МЕТhod of determining for a given restricted lorentz transformation matrix, which can always be factorised as the product of a boost matrix and a purely spatial rotation matrix, the angle and the axis of the rotation matrix

Abstract

Method of determining for a given restricted Lorentz transformation matrix, which can always be factorised as the product of a boost matrix and a 4-dim. rotation matrix, the angle ϑ and the axis ω̂ of the 3-dim. rotation matrix R contained in said 4-dim. rotation matrix. For example, if the first component of the first column of a restricted Lorentz transformation matrix is denoted with u 0 , the trace of the restricted Lorentz transformation matrix is denoted with P and the trace of the squared restricted Lorentz transformation matrix is denoted with Q, then the angle ϑ can be calculated by (I). Several similarly simple formulas are given for the angle ϑ, for the axis ω̂ and for said 3-dim. rotation matrix R.

Description

METHOD OF DETERMINING FOR A GIVEN RESTRICTED LORENTZ TRANSFORMATION MATRIX, WHICH CAN ALWAYS BE FACTORISED AS THE PRODUCT OF A BOOST MATRIX AND A PURELY SPATIAL ROTATION MATRIX, THE ANGLE AND THE AXIS OF THE ROTATION MATRIXFIELD OF THE INVENTION

[0001] This invention relates generally to the field of special relativity.

[0002] This invention relates in particular to the manipulation of matrices representing restricted Lorentz transformations [6, chapter 6 on p.167, in particular subchapter "6.3.3. Restricted Lorentz Group" on p.174] in four-dimensional Minkowski space.

[0003] In this application we use for the Minkowski metric the convention^-1 0 0 0^0 1 0 00 0 1 0Greek indices run from 0 to 3, latin indices run from 1 to 3. Capital letters L, R and B denote contravariant tensors of rank 2 with matrix componentsLμν, Rμν,. The products Try, RT] and BT] denote mixed tensors of rank 2 with matrix componentsThe products T]LT], T]RT] and T]BT] denote covariant tensors of rank 2 with matrix components L, R, B.

[0004] A point • between four-vectors denotes always the Minkowski pseudo scalar product (a • u = aTTu =a point • between three-dimensional vectors denotes always a normal Euclidian scalar product (a • u = aTu) and a point • between a matrix and another matrix or a vector denotes always matrix multiplication, however the latter point is most often omitted. Throughout the application documents three-dimensional quantities are always denoted by bold symbols like r,u,a,w, R, L, B in contrast to the associated four-dimensional quantities r, u, a, 7?, L, B. Three-dimensional vectors with a hat " such as & denote always unit vectors.

[0005] We define the signum function sgn byx M- — - = — if x 0sgn[rr]:= < 1^1xG {1, — 1} if x = 0<From this definition followssgn[rr] = — — \ / x G R O sgn2[rr] = 1 \ / x G Rsgn[rr]andsgn[rr] |rr| = x / x G R sgn[rr]rr = \x | / x G RNote that sgn[0] = ±1 in our definition, while in a more common definition of the signum function sgn[0] = 0 applies.BACKGROUND OF THE INVENTION

[0006] Textbooks on special relativity like [6, equ.(4.42) on p.108, equ.(4.41) on p.107, equ.(4.19) on p.102, example 4.1 on p.96, 97, equ.(2.7)-(2.9) on p.32, equ.(2.12)on p.35 and remark 2.9 on p.36] or [10, equ. (6.6) on p.164 and equ. (4.23) on p.118] or [9, equ. (11.36) on p.532 and equ. (11.17) on p.525] or [7, equ. (17.36) on p.373] disclose that if a massive particle is moving with constant speed s = (s1, s2, s3)Trelative to a frame at rest with time coordinate t, then the particle is represented in 4-dimensional Minkowski space by the timelike worldlinewith T being the proper time of the particle and with c denoting the speed of light. In line with [6, equ. 2.12 on p.35] we define the four-velocity u of the particle bysuch thatu• U = UTT]U = u^u11= = — 1 (3)

[0007] Other authors like [10, equ. (6.6) on p.164], [9, equ. (11.36) on p.532], [7, line between equ. (17.30) and equ. (17.31) on p.372 or equ. (17.36) on p.373] or [1, dv p.819, left col., paragraph preceding equ. (5)] define the four-velocity by u = dr / dτ, in which case equ. (3) assumes the form u • u = — c2[10, Exercise 6.2.1. on p.164], [7, equ. (17.20) on p.369] and [1, p.819, left col., paragraph following equ. (5)]. Also in the latter case equ. (3) remains valid if one chooses natural units with c = 1.

[0008] A Lorentz transformation matrix Lη = (Lμν) is a 4x4-matrix representinga mixed tensor, which fulfills the four following equivalent conditions ([6, p.171]):(Lηb) · (Lηc) = (Lηb)Tη(Lηc) = bTηc = b • c for all four-vectors b, c (4)⇔ (Lη)Tη(Lη) = η (5) ⇔ (Lη)−1= η(Lη)Tη (6)⇔ (7)

[0009] A Lorentz transformation matrix, which transforms coordinates in a right handed Minkowski-orthogonal basis moving along a straight timelike worldline as defined in equ.(l) into coordinates in a right-handed Minkowski-orthogonal basis at rest, has to be a restricted Lorentz transformation matrix, i.e. it has to obey additionally the two following conditions ([6, equ.(6.18) on p.174]):det (Lη) = 1 and L00= u0≥ 1The conditions of equ.(8) exclude time and space inversions and ensure that the particle is moving in the future direction [6, p.16]. Since such a restricted Lorentz transformation must transform the four velocity (1 0 0 0)Tofanobserver at rest in the moving coordinate system into the four velocity (u° u1u2u3)Tof the moving coordinate system, the first column vector of a restricted Lorentz transformation matrix Lq is always equal to the four velocity u = (u0u)T, such that in particular L°o= u° applies.

[0010] Examples of restricted Lorentz transformations are purely spatial rotations(Rη)T= (Rη)−1⇔ RT= R−1and det(Rη) = 1 ⇔ det R = 1 (10)and Lorentz boosts, which are also called pure Lorentz transformations or special Lorentz transformations (some authors denote only boosts as Lorentz transformations and use different expressions for other transformations of the Lorentz group or of the restricted Lorentz group). The matrix of a Lorentz boost can be brought in the following form ([5, equ.(3-46) on p.66 and equ.(3-33) on p.53], [6, equ.(6.72) on p.198]):Bη = ... with 1 := ... (11)(Bη)−1= ... (12)— u — u equ.(2) u2Tr(B) = 1 •') I oTr(Bη) = u0+ Tr(B) = 2(u0+ 1) (13) Other authors define the boost and the inverse boost (which is the boost for u → −u) exactly the other way around ([10, equ.(1.45) on p.25], [9, equ. (11.98) on p.547]).

[0011] It is further known from textbooks ([6, chapter "6.5. Polar Decomposition" on p.191-193]), that any restricted Lorentz transformation matrix LT] can be written in a unique way as the product of a boost matrix BT] and a rotation matrix RT] Lη = (Bη)(Rη) (14)

[0012] Thus for a given restricted Lorentz transformation matrix LT] and its associated Lorentz boost BT] the rotation matrix RT] can be calculated byRη = (Bη)−1(Lη)

[0013] From the four dimensional rotation matrix RT] then the three dimensional rotation matrix R can be extracted by equ.(9).

[0014] As shown for example in document [8, p.393] or [13, para. 0016] every three dimensional rotation matrix R can be brought in the following standard form using the Rodrigues formula ([8, p.393], [13, equ.(18), (19)]), which defines an arbitrary rotation matrix R in terms of the rotation angle d and the unit vector pointing in the direction of the rotation axis:R(ϑ, ω̂) = (sinϑ)[ω̂]×+ 1 + (1 − cosϑ)([ω̂]×)2== (sinϑ)[ω̂]×+ (1 − cosϑ)ω̂ω̂T+ cosϑ·1with the unit matrix 1 being defined in equ.(11) and with skew symmetric matrix [a]xassociated with a three-dimensional vector a being defined by(18)with b being an arbitrary three-dimensional vector. From this follows according to [8, fact 4.12.1. i) on p.384][a]2×= aaT− a21 (19) with 1 being the three dimensional unity matrix defined in equ.(11). For GJ thus the following applies:[ω̂]2×= −1-u>2 £qFrom this relation follows the equivalence of equ.(16) and (17).

[0015] From equ.(17) one can see that the trace of a general rotation is given by / ~ 2, ' 2. ~ 2\ Tr(R) = Tr[...] = (1 − cosϑ)(ω̂12+ ω̂22+ ω̂32) + 3cosϑ = 1 + 2cosϑ (20) equ.(9)⇒ Tr(Rη) = 1 + Tr (R) = 2 (1 + cosϑ) (21)

[0016] The antisymmetric part (sinϑ)[ω̂]×= ½(R − RT) of equ.(17) provides an easy way to determine the rotation angle d and the rotation axis GJ of an arbitrary rotation matrix R except if the antisymmetric part turns out to be equal to [0]×. In the latter case the rotation angle is either d = 0 and the rotation axis undetermined (and the rotation matrix the unity matrix R = 1) or the rotation angle is d = TT and equ.(17)ϑ = π ⇒ R = 2ω̂ω̂T− 1 ⇔ ω̂ω̂T= ½(R + 1) (22)applies, such that the absolute values of the three components of the rotation axis GJ can be determined as the three square roots of the three diagonal elements of the matrix - (R + 1) and the relative signs of the three components can be determined from the off-diagonal elements of the matrix - (R + 1).

[0017] Equ.(20) shows that another way to determine the rotation angle d of an arbitrary rotation matrix R is to evaluate the trace of the rotation matrix, which requires only the symmetric part.

[0018] IIf cosϑ ≠ ±1 ⇔ ϑ ∉ nπℤ ⇔ sinϑ ≠ 0, for a general rotation R(d, >) as defined in equ.(16), (17) alwaysR(ϑ,ω̂) = R(−ϑ, −ω̂) applies, i.e. the sign of the rotation axis G depends on the sign of d. If cosϑ = 1 ⇔ ϑ ∈ 2πℤ, the rotation axis GJ of said three dimensional matrix R is trivially not well defined. If cosϑ = −1 ⇔ ϑ ∈ 2πℤ + π, the rotation axis GJ of said three dimensional matrix R(d, >)is well defined only up to the factor ±1, since for a rotation generallyR(π, ω̂) = R(−π, −ω̂) = R(−π, ω̂) = R(−π, −ω̂) applies.

[0019] From the above one can see that in order to calculate for a given restricted Lorentz transformation matrix LT] and its associated boost matrix BT] the rotation angle d and the rotation axis CJ of the three dimensional matrix R contained in four dimensional rotation matrix Rη = (Bη)−1(Lη) a sequence of nontrivial steps is required, which are often performed with the help of computer algebra software.SUMMARY OF THE INVENTION

[0020] A problem to be solved by the current invention is to provide simple closed form expressions for said angle d and said rotation axisin terms of components of a given restricted Lorentz transformation matrix LT].

