An exact derivation of the Thomas precession rate using the Lorentz transformation
11 An exact derivation of the Thomas precession rate using the Lorentz transformation
Masud Mansuripur
James C. Wyant College of Optical Sciences, The University of Arizona, Tucson, Arizona 85721 [Published in the
Proceedings of SPIE , Spintronics XIII, 114703U (2020); doi: 10.1117/12.2569025]
Abstract . Using the standard formalism of Lorentz transformation of the special theory of relativity, we derive the exact expression of the Thomas precession rate for an electron in a classical circular orbit around the nucleus of a hydrogen-like atom.
1. Introduction . This paper is a follow-up to a recent article in which we discussed the origin of the spin-orbit interaction based on a semi-classical model of hydrogen-like atoms without invoking the Thomas precession in the rest-frame of the electron. We argued that the Thomas precession leads to a relativistic version of the Coriolis force of Newtonian mechanics, which cannot possibly affect the energy levels of the atom, since it is a fictitious force 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 rotation matrices in three-dimensional space as well as the associated notion of the generator matrix. The results of Sec.2 are then extended in Sec.3 to derive boost matrices and their generators in Minkowskiβs four-dimensional spacetime. The 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 speed 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 Β½ factor in the spin-orbit coupling energy of hydrogen-like atoms.
2. Rotation in three-dimensional space . Consider the vector ππ = ( π₯π₯ , π¦π¦ , π§π§ ) ππ in a three-dimensional Cartesian coordinate system specified by the unit vectors ( πποΏ½ , πποΏ½ , πποΏ½ ) . In a different Cartesian system ( πποΏ½ β² , πποΏ½ β² , πποΏ½ β² ) , which shares the same origin with ( πποΏ½ , πποΏ½ , πποΏ½ ) but is rotated relative to it, the same vector may be expressed as ππ β² = ( π₯π₯ β² , π¦π¦ β² , π§π§ β² ) ππ , where οΏ½π₯π₯ β² π¦π¦ β² π§π§ β² οΏ½ = οΏ½ππ ππ ππ ππ ππ ππ ππ ππ ππ οΏ½ οΏ½π₯π₯π¦π¦π§π§οΏ½ . (1) In short, ππ β² = π΄π΄ππ , with the matrix π΄π΄ specifying a linear transformation from ( πποΏ½ , πποΏ½ , πποΏ½ ) to the ( πποΏ½ β² , πποΏ½ β² , πποΏ½ β² ) system. In a coordinate rotation, the length of ππ must remain intact, that is, ππ β²ππ ππ β² = ππ ππ π΄π΄ ππ π΄π΄ππ . Since the same relation holds for all vectors ππ , we conclude that π΄π΄ ππ π΄π΄ = πΌπΌ . The matrix π΄π΄ possessing this important property is said to be unitary . It is thus seen that the inverse of π΄π΄ is equal to its transpose π΄π΄ ππ , and since the inverse of a square matrix is unique, we must also have π΄π΄π΄π΄ ππ = πΌπΌ . The vectors ( ππ , ππ , ππ ) , ( ππ , ππ , ππ ) , and ( ππ , ππ , ππ ) thus form the orthonormal bases of the rotated coordinate system. Note from Eq.(1) that π₯π₯ β² is the projection of ( π₯π₯ , π¦π¦ , π§π§ ) onto ( ππ , ππ , ππ ) , while π¦π¦ β² is the projection of ( π₯π₯ , π¦π¦ , π§π§ ) onto ( ππ , ππ , ππ ) , and π§π§ β² is the projection of ( π₯π₯ , π¦π¦ , π§π§ ) onto ( ππ , ππ , ππ ) , thus confirming that the rows of π΄π΄ form the basis-vectors of the rotated system ( πποΏ½ β² , πποΏ½ β² , πποΏ½ β² ) . A right-handed rotation around the π₯π₯ -axis through an angle ππ is represented by the linear transformation ππ β² = π π π₯π₯ ( ππ ) ππ , where π π π₯π₯ ( ππ ) = οΏ½ ππ sin ππ β sin ππ cos πποΏ½ . (2) Differentiating π π π₯π₯ ( ππ ) with respect to ππ , we find ddππ π π π₯π₯ ( ππ ) = οΏ½ β sin ππ cos ππ β cos ππ β sin πποΏ½ = οΏ½ β οΏ½ οΏ½ ππ sin ππ β sin ππ cos πποΏ½ = ππ π₯π₯ π π π₯π₯ ( ππ ) . (3) Here, the matrix ππ π₯π₯ is the so-called generator of rotations around the π₯π₯ -axis. The rotation matrix π π π₯π₯ ( ππ ) may thus be written as π π π₯π₯ ( ππ ) = exp( ππ π₯π₯ ππ ) . (4) To verify the above identity, note that ππ π₯π₯2 = οΏ½ β β οΏ½ and ππ π₯π₯3 = βππ π₯π₯ , and that, therefore, exp( ππ π₯π₯ ππ ) = πΌπΌ + ππ π₯π₯ ππ + ππ π₯π₯2 ππ ! + ππ π₯π₯3 ππ ! + ππ π₯π₯4 ππ ! + β― = πΌπΌ + οΏ½ππ β ππ ! + ππ ! β ππ ! + β― οΏ½ ππ π₯π₯ + οΏ½ ππ ! β ππ ! + ππ ! β β― οΏ½ ππ π₯π₯2 = οΏ½ ππ sin ππ β sin ππ cos πποΏ½ . (5) In general, a sequence of three right-handed rotations through the angles ππ π₯π₯ , ππ π¦π¦ , ππ π§π§ , first around the π₯π₯ -axis, then around the π¦π¦ β² -axis, and finally around the π§π§ β³ -axis, can bring the original ( πποΏ½ , πποΏ½ , πποΏ½ ) coordinate system to any desired ( πποΏ½ β΄ , πποΏ½ β΄ , πποΏ½ β΄ ) system that shares its origin with ( πποΏ½ , πποΏ½ , πποΏ½ ) . (To see the generality of this scheme, note that any arbitrary final system can be realigned with the initial system upon reversing the sequence of three rotations.) Thus, the corresponding rotation matrix is π π οΏ½ππ π₯π₯ , ππ π¦π¦ , ππ π§π§ οΏ½ = π π π§π§ ( ππ π§π§ ) π π π¦π¦ ( ππ π¦π¦ ) π π π₯π₯ ( ππ π₯π₯ ) = οΏ½ cos ππ π§π§ sin ππ π§π§ β sin ππ π§π§ cos ππ π§π§
