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Home , Lorentz transformation, Rapidity, Thomas precession

An exact derivation of the Thomas rate using the Masud Mansuripur James C. Wyant College of Optical Sciences, The University of Arizona, Tucson, Arizona 85721 [Published in the Proceedings of SPIE 11470, Spintronics XIII, 114703U (2020); doi: 10.1117/12.2569025] Abstract. Using the standard formalism of Lorentz transformation of the special , we derive the exact expression of the rate for an in a classical circular orbit around the nucleus of a hydrogen-like . 1. Introduction. This paper is a follow-up to a recent article1 in which we discussed the origin of the -orbit interaction based on a semi-classical model of hydrogen-like without invoking the Thomas precession2-10 in the rest-frame of the electron. We argued that the Thomas precession leads to a relativistic version of the Coriolis force of Newtonian , which cannot possibly affect the levels of the atom, since it is a that appears only in the rotating frame where the electron is ensconced. In the present paper we derive the exact expression of the Thomas precession rate and show its close connection to the rotating frame phenomenon that gives rise to the Coriolis force. The organization of the paper is as follows. In Sec.2 we describe the formalism of matrices in three-dimensional as well as the associated notion of the generator . The results of Sec.2 are then extended in Sec.3 to derive boost matrices and their generators in Minkowski’s four-dimensional . The 4 × 4 Lorentz transformation matrix for a boost along an arbitrary direction in space is subsequently derived and analyzed in Sec.4. An alternative derivation of the same boost matrix is the subject of Sec.5. In Sec.6, we use a sequence of boost and rotation matrices to derive the general formula for the Lorentz transformation from an inertial frame moving along one direction to a second inertial frame moving with the same along a different direction in space. The final result is subsequently used in Sec.7 to arrive at the desired Thomas precession rate and to explain how this precession supposedly accounts for a missing ½ factor in the spin-orbit coupling energy of hydrogen-like atoms. 2. Rotation in three-dimensional space. Consider the vector = ( , , ) in a three-dimensional Cartesian specified by the unit vectors ( , , ). In a different𝑇𝑇 Cartesian system ( , , ), which shares the same origin with ( , , ) but is rotated𝑉𝑉 𝑥𝑥 relative𝑦𝑦 𝑧𝑧 to it, the same vector may′ be′ expressed′ as = ( , , ) , where 𝒙𝒙� 𝒚𝒚� 𝒛𝒛� 𝒙𝒙� 𝒚𝒚� 𝒛𝒛� ′ ′ ′ ′ 𝑇𝑇 𝒙𝒙� 𝒚𝒚� 𝒛𝒛� 𝑉𝑉 𝑥𝑥 𝑦𝑦 𝑧𝑧 ′ = 11 12 13 . (1) 𝑥𝑥′ 𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑥𝑥 21 22 23 �𝑦𝑦′ � �𝑎𝑎 𝑎𝑎 𝑎𝑎 � �𝑦𝑦� In short, = , with the 3 × 3 matrix31 specify32 33ing a linear transformation from ( , , ) to 𝑧𝑧 𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑧𝑧 the ( , , ) system.′ In a coordinate rotation, the length of must remain intact, that is, = ′ .′ Since′ 𝑉𝑉 the𝐴𝐴 same𝐴𝐴 relation holds for all𝐴𝐴 vectors , we conclude that = . The𝒙𝒙� matrix𝒚𝒚�′𝑇𝑇𝒛𝒛� ′ possessing𝑇𝑇 𝑇𝑇𝒙𝒙� 𝒚𝒚� 𝒛𝒛�this important property is said to be unitary. It is thus𝑉𝑉 seen that the𝑇𝑇 inverse of 𝑉𝑉 is equal𝑉𝑉 to𝑉𝑉 its𝐴𝐴 𝐴𝐴𝐴𝐴 , and since the inverse of a square matrix𝑉𝑉 is unique, we must𝐴𝐴 𝐴𝐴 also𝐼𝐼 have = 𝐴𝐴. The vectors ( ,𝑇𝑇 , ), ( , , ), and ( , , ) thus form the orthonormal𝐴𝐴 bases𝑇𝑇 of the rotated coordinate𝐴𝐴 system. Note from Eq.(1) that is the projection of ( , 𝐴𝐴, 𝐴𝐴) onto𝐼𝐼 11 12 13 21 22 23 31 32 33 ( , , )𝑎𝑎, while𝑎𝑎 𝑎𝑎 is the𝑎𝑎 projection𝑎𝑎 𝑎𝑎 of ( , ,𝑎𝑎) onto𝑎𝑎 (𝑎𝑎 ′ , , ), and is the projection of ( , , ) onto ( , ′, ), thus confirming that the rows𝑥𝑥 of form the ′ -vectors𝑥𝑥 𝑦𝑦 𝑧𝑧 of the 11 12 13 21 22 23 rotated𝑎𝑎 𝑎𝑎 system𝑎𝑎 ( , ,𝑦𝑦 ). 𝑥𝑥 𝑦𝑦 𝑧𝑧 𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑧𝑧 31 32 33 𝑥𝑥 𝑦𝑦 𝑧𝑧 𝑎𝑎 ′ 𝑎𝑎′ ′ 𝑎𝑎 𝐴𝐴 𝒙𝒙� 𝒚𝒚� 𝒛𝒛� 1 A right-handed rotation around the -axis through an angle is represented by the linear transformation = ( ) , where ′ 𝑥𝑥1 0 0 𝜃𝜃 𝑉𝑉 𝑅𝑅𝑥𝑥 𝜃𝜃 𝑉𝑉 ( ) = 0 cos sin . (2) 0 sin cos 𝑅𝑅𝑥𝑥 𝜃𝜃 � � Differentiating ( ) with respect to , we find 𝜃𝜃 𝜃𝜃 − 𝜃𝜃 𝜃𝜃 𝑅𝑅𝑥𝑥 0𝜃𝜃 0 0 𝜃𝜃 0 0 0 1 0 0 ( ) = 0 sin cos = 0 0 +1 0 cos sin = ( ). (3) d 0 cos sin 0 1 0 0 sin cos d𝜃𝜃 𝑅𝑅𝑥𝑥 𝜃𝜃 � − 𝜃𝜃 𝜃𝜃 � � � � 𝜃𝜃 𝜃𝜃� 𝑆𝑆𝑥𝑥𝑅𝑅𝑥𝑥 𝜃𝜃 Here, the 3 × 3 matrix− 𝜃𝜃 is− the 𝜃𝜃so-called generator− of rotations− 𝜃𝜃 around𝜃𝜃 the -axis. The ( ) may thus be written as 𝑆𝑆𝑥𝑥 𝑥𝑥 ( ) = exp( ) 𝑅𝑅𝑥𝑥 𝜃𝜃 . (4) 𝑅𝑅𝑥𝑥 𝜃𝜃 0 0 𝑆𝑆𝑥𝑥𝜃𝜃0 To verify the above identity, note that = 0 1 0 and = , and that, therefore, 2 0 0 1 3 𝑆𝑆𝑥𝑥 � − � 𝑆𝑆𝑥𝑥 −𝑆𝑆𝑥𝑥 exp( ) = + + + +− + 2! 3! 4! 2 𝜃𝜃 3 𝜃𝜃 4 𝜃𝜃 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑆𝑆 𝜃𝜃 = 𝐼𝐼 + 𝑆𝑆 𝜃𝜃 𝑆𝑆+2 𝑆𝑆 3+ 𝑆𝑆 4 + ⋯ + 3! 5! 7! 2! 4! 6! 𝜃𝜃 𝜃𝜃 𝜃𝜃 𝜃𝜃 𝜃𝜃 𝜃𝜃 2 𝐼𝐼 1 �𝜃𝜃 −0 3 5 0− 7 ⋯ � 𝑆𝑆𝑥𝑥 � 2 − 4 6 − ⋯ � 𝑆𝑆𝑥𝑥 = 0 cos sin . (5) 0 sin cos � 𝜃𝜃 𝜃𝜃� , , In general, a sequence of three− right𝜃𝜃-handed𝜃𝜃 rotations through the angles , first around the -axis, then around the -axis, and finally around the -axis, can bring the original ( , , ) 𝑥𝑥 𝑦𝑦 𝑧𝑧 coordinate system to any desired′ ( , , ) system that shares″ its origin with𝜃𝜃 𝜃𝜃 ( 𝜃𝜃, , ). (To see the generality𝑥𝑥 of this scheme𝑦𝑦, note that‴ any‴ ‴arbitrary final system𝑧𝑧 can be realigned with the𝒙𝒙� initial𝒚𝒚� 𝒛𝒛� system upon reversing the sequence𝒙𝒙� of three𝒚𝒚� 𝒛𝒛� rotations.) Thus, the corresponding rotation𝒙𝒙� 𝒚𝒚� 𝒛𝒛matrix� is , , = ( ) ( ) ( )