[0021] In order to solve said problem, the following notions are introduced. An arbitrary Lorentz transformation matrix can always be partitioned in the following wayLr] =which defines the three dimensional vector v and the three dimensional matrix L. As shown in annex 30 (and in the prior art, see for example [14, annex 12] or [13, annex 2.1.]), always v2= u2applies. The three dimensional vector x is defined byMx:= |with skew symmetric matrix [a]xassociated with an arbitrary three-dimensional vector a being defined in equ.(18) above.

[0022] In order to solve said problem, additionally the trace of LT] and the trace of ( T;)2are denoted as follows:P := Tr(Lη) = u0+ Tr(L) (trace of plain matrix) Q := Tr[(Lη)2] (trace of squared matrix)

[0023] According to the invention as defined in claim 1 the rotation angle d of said three dimensional matrix R contained in four dimensional rotation matrix Rη = (Bη)−1(Lη) can be expressed in terms of u°, P, Q, u, v and x-

[0024] As defined in claim 2 the cosinus cos'd of the rotation angle d can for example be expressed in the following surprisingly simple ways:cosϑ = ½[P − uTv / (u0+1) − u0− 1] (25) cosϑ = (P+2)2−Q / 8(u0+1) − 1 (26)cosϑ = ûTv̂ − [Q−(P−2)2] / 8(u0−1) (27)8(u° - 1) Also the two following equations can be used to determine cos'd, although each equation contains a sign ambiguity, since these ambiguities can always be resolved by evaluating both equations unless u = v or u2= 0:cosϑ = ½{ûTv̂ − 1 ± √[(1 + ûTv̂)2− [(û + v̂)Tχ]2]} if u2≠ 0 (28)

[0025] It has to be emphasised that the above list of equations for cos'd is by no means exhaustive. Further equations for cos'd can for example be constructed by forming the average of two or more of equ.(25)-(29). It is for example shown in annex 18 that forming the average of equ.(26) and (27) yieldscosϑ = ½{ûTv̂ − 1 + 1 / (4u2)[u0(P2− Q + 4) − 4P]} (30)2 I 4u2 L JJComparing this result with equ.(28) one can see that the sign ambiguity ± in equ.(28) can be resolved to ± = sgn[u0(P2− Q + 4) − 4P].

[0026] According to the invention as defined in claim 1 also co can be expressed in terms of u0, P, Q, u, v and χ. It will be shown in the detailed description that the four 3-dim. vectors u + v, u x v, x and to are always lying in the plane perpendicular to u — v. Thus co can be expressed as a linear combination of any two linearly independent vectors of the three vectors u + v, u x v and x- More specifically as defined in claim 3 co can be expressed for example in the following surprisingly simple ways:r nll / l u X V \ “ = sgn [szn$\ x + 7 ^—7 (31)√(1 − cos2ϑ) · (... 2u0+ 1) if cosϑ ≠ ±1, ω̂ = sgn[sinϑ] · sgn[χT(u+v) / 2]·√[Q−(P−2)2]·(u+v) / 2 + √[(P+2)2−Q]·χ / (√2·√(1−cosϑ)·√(u0+1)·P) (32)√2·√(1−cosϑ)·√(u0+1)·P if F 0 and cosd 1,ω̂ = sgn[sinϑ] · sgn[χTu]·[√(uTv−u2cosϑ)·(u+v) + 2(u0+1)√(1+cosϑ)·χ] / [P(u0+1)√(1−cosϑ)] (33)P(u0+ 1)√(1 − cosϑ)if P ≠ 0 and cosϑ ≠ 1 orω̂ = sgn[sinϑ] · ... (34)(u2+ uTv)y / l — cosd if v — u and cost) 1. In the above equations for the direction co the expression cosϑ can always be replaced by one of the right hand sides of equ.(25)-(30) and the sign of d and thus of sinϑ can be chosen arbitrarily.

[0027] If one of the above equ.(31)-(34) foris not well defined, because the denominator is equal to zero, then - as shown in the detailed description - always one of the following three cases applies:1. If cos'd = 1 => d G 2TTZ, the rotation matrices Rr] and R are unity matrices and the restricted Lorentz transformation matrix LT] under consideration is a boost matrix BT]. In this case the direction J is thus not well defined.2. If u2= v2= 0, the restricted Lorentz transformation matrix LT under consideration is a rotation matrix RT] and thus L is equal to R. In this case the antisymmetric part of equ.(17) or equ.(22) can be used to determine the rotation axis3. If cosd = — 1 => d 6 2TTZ + TT the rotation axis J is well defined only up to the factor ±1 for reasons given in the paragraph below the paragraph containing equ.(22) above and can be derived from the following equation:, uvT\=2 (35)Thus similar to equ.(22) above the absolute values of the three components of the rotation axis J can in this case be determined as the three square roots of the three 1 ( uv1\ diagonal elements of the matrix - I L + 1 -Qj ) and, re^a^ve signs°f thethree components can be determined from the off-diagonal elements of said matrix.1 / uu^ \ If additionally v = — u applies, equ.(35) simplifies to = - I L + 1 — - - j. If2 \ u “I- 11 additionally v = u = 0 applies, then also Lrj = R and R = L applies and equ.(35) simplifies even to <jd<jdT= - (L + 1) = - (R + 1) as in equ.(22) above.

[0028] The above list of equations for J is by no means exhaustive. Further equations for J can for example be constructed by forming the average of two or more of equ.(31)-(34).

[0029] With the rotation angle d and the rotation axisbeing given the rotation matrix R can be calculated using equ.(16) or (17) above. If one is not particularly interested in the angle d and the axisbut only in the rotation matrix R, then according to the invention as defined in claim 4 a surprisingly simple method to determine the rotation matrix R is given by the following equation:UV TR=L- ^ (36)

[0030] The calculation of the trace Q in the above equations can be simplified by using the following formula / u + v\2Q = Tr[(Lri)2] = 4 (37)I 2 / and the trace P can be expressed as- y2- if u ± v (38)— u2which enables further variations of the above formulas for cos'd and

[0031] It is shown in the detailed description that the three dimensional matrix R contained in four dimensional rotation matrix Rη =(LT / ) fulfills the condition v = RTU -tv u = Rv and that the axis in equ.(34) is thus a special case of the axis of a general rotation imaging vector v into vector u. According to the invention as defined in claim 6 the most general rotation R with rotation angle d, which images a 3-dim. vector a into a vector b, i.e. b = Ra, with a2= b20 A a — b must have the axis±y / aTb — cosd(a + b) + sgn [smd] y / 1 + cosd(a x b)CJ = - 7 — — - (39)(1 + aTb)y / l ~ cosd

[0032] Article

[0011] entitled " On the rotations taking one vector into another" derives also a rotation matrix which takes one vector into another, but the rotation matrix isnot given in dependence of the rotation angle as defined in claim 5, but in dependence of an arbitrary real number ([11, equ.(3.10) on p.617]). A form of the rotation matrix R in the special case that the angle d is chosen to be equal to TT is disclosed in [11, case II in the lower half of p.615]. A form of the rotation matrix R in the special case that the angle d is chosen to be equal to the angle between vectors a and b is disclosed in [3, Lemma 4.1 on p.10] and [2, equ.(18), (20), (73) and (76) and in particular the third line below equ.(21)].

[0033] The reader might be tempted to check, whether the above equations provide the correct results in the simple cases that the restricted Lorentz transformation matrix LT] under consideration is a Lorentz boost matrix BT] or a purely spatial rotation matrix RT]. This is shown in the detailed description.BRIEF DESCRIPTION OF THE DRAWINGS

[0034] The only figure 1 shows in the P, Q-plane the area of possible P, Q-pairs of restricted Lorentz transformation matrices LT] for speeds below the speed of light, i.e with u2< 1.DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENTSProof of equ.(25), (29), (31), (35) and (36)

[0035] Using the partitioned form of a general restricted Lorentz transformation matrix LT] given in equ.(23), of a boost matrix BT] given in equ.(11) and of a purely spatial rotation matrix RT] given in equ.(9) the fact that every restricted Lorentz transformation matrix can be factorised as a product of a boost matrix and a purelyspatial rotation matrix as shown in equ.(14) can be expressed in the following way:

[0036] From equ.(40) one can see that for every restricted Lorentz transformation v = RTU applies, which implies that for every restricted Lorentz transformation v2= u2applies (as already shown in annex 30). From equ.(40) one can further conclude thatThis proves equ.(36).

[0037] Calculating the angle d of rotation R by means of equ.(20) yields equ.(20) equ.(36)co-stf A | [(7V (R) - 1]) A 1 / uv1\ Tr (L) - Tr — - - 1V\u° + 1 / / / ' + 1 This proves equ.(25).

[0038] Calculating the antisymmetric part of R taking equ.(17) into account yields equ.(36), T T ^ (Smd>]x= -(R - RT) +[8, fact 4.12.1 xi) on p.384] and equ.(24)1M x + - / • - 1 U X V

[0039] From this follows firstly / 1 U X V \sin2G =\X +2u° + 1 / which proves equ.(29), and secondly1 / 1 U X V \Gu = if sinG 0sind \ + 2 u° + 1 / which proves equ.(31).

[0040] In case of V 6 2TTZ, GJ is trivially not well defined. In case of d 6 2TTZ + TT equ.(17) and equ.(36) yieldUVR = 2GuGuT— 1 => L = 2GJGJT— I d — - —~ ~ T 1 uvT\<=> LULU L + l -2 U° + 1 / This proves equ.(35).

[0041] In this case the absolute values |a | of the components of Gu can be directly calculated from the diagonal elements of the right hand matrix= 2 " LLii+1“ he “I- ± _V / =2’3and for nonzero | >j|, | > | the relative signs can be determined from the off-diagonal components of the right hand matrix:pry 7W— = sgn - Vi jL u° + 1 JProof of equ.(26), (27), (28), (32), (33) and (38)

[0042] As already shown in equ.(40) above the factorization of equ. (14) can be expressed in the following way:equ. (11), equ. (9) / o\U

[0043] Using equ. (17) one can now determine L, v and quantities derived therefrom in terms of u,annex 1 ( ' d L = (swi$)[w]x+ (1 — cosd)u)a>T+ cosdl — (smd)U XT T / T\ o + ( 1 — COSV) U U? — - P cosv—U-Uv' U° + 1 U° + 1 annex 2v = -(sin^)w x u + (1 — cosd) u) + cosduequ. (44) u + v i -(sW)w x u + (1 — cosd) + (1 + cosd) u(1 — cos'd) (^Tu) + 1 + cosdu — v = (swi$)w x u — (1 — cosd) (a)Tu) + (1 — cosd) u equ. (45) 1 u — (1 — cosid) annex 4 ux v = — (swi$)u2w + (smd) u + (cosd — 1) <i x uannex 51 sired 1 (1 — cos X w - - (51)2 annex 61 — cos'd x24annex 7, equ.(31) 1 u x V I „. X + ^ o, 1 =sm^ 53 2 MU+ 1 annex 8 equ.(48)XT(u - v) 0 = (u + vf (u - v) = (u x vf (u - v) &T(u - v)(54) equ.(54) equ.(53) T 1 T r / u + v\ J. z^T\. Xu= Xv =X I — 2 — J=v1’u / sm^ (55) annex 9 P = (cosd + 1) (M° + 1) + (1 — cos'd)(u° — 1) > 0 (56) annex 32equ.(49) XT(u x v) i (1 — cos'd) — I F i ~ \ 2 / F (57) annex 10 Q = (F - 2)2+ 8 (1 - cos'd) (u° - 1) (<^Tu)2> 0 (58)> 0 annex 10 Q 1 (F + 2)2- 8(cosd + 1)(M° + 1) (59)> 0 annex 10 Q = (F — A / 8)2+ 4 + a?(u,d;,d) (60)< 0 if u2< 1

[0044] Equ.(56), (58) and (59) can be combined in the following way(61)

[0045] Equ.(56), (58), (60) and (59) limit the area of possible (F, Q)-pairs in the P, Q-plane as shown in Fig.l. With the above equations it is easy to show (annex 13) thatequ.(55)21 [(P + 2)2- Q\ 1 [Q - (P - 2)2] = X + (++)] (62) oL JoL J K / L \ 2 / _ This relation is interesting with respect to Fig.l, since one can see that the P- Q-point of a restricted Lorentz transformation lies on the border curves Q = (F — 2)2or Q = (F + 2)2if and only if xT( — ± ) = XTu =x'N =0 applies. The conditionin terms of d, il’.u for the F, Q-point of a restricted Lorentz transformation matrix LT to he on the border curve Q = (F — 2)2(on the border curve Q = (F + 2)2) can be seen from equ.(58) (equ.(59)), while the condition to he either on curve Q = (F — 2)2or on curve Q = (F + 2)2can be seen from equ.(62).

