� 𝜕𝜕𝐸𝐸𝑥𝑥 𝜕𝜕𝐸𝐸𝑧𝑧 3 − 4𝜋𝜋 ∭−∞ � 𝑟𝑟 � ∙ � 𝜕𝜕(𝜕𝜕 ) 𝒙𝒙� 𝜕𝜕(𝜕𝜕 ) 𝒛𝒛�� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = ext + ext 𝜀𝜀0𝑉𝑉𝑚𝑚0 ∞ 𝑧𝑧𝒙𝒙�−𝑥𝑥𝒛𝒛� 𝜕𝜕𝐸𝐸𝑧𝑧 𝜕𝜕𝐸𝐸𝑧𝑧 3 ( ) = − 4𝜋𝜋 ∭−∞ � 𝑟𝑟 � ∙ � 𝜕𝜕𝜕𝜕 𝒙𝒙� (𝜕𝜕𝜕𝜕, 𝒛𝒛�� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑, ) = 0. (78) 𝜀𝜀0𝑉𝑉𝑚𝑚0 ∞ 𝜕𝜕 𝑧𝑧 𝜕𝜕 𝑥𝑥 ext −1 3 3 We must now add4𝜋𝜋 to ∭the− ∞above�𝜕𝜕𝜕𝜕 � result𝑟𝑟 � − the𝜕𝜕𝜕𝜕 � correction𝑟𝑟 �� 𝐸𝐸𝑧𝑧 due𝑥𝑥 𝛾𝛾 to the𝑦𝑦 𝑧𝑧 internal𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 E-field of the induced electric dipole = . This correction arises from the difference between the internal B- field of the magnetic dipole and its corresponding -field, which is just the magnetization 0 0 of the dipole. In𝒑𝒑 the 𝜀𝜀spherical𝑉𝑉𝑚𝑚 𝒙𝒙� dipole model, the correction amounts to an additional = 0 in the region < , whose contribution to ( =𝜇𝜇0)𝐻𝐻 is readily seen to be ( )(0). In the𝑀𝑀 end, we recover′ the standard formula for the energy of an electric dipole immersedext𝑬𝑬 in− a𝛾𝛾 static𝛾𝛾𝛾𝛾𝒙𝒙� external E-field,𝑟𝑟 even𝑅𝑅 though the field distribution∆ℰ 𝑡𝑡 of a relativistically-induced−𝒑𝒑 ∙ 𝑬𝑬 electric dipole turned out to be quite different from that of an ordinary electric dipole.

6.2. Magnetic dipole in an external magnetic field. It is instructive to examine the similarities and differences between an electric dipole immersed in an external electric field, and a magnetic dipole immersed in an external magnetic field. One major difference is that, while in the preceding derivations involving a static E-field, we could use the identity ( ) z × ( ) = 0, the corresponding identity cannot be invoked for ( ) magneticext fields, where × ( ) = ( ). As a simple example, + + + + 𝜵𝜵consider𝑬𝑬 a small,𝒓𝒓 uniformly-magnetizedext spherical particle (i.e., a magnetic free + + dipole) placed inside a pair𝜵𝜵 of𝑯𝑯 infinitely𝒓𝒓 -long,𝑱𝑱 uniformly𝒓𝒓 -charged + + − − − − concentric cylinders whose radii are nearly equal and have equal but − − opposite surface-charge-densities, as shown in Fig.7. − − − Fig.7. A small spherical magnetic dipole = inside a pair of concentric, − infinitely-long, uniformly-charged cylinders. The electric charge-densities of the two − − − − 0 cylinders are equal in magnitude and opposite𝒎𝒎 in𝑚𝑚 sign.𝒛𝒛� When the cylinders are spun in opposite directions, a uniform magnetic field ( ) = is created inside the cylinders.