𝑅𝑅�𝜃𝜃𝑥𝑥 𝜃𝜃𝑦𝑦 𝜃𝜃𝑧𝑧� 𝑅𝑅𝑧𝑧 cos𝜃𝜃𝑧𝑧 𝑅𝑅𝑦𝑦 𝜃𝜃𝑦𝑦sin𝑅𝑅𝑥𝑥 𝜃𝜃𝑥𝑥 0 cos 0 sin 1 0 0 = 0 1 0 0 cos sin sin cos 0 𝑦𝑦 𝑦𝑦 . (6) 0𝜃𝜃𝑧𝑧 0𝜃𝜃𝑧𝑧 1 sin 𝜃𝜃 0 −cos 𝜃𝜃 0 sin cos �− 𝜃𝜃𝑧𝑧 𝜃𝜃𝑧𝑧 � � � � 𝜃𝜃𝑥𝑥 𝜃𝜃𝑥𝑥 � Since the above rotation matrices do not commute𝜃𝜃𝑦𝑦 with𝜃𝜃𝑦𝑦 each other,− the𝜃𝜃𝑥𝑥 order 𝜃𝜃of𝑥𝑥 multiplication in Eq.(6) is important and should not be altered. The generator matrices for the above rotations are 0 0 0 0 0 1 0 +1 0 = 0 0 +1 , = 0 0 0 , = 1 0 0 . (7) 0 1 0 +1 0 −0 0 0 0 𝑆𝑆𝑥𝑥 � � 𝑆𝑆𝑦𝑦 � � 𝑆𝑆𝑧𝑧 �− � One may write −, , = exp( ) exp( ) exp( ), with the caveat that, since , , do not commute𝑥𝑥 𝑦𝑦with𝑧𝑧 each other,𝑧𝑧 𝑧𝑧 it is not𝑦𝑦 𝑦𝑦permissible𝑥𝑥 𝑥𝑥 to write the product of the exponentials as exp( 𝑅𝑅�𝜃𝜃). 𝜃𝜃 𝜃𝜃 � 𝜃𝜃 𝑆𝑆 𝜃𝜃 𝑆𝑆 𝜃𝜃 𝑆𝑆 𝑆𝑆𝑥𝑥 𝑆𝑆𝑦𝑦 𝑆𝑆𝑧𝑧 𝜽𝜽 ∙ 𝑺𝑺

2 3. Boost (from one inertial frame to another) as a coordinate rotation in . In the special theory of relativity, where and space are treated on an equal footing, a general coordinate transformation (involving a coordinate rotation as well as a boost) is written

= 00 01 02 03 . (8) 𝑐𝑐𝑐𝑐′′ 𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑐𝑐𝑐𝑐 10 11 12 13 𝑥𝑥′ 𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑥𝑥 � � � 20 21 22 23� � � 𝑦𝑦 𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑦𝑦 ′ 30 31 32 33 The 4-vector = ( , 𝑧𝑧, , ) is𝑎𝑎 transformed𝑎𝑎 𝑎𝑎 between𝑎𝑎 primed𝑧𝑧 and unprimed reference frames by multiplication with a 4 × 4 matrix𝑇𝑇 . The metric of the flat space-time of special relativity is 𝑉𝑉 𝑐𝑐𝑐𝑐 𝑥𝑥 𝑦𝑦 𝑧𝑧 𝐴𝐴 1 0 0 0 0 1 0 0 = . (9) −0 0 1 0 𝑔𝑔 � 0 0 0 1� Since the Lorentz transformation preserves the norm of the 4-vector , the matrix must satisfy the identity = . 𝑇𝑇 Suppose the ( , , ) 𝑇𝑇coordinate axes of the reference frame𝑉𝑉 𝑔𝑔𝑔𝑔 (the so-called laboratory𝑉𝑉 frame)𝐴𝐴 are parallel to the corresponding𝐴𝐴 𝑔𝑔𝑔𝑔 axes𝑔𝑔 of the ( , , ) system, often referred to as the rest-frame . The latter frame (𝒙𝒙� )𝒚𝒚� moves𝒛𝒛� relative to the former′ ′ (′ ) at constant𝑆𝑆 along the positive - axis, while at = ′ = 0 the origins of the 𝒙𝒙�two𝒚𝒚� coordinate𝒛𝒛� systems coincide. Using the standard𝑆𝑆′ notation tanh( ) =𝑆𝑆 ′ = and cosh( ) = = 1 𝑆𝑆 1 , where is the𝑉𝑉 so-called and𝑥𝑥 𝑡𝑡 𝑡𝑡 is the speed of in vacuum, the pure Lorentz boost from2 to may be written as follows: 𝛼𝛼 𝛽𝛽 𝑉𝑉⁄𝑐𝑐 𝛼𝛼 𝛾𝛾 ⁄� − 𝛽𝛽 ′ 𝛼𝛼 𝑐𝑐 cosh sinh 0 0𝑆𝑆 𝑆𝑆 sinh cosh 0 0 ′ = . (10) 𝑐𝑐𝑡𝑡′ 0 𝛼𝛼 − 0 𝛼𝛼 1 0 𝑐𝑐𝑐𝑐 𝑥𝑥 − 𝛼𝛼 𝛼𝛼 𝑥𝑥 � ′ � � 0 0 0 1� � � 𝑦𝑦′ 𝑦𝑦 Differentiation with respect to𝑧𝑧 of the above transformation matrix𝑧𝑧, denoted by ( ), yields

𝛼𝛼cosh sinh 0 0 sinh cosh 𝛬𝛬𝑥𝑥 0𝛼𝛼 0 sinh cosh 0 0 cosh sinh 0 0 ( ) = = 𝛼𝛼 − 𝛼𝛼 𝛼𝛼 − 𝛼𝛼 d d 0 0 1 0 0 0 0 0 − 𝛼𝛼 𝛼𝛼 − 𝛼𝛼 𝛼𝛼 d𝛼𝛼 𝛬𝛬𝑥𝑥 𝛼𝛼 d𝛼𝛼 � 0 0 0 1� � 0 0 0 0� 0 1 0 0 cosh sinh 0 0 1 0 0 0 sinh cosh 0 0 = 0 −0 0 0 0 𝛼𝛼 − 0 𝛼𝛼 1 0 − − 𝛼𝛼 𝛼𝛼 � 0 0 0 0� � 0 0 0 1� 0 1 0 0 1 0 0 0 = ( ) = ( ). (11) 0 0 0 0 − �0 0 0 0� 𝛬𝛬𝑥𝑥 𝛼𝛼 −𝐾𝐾𝑥𝑥𝛬𝛬𝑥𝑥 𝛼𝛼

