Lorentz Transform of an Arbitrary Force Field on a Particle in its Rest Frame using the Hamilton-Lagrangian Formalism

Transcript

1 B/IR#15-01 September, 2015; updated June, 2016 Lorentz Transform of an Arbitrary Force Field on a Particle in its Rest Frame using the Hamilton-Lagrangian Formalism C. Tschalär Abstract: A formalism is described for deriving the Lagrangian, Hamiltonian, canonical momentum of a particle interacting with an arbitrary force field, based on the interaction energy in the rest frame of the particle. Transformation to the laboratory frame is described the resulting force on the particle derived. I

2 II

3 Lorentz Transform of an Arbitrary Force Field Acting on a Particle in its Rest Frame, using the Hamilton-Lagrangian Formalism C. Tschalär June 24, Rest Frame 1.1 Lagrangian In the rest frame of a particle of mass m, its Lagrangian is the difference between its kinetic energy the sum of its restmass energy its energy of interaction with the external field 1) L = E kin (mc 2 + E field). (1) In the particle s true rest frame (TRF; double-starred quantities), its kinetic energy is zero so that L = mc 2 ϕ (2) where ϕ is the interaction energy of the particle with the external field in the TRF. In order to describe the velocity dependence of the Lagrangian for very small velocities, we define a fixed rest frame (FRF; starred quantities) which moves with a fixed velocity c β 0 with respect to the laboratory frame (LF). The TRF moves at an infinitesimal velocity c β with respect to the FRF with velocity c β with respect to the LF. The FRF Lagrangian L is related to L by L = L / (3) where 1/ 1 β 2, since L is a Lorentz invariant 2). Thus, for a particle of mass m, we have L = mc 2 / ϕ / (4) where the first part, L µ = mc 2 /, represents the particle s motion mass energy, the second part, L ϕ = ϕ (t, x, β ), the particle-field interaction, where we have defined ϕ (t, x, β ) ϕ /. 1

4 1 REST FRAME Canonical Momentum The canonical momentum P = P µ + P ϕ conjugate to the coordinates x associated with a Lagrangian L is defined 1) as the partial derivative of the Lagrangian with respect to the particle s velocity c β. Its parts cp µ,k = L µ/ β k = β kmc 2 (5) cp ϕ,k = L ϕ/ β k = ϕ / β k, (6) without being canonical momenta in themselves, represent the particle motion the external force field, respectively. The Taylor expansion of ϕ in β about β = 0 is where ϕ (t, x, β ) = ϕ 0(t, x ) cβ mp ϕ0,m 1 2 β mβ l Q lm + O(β 3 ) (7) ϕ 0 ϕ (t, x, β = 0) = ϕ ( β = 0) (8) ϕ cpϕ0,m ; Q 2 ϕ βm lm (9) β =0 βm β l we follow the convention of summing over double indices. Thus we obtain the following expressions for the field part of the canonical momentum its derivative with respect to β in the FRF: 1.3 Hamiltonian cp ϕ,k(t, x, β ) = cp ϕ0,k(t, x ) + β l Q lk(t, x ) + O( β 2 ); (10) c P ϕ,k/ β n = Q nk. (11) The Hamiltonian H is defined as the sum of the kinetic interaction energy 1). It is related to the Lagrangian the canonical momentum by H = c β P L separates like the Lagrangian into the parts H µ associated with particle motion H ϕ with the external field: H µ = c β P µ L µ = mc 2 β mβ m + mc 2 / = mc 2 ; (12) H ϕ = c β P ϕ L ϕ = ϕ 0(t, x ) β mβ l Q lm(t, x ). (13) The Hamiltonian, expressed as a function of t, x, β of t, x, P is therefore H (t, x, β ) = Hµ + Hϕ = mc 2 + ϕ 0(t, x ) β mβl Q lk(t, x ) ; (14) H (t, x, P ) = m 2 c 4 + c 2 P P ϕ(t, x ) 2 + ϕ 0(t, x ) β mβl Q lk(t, x ). (15) β =0

