The Magnetic Force as a Kinematical Consequence of the Thomas Precession
aa r X i v : . [ phy s i c s . c l a ss - ph ] A p r The Magnetic Force as a Consequence of the Thomas Precession
David C. Lush [email protected] (Dated: April 23, 2019)The requirements imposed by relativistic covariance on the physical description of two interactingclassical charged particles are investigated. Because rotational pseudoforces cannot be caused byThomas precession, kinematical considerations demand the presence of compensatory forces whenThomas precession of an inertial reference frame is observed. The magnetic force on a moving chargeis apparently one such force, where Thomas precession of the laboratory frame is seen by an observerco-moving with the charge. Thus, no acceleration of the field source charge is required to cause thenecessary Thomas precession, consistent with the known properties of the magnetic interaction.However, when the field source charge is accelerating, an additional magnetic-like force is expected.Other forces corresponding to the Euler and centrifugal rotational pseudoforces are also predicted bythis line of reasoning. The plausibility that an anti-centrifugal force of the Thomas precession mayaccount for the binding of quarks into nucleons is investigated. The similarity of the magnetic forceon a relativistically-moving charge in the radiative magnetic field of a nearby Coulomb-acceleratingcharge to the predicted anticentrifugal force of the Thomas precession is shown.
PACS numbers: 03.30.+p, 41.20.-q, 45.05.+x, 45.20.da
I. INTRODUCTION
The rest frames of charged particles interacting elec-trodynamically are known to rotate relative to the labo-ratory inertial reference frame, as well as mutually, dueto the Thomas precession [1]. Kinematics requires thatwhen a reference frame rotates relative to an inertial ref-erence frame, rotational pseudoforces must be present inthe rotating frame. However, it is clear that Thomas pre-cession cannot cause rotational pseudoforces in general,because it depends only on the relative motion betweentwo reference frames, and the Thomas precessing framemay be an inertial frame if the observer is accelerating.Thus, for example, since there can be no Coriolis forcein the rest frame of a field-source charged particle that isonly translating with constant velocity, an observer co-moving with a charged test particle that is translatingrelative to the field-source particle, and cross-acceleratingrelative to the translation, must infer the presence offorces that compensate for the lack of Coriolis force theobserver expects due to the observed Thomas precession.It is contended in the present contribution that thenecessity of an observer moving non-inertially to inferpseudoforce-compensating forces implies even in iner-tial frames the existence of related forces, that will beherein referred to as anti-pseudoforces, the most obviousof which can be recognized as the magnetic part of theLorentz force. The formal similarity of the magnetic forceto a Coriolis force is viewed as a direct consequence of itarising as an anti-Coriolis force. Similar reasoning whenextended to centrifugal and Euler pseudoforces impliesthe existence of other tangible forces. These may pro-vide relativistic kinematical bases for the strong and weakforces. Also, when a magnetic field source charge is accel-erating as well as translating, an additional magnetic-likeforce must be expected. It is shown further that under highly relativistic and short range conditions, the force ona charged particle due to the magnetic acceleration fieldcan behave similarly to predicted anticentrifugal force.That is, the magnetic field of an accelerating charge cancause a force between two charges that is attractive inde-pendent of the relative polarity of the charges, and canovercome Coulombic repulsion at sub-nucleonic scale inthe highly relativistic limit.
II. SIMILARITY OF THE MAGNETIC FORCETO A CORIOLIS FORCE
In this section the approximate Lorentz force on acharged test particle interacting electromagnetically withanother charged particle is determined and the magneticpart of the interaction identified, and then it is shownhow the magnetic force may be interpreted as an anti-Coriolis force. The anti-Coriolis force acts in any inertialreference frame where both the field source particle andtest particle are moving.
A. Interaction of Two Charged Particles, WhereOne Particle is Non-Accelerating
The relativistic law of inertia for a massive particle ofmomentum P and rest mass m , acted on by a force F is F = d P dt = ddt [ γm v ] = ˙ γm v + γm a , (1)where v is the particle velocity and a ≡ d v /dt ≡ ˙ v itsacceleration, and γ ≡ / p (1 − ( v/c ) with v ≡ | v | and cthe speed of light. The acceleration due to F is thus a = (cid:20) γm (cid:21) [ F − ˙ γm v ] = (cid:20) γm (cid:21) h F − γ ( β · ˙ β ) m v i , (2)where β ≡ v /c (and using the well-known identity that˙ γ = γ ( β · ˙ β )). The force of interest is the Lorentz force on a moving charged particle in the electromag-netic field caused by another moving charged particle. Inthis case the electromagnetic field is exactly described bythe Li´enard-Wiechert field expressions [2]. The Li´enard-Wiechert field expressions in three-vector notation are E ( r , t ) = q " n − β γ (1 − β · n ) R ret + qc n × (cid:16) ( n − β ) × ˙ β (cid:17) (1 − β · n ) R ret (3)and B ( r , t ) = [ n × E ] ret , (4)where, if r is the