Lorentz force rotor formulation.

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Lorntz forc rotor formulation. Ptr Joot ptr.joot@gmail.com March 18, 2009. Last Rvision: Dat : 2009/03/2321 : 19 : 46 Contnts 1 Motivation. 1 2 In trms of GA. 1 2.1 Omga bivctor............................ 4 2.1.1 Vrify rotation form..................... 4 2.1.2 Th rotation bivctor..................... 5 2.2 Omga bivctor for boost...................... 5 3 Tnsor variation of th Rotor Lorntz forc rsult. 5 3.1 Tnsor stup.............................. 6 3.2 Lab fram vlocity of particl in tnsor form........... 7 3.3 Lorntz forc in tnsor form..................... 7 3.4 Evolution of Lab fram vctor.................... 8 4 Gaug transformation for spin. 9 1 Motivation. Both [Baylis t al.(2007)baylis, Cabrra, and Kslica] and [Doran and Lasnby(2003)] covr rotor formulations of th Lorntz forc quation. Work through som of this on my own to bttr undrstand it. 2 In trms of GA. An activ Lorntz transformation can b usd to translat from th rst fram of a particl with worldlin x to an obsrvr fram, as in y = Λx Λ (1) 1

Hr Lorntz transformation is usd in th gnral sns, and can includ both spatial rotation and boost ffcts, but satisfis Λ Λ = 1. Taking propr tim drivativs w hav ẏ = Λx Λ + Λx Λ = Λ ( Λ Λ ) x Λ + Λx ( ΛΛ) Λ Sinc ΛΛ = Λ Λ = 1 w also hav Hr s whr a bivctor variabl 0 = Λ Λ + Λ Λ 0 = Λ Λ + ΛΛ Ω/2 = Λ Λ (2) is introducd, from which w hav ΛΛ = Ω/2, and ẏ = 1 2 ( ΛΩx Λ ΛxΩ Λ ) Or ΛẏΛ = 1 (Ωx xω) 2 Th inclusion of th factor of two in th dfinition of Ω was chating, so that w gt th bivctor vctor dot product abov. Prsuming Ω is rally a bivctor (rturn to this in a bit), w thn hav ΛẏΛ = Ω x (3) W can xprss th tim volution of y using this as a stpping ston, sinc w hav ΛyΛ = x 2

Using this w hav 0 = ΛẏΛ Ω x 1 = ΛẏΛ Ωx 1 = ΛẏΛ Ω ΛyΛ 1 = ( Λẏ ΛΛΩ Λy ) Λ 1 = Λ ( ẏ ΛΩ Λy ) Λ 1 So w hav th complt tim volution of our obsrvr fram worldlin for th particl, as a sort of an ignvalu quation for th propr tim diffrntial oprator ẏ = ( ΛΩ Λ ) y = ( 2 Λ Λ ) y Now, what Baylis did in his lctur, and what Doran/Lasnby did as wll in th txt (but I didn t undrstand it thn whn I rad it th first tim) was to idntify this tim volution in trms of Lorntz transform chang with th Lorntz forc. Rcall that th Lorntz forc quation is v = mc F v (4) whr F = E + icb, lik Λ Λ is also a bivctor. If w writ th vlocity worldlin of th particl in th lab fram in trms of th rst fram particl worldlin as v = Λctγ 0 Λ Thn for th fild F obsrvd in th lab fram w ar lft with a diffrntial quation 2 Λ Λ = F/mc for th Lorntz transformation that producs th obsrvd motion of th particl givn th fild that acts on it Λ = FΛ (5) 2mc Okay, good. I undrstand now wll nough what thy v don to rproduc th nd rsult (with th xcption of my rsult including a factor of c sinc thy v workd with c = 1). 3

2.1 Omga bivctor. It s bn assumd abov that Ω = 2 Λ Λ is a bivctor. On way to confirm this is by xamining th grads of this product. Two bivctors, not nccssarily rlatd can only hav grads 0, 2, and 4. Bcaus Ω = Ω, as sn abov, it can hav no grad 0 or grad 4 parts. Whil this is a powrful way to vrify th bivctor natur of this objct it is fairly abstract. To gt a bttr fl for this, lt s considr this objct in dtail for a purly spatial rotation, such as R θ (x) = Λx Λ Λ = xp( inθ/2) = cos(θ/2) in sin(θ/2) whr n is a spatial unit bivctor, n 2 = 1, in th span of {σ k = γ k γ 0 }. 2.1.1 Vrify rotation form. To vrify that this has th appropriat action, by linarily two two cass must b considrd. First is th action on n or th componnts of any vctor in this dirction. R θ (n) = Λn Λ = (cos(θ/2) in sin(θ/2)) n Λ = n (cos(θ/2) in sin(θ/2)) Λ = nλ Λ = n Th rotation oprator dos not chang any vctor colinar with th axis of rotation (th normal). For a vctor m that is prpndicular to axis of rotation n (i: 2(m n) = mn + nm = 0), w hav R θ (m) = Λm Λ = (cos(θ/2) in sin(θ/2)) m Λ = (m cos(θ/2) i(nm) sin(θ/2)) Λ = (m cos(θ/2) + i(mn) sin(θ/2)) Λ = m( Λ) 2 = m xp(inθ) This is a rotation of th vctor m that lis in th in plan by θ as dsird. 4