3 The 4 × 4 matrix is thus seen to be the generator of pure boost along the -axis. Consequently, the boost may be expressed as ( ) = exp( ). Similarly, the generator 𝑥𝑥 matrices for pure boosts along𝐾𝐾 the - and -axes are 𝑥𝑥 𝛬𝛬𝑥𝑥 𝛼𝛼 −𝛼𝛼𝛼𝛼𝑥𝑥 0 0 1 0 0 0 0 1 0 0𝑦𝑦 0 0𝑧𝑧 0 0 0 0 = , = . (12) 1 0 0 0 0 0 0 0 𝐾𝐾𝑦𝑦 �0 0 0 0� 𝐾𝐾𝑧𝑧 �1 0 0 0� A sequence of three successive boosts, first from to along the -axis, then from to along the -axis, and finally from to along the -axis,′ is described by the Lorentz ′matrix″ ( , , ′) = ( ) ( ) ( )″= exp‴( ) exp𝑆𝑆 (″ 𝑆𝑆 ) exp( 𝑥𝑥 ). Considering𝑆𝑆 that𝑆𝑆 𝑦𝑦 , , 𝑆𝑆 𝑆𝑆 𝑧𝑧 the matrices𝑥𝑥 𝑦𝑦 𝑧𝑧 𝑧𝑧 𝑧𝑧 do𝑦𝑦 not𝑦𝑦 commute𝑥𝑥 𝑥𝑥 with each𝑧𝑧 𝑧𝑧 other, one𝑦𝑦 should𝑦𝑦 avoid𝑥𝑥 𝑥𝑥the temptation to write 𝛬𝛬(𝛼𝛼 , 𝛼𝛼 , 𝛼𝛼 ) as 𝛬𝛬exp𝛼𝛼( 𝛬𝛬 𝛼𝛼). 𝛬𝛬 𝛼𝛼 −𝛼𝛼 𝐾𝐾 −𝛼𝛼 𝐾𝐾 −𝛼𝛼 𝐾𝐾 𝐾𝐾𝑥𝑥 𝐾𝐾𝑦𝑦 𝐾𝐾𝑧𝑧 4.𝛬𝛬 𝛼𝛼Boost𝑥𝑥 𝛼𝛼𝑦𝑦 in𝛼𝛼𝑧𝑧 an arbitrary−𝜶𝜶 ∙ 𝑲𝑲direction. Let and be two reference frames whose origins overlap at = = 0. The frame moves with velocity ′= + + relative to the laboratory frame . The′ rapidity and the′ polar coordinates𝑆𝑆 ( 𝑆𝑆, ) of are given by tanh( ) = = , = 𝑥𝑥 𝑦𝑦 𝑧𝑧 𝑡𝑡 sin𝑡𝑡 cos , = sin𝑆𝑆 sin , and = 𝑽𝑽cos 𝑉𝑉. 𝒙𝒙�A Lorentz𝑉𝑉 𝒚𝒚� 𝑉𝑉transform𝒛𝒛� ation from to thus 𝑥𝑥 involves𝑆𝑆 the following𝛼𝛼 sequence of steps: 𝜃𝜃 𝜙𝜙 𝑽𝑽 𝛼𝛼 𝛽𝛽 𝑉𝑉⁄𝑐𝑐 ′𝛽𝛽 𝛽𝛽 𝜃𝜃 𝜙𝜙 𝛽𝛽𝑦𝑦 𝛽𝛽 𝜃𝜃 𝜙𝜙 𝛽𝛽𝑧𝑧 𝛽𝛽 𝜃𝜃 𝑆𝑆 𝑆𝑆 i) Rotation around through the angle . ii) Rotation around through the angle . iii) Boost along with𝑧𝑧 rapidity . 𝜙𝜙 iv) Rotation around 𝑦𝑦 through the angle𝜃𝜃 . v) Rotation around𝑧𝑧 ′ through the𝛼𝛼 angle . 𝑦𝑦′ −𝜃𝜃 The matrix product representing the above transformation is readily evaluated as follows: 𝑧𝑧 −𝜙𝜙 1 0 0 0 1 0 0 0 ( ) 0 cos sin 0 0 cos 0 sin ( ) = ( ) ( ) ( ) ( ) ( ) = 0 sin cos 0 0 0 1 0 boost 𝜙𝜙 − 𝜙𝜙 𝜃𝜃 𝜃𝜃 𝛬𝛬 𝜷𝜷 𝑅𝑅𝑧𝑧 −𝜙𝜙 𝑅𝑅𝑦𝑦 −𝜃𝜃 𝐴𝐴𝑧𝑧 𝛼𝛼 𝑅𝑅𝑦𝑦 𝜃𝜃 𝑅𝑅𝑧𝑧 𝜙𝜙 �0 0 0 1� �0 sin 0 cos � 𝜙𝜙 𝜙𝜙 cosh 0 0 sinh 1 0 0 0 1 0 −0 𝜃𝜃0 𝜃𝜃 0 1 0 0 0 cos 0 sin 0 cos sin 0 × 0 𝛼𝛼 0 1 − 0 𝛼𝛼 0 0 1 0 0 sin cos 0 𝜃𝜃 − 𝜃𝜃 𝜙𝜙 𝜙𝜙 � sinh 0 0 cosh � �0 sin 0 cos � �0 0 0 1� − 𝜙𝜙 𝜙𝜙 cosh− 𝛼𝛼 sinh sin 𝛼𝛼cos 𝜃𝜃 sinh sin𝜃𝜃 sin sinh cos ( ) ( ) ( ) sinh sin𝛼𝛼 cos 1 + cosh− 𝛼𝛼 1 sin𝜃𝜃 𝜙𝜙cos cosh− 1 𝛼𝛼sin 𝜃𝜃sin 𝜙𝜙cos cosh − 1 sin𝛼𝛼 cos𝜃𝜃 cos = sinh sin sin (cosh 1) sin sin2 cos2 1 + (cosh 12) sin sin (cosh 1) sin cos sin ⎛− 𝛼𝛼 𝜃𝜃 𝜙𝜙 𝛼𝛼 − 𝜃𝜃 𝜙𝜙 𝛼𝛼 − 𝜃𝜃 𝜙𝜙 𝜙𝜙 𝛼𝛼 − 𝜃𝜃 𝜃𝜃 𝜙𝜙⎞ ⎜ ( ) 2 ( ) 2 2 ( ) ⎟ ⎜− sinh𝛼𝛼 cos𝜃𝜃 𝜙𝜙 cosh𝛼𝛼 − 1 sin 𝜃𝜃cos 𝜙𝜙cos 𝜙𝜙 cosh 𝛼𝛼1−sin cos𝜃𝜃 sin 𝜙𝜙 1 +𝛼𝛼 cosh− 𝜃𝜃1 cos𝜃𝜃 𝜙𝜙⎟ 2 ⎝ − 𝛼𝛼 𝜃𝜃 𝛼𝛼 − 𝜃𝜃 𝜃𝜃 𝜙𝜙 𝛼𝛼 − 𝜃𝜃 𝜃𝜃 𝜙𝜙 𝛼𝛼 − 𝜃𝜃 ⎠

( ) ( ) ( ) 𝛾𝛾 1 + −𝛾𝛾1𝛽𝛽𝑥𝑥 1−𝛾𝛾𝛽𝛽𝑦𝑦 1−𝛾𝛾𝛽𝛽𝑧𝑧 = . (13) ( 1) 2 2 1 + ( 1) 2 ( 1) 2 ⎛−𝛾𝛾𝛽𝛽𝑥𝑥 𝛾𝛾 − 𝛽𝛽𝑥𝑥 ⁄𝛽𝛽 𝛾𝛾 − 𝛽𝛽𝑥𝑥𝛽𝛽𝑦𝑦⁄𝛽𝛽 𝛾𝛾 − 𝛽𝛽𝑥𝑥𝛽𝛽𝑧𝑧⁄𝛽𝛽 ⎞ ⎜ ( ) 2 ( ) 2 2 ( ) 2 ⎟ ⎜−𝛾𝛾𝛽𝛽𝑦𝑦 𝛾𝛾 − 1 𝛽𝛽𝑥𝑥𝛽𝛽𝑦𝑦⁄𝛽𝛽 𝛾𝛾1− 𝛽𝛽𝑦𝑦⁄𝛽𝛽 1 +𝛾𝛾 − 𝛽𝛽1𝑦𝑦𝛽𝛽𝑧𝑧⁄𝛽𝛽 ⎟ 2 2 2 2 ⎝−𝛾𝛾𝛽𝛽𝑧𝑧 𝛾𝛾 − 𝛽𝛽𝑥𝑥𝛽𝛽𝑧𝑧⁄𝛽𝛽 𝛾𝛾 − 𝛽𝛽𝑦𝑦𝛽𝛽𝑧𝑧⁄𝛽𝛽 𝛾𝛾 − 𝛽𝛽𝑧𝑧 ⁄𝛽𝛽 ⎠