0 The cylinders are initially at rest,𝑯𝑯 𝒓𝒓but, 𝐻𝐻at𝒛𝒛� = 0, they begin to rotate slowly in opposite directions. A uniform magnetic field thus builds up inside the cylinders, immersing the magnetic dipole in an external magnetic field𝑡𝑡 . Let us first analyze the system in accordance 0 with the Einstein-Laub formulation. Here𝐻𝐻 𝒛𝒛� no energy is exchanged between the dipole and the 𝑚𝑚0𝒛𝒛�

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cylinders as the latter spin up. Moreover, the H-field inside the cylinders, being the sum of the dipole field and the uniform field set up by the rotating cylinders, does not make a net contribution to the energy (via a cross-term) upon integrating the H-field energy-density over the interior volume of the cylinders. In other words, given the dipolar field ( ) , ( ) = 𝑚𝑚0 2 cos 𝜃𝜃𝒓𝒓� + sin 𝜃𝜃𝜽𝜽� 3 (79) 4𝜋𝜋𝜇𝜇0, 𝑟𝑟 𝑟𝑟 <≥ 𝑅𝑅, 𝑯𝑯 𝒓𝒓 � 𝑚𝑚0𝒛𝒛� 3 the total energy-density inside the cylinders− 4𝜋𝜋𝜇𝜇0 𝑅𝑅will be 𝑟𝑟 𝑅𝑅 ( ) = ½ | ( )| = ½ ( )( ) + ( ) 2 ext 2 ℰ 𝒓𝒓 = ½𝜇𝜇0 𝑯𝑯total+ ½𝒓𝒓 | ( )|𝜇𝜇0+�𝑯𝑯 𝒓𝒓 ( ).𝑯𝑯 𝒓𝒓 � (80) 2 2 Now, the third term on the right𝜇𝜇0𝐻𝐻0-hand-𝜇𝜇side0 𝑯𝑯 of𝒓𝒓 Eq.(80)𝜇𝜇0 𝐻𝐻integrates0𝐻𝐻𝑧𝑧 𝒓𝒓 to zero over the volume of the cylinder. An easy way to see this is to note that, for a point-dipole , the -field is

( ) = , 𝑚𝑚0𝒛𝒛� 𝐻𝐻 (81) 𝑚𝑚0 𝜕𝜕 𝒓𝒓 3 whose integral over from to +𝑯𝑯 𝒓𝒓 ends− up4𝜋𝜋 being𝜇𝜇0 𝜕𝜕𝑧𝑧 � zero.𝑟𝑟 � The question remains, however, as to what happens to the dipole’s energy in the external field , because surely there will be a torque acting on the 𝑧𝑧dipole −if∞ it is turned∞ away from the z-axis, and the action of this torque on 0 the rotating dipole should change its rotational kinetic energy𝐻𝐻 𝒛𝒛� in exactly the same way that an external E-field exchanges energy with a rotating electric dipole. To answer the above question, imagine the magnetic dipole losing its magnetization, say, as a result of warming up slowly to beyond its Curie . The changing magnetization produces an E-field in the surrounding space in accordance with × = , which acts on the rotating cylinders to speed them up, if the initial magnetization is pointing up — or to slow 𝑡𝑡 them down, if the initial magnetization is downward. The kinetic𝜵𝜵 energy𝑬𝑬 −thus𝜕𝜕 𝑩𝑩 imparted to the cylinders can be shown to be exactly equal to ± . The energy of a magnetic dipole in an external H-field is thus seen to be = ( )(0), as expected from an argument based on 0 0 the torque exerted by the external H-field on theext dipole.𝑚𝑚 𝐻𝐻 In contrast to the case of an external E- 0 field acting on an electric dipole, however,ℰ −𝒎𝒎 the∙ 𝑯𝑯 energy associated with a magnetic dipole is not exchanged with the surrounding H-field but, rather, is exchanged with the source of the external H-field, which, in the present example, is the rotating pair of cylinders. In the Lorentz formulation, the magnetic field energy-density is ( , ) = ½ | ( , )| . In ( ) this case the stored energy-density in the cross-term, namely, ( ) ( )−,1 integrates2 to 0 . However, the bound current-density associated with the magnetic−1 ℰext𝒓𝒓 𝑡𝑡dipole,𝜇𝜇 i.e.,𝑩𝑩 𝒓𝒓 𝑡𝑡 × , 0 now exchanges energy with the E-field produced by the cylinders𝜇𝜇 𝑩𝑩as they𝒓𝒓 spin∙ 𝑩𝑩 𝒓𝒓up from−1 rest to 0 0 0 𝑚𝑚their𝐻𝐻 final . (Note: In the Einstein-Laub formulation, the bound current𝜇𝜇 𝜵𝜵of the𝑴𝑴 magnetic dipole does not participate in the exchange of energy with the fields.) The energy thus extracted from the bound current during the spin-up process turns out to be exactly equal to . Therefore, no net energy is given by the spinning cylinders to the dipole and its B-field during spin-up. Once again, any energy exchanged in consequence of a rotation of the magnetic 0 0 𝑚𝑚dipole𝐻𝐻 will be between the dipole and the spinning cylinders.