00 0 1 οΏ½ οΏ½ cos ππ π¦π¦ β sin ππ π¦π¦ ππ π¦π¦ ππ π¦π¦ οΏ½ οΏ½ ππ π₯π₯ sin ππ π₯π₯ β sin ππ π₯π₯ cos ππ π₯π₯ οΏ½ . (6) Since the above rotation matrices do not commute with each other, the order of multiplication in Eq.(6) is important and should not be altered. The generator matrices for the above rotations are ππ π₯π₯ = οΏ½ β οΏ½ , ππ π¦π¦ = οΏ½ β
10 0 0+1 0 0 οΏ½ , ππ π§π§ = οΏ½ β οΏ½ . (7) One may write π π οΏ½ππ π₯π₯ , ππ π¦π¦ , ππ π§π§ οΏ½ = exp( ππ π§π§ ππ π§π§ ) exp( ππ π¦π¦ ππ π¦π¦ ) exp( ππ π₯π₯ ππ π₯π₯ ) , with the caveat that, since ππ π₯π₯ , ππ π¦π¦ , ππ π§π§ do not commute with each other, it is not permissible to write the product of the exponentials as exp( π½π½ β πΊπΊ ) .
3. Boost (from one inertial frame to another) as a coordinate rotation in special relativity . In the special theory of relativity, where time and space are treated on an equal footing, a general coordinate transformation (involving a coordinate rotation as well as a boost) is written οΏ½ππππβ²π₯π₯ β² π¦π¦ β² π§π§ β² οΏ½ = οΏ½ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ οΏ½ οΏ½πππππ₯π₯π¦π¦π§π§ οΏ½ . (8) The 4-vector ππ = ( ππππ , π₯π₯ , π¦π¦ , π§π§ ) ππ is transformed between primed and unprimed reference frames by multiplication with a matrix π΄π΄ . The metric of the flat space-time of special relativity is ππ = οΏ½β οΏ½ . (9) Since the Lorentz transformation preserves the norm ππ ππ ππππ of the 4-vector ππ , the matrix π΄π΄ must satisfy the identity π΄π΄ ππ πππ΄π΄ = ππ . Suppose the ( πποΏ½ , πποΏ½ , πποΏ½ ) coordinate axes of the reference frame ππ (the so-called laboratory frame) are parallel to the corresponding axes of the ( πποΏ½ β² , πποΏ½ β² , πποΏ½ β² ) system, often referred to as the rest-frame ππβ² . The latter frame ( ππ β² ) moves relative to the former ( ππ ) at constant velocity ππ along the positive π₯π₯ -axis, while at ππ = ππ β² = 0 the origins of the two coordinate systems coincide. Using the standard notation tanh( πΌπΌ ) = π½π½ = ππ ππβ and cosh( πΌπΌ ) = πΎπΎ = 1 οΏ½ β π½π½ β , where πΌπΌ is the so-called rapidity and ππ is the speed of light in vacuum, the pure Lorentz boost from ππ to ππ β² may be written as follows: οΏ½ππππ β² π₯π₯ β² π¦π¦ β² π§π§ β² οΏ½ = οΏ½ cosh πΌπΌ β sinh πΌπΌ β sinh πΌπΌ cosh πΌπΌ οΏ½ οΏ½πππππ₯π₯π¦π¦π§π§ οΏ½ . (10) Differentiation with respect to πΌπΌ of the above transformation matrix, denoted by π¬π¬ π₯π₯ ( πΌπΌ ) , yields ddπΌπΌ π¬π¬ π₯π₯ ( πΌπΌ ) = ddπΌπΌ οΏ½ cosh πΌπΌ β sinh πΌπΌ β sinh πΌπΌ cosh πΌπΌ οΏ½ = οΏ½ sinh πΌπΌ β cosh πΌπΌ β cosh πΌπΌ sinh πΌπΌ οΏ½ = οΏ½ β β οΏ½ οΏ½ cosh πΌπΌ β sinh πΌπΌ β sinh πΌπΌ cosh πΌπΌ οΏ½ = β οΏ½ οΏ½ π¬π¬ π₯π₯ ( πΌπΌ ) = βπΎπΎ π₯π₯ π¬π¬ π₯π₯ ( πΌπΌ ) . (11) The matrix πΎπΎ π₯π₯ is thus seen to be the generator of pure boost along the π₯π₯ -axis. Consequently, the boost may be expressed as π¬π¬ π₯π₯ ( πΌπΌ ) = exp( βπΌπΌπΎπΎ π₯π₯ ) . Similarly, the generator matrices for pure boosts along the π¦π¦ - and π§π§ -axes are πΎπΎ π¦π¦ = οΏ½ οΏ½ , πΎπΎ π§π§ = οΏ½ οΏ½ . (12) A sequence of three successive boosts, first from ππ to ππ β² along the π₯π₯ -axis, then from ππ β² to ππ β³ along the π¦π¦ β² -axis, and finally from ππ β³ to ππ β΄ along the π§π§ β³ -axis, is described by the Lorentz matrix π¬π¬ ( πΌπΌ π₯π₯ , πΌπΌ π¦π¦ , πΌπΌ π§π§ ) = π¬π¬ π§π§ ( πΌπΌ π§π§ ) π¬π¬ π¦π¦ ( πΌπΌ π¦π¦ ) π¬π¬ π₯π₯ ( πΌπΌ π₯π₯ ) = exp( βπΌπΌ π§π§ πΎπΎ π§π§ ) exp( βπΌπΌ π¦π¦ πΎπΎ π¦π¦ ) exp( βπΌπΌ π₯π₯ πΎπΎ π₯π₯ ) . Considering that the matrices πΎπΎ π₯π₯ , πΎπΎ π¦π¦ , πΎπΎ π§π§ do not commute with each other, one should avoid the temptation to write π¬π¬ ( πΌπΌ π₯π₯ , πΌπΌ π¦π¦ , πΌπΌ π§π§ ) as exp( βπΆπΆ β π²π² ) .