4 Note that the pure boost matrix ( ) given by Eq.(13) is symmetric. The inverse of this matrix transforms the space-time coordinates of back to . This, however, is equivalent to switching the sign of , which flips the signs of 𝛬𝛬, 𝜷𝜷, ′ . The inverse of ( ) of Eq.(13) is thus obtained simply 𝑆𝑆 𝑆𝑆 by reversing the sign of . 𝑥𝑥 𝑦𝑦 𝑧𝑧 The𝑽𝑽 above definition of a pure𝛽𝛽 boost𝛽𝛽 𝛽𝛽 might lead one to𝛬𝛬 believe𝜷𝜷 that the ( , , ) axes of are parallel to the corresponding𝜷𝜷 axes ( , , ) of . This, however, is not the case, as can be readily seen by examining the simple example′ where′ ′ =′ + = ( )(cos 𝒙𝒙�+𝒚𝒚�sin𝒛𝒛� ). At 𝑆𝑆= 0, 𝒙𝒙� 𝒚𝒚� 𝒛𝒛� 𝑆𝑆 ( , , ) when the origins of the two coordinate systems overlap,𝑥𝑥 𝑦𝑦 let the point in the laboratory frame correspond to the point ( , 0, 0) located𝜷𝜷 𝛽𝛽 on𝒙𝒙� the𝛽𝛽 𝒚𝒚� -axis𝑉𝑉 ⁄of𝑐𝑐 the rest𝜙𝜙 𝒙𝒙�-frame 𝜙𝜙 𝒚𝒚�. A Lorentz𝑡𝑡 transformation from to yields ′ ′ 𝑥𝑥 𝑦𝑦 𝑧𝑧 ′ 𝑆𝑆 ′ 𝑥𝑥 𝑥𝑥 𝑆𝑆 𝑆𝑆 𝑆𝑆 0 0 1 + ( 1) ( 1) 0 ′ = 𝛾𝛾 −𝛾𝛾𝛽𝛽𝑥𝑥 −𝛾𝛾𝛽𝛽𝑦𝑦 . (14) 𝑐𝑐0𝑡𝑡′ 2 2 2 𝑥𝑥 ( 1) 𝑥𝑥 1 + ( 1𝑥𝑥) 𝑦𝑦 0 𝑥𝑥0 ⎛−𝛾𝛾𝛽𝛽 𝛾𝛾 − 𝛽𝛽 ⁄𝛽𝛽 𝛾𝛾 − 𝛽𝛽 𝛽𝛽 ⁄𝛽𝛽 ⎞ 𝑥𝑥 � � ⎜ 0 0 2 0 2 2 1⎟ � � −𝛾𝛾𝛽𝛽𝑦𝑦 𝛾𝛾 − 𝛽𝛽𝑥𝑥𝛽𝛽𝑦𝑦⁄𝛽𝛽 𝛾𝛾 − 𝛽𝛽𝑦𝑦⁄𝛽𝛽 𝑦𝑦 𝑧𝑧 Consequently, ( , , ⎝) = 1 + ( 1)( ) , ( 1) , 0 (⎠ ), indicating that the 2 2 ′ -axis, when viewed from the laboratory frame,𝑦𝑦 appears to be𝑥𝑥 rotated𝑦𝑦 clockwise through an angle ′ , where 𝑥𝑥 𝑦𝑦 𝑧𝑧 � 𝛾𝛾 − 𝛽𝛽 ⁄𝛽𝛽 − 𝛾𝛾 − 𝛽𝛽 𝛽𝛽 ⁄𝛽𝛽 � 𝑥𝑥 ⁄𝛾𝛾 𝑥𝑥 ′ ( ) ( ) 𝑥𝑥 tan = | | = = 𝜓𝜓 ( )( 2) ( ) . (15) 𝛾𝛾−1 𝛽𝛽𝑥𝑥𝛽𝛽𝑦𝑦⁄𝛽𝛽 ′ 𝛾𝛾−1 sin 𝜙𝜙 cos 𝜙𝜙 2 2 Similarly, by finding 𝜓𝜓the𝑥𝑥 coordinates𝑦𝑦⁄𝑥𝑥 1 +( 𝛾𝛾,−1, 𝛽𝛽)𝑦𝑦 ⁄of𝛽𝛽 a point1+ 𝛾𝛾in−1 thesin laboratory𝜙𝜙 frame at = 0 corresponding to the point (0, , 0) in the rest-frame , we find the apparent counterclockwise rotation angle of the -axis as′ seen from𝑥𝑥 the𝑦𝑦 𝑧𝑧laboratory′ frame to be 𝑆𝑆 𝑡𝑡 ′ ′ 𝑦𝑦 𝑆𝑆 𝑦𝑦 ( ) ( ) 𝜓𝜓 tan𝑦𝑦 = | | = = ( )( 2) ( ) . (16) 𝛾𝛾−1 𝛽𝛽𝑥𝑥𝛽𝛽𝑦𝑦⁄𝛽𝛽 ′ 𝛾𝛾−1 sin 𝜙𝜙 cos 𝜙𝜙 𝑦𝑦 2 2 Note that the angles 𝜓𝜓 and 𝑥𝑥⁄𝑦𝑦 are unequal1+ 𝛾𝛾−1 𝛽𝛽and𝑥𝑥⁄𝛽𝛽 also in1+ opposite𝛾𝛾−1 cos directions,𝜙𝜙 which means that, viewed from the frame, ′the ′ frame appears to be deformed, not rotated. 𝑥𝑥 𝑦𝑦 The pure boost by 𝜓𝜓 , embodied𝜓𝜓′ ′ ′ in the transformation matrix ( ) of Eq.(13), may be expressed in more𝑥𝑥 𝑥𝑥𝑥𝑥compact form𝑥𝑥 as𝑦𝑦 𝑧𝑧 𝑽𝑽 𝛬𝛬 𝜷𝜷 = ( ), (17a)

= + [𝑐𝑐𝑐𝑐( ′ 1𝛾𝛾) 𝑐𝑐𝑐𝑐 −](𝜷𝜷 ∙ 𝒓𝒓 ) . (17b) ′ 2 Writing = + 𝒓𝒓, with𝒓𝒓 the 𝛾𝛾subscripts− ⁄𝛽𝛽 𝜷𝜷and∙ 𝒓𝒓 𝜷𝜷 referring,− 𝛾𝛾𝜷𝜷𝑐𝑐𝑐𝑐 respectively, to the parallel and perpendicular components of relative to = , Eq.(17) may be further simplified, yielding 𝒓𝒓 𝒓𝒓∥ 𝒓𝒓⊥ ∥ ⊥ 𝒓𝒓 = [𝜷𝜷 (𝑽𝑽⁄𝑐𝑐 ) ], (18a) 2 𝑡𝑡′ =𝛾𝛾 (𝑡𝑡 − 𝑉𝑉⁄𝑐𝑐) +𝑟𝑟∥ . (18b) ′ 5. Alternative treatment of boost 𝒓𝒓in an𝛾𝛾 arbitrary𝒓𝒓∥ − 𝑽𝑽𝑡𝑡 direction𝒓𝒓⊥ . An alternative derivation of Eq.(13) starts from the assumption that the direction of is fixed, but that its magnitude changes incrementally, from to + d . Recalling that tanh( ) = and associating the rapidity with 𝑽𝑽 𝑉𝑉 𝑉𝑉 𝑉𝑉 𝛼𝛼 𝑉𝑉⁄𝑐𝑐 𝛼𝛼