5 2 LORENTZ BOOST TO THE LABORATORY FRAME 3 2 Lorentz Boost to the Laboratory Frame Since the product L is invariant under Lorentz transformations 2), the Lagrangian L, the canonical momentum P, the Hamiltonian H in the LF are 1) L = ( /)L ; cp k = c L/ β k ; H = c P β L. (16) The particle motion parts of the LF Lagrangian L µ, of the canonical LF momentum cp µ,k L µ / β k, of the LF Hamiltonian H µ = c P µ β Lµ are L µ = mc 2 /; c P µ = mc 2 β; H µ = mc 2. (17) The field interaction part of the LF Lagrangian L ϕ is L ϕ = ϕ = ϕ 0 + cp ϕ0,mβ m β mβl Q lm + O(β 3 ) (18) which, according to eqns. (120) (121), becomes L ϕ = 0 ( β β 0 1)ϕ 0 +c P ϕ0 β +φ( β 0 β)(c P ϕ0 β0 ) 0 (c P ϕ0 β 0 ) β mβ l Q lm +O(β 3 ) (19) where φ 0 /( 0 + 1). The corresponding part of the LF momentum P ϕ,k = L ϕ / β k is then, using eqn. (124), cp ϕ,k = 0 β 0,k ϕ 0+cP ϕ0,k+ 0 φβ 0,k (c P ϕ0 β 0 )+δ mk + 0 φβ 0,k β 0,m β l Q lm+ 0 2 β 0,kβ mβ l Q lm +O(β 3 ) the field part of the LF Hamilronian H ϕ is (20) H ϕ = 0 ϕ (c P ϕ0 β 0 ) + 0 β 0,m β l Q lm β mβ l Q lm + O(β 3 ). (21) If we rewrite eqns. (20) (21) as cp ϕ,k = cp ϕ0,k + β l Q kl + 0 φβ 0k β 0m (cp ϕ0,m + β l Q ml) + 0 β 0k (ϕ β mβ l Q lm) (22) we find that H ϕ = 0 (ϕ β mβ l Q lm) + 0 β 0m (cp ϕ0,m + β l Q ml), (23) c P ϕ = c P ϕ + 0 φ β o ( β 0 P ϕ ) + 0 β0 H ϕ + O(β 3 ) (24) H ϕ = 0 H ϕ + 0 ( β 0 P ϕ ) + O(β 3 ). (25) Thus, the Lorentz transformation of the FRF energy-momentum 4-vector (H ; c P ) to the LF energy-momentum 4-vector (H; c P ) is a canonical transformation i.e. the FRF Hamiltonian H its conjugate FRF momentum P transform into the LF Hamiltonian H its

6 3 THOMAS EFFECT 4 conjugate LF momentum P up to orders of β 2. It is shown in Appendix 7.2 that this holds true for all orders of β. Finally, using the definitions c P ϕ0 P ϕ ( β = 0); H ϕ0 H ϕ ( β = 0), (26) the total LF Hamiltonian as a function of t, x, β of t, x, P is H(t, x, β) = H µ + H ϕ = mc 2 + H ϕ0 + 0 β 0,m βl Q lm + O(β 2 ) (27) H(t, x, P ) = m 2 c 4 + c 2 ( P P ϕ ) 2 + H ϕ0 + 0 β 0,m βl Q lm + O(β 2 ) (28) cp k = mc 2 β k + cp ϕ0,k + δ km + 0 φβ 0,m β l Q lm + O(β 2 ) (29) c P k / β n = mc 2 (β k) β n + 0 δ km + 0 φβ 0,m δ ln + 0 φβ 0,n β 0,l Q lm + O(β ) (30) according to eqn. (123). 3 Thomas Effect The Lorentz boost describes the relation between two frames at constant relative velocity. If however the frames are accelerated, they also rotate with respect to each other by the well-known Thomas rotation 3) described in Appendix 7.3. While a Lorentz boost from the LF to the TRF by β or from the LF to the FRF by β 0 does not rotate the 3-dimensional coordinates, a subsequent Lorentz boost from the FRF to the TRF by β results in a rest frame RTRF whose spatial coordinates are rotated with respect to the spatial coordinates in the non-rotated TRF (NRTRF) boosted directly by β from the LF. In consequence, a physical vector V in the NRTRF appears rotated with respect to its appearence V in the RTRF so that, according to eqn. (150), β =0 V = V φ V ( β β 0 ) + O( β 2 ). (31) Therefore, the derivative of V with respect to β is V V = φ V βk β (ˆk V β k 0 ) = βk β =0 β =0 φˆk( β 0 V ) β 0 k (32) where ˆk is the unit vector in k-direction. This spacial rotation from the RTRF to the NRTRF affects the representation of all vectors in the TRF, in particular those making up the external field ϕ. However, if ϕ is a function of a scalar or of the scalar product of two vectors V W, it may be expressed in either the rotated or the non-rotated vector since V W = V W φ V β ( β 0W ) β 0 ( β W ) + W β ( β 0V ) β 0 ( β V ) (33) where the expression in the square parentheses is zero. V