displacement from a field-source chargedparticle at the retarded time t ′ ≡ t − R/c to a field point attime t , then R ≡ | r | , and n = r /R . Also β ≡ v /c , where v is the field-source particle velocity. The subscript “ret”refers to that the quantity in the brackets is evaluated atthe retarded time.The first term on the right hand side of Eq. (3) is calledthe velocity, or non-radiative, electric field. The secondis called the acceleration or radiative electric field. In thepresent application, with the electromagnetic field causedby a non-accelerating charge, the acceleration fields van-ish identically. It can also be seen by inspection that themagnitude of the field difference from the Coulomb fieldof the particle (that is, the electric field in the rest frameof the particle) due to motion is small when β << v/c ) compared to the Coulomb force, it will be suffi-cient to represent all forces and fields only to this order.From (3) and with the source particle non-accelerating,the electric field is exactly E ( r , t ) = q s " n − β s γ s (1 − β s · n ) R ret (5)and the magnetic field is approximated to order ( v/c ) as B ( r , t ) ≈ q s R [ β s × n ] . (6)Let r s and r t represent position vectors to an elec-tromagnetic field source particle of mass m s , and a testparticle of mass m t , in an inertial reference frame (IRF)that will be referred to herein as the laboratory frame,and R ≡ | r | ≡ | r t − r s | . The velocities of both thesource and test particles are assumed nonvanishing inthe laboratory frame. It is also assumed for simplicitythat the particles have no intrinsic magnetic moments. The Lorentz force on the test particle in the laboratoryframe is then F = q t v t c × B + q t E . (7)For the uniformly translating field source particle, itis straightforward to evaluate the retardation effects ex-plicitly and rewrite Eq. (7) in terms of non-retardedquantites. The result accurate to order ( v/c ) is F ≈ q s q t R [ β t × [ β s × n ]]+ q s q t n (cid:0) − β s · n ) / β s / (cid:1) R . (8)If the force is considered as consisting of electric andmagnetic parts so that F ≡ F electric + F magnetic , it willbe useful to further consider the electric part of the forceas consisting of a Coulomb force plus additional termsthat are at order β and higher powers of β , where theCoulomb part of the electric force is F Coul ≡ q t q s R ( r t − r s ) = q t q s r R . (9)The acceleration of the test particle due to theCoulomb force, using Eq. (2), is a Coul = (cid:20) γ t m t (cid:21) h F Coul − γ t ( β t · ˙ β t ) m t v t i , (10)which to order ( v/c ) is a Coul ≈ F Coul γ t m t − ( β t · ˙ β t ) v t , (11)The acceleration of the test particle due to theCoulomb force, neglecting the relativistic terms from Eq.(11) that are of order β and higher, is then a Coul ≈ q t q s r m t R . (12)If we let F ≡ F electric + F magnetic with F magnetic = q t β t × B , (13)then the magnetic force can be written using Eq. (6) as F magnetic ≈ h q t q s R i [ β t × ( β s × n )] = − m t ω × v t , (14)with ω = (cid:20) q t q s cm t R (cid:21) [ β s × n ] ≈ (cid:20) c (cid:21) [ v s × a Coul ] (15)where a Coul is the Coulomb acceleration of the test par-ticle as given by Eq. (12).Now, Eq. (14) shows that the magnetic force on thetest particle is formally identical to a Coriolis force thatwould be present if the reference frame of the descriptionwas rotating with angular velocity as given by Eq. (15).Furthermore, the above angular velocity expression canbe related to the expression for the Thomas precession[2] in the limit of small v/c , of the rest frame of a particlewith velocity v and acceleration a relative to the observerseeing the Thomas precession. That is, ω T = γ γ + 1 a × v c ≈ a × v c . (16)The acceleration of coordinate axes that are fixed inthe laboratory frame, relative to the observer co-movingwith the test particle, is simply the opposite of the testparticle acceleration, since the laboratory frame is non-accelerating. Similarly, the relative velocity to the sameobserver of an object fixed in the laboratory frame is theopposite of the test particle velocity as observed fromthe laboratory frame. However, the sign of the Thomasprecession must be inverted to account for the observerbeing in the non-inertial frame. This results in an angu-lar velocity of Thomas precession of the laboratory framecoordinate axes seen by the test particle co-moving ob-server of ω ′ T ≡ − ω T ≈ − a t × v t c , (17)which is not equal to the expected angular velocity ac-cording to Eq. (15). In order to derive the magneticforce from Thomas precession seen from the test parti-cle, it will be necessary to consider not just the Thomasprecession of the laboratory frame, but also that of thefield source particle rest frame, in both cases as seen bythe test particle co-moving observer. B. The Magnetic Force as a Coriolis Effect of theThomas Precession