5 a vector of length in the direction of = , we will have (d , d , d ) = ( , , )d . Invoking Eqs.(11) and (12), we may write 𝛼𝛼 𝜷𝜷� 𝜷𝜷⁄𝛽𝛽 𝛼𝛼𝑥𝑥 𝛼𝛼𝑦𝑦 𝛼𝛼𝑧𝑧 𝛽𝛽̂𝑥𝑥 𝛽𝛽̂𝑦𝑦 𝛽𝛽̂𝑧𝑧 𝛼𝛼 0 0 0 0 ( ) = −𝛽𝛽̂𝑥𝑥 −𝛽𝛽̂𝑦𝑦 −𝛽𝛽̂𝑧𝑧 ( ) = ( ) ( ). (19) d 𝑥𝑥 0 0 0 � ⎛−𝛽𝛽̂ ⎞ � � d𝛼𝛼 𝛬𝛬𝜷𝜷 𝛼𝛼 0 0 0 𝛬𝛬𝜷𝜷 𝛼𝛼 −𝜷𝜷� ∙ 𝑲𝑲 𝛬𝛬𝜷𝜷 𝛼𝛼 ⎜−𝛽𝛽̂𝑦𝑦 ⎟ Consequently, ( ) = exp⎝−𝛽𝛽[̂𝑧𝑧 ( ) ]. Note that⎠ the incremental nature of the boost (i.e., from to + while keeping constant) has made it possible in this instance to combine the 𝛬𝛬𝜷𝜷� 𝛼𝛼 − 𝜷𝜷� ∙ 𝑲𝑲 𝛼𝛼 three exponentials into a single entity, exp[ ( ) ], without due consideration for the sequential � order 𝑉𝑉of individual𝑉𝑉 𝑑𝑑𝑑𝑑 boosts along ,𝜷𝜷 , and .† Proceeding to evaluate exp[ ( ) ] via its Taylor � series expansion, we find − 𝜷𝜷 ∙ 𝑲𝑲 𝛼𝛼 𝒙𝒙� 𝒚𝒚� 𝒛𝒛� − 𝜷𝜷� ∙ 𝑲𝑲 𝛼𝛼 0 0 1 0 0 0 0 0𝑥𝑥 0𝑦𝑦 0𝑧𝑧 0𝑥𝑥 0𝑦𝑦 0𝑧𝑧 ( ) = 𝛽𝛽̂ 𝛽𝛽̂ 𝛽𝛽̂ 𝛽𝛽̂ 𝛽𝛽̂ 𝛽𝛽̂ = 2 . (20) 0 0 0 0 0 0 0 ̂𝑥𝑥 ̂𝑥𝑥 ̂𝑦𝑦 ̂𝑥𝑥 ̂𝑧𝑧 2 ⎛𝛽𝛽̂𝑥𝑥 ⎞ ⎛𝛽𝛽̂𝑥𝑥 ⎞ ⎛ 𝛽𝛽 𝛽𝛽 𝛽𝛽 𝛽𝛽 𝛽𝛽 ⎞ 𝜷𝜷� ∙ 𝑲𝑲 0 2 ⎜ ̂𝑦𝑦 0 0 0 ⎟ ⎜ ̂𝑦𝑦 0 0 0 ⎟ ⎜ 𝛽𝛽̂𝑥𝑥𝛽𝛽̂𝑦𝑦 𝛽𝛽̂𝑦𝑦 𝛽𝛽̂𝑦𝑦𝛽𝛽̂𝑧𝑧⎟ 𝛽𝛽 𝛽𝛽 2 ̂𝑥𝑥 ̂𝑧𝑧 ̂𝑦𝑦 ̂𝑧𝑧 ̂𝑧𝑧 ⎝𝛽𝛽0̂𝑧𝑧 ⎠ ⎝1𝛽𝛽̂𝑧𝑧 0 0 ⎠ 0 ⎝ 𝛽𝛽 𝛽𝛽0 𝛽𝛽 𝛽𝛽 𝛽𝛽 ⎠ 0 0𝑥𝑥 0𝑦𝑦 0𝑧𝑧 0𝑥𝑥 0𝑦𝑦 0𝑧𝑧 ( ) = 𝛽𝛽̂ 𝛽𝛽̂ 𝛽𝛽̂ 2 = 𝛽𝛽̂ 𝛽𝛽̂ 𝛽𝛽̂ . (21) 0 0 0 0 ̂𝑥𝑥 ̂𝑥𝑥 ̂𝑦𝑦 ̂𝑥𝑥 ̂𝑧𝑧 0 0 0 3 ⎛𝛽𝛽̂𝑥𝑥 ⎞ ⎛ 𝛽𝛽 𝛽𝛽 𝛽𝛽 𝛽𝛽 𝛽𝛽 ⎞ ⎛𝛽𝛽̂𝑥𝑥 ⎞ 𝜷𝜷� ∙ 𝑲𝑲 0 2 ⎜ ̂𝑦𝑦 0 0 0 ⎟ ⎜ 𝛽𝛽̂𝑥𝑥𝛽𝛽̂𝑦𝑦 𝛽𝛽̂𝑦𝑦 𝛽𝛽̂𝑦𝑦𝛽𝛽̂𝑧𝑧⎟ ⎜ ̂𝑦𝑦 0 0 0 ⎟ 𝛽𝛽 2 𝛽𝛽 ̂𝑥𝑥 ̂𝑧𝑧 ̂𝑦𝑦 ̂𝑧𝑧 ̂𝑧𝑧 ⎝𝛽𝛽̂𝑧𝑧 ⎠ ⎝ 𝛽𝛽 𝛽𝛽 𝛽𝛽 𝛽𝛽 𝛽𝛽 ⎠ ⎝𝛽𝛽̂𝑧𝑧 ⎠ exp[ ( ) ] = ( ) + ( ) ( ) + ( ) ! 2! 3! 4! 𝛼𝛼 𝛼𝛼 2 𝛼𝛼 3 𝛼𝛼 4 − 𝜷𝜷� ∙ 𝑲𝑲 𝛼𝛼 𝐼𝐼 − 1 𝜷𝜷� ∙ 𝑲𝑲 2 𝜷𝜷� ∙ 𝑲𝑲 − 3 𝜷𝜷� ∙ 𝑲𝑲 4 𝜷𝜷� ∙ 𝑲𝑲 − ⋯ = ( + + + )( ) + ( + + + )( ) ! 3! 5! 2! 4! 6! 𝛼𝛼 𝛼𝛼 𝛼𝛼 𝛼𝛼 𝛼𝛼 𝛼𝛼 2 1 3 5 � 2 4 6 � = 𝐼𝐼 − (sinh )( )⋯+ (cosh𝜷𝜷 ∙ 𝑲𝑲 1)( ) ⋯ 𝜷𝜷 ∙ 𝑲𝑲 2 = 𝐼𝐼 − 𝛼𝛼+ [𝜷𝜷�( ∙ 𝑲𝑲 1) ]( 𝛼𝛼 − ) 𝜷𝜷� ∙ 𝑲𝑲 2 2 𝐼𝐼 − 𝛾𝛾𝜷𝜷 ∙ 𝑲𝑲 𝛾𝛾 − ⁄𝛽𝛽 𝜷𝜷 ∙ 𝑲𝑲 2 2 2 2 𝛾𝛾 1 + ( −𝛾𝛾𝛽𝛽1)𝑥𝑥 ( 1−)𝛾𝛾𝛽𝛽𝑦𝑦 ( 1−)𝛾𝛾𝛽𝛽𝑧𝑧 = . (22) ( 1) 2 1 + ( 1) 2 2 ( 1) 2 ⎛−𝛾𝛾𝛽𝛽𝑥𝑥 𝛾𝛾 − 𝛽𝛽𝑥𝑥⁄𝛽𝛽 𝛾𝛾 − 𝛽𝛽𝑥𝑥𝛽𝛽𝑦𝑦⁄𝛽𝛽 𝛾𝛾 − 𝛽𝛽𝑥𝑥𝛽𝛽𝑧𝑧⁄𝛽𝛽 ⎞ 2 2 2 2 ⎜−𝛾𝛾𝛽𝛽𝑦𝑦 (𝛾𝛾 − 1)𝛽𝛽𝑥𝑥𝛽𝛽𝑦𝑦⁄𝛽𝛽 ( 𝛾𝛾1−) 𝛽𝛽𝑦𝑦⁄𝛽𝛽 1 +𝛾𝛾 −( 𝛽𝛽1𝑦𝑦)𝛽𝛽𝑧𝑧⁄𝛽𝛽 ⎟ 𝑧𝑧 𝑥𝑥 𝑧𝑧 𝑦𝑦 𝑧𝑧 𝑧𝑧 The above matrix ⎝is− seen𝛾𝛾𝛽𝛽 to be𝛾𝛾 −identical𝛽𝛽 𝛽𝛽 ⁄with𝛽𝛽 that𝛾𝛾 obtained− 𝛽𝛽 𝛽𝛽 in⁄ 𝛽𝛽Eq.(13), which𝛾𝛾 − was𝛽𝛽 ⁄derived𝛽𝛽 ⎠ using a combination of boost and rotations.

†If there is any doubt as to the validity of Eq.(19), one should consider deriving it directly from Eq.(13). Of course, we are taking the opposite path here, attempting to derive Eq.(13) from Eq.(19).