[0046] With the above equations it is now possible to prove equ.(26), (27) and (28) for cos'd and equ.(32) and (33) forequ.(26) is equivalent to equ.(58), equ.(27) is proven in annex 31 and equ.(28) is proven in annex 33. Equ.(32) is proven in annex 14 and equ.(33) is proven in annex 32.Proof of equ.(34) and (39)

[0047] Annex 34 proves that the most general rotation R with rotation angle d 2TTZ, which images a 3-dim. vector a into a vector b, i.e. b = Ra, with a2= b27^ 0 Aa — b must have the axis defined in equ.(39), namely±y / aTb — cosd(a + b) + sgn [smd] y / 1 + cos'd (a x b)(1 + aTb)y / l — cos'dIf a2= b2= 0, every rotation R trivially fulfills the condition b = Ra and if a2= b27^ 0 A a = — b, then d G 2TTZ + TT and the axis CJ is an arbitrary vector perpendicular to both a and b.

[0048] The angle can always be chosen to be TT and - as shown in annex 34 - in this case the rotation axis as defined in equ.(39) assumes the formCJ = ±a + b (63) and the rotation matrix assumes the formR = 2 a + b a + b — 1 (64) It is immediately clear from geometric considerations that a rotation by x around the axis ±a + b images a into b and vice versa. This special case is disclosed in [11, case II in the lower half of p.615; note that in case II in the upper half of p.615 the equ.R = — 13 + cc' contains a typing error and should read R = — 13 + 2cczas correctly stated in case II in the lower half of p.615].

[0049] The angle d can also always be chosen to be equal to the angle between a and b and in this case the rotation axis as defined in equ. (39) assumes the form J = sgn [smd] a x b (65) as shown in annex 34. It is immediately clear from geometric considerations that a rotation by the angle between a and b around the axis a x b images a into b or vice versa depending on the sign of the angle. The rotation matrix assumes in this case -as likewise shown in annex 34 - the formR = 1 + b̂âT− âb̂T+ 1 / (1+âTb̂) (b̂âT− âb̂T)^ (66)1 + aTb '7This special case is disclosed in [3, Lemma 4.1 on p.10] and [2, equ. (18), (20), (73) and (76) and in particular the third line below equ. (21)].

[0050] From equ. (40) one can see that v = RTu => u = Rv applies. Inserting this in equ. (39) yields±y / vTu — cosd(v + u) + sgn [smd] y / 1 + cosd(v x u)(1 + vTu)y / l — cosd±y / v^ u — u2cosd(v + u) + sgn [smd] y / 1 + cosd(v x u)(u2+ vTu)y / l — cos'dif 'd 2TTZ A u2= v20 A u —v. In this case there is a sign ambiguity, but for a given restricted Lorentz transformation J must be uniquely defined. In order to determine the correct sign, in annex 35 equ.(44), (46), (47) and (50) are inserted into equ.(67) and it is demanded that the result is equal to J. The result is that±= sgn [XTR] sgn[smd] (68)must apply. Inserting this in equ.(67) yieldssgn [xTu| sgn[smd]y / vTu — u2cosd(v + u) + sgn [smd] y / 1 + cosd(v x u)(u2+ vTu)y / l — cos'dsgn A / UTV — u2cosd(u + v) — y / 1 + cos'd (u x v)(u2+ uTv)y / l — cos'dThis proves equ.(34).Discussion of the cases, in which the denominators of equ.(31)-(34) for CJ vanish

[0051] In case of cos'd = 1 => 'd E 2TTZ, J is trivially not well defined.

[0052] In case of P = 0 the following applies: from Fig.l one can see thatP = 0 => Q = 4From equ.(56) one can easily derive the conditions, which must be fulfilled by the quantities u, for P = 0:P = 0 O [cosd = — 1 A = 0 V u2= ojj (69)In annex 15 it is shown that the condition P = 0 can also be formulated in terms of u, v, L in the following way:P = 0 w u = -v A L = L' 7 1 (70)Thus one can see from a glance on LT, whether P = 0 applies or not without explicitly calculating P.

[0053] One can see from equ.(69) and (70) that in case of P = 0 always cos'd = — 1 applies (from equ.(70) this can be seen due to v = RTu => u = Rv proven in equ.(40)). Thus in those cases with cos'd1 in which the denominator of equ.(32) or equ.(33) for GJ is equal to zero always equ.(35) can be used to determine GJ with 1 ( uur\ equ.(35) simplifying to GJGJT= - I L + 1 H — - j if v2= u20 and even to 2 \ Ut “I- 11 GJGJT= - (L + 1) if v2= u2= 0.

[0054] In equ.(34) in those cases with cos'd1 the denominator is zero if u = — v applies. If u2= v20, necessarily cos'd = — 1 applies (due to v = RTu u = Rv 1 ( uur\ proven in equ.(40)), such that equ.(35) simplified to GJGJT= - L + 1 H — - can2 ( / / ' + 1 j be used to determineIf u2= v2= 0, the restricted Lorentz transformation matrix LT] under consideration is a rotation matrix RT] and thus L is equal to R. In this case the antisymmetric part of equ.(17) or equ.(22) can be used to determine the rotation axis GJ.

[0055] In equ.(31) in those cases with cos'd1 the denominator is zero if cos'd = — 1 applies, such that equ.(35) can be used to determine GJ..

[0056] We have thus shown that if in one of equ.(31)-(34) the denominator is equal to zero, then either cos'd = 1 (u> not well defined) or cos'd = — 1 (equ.(35) can be used to determine G>) or L = R (the antisymmetric part of equ.(17) or equ.(22) can be used to determine the rotation axis G>) applies as stated in the summary of the invention.Proof of equ.(37) and (38) and of other useful relations

[0057] The calculation of Q is rather simple using equ.(37) above, which readsQ = Tr[(Lri)2] = 4This equation and the following equation for PP = Tr^Lrj) = 2 < u' (71)are proven in annex 11 using only the symmetries of a general restricted Lorentz transformation matrix. While equ.(37) can simplify the calculation of Q = Tr [(LT / )2], equ.(71) is less helpful, since firstly it is easier to calculate trace Tr(Lη) directly from Lη and secondly we found no simple condition to decide, which of the two signs apply: for example for boosts ± = + applies, while for pure spatial rotations ± = sgn[cosϑ] applies as shown in annex 11.

[0058] Equ.(38) for the trace P follows because of (u — v)2= 2(u2— uTv) directly from equ.(57).u + v

[0059] P can also be expressed only in terms of - and x:u + v\22+ 2 - x2±8= |P=: Q+4 see equ.(37)2— (Q+4)| see equ.(116) with± = sgn [P2- (Q + 4)]as proven in annex 23. The line P2— (Q + 4) = 0<=> Q(P) = P2— 4 separates in Fig.1 those (P, Q) values, for which ± = + applies from those, for which ± = — applies. This line comprises the end points (P, Q) = (2,0) and (P, Q) = (√8, 4).

[0060] There are thus three equations for P, namely equ.(38), (71) and (72).

[0061] From u2= v2follows by elementary geometrical considerations that vectors u + V u — V— - — and u x v are lying in a plane perpendicular to — - —. From equ. (54), (31) and u T v (32) one can see that x and Gi are also lying in the plane spanned by vectors — - —and u x v. It is thus possible to express any vector in this plane and in particular Gi as a linear combination of any two other linearly independent vectors in this plane. For Gi this is shown in equ.(31)-(34) and for u x v it is shown in annex 16 that annex 16.J 4 u + v\T / u + v\ u + v\2(73)p —)XI 2 / —)XFrom the above follows further that the cross product of any two linearly independent u + vvectors lying in the plane spanned by vectors — - — and u x v must be a multiple of u — v. In annex 17 it is for example proven thatu - v = — [x x (u + v)] =Application of equ. (25) - (38) in case of boosts and rotations

[0062] That the above equ. (25) - (38) provide the correct results in case of a boost is shown hereinafter. In this case the following relations apply:v = uX = oequ. (13)P2 equ. (11) Lo u annex 37 2u2+ 1 2 / / u (Bltf A uu u 2u°u 1 + 2uu2=> Q = Tr[(Bη)2] = 2u2+ 1 + 3 + 2u2= 4u2+ 4 =Alternatively trace Q can be determined via equ. (37):equ.(37) / u + v\2Q = 4 + 1 - X = 4 (u2+ 1) = 4 (u0)2I 2 / 2

[0063] Using these relations it is easy to show that cos'd is equal to 1:eQU^26)(F + 2)2- Q > i=(2u° + 2 + 2)2- 4 (u0)28(u° + 1) 8(u° + 1)4 (u0)2— (2u° + 2 — 2)28(z / ° - 1) equ. (28) -cos'd i - < uTv — 1 ± y (1 ± uTv)2— [(u + v)Tx]? = ±-\ / (1 + I)2= ±1 equ. (29 ) - „ 1 / / 1 u x v \ cosd = =Ei / l — f x + - o, = ±1V \ 2 w" + 1 ) equ. (30) cos'd i { uTv — I d - - |u°(F2— Q + 4) — 4F]2 I 4u2 L J2 (t? - 1) +1+ 1 - - 2] - 1

[0064] This is one of those cases, in which a combination of equ. (28) and equ. (29) is not sufficient to determine cos'd, but equ. (28) can be used alone if one uses ± = sgn u°(F2— Q + 4) — 4F =+ as explained in the paragraph below equ. (30).

[0065] In case of cos'd = 1 equ.(31)-(34) are not well defined, since co is arbitrary on the unit sphere.