7 4 EXTERNAL FORCES 5 4 External Forces The force F exerted on a particle by an external field is defined as the total derivative of the motion-related momentum P µ with respect to time. Since the total time derivative of the canonical momentum is given by the Lagrange equation 1) we find F k dp µ,k dp k L = dp ϕ,k x k β,t Using eqn. (13), we obtain Hϕ Pϕ,m F k = + cβ m x k x k or where t β F = c β ( P ϕ ) L = x k L = x k t β P ϕ t β,t β,t = dp µ,k Pϕ,k t Pϕ,k cβ m x m tβ x β + dp ϕ,k Pϕ,k x β x m Pϕ,k H P ϕ ϕ β m t t x, (34) dβ m t β x β dx m Pϕ,k Pϕ,k β m β m = mc d( β) t x t x dβ m dβ m. (35) (36). (37) c P ϕ,k / β n = 0 δ km + 0 φβ 0,k β 0,m δ ln + 0 φβ 0,n β 0,l Q lm (38) It may be noteworthy that, if a particle traverses a localized force field whose value vanishes outside a given boundary, its momentum P µ is changed only by the L-term since d P ϕ / is a total time differential whose integral across the force field is zero. 5 Examples 5.1 Electro-Magnetic Potential The TRF interaction energy for a particle with a charge e in an electro-magnetic field potential Φ is ϕ = eφ the field part of the FRF Lagrangian is L em = eφ / = eφ 0(t, x ) + ec β A 0(t, x ) (39) all quatities Q lm higher derivatives with respect to β are zero since (Φ ; c A ) is a 4-vector thus Φ (t, x, β ) = Φ 0(t, x ) c β A 0(t, x ) (40) where A is the electro-magnetic vector potential in the FRF Φ 0 Φ ( β = 0) A 0 A ( β = 0). The field part P em of the canonical momentum is therefore Pem,k = 1 L em = ea 0,k (41) c β k

8 5 EXAMPLES 6 the field part of the RF Hamiltonian is Hem = eφ 0. In the LF, the field part L em of the Lagrangian is, according to eqns. (19) (41), L em = e 0 ( β β 0 1)Φ 0 + ca 0β + 0 φ( β 0β)(c A 0β0 ) 0 (ca 0β 0 ) eφ/ (42) the field parts H em P em of the Hamiltonian the canonical momentum are H em = e 0 Φ 0 + ec 0β0 A eφ 0 (t, x) ; (43) P em = e A + φ 0β0 ( β 0A ) + 0β0 Φ 0/c ea(t, x) (44) according to eqns. (25) to (26). The resulting electro-magnetic force F em on the particle is, according to eqn. (37), F em = ecβ ( A) A e e Φ t = e(cβ B) + ee (45) which is the Lorentz force on a particle of charge e in an electro-magnetic field ( E; B). 5.2 Stern-Gerlach-Thomas Spin Interaction The spin s of a particle does not change in magnitude therefore does not change at all in a Lorentz boost from the LF to the NRTRF so that the NRTRF spin vector s = s. In the FRF however, the spin vector s, where s ( β = 0) = s, is rotating with β according to eqn. (31) so that x s = s φ s ( β β 0 ) = s φ s ( β β 0 ) + O( β 2 ). (46) Thus the change of s with time t in the FRF is d s = s t + s dβ k β (47) k, since s / βk = 0, d s = s t s ( d β β 0 )φ (48) The first term in eqn. (48) describes the so-called Stern-Gerlach spin precession in the NRTRF electromagnetic field ( B, E ) at fixed β 3), i.e. s t = Ω SG s (49) where, for a particle with charge e, mass m, a gyro-magnetic factor g, SG( x, β, t ) = e g B m 2 ( x, β, t ). (50) Ω