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7. Concluding remarks. While the conservation laws of EM energy and linear and angular momenta, being direct consequences of the energy-momentum stress tensor formulation(s) of classical electrodynamics [1], are universal, their verification in specific situations may or may not be straightforward. In this paper, we have examined a few non-trivial EM systems which are sufficiently simple to be amenable to exact solutions of Maxwell’s equations in conjunction with the electrodynamic laws of force, torque, energy, and momentum. It is well known that a light pulse reflecting from a mirror or passing through an absorptive or transparent object exchanges its EM energy and momentum with the material medium. The source of the radiation does not participate in these exchanges, since the light pulse, once detached from its source, is essentially autonomous. The same, however, cannot be said about exchanges that take place between static (or quasi-static) fields and material media. As the examples of the preceding sections have amply demonstrated, the (undetached) sources of the EM fields in such situations could actively participate in the exchange of energy and momentum. When magnetic materials interact with an external electric field, the standard (i.e., Lorentz) formulation of classical electrodynamics could deviate from an alternative formulation of force, torque, energy and momentum first proposed by A. Einstein and J. Laub in 1908 [33]. In such cases we have analyzed the system under consideration from both perspectives, so that the reader might appreciate the similarities and differences of the two formulations. There exist other formulations of classical electrodynamics (e.g., Chu, Minkowski, Abraham), each built upon the foundation of Maxwell’s macroscopic equations, but deviating from the standard formulation in their structure of the corresponding energy-momentum stress tensor [34]. We have chosen to stay away from these alternative formulations, as we believe them to be phenomenological rather than rooted in realistic microscopic models of interaction between matter and EM fields.

Appendix A Below we list (or evaluate) the various integrals and identities used throughout the paper.