[0066] That equ.(36) provides the correct result in case of a boost can be seen easily:equ.(11)T TRTuv1 -j,uu uvu° + 1u° + 1 u° + 1The rotation matrix linking a boost to a boost is trivially the unity matrix.

[0067] That the above equ.(25) - (38) provide the correct results in case of a purely spatial rotation RT] as defined in equ.(9) is shown in the following. In this case the following relations apply:v = u = 0equ.(17)X — (sin??)wequ.(21)F = 2 (1 + cos??)R2= cos (2??) 1 + sin (2??) [co]x+ [1 — cos(2??)]=s Tr(R2) = 3cos (2??) + [1 — cos(2??)] co2= 1 + 2cos (2??)=⇒ Q = Tr[(Rη)2] = 2 + 2cos(2ϑ) = 2 + 2(2cos2ϑ — 1) = 4cos2ϑAlternatively trace Q can be determined via equ.(37):equ.(37) / u + v\2Q = 4[(u+v / 2)2+ 1 − χ2] = 4(1 − sin2ϑ) = 4cos2ϑ(F + 2)2— Q = (2 + 2cos'& + 2)2— 4cos2'$ = 4 [(cos?? + 2)2— cos2??] = 16 (cos?? + 1)Q - (F - 2)2= 4cos2?? - (2 + 2cos?? - 2)2= 0

[0068] Using these relations it is easy to show by means of equ.(25)-(26) and equ.(29) that d is equal to the rotation angle:| [2 (1 + cost?) — 1 — 1] = cost? equ.(26)16 (cost? + 1) cost? i 1 = cost?8(1 + 1) Equ.(27) and equ.(28) are not well defined, since u, v are not well defined due to u = v = 0.equ.(29) I - - -1 u x 2cost? i ±1 / 1 — III - 7 - I = ±A / 1 — sm2t? = ±cost?V \ 2 w° + l / Equ.(30) is likewise not well defined, since u, v are not well defined due to u = v = 0.This is one of the cases, in which a combination of equ.(28) and equ.(29) is not sufficient to determine cos'd.

[0069] Using the above relations it is further easy to show by means of equ.(31)-(33) that co points in the direction of the rotation axis::equ.(31), 1 r • oi 1 ( 1 u X V \ r • O1 > - co = sgn |sm+9 0X + o01 1V 1 —= s&n1 •=(76)cos+7 \ 2uu+ l / swif cost? 7^ ±1equ^(32) v / 16 (cost? + l)(smt?) co = sgn [smt?] J— - — - - - (77)y / 2y / l — cost? y / l ± 12 (1 + cost?) if P 7^ 0 and cost? 7^ 1j. 2(1 + 1) y / 1 ± cost?smt? = sgn smt? — - — - - —.: (78)2 (1 + cost?) (1 + 1) y / 1 — cost? if P 7^ 0 and cos'd 7^ 1. Equ.(34) is not well defined in this case because v = — u = 0.

[0070] That equ.(36) provides the correct result in case of a purely spatial rotation can be seen easily:equ.(9)TuvR = L — _L | > _ = R77° + 1The rotation matrix linking a rotation matrix to the corresponding boost matrix, which is the unity matrix, is trivially the rotation matrix itself.

[0071] In annex 22 it is further shown that equ.(32)-(34) forprovide also in case of cosϑ = −1 ∧ u2≠ 0 ∧ ûTω̂ ≠ 0 the correct result ω̂ = ± u + v (note that in case of cosϑ = — 1 both directions of the axisyield the same rotation).Application of equ.(25) - (38) to the product of two boosts in order to determine angle, axis and matrix of the Wigner rotation (Thomas rotation)

[0072] The above equ.(25)-(30) for cos'd and equ.(31)-(34) for J can for example be used to determine the angle and the rotation axis of the Wigner rotation, which is also known as Thomas rotation (not to be confused with the Thomas precession). The starting point for the calculation of the Wigner rotation is the restricted Lorentz transformation matrix L.oi obtained by multiplication to two arbitrary Lorentz boosts:UZ \ annex 24 \ Ui!+tI§ + 1 / The Wigner rotation (Thomas rotation) is now the rotation linking this restricted Lorentz transformation to its boost.

[0073] The Wigner angle (Thomas angle) d3and the Wigner axis (Thomas axis) >3can now be determined using the above formulae. For doing so we need the following relations:equ.(79)u 0, = t 0 0+ U, U2(80)annex 25 t H+ M° + U° + 1)(U° - 1) U3X v3x u2(81)M + 1) M + 1)U1> 0annex 26 equ.(81) t 1 + U® +u3 + 1 t 1 U3X v3(82) 2 M + 1) + i)U1 x U22 U3 - Ianne J_x 27 ( zu 0? +, v% 0 + 1 0 +, - I1 \)22H + 1) H + 1)ex 28, 0, 0, 0, ix2 equ.(84) 1 (W + ^2 + ^3 + 1) >11p3 (u? + 1) H + 1) («3 + 1)u3 + 1annex 29 2 equ.(84)(u? + 11% + 11% + I)2Qi = (A - 2)2(86)H + i) H + i)=> Q3- (A - 2)2= 0 A (P3+ 2)2Q3 = 8P3 (87)

[0074] With these relations it is easy to show by means of equ.(25)-equ.(30) thatd is equal to the angle between u3and v3:equ.(25) -i annex 19u| v3COsd3- P3- 23U3V3 (88) equ.(26) equ.(87) equ.(85)(P3+ 2)2- Q,3P3 cos031 = - 1 = u3v3equ.(27) a (P9x2 equ.(87). T. q<3 - (P3 - 2) J. cos,03U, V3- —p, -; - = U, V33 38H - 1) equ.(28)1 cos,032 equ.(82)equ.(29)1\ 2 annex 201 u3x v3\ j_ COsd3(89)2 «3 + 1 / equ.(30) annex 191 cos,03u3v3(90)2 Annex 20 discloses likewise that|sm$3| = |u3x v3| =

[0075] Because of equ.(85) these are the well known forms of cosϑ3and sinϑ3of the Wigner rotation angle (Thomas rotation angle), see [13, equ.(30) and (32) on p.13] or [6, equ.(6.115) on p.213]. From cosϑ3= û3Tv̂3and from the fact that the rotation R images v3on u3one can further conclude that ω̂3= — sgn [sinϑ3] û3× v̂3, from which follows according to equ.(81) ω̂3= — sgn [sinϑ3] û1× u2, which is the well known direction of the axis of the Wigner rotation (Thomas rotation), see [13, equ.(28) on p.12 and paragraph 0028 on pages 12,13] or [6, paragraph below equ.(6.112) on p.213]. Note that if one chooses (as is done in both of [6, equ.(6.118)-(6.120) in combinationwith equ.(6.122)] and [13, equ.(30) on p.13]) the sign of smd3negative, thenω̂3= û1× u2That the same result is obtained by equ.(31)-(34) is shown in annex 21.

[0076] The complete Wigner matrix can be calculated using equ.(36):WvlR, — L3 —«3 +1equ.(79)U1U^+u?u2+ ufu2+ 1annex 36 i o 11 T_1~U2 _,U1U1(u? + 1) + ufu2+ 1)! T 1 — U1+ U2U2H + i)(<w + uru2+ ij+T2ufu2+ (u? + 1) («2 + 1)U1U2(u? + 1) + 1) + ufu2+ 1) T 1=U2U1u?u2+ ufu2+ 1[13, annex 5] 1(93)equ 2) 71 + 72 + 7172 + 7172^7^ +1_L 1 _ c 72s±27181 2 / T / \ / \ I Si s c (71 + 1) (72 + 1) 712c 72 + 7172^- _ c273 1 71s! 72S2]2(94), \ / \ / Si s2c c Jx(71 + 1) (72 + 1) 7172 + 7172^-_ c273For the last step we used the relation Ui = for i = 1,2,3 fromequ. (2) above.

[0077] Equ.(92) is the form of the Wigner rotation obtained in [1, equ.(30) on p.820] and in [13, equ.(27) on p.12]. Equ.(93) is the form of the Wigner rotation obtained in [13, equ.(28) on p.12]. Finally in the form of equ.(94) the Wigner rotation matrix was obtained in [4, equ. (4), (14), (15a), (15b), (16)] and in [13, equ.(29) on p.12], Note that the transformation in annex 36 is significantly simpler than the corresponding transformation in [13, annex 3].Annex 1: Proof of equ.(43)

[0078] We start with the left hand side of equ.(43):equ.(42)L =equ.(17)zT \ = ( 1 + u i j { (smd) [w]x+ (1 — cos'd)odCjT+ cos'd!= (sind) [d>]x+ (1 — cos'd)tia>T+ cosdl + uu^ 4 — [(smd)[d;]x + (1 — cos'd)&&T+ cosdl] == (swi$)[w]x+ (1 — cos,d)<jj<jjT+ cosdl += (swi$)[w]x+ (1 — cos,d)<jj<jjT+ cosdl + u (a? x u uu (sin / d) + co sv— —This is the right hand side of equ.(43).Annex 2: proof of equ.(44)

[0079] We start with the left hand side of equ.(44):equ.(17)v i RTU = [(sW)[w]x+ (1 — cos'd)<jjCjT+ cosdl] u == (smd)[a>]x + (1 — cos'd)&&T+ cosdl] u == -(swi$)w x u + (1 — cos'd) (c<)Tu) & + cosduThis is the right hand side of equ.(44).Annex 3: proof of equ.(57)

[0080] We start with the left hand side of equ.(57):equ.(50),(51)1 o / 0 \ ~1 sin^ / \14 -C0S^ (uT(20. -sinv (u + 1 a? - - - (a? u u H - - - - - -co x u 2 V / 2 u° + M / 2 a0+ 1 -(sffl??)u2w + (smd) (^Tuj u + (cos'd — 1) co x u1 S / Ild \2 / .2\ 1sm$ ( \ 2 / • ox Z- 7 \ - - - u? u — (sznv u - - - a? u u (sznv) a? u + 2 u° + M ' 2 M° + Mv) 1 (1 - cos'd) ( U' CJ ),X+- - — (co x u) (cos'd - 1) (a, u) = 2 Mu+ 1 -7= — -sin2id (u° + 1) u2+ -sin2id (u° + 1) (4)Tu) +4 44 (*T4+o - >) 444 (^4 h° + o h° - 1) +1 (1 — cos'd)2,,2 / , 7 \2- - - (a? x u) (co1u) = 2 / / ° + 17' = -sm2d (u° + 1) (£4U) — u2+ 2 V ) k )(1 — cos'd)2equ.(56) 1This is the right hand side of equ.(57).Annex 4: proof of equ. (50)

[0081] We start with the left hand side of equ. (50):U X V =equ. (44)= u x |-(swi$)w x u + (1 — cos'd) (c<)Tu) & + cos$u| = — (smd)u x (d> x u) 4- (1 — cos'd) (c<)Tu)u xco = = — (sid'd) (u2u) — (d / uj U) + (1 — cos'd) u x a) == (sired) (c<)Tu)u —(sired)u2cb + (1 — cos'd) (c<)Tu)ux co This is the right hand side of equ. (50).Annex 5: proof of equ.(51)

[0082] Preliminary calculations:[8, fact 4.12.1 xi) on p.384](95) uw UCJ — CJU uo?T+ am (96)