In accordance with Eq. (16), but with a sign inversionto account for the difference in observer, the test par-ticle co-moving observer sees any Cartesian coordinateaxes that are fixed in the laboratory frame as Thomasprecessing with angular velocity ω l ≈ − a t × v t c . (18) Similarly, the angular velocity of the Thomas preces-sion of the source particle rest frame seen by the testparticle co-moving observer is ω s ≈ − a t × ( v t − v s )) c . (19)The relative angular velocity of the laboratory framecompared to the field-source particle rest frame is then ω r = ω l − ω s ≈ − a t × v s c . (20)The observer co-moving with the test particle thus seesthe laboratory frame as rotating with angular velocity ω r with respect to the field source particle rest frame.Although the test particle co-moving observer sees bothframes as rotating, it is the relative rotation that de-termines the kinematical relationship between them. Ifthe law of motion is known in either of the two frames,it can be determined in the other using standard kine-matics. If the source rest frame is taken as the non-rotating frame, then the test particle co-moving observerpredicts that the lab frame equation of motion must bethe source frame equation plus the Coriolis, Euler, andcentrifugal rotational pseudoforces. These are viewed asanti-pseudoforces since the source frame is rotating witha larger-magnitude angular velocity than the lab frame,from the point of view of the test particle rest frame ob-server, yet the electromagnetic interaction in the sourceparticle rest frame is perfectly radial in general and soapparently lacking in any rotational pseudoforces. (Thisis only to be expected, since an observer co-moving withthe uniformly-translating field source particle experiencesno Thomas precession.)The expected Coriolis force in the lab frame relative tothe source particle rest frame, as observed from the testparticle rest frame is F Coriolis = − m t ω r × v t , (21)or, with the relative angular velocity of the Thomas pre-cession between the source and laboratory frames givenby Eq. (20), F Coriolis = m t c [ a t × v s ] × v t . (22)Approximating the test particle acceleration as thatdue to the Coulomb force due to the source particle ac-cording to Eq. (12) obtains F Coriolis = q t q s R [ β t × [ β s × n ]] . (23)Comparing with Eq. (14), it is apparent that F Coriolis = F magnetic . (24)The magnetic force on the test particle is thus inter-pretable as a Coriolis force caused by Thomas precession.It seems worth remarking that although the Coriolisand other inertial pseudoforces generally are directly pro-portional to the mass of the object on which they appearto act, the magnetic component of the Coriolis force man-ifesting here does not depend on the test particle mass.The amount of Thomas precession seen by the test par-ticle rest frame observer is inversely proportional to thetest particle mass, which has directly canceled the massfactor that would be otherwise present. Such cancela-tion of the mass factor is of course essential to admittingthe possibilty that the magnetic force is due to a Cori-olis effect. However, the argument presented here seemsto also be applicable to other situations where such can-celation cannot occur. Specifically, had the field-sourceparticle been allowed to freely accelerate in the Coulombfield of the test particle, then this acceleration would havecontributed to the Thomas precession of the source par-ticle rest frame seen by the observer in the test particlerest frame, with amount of additional Thomas precessiondepending inversely on the mass of field-source particle.Thus, an additional Coriolis force component would bepresent that would be proportional to the ratio of the testparticle to source particle masses. In the case of interac-tions between equal-mass free particles, this would seemto result in a doubling of the strength of the magneticinteraction.The magnetic force has been related by other authorsto a Coriolis force [3], as well as to an anti-Coriolis force[4]. However, those authors do not infer the existenceof anti-Euler or anti-centrifugal forces. The apparent in-completeness of the Lorentz force has also been notedpreviously [5]. III. STRENGTH OF THE ANTI-CENTRIFUGALFORCE COMPARED TO COULOMBREPULSION
The same arguments that lead to expectation of ananti-Coriolis force lead also to expectation of an anti-centrifugal force, that can be given notionally as F anticentrifugal ≡ F a.c. ≡ γm ω T × ( ω T × r ) , (25) with (using again the formula for the angular velocityof the Thomas precession given in [2], but here withoutspecialization to small v/c ), ω T = − ( γ −
1) [ v × a ] v ≈ − γ [ v × a ] v = − γ [ β × a ] cβ (26)for β approaching unity. (The relativistic version of thecentrifugal force on which Eq. (25) is based, that includesthe leading Lorentz factor, γ , is derived in [6].) The anti-centrifugal force according to Eq. (25) on one particledue to the other is thus, F a.c. = γ m c β γ [ β × a ] × ([ β × a ] × r ) . (27)where r ≡ r − r , and where the two particles arenow distinguished by the subscripts 1 and 2, since thefield-source particle is no longer constrained to be non-accelerating, and in fact is accelerating under the influ-ence of the Coulomb field of particle 2. Therefore thenotion of a test particle that doesn’t influence the field-source particle motion must be abandoned, and the sub-scripts s and t have been replaced by 1 and 2.For Coulomb attraction or repulsion in the case of mu-tual circular motion of the two charged particles and ne-glecting effects of retardation, Eq. (2) gives the accelera-tion of one of the particles in the laboratory frame due tothe velocity electric field according to the exact electricfield expression of Eq. (5) of the other as a = (cid:20) γ m (cid:21) q q r γ R . (28)In the approximation valid in the highly relativisticcase where β and β approach unity and for circularmotion neglecting retardation, and assuming the acceler-ation of particle 1 is Coulombic (but retaining the inverse γ factor in the acceleration according the exact electricfield expression of Eq. (5)), the expected magnitude ofthe anti-centrifugal force acting on particle 2 is thus (cid:12)(cid:12) F a.c. (cid:12)(cid:12) ≈ γ m c (cid:12)(cid:12) a (cid:12)(cid:12) R ≈ γ m c q q γ γ m γ R = m c q q m γ R . (29)If it is assumed the two particles are of like chargepolarity with charge magnitude of the order of the ele-mentary charge, and repulsed by Coulomb repulsion, yetcircularly orbiting each other with laboratory-frame ve-locities near the speed of light, then it will be a simplematter to calculate approximately ( i.e. , neglecting retar-dation) at what orbital radius the anticentrifugal force will overcome Coulomb repulsion. This orbital radiuscan then be compared with the measured size of a nu-cleon such as the proton.Equating the anticentrifugal force of Eq. (29) and themagnitude of the electric force based on the electric fieldof Eq. (5) but neglecting retardation obtains, for equalmass particles and asumming | q | = | q | = e (with e thefundamental charge magnitude) and γ = γ ≡ γ , andusing the proton mass for γm , R ≈ e γmc ≈ − cm . (30)This orbital radius or (strictly) inter-particle separa-tion is about two orders of magnitude less than the esti-mated size of the proton [7]. Thus the present analysiswould seem to indicate that the expected anticentrifu-gal force of the Thomas precession is too weak to po-tentially account for the fundamental strong force thatbinds quarks into nucleons. However, it would seem avalid candidate for the needed force that could bind con-jectured constituent particles of quarks and leptons [8].Furthermore, it is plausible that a force of relativistickinematical origin such as the anti-centrifugal force of theThomas precession would not have the same binding en-ergy and relativistic mass equivalency usually expected,as needed to bind such constituent particles ( e.g., “pre-ons”) without exceeding the known masses of the leptonsand quarks.Although the Thomas precession is widely creditedwith accounting for the spin-orbit coupling being half itsotherwise classically expected value, the original analysis [1] was performed prior to recognition of the need to ac-count for the “hidden” relativistic momentum of a mag-netic dipole in an electric field. Accounting for hiddenmomentum can be seen to directly affect this conclusion[9]. Perhaps further re-examination of the relationship ofthe Thomas precession to energy is warranted. IV. THE STRONG FORCE OFMAXWELL-LORENTZ ELECTRODYNAMICS
It is worth noting that a force similar to the expectedanti-centrifugal force of the Thomas precession can befound in relativistic electrodynamics. If retardation ef-fects can be neglected, conventional electrodynamics pre-dicts that two charges closely approaching each other atrelativistic relative velocity experience a mutually attrac-tive force that is independendent of relative polarity andsimilar in strength to the predicted anti-centrifugal forcedue to Thomas precession.Based on the Li´enard-Wiechert field expressions pro-vided above as Eqs. (3) and (4), the magnetic force onparticle 2 due to the acceleration field of particle 1 isgenerally F m = q v c × B = q β × n × q c n × (cid:16) ( n − β ) × ˙ β (cid:17) (1 − β · n ) R ret , (31)where here n ≡ r /R ≡ ( r − r ) /R with R ≡ | r − r | .The acceleration of particle 1 in the electric velocity fieldof particle 2 is found from Eqs. (2) and (3), neglecting re-tardation and assuming mutually circular orbital motionof particle 1 around particle 2, as a ≈ q q γ m γ R [ r ] ≡ − q q γ m γ R [ n ] , (32)or, if ˙ β = a /c ≈ − a n /c with a ≡ | a | , and stillneglecting retardation, F magnetic ≈ β × [ n × [ n × ( β × n )]] q q c R q q γ m γ R . (33)Now assume mutual circular motion and let ˆ x , ˆ y , ˆ z rep-resent orthogonal unit vectors in a right-handed coordi-nate system, and align ˆ x with n and ˆ y with v . Then forcircular motion with β = − β and | β | = | β | ≈
1, andneglecting retardation, Eq. (33) for the magnetic force onparticle 2 due to its motion in the magnetic accelerationfield of particle 1 becomes F magnetic ≈ ( − n ) q q γ m c γ R . (34) Recalling that n is directed towards particle 2 fromparticle 1, it is apparent that this force is attractive forlike charges, as well as for opposite. Also, comparingEq. (34) with Eq. (29), it is apparent that, for mutualcircular motion and neglecting retardation, F magnetic ≈ γ m γ m F anticentrifugal . (35)In the case of equal mass particles with γ = γ sothat F magnetic ≈ F anticentrifugal , it is apparent that un-der conditions as stated the magnetic force is approx-imately equal to the predicted anti-centrifugal force ofthe Thomas precession. V. CONCLUDING REMARKS
It has been proposed that relativistic kinematic con-siderations necessitate that the Thomas precession givesrise to certain forces that resemble the rotational pseud-oforces known as the Coriolis, Euler, and centrifugalforces. It has been shown to order v /c that the pre-dicted anti-Coriolis force of the Thomas precession ac-counts for the existence of the magnetic force. The pos-sible correspondence of the strong nuclear force to ananti-centrifugal force of the Thomas procession was in-vestigated. Appendix A: Two-Particle Interaction Seen by anObserver Co-Moving with the Test Particle