( ) + ( ) = 2( / ) ( ) (Gradshteyn & Ryzhik* 8.471-1) (A1)

𝑚𝑚−1 𝑚𝑚+1 𝑚𝑚 𝐽𝐽 (𝑥𝑥) 𝐽𝐽 (𝑥𝑥) = 2 𝑚𝑚( 𝑥𝑥) 𝐽𝐽 𝑥𝑥 (Gradshteyn & Ryzhik 8.471-2) (A2) ′ 𝑚𝑚−1 𝑚𝑚+1 𝑚𝑚 𝐽𝐽( / )𝑥𝑥 −( 𝐽𝐽) 𝑥𝑥 ( ) =𝐽𝐽 𝑥𝑥 ( ) (Gradshteyn & Ryzhik 8.472-2) (A3) ′ 𝑚𝑚 𝑚𝑚 𝑚𝑚+1 𝑚𝑚 𝑥𝑥( 𝐽𝐽) 𝑥𝑥 =−½𝐽𝐽 𝑥𝑥[ ( 𝐽𝐽) 𝑥𝑥 ( ) ( )] (Gradshteyn & Ryzhik 5.54-2) (A4) 2 2 2 ∫ 𝑥𝑥𝐽𝐽𝑚𝑚 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑥𝑥 𝐽𝐽𝑚𝑚 𝑥𝑥 − 𝐽𝐽𝑚𝑚−1 𝑥𝑥 𝐽𝐽𝑚𝑚+1 𝑥𝑥 ( ) ( ) = ½ ( ) ( ) = ½ ( ) ( ). (A5) 0 0 𝑥𝑥 2 ′ 2 2 𝑥𝑥 2 2 0 𝑚𝑚 𝑚𝑚 0 𝑚𝑚 0 0 𝑚𝑚 0 𝑚𝑚−1 0 𝑚𝑚+1 0 ∫ 𝑥𝑥 𝐽𝐽( /𝑥𝑥 )𝐽𝐽 𝑥𝑥(𝑑𝑑𝑑𝑑) + 𝑥𝑥( 𝐽𝐽) 𝑥𝑥 − ∫ 𝑥𝑥𝐽𝐽 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑥𝑥 𝐽𝐽 𝑥𝑥 𝐽𝐽 𝑥𝑥 0 𝑥𝑥 2 2 ′ 2 ∫0 𝑥𝑥� 𝑚𝑚 𝑥𝑥 =𝐽𝐽𝑚𝑚 𝑥𝑥{ [( 𝐽𝐽𝑚𝑚/ )𝑥𝑥 �(𝑑𝑑𝑑𝑑) ( )] + 2 ( ) ( )} 0 𝑥𝑥 ′ 2 ′ = ∫0 [𝑥𝑥 𝑚𝑚 𝑥𝑥( )𝐽𝐽𝑚𝑚+ 2𝑥𝑥 − 𝐽𝐽(𝑚𝑚 )𝑥𝑥 ( )] 𝑚𝑚 𝐽𝐽𝑚𝑚 𝑥𝑥 𝐽𝐽𝑚𝑚 𝑥𝑥 𝑑𝑑𝑑𝑑 0 𝑥𝑥 2 ′ = ½∫0 𝑥𝑥{ 𝐽𝐽𝑚𝑚+1(𝑥𝑥 ) 𝑚𝑚𝐽𝐽(𝑚𝑚 )𝑥𝑥 𝐽𝐽𝑚𝑚 (𝑥𝑥 )𝑑𝑑𝑑𝑑+ 2 [ ( )/ ] }. (A6) 2 2 2 0 𝑚𝑚+1 0 𝑚𝑚 0 𝑚𝑚+2 0 𝑚𝑚 0 0 *I. S. Gradshteyn and I.𝑥𝑥M. Ryzhik,𝐽𝐽 𝑥𝑥“Table− of𝐽𝐽 Integrals,𝑥𝑥 𝐽𝐽 Series,𝑥𝑥 and Products𝑚𝑚 𝐽𝐽 𝑥𝑥,” 7th 𝑥𝑥Edition, Academic Press (2007).