[0083] Actual proof of equ.(51):equ.(43),(95),(96)1 smPr / ^,, 1 (1 - cosP) (uT& = (smP)[a?]x— T: w x u x uvl - - - - -:2u° + ltv J Jx2 u° + l 1 1 (1 — cosd) (uTo? ) => x = (smP)o? - - - (a? x u) x u 4 - - - - - -& x u =A!2 u° + 1v!2 u° + 11 sind 1 (1 — COS’2 1 sind r2, Z. T \ i 1 (1 - cos'd) (uTa?) sznd)cv 4 - - - u a? — I a? u u 4 - - - - - -w x u =J2 u° + 1 L V 7 1 ^ 2 u° + 1 1 sind2- 1 smP.. 1 (1 — cosd) (uTa?) smv w 4 - - - u w - - - ar u ) u 4 - 7. - - - x u =V J2 u° + 1 2 u° + 1 / 2 u° + 1. _ 1.n fo ~ 1 sind nT\ 1 (1 - cosd) (uTo>) sznd )oj 4 — sznd \u — l a? - - - a? u u 4 - - - - - -OJ x u =72 V / 2 u° + 1 V J 2 u° + 1 1 o / 0 \ ~1 sin^ / \ 1 t1“C0S^ (uT(20. = -sind (u + 1 a? - - - a? u u 4 - T - - - -OJ x u = 2 V / 2 u° + 1 V J 2 u° + 1|(u° — 1)(1 — cos-The penultimate line proves equ.(51).Annex 6: proof of equ.(52)

[0084] We start with the left hand side of equ.(52):equ.(51)1 / . 2 2-sm2d (£TU) + 4 v / ( \ 2 1 sm2d -| - 9 LU u I u -I - 5 - — ( < JJ X U 4 (Mo + I)274 (u° + I)2 V1 (1 — cos'd)2(uTc<)[sm2d (— u° — 3) + (1 — 2cosd + co.s2d)( / / ° — 1)] +(1 — cosd)2 / X 4- 5— u? u =...(«° + 1)2'

[0085] In order to continue the transformation we need the following relation:sm2d (— u° — 3) + (1 — 2cosd + cos2d)(u° — 1) == u°[— sm2d + 1 — ‘loos'd + cos2d] — (1 — loos'd + cost'd) — 3sin2'd = [— (1 — cos2d) + 1 — loos'd + cos2d] — 1 + loos'd — cost'd — 3(1 — cost'd) = = u°[— loos'd + 2cos2d] — 1 + loos'd — cost'd — 3 + 3cos2d = = u° loos'd (cos'd — 1) + loos'd — 4 + 2cos2d = = 2 |u0cos2d — u^cosd + cos'd — 2 + cos2d| == 2 [(U° + 1 cost'd + loos'd — (u° + 1) cos'd — 2] = = 2 [(u° + 1 cos'd (cos'd — 1) + 2 (cos'd — 1)] == 2 (cos'd — 1) [( ° + 1 cos'd + 2]

[0086] Now we can continue the transformation: / T\2,. 2 (£> u)r / .n(1 — cos2d) (u° + 1 ) + — —1 (cos'd — 1) (u° + 1 ) cos'd + 2 +v 7uu+ 1LV 7 J— (1 — cos'd) (u°This is the right hand side of equ.(52).Annex 7: second proof of equ.(53) (for the first proof see proof of equ.(31) in the paragraph below equ.(41))

[0087] We start with the left hand side of equ.(53)equ.(50),(51)1 u x vX +2 / / ° + 11 o / o \ ~1 Slll,} / \ 1 “cos^ (uT(20. = -sinid (u + 1 u? - - - (a? u u H - - - - - -eu x u+ 2 V / 2 u° + 1 V J 2 / 7° + 1 > +0+ 1) [—(s^n^)u2(^ + (sid'd) (^Tu)u+ (cos'd — 1) (^Tu) & x u| ==- sind (u° + 1) a> + >Q- — [— (sid'd) (u° + 1) (u° — 1) = sid'dw(u “I- 1 ) This is the right hand side of equ.(53).Annex 8: proof of equ.(54)

[0088] We start with the left hand side of equ.(54):equ.(51),(48)1.. / o. 1 sidd / \ 1 (1- - cos'd) (uT&) -sid'd lu + 1 a? - - - l a? u u H - - - - - -cu x u(sid’d)eb x u — (1 — cosd) (o?Tuj u? + (1 — cosd) u COS' 1 Sid'd / - T \3I a\ / - T \ 2 H - - - ( I — cos'd) co U - - - (I — cos'd) leu u u + 2 M° + r ' 2 u° + IV' I (I - cos'd) (uTo?)+2 -x u) =1 / -T \3 1 SUI^ / -T \ 2=2^n(1“C0S^ “ 2^n “c->rUJU +This is the right hand side of equ.(54).Annex 9: proof of equ.(56)

[0089] First we determine Tr(L):equ.(43)2Tr(L) = (1 — cosd)u>2+ 3cosd + (1 — cosd) (uTu)') — - 1- cos'd— — =-7Z / U+ 1 / / ' + 1 = 1 — cosd + 3cosd + (1 — cosd) {uT& J (u° — 1 J + cosd (u° — 1) =( rr~i \ 2 / rj \ rj u w l ( M — 1 ) + u cosd = = 1 + cosd [u® + lj + (1 — cosd) (uTu>j (u° — 1J (97)

[0090] Now we can determine P = Tr(Lη):equ.(23) equ.(97) ⇒ P = u° + Tr(L) = {cosd + 1) (u° + 1) + (1 — cosd) (u° — 1)This proves equ.(56).Annex 10: proof of equ.(58), (59) and (60)

[0091] In equ.(106) of annex 11 the following relation is provenQ = Tr[(Lη)2] = 2 (u2+ u' v + 2 - 2χ2)and we use this relation in the following to calculate Q in terms of u, uyd. To do so we need two preliminary calculations.

[0092] The first preliminary calculation is as follows:u2+ vTu + 2 - 2χ2=equ.(52)1 2(1 — cos'd} + cosidii2+ 2+ 1 _ I / \ 2 / \ 2 r 1 + (—2) - - - < (1 + cos-d} (u° + 1) — 2 (u° ~ 1) [cosd(u° + 1) + 2] +— (1 — cosd) (iz° / \ 2 = u2— [cos-d — 1) + cosdu2+ 2+oo _ 1 I / \ 9 / \ 2 r i H - - - < (1 + cosd) (u° + 1) — 2 (U° — 1) [cosd(u° + 1) + 2] +4 — (1 — cos-d} [u° / \ 2 = [cos-d + l)u2— [cos-d — l)u2(d;Tu) + 2+_f-C0S- J (i -|_ cos-d} (u° + 1) — 2 [U° ~ 1) [cosd(u° + 1) + 2] +— (1 — cos-d} [u° - 1)2(d / u)4= {cos'd + l)u2+ 2+-\-C0S- J (i -|- cos'd) (u° + 1) — 2 (IZ° — 1) [COS#(M° + 1) + 2^ += {cos'd + l)u2+ 2++CQS^ — -{(1C0S^ (M° 1) ~ (u° ~ 1) [C0S^(M° + 1) + 2] +— 2{u° + l)(iz° — 1) (coTu) — (1 — cos'd) (M° — I)2(k>Tu) * == {cos'd + l)u2+ 2+ COSV — 1Z 1o\ / 0, \2H - < (I + cos'd) \ u° + I J — 24 — (I — cos'd) (iz°= {cos'd + l)u2+ 2 +CQS^ — -(l + cos'd) (u° + l) +_2C0S^ — I (M°—l) [(cos^ + l)( / / 0+ I) + 2] + cos'd - l o 2 - - (1 — cos'd) (w — 1) {ar uj == {cos'd + l)u2+ 2 H - - - (1 + cos'd) (u° + 1) ++2- — c°S$ (coTu) {U° — 1) [(cos$ + l)(iz° + 1) + 2] ++ | (1 — cos'd)2{u° — I)2(^ / 7u) = I I 2 = - < 2(cos# + l)u2+ 4 + {cos'd — 1) (1 + cos'd) (u° + 1) ++2 (1 — cos'd) {u° ~ 1) [(cos# + l)(iz° + 1) + 2] ++ (1 — cos'd)2(u° - I)2(£TU)4(98)

[0093] The second preliminary calculation is as follows:2(cos'& + l)u2+ 4 + (cos-# — 1) (1 + cos#) (u° + 1 == 2(cos-# + l)(u° + l)(u° — 1) + 4 + (cos2^ — 1) (u° + 1) == 2(cos-# + l)[(-u°)2— 1] + 4 + [cos2^ — 1) («0)2+ 2u° + 1= 2[(M°)2COS'# + (u0)2— cos-# — 1] + 4++cos2-# (M0)2+ 2U° + 1 («0)2+ 2u° + 1= 2(M°)2COS'# + 2(u0)2— 2cos-# — 2 + 4+ / \ 2 — \ u° j — 2u° — 1 + cos2-# (M0)2+ 2U° + 1r / \ 2 = 2(U°)2COS'# + (u0)2— 2cos'd — 2u° + 1 + cos2-# \u°) + 2u° + 1 == 2(U°)2COS'# + (u0)2— 2cos'd + 2u°(cos2'# — 1) + 1 + cos2-# + cos2-# == (cos2# + 2cos-# + 1) + 2u°(cos2'# — 1) + cos2-# — 2cos-# + 1 == [(cos'# + l)(u° + 1) - 2]2(99)

[0094] After these preliminary calculations Q can be determined in terms of u, co, ■#:equ.(106) equ.(98)Q = Tr[(Lrf] = 2 (u2+ uTv + 2 - 2χ2) = 2(cos-# + l)u2+ 4++ (cos-# — 1) (1+cos#) (u° + +2 (1 — COS'#) (u° — 1) [(cos'# + l)(u° + 1) + 2] ++ (1 — COS'#)2(u° — I)2(cOTft) =equ.(99)= [(cos'# + l)(u° + 1) — 2] ++2 (1 — cos-#) (coTu) (w° — 1) [(cos-# + l)(u° + 1) + 2] ++(1 — COS'#)2(u° — I)2= •••This expression can be further simplified in two ways.

[0095] The first way is as follows:... = [(cost? + l)(u° + 1) — 2] ++2(1 — cos(1 — cos'd) (w° — 1) + [(cost? + l)(iz° + 1) — 2] ++2 [(cost? + l)(u° + 1) - 2] / \ 2 (cost? + l)(iz° + 1) — 2 + (1 — cost?) (coTu) (w° 1)equ.(56)1 / \ 28 (1 — cos'd) (coTu) (w° — 1) + (F — 2)2This proves equ (58).

[0096] The second way is as follows:... = —8(cosid + l)(u° + 1) + [(cost? + l)(u° + 1) + 2] ++2 (1 — cost?) (coTu) (w° — 1) [(cost? + l)(iz° + 1) + 2] ++ (1 — cosd)2(u° — I)2(coTu) = / \ 2 1 2 = — 8(cost? + l)(iz° + 1) + (cos'd + l)(u° + l) + 2 + (l — cos'd) (u°(100) equ. (56) _= -8(cosd + l)(u° + 1) + (2 + F)2This proves firstly equ.(59).