9 5 EXAMPLES 7 The second term is related to the Thomas precession s c(dβ / ) where Ω T ( x, β, t ) Ω T For β = 0, where the NRTRF becomes the FRF, where s = s, caused by the acceleration + 1 (d β β). (51) B = B, Ω T ( x, β = 0, t ) = Ω T = φ( d β β 0 ), (52) SG( x, β = 0, t ) = Ω SG = e g B m 2 ( x, t ), (53) Ω the total spin precession in the FRF is d s = ( Ω SG + Ω T ) s (54) which, according to ref. 3, results in a NRTRF field interaction energy ϕ at β = 0 of ϕ ( x, β = 0, t ) = s ( Ω SG + Ω T ) (55) where s Ω SG is the Stern-Gerlach energy. If we infinitesimally change the particle velocity by a fixed amount cβ, the NRTRF, in which the particle velocity is zero which does not rotate with respect to the FRF, differs infinitesimally from the FRF: The vectors Ω SG Ω T become Ω SG Ω T the NRTRF field interaction energy ϕ is ϕ = s ( Ω SG + Ω T ). (56) The corresponding field part of the Lagrangian in the NRTRF is then L SGT = s ( Ω SG + Ω T ). (57) The field part of the FRF Lagrangian L SGT is, according to eqn. (3), where the precession angular velocities are Ω where, according to eqn. (31), L SGT = 1 s ( Ω SG + Ω T ) (58) SG = e g B m 2, Ω T = + 1 ( β dβ / ) (59) B = B β E /c φ B ( β β 0 ) + O( β 2 ) (60) B is the magnetic field in the FRF at β = 0.

10 5 EXAMPLES 8 Thus, the components of L SGT = L SG + L T are L SG = µ s B + β Σ + φ( β B )( β 0 s) φ( β s)( β 0 B ) + O( β 2 ) (61) L T = ( + 1) s ( β d β ) = ( + 1) β ( β s) (62) where µ (e/m)(g/2), β β /, Σ s E /c. The resulting field parts of the canonical momentum P SGT = P SG + P T in the FRF are c P SG = µ Σ + φ B ( β 0 s) φ s( β 0 B ) (63) c P T,k = according to eqn. (6) using the relations ( +1 ) β m = ˆk ( β s) + φβ ( β s) (64) βk 3 ( + 1) 2 β m ; ( β +1 ) β k β =0 The corresponding field part of the FRF Hamiltonian H SGT is H SGT = c β P SGT L SGT = µ( s B )+ or, according to eqns. (118) (119), HSGT = µ( s B ) + φβ 0 (β k β β ) s βk β β ( β s)+φβ k ( + 1) = ˆk (65) β ( s β) (66) β k = µ( s B ) + φ(βk β βk β ) ( s β 0 + O( β 2 ) (67) In terms of the velocity-independent LF fields B(t, x) E(t, x), where B = B ( β E/c) E = E + (c β B) ( β B) β, (68) ( β E) β, (69) Σ s E/c, the field part of the LF Lagrangian L SGT = ( /)L SGT is thus see eqns. (58) (59): L SGT = µ ( s B) + ( β Σ) + 1 ( s β)( B β) + 1 β( + 1 β s) (70)

11 5 EXAMPLES 9 or L SGT = s Ω where the LF spin precession frequency Ω is Ω = µ B + ( E/c β) + 1 β( B β) ( β β ) (71) The corresponding field parts of the LF momenta P SGT,k = L SGT / β k = P SG,k + P T,k are therefore cp SG = µ Σ 3 β ( + 1) ( β s)( β B) s( βb) + B( β s) (72) cp T,k = 3 ( + 1) β 2 k β ( β s) ( β s) k + 1 β + 1 β s β k (73) the field part of the Hamiltonian H SGT = cβ P SGT L SGT is H SGT = µ ( s B) ( β s)( β B) ( s β) β β k ( 1) β β k. (74) The corresponding LF force F SGT on the particle is, according to eqns. (17) (35), F SGT = L SGT ( β 0 )c PSGT PSGT β m P SGT / β m = mc( β) (75) where the dot indicates the partial derivative with respect to time. The time derivative d s/ = Ω s does not contribute to P SGT / t since P SGT,k = ( s Ω) = s Ω t β k t β k t + Ω ( s Ω) = 2 Ω s, (76) β k t according to eqns. (16) (72), so that the spin s may be considered a constant in the equation (75) for the SGT force Example of a Charged Particle Accelerated by the Lorentz Force A particle with charge Ze in an electro-magnetic field ( E, B) is subject to the Lorentz force Zec Λ where Λ E/c + β B (77) Assuming that the Lorentz force is much larger than the SGT force, i.e. ZecB average e mc sḃaverage (78) or Z mc2 MeV fr Hz s h where fr is the oscillating frequency of B, the resulting acceleration in the TRF is (79) β = ν c E = ν( Λ β ɛ) (80)