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[ ( ) ( ) 2 ( ) ( )] 𝑥𝑥0 2 2 ′ ∫0 𝑥𝑥𝐽𝐽=𝑚𝑚 ½𝑥𝑥 −{ 𝑥𝑥𝐽𝐽𝑚𝑚(+1)𝑥𝑥 − 𝑚𝑚( 𝐽𝐽𝑚𝑚) 𝑥𝑥 𝐽𝐽𝑚𝑚( 𝑥𝑥) 𝑑𝑑𝑑𝑑 ( ) + ( ) ( ) 2 [ ( )/ ] } 2{ 2 ( )[ ( ) ( )] 2 ( )[ ( ) ( )] [ ( ) 2 ] } = ½𝑥𝑥0 𝐽𝐽𝑚𝑚 𝑥𝑥0 − 𝐽𝐽𝑚𝑚−1 +𝑥𝑥0 𝐽𝐽𝑚𝑚+1 𝑥𝑥0 − 𝐽𝐽𝑚𝑚+1 𝑥𝑥0 𝐽𝐽𝑚𝑚 𝑥𝑥0 𝐽𝐽+𝑚𝑚+2 𝑥𝑥0 − 𝑚𝑚2𝐽𝐽𝑚𝑚 𝑥𝑥0 𝑥𝑥0/ {(2 ) ( ) ( ) ( ) ( ) [ ( ) ] } 2 = 𝑥𝑥0 𝐽𝐽𝑚𝑚+ 1𝑥𝑥0 𝐽𝐽𝑚𝑚 𝑥𝑥0 𝐽𝐽𝑚𝑚+2/ 𝑥𝑥0 − 𝐽𝐽𝑚𝑚+1 𝑥𝑥0 𝐽𝐽𝑚𝑚−1 /𝑥𝑥0 𝐽𝐽𝑚𝑚+1 𝑥𝑥0 /− 𝑚𝑚 𝐽𝐽𝑚𝑚 𝑥𝑥0 𝑥𝑥0 2{ ( ) ( ) [ ( ) ] } 2 = 𝑥𝑥0 𝑚𝑚 𝐽𝐽𝑚𝑚 𝑥𝑥0 𝐽𝐽𝑚𝑚/+1 𝑥𝑥0 𝑥𝑥0 − 𝑚𝑚𝐽𝐽/𝑚𝑚+1 𝑥𝑥 0 𝐽𝐽𝑚𝑚 𝑥𝑥0 𝑥𝑥0 − 𝑚𝑚 𝐽𝐽𝑚𝑚 𝑥𝑥0 𝑥𝑥0 [ 2 ( ) ( )] ( ) 2 = 𝑥𝑥0 𝐽𝐽𝑚𝑚 𝑥𝑥0 𝐽𝐽𝑚𝑚+1 𝑥𝑥0 𝑥𝑥0 − 𝑚𝑚 𝐽𝐽𝑚𝑚 𝑥𝑥. 0 𝑥𝑥0 (A7) 𝑥𝑥0 𝐽𝐽𝑚𝑚+1 𝑥𝑥0 − 𝑚𝑚𝐽𝐽𝑚𝑚 𝑥𝑥0 𝐽𝐽𝑚𝑚 𝑥𝑥0 = ln + + (Gradshteyn & Ryzhik 2.261) (A8) 𝑑𝑑𝑑𝑑 2 2 � √𝑎𝑎+𝑥𝑥 �√𝑎𝑎 𝑥𝑥 𝑥𝑥� = (Gradshteyn & Ryzhik 2.271-5) ( ) (A9) 𝑑𝑑𝑑𝑑 𝑥𝑥 2 3⁄2 2 � 𝑎𝑎+𝑥𝑥 𝑎𝑎√𝑎𝑎+𝑥𝑥 = (Gradshteyn & Ryzhik 2.271-6) (A10) ( ) ( 3 ) 𝑑𝑑𝑑𝑑 1 𝑥𝑥 𝑥𝑥 2 5⁄2 2 2 2 3⁄2 � 𝑎𝑎+𝑥𝑥 𝑎𝑎 �√𝑎𝑎+𝑥𝑥 − 3 𝑎𝑎+𝑥𝑥 � 1 = (1 ) = (Gradshteyn & Ryzhik 2.262-2) (A11) 1 1 2 2 3⁄2 �0 𝑥𝑥√ − 𝑥𝑥 𝑑𝑑𝑑𝑑 −⅓ − 𝑥𝑥 �0 ⅓ , ( < 1) cos ln(1 2 cos + ) = (Gradshteyn & Ryzhik 4.397-6) (A12) 2 𝜋𝜋 , ( > 1) 2 −𝜋𝜋𝜋𝜋 𝑎𝑎 ∫0 𝑥𝑥 − 𝑎𝑎 𝑥𝑥 𝑎𝑎 𝑑𝑑𝑑𝑑 � 2 ⁄ Appendix− 𝜋𝜋 𝑎𝑎 𝑎𝑎 B An alternative method of computing the Livens momentum for the system of Fig.3 begins by expressing the bound current-density associated with the magnetization distribution as ( , ) = × ( , ), followed by invoking the Biot-Savart law of magnetostatics, as follows: −1 𝑱𝑱bound 𝒓𝒓 𝑡𝑡 𝜇𝜇0 𝜵𝜵 𝑴𝑴 𝒓𝒓 𝑡𝑡