4. Boost in an arbitrary direction . Let ππ and ππ β² be two reference frames whose origins overlap at ππ = ππ β² = 0 . The frame ππ β² moves with velocity π½π½ = ππ π₯π₯ πποΏ½ + ππ π¦π¦ πποΏ½ + ππ π§π§ πποΏ½ relative to the laboratory frame ππ . The rapidity πΌπΌ and the polar coordinates ( ππ , ππ ) of π½π½ are given by tanh( πΌπΌ ) = π½π½ = ππ ππβ , π½π½ π₯π₯ = π½π½ sin ππ cos ππ , π½π½ π¦π¦ = π½π½ sin ππ sin ππ , and π½π½ π§π§ = π½π½ cos ππ . A Lorentz transformation from ππ to ππ β² thus involves the following sequence of steps: i) Rotation around π§π§ through the angle ππ . ii) Rotation around π¦π¦ through the angle ππ . iii) Boost along π§π§ with rapidity πΌπΌ . iv) Rotation around π¦π¦ β² through the angle βππ . v) Rotation around π§π§ β² through the angle βππ . The matrix product representing the above transformation is readily evaluated as follows: π¬π¬ ( π·π· ) = π π π§π§ ( βππ ) π π π¦π¦ ( βππ ) π΄π΄ π§π§ ( boost ) ( πΌπΌ ) π π π¦π¦ ( ππ ) π π π§π§ ( ππ ) = οΏ½ ππ β sin ππ
00 sin ππ cos ππ
00 0 0 1 οΏ½ οΏ½ ππ ππ β sin ππ πποΏ½ Γ οΏ½ cosh πΌπΌ β sinh πΌπΌ β sinh πΌπΌ πΌπΌ οΏ½ οΏ½ ππ β sin ππ ππ ππ οΏ½ οΏ½ ππ sin ππ β sin ππ cos ππ
00 0 0 1 οΏ½ = ββββ cosh πΌπΌ β sinh πΌπΌ sin ππ cos ππ β sinh πΌπΌ sin ππ sin ππ β sinh πΌπΌ cos ππβ sinh πΌπΌ sin ππ cos ππ πΌπΌ β
1) sin ππ cos ππ (cosh πΌπΌ β
1) sin ππ sin ππ cos ππ (cosh πΌπΌ β
1) sin ππ cos ππ cos ππβ sinh πΌπΌ sin ππ sin ππ (cosh πΌπΌ β
1) sin ππ sin ππ cos ππ πΌπΌ β
1) sin ππ sin ππ (cosh πΌπΌ β
1) sin ππ cos ππ sin ππβ sinh πΌπΌ cos ππ (cosh πΌπΌ β
1) sin ππ cos ππ cos ππ (cosh πΌπΌ β
1) sin ππ cos ππ sin ππ πΌπΌ β
1) cos ππ β βββ = ββββ πΎπΎ βπΎπΎπ½π½ π₯π₯ βπΎπΎπ½π½ π¦π¦ βπΎπΎπ½π½ π§π§ βπΎπΎπ½π½ π₯π₯ πΎπΎ β π½π½ π₯π₯2 π½π½ β ( πΎπΎ β π½π½ π₯π₯ π½π½ π¦π¦ π½π½ β ( πΎπΎ β π½π½ π₯π₯ π½π½ π§π§ π½π½ ββπΎπΎπ½π½ π¦π¦ ( πΎπΎ β π½π½ π₯π₯ π½π½ π¦π¦ π½π½ β πΎπΎ β π½π½ π¦π¦2 π½π½ β ( πΎπΎ β π½π½ π¦π¦ π½π½ π§π§ π½π½ ββπΎπΎπ½π½ π§π§ ( πΎπΎ β π½π½ π₯π₯ π½π½ π§π§ π½π½ β ( πΎπΎ β π½π½ π¦π¦ π½π½ π§π§ π½π½ β πΎπΎ β π½π½ π§π§2 π½π½ β β βββ . (13) Note that the pure boost matrix π¬π¬ ( π·π· ) given by Eq.(13) is symmetric. The inverse of this matrix transforms the space-time coordinates of ππ β² back to ππ . This, however, is equivalent to switching the sign of π½π½ , which flips the signs of π½π½ π₯π₯ , π½π½ π¦π¦ , π½π½ π§π§ . The inverse of π¬π¬ ( π·π· ) of Eq.(13) is thus obtained simply by reversing the sign of π·π· . The above definition of a pure boost might lead one to believe that the ( πποΏ½ , πποΏ½ , πποΏ½ ) axes of ππ are parallel to the corresponding axes ( πποΏ½ β² , πποΏ½ β² , πποΏ½ β² ) of ππ β² . This, however, is not the case, as can be readily seen by examining the simple example where π·π· = π½π½ π₯π₯ πποΏ½ + π½π½ π¦π¦ πποΏ½ = ( ππ ππβ )(cos ππ πποΏ½ + sin ππ πποΏ½ ) . At ππ = 0 , when the origins of the two coordinate systems overlap, let the point ( π₯π₯ , π¦π¦ , π§π§ ) in the laboratory frame ππ correspond to the point ( π₯π₯ β² , 0, 0) located on the π₯π₯ β² -axis of the rest-frame ππ β² . A Lorentz transformation from ππ to ππ β² yields οΏ½ππππ β² π₯π₯ β² οΏ½ = βββ πΎπΎ βπΎπΎπ½π½ π₯π₯ βπΎπΎπ½π½ π¦π¦ βπΎπΎπ½π½ π₯π₯ πΎπΎ β π½π½ π₯π₯2 π½π½ β ( πΎπΎ β π½π½ π₯π₯ π½π½ π¦π¦ π½π½ β βπΎπΎπ½π½ π¦π¦ ( πΎπΎ β π½π½ π₯π₯ π½π½ π¦π¦ π½π½ β πΎπΎ β π½π½ π¦π¦2 π½π½ β