12 5 EXAMPLES 10 where ν Ze/m ɛ ( β E)/c. From eqns. (70), (77), (80), we thus obtain the well-known 4) LF Lagrangian for the spin motion of a particle of charge Ze, mass m, gyro-magnetic factor g in an electromagnetic field ( E, B): L SGT = µ ( s B) + ( β Σ) + 1 ( s β)( B β) + ν { Λ + 1 β( + 1 β E/c)} s β(81) = µ ν 1 ( s B) + µ ν ( β + 1 Σ) (µ ν) + 1 ( β s)( β B). (82) In the LF, the particle momentum change is where β β/. Thus d(mc β)/ = Zec Λ; d( β)/ = 3 ( β β) β + β = ν Λ (83) ( β β) = ν( Λ β) 3 = νɛ 3 ; β = ν ( Λ ɛ β). (84) According to eqn. (75), the Stern-Gerlach part of F SGT is From eqns. (70) (72), we obtain F SG = L SG c( β ) P SG PSG β m P SG β m. (85) 1 L µ SG = ( s B) + ( β Σ) + 1 ( β s) ( β B); (86) c µ ( β ) P SG = ( β ) Σ 3 β ( + 1) ( β s)( β )( β B) s( β )( β B) + ( β s)( β ) B;(87) c µ P SG = Σ 3 β ( + 1) 2 ( β s)( β B) + 1 s( β B) + ( β s) B (88), using the relations we find c β m P SG = µν β m β m / β m = 3 ( β β) = νɛ ; βm β/ β m = β = (ν/)( Λ ɛ β), (89) 2 3 ɛβ( β s)( β B) 2 Λ( (+1) 3 (+1) β s)( β B) + β( Λ s)( β B) + ( β s)( Λ B) s( Λ B) + B( Λ s) + (+1) 2 ɛ s( β B) + B( β s) (90) If, for simplicity, we assume that β point in the x-3 direction, i.e. β = (0; 0; β), we obtain the force parallel to β,. 1 µ F SG, = ( s B) + ( 1) s 3 B 3 1 Σ c + 1 βs c 3 Ḃ ν 1 E s c +1 c 3B 3 + s +1 ( Λ B) + B ( Λ s). (91)

13 5 EXAMPLES 11 the force perpedicular to β 1 F µ SG, = ( s B) 1 s 3B 3 + β( Σ 3 3Σ ) 1 c + β ( s (+1)c Ḃ3 + s 3 B ) + ν c The Thomas rotation part of the SGT force is 1 +1 Λ s 3 B s ( Λ B) + B ( Λ s) 1 Σ + 1 3( s B 3 + s 3 B )+ E (+1) 3 s B 3 + B s 3 (92) From eqns. (70) (73), we find F T = L T c( β ) P T PT β m P T β m. (93) 1 L ν T = c ν ( β ) P T = c ν P T = + 1 β( Λ s) = ( + 1 β B)( β s) β 2 ( s B) β Σ ; (94) 3 β( ( + 1) β )( Λ s) β + ( β ) Λ s + ( β ) B ( s β) (95) 3 β( ( + 1) Λ s) β + ( Λ s) + B ( s β) ; (96) c β m ν 2 P T β m = 2 ( 1) ɛβ β( Λ s) + 2 Λ β( Λ s) + ɛ ( Λ s)+ (+1) 3 (+1) 2 (+1) β β( Λ s) + 1 ( Λ B) s + B ( s Λ) (+1) 2 +1 ɛ2 β{ β ( β B) s} ɛ ( β B) s ɛ B ( s β). (+1) 2 +1 (+1) 2 (97) For β = (0; 0; β), the Thomas forces become 1 ν F T, = + ν c ( 1) β +1 3Σ + Σ 2 (+1)c + ( 1) ( s B ) + β ( s c B )+ 2ɛ E Σ (+1) 2 + ( s B) 1 ( s c +1 B ) B +1 ( Λ s) 1 s +1 ( Λ B) (98) + ν c 1F ν T, = 2 β β( +1 3Σ Σ 3 ) + 1 Σ c β (s c 3 (+1) 2 Λ Σ 3 + B + s Ḃ 3 ) 1 ( s B ) + 3 (s 3B + s B 3 )+ ɛ Σ (+1) 2 + Λ ( s B) 1 s +1 3B E 3 (+1) c (s 3B + s B 3 ) 1 B +1 ( s Λ) + s ( B Λ) (99) The total SGT force on the spin therefore is. F SGT, = + ν c ν ( 1) β +1 3Σ + ν 2 µ 1 Σ +1 c + ν( 1) ( s B) + (µ ν)( 1) s 3 B 3 + +µ ( s B) + ν β ( s c B) + µ ν β(s c 3 Ḃ 3 )+ ν 2ɛ (+1) 2 Σ + ν 2 +1 E c ( s B) + µ ν +1 { B ( s Λ) ( 1) E c (s 3B 3 )} + µ ν +1 s ( B Λ) (100)