6 6. The Thomas Precession. Consider the laboratory frame , a first rest-frame moving within with (normalized) velocity = , and a second rest-frame , also moving ′within but with 𝑆𝑆 𝑆𝑆 𝑆𝑆 (normalized) velocity = 1 . Note1 that and share the same″ magnitude , differing only in 𝜷𝜷 𝛽𝛽𝜷𝜷� 𝑆𝑆 𝑆𝑆 their directions. The three2 coordinate2 systems1 share 2a common -axis and their origins coincide at = = = 0, although𝜷𝜷 their𝛽𝛽𝜷𝜷� and axes𝜷𝜷 are oriented𝜷𝜷 differently. In the case 𝛽𝛽of , the -axis is 𝑧𝑧 aligned′ with″ , while in the case of , the -axis is aligned with . Let , =′ (cos , ) + 𝑡𝑡 𝑡𝑡 𝑡𝑡 𝑥𝑥 𝑦𝑦 𝑆𝑆 𝑥𝑥′ (sin , ) . A Lorentz1 transformation from″ to thus involves the following2 sequence1 2 of steps:1 2 𝜷𝜷� 𝑆𝑆 ′ 𝑥𝑥″ ″ 𝜷𝜷� 𝜷𝜷� 𝜃𝜃 𝒙𝒙� 1 2 i) Boost𝜃𝜃 𝒚𝒚� from to along the -axis. 𝑆𝑆 𝑆𝑆 ii) Rotation in ′around the -axis′ through the angle . 𝑆𝑆 𝑆𝑆 𝑥𝑥 iii) Rotation in around the -axis through the angle + 1. 𝑆𝑆 𝑧𝑧 −𝜃𝜃 iv) Boost from to along the -axis. 2 𝑆𝑆 𝑧𝑧 ″ 𝜃𝜃 Steps (ii) and𝑆𝑆 (iii)𝑆𝑆″, of course,𝑥𝑥 may be combined into a single rotation around the -axis through = . The matrix representing the transformation from to is thus given by ′ ″ 𝑧𝑧 2 1 𝜃𝜃 𝜃𝜃 − 𝜃𝜃 0 0 1 0 0 𝑆𝑆 0 𝑆𝑆 0 0 0 0 0 cos sin 0 0 0 ( , ) = 0𝛾𝛾 −0𝛾𝛾𝛾𝛾 1 0 0 sin cos 0 𝛾𝛾0 𝛾𝛾0𝛽𝛽 1 0 −𝛾𝛾𝛾𝛾 𝛾𝛾 𝜃𝜃 𝜃𝜃 𝛾𝛾𝛽𝛽 𝛾𝛾 𝛬𝛬 𝛽𝛽 𝜃𝜃 � 0 0 0 1� �0 0 0 1� � 0 0 0 1� − 𝜃𝜃 𝜃𝜃 (1 cos ) (1 cos ) sin 0 2 (1 2 cos ) 2(cos ) sin 0 = 𝛾𝛾 − 𝛽𝛽 𝜃𝜃 𝛾𝛾 𝛽𝛽 − 𝜃𝜃 −𝛾𝛾𝛾𝛾 𝜃𝜃 2 2 2 . (23) ⎛−𝛾𝛾 𝛽𝛽 sin− 𝜃𝜃 𝛾𝛾 sin𝜃𝜃 − 𝛽𝛽 𝛾𝛾cos 𝜃𝜃 0⎞ ⎜ ⎟ −𝛾𝛾𝛽𝛽0 𝜃𝜃 −𝛾𝛾 0 𝜃𝜃 0 𝜃𝜃 1 Note that ( , ⎝) in Eq.(23) is not a , implying that it⎠ does not represent a pure boost but, rather, a boost followed by a coordinate rotation. Writing ( , ) = ( ) ( )( ), where is 𝛬𝛬the𝛽𝛽 (as𝜃𝜃 yet unknown) boost velocity (from to ), and is the (as yet boostunknown′ ) 𝑧𝑧 rotation angle′ within , we may write ′ ″ 𝛬𝛬 𝛽𝛽 𝜃𝜃 𝑅𝑅 𝜓𝜓 𝐴𝐴 𝜷𝜷 𝜷𝜷 𝑆𝑆 𝑆𝑆 𝜓𝜓 ( ) ( ) = ( 𝑆𝑆″) ( , ) boost ′ 𝐴𝐴 𝜷𝜷 𝑅𝑅𝑧𝑧1−𝜓𝜓 0𝛬𝛬 𝛽𝛽 𝜃𝜃 0 0 (1 cos ) (1 cos ) sin 0 0 cos sin 0 2 (1 2 cos ) 2 (cos ) sin 0 = 𝛾𝛾 − 𝛽𝛽 𝜃𝜃 𝛾𝛾 𝛽𝛽 − 𝜃𝜃 −𝛾𝛾𝛾𝛾 𝜃𝜃 0 sin cos 0 2 sin 2 sin 2 cos 0 ⎛ 𝜓𝜓 − 𝜓𝜓 ⎞ ⎛−𝛾𝛾 𝛽𝛽 − 𝜃𝜃 𝛾𝛾 𝜃𝜃 − 𝛽𝛽 𝛾𝛾 𝜃𝜃 ⎞ ⎜0 0 0 1⎟ ⎜ ⎟ 𝜓𝜓 𝜓𝜓 −𝛾𝛾𝛽𝛽0 𝜃𝜃 −𝛾𝛾 0 𝜃𝜃 0 𝜃𝜃 1

(⎝1 cos ) ⎠ ⎝ (1 cos ) sin 0 ⎠ (1 cos2 ) cos2 + sin sin (cos 2 ) cos + sin sin sin cos cos sin 0 = 𝛾𝛾 − 𝛽𝛽 𝜃𝜃 𝛾𝛾 𝛽𝛽 − 𝜃𝜃 −𝛾𝛾𝛾𝛾 𝜃𝜃 . (24) 2 ( ) 2( 2) ⎛−𝛾𝛾 𝛽𝛽 1 − cos 𝜃𝜃 sin 𝜓𝜓 𝛾𝛾𝛽𝛽sin 𝜃𝜃cos 𝜓𝜓 𝛾𝛾 cos 𝜃𝜃 − 𝛽𝛽 sin 𝜓𝜓 𝛾𝛾sin 𝜃𝜃cos 𝜓𝜓 𝛾𝛾 sin 𝜃𝜃 sin 𝜓𝜓+−cos 𝜃𝜃cos 𝜓𝜓 0 ⎞ ⎜ 2 2 2 ⎟ −𝛾𝛾 𝛽𝛽 − 𝜃𝜃 0 𝜓𝜓 − 𝛾𝛾𝛽𝛽 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 − 𝛽𝛽 0𝜓𝜓 − 𝛾𝛾 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓0 𝜃𝜃 𝜓𝜓 1 ⎝ ⎠

7 This pure boost must have a symmetric matrix. The acceptable value of is thus found by ensuring the of the matrix in Eq.(24). There are three constraints on the matrix, but they all lead to the same value of , as shown below. 𝜓𝜓

1st constraint: (1 𝜓𝜓cos ) cos + sin sin = (1 cos ) sin2 sin = (1 cos )(1 + cos ) 2 −𝛾𝛾 𝛽𝛽 − 𝜃𝜃 𝜓𝜓 𝛾𝛾𝛽𝛽 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝛽𝛽 − 𝜃𝜃 sin(½ ) cos(½ ) sin(½ ) cos(½ ) = sin (½ ) cos (½ ) → 𝜃𝜃 𝜓𝜓 𝛾𝛾 − 𝜃𝜃 𝜓𝜓 tan(½ ) = tan(½ ). 2 2 (25a) → 𝜃𝜃 𝜃𝜃 𝜓𝜓 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓

nd → 𝜓𝜓 𝛾𝛾 𝜃𝜃 2 constraint: (1 cos ) sin sin cos = sin (12 cos ) sin = sin (1 cos ) tan(½ ) = tan(½ ) −𝛾𝛾 𝛽𝛽 − 𝜃𝜃 𝜓𝜓 − 𝛾𝛾𝛽𝛽 𝜃𝜃 𝜓𝜓 −𝛾𝛾𝛾𝛾 𝜃𝜃 . (25b)

rd → 𝛾𝛾 − 𝜃𝜃 𝜓𝜓 𝜃𝜃 − 𝜓𝜓 → 𝜓𝜓 𝛾𝛾 𝜃𝜃 3 constraint: (cos ) sin sin cos = sin cos cos sin ( 2 + 1) cos 2 sin sin = 2 sin cos 𝛾𝛾 𝜃𝜃 − 𝛽𝛽 𝜓𝜓 − 𝛾𝛾 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 − 𝜃𝜃 𝜓𝜓 ( 2 + 1) cos sin ( 2 2 1) sin = 2 sin cos → 𝛾𝛾 𝜃𝜃 𝜓𝜓 − 𝛾𝛾 𝛽𝛽 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 [1 2+ cos (1 cos 2)] tan = 2 sin cos → 𝛾𝛾 𝜃𝜃 𝜓𝜓 − 𝛾𝛾 − 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 [cos (½ ) 2 sin (½ )] tan = 2 sin(½ ) cos(½ ) → 𝜃𝜃 − 𝛾𝛾 − 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 [1 2 tan (½2 )] tan2 (½ ) = tan(½ ) [1 tan (½ )] → 𝜃𝜃 − 𝛾𝛾 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜃𝜃 [tan(½2 ) 2 tan(½ )][1 + tan(½ ) tan(½ )] 2= 0 → − 𝛾𝛾 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 − 𝜓𝜓 tan(½ ) = tan(½ ). (25c) → 𝜓𝜓 − 𝛾𝛾 𝜃𝜃 𝛾𝛾 𝜃𝜃 𝜓𝜓 With the rotation→ angle 𝜓𝜓thus determined𝛾𝛾 𝜃𝜃 , the boost matrix of Eq.(24) becomes