00 0 0 1 β ββ οΏ½ π₯π₯π¦π¦ π§π§ οΏ½ . (14) Consequently, ( π₯π₯ , π¦π¦ , π§π§ ) = οΏ½ πΎπΎ β π½π½ π¦π¦ π½π½β ) , β ( πΎπΎ β π½π½ π₯π₯ π½π½ π¦π¦ π½π½ β , 0 οΏ½ ( π₯π₯ β² πΎπΎβ ) , indicating that the π₯π₯ β² -axis, when viewed from the laboratory frame, appears to be rotated clockwise through an angle ππ π₯π₯ β² , where tan ππ π₯π₯ β² = | π¦π¦ π₯π₯β | = ( πΎπΎβ1 ) π½π½ π₯π₯ π½π½ π¦π¦ π½π½ β1+ ( πΎπΎβ1 )( π½π½ π¦π¦ π½π½β ) = ( πΎπΎβ1 ) sin ππ cos ππ1+ ( πΎπΎβ1 ) sin ππ . (15) Similarly, by finding the coordinates ( π₯π₯ , π¦π¦ , π§π§ ) of a point in the laboratory frame ππ at ππ = 0 corresponding to the point (0, π¦π¦ β² , 0) in the rest-frame ππ β² , we find the apparent counterclockwise rotation angle ππ π¦π¦ β² of the π¦π¦ β² -axis as seen from the laboratory frame to be tan ππ π¦π¦ β² = | π₯π₯ π¦π¦β | = ( πΎπΎβ1 ) π½π½ π₯π₯ π½π½ π¦π¦ π½π½ β1+ ( πΎπΎβ1 )( π½π½ π₯π₯ π½π½β ) = ( πΎπΎβ1 ) sin ππ cos ππ1+ ( πΎπΎβ1 ) cos ππ . (16) Note that the angles ππ π₯π₯ β² and ππ π¦π¦ β² are unequal and also in opposite directions, which means that, viewed from the π₯π₯π¦π¦π§π§ frame, the π₯π₯ β² π¦π¦ β² π§π§ β² frame appears to be deformed, not rotated. The pure boost by π½π½ , embodied in the transformation matrix π¬π¬ ( π·π· ) of Eq.(13), may be expressed in more compact form as ππππβ² = πΎπΎ ( ππππ β π·π· β ππ ) , (17a) ππ β² = ππ + [( πΎπΎ β π½π½ β ]( π·π· β ππ ) π·π· β πΎπΎπ·π·ππππ . (17b) Writing ππ = ππ β₯ + ππ β₯ , with the subscripts β₯ and β₯ referring, respectively, to the parallel and perpendicular components of ππ relative to π·π· = π½π½ ππβ , Eq.(17) may be further simplified, yielding ππβ² = πΎπΎ [ ππ β ( ππ ππ β ) ππ β₯ ] , (18a) ππ β² = πΎπΎ ( ππ β₯ β π½π½ππ ) + ππ β₯ . (18b)
5. Alternative treatment of boost in an arbitrary direction . An alternative derivation of Eq.(13) starts from the assumption that the direction of π½π½ is fixed, but that its magnitude changes incrementally, from ππ to ππ + d ππ . Recalling that tanh( πΌπΌ ) = ππ ππβ and associating the rapidity πΌπΌ with a vector of length πΌπΌ in the direction of π·π·οΏ½ = π·π· π½π½β , we will have (d πΌπΌ π₯π₯ , d πΌπΌ π¦π¦ , d πΌπΌ π§π§ ) = ( π½π½Μ π₯π₯ , π½π½Μ π¦π¦ , π½π½Μ π§π§ )d πΌπΌ . Invoking Eqs.(11) and (12), we may write ddπΌπΌ π¬π¬ π·π·οΏ½ ( πΌπΌ ) = βββ βπ½π½Μ π₯π₯ βπ½π½Μ π¦π¦ βπ½π½Μ π§π§ βπ½π½Μ π₯π₯ βπ½π½Μ π¦π¦ βπ½π½Μ π§π§ β ββ π¬π¬ π·π·οΏ½ ( πΌπΌ ) = ( βπ·π·οΏ½ β π²π² ) π¬π¬ π·π·οΏ½ ( πΌπΌ ) . (19) Consequently, π¬π¬ π·π·οΏ½ ( πΌπΌ ) = exp[ β ( π·π·οΏ½ β π²π² ) πΌπΌ ] . Note that the incremental nature of the boost (i.e., from ππ to ππ + ππππ while keeping π·π·οΏ½ constant) has made it possible in this instance to combine the three exponentials into a single entity, exp[ β ( π·π·οΏ½ β π²π² ) πΌπΌ ] , without due consideration for the sequential order of individual boosts along πποΏ½ , πποΏ½ , and πποΏ½ . β Proceeding to evaluate exp[ β ( π·π·οΏ½ β π²π² ) πΌπΌ ] via