14 5 EXAMPLES 12 F SGT, = (µ ν )β +1 Σ 3 β 3Σ 1 Σ c + (µ ν 1 ) ( s B)+ 1 +(µ ν) { 3( s B 3 + s 3B ) s 3 B 3 } + β s (+1)c Ḃ3 + s 3 B + + ν2 Λ 2 β Σ c (+1) ɛ Σ (+1) 2 + Λ ( s B) + + ν(µ ν) 1 s c +1 ( Λ B) + B ( Λ s) + 1 Λ +1 s 3 B 3 E 3 ( s c B 3 + s 3B ). (101) or For large values of, these forces approach ν 3Σ + 1 Σ c + ( s β B) + ( s c B) + (µ ν){ β s 3 B 3 + s c 3Ḃ3} F SGT, 2ν 3Σ ν+µ Σ c + (µ ν) ( s B) (µ ν) (s 3 B 3 )+ + ν (µ ν) B ( s E Λ) s c c 3B 3 + νµ s c ( B Λ) F SGT, β d F SGT, νσ + ν( s B) + (µ ν)s 3 B 3 (102) (µ + ν) d Σ + (µ ν) ( s B ) + Σ ν c (µ ν) B ( s Λ) E c s 3B 3 + νµ c s ( B Λ). (103) (µ ν) Σ 3 3Σ 1 Σ c + ( s B) + 3 ( s B 3 + s 3B ) (s 3 B 3 ) + + µ ν ( s c Ḃ 3 + s 3 B ) + ν2 Λ c (Σ 3 + s B) + ν (µ ν) Λ c s 3 B 3 = = (µ ν) (Σ 3 + s B ) + d ( s B 3 + s 3B Σ ) + νc Λ νσ 3 + ν( s B ) + µs 3 B 3 (104) In the FRF, F SGT is obtained by letting β approach zero: ν 1 F SGT = 2 µ c Σ + µ ( s B ) + ν2 c 2 E ( s B ) + ν 2c 2 (µ ν) B ( s E ) + s( B E ) (105) since Λ = E /c; ɛ = 0; = 1. The SGT-force is composed of two parts, one proportional to ν µ (which are of the same order of magnitude) which depends on space time variations of the electromagnetic field, another proportional to ν 2 νµ depending on the field itself. Since ν = Ze/m for charged particles is largest for electrons, the ratio of the two parts of the force is larger by more than two orders of magnitude than for the next heavier charged particle, the muon. At the same time, µ ν is about three orders of magnitude smaller than ν or µ for electrons muons. It is useful therefore to list the components of the SGT-force for electrons muons neglecting the (µ ν)-terms: F SGT, ;el. ν ( 1) β +1 3Σ Σ (+1)c + ( s B) + β ( s c B) + + ν2 2ɛ Σ c (+1) E ( s B) + 1 s (+1) 2 c +1 ( B Λ) (106)