An observer co-moving with the test particle sees aThomas precession of the field-source particle rest frame.Also, in the source particle rest frame there is no mag-netic force acting on the test particle, only the velocity-independent Coulomb force.At a succession of instants of time, an observer co-moving with the source particle can calculate the fieldsource particle position relative to the test particle, asthe source particle relative position would be expressedin the rest frame of the test particle.
1. Change of Field Source Particle Position asMeasured by an Observer Co-Moving with the TestParticle
Let η ( ζ ) , µ ( ζ ) represent the test particle position andvelocity measured by an observer in the source particlerest frame (an inertial reference frame, since the fieldsource particle is restricted in the present analysis tomoving uniformly with respect to the inertial laboratoryreference frame), as a function of time, ζ , in the sourceparticle rest frame, and ρ s ( τ ) , ν s ( τ ) , α s ( τ ) represent thesource particle position, velocity, and acceleration rela-tive to the test particle, as measured by an observer co-moving with the test particle, as a function of the testparticle proper time τ .Also let σ s ( ξ ) , λ s ( ξ ) represent the source particle po-sition and velocity in an inertial reference frame momen-tarily co-moving with the test particle at time ζ = ζ ,where ξ is the source particle time coordinate in this in-ertial reference frame, which will be referred to as thetest particle momentary rest frame. The space origin ofthe field source particle rest frame coordinate system isat the field source particle, and the observer co-movingand co-located with the test particle chooses a coordinatesystem for the inertial reference frame moving with thetest particle at its proper time τ = τ and space originat the test particle.Similarly, let η p ( ζ ) , µ p ( ζ ) and σ p ( ξ ) , λ p ( ξ ) representthe position and velocity of an arbitrarily located par-ticle in the source particle rest frame and test particlemomentary rest frame coordinate systems, respectively.The arbitrary particle position in the test particle mo-mentary rest frame can be expressed accurately to order v /c (see Appendix D) in terms of test particle momen-tary rest frame at ζ = ζ quantities as σ p ( ζ ) ≈ η p − η + ( β · ( η p − η )) β / − ( ζ − ζ ) µ with β ≡ µ /c , and γ ≡ (1 − ( µ/c ) ) − / , and β ≡ β ( t = t ), and γ ≡ γ ( ζ = ζ ). In the case that the arbitraryparticle is the source particle, which is at the origin of itsown rest frame here, then σ s ( ζ ) ≈ − η − ( β · η ) β / − ( ζ − ζ ) µ . The space origin of the system ( σ , ξ ) coincides withthe test particle at τ = τ , which corresponds to ξ = 0.Then the arbitrary particle time coordinate according tothe observer co-moving with the test particle i ξ ( ζ ) = ( ζ − ζ ) γ − γ β · ( η p ( ζ ) − η ) /c. Thus dξdζ = γ − γ β · µ p ( ζ ) /c. Since the source particle is by definition stationary inthe source particle rest frame, for the field source particle dξdζ = γ , (A1)and for the test particle dξdζ (cid:12)(cid:12) ( ζ = ζ ) = γ − γ β · µ ( ζ ) /c ≈ γ − . (A2)Also, let σ s ≡ σ s ( ζ = ζ ) ≈ − η − ( β · η ) β / η ≡ η ( ζ = ζ ), and ξ ≡ ξ ( ζ = ζ ) = δζγ − γ β · ( η δ − η ) /c. For the source particle this becomes ξ ≡ ξ ( ζ = ζ ) = γ β · η /c. (A3)The arbitrary particle position in an inertial coordinatesystem momentarily co-moving with the test particle attime ζ + δζ can be expressed (accurately to order v /c )as σ p ′ ≈ η p − η δ + ( β δ · ( η p − η δ )) β δ / − ( ζ − ( ζ + δζ )) µ δ with η δ ≡ η ( ζ = ζ + δζ ), and with β δ ≡ β ( ζ = ζ + δζ ),and γ δ ≡ γ ( ζ = ζ + δζ ). For the source particle thisbecomes σ s ′ ≈ − η δ − ( β δ · η δ ) β δ / − ( ζ − ( ζ + δζ )) µ δ . (A4)Similarly, let ξ ′ = ( ζ − ( ζ + δζ )) γ δ − γ δ β δ · ( η − η δ ) /c. (A5)Then σ sδ ≡ σ s ′ ( ζ ′ = ζ ′ ) ≡ σ s ′ ( ζ = ζ + δζ ) ≈ − η δ − ( β δ · η δ ) β δ / η δ ≡ η ( ζ = ζ + δζ ), and ξ δ ≡ ξ ′ ( ζ ′ = ζ ′ ) ≡ ξ ′ ( ζ = ζ + δζ ) = − γ δ β δ · ( η pδ − η δ ) /c (A7)or, in the case of the arbitrary particle being the field source particle so that η pδ = η sδ ≡ ξ δ ≡ ξ ′ ( ζ = ζ + δζ ) = γ δ β δ · η δ /c (A8)with η δ ≡ η ( ζ = ζ + δζ ) ≈ η + δζ µ (A9)where µ is the test particle velocity relative to the source particle (measured by a source particle co-moving observer).Now let δ ρ s ≡ σ s ′ ( ζ + δζ ) − σ ( ζ ) ≡ σ sδ − σ s . Then, δ ρ s ≈ − η δ − ( β δ · η δ ) β δ / η + ( β · η ) β / . Replacing η δ by η + δζ µ and β δ by β + δζ ˙ β and reducing obtains δ ρ s ≈ − δζ µ − δζ ( β · η ) ˙ β / − δζ (cid:16) ˙ β · η (cid:17) β / − δζ ( β · µ ) β / δζ → δ ρ s δζ ≡ d ρ s dζ ≈ − µ − ( β · η ) ˙ β / − (cid:16) ˙ β · η (cid:17) β / − ( β · µ ) β / β · µ ) / β = (1 + β / µ ≈ γ µ , d ρ s dζ ≡ ν s ≈ − γ µ − ( β · η ) ˙ β / − (cid:16) ˙ β · η (cid:17) β /
2. Rate of Change of Field Source Particle TimeCoordinate as Measured by an Observer Co-Movingwith the Test Particle