its Taylor series expansion, we find ( π·π·οΏ½ β π²π² ) = βββ π½π½Μ π₯π₯ π½π½Μ π¦π¦ π½π½Μ π§π§ π½π½Μ π₯π₯ π½π½Μ π¦π¦ π½π½Μ π§π§ β ββ βββ π½π½Μ π₯π₯ π½π½Μ π¦π¦ π½π½Μ π§π§ π½π½Μ π₯π₯ π½π½Μ π¦π¦ π½π½Μ π§π§ β ββ = βββ π½π½Μ π₯π₯2 π½π½Μ π₯π₯ π½π½Μ π¦π¦ π½π½Μ π₯π₯ π½π½Μ π§π§ π½π½Μ π₯π₯ π½π½Μ π¦π¦ π½π½Μ π¦π¦2 π½π½Μ π¦π¦ π½π½Μ π§π§ π½π½Μ π₯π₯ π½π½Μ π§π§ π½π½Μ π¦π¦ π½π½Μ π§π§ π½π½Μ π§π§2 β ββ . (20) ( π·π·οΏ½ β π²π² ) = βββ π½π½Μ π₯π₯ π½π½Μ π¦π¦ π½π½Μ π§π§ π½π½Μ π₯π₯ π½π½Μ π¦π¦ π½π½Μ π§π§ β ββ βββ π½π½Μ π₯π₯2 π½π½Μ π₯π₯ π½π½Μ π¦π¦ π½π½Μ π₯π₯ π½π½Μ π§π§ π½π½Μ π₯π₯ π½π½Μ π¦π¦ π½π½Μ π¦π¦2 π½π½Μ π¦π¦ π½π½Μ π§π§ π½π½Μ π₯π₯ π½π½Μ π§π§ π½π½Μ π¦π¦ π½π½Μ π§π§ π½π½Μ π§π§2 β ββ = βββ π½π½Μ π₯π₯ π½π½Μ π¦π¦ π½π½Μ π§π§ π½π½Μ π₯π₯ π½π½Μ π¦π¦ π½π½Μ π§π§ β ββ . (21) exp[ β ( π·π·οΏ½ β π²π² ) πΌπΌ ] = πΌπΌ β πΌπΌ1 ! ( π·π·οΏ½ β π²π² ) + πΌπΌ ! ( π·π·οΏ½ β π²π² ) β πΌπΌ ! ( π·π·οΏ½ β π²π² ) + πΌπΌ ! ( π·π·οΏ½ β π²π² ) β β― = πΌπΌ β ( πΌπΌ1 ! + πΌπΌ ! + πΌπΌ ! + β― )( π·π·οΏ½ β π²π² ) + ( πΌπΌ ! + πΌπΌ ! + πΌπΌ ! + β― )( π·π·οΏ½ β π²π² ) = πΌπΌ β (sinh πΌπΌ )( π·π·οΏ½ β π²π² ) + (cosh πΌπΌ β π·π·οΏ½ β π²π² ) = πΌπΌ β πΎπΎπ·π· β π²π² + [( πΎπΎ β π½π½ β ]( π·π· β π²π² ) = ββββ πΎπΎ βπΎπΎπ½π½ π₯π₯ βπΎπΎπ½π½ π¦π¦ βπΎπΎπ½π½ π§π§ βπΎπΎπ½π½ π₯π₯ ( πΎπΎ β ) π½π½ π₯π₯ π½π½ β ( πΎπΎ β ) π½π½ π₯π₯ π½π½ π¦π¦ π½π½ β ( πΎπΎ β ) π½π½ π₯π₯ π½π½ π§π§ π½π½ β βπΎπΎπ½π½ π¦π¦ ( πΎπΎ β ) π½π½ π₯π₯ π½π½ π¦π¦ π½π½ β ( πΎπΎ β ) π½π½ π¦π¦ π½π½ β ( πΎπΎ β ) π½π½ π¦π¦ π½π½ π§π§ π½π½ β βπΎπΎπ½π½ π§π§ ( πΎπΎ β ) π½π½ π₯π₯ π½π½ π§π§ π½π½ β ( πΎπΎ β ) π½π½ π¦π¦ π½π½ π§π§ π½π½ β ( πΎπΎ β ) π½π½ π§π§ π½π½ β β βββ . (22) The above matrix is seen to be identical with that obtained in Eq.(13), which was derived using a combination of boost and rotations. β 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. The Thomas Precession . Consider the laboratory frame ππ , a first rest-frame ππ β² moving within ππ with (normalized) velocity π·π· = π½π½π·π·οΏ½ , and a second rest-frame ππ β³ , also moving within ππ but with (normalized) velocity π·π· = π½π½π·π·οΏ½ . Note that π·π· and π·π· share the same magnitude π½π½ , differing only in their directions. The three coordinate systems share a common π§π§ -axis and their origins coincide at ππ = ππ β² = ππ β³ = 0 , although their π₯π₯ and π¦π¦ axes are oriented differently. In the case of ππ β² , the π₯π₯β² -axis is aligned with π·π·οΏ½ , while in the case of ππ β³ , the π₯π₯β³ -axis is aligned with π·π·οΏ½ . Let π·π·οΏ½ , = (cos ππ , ) πποΏ½ +(sin ππ , ) πποΏ½ . A Lorentz transformation from ππ β² to ππ β³ thus involves the following sequence of steps: i) Boost from ππ β² to ππ along the π₯π₯ β² -axis. ii) Rotation in ππ around the π§π§ -axis through the angle βππ . iii) Rotation in ππ around the π§π§ -axis through the angle + ππ . iv) Boost from ππ to ππβ³ along the π₯π₯ β³ -axis. Steps (ii) and (iii), of course, may be combined into a single rotation around the π§π§ -axis through ππ = ππ β ππ . The matrix representing the transformation from ππ β² to ππ β³ is thus given by π¬π¬ ( π½π½ , ππ ) = οΏ½ πΎπΎ βπΎπΎπ½π½ βπΎπΎπ½π½ πΎπΎ οΏ½ οΏ½ ππ sin ππ β sin ππ cos ππ