(1 𝜓𝜓cos ) (1 cos ) sin 0 2 (1 2cos ) (cos 2 ) cos + sin sin sin cos cos sin 0 ( )( ) = 𝛾𝛾 − 𝛽𝛽 𝜃𝜃 𝛾𝛾 𝛽𝛽 − 𝜃𝜃 −𝛾𝛾𝛾𝛾 𝜃𝜃 2 2 2 . (26) boost ′ ⎛ 𝛾𝛾 𝛽𝛽 −sin 𝜃𝜃 𝛾𝛾 sin𝜃𝜃 − 𝛽𝛽cos 𝜓𝜓cos𝛾𝛾 sin𝜃𝜃 𝜓𝜓 𝛾𝛾 sin 𝜃𝜃 sin 𝜓𝜓+−cos 𝜃𝜃cos 𝜓𝜓 0⎞ 𝐴𝐴 𝜷𝜷 ⎜ ⎟ −𝛾𝛾𝛾𝛾0 𝜃𝜃 𝛾𝛾 𝜃𝜃 𝜓𝜓0− 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓0 𝜃𝜃 𝜓𝜓 1 Comparison with⎝ Eq.(22) shows that the following equalities must hold ⎠ = (1 cos ), (27a) ′ 2 2 𝛾𝛾 𝛾𝛾= − 𝛽𝛽(1 cos𝜃𝜃 ), (27b) ′ ′ 2 𝛾𝛾 𝛽𝛽𝑥𝑥 = −𝛾𝛾sin𝛽𝛽 ,− 𝜃𝜃 (27c) ′ ′ 𝛾𝛾 𝛽𝛽=𝑦𝑦 0, 𝛾𝛾𝛾𝛾 𝜃𝜃 (27d) ′ 1𝛽𝛽𝑧𝑧+ ( 1)( ) = (cos ) cos + sin sin , (27e) ′ ′ ′ 2 2 2 ( 𝛾𝛾1)− 𝛽𝛽𝑥𝑥⁄𝛽𝛽= sin𝛾𝛾 cos 𝜃𝜃 − 𝛽𝛽cos sin𝜓𝜓 , 𝛾𝛾 𝜃𝜃 𝜓𝜓 (27f) ′ ′ ′ ′2 1𝛾𝛾+−( 𝛽𝛽𝑥𝑥1𝛽𝛽)𝑦𝑦(⁄𝛽𝛽 ) 𝛾𝛾= 𝜃𝜃sin 𝜓𝜓sin− + cos𝜃𝜃 𝜓𝜓cos . (27g) ′ ′ ′ 2 𝛾𝛾 − 𝛽𝛽𝑦𝑦⁄𝛽𝛽 𝛾𝛾 𝜃𝜃 𝜓𝜓 𝜃𝜃 𝜓𝜓 8 The boost velocity from to is thus found from Eqs.(27a)-(27d) as follows: ′ ″ ( ) 𝑆𝑆 𝑆𝑆 = . (28) 2 ( ) (½ ) ′ 𝛾𝛾 𝛽𝛽 cos 𝜃𝜃−1 𝒙𝒙� + 𝛾𝛾𝛾𝛾 sin 𝜃𝜃𝒚𝒚� 2 2 Note that, at non-relativistic ,𝜷𝜷 1+ 2 𝛾𝛾 −1 sin 𝜃𝜃= (cos 1) + ( sin ) . To confirm that of Eq.(28) is indeed consistent with ′of Eq.(27a), that is, = 1/′ 1 , we′ write 𝜷𝜷 ≅ 𝜷𝜷2 − 𝜷𝜷1 𝛽𝛽 𝜃𝜃 − 𝒙𝒙� 𝛽𝛽 𝜃𝜃 𝒚𝒚� ′ ′ ′ ′2 ( ) 1𝜷𝜷 = 1 + + = 1 𝛾𝛾 𝛾𝛾= � − 𝛽𝛽= (1 ) 4 2 ( 2 2 )2 2 ( ) . (29) ′2 ′2 ′2 ′2 𝛾𝛾 𝛽𝛽 1−cos 𝜃𝜃 +𝛾𝛾 𝛽𝛽 sin 𝜃𝜃 1 ′ 2 𝑥𝑥 𝑦𝑦 𝑧𝑧 4 2 2 4 2 2 To confirm− 𝛽𝛽 the remaining− �𝛽𝛽 𝛽𝛽 identities𝛽𝛽 � in Eq.(2− 7), we𝛾𝛾 1first−𝛽𝛽 cosprove𝜃𝜃 the following𝛾𝛾 1−𝛽𝛽 cosrelations:𝜃𝜃 ⁄𝛾𝛾 1 = (1 cos ) 1 = ( 1)(1 cos ) = 2( 1) sin (½ ), (30a) ′ 2 2 2 2 2 (½ ) (𝛾𝛾 − ) 𝛾𝛾= 1 +−(𝛽𝛽 )𝜃𝜃 =− 1 + 𝛾𝛾 − −= 1 +𝜃𝜃 𝛾𝛾 −= 1 + 𝜃𝜃 = , (30b) ( ) 2 (½ ) 2 (½ ) (½ ) ′ ′ 2 ′ ′ 2 sin 𝜃𝜃 cos 𝜃𝜃 1 1 𝑥𝑥 𝑦𝑦 𝑥𝑥 2 2 𝛽𝛽 ⁄𝛽𝛽 𝛽𝛽 ⁄𝛽𝛽 �𝛾𝛾(1−cos 𝜃𝜃)� �𝛾𝛾 sin(½𝜃𝜃)� tan 𝜓𝜓 sin 𝜓𝜓 ( ) = 1 + ( ) = 1 + = 1 + = 1 + tan (½ ) = , (30c) 2 (½ ) 2 (½ ) ′ ′ 2 ′ ′ 2 𝛾𝛾 1−cos 𝜃𝜃 𝛾𝛾 sin 𝜃𝜃 2 1 𝑦𝑦 𝑥𝑥 𝑦𝑦 2 𝛽𝛽 ⁄(𝛽𝛽 ) = ( 𝛽𝛽 ⁄𝛽𝛽) + ( �) =sin 𝜃𝜃tan(�½ ) � cos 𝜃𝜃=� 𝜓𝜓 cos 𝜓𝜓 (½ ) . (30d) ′2 ′ ′ ′ ′ ′ ′ 1 2 Substitution𝛽𝛽 � 𝛽𝛽𝑥𝑥𝛽𝛽𝑦𝑦 into 𝛽𝛽Eqs.(2𝑥𝑥⁄𝛽𝛽𝑦𝑦 7e)-𝛽𝛽(2𝑦𝑦⁄7𝛽𝛽g)𝑥𝑥 now− yields 𝜓𝜓 − tan 𝜓𝜓 − sin 𝜓𝜓