( , ) = ( , ) × . (B1) ∞ | ′| 𝜇𝜇0 ′ 𝒓𝒓 − 𝒓𝒓 ′ ′ ′ ′ 3 4𝜋𝜋 bound 𝒓𝒓 − 𝒓𝒓 Note that the Bio𝑩𝑩-Savart𝒓𝒓 𝑡𝑡 law �ignores−∞ 𝑱𝑱 the time𝒓𝒓 -dependence𝑡𝑡 𝑑𝑑of𝑥𝑥 𝑑𝑑𝑦𝑦( )𝑑𝑑 𝑧𝑧by ignoring the effects of retardation. Integration over the volume of space between the parallel plates of Fig.3 then yields ( ) 𝒎𝒎 𝑡𝑡 ( ) = ( , ) × ( , ) × ( × ) = ( ) ( ) Livens ∞ 𝑒𝑒𝑒𝑒 0 −∞ 𝒑𝒑 = 𝑡𝑡 𝜀𝜀 ∭ 𝑬𝑬 𝒓𝒓 𝑡𝑡 𝑩𝑩 𝒓𝒓 𝑡𝑡×𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑[ × ( , )𝑨𝑨] × 𝑩𝑩 𝑪𝑪 𝑨𝑨 ∙ 𝑪𝑪 𝑩𝑩 − 𝑨𝑨 ∙ 𝑩𝑩 𝑪𝑪 ∞ | ′| 𝜎𝜎0 +𝑑𝑑⁄2 ∞ ∞ ′ 𝒓𝒓 − 𝒓𝒓 ′ ′ ′ ′ 3 4𝜋𝜋 𝑥𝑥=−𝑑𝑑⁄2 𝑦𝑦=−∞ 𝑧𝑧=−∞ 𝒓𝒓 − 𝒓𝒓 �−∞ ∫ ∫ ∫ 𝒙𝒙� � 𝜵𝜵 𝑴𝑴 𝒓𝒓 𝑡𝑡 � 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧 = [ × ( , )] ∞ | ′| 𝜎𝜎0 ′ +𝑑𝑑⁄2 ∞ ∞ 𝑥𝑥 − 𝑥𝑥 ′ ′ ′ ′ 3 4𝜋𝜋 𝑥𝑥=−𝑑𝑑⁄2 𝑦𝑦=−∞ 𝑧𝑧=−∞ 𝒓𝒓 − 𝒓𝒓 �−∞ 𝜵𝜵 𝑴𝑴 𝒓𝒓 𝑡𝑡 ∫ ∫ ∫ 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧 [ × ( , )] ∞ | ′| 𝜎𝜎0 ′ +𝑑𝑑⁄2 ∞ ∞ 𝒓𝒓 − 𝒓𝒓 ′ ′ ′ ′ 3 4𝜋𝜋 𝑥𝑥=−𝑑𝑑⁄2 𝑦𝑦=−∞ 𝑧𝑧=−∞ 𝒓𝒓 − 𝒓𝒓 − �−∞ 𝒙𝒙� ∙ 𝜵𝜵 𝑴𝑴 𝒓𝒓 𝑡𝑡 ∫ ∫ ( ∫ ) 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧 = [ × ( , )] ∞ [( ) ′ ] 𝜎𝜎0 ′ +𝑑𝑑⁄2 ∞ 𝑥𝑥 − 𝑥𝑥 𝜌𝜌 ′ ′ ′ ′ 2 2 3⁄2 2 𝑥𝑥=−𝑑𝑑⁄2 𝜌𝜌=0 𝑥𝑥 − 𝑥𝑥 + 𝜌𝜌 �−∞ 𝜵𝜵 𝑴𝑴 𝒓𝒓 𝑡𝑡 ∫ ∫ 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧 22