2.1.2 Th rotation bivctor. W want drivativs of th Λ objct. Λ = θ ( sin(θ/2) in cos(θ/2)) iṅ cos(θ/2) 2 = in θ (in sin(θ/2) cos(θ/2)) iṅ cos(θ/2) 2 = 1 2 xp( inθ/2)in θ iṅ cos(θ/2) So w hav Ω = 2 Λ Λ = in θ 2 xp(inθ/2)iṅ cos(θ/2) = in θ 2 cos(θ/2) (cos(θ/2) in sin(θ/2)) iṅ = in θ 2 cos(θ/2) (cos(θ/2)iṅ + nṅ sin(θ/2)) Sinc n ṅ = 0, w hav nṅ = n ṅ, and sur nough all th trms ar bivctors. Spcifically w hav Ω = θ(in) (1 + cos θ)(iṅ) sin θ(n ṅ) 2.2 Omga bivctor for boost. TODO. 3 Tnsor variation of th Rotor Lorntz forc rsult. Thr isn t anything in th initial Lorntz forc rotor rsult that intrinsically rquirs gomtric algbra. At last until on actually wants to xprss th Lorntz transformation consisly in trms of half angl or boost rapidity xponntials. In fact th logic abov is not much diffrnt than th approach usd in [Tong()] for rigid body motion. Lt s try this in matrix or tnsor form and s how it looks. 5

3.1 Tnsor stup. Bfor anything ls som notation for th tnsor work must b stablishd. Similar to 1 writ a Lorntz transformd vctor as a linar transformation. Sinc w want only th matrix of this linar transformation with rspct to a spcific obsrvr fram, th dtails of th transformation can b omittd for now. Writ y = L(x) (6) and introduc an orthonormal fram {γ }, and th corrsponding rciprocal fram {γ }, whr γ γ ν = δ ν. In this basis, th rlationship btwn th vctors bcoms y γ = L(x ν γ ν ) = x ν L(γ ν ) Or y = x ν L(γ ν ) γ Th matrix of th linar transformation can now b writtn as Λ ν = L(γ ν ) γ (7) and this can now b usd to xprss th coordinat transformation in abstract indx notation y = x ν Λ ν (8) Similarily, for th invrs transformation, w can writ x = L 1 (y) (9) Π ν = L 1 (γ ν ) γ (10) x = y ν Π ν (11) I v sn this xprssd using primd indxs and th sam symbol Λ usd for both th forward and invrs transformation... lacking skill in tricky indx manipulation I v avoidd such a notation bcaus I ll probably gt it wrong. Instad diffrnt symbols for th two diffrnt matrixs will b usd hr and Π was pickd for th invrs rathr arbitrarily. 6

With substitution y = x ν Λ ν = (y α Π α ν )Λ ν x = y ν Π ν = (x α Λ α ν )Π ν th pair of xplicit invrs rlationships btwn th two matrixs can b rad off as δ α = Π α ν Λ ν = Λ α ν Π ν (12) 3.2 Lab fram vlocity of particl in tnsor form. In tnsor form w want to xprss th worldlin of th particl in th lab fram coordinats. That is v = L(ctγ 0 ) = L(x 0 γ 0 ) = x 0 L(γ 0 ) Or v = x 0 L(γ 0 ) γ = x 0 Λ 0 3.3 Lorntz forc in tnsor form. Th Lorntz forc quation 4 in tnsor form will also b ndd. Th bivctor F is So w can writ F = 1 2 F νγ γ ν F v = 1 2 F ν(γ γ ν ) γ α v α = 1 2 F ν(γ δ ν α γ ν δ α)v α = 1 2 (vα F α γ v α F αν γ ν ) 7