15 6 REFERENCES 13 where, for 1, F SGT, ;el. ν F SGT, ;el. ν 3 Σ + 1 c β +1 Σ 3 3Σ 1 (+1)c + ν2 Λ { 2 β Σ c (+1) ( s B)} + Σ + 1 ( s B) + ɛ (107) Σ (+1) 2 Σ + ( s B) β + c ( s B) 2ν 3Σ + 1 Σ + ν2 c c s ( B Λ) (108) F SGT, ;el. ν2 Λ c Σ 3 + ( s B), (109) where the FRF value of F SGT is F SGT = ν Σ + ν 2c ( s B ) + ν2 E c ( s B ). (110) 2 For a particle traversing a localized force field as mentioned at the end of chapter 4, the net SGT force F SGT,net = L SGT affecting the total momentum change integrated over the traversal is, according to eqns. (81), (86), (94), F SGT,net = µ ν 1 ( s B) + µ ν ( β + 1 Σ) + 1 (µ ν)( β s) ( β B) (111) which, for large, becomes approximately proportional to µ ν is therefore suppressed by 1/ for electrons down to a minimal value comparable to the value for protons. Furthermore, the energy change de 2 imparted by the SGT-force to an incident particle of energy E = mc 2 momentum p = mc β is de 2 = 2E de = 2c 2 p d p = 2mc 2 c β L SGT. (112) Since the time integral of dl SGT / = ( L SGT / t) + ( L SGT cβ) over the particle path through a localized field is zero, we find that the total energy change E over the path is given by the relation (with E 0 the incident particle energy) de 2 2E 0 E = 2mc 2 cβ L SGT ; E ( L SGT / t) (113) path path if the velocity factor stays approximately constant, neglecting terms like ( E) 2 ( p) 2 quadratic in L SGT. 6 References 1) P.M. Morse H. Feshbach; Methods of Theoretical Physics. McGraw-Hill, 1953, pp ) ibid.; p ) J.D. Jackson; Classical Electrodynamics, Wiley & Sons, 1975, ch ) A.A. Pomeranskii J.B. Khriplovich, Journal of Experimental Theoretical Physics; Vol. 86, No.5, path

16 7 APPENDIX 14 7 Appendix 7.1 Lorentz Transform of the Space-Time 4-Vector Velocity Beginning in frame 1 with the infinitesimal space-time 4-vector (c ; dx ) the velocity cβ = dx / in this frame, we Lorentz-boost this 4-vector by a velocity cβ 0 into frame 2 to obtain the boosted 4-vector (c; dx) the velocity cβ = dx/ relative to frame 2, where c = 0 c + 0β0 dx ; dx = dx + φ 0β0 ( β 0 dx ) + 0β0 c (114) so that β = β β +1 0 ( β 0 β ) + 0β0 0 (1 + β 0 β ; β β = + φ 0β0 ( β 0 β) 0β0 ) 0 (1 β 0 β). (115) From these relations between frame-1 frame-2 velocities, a number of other useful relations may be derived, such as: β 0 β β = 0 ( β + β 0 ) 1 + β 0 β ; β0 β β = 0 ( β β 0 ) 1 β 0 β (116) 1 = 2 0(1 + β 0β ) 2 β 2 φ( )( β 0β ) β 0 2φ0 ( β 0β ) 2 2 0β0 β 2(0 2 0 ) β 0β 2 0(1 2 + β 0β ) 2 (117) or so that 2 = 2 0(1 + β 0 β ) 2 1 β β 0 β φ ( β 0 β ) 2 (118) = 2 2 0(1 + β 0 β ) 2, (119) = 0 (1 + β 0 β ); = 0 (1 β 0 β); 2 0 (1 + β 0 β )(1 β 0 β) = 1 (120) β = β + φ 0 β0 ( β 0 β ) + 0 β0 ; β = β + φ0 β0 ( β 0 β) 0 β0. (121) The partial derivatives of β with respect to β are β l β k = δ kl φ 2 0 β 0l β 0k (1 + β 0β ) β 0k βl + φ 0 β 0l ( β 0β ) + 0 β 0l (122) = δ kl 2 2 0β 0k φβ 0l + β l. (123)

17 7 APPENDIX 15 Inversely, by replacing β with β β 0 with β 0 (see eqn. (123)), we find β l β k = δ kl 2 2 0β 0k φβ 0l β l = δ kl + 0 φβ 0k β 0l + 0 β 0k β l. (124) For β 0 (rest frame), these derivatives become β l β k = δ kl 0 β 0kβ 0l ; β l β k = 0 δ kl + φ 2 0β 0k β 0l. (125) If β 0 points in the x 3 -direction ( β 0 = 0; 0; β 0 ), we obtain β l β k = δ kl 2 = δ kl 0β 2 0 δ k3 φβ 0 δ 3l + βl (126) 1 ( 0 1)δ k3 2 0β 2 0 βl δ k3 (127) which, for β 0l = β 0 δ 3l β 0, becomes β l = δ kl δ k3 0 0 β k. (128) 7.2 Canonical Property of the Lorentz Transform of the Energy- Momentum 4-Vector (H ; c P ) The space component P of the energy-momentum 4-vector (H ; c P ) defined in Section 2 is the canonically conjugate momentum of the Hamiltonian H is defined by the Hamilton canonical equations 1) H P k x,t = cβ k; H x k = dp k P,t. (129) Alternatively, we may define the associated Lagrangian L as The partial derivative with respect to βk is L P = cpk + cβ m m x,t β k Since, according to eqn. (129), H H P m β k x,t = P m x,t β k L c β P H. (130) x,t + H x m β k P,t x,t x m β k H β k x,t + x,t. (131) H t x, P t β k x,t = P m cβ βk m (132)