00 0 0 1 οΏ½ οΏ½ πΎπΎ πΎπΎπ½π½ πΎπΎπ½π½ πΎπΎ οΏ½ = ββββπΎπΎ (1 β π½π½ cos ππ ) πΎπΎ π½π½ (1 β cos ππ ) βπΎπΎπ½π½ sin ππ βπΎπΎ π½π½ (1 β cos ππ ) πΎπΎ (cos ππ β π½π½ ) πΎπΎ sin ππ βπΎπΎπ½π½ sin ππ βπΎπΎ sin ππ cos ππ
00 0 0 1 β βββ . (23) Note that π¬π¬ ( π½π½ , ππ ) in Eq.(23) is not a symmetric matrix, implying that it does not represent a pure boost but, rather, a boost followed by a coordinate rotation. Writing π¬π¬ ( π½π½ , ππ ) = π π π§π§ ( ππ ) π΄π΄ ( boost ) ( π·π· β² ) , where π·π· β² is the (as yet unknown) boost velocity (from ππ β² to ππ β³ ), and ππ is the (as yet unknown) rotation angle within ππβ³ , we may write π΄π΄ ( boost ) ( π·π· β² ) = π π π§π§ ( βππ ) π¬π¬ ( π½π½ , ππ ) = βββ ππ β sin ππ
00 sin ππ cos ππ
00 0 0 1 β ββ ββββπΎπΎ (1 β π½π½ cos ππ ) πΎπΎ π½π½ (1 β cos ππ ) βπΎπΎπ½π½ sin ππ βπΎπΎ π½π½ (1 β cos ππ ) πΎπΎ (cos ππ β π½π½ ) πΎπΎ sin ππ βπΎπΎπ½π½ sin ππ βπΎπΎ sin ππ cos ππ
00 0 0 1 β βββ = ββββ πΎπΎ (1 β π½π½ cos ππ ) πΎπΎ π½π½ (1 β cos ππ ) βπΎπΎπ½π½ sin ππ βπΎπΎ π½π½ (1 β cos ππ ) cos ππ + πΎπΎπ½π½ sin ππ sin ππ πΎπΎ (cos ππ β π½π½ ) cos ππ + πΎπΎ sin ππ sin ππ πΎπΎ sin ππ cos ππ β cos ππ sin ππ βπΎπΎ π½π½ (1 β cos ππ ) sin ππ β πΎπΎπ½π½ sin ππ cos ππ πΎπΎ (cos ππ β π½π½ ) sin ππ β πΎπΎ sin ππ cos ππ πΎπΎ sin ππ sin ππ + cos ππ cos ππ β βββ . (24) This pure boost must have a symmetric matrix. The acceptable value of ππ is thus found by ensuring the symmetry of the matrix in Eq.(24). There are three constraints on the matrix, but they all lead to the same value of ππ , as shown below. 1 st constraint: βπΎπΎ π½π½ (1 β cos ππ ) cos ππ + πΎπΎπ½π½ sin ππ sin ππ = πΎπΎ π½π½ (1 β cos ππ ) β sin ππ sin ππ = πΎπΎ (1 β cos ππ )(1 + cos ππ ) β sin(Β½ ππ ) cos(Β½ ππ ) sin(Β½ ππ ) cos(Β½ ππ ) = πΎπΎ sin (Β½ ππ ) cos (Β½ ππ ) β tan(Β½ ππ ) = πΎπΎ tan(Β½ ππ ) . (25a) 2 nd constraint: βπΎπΎ π½π½ (1 β cos ππ ) sin ππ β πΎπΎπ½π½ sin ππ cos ππ = βπΎπΎπ½π½ sin ππ β πΎπΎ (1 β cos ππ ) sin ππ = sin ππ (1 β cos ππ ) β tan(Β½ ππ ) = πΎπΎ tan(Β½ ππ ) . (25b) 3 rd constraint: πΎπΎ (cos ππ β π½π½ ) sin ππ β πΎπΎ sin ππ cos ππ = πΎπΎ sin ππ cos ππ β cos ππ sin ππ β ( πΎπΎ + 1) cos ππ sin ππ β πΎπΎ π½π½ sin ππ = 2 πΎπΎ sin ππ cos ππ β ( πΎπΎ + 1) cos ππ sin ππ β ( πΎπΎ β
1) sin ππ = 2 πΎπΎ sin ππ cos ππ β [1 + cos ππ β πΎπΎ (1 β cos ππ )] tan ππ = 2 πΎπΎ sin ππ cos ππ β [cos (Β½ ππ ) β πΎπΎ sin (Β½ ππ )] tan ππ = 2 πΎπΎ sin(Β½ ππ ) cos(Β½ ππ ) β [1 β πΎπΎ tan (Β½ ππ )] tan(Β½ ππ ) = πΎπΎ tan(Β½ ππ ) [1 β tan (Β½ ππ )] β [tan(Β½ ππ ) β πΎπΎ tan(Β½ ππ )][1 + πΎπΎ tan(Β½ ππ ) tan(Β½ ππ )] = 0 β tan(Β½ ππ ) = πΎπΎ tan(Β½ ππ ) . (25c) With the rotation angle ππ thus determined, the boost matrix of Eq.(24) becomes π΄π΄ ( boost ) ( π·π· β² ) = ββββπΎπΎ (1 β π½π½ cos ππ ) πΎπΎ π½π½ (1 β cos ππ ) βπΎπΎπ½π½ sin ππ πΎπΎ π½π½ (1 β cos ππ ) πΎπΎ (cos ππ β π½π½ ) cos ππ + πΎπΎ sin ππ sin ππ πΎπΎ sin ππ cos ππ β cos ππ sin ππ βπΎπΎπ½π½ sin ππ πΎπΎ sin ππ cos ππ β cos ππ sin ππ πΎπΎ sin ππ sin ππ + cos ππ cos ππ