[0097] This can secondly be further transformed into the following expression annex 121(P — A / 8)2+ 4 + x with x < 0 if u° < A / 2which proves equ.(60).Annex 11: proof of equ.(37) and (71)

[0098] If an arbitrary restricted Lorentz transformation matrix LT] is partitioned as defined in equ.(23), then the square ( T;)2assumes the form(i’l)2= (101)and the traces P = Tr Dq) and Q = Tr[(L7y)2] can always be expressed in terms of u, v and x (defined by equ.(24)) as shown in the following.

[0099] 1. Calculation of Tr(L2) and Tr[(Lr)2]:= LS:= Lt= [x]x L2= L2+ LsLa+ LaLs +L2= L2+ L2+ LsLa- (LsLaf antisymmetric Tr(L2) = Tr(L2+ L2) (102) equ.(124) LLT= (Ls+ La) (Ls- La) = L2- LsLa+ LaLs— L2= 1 + uuTequ.(125) LTL = (Ls- La) (Ls+ La) = L2+ LsLa- LaLs- L21 + vvTSummation of the last two equations yields=> L2- L2= i (1 + uuT+ 1 + vvTjL2= 1 + | (uu' +vv7) + L2(103)equ.(24) equ.(19)La - Mx - XX' ~ X21= Tr(L2) = x2- 3X2= -2x2(104) equ.(102) equ.(103)Tr(L2) J= Tr(L2+ L2) J= Tr 1 + | (uuT+ vvT) + L2+ L2equ. (104) = 3 + | (u2+ v2) + 2 (-2*2) = equ. (121) i 3 + | (u2+ u2) - 4x2= 3 + u2- 4x2(105) equ.(101),(105)equ. (2)= u2+ 1 + vTu + uTv + 3 + u2- 4x2== 2UTV + 4 + 2u2— 4^2= 2 (uTv + 2 + u2— 2^2) = (u + v)2+ 4 — 4^2= (106)(u - v)2+ 4 - 4x2+ 4uTvThis is equ. (37).

[0100] 2. Calculationof[8, equ. (3.8.22) on p.301] equ. (124).TL = (dct L)L1= (det L)LT[ 1 - a / Vz. equ. (123)z,,T. / T T■, mr \ I.,T. / T TuuvuJ= (detL) L - L j = (detL) L ~equ. (126).T._L O / T TVU) _?.0T T— u l L — - I — d L — V LrL\WU / Tr(l / ) = u°Tr(L) - vTu (107)

[0101] 3. Calculation of Tr(L) and Tr^Lrj)According to [8, fact 6.9.2.ii)] the following holds:Tr(A) = ½[(Tr())2- Tr(2)]equ.(107),(105) u°Tr() -T= |[(7V(L))2- (3 + u2- 4x2)] O 0 = (Tr(L))2- 2u°Tr(L) + 2vTu - 3 - u2+ 4*22u° ± i / 4(u0)2— 4 (2vTu — 3 — u2+ 4x2) => Tr(L) = - - - - = = u° ± y (770)2— (2vTu — 3 — u2+ 4x2) = u° ± y / u2+ 1 — 2vTu + 3 + u2— 4x2= = u° ± √(2u2+ 4 — 2T— 4χ2) = u° ± √2√(u2—T+ 2 — 2χ2) == u° ± √( — )2+ 4 — 4χ2(108) equ.(23) ⇒ P = Tr(Lη) ≜ i u° + Tr(L) = 2u° ± \ / 2 \ / u2- v' u + 2 - 2χ2== 2u° ± y / (u — v)2+ 4 — 4x2= 2u° ± y 2u2— 2vTu + 4 — 4x2== 2u° ± y / 4u2— 2u2— 2vTu + 4 — 4x2= = 2u° ± y 4u2— (2u2+ 2vTu) + 4 — 4x2= = 2u° ± √(4(u2+ 1) — ( + )2— 4χ2) = 2u° ± √(4(u0)2— [( + )2+ 4χ2]) =This is equ.(71). The ± in this formula for P = Tr{Lg) is given by± = sgn(P — 2u°)Using equ.(61) this can be reformulated into±1 = sgn[(cosϑ + 1)(u° + 1) + (1 — cosϑ)(u° — 1)(ûTω̂)2— 2u°]= sgn[cosϑ(u° + 1) + u° + 1 + (1 — cosϑ)(u° — 1)(ûTω̂)2— 2u°]= sgn[u°cosϑ + cosϑ — u° + 1 + (1 — cosϑ)(u° — 1)(ûTω̂)2]Thus one can see that for boosts cos'd = 1) the correct sign is ± = sgn{2) = + and for pure rotations {u° = 1) the correct sign is ± = sgn(2co d = sgn{cos'd).Annex 12: proof of equ.(60)

[0102] We start with the last relation of annex 10, which defines the quantity x:-8(cosd + l)(u° + 1) + (2 + F)2= (F - y / 8)2+ 4 + xO —8(cosd + l)(u° + 1) + 4 + 4F + F2= F2- 2\ / 8P + 8 + 4 + xO -8(cosd + l)(u° + 1) + 4F = -4y / 2F + 8 + xO —8(cosd + l)(u° + 1) + 4F + 4\ / 2F - 8 = xO —2(005-# + l)(u° + 1) + F(1 + y / 2) - 2 = |equ.(56)— 2(cos-# + l)(u° + 1) + / \ / \ / \ 2 (cos-# + 1) [u° + lj + (1 — COS'#) [u° — lj (uTCOJ (l+ x / 2) - 2 = ^=> (1 + A / 2 — 2)(cos-# + l)(u° + 1) + (1 — cos-#) \u° — Ij (uTcoj (1 + A / 2) — 2 = —(A / 2 — l)(cos# + l)(iz° + 1) + (1 — cos-#) (u° — 1J (uTcoj (1 + A / 2) — 2 = —

[0103] Using this form of x we now show that x ≤ 0 if u° ≤ √2 ⇔ u2≤ 1:ZY»,, \ 2x / 4 + 2 = (√2 — 1)(cosϑ + 1)(u° + 1) + (1 — cosϑ)(u° — 1)(ûTω̂)2(1 + √2) ≤ / ■, \ 2(uTco < 1≤ (√2 — 1)(cosϑ + 1)(u° + 1) + (1 — cosϑ)(u° — 1)(1 + √2) = = (√2 — 1)(u°cosϑ + cosϑ + u° + 1) + (—u°cosϑ + cosϑ + u° — 1)(1 + √2) == — u°cosd — cosd — u° — 1 — u°cosd + cos# + u° — 1 ++ (√2)(u°cosϑ + cosϑ + u° + 1 — u°cosϑ + cosϑ + u° — 1) ==2[(—u°cosϑ — 1) + (√2)(cosϑ + u°)] =u°(√2 — cosϑ) + √2cosϑ — 1≥ 0, u° ≤ √2 ≤ 2[√2(√2 — cosϑ) + √2cosϑ — 1] = 2 ⇒ x ≤ 0Annex 13: proof of equ.(62)

[0104] We start with the left hand side of equ.(62):equ.(61) 1 / 8[(P + 2)2— Q] · 1 / 8[Q — (P — 2)2] ≜0L J0L J= (1 + cosϑ)(1 + u°)(1 — cosϑ)(ûTω̂)2(u° — 1) =+ iz°) (1 — cosd) (v° — 1) == (u° + l)(iz° — 1) (1 — cost'd) == u2sin2ϑ(ω̂Tû)2= sin2ϑ(ω̂Tu)2This is the right hand side of equ.(62).Annex 14: proof of equ.(32)

[0105] Hereinafter we use the following abbreviation:equ.(55)[±] := sgn[χT((u + v) / 2)][±]:= sgn (109)X[~

[0106] With this abbreviation we make the following preliminary calculation:[±]√(1 / 8[Q — (P — 2)2])(u+v) / 2 + √(1 / 8[(P + 2)2— Q])χ = (110) V o 2 V o equ.(61) ≜ [±]|ω̂Tû|√((1 — cosϑ)(u° — 1))·· 1 / 2[—(cosϑ — 1)(ω̂Tu)ω̂ + (1 + cosϑ)u — (sinϑ)ω̂ × u] ++y / (1 + cosft)(u° + 1)- 1 o / 0 \ ~1 Sl ll,} / \ 1 “C0S^ (uT(20. -sinv \u + 1 a? - - - (a? u ul - - - - - -w x u 2 V / 2 u° + M / 2 u° + 1 1[±]|ω̂Tû|√((1 — cosϑ)(u° — 1))(1 — cosϑ)(ω̂Tu) ++ √((1 + cosϑ)(u° + 1)) · sinϑ(u° + 1)[±]|ω̂Tû|√((1 — cosϑ)(u° — 1))(1 + cosϑ) +— √((1 + cosϑ)(u° + 1)) · sinϑ / (u° + 1) · (ω̂Tu)[±]|ω̂Tû|√((1 — cosϑ)(u° — 1))(sinϑ) ++ [±] sgn[ω̂Tû]√((1 — cosϑ)(u° — 1))(1 — cosϑ)√(u2) ++ y (1 + cos$)(u° + 1) sgn(sm$)-y(l — cos$) (1 + cos$) (u° + 1) + -1u u? u [±] sgn [^Tu] y^(l — cos'd) (u° — 1) (1 + cosd) + 2 V / — y^(l + COS'Z?)(M° + 1) sgn(smd)^ (1 — cos'd) (1 + cosd) o _|_ ^ ++- 1u.? x u / a.?r u.\ • 2 V ) [±] sgn [a>Tu] ^(7 — cos'd) (u° — 1) sgn(sin'd)^ (1 — cos'd) (1 + 003$) += -d>-y / l — cos$ [±] faiTu') sgn [a)Tu] \ / u° — 1(1 — cos$)Vrf+ 2— u (1 + cos$) A / 1 — cos$sgn(sin$)- r— - Vu2Az° - 1 - / A / M° + 1 1 cos'd) A / 1 + cos$- ~2 u2■ [±] sgn(sin$) sgn A / M° — 1 — A / 1 — cos'd [±] sgn A / M° — 1(1 — cos'd) A / U2+ +A / M° + 1 sgn(sm$)(l + cos'd) (u° + 1 1 / \ 2 -u (CJ ' UJ A / 1 + cos'd x / 1 — cos$ sgn(sin$)-1 / \ 2- > x u (CJ7U A / 1 — cosd A / 1 + cosd [(sgn(sm )]2• 1 [±] sgn(sind) sgn J (M° — 1) (iz° + 1) — A / U^ \ / zz° + 1 1 A / 1 — cosd [±] sgn [u)Tu] \ / u° — 1(1 — cosd) A / UA +A / M° + 1 sgn(sm$)(l + cosd) (u° + 1 — u (u>Tu) Vl + cosdVl - cos2d 2 V ' sSimnd^ A.-u> x u (&Tu') A / 1 — COS$A / 1 — cos2d fy-—, 2 V i \\sind\J y / u°r,, sind | sznd | 1 A / 1 — cosd [±] sgn \ / u° — 1(1 — cosd)Vu2+- u (u^u) A / 1 + cosd |sm$| sgn(sm^). — [±] sgn(sind) sgn [a)Tu| — 1 2 ' ' difl 1 (u^u) A / 1 — cosd |sm$| [sgn(sm$)]2• ~2 \ ' A A / 0 -L. 1 • [±] sgn(sm$) sgn [u)Tu| — 1 1 = —co U° — 1(1 — COS' 21 \. „ A / 1 + cos'd + -u a? u sin / d —. [±] sgn(sm ) sgn [^7fi] — 1 2 V71, ( -T \ ■ 7 • QA l - cos'd —OJ x u w u) Sind sgn szn'd) — [±] sgn sin'd) sgn u)Tu — 1 2 -7\ / u° + 1V 1 — cos'd [±] sgn [a)Tu] \ / u° — 1(1 — cos'd)+Vu° + 1 sgn(sm$)(l + cos'd) (u° + 11 / ~ T \ • O A / 1 + cos'd H — u a? u si / nv —. [±] sgn(sin'd) sgn — 12 V7x4oTT [±] equ.(109)1, / . T \ •Q / • Q\ l - cos'd - x u w uj sift'd sgn(sz ) — [±] sgn sin'd) sgn u)Tu — 1 2 -7v u° + 11 A / 1 — cos'd [±] sgn [a)Tu] \ / u° — 1(1 — cos'd) A / UM-+V z / ° + 1 sgn(sm$)(l + cos'd) (u° + 1equ.(109)1= — u>x / l — cos'd sgnfsin'd) sgn2(u)Tu] (u)Tu] \ / u° — 1(1 — cos'd) A / U2+ 2+A / M° + 1 sgn(sm$)(l + cos'd) (u° + 1= — u>x / l — cos'd sgn(sin'd) aiTu') A / M° — 1(1 — cos'd) Vu^++Vu° + 1(1 + cos1 x / 1 — cos'd l)(iz° + 1)(1 — cos'd)Vi^+ 2 \ / M0+ 1+ (M° + 1)(1 + cos'd) (u° + 1)1 \ / l — COS'# / \ 2 / \ 2= / o (coTu) u2(l — cos-d) + (u° + 1) (1 + COS'#) 2 Vuu+, W 1s8n(sm'1 Vl — COS'# - 2“ + 1BgD(3Mi / \ 2 / \ 2 (coTuj (u° — l)(iz° + 1)(1 — COS'#) + (u° + Ij (1 + COS'#)= -co x / 1 — cosdy / u0+ 1 sgn(sm#) • / \ 2 / \ (coTu) (u° — 1)(1 — COS'#) + (u° + lj (1 + COS'#)equ.(56) 1 sgn(sm#) - x / 1 — cosdy / u0+ IFco (H l)