( ) [ × ( , )] ∞ [( ) ′ ] 𝜎𝜎0 ′ +𝑑𝑑⁄2 ∞ 𝑥𝑥 − 𝑥𝑥 𝜌𝜌 ′ ′ ′ ′ 2 2 3⁄2 Terms containing 2 x component of𝑥𝑥 curl𝑥𝑥= −𝑑𝑑⁄2 𝜌𝜌=0 𝑥𝑥−𝑥𝑥 + 𝜌𝜌 − �−∞ 𝜵𝜵 𝑴𝑴 𝒓𝒓 𝑡𝑡 ∫ ∫ 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 𝑑𝑑 and𝑦𝑦 𝑑𝑑 𝑧𝑧 vanish ( ) = [ × ( , )] , ∞ ( ′ ) 𝒚𝒚� 𝒛𝒛� 𝜎𝜎0 ′ +𝑑𝑑⁄2 𝑥𝑥 − 𝑥𝑥 ′ ′ ′ y,z components𝑦𝑦 𝑧𝑧of curl ′ 2 2 𝑥𝑥=−𝑑𝑑⁄2 � 𝑥𝑥 − 𝑥𝑥 �−∞ 𝜵𝜵 𝑴𝑴 𝒓𝒓 𝑡𝑡 ∫ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧 = [ × ( , )] , | | 0 ∞ 𝜎𝜎 ′ ′ +𝑑𝑑⁄2 ′ ′ ′ 2 �−∞ 𝜵𝜵 ( 𝑴𝑴 𝒓𝒓 𝑡𝑡 𝑦𝑦 𝑧𝑧� 𝑥𝑥)− 𝑥𝑥 (𝑥𝑥=−𝑑𝑑⁄2�𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑)𝑧𝑧 = + Integration by parts ∞ ′ ′ ′ ′ ′ ′ ′ ′ = −𝜎𝜎0 �−∞� 𝜕𝜕𝑧𝑧 𝑀𝑀𝑥𝑥 − 𝜕𝜕𝑥𝑥 𝑀𝑀𝑧𝑧 𝒚𝒚� 𝜕𝜕𝑥𝑥 𝑀𝑀𝑦𝑦=− 𝜕𝜕𝑦𝑦 𝑀𝑀𝑥𝑥 𝒛𝒛��𝑥𝑥 𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧 ∞ ∞ ′ ′ ′ ′ ′ ′ ′ ′ = 𝜎𝜎0 �×−∞ 𝑥𝑥 𝜕𝜕𝑥𝑥 �(𝑀𝑀𝑧𝑧,𝒚𝒚�)− 𝑀𝑀𝑦𝑦𝒛𝒛��𝑑𝑑𝑥𝑥 𝑑𝑑=𝑦𝑦 𝑑𝑑𝑧𝑧 ×−𝜎𝜎(0 )�. −∞�𝑀𝑀𝑧𝑧𝒚𝒚� − 𝑀𝑀𝑦𝑦𝒛𝒛��𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧 (B2) ∞ ′ ′ ′ ′ 0 0 0 The above𝜎𝜎 𝒙𝒙� result∭−∞ is𝑴𝑴 the𝒓𝒓 same𝑡𝑡 𝑑𝑑𝑥𝑥 as𝑑𝑑 𝑦𝑦that𝑑𝑑𝑧𝑧 used𝜀𝜀 in𝑬𝑬 Eq.(𝒎𝒎53).𝑡𝑡 While the Abraham momentum of the EM field in the system of Fig.3 is equal to zero, the Livens momentum has a contribution arising from the interaction between the external -field and the magnetic dipole moment of the particle.

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