It was found above that the arbitrarily located particletime coordinate in the test particle momentary rest framewith origin at the test particle position at time ζ = ζ in the field source particle rest frame, corresponding totime ζ = ζ in the field source particle rest frame, can beexpressed as ξ ≡ ξ ( ζ = ζ ) = δζγ − γ β · ( η pδ − η ) /c. For the source particle this becomes ξ = γ β · η /c. (A11)Suppose temporarily that the test particle is not accel-erating, and suppose that at a slightly later time ζ + δζ we perform a boost so that ξ ′ ( ζ ) = ( ζ − ( ζ + δζ )) γ − γ β · ( η p ( ζ + δζ ) − η δ ) /c. For the arbitrary particle being the source particle,which is staionary at the origin of its rest frame coor-dinates, this becomes ξ ′ ( ζ ) = ( ζ − ( ζ + δζ )) γ + γ β · η δ /c. (A12)At ζ = ζ + δζ , ξ ′ ( ζ + δζ ) = γ β · η δ /c. (A13)In the above analysis, the change of the source particleposition was found between two inertial coordinate sys-tems with space origins at the test particle, and where theaccelerating test particle is momentarily at rest, at twosuccessive times separated by a small time interval of δζ in the source particle rest frame. In order to determine arate of change of the source particle position relative tothe test particle as measured by the observer co-movingwith the test particle, it is necessary to determine thecorresponding interval of time that elapses between thetwo successive positions, according to the test particle co-moving observer. However, in the limiting case that thetest particle is not accelerating, the difference betweenthe two time coordinates given by Eqs. (A11) and (A13)vanishes. It is thus needed to find the translation of timethat will place the time of the successive snapshots of po-sition consistently with the elapsed test particle propertime. That is, it is necessary to find a time translation∆ and a translated time coordinate ξ ′′ ( ζ ) = ξ ′ ( ζ ) + ∆,such that ξ ′′ ( ζ + δζ ) = ξ ( ζ + δζ ) = γ δζ ≡ δτ , for thecase of a non-accelerating test particle. So, let∆ = ξ ( ζ + δζ ) − ξ ′ ( ζ + δζ ) (A14)or ∆ = δζγ + γ β · η /c − γ β · η δ /c. (A15)Substitution for η δ ≈ η + δζ µ and reducing obtainsthat ∆ ≈ δζ ( γ − β · µ /c ) (A16)and so ξ ′′ ( ζ ) ≈ ( ζ − ζ ) γ + β · η /c (A17)which becomes, for the case of an accelerating test par-ticle ξ ′′ ( ζ ) ≈ ( ζ − ζ ) γ δ + β δ · η /c. (A18)Allowing for test particle acceleration, let δτ ≡ ξ ′′ ( ζ + δζ ) − ξ s ( ζ ) ≈ δζγ δ + β δ · η /c − β · η /c or δτ ≈ δζγ + δζ ˙ β · η /c (A19) or (dropping the subscript zero because the time ζ isarbitrary), δτ ≈ δζγ + δζ ˙ β · η /c (A20)so lim δζ → δτδζ ≡ dτdζ ≈ γ + ˙ β · η /c (A21)and dζdτ ≈ γ − − ˙ β · η /c. (A22)
3. Rate of Change of Field Source Particle Positionas Measured by an Observer Co-Moving with theTest Particle
The source particle velocity as measured by the testparticle co-moving observer can be evaluated as ν s ≡ d ρ s dτ = d ρ s dζ dζdτ . (A23)Using results from above, then, ν s ≈ − h γ µ + (cid:16) ˙ β · η (cid:17) β / β · η ) ˙ β / i h γ − − ˙ β · η /c i (A24)or ν s ≈ − µ + (cid:16) ˙ β · η (cid:17) β / − ( β · η ) ˙ β / . (A25)If the test particle rest frame is taken as the non-rotating frame, then the (relative) velocity in that frameis expected to be expressible as d ρ dτ ≡ ν = v + ω × r (A26)with ω × r = − (cid:16) ˙ β · r (cid:17) β / β · r ) ˙ β / , so with the vector identity a × ( b × c ) = ( c · a ) b − ( b · a ) c ,with a = r , b = β , c = ˙ β , ω ≈ (cid:16) β × ˙ β (cid:17) / − a × v c = − ω T ≡ ω ′ T . (A27)Comparing this result with Eq. (16), it appears to beopposite in sign. However, the sign difference is easily accounted for by noting that the accelerated frame istaken as non-rotating here. As seen from the laboratoryframe, the accelerated frame rotates oppositely. That is,if (Jackson [2] Eq. (11.107), for a general vector G ) (cid:18) d G dt (cid:19) nonrot = (cid:18) d G dt (cid:19) rest frame + ω T × G or, more generally, (cid:18) d G dt (cid:19) nonrot = (cid:18) d G dt (cid:19) rot + ω × G . So if (cid:18) d G dt (cid:19) laboratory = (cid:18) d G dt (cid:19) rest frame + ω × G then (cid:18) d G dt (cid:19) rest frame = (cid:18) d G dt (cid:19) laboratory + ( − ω ) × G . Interchanging the designation of which reference frameis taken as the rotating frame thus inverts the sign of theassociated angular velocity.
Appendix B: The Magnetic Force as a PurelyRelativistic-Kinematic Effect of CoulombAcceleration
In this section, it is shown explicitly that presence ofa magnetic force on a test charge involves necessarily aCoulombic acceleration of the test particle, in at leastone inertial reference frame. The argument is based onthe interaction of two charged particles, but because oneof them is constrained to be non-accelerating, the ar-gument extends by the principle of linear superpositionto the case of a magnetic field generated by a current-carrying neutral wire. If, in addition to relatively accel-erating, the field-source particle and the test particle arealso relatively translating with a component transverseto the acceleration, the field-source particle rest frame isthen necessarily Thomas precessing with respect to anobserver co-moving with the test particle. This is pre-cisely the condition under which a magnetic force is seento act on the test particle by an observer in the labo-ratory frame. Therefore, the magnetic force can alwaysbe associated with Thomas precession of the source par-ticle rest frame from the point of view of an observerco-moving with the test particle.
1. Implications of the Relativistic Law of Inertia