[0107] We have thus proven thatsgn [χT / 2] √(1 / 8)[Q-(P-2)2] / 2 + √(1 / 8)[(P+2)2-Q]χ == sgn[sinϑ] · (1 / 2)√(1-cosϑ)√(u0+1) P ω̂⟺ ω̂ = sgn[sinϑ]x / 2x / l — cosd / u0+ IFwhich is identical to equ.(32).Annex 15: proof of statement (70)

[0108] Statement (70) reads:P = 0 ⟺ [u = -v ∧ L = LT≠ 1]

[0109] Proof of directionequ.(69)P = 0 ⇒ ϑ ∈ (2ℤ+1)π ∧ [ûTω̂ = 0 ∨ u2= 0]equ.(42) if ûTω̂ = 0= 0 = −u if u2= 0equ. (51)⇒ χ = 0 ⟺ L = LTTr(L) = 1 − (u0+1) = −u0≠ 3 = Tr(1) ⇒ L ≠ 1

[0110] Proof of direction " ="equ.(42) equ.(69)1. v = RTu = −u ≠ 0 ⇒ [ϑ ∈ (2ℤ+1)π ∧ ω̂Tû = 0] ⇒ P = 0 equ.(42) equ.(51) 2. v = RTu = −u = 0 ⇒ u2= 0 ⇒ χ = sinϑω̂2ω̂ω̂T− 1 if ϑ ∈ (2ℤ+1)πu = −v = 0 ∧ χ = 0 ∧ L ≠ 1 ⇒ ϑ ∈ (2ℤ+1)π ⇒ P = 0 ⟺ L = LTbination of 1. and 2. proves directionAnnex 16: proof of equ.(73)

[0111] We start with the expression in curly braces of equ.(73):[(u+v) / 2]Tχ [(u+v) / 2] − [(u+v) / 2]2= (equ.(47),equ.(55)) (ω̂Tu) sinϑ (u+v) / 2 + equ.(46),equ.(51) −(1 / 2)u2[(1−cosϑ)(ω̂Tû)2+ 1 + cosϑ] x = 2 1 o / 0 \. 1 sid'd,T\ 1 (1 - cos'd} (uTu>) -sired \u + 1 a? - - - (a? u u H - - - - - -cv x u 2 V / 2 u° + M / 2 u° + 11 Ysind (^Tu) {(1 + cos'd) | + |u2(1 — cosd) / + 1 + cosd*T \ / • 2 1 2, J 1 (1 — COS'#) co u) < — (sin d)- — -; u2 r(1 — cos'd) (co u) + 1 + cos'd - - - - - — ".. U “I- -L 2 I / \ ( T \ 1 = — sm$u < (cosd — 1) uj -1 " 2 +- (1 — cos'd) + 1 + cos'd (u° - 1 2(1 - cos2#)| +1 — cosd 1 ^(M° + l)(iz° — 1) (1 — cos'd) (k>Tu) + 1 + cos'd — - - >co x u M° + l J 2 | / . / ' / \ 2 X = — sindu < (cosd — 1) (^co u) -+ - (1 — COS'1h 0 2+— (1 — COS#)(1 + 005-#)^+1 — (u° — 1) (1 — cosd) (^Tu) + 1 + cosd (1 — cosd) 4 -1SZTIVU 2 4 cosd) + (ucosd) + 1 + cosd−sinϑu2 / 4 + (1 / 4)sinϑ{(1+cosϑ)(2+u0−1) + (u0−1)(1−cosϑ)}u+(ω̂Tu)(1−cosϑ){(1+cosϑ)(2+u0−1) + (u0−1)(1−cosϑ)} u×ω̂ =cosd) (u° — 1)(ω̂Tu)(1−cosϑ) u × ω̂ = (P / 4) u × v

[0112] We have thus proven4 2This is equ.(73).Annex 17: Proof of equ.(74)

[0113] Proof of the first part of equ.(74)equ.(51),equ.(46)1 o / o \ ~1 Slll,} / \ 1 “COS^ (H7^). -simd \u + 1 co - - - (co u u 4 - - - - - -w x u 2 V / 2 / / ° + 1 v J 2 a0+ 1 coTu ) co + (1 + cost?) u — (stnt?)co x u / \ 2 1 (cost? — l)smt? OsT \2zv 1 (1 “ cos'd)2(uTco) CO U I u x co H - - - - - — (co X u) X co +2 u° + 1 ) 2 u° + 1V 71 ox - a / o 1 (1 - cosP) (1 + cosP) (uT^)+ - ( 1 + cosv) sinv (u + 1 co x u 4 - - - - - - co x u x u+ 2v' J 2 77° + 1v’1. 2 o / 0 A « x 1 so / 24 / T\.. — sm id (u + 1 co x (co x u) 4 - - - co u u x (co x u) = 2 V ’ 2 u° + 1 ) ’ 1 (1 — cosd) sind t?) sind (u° 4- 1) & X u+ 2 u° + 1 1 (1 — cos'd)2(uTco co x (co x u) + 2 u° + 1 1 (1 - cos2P) (u'co) 1 sin^^ - - - - H - a) U X (co X u) = 2 u° + 1 2 u° + M 1 (1 — cost?) u2= sind u 2 M° + 12 1 (1 — COS’ <jJ — 2 u{1 (1 — cos 1 sm2P u2a> 2 1 (1 — cos'd) u2= sind 2 M° + 1 1 (1 — cos'd)22 u° + 1(1 / 2) sin2ϑ (uTω̂) / (u0+1) · u2ω̂= −sinϑ(1−cosϑ)(u0+1) cosϑ ω̂ × u1 (1 — cos'd) (u° + l)(iz° cos 2• (1 — cosd)= (P / 2)(u−v)This proves the first part of equ.(74).

[0114] The proof of the other parts of equ.(74) is as follows. From equ.(73) follow the relations / u + v\X xI 2 /

[0115] Combining these equations with the first part of equ.(74) provides the following further relations, which prove the remaining parts of equ.(74):1 (u - v) =2v J

[0116] Note that the identity between the first and the last term in the preceding equation follows by elementary geometry from u2= v2(equ.(121)):= Z.2 {[(u + v)Tv] u - [(u + v)Tu] v} =(u + v)k L J L J J1 r / T 2\ / T 2\ 1=(u2+ 2uTv + v2) v + v ) u - (u v + u ) v] =equ.(121)— 7 7 - T--(uTV + u2>) (u — v) = - (u — v)(2u2+ 2uTv) V Jv 727Annex 18: proof of equ.(30)

[0117] The average of equ.(26) and (27) yields1 [ (F + 2)2- Q Q - (F - 2)2] cos'd = 1 + uTv - 2 8(u° + 1) 8(u° - 1) J= - 1 + A [[( + 2)2- <?] p - 1) - [Q - (P - 2)2] P + 1)] } = 2 I oilL L J L J JJ = Uu'v - 1 + [(F + 2)2(«° - 1) - Q(u° - 1) + (F - 2)2(«° + 1) - Q(u° + 1)] } = 2 I 8u2 L JJ = | |uTv - 1 + [[(F + 2)2+ (F - 2)2] u° + (F - 2)2- (F + 2)2- 2Qu°] | == i (uTv - 1 + -i- [2 / / ° (F2+ 4) + 2 (-4F) - 2Qu°] | = 2 I olrL JJ = | {uTv - 1 + -L p(F2- Q+4) - 4F] |2 I 4uz L J) This is the right hand side of equ.(30).Annex 19: proof of equ.(88) and equ.(90)

[0118] Starting from equ.(25) we calculate cos'd? / equ.(85)- | KV3(U° + 1) - - l)ufv3] = il! v.This proves equ.(88).

[0119] Now we calculate cos'd? starting from equ.(30):equ.(30), costf3= | <ju|v3 - 1 + [u°(P32- Q3+ 4) - 4P3This proves equ.(90).Annex 20: proof of equ.(91) and equ.(89)

[0120] We start from equ.(41):(112)equ.(81)(113)equ.(112) equ.(113) |sm-z93| = |u3x v31 =This proves equ.(91).