1 + ( 1)( ) = (cos ) cos + sin sin 1 + 2( ′ 1) sin′ (′½2 ) sin2 (½ ) = 2 (cos ) cos + sin sin 𝛾𝛾 − 𝛽𝛽𝑥𝑥⁄𝛽𝛽 𝛾𝛾 𝜃𝜃 − 𝛽𝛽 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 1 + ½( 2 1)(1 2 cos )(12 cos ) 2 cos cos2 + ( 1) cos = sin sin → 𝛾𝛾 − 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 − 𝛽𝛽 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 ( + 1)(21 cos cos ) ( 1)(cos 2 cos ) = 2 sin2 sin → 𝛾𝛾 − − 𝜃𝜃 − 𝜓𝜓 − 𝛾𝛾 𝜃𝜃 𝜓𝜓 𝛾𝛾 − 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 2(1 cos )(1 + cos ) + (1 2+ cos )(1 cos ) = 2 sin sin → 𝛾𝛾 − 𝜃𝜃 𝜓𝜓 − 𝛾𝛾 − 𝜃𝜃 − 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 2 ( ) ( ) ( ) ( ) ( ) ( ) → 𝛾𝛾 tan− ½ 𝜃𝜃 + tan ½𝜓𝜓 = 2 tan ½𝜃𝜃 tan− ½ 𝜓𝜓 2 tan𝛾𝛾 𝜃𝜃½ 𝜓𝜓= 2 tan ½ . (31a) 2 2 2 2 2 → 𝛾𝛾 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 → 𝜓𝜓 𝜓𝜓 ( 1) = sin cos cos sin ′ ′ ′ ′2 𝛾𝛾( − 1𝛽𝛽)𝑥𝑥sin𝛽𝛽𝑦𝑦⁄(𝛽𝛽½ ) sin𝛾𝛾 𝜃𝜃= sin𝜓𝜓 −cos 𝜃𝜃 cos𝜓𝜓 sin [cos2 (½ ) 2 sin (½ )] sin = sin cos → − 𝛾𝛾 − 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 − 𝜃𝜃 𝜓𝜓 [1 2 tan (½2 )] tan2 = 2 tan(½ ) → 𝜃𝜃 − 𝛾𝛾 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 [ 2 ( 2 )] ( ) → 1 − 𝛾𝛾tan ½ 𝜃𝜃tan =𝜓𝜓 2 tan𝛾𝛾 ½ 𝜃𝜃 tan = tan . (31b) 2 → − 𝜓𝜓 𝜓𝜓 𝜓𝜓 → 𝜓𝜓 𝜓𝜓 1 + ( 1)( ) = sin sin + cos cos ′ ′ ′ 2 1 + 2(𝛾𝛾 − 1)𝛽𝛽sin𝑦𝑦⁄𝛽𝛽(½ ) cos𝛾𝛾 (½𝜃𝜃 ) =𝜓𝜓 sin 𝜃𝜃sin 𝜓𝜓+ cos cos 1 + ½( 2 1)(1 2 cos )(12+ cos ) = sin sin + cos cos → 𝛾𝛾 − 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 𝜃𝜃 𝜓𝜓 ½( + 12)(1 cos cos ) ½( 1)(cos cos ) = sin sin → 𝛾𝛾 − − 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 𝜃𝜃 𝜓𝜓 (12 cos )(1 + cos ) + (1 + cos2 )(1 cos ) = 2 sin sin → 𝛾𝛾 − 𝜃𝜃 𝜓𝜓 − 𝛾𝛾 − 𝜃𝜃 − 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 2 ( ) ( ) ( ) ( ) ( ) ( ) → 𝛾𝛾 tan− ½ 𝜃𝜃 + tan ½𝜓𝜓 = 2 tan ½𝜃𝜃 tan− ½ 𝜓𝜓 2 tan𝛾𝛾 𝜃𝜃½ 𝜓𝜓= 2 tan ½ . (31c) 2 2 2 2 2 → 𝛾𝛾 𝜃𝜃 𝜓𝜓 𝛾𝛾 𝜃𝜃 𝜓𝜓 → 𝜓𝜓 𝜓𝜓 9 This completes the proof that is indeed reached from after a pure boost by of Eq.(28), followed by a rotation around the -axis″ through the angle , where′ tan(½ ) = tan(½′ ). 𝑆𝑆 𝑆𝑆 𝜷𝜷 7. Discussion. If happens to 𝑧𝑧be small, then in going𝜓𝜓 from to 𝜓𝜓, the 𝛾𝛾coordinate𝜃𝜃 system undergoes a rotation through the angle , which indicates that′ a rotation″ through the small angle in the laboratory𝜃𝜃 frame will be seen by an observer in𝑆𝑆 the 𝑆𝑆rest-frame as a rotation through the larger angle . If a 𝜓𝜓 ≅ at𝛾𝛾 rest𝛾𝛾 in , with its spin axis aligned, say,′ with the - axis, suddenly𝜃𝜃 jumps to , its spin𝑆𝑆 axis relative to the′ laboratory frame will remain𝑆𝑆 unchanged′ (because no acts𝛾𝛾 𝛾𝛾on″ the gyroscope). However,𝑆𝑆 because of the coordinate rotation, the𝑥𝑥 gyroscope’s axis in its new𝑆𝑆 rest-frame appears to have rotated through an angle of . While a rotation through may be″ ascribed to the Coriolis torque acting on the gyroscope in its (steadily rotating) rest-frame, the extra𝑆𝑆 bit of rotation, ( 1) , is a purely relativistic−𝛾𝛾𝛾𝛾 effect commonly associated with the− 𝜃𝜃name of Llewellyn Thomas, who used it in conjunction with the Bohr model to account for a missing ½ factor in the spin−-orbit𝛾𝛾 − coupling𝜃𝜃 energy of the hydrogen atom.2,3 If the rate-of-change of with the (i.e., time measured in the rest-frame) is denoted by d d , then, taking time-dilation into account, the frequency of the Thomas 𝜃𝜃 𝜏𝜏 precession around the -axis (observed within the rest-frame of the gyroscope) may𝑇𝑇 be written as 𝜃𝜃⁄ 𝜏𝜏 𝛺𝛺 ( ) ( ) = ( 1) = = = 𝑧𝑧 2 3 −2 3 2 . (32) d𝜃𝜃 𝛾𝛾�𝛾𝛾 −1� d𝜃𝜃 𝛾𝛾 1−𝛾𝛾 d𝜃𝜃 𝛾𝛾 𝛽𝛽 d𝜃𝜃 𝑡𝑡 𝑇𝑇 Here, d (𝛺𝛺 ) d− 𝛾𝛾is 𝛾𝛾the− timed𝑡𝑡-rate−-of-𝛾𝛾change+1 d𝑡𝑡 of − as𝛾𝛾 +measured1 d𝑡𝑡 −in� the𝛾𝛾+1 �laboratoryd𝑡𝑡 frame. The Thomas precession is claimed to be responsible for a factor of 2 reduction of the spin-orbit coupling energy according𝜃𝜃 𝑡𝑡 ⁄ to𝑡𝑡 the semi-classical model of hydrogen𝜃𝜃 -like atoms. Seen from the electron’s rest frame, the positively-charged appears to rotate around the electron and, therefore, produce, at the location of the electron, a constant magnetic that is perpendicular to the orbital plane of the electron. This should cause the intrinsic magnetic moment of the electron to precess around the -axis at a rate that can be readily computed as a function of the𝑥𝑥𝑥𝑥 electron’s -factor, its orbital radius, and its around the nucleus — for a detailed analysis see Jackson’s textbook𝑧𝑧8 or Appendix 𝛺𝛺A of Ref.[1]. It turns out that ½ , which is 𝑔𝑔 why Thomas and his contemporaries believed that the magnetic field of the nucleus𝑇𝑇 (as seen in the electron’s rest frame) is only half as effective as it would be in the absence𝛺𝛺 ≅ −of the𝛺𝛺 Thomas precession. And this is how the semi-classical understanding of the spin-orbit coupling energy in hydrogen-like atoms was supposedly reconciled with the relevant experimental observations.

References 1. G. Spavieri and M. Mansuripur, “Origin of the Spin-Orbit Interaction,” Physica Scripta 90, 085501 (2015)] 2. L. H. Thomas, “The of the Spinning Electron,” Nature 117, 514 (1926). 3. L. H. Thomas, “The of an Electron with an Axis,” Phil. Mag. 3, 1-22 (1927). 4. W. H. Furry, “Lorentz Transformation and the Thomas Precession,” Am. J. Phys. 23, 517-525 (1955). 5. D. Shelupsky, “Derivation of the Thomas Precession Formula,” Am. J. Phys. 35, 650-651 (1967). 6. G. P. Fisher, “The Thomas Precession,” Am. J. Phys. 40, 1772-1781 (1972). 7. R. A. Muller, “Thomas precession: Where is the torque?” Am. J. Phys. 60, 313-317 (1992). 8. J. D. Jackson, Classical Electrodynamics, 3rd edition, sections 11-8 and 11-11, Wiley, New York (1999). 9. H. Kroemer, “The Thomas precession factor in spin-orbit interaction,” Am. J. Phys. 72, 51-52 (2004). 10. K. Rębilas, “Thomas precession and torque,” Am. J. Phys. 83, 199-204 (2015).

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