The relativistic law of inertia for the test particle inthe field source particle rest frame (which is an inertialframe here since the source particle is non-accelerating)may be written as (cid:18) d [ γ t v t ] dτ (cid:19) (srf) = F (srf) m t , (B1)where the parenthetic superscript (srf) indicates quanti-ties defined in the field-source particle rest frame. Thedemonstration will proceed by rewriting Eq. (B1) interms of laboratory frame quantities. To facilitate eval- uation of the left hand side of Eq. (B1) in terms oflab-frame quanitities, consider (based on the Lorentztransformation as given in Appendix B below) that thetime component of the four-vector displacement from thesource to the test particle Lorentz transforms from a 4-coordinate system in the laboratory reference frame toone in the field source particle frame, where the originsof both systems are the same event, as ct (srf) ≡ cτ = γ s ( ct − β s (ˆ v s · r t )) (B2)where t is the time coordinate, in the laboratory frame,of the test particle when it is at the position r t . v s isthe field-source particle velocity in the laboratory frame.Thus τ = γ s ( t − ( β s · r t ) /c ) , (B3)and so (and with the source particle non-accelerating) dτdt = γ s (1 − β s · ˙ r t /c ) = γ s (1 − β s · β t ) . (B4)For the field source particle moving at constant veloc-ity, then, d (cid:16) P (srf) (cid:17) dt = d (cid:16) P (srf) (cid:17) dτ dτdt = d (cid:16) P (srf) (cid:17) dτ [ γ s (1 − β s · β t )] . (B5)Thus, d (cid:16) P (srf) (cid:17) dτ = d (cid:16) P (srf) (cid:17) dt [ γ s (1 − β s · β t )] − , (B6)and so d (cid:16) [ γ t v t ] (srf) (cid:17) dt = F (srf) m t [ γ s (1 − β s · β t )] . (B7)Expanding the left hand side of Eq. (B7) obtains (cid:16) [ γ t ] (srf) (cid:17) d (cid:16) [ v t ] (srf) (cid:17) dt + (cid:16) [ v t ] (srf) (cid:17) d (cid:16) [ γ t ] (srf) (cid:17) dt = F (srf) m t [ γ s (1 − β s · β t )] , (B8)where (to order β ) ( v t ) (srf) ≈ (1 + β t · β s ) v − ( β s / v t + ( β s · v t ) β s / , (B9)so (and with the source particle non-accelerating) d (cid:16) [ v t ] (srf) (cid:17) dt ≈ − ( β s / a t + [ β s · a t ] β s / β t · β s ) v + (1 + β t · β s ) a . (B10)Also, ( γ t ) (srf) = γ s γ t (1 − β t · β s ) , (B11)0so, to order β , (cid:18) dγ t dt (cid:19) (srf) ≈ h γ s ˙ γ t (1 − β t · β s ) − γ s γ t (cid:16) ˙ β t · β s (cid:17)i . (B12)With ˙ γ = γ β · ˙ β ≈ β · ˙ β and keeping only to order β , (cid:18) dγ t dt (cid:19) (srf) ≈ h(cid:16) β t · ˙ β t (cid:17) − (cid:16) ˙ β t · β s (cid:17)i = (cid:16) β · ˙ β t (cid:17) , (B13)so ( γ t ) (srf) (cid:18) d [ v t ] dt (cid:19) (srf) + v (cid:16) β · ˙ β t (cid:17) ≈ F (srf) m t [ γ s (1 − β s · β t )] . (B14)Substituting using (B10),( γ t ) (srf) (cid:16) ( β s / a t + [( β s · a t )] β s / β t · β s ) v + (1 + β t · β s ) a (cid:17) + v (cid:16) β · ˙ β (cid:17) ≈ F (srf) m t [ γ s (1 − β s · β t )] (B15)which reduces straightforwardly to − ( β s / a t + [( β s · a t )] β s / γ t ) (srf) a t + ( β t · β s ) a + v h(cid:16) β t · ˙ β t (cid:17)i ≈ γ s F (srf) m t [1 − β s · β t ] , (B16)(and since the field source particle is non-accelerating so that a ≡ a t ). Also have (from Eq. (2)) that a t = (cid:20) γ t m t (cid:21) h F t − γ t ( β t · ˙ β t ) m t v t i , (B17)so − ( β s / a t + [( β s · a t )] β s / (cid:20) ( γ t ) (srf) γ t m t (cid:21) h F t − γ t ( β t · ˙ β t ) m t v t i + ( β t · β s ) a + v h(cid:16) β t · ˙ β t (cid:17)i ≈ γ s F (srf) m t [1 − β s · β t ] , (B18)or − ( β s / a t + [( β s · a t )] β s / (cid:20) ( γ t ) (srf) γ t m t (cid:21) F t + ( β t · β s ) a − v s h(cid:16) β t · ˙ β t (cid:17)i ≈ γ s F (srf) m t [1 − β s · β t ] . (B19)The electric field in the source rest frame due to the stationary source particle is E (srf) = q s h n R i (srf) , (B20)so substituting q t E (srf) for F (srf) obtains − ( β s / a t + [( β s · a t )] β s / (cid:20) ( γ t ) (srf) γ t m t (cid:21) F t + ( β t · β s ) a − v s h(cid:16) β t · ˙ β t (cid:17)i ≈ γ s m t h q s q t n R i (srf) [1 − β s · β t ] . (B21)It is shown below that n (srf) (cid:2) R (srf) (cid:3) = " n − γ s β s − (1 − γ s )(ˆ v s · n )ˆ v s γ s R [1 − ( β s · n )] (lab)ret , (B22)so, to order β s , n (srf) (cid:2) R (srf) (cid:3) ≈ " n − β s γ s R [1 − ( β s · n )] (lab)ret + (cid:20) ( β s · n ) β s / R (cid:21) . (B23)1Substituting Eq. (B23) into Eq. (B21): − ( β s / a t + [( β s · a t )] β s / (cid:20) ( γ t ) (srf) γ t m t (cid:21) F t + ( β t · β s ) a − v s h(cid:16) β t · ˙ β t (cid:17)i ≈ γ s q s q t m t " n − β s γ s R [1 − ( β s · n )] (lab)ret + (cid:20) ( β s · n ) β s / R (cid:21) [1 − β s · β t ] , (B24)or (suppressing notation indicating retardation and since all quantities are assumed defined in the laboratory frameunless indicated otherwise), − ( β s / a t + (cid:20) ( γ t ) (srf) γ t m t (cid:21) F t + ( β t · β s ) a − v s h(cid:16) β t · ˙ β t (cid:17)i ≈ γ s q s q t m t " n − β s γ s R [1 − ( β s · n )] [1 − β s · β t ] . (B25)Substituting for ( γ t ) (srf) = γ s γ t (1 − β t · β s ), − ( β s / a t + (cid:20) γ s (1 − β t · β s ) m t (cid:21) F t + ( β t · β s ) a − v s h(cid:16) β t · ˙ β t (cid:17)i ≈ γ s q s q t m t " n − β s γ s R [1 − ( β s · n )] [1 − β s · β t ] . With (for the non-accelerating field-source particle) a ≈ F t /m t , − ( β s / a t + (cid:20) γ s m t (cid:21) F t − v s h(cid:16) β t · ˙ β t (cid:17)i ≈ γ s q s q t m t " n − β s γ s R [1 − ( β s · n )] [1 − β s · β t ] , (B26)or (since γ s ≈ β s / (cid:20) m t (cid:21) F t − v s h(cid:16) β t · ˙ β t (cid:17)i ≈ γ s q s q t m t " n − β s γ s R [1 − ( β s · n )] [1 − β s · β t ] , (B27)or F t ≈ q s q t γ s " n − β s R [1 − ( β s · n )] − h q s q t R i [( β s · β t ) n − ( β t · n ) β s ] . (B28)With the vector identity a × ( b × c ) = ( c · a ) b − ( b · a ) c , F t ≈ q s q t γ s " n − β s R [1 − ( β s · n )] − h q s q t R i [ β t × ( n × β s )] , or F t ≈ q s q t γ s " n − β s R [1 − ( β s · n )] + q t β t × B , (B29)where B ≈ h q s R i [ β s × n ] . (B30) The above analysis demonstrates that the magneticforce can be interpreted as a relativistic-kinematic con-sequence of the acceleration of the test particle by theCoulomb force, and the relative motion of the field-sourceand test particles. Appendix C: Lorentz transformation, to the fieldsource particle rest frame, of the field sourceparticle to test particle null displacement, and theresulting electric field