18 7 APPENDIX 16 so that we obtain the relation Furthermore, since H H x k β,t = x k c β m + P,t we find the Lagrange relation L x m P m β k H P m β,t = cβ m x,t L β k x,t P m x k H β k x,t = 0, (133) = cp k. (134) P m x k β,t = dp k β,t H x k + cβ m = dp k β,t P m x k β,t, (135). (136) Therefore, the Lagrangian relations (134) (136) are equivalent to the Hamiltonian canonical equations (129). The Lorentz-boosted Lagrangian of eqn. (16) is, in terms of the frame-1 Hamiltonian H its canonical frame-1 momentum P, L = L = β mcp m H. (137) The corresponding canonical frame-2 momentum is therefore L = ( ) β k β k β m cpm ( ) H c P + β k β m m βl H β l (138) βl β k where the term in square parentheses is zero according to eqn. (132). Therefore, using eqns. (120) (121), we obtain or L β k = β m + φ 0 β 0,m ( β β 0β) 0 β 0,m cpm k β k 0 (1 β β 0 )H (139) L β k = cp k + φ 0 β 0,k ( β 0 c P ) + 0 β 0,k H = cp k. (140) The frame-2 Hamiltonian H is, according to eqn. (16), or H = β k cp k + φ 0( β 0 β)( β0 c P ) + 0 ( β 0 β)h c β P φ 0 ( β 0 β)( β0 c P ) + 0 ( β 0 P ) + 0 (1 β β 0 )H (141) H = 0 H + 0 ( β 0 c P ) (142) The frame-1 Hamiltonian H its canonically conjugate frame-1 momentum P transform, according to eqns. (140) (142), to the Lorentz-boosted frame-2 Hamiltonian H its canonically conjugate momentum P like the components of an energy-momentum 4-vector (H; c P ). The Lorentz boost is therefore a canonical transformation..

19 7 APPENDIX Thomas Effect of an Infinitesimal Velocity on Vectors in a Lorentz-Boosted Rest Frame A Lorentz boost A( β 0 ) of a 4-vector v by a velocity c β 0 from the laboratory frame (LF) results in a 4-vector v 0 = A( β 0 )v. After a further boost A( β ) by an infinitesimal velocity c β, we obtain a 4-vector ṽ : ṽ = A( β )A( β 0 )v. (143) A direct Lorentz boost of v to a frame with the same velocity relative to the LF as the ṽ -frame results in a 4-vector v : v = A( β 0 + δ β)v (144) with v 0 = ṽ 0. The relation between ṽ v can be expressed as ṽ = Ãv where If β 0 points in the x 3 -direction, we find β 0 A( β 0 ) = β neglecting terms of order βk 2, so that Since A( β )A( β 0 ) = A( β 0 + δ β) = à = A( β )A( β 0 )A( β 0 δ β). (145) ; A( β ) = β 0 β3 β1 β2 0 β β 0 0 β3 0 β 0 β1 0 β β 0 β2 0 β 0 0 β β 0 β3 1 β 1 β 2 β 3 β β β β 3 0 δβ 3 0 δβ 1 0 δβ 2 0 β 0 0δβ δβ φ 0 β 0 δβ 1 0 δβ φ 0 β 0 δβ 2 0 β 0 0δβ β 3 0 δβ 3 (146) + O( β 2 ). (147) + O( β 2 ), (148) we require 0 δβ 1 = β1; 0 δβ 2 = β2; 0δβ 2 3 = β3 to satisfy the condition v0 = ṽ0 obtain à = φβ 0 β φβ 0 β2 + O( β 2 ). (149) 0 φβ 0 β1 φβ 0 β2 1 Therefore, à represents the sum of two rotations around the x 1- x 2 -axes by the infinitesimal angles φβ 0 β2 φβ 0 β1, respectively, from the vector v to the vector v. This may be summarized for arbitrary directions of β 0 by the relation v = v + φ v ( β β 0 ) + O( β 2 ) (150) which represents a rotation of v around the β β 0 -axis by an infinitesimal angle φ β β 0.