00 0 0 1 β βββ . (26) Comparison with Eq.(22) shows that the following equalities must hold πΎπΎ β² = πΎπΎ (1 β π½π½ cos ππ ) , (27a) πΎπΎ β² π½π½ π₯π₯β² = βπΎπΎ π½π½ (1 β cos ππ ) , (27b) πΎπΎ β² π½π½ π¦π¦β² = πΎπΎπ½π½ sin ππ , (27c) π½π½ π§π§β² = 0 , (27d) πΎπΎ β² β π½π½ π₯π₯β² π½π½ β² β ) = πΎπΎ (cos ππ β π½π½ ) cos ππ + πΎπΎ sin ππ sin ππ , (27e) ( πΎπΎ β² β π½π½ π₯π₯β² π½π½ π¦π¦β² π½π½ β²2 β = πΎπΎ sin ππ cos ππ β cos ππ sin ππ , (27f) πΎπΎ β² β π½π½ π¦π¦β² π½π½ β² β ) = πΎπΎ sin ππ sin ππ + cos ππ cos ππ . (27g) The boost velocity from ππ β² to ππ β³ is thus found from Eqs.(27a) -(27d) as follows: π·π· β² = πΎπΎ π½π½ ( cos ππβ1 ) πποΏ½ + πΎπΎπ½π½ sin πππποΏ½1+2 ( πΎπΎ β1 ) sin (Β½ ππ ) . (28) Note that, at non-relativistic velocities, π·π· β² β π·π· β π·π· = π½π½ (cos ππ β πποΏ½ β² + ( π½π½ sin ππ ) πποΏ½ β² . To confirm that π·π· β² of Eq.(28) is indeed consistent with πΎπΎ β² of Eq.(27a), that is, πΎπΎ β² = 1/ οΏ½ β π½π½ β²2 , we write β π½π½ β²2 = 1 β οΏ½π½π½ π₯π₯β²2 + π½π½ π¦π¦β²2 + π½π½ π§π§β²2 οΏ½ = 1 β πΎπΎ π½π½ ( ) +πΎπΎ π½π½ sin πππΎπΎ ( cos ππ ) = ( cos ππ ) = (1 πΎπΎ β² β ) . (29) To confirm the remaining identities in Eq.(27), we first prove the following relations: πΎπΎ β² β πΎπΎ (1 β π½π½ cos ππ ) β πΎπΎ β β cos ππ ) = 2( πΎπΎ β
1) sin (Β½ ππ ) , (30a) ( π½π½ β² π½π½ π₯π₯β² β ) = 1 + ( π½π½ π¦π¦β² π½π½ π₯π₯β² β ) = 1 + οΏ½ sin πππΎπΎ ( ) οΏ½ = 1 + οΏ½ cos (Β½ ππ ) πΎπΎ sin (Β½ ππ ) οΏ½ = 1 + (Β½ ππ ) = (Β½ ππ ) , (30b) ( π½π½ β² π½π½ π¦π¦β² β ) = 1 + ( π½π½ π₯π₯β² π½π½ π¦π¦β² β ) = 1 + οΏ½ πΎπΎ ( ) sin ππ οΏ½ = 1 + οΏ½ πΎπΎ sin (Β½ ππ ) cos (Β½ ππ ) οΏ½ = 1 + tan (Β½ ππ ) = (Β½ ππ ) , (30c) π½π½ β²2 ( π½π½ π₯π₯β² π½π½ π¦π¦β² ) οΏ½ = ( π½π½ π₯π₯β² π½π½ π¦π¦β² β ) + ( π½π½ π¦π¦β² π½π½ π₯π₯β² β ) = β tan(Β½ ππ ) β (Β½ ππ ) = β . (30d) Substitution into Eqs.(27e) -(27g) now yields πΎπΎ β² β π½π½ π₯π₯β² π½π½ β² β ) = πΎπΎ (cos ππ β π½π½ ) cos ππ + πΎπΎ sin ππ sin ππ β πΎπΎ β
1) sin (Β½ ππ ) sin (Β½ ππ ) = πΎπΎ (cos ππ β π½π½ ) cos ππ + πΎπΎ sin ππ sin ππ β πΎπΎ β β cos ππ )(1 β cos ππ ) β πΎπΎ cos ππ cos ππ + ( πΎπΎ β
1) cos ππ = πΎπΎ sin ππ sin ππ β ( πΎπΎ + 1)(1 β cos ππ cos ππ ) β ( πΎπΎ β ππ β cos ππ ) = 2 πΎπΎ sin ππ sin ππ β πΎπΎ (1 β cos ππ )(1 + cos ππ ) + (1 + cos ππ )(1 β cos ππ ) = 2 πΎπΎ sin ππ sin ππ β πΎπΎ tan (Β½ ππ ) + tan (Β½ ππ ) = 2 πΎπΎ tan(Β½ ππ ) tan(Β½ ππ ) β (Β½ ππ ) = 2 tan (Β½ ππ ) . (31a) ( πΎπΎ β² β π½π½ π₯π₯β² π½π½ π¦π¦β² π½π½ β²2 β = πΎπΎ sin ππ cos ππ β cos ππ sin ππ β β ( πΎπΎ β
1) sin (Β½ ππ ) sin ππ = πΎπΎ sin ππ cos ππ β cos ππ sin ππ β [cos (Β½ ππ ) β πΎπΎ sin (Β½ ππ )] sin ππ = πΎπΎ sin ππ cos ππ β [1 β πΎπΎ tan (Β½ ππ )] tan ππ = 2 πΎπΎ tan(Β½ ππ ) β [1 β tan (Β½ ππ )] tan ππ = 2 tan(Β½ ππ ) β tan ππ = tan ππ . (31b) πΎπΎ β² β π½π½ π¦π¦β² π½π½ β² β ) = πΎπΎ sin ππ sin ππ + cos ππ cos ππ β πΎπΎ β
1) sin (Β½ ππ ) cos (Β½ ππ ) = πΎπΎ sin ππ sin ππ + cos ππ cos ππ β πΎπΎ β β cos ππ )(1 + cos ππ ) = πΎπΎ sin ππ sin ππ + cos ππ cos ππ β Β½( πΎπΎ + 1)(1 β cos ππ cos ππ ) β Β½( πΎπΎ β ππ β cos ππ ) = πΎπΎ sin ππ sin ππ β πΎπΎ (1 β cos ππ )(1 + cos ππ ) + (1 + cos ππ )(1 β cos ππ ) = 2 πΎπΎ sin ππ sin ππ β πΎπΎ tan (Β½ ππ ) + tan (Β½ ππ ) = 2 πΎπΎ tan(Β½ ππ ) tan(Β½ ππ ) β (Β½ ππ ) = 2 tan (Β½ ππ ) . (31c) 0 This completes the proof that ππ β³ is indeed reached from ππ β² after a pure boost by π·π· β² of Eq.(28), followed by a rotation around the π§π§ -axis through the angle ππ , where tan(Β½ ππ ) = πΎπΎ tan(Β½ ππ ) .