The Lorentz transformation for a general pure boost is[2] A boost ( β ) = γ − γβ − γβ − γβ − γβ ( γ − β β ( γ − β β β ( γ − β β β − γβ γ − β β β ( γ − β β ( γ − β β β − γβ γ − β β β ( γ − β β β ( γ − β β (C1)2Suppose we identify the 1 direction as the direction of the test particle velocity in the field-source particle rest frame(where the field source particle is non-accelerating here), then A boost ( β ) = γ − γβ − γβ ( γ − β β = γ − γβ − γβ γ (C2)where β = β is the velocity of the boost here, and γ = 1 / p − β . So, A boost ( β ) = γ − γβ − γβ γ (C3)The four-position is generally R ≡ ( ct, r ) , (C4)where r is the 3-vector position. The test particle four-position is then R t = ( ct, r t ) . (C5)The source particle four-position at the retarded timeand space position of the source particle relative to thetest particle position at time t is then R s = ( c ( t − R/c ) , r s ( t − R/c )) . (C6)The null four-displacement from the field source parti-cle to the test particle (to represent the field point beingat the test particle) is then R ≡ R t − R s = ( R, r ) . (C7)The field source particle to test particle displacementcan be put into components parallel and perpendicular tothe test particle velocity v relative to the source particleas R = ( R, (ˆ v · r )ˆ v + ( r − (ˆ v · r )ˆ v )) . (C8) The test particle four-position transformed from theinertial laboratory frame to the (inertial, for the non-accelerating) field-source particle rest frame, or sourcerest frame (srf) is thus R (srf) ≡ ( R (srf) , r (srf) ) , (C9)with R (srf) = γR − γβ (ˆ v · r ) , (C10)and r (srf) = − γβR ˆ v + γ (ˆ v · r )ˆ v + ( r − (ˆ v · r )ˆ v ) , (C11)or r (srf) = r − γR β + ( γ − v · r )ˆ v . (C12)The field source particle to test particle separation inthe source particle rest frame is thus r (srf) = r − (1 − γ )(ˆ v · r )ˆ v − γβR ˆ v (C13)Also, R (srf) = [( r − (1 − γ )(ˆ v · r )ˆ v − γβR ˆ v ) · ( r − (1 − γ )(ˆ v · r )ˆ v − γβR ˆ v )] / , (C14)which reduces to R (srf) = R (cid:2) − (cid:2) − γ (cid:3) (ˆ v · n ) − v · n ) γ β + γ β (cid:3) / , (C15)so n (srf) = r − (1 − γ )(ˆ v · r )ˆ v − γβR ˆ v R [1 − [1 − γ ] (ˆ v · n ) − v · n ) γ β + γ β ] / , (C16)or, n (srf) = n − γ β − (1 − γ )(ˆ v · n )ˆ v [1 − [1 − γ ] (ˆ v · n ) − v · n ) γ β + γ β ] / . (C17)3Using results above, γ n (srf) (cid:2) R (srf) (cid:3) = γ n − γ β − ( γ − γ )(ˆ v · n )ˆ v R [1 − [1 − γ ] (ˆ v · n ) − v · n ) γ β + γ β ] / , (C18)which can be re-arranged to obtain γ n (srf) (cid:2) R (srf) (cid:3) = n − γ β − (1 − γ )(ˆ v · n )ˆ v γ s R [1 − ( β · n )] . (C19)Eq. (C19) is the needed result for the analysis of Appendix A. It is perhaps worth noting that it can be furtheremployed to obtain exactly the non-radiative electric field in the lab frame as E q = γ n (srf) (cid:2) R (srf) (cid:3) − γ γ + 1 (cid:2)(cid:0) β · n (srf) (cid:1) β (cid:3)(cid:2) R (srf) (cid:3) = n − γ β − (1 − γ )(ˆ v · n )ˆ v γ R [1 − ( β · n )] − γ γ + 1 (cid:2)(cid:0) β · n (srf) (cid:1) β (cid:3)(cid:2) R (srf) (cid:3) , (C20)which under further algebraic manipulations becomes E q = n − β γ R [1 − ( β · n )] . (C21) Appendix D: Lorentz Transformation of anArbitrary Four-Position from an Inertial Frame tothe Test Particle Momentary Rest Frame, with aTranslation of Origin
Consider an arbitrarily located particle position ( r p ( t )expressed in an inertial reference frame coordinate sys-tem ( x , x , x , t ), where t is the time coordinate, withspace origin fixed with respect to the laboratory frameobserver, and an inertial reference frame coordinate sys-tem ( σ , σ , σ , ξ ) with space origin momentarily co-moving with the test particle at t = t , and space originco-located with the field source particle at t = t . Thearbitrarily located particle four-position in the lab frameis then R p = ( ct, r p ( t ) = ( ct, r p ) . (D1) Suppose the test particle position as a function of timeis r ( t ). The arbitrary four position with space part di-vided into components parallel and perpendicular to thetest particle velocity v ( t ) = d r ( t ) /dt at time t = t , v ,is then R p = ( ct, (ˆ v · r p )ˆ v + ( r p − (ˆ v · r p )ˆ v )) . (D2)The arbitrary four position transformed to the test par-ticle momentary rest frame (tmrf), relative to the testparticle located at the position r ( t = t ) ≡ r is P p ≡ R (tmrf) ≡ ( cξ, σ p ) (D3)with ξ = ξ ( t, r p ( t )) = γ ( t − t ) − γ β (ˆ v · ( r p ( t ) − r )) /c with β ≡ v /c , and σ p = − γ β c ( t − t )ˆ v + γ (ˆ v · ( r p ( t ) − r ))ˆ v + ( r p ( t ) − r ) − (ˆ v · ( r p ( t ) − r ))ˆ v ) , or σ p = − γ ( t − t ) v + ( γ − v · ( r p ( t ) − r ))ˆ v + ( r p ( t ) − r ) , (D4)or, with γ ≈ β / σ p ≈ − γ ( t − t ) v + β (ˆ v · ( r p ( t ) − r ))ˆ v / r p ( t ) − r ) , (D5)or, σ p ≈ r p ( t ) − r + ( β · ( r p ( t ) − r )) β / − ( t − t ) v . (D6) [1] L. H. Thomas, “The Kinematics of an Electron with anAxis,” Philos. Mag. 3, 1-22 (1927) [2] J. D. Jackson, Classical Electrodynamics,
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