And v σ = mc (F v) γ σ = 2mc (vα F α γ v α F αν γ ν ) γ σ = 2mc vα (F σα F ασ ) = mc vα F σα Or v σ = 3.4 Evolution of Lab fram vctor. mc vα F σ α (13) Givn a lab fram vctor with all th (propr) tim volution xprssd via th Lorntz transformation y = x ν Λ ν w want to calculat th drivativs as in th GA procdur ẏ = x ν Λ ν = x α δ α ν Λ ν = x α Λ α β Π β ν Λ ν With y = v, this is v σ = v α Π α ν Λ σ ν = v α mc Fσ α So w can mak th idntification of th bivctor fild with th Lorntz transformation matrix Π α ν Λ σ ν = mc Fσ α 8

With an additional summation to invrt w hav Λ β α Π α ν Λ σ ν = Λ β α mc Fσ α This lavs a tnsor diffrntial quation that will provid th complt tim volution of th lab fram worldlin for th particl in th fild Λ ν = mc Λ α F ν α (14) This is th quivalnt of th GA quation 5. Howvr, whil th GA quation is dirctly intgrabl for constant F, how to do this in th quivalnt tnsor formulation is not so clar. Want to rvisit this, and try to prform this intgral in both forms, idally for both th simplr constant fild cas, as wll as for a mor gnral fild. Evn bttr would b to b abl to xprss F in trms of th currnt dnsity vctor, and thn trat th propr intraction of two chargd particls. 4 Gaug transformation for spin. In th Baylis articl 5 is transformd as Λ Λ ω0 xp( i 3 ω 0 τ). Using this w hav Λ d dτ (Λ ω 0 xp( i 3 ω 0 τ)) = Λ ω0 xp( i 3 ω 0 τ) Λ ω0 (i 3 ω 0 ) xp( i 3 ω 0 τ) For th transformd 5 this givs Λ ω0 xp( i 3 ω 0 τ) Λ ω0 (i 3 ω 0 ) xp( i 3 ω 0 τ) = Cancling th xponntials, and shuffling Λ ω0 = 2mc FΛ ω 0 xp( i 3 ω 0 τ) 2mc FΛ ω 0 + Λ ω0 (i 3 ω 0 ) (15) How dos h commut th i 3 trm with th Lorntz transform? about instad transforming as Λ xp( i 3 ω 0 τ)λ ω0. Using this w hav How Λ d dτ (xp( i 3ω 0 τ)λ ω0 ) = xp( i 3 ω 0 τ) Λ ω0 (i 3 ω 0 ) xp( i 3 ω 0 τ)λ ω0 9

thn, th transformd 5 givs xp( i 3 ω 0 τ) Λ ω0 (i 3 ω 0 ) xp( i 3 ω 0 τ)λ ω0 = 2mc F xp( i 3ω 0 τ)λ ω0 Multiplying by th invrs xponntial, and shuffling, noting that xp(i 3 α) commuts with i 3, w hav Λ ω0 = (i 3 ω 0 )Λ ω0 + 2mc xp(i 3ω 0 τ)f xp( i 3 ω 0 τ)λ ω0 = ( ) 2mc (i 3 ω 0 ) + xp(i 3 ω 0 τ)f xp( i 3 ω 0 τ) 2mc So, if on writs F ω0 = xp(i 3 ω 0 τ)f xp( i 3 ω 0 τ), thn th transformd diffrntial quation for th Lorntz transformation taks th form Λ ω0 Λ ω0 = ( ) 2mc (i 3 ω 0 ) + F ω0 Λ ω0 2mc This is closr to Baylis s quation 31. Dropping ω 0 subscripts this is Λ = ( ) 2mc (i 3 ω 0 ) + F Λ 2mc A phas chang in th Lorntz transformation rotor has introducd an additional trm, on that Baylis appars to idntify with th spin vctor S. My way of gtting thr sms fishy, so I think that I m missing somthing. Ah, I s. If w go back to 15, thn with S = Λ ω0 (i 3 ) Λ ω0 (an application of a Lorntz transform to th unit bivctor for th 2 3 plan), on has Rfrncs Λ ω0 = 1 2 ( ) mc F + 2ω 0S Λ ω0 [Baylis t al.(2007)baylis, Cabrra, and Kslica] W. E. Baylis, R. Cabrra, and D. Kslica. Quantum/classical intrfac: Frmion spin, 2007. URL http: //www.citbas.org/abstract?id=oai:arxiv.org:0710.3144. [Doran and Lasnby(2003)] C. Doran and A.N. Lasnby. Gomtric algbra for physicists. Cambridg Univrsity Prss Nw York, 2003. [Tong()] Dr. David Tong. Classical mchanics. http://www.damtp.cam.ac. uk/usr/tong/dynamics.htm. 10