7. Discussion . If ππ happens to be small, then in going from ππ β² to ππ β³ , the coordinate system undergoes a rotation through the angle ππ β πΎπΎππ , which indicates that a rotation through the small angle ππ in the laboratory frame ππ will be seen by an observer in the rest-frame ππ β² as a rotation through the larger angle πΎπΎππ . If a gyroscope at rest in ππ β² , with its spin axis aligned, say, with the π₯π₯ β² -axis, suddenly jumps to ππ β³ , its spin axis relative to the laboratory frame will remain unchanged (because no torque acts on the gyroscope). However, because of the coordinate rotation, the gyroscopeβs axis in its new rest-frame ππ β³ appears to have rotated through an angle of βπΎπΎππ . While a rotation through βππ may be ascribed to the Coriolis torque acting on the gyroscope in its (steadily rotating) rest-frame, the extra bit of rotation, β ( πΎπΎ β ππ , is a purely relativistic effect commonly associated with the name of Llewellyn Thomas, who used it in conjunction with the Bohr model to account for a missing Β½ factor in the spin-orbit coupling energy of the hydrogen atom. If the rate-of-change of ππ with the proper time ππ (i.e., time measured in the rest-frame) is denoted by d ππ d ππβ , then, taking time-dilation into account, the frequency πΊπΊ ππ of the Thomas precession around the π§π§ -axis (observed within the rest-frame of the gyroscope) may be written as πΊπΊ ππ = βπΎπΎ ( πΎπΎ β dππdπ‘π‘ = β πΎπΎοΏ½πΎπΎ β1οΏ½πΎπΎ+1 dππdπ‘π‘ = β πΎπΎ ( β2 ) πΎπΎ+1 dππdπ‘π‘ = β οΏ½ πΎπΎ π½π½ πΎπΎ+1 οΏ½ dππ ( π‘π‘ ) dπ‘π‘ . (32) Here, d ππ ( ππ ) d ππβ is the time-rate-of-change of ππ as measured in the laboratory frame. The Thomas precession is claimed to be responsible for a factor of reduction of the spin-orbit coupling energy according to the semi-classical model of hydrogen-like atoms. Seen from the electronβs rest frame, the positively-charged atomic nucleus appears to rotate around the electron and, therefore, produce, at the location of the electron, a constant magnetic field that is perpendicular to the π₯π₯π¦π¦ orbital plane of the electron. This magnetic field should cause the intrinsic magnetic moment of the electron to precess around the π§π§ -axis at a rate πΊπΊ that can be readily computed as a function of the electronβs ππ -factor, its orbital radius, and its angular velocity around the nucleus β for a detailed analysis see Jacksonβs textbook or Appendix A of Ref.[1]. It turns out that πΊπΊ ππ β β Β½ πΊπΊ , which is why Thomas and his contemporaries believed that the magnetic field of the nucleus (as seen in the electronβs rest frame) is only half as effective as it would be in the absence of the 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 G. Spavieri and M. Mansuripur, βOrigin of the Spin-Orbit Interaction,β
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