[0121] We start from equ.(29):equ.(112)= ±^1 - (u3X V3)2= ± |ug V3| = ±Ug V3This is equ.(89).Annex 21: calculation of the Wigner axis (Thomas axis) using equ.(31) - (34)

[0122] Evaluating equ.(31) we get:equ.(31)_ 1 r n i 1 / 1 U3 X V3 \ = sgn [szn$3\2 QX3+ 0, 1 =Vl — cos2v3\ 2 u3+ 1 )equ.(112) equ.(113)= — sgn[s3]u3x v3= - sgn[sm^3]ui x u2

[0123] Evaluating equ.(32) we get:equ.(32)U3> > sgn > = sgn [sm3]> >V2\ / l - cosd3 / u + 1P3equ.^(87)= sgn [smtf3] — >3x3=\ / 2Vl - cos'd3\jul + 1P32= sgn [switf3] > X3=Vl - cos^3\juQ3+ 1VAequ.(82).. equ.(85),(25)-(30) 1 r • 0 i2 / 1 u3x v3\= sgn smv3-, - - „ - =VI - eostl3. JuS, + 1775 v2-1 / = sgn [smV] >,z / — — ( -uf >^o^3) =0. - u^v3yu[] + 1^ (u^v3+ 1) (u% + 1) \u3 1 Jr • 0 i—(M3 + 1) / \ = sgn sm-zV],, — (u3x v3) =V1- ^3^3 + i (u^v3+ 1) H + 1)= sgn [sW3] >1= (u3x v3) = sgn [sin^3] >1= (u3x v3) =^1 - (u^v3) V (u3x v3)equ.(81)= — sgn [sm3] u3x v3= — sgn [sin^] u, x u2

[0124] Evaluating equ. (33) we get:equ.(33)<-^3sgn [ ju3] - u3cosid3(U3+ v3) + 2( / z° + 1)^ / 1 + cosi?3x3= sgn [smi?3]P3H + 1) \ / l - COS1?3 equ.(83),(25)-(30)sgn [sm^3] - - —.A M + 1) yl - ufv3 equ.(85)= sgn [sm$3]equ.(82)1 U3X v3i sgn [sm$3]2 1 / 0 - 1 U3 X V3= — sgn [smi?3]1 / 0 — 11 - uf v3= - sgn [smi?3] -U3 X V3= = - sgn [smi?3]U3 x V3u2^! - (u v3) V (u3x v3)2equ.(81)= — sgn [smi?3] u3x v3= — sgn[smi?]ui x U2

[0125] Evaluating equ.(34) we get:equ.(34)sgn [X3U3] VU3V3 - u|cos$3(u3+ v3) - VI + COS$3(u3X v3) = sgn[smi?3] -— — — - — = - = (ug + u3v3) l - cos'dequ.(83),(25)-(30)= sgn[sm$3]. -^1 + u^v3(u3x v3) u3x v3— sgnta -1— — sgn SZ721 / 3 —, = —, =( 1 + u v3) ^1 - u v3. + u v3. - u v3r • o iu3x v3,. „, U = sgn s3X V - [ m-fl3= = - sgn[sinv3\3 / i IAT* \2(A x, * \2equ.(81)= — sgn[sm3]u3x v3= — sgn[sm3]ui x u2Annex 22: this annex proves that the calculation of the rotation axis in case of cos'd = -1 A u20 A uvc5 0 using equ. (32) - (34) yields always ±u + v (equ. (31) cannot be used in case of cos'd = — 1)

[0126] From equ. (46) one can see in this case thatu + v = 2 (u>Tu) db => (u + v)2= 4u2(<^Tu)2= 4(u° + l)(u° - 1) (<^Tu)2applies, from equ. (56) one can see thatF = 2(u° - 1) (u>Tu)2=> STv = >U +v = (114)'7^2(1 / 0 + 1)Fapplies, from equ. (55) one can see that< TZ rrt rp f U |~ \7 \Xu = x v = x I ) = 0applies, from equ. (59) one can see that(F + 2)2- Q = 0applies and from equ. (61) one can see thatP = 1 [Q - (P - 2)2] (115) 0L Japplies.

[0127] Inserting the preceding results in equ. (32)- (34) for db yields the following:> > equ.(32)sg“ XT-4— \ / <2 - (I3- 2)2>^i^ + < J(P + 2)2- Qx sgn [smd] - - - - - >, > - _ - y / 2y / l ~ cos'd \ / u^ + IF > sgn [0] > equ. (115) = sgn [0] > -. — = - - — = ± — u + v X / 2X / 2X4OT1F 4y / / / 0+ IF equ. (114) ± ~ + v =y / 2y4oTTy / Fequ.(33)— iPcos'd (u + v) + 2(M° + 1)\ / 1 + cos^x i sgn P (u° + 1) Vl — cos'd.. sgn [0] V uTv + u2(u + v)= Sgll [01- P(u° + l) \ / 2 P (M° + 1) \ / 2 n:, \ equ.(114)equ.(34)= sgn[sm$] (u2+ uTv)Vl — cos'd sgn [0] V uTv + u2(u + v) u + v = sgn[0] -L—Tv)V2Annex 23: proof of equ.(72)

[0128] To prove equ.(72) we make the following transformation:F2= Q + 4 + F2- (Q + 4) == Q + 4 + sgn [F2- (Q + 4)] |F2- (Q + 4) | == Q + 4 + sgn [F2- (Q + 4)] 7[F2- (Q + 4)]2=T2[F2— (Q + 4)]| =...

[0129] To continue the transformation we need the following relation: / u + v\29r7 / u + v \i2 / - X +1- [x = equ.(37),(55) - ^Q - ^ [(P + 2)2- Q] [Q - (F - 2)2] == {-P4+ 2(Q + 4)F2- (Q - 4)2} == ~¥ 64A (1“P4 + + 4)p2- W -4)2-16(?}J= = {-P4+ 2(Q + 4)F2- < Q2- 8Q + 16) - 16Q} = 641-7 J= -^ {-F4+ 2(Q + 4)F2- (Q2+ 8Q + 16) } = 641 v 7 J1 ( 1 1 = - {F4- 2(Q + 4)F2+ (Q +4)2} = - [F2- (Q + 4)] (116)641 J18L JJ

[0130] Now we can continue the transformation:= 4=Q+4 see equ.(37) = |P2— (Q+4) | see equ.(H6) with± = sgn [F2- (Q + 4)]This proves equation (72).Annex 24: proof of equ.(79)

[0131] We start with the left hand side of equ.(79):U1This is the right hand side of equ.(79).Annex 25: proof of equ.(81)

[0132] From equ.(79) the following relation can be read:equ.(79)r..j. / nuf u2\ u3x v3= ui [ u2+ + u2\ Ml + 1 j (nT \ / T \Ut U2 \ I n Ut U2 \U2 + 0, 1XU2pl + 0, 1 + U2X Ui =Ml + 1 / \ U2+ 1 / Ui X U2

[0133] Next we transform the factor in front of Ui x U2:

[0134] To continue the transformation, we need the following identity:[ufu2+ (u®u2+ u® + / / J + 1)] [ufu2- (1 - M1M2)] == [ufu2+ (M? M2+U1 +U2 + 1)] ufu2+— [u^U2 + + ^1 + ^2 + 1)] ++ [ufu2+ (u? U2+ U? + U°2+ 1)] =( T ( 0 0 0 0 1 \ T uxU2I + I u1u2+ u±+ u2+ 1 ) UL U2+— U^U2— +U1 +u2 + 1) ++ufu2M<j)M2 + (M1M2 +U1 +U2 + 1)M1M2== (ufu2) + (21Z(j)M2+ «? + «2) ufu2++ (uyUz + U? + U2+ 1) (uyUz — 1)

[0135] Now we can continue the transformation:> (ufu2+ u?u2+ u? + u2+ l)(ufu2+ u?u2- 1)(u? + 1) H +!)u3X v3=This proves equ.(81).Annex 26: proof of equ.(82)

[0136] We start with the definition of x in equ.(24):equ.(24) equ.(79)T T[X3]x - | (L3 - Lf) = | UiU2— U2UXequ.(79)A 1 M? + / / J +u3 + 1 ~~~7, - —~7. -? Ui X U22 H + 1) H + 1) This proves equ.(82).Annex 27: proof of equ.(84)

[0137] From equ.(79) one can read the trace P3 as follows:This proves equ.(84).Annex 28: proof of equ.(85)

[0138] In annex 28a it is shown that / \ 2 T / rr \ Q Q U3V3 = lu1u2l + u2U| + + / / J (u? U2+ ufu2- 1) (117)Further it is immediately apparent thatequ.(2) equ.(79)u| = U3 u3= (1Z3 + 1)(«3 - 1) = (u'j’u® + ufu2+ i)(u(j)u2+ ufu2- 1)

[0139] With these results it is easy to show thatU^V3+ U. TU3+ + + +j)=Annex 28a: proof of equ.(117)

[0140] We start with a transformation of the left hand side of equ.(117):+ (M2«1 + +U1 ~ 1)U^U1 ++ (M2«1 + +M1 1) (u°2 +ui) (118)

[0141] We continue with a transformation of the right hand side of equ.(117):1(u'j’u!] + ufu2- 1) = M + 1) M + 1)+ (up / J + M? + «2—1)ufu2 + («1 + U2) (U1U2 ~ 1)=1 «2 M + 1) M + 1)+ («2W(j)+ «2 +U1 ~ 1)UIU1 + (119) Since expressions (118) and (119) are identical, equ.(117) is proven.Annex 29: proof of equ.(86)

[0142] In order to prove equ.(86) we need the following identity:

[0143] Now we can prove equ.(86) by starting with its left hand side:

[0144] In order to continue the transformation, we need the following identity:U1 (U2 + 1) + 2ufu2(u® + 1) (u® + 1) + U2 (u® + 1) — U^U2 + (ufu2) == [(-aS] + 1) (Mi + 1)] + 2ufu2(u® + 1) (u® + 1) + (ufu2) + — [(M® + 1) (U1 + 1)1 +U1 (U2 + 1) +U2 (U1 + 1)— ulu2== [(«2 + 1) (M1 + 1) +UfU2] + - [(u» + 1) (u? + l)]2+ («? + 1) («? - 1) («» + l)2++ (M2 + 1) (M2 - 1) + l)2- + 1) - 1) (M2 + 1) (M2 - 1) =M? — 1 — M? —u2 ~= [(«2 + 1) (M1 + 1) +UfU2] +2(u2+ 1) (u? + 1) [- (u2+ 1) + (M2 - 1)] == [(«2 + 1) (M1 + 1) +UfU2] - 4 («2+ 1) (u® + 1) = equ.(80)= f \l / jL* + «2 +u3 O + 1) / —4 ( \u2 + 1)! ( \u® A + 1)!

[0145] Now we can continue the transformation:This is the right hand side of equ.(86).Annex 30: Partitioning of a restricted Lorentz transformation matrix

[0146] A restricted Lorentz transformation matrix LT] can trivially always be partitioned in the following way:uu L

[0147] From conditions (4), (6), (8) and (2) then the following relations can be derived:equ.(6) u — u M-1i 'q(L'q')T,q =L2 1 = LT / (LT / )1M°U — Lv — uuT+ LL:0)2- u21 = (LT / )XLT / =,0x2 2(121)xu (122) 2.2. — lTLr u (123) uu-uuT+ LLT(124) [8, fact (3.21.1) on p.351], -i uu ' A L1= L' l 1 (3,0A2(125)equ.(124),(125)Commutator: [L, LT] = LLT— LTL = uu

[0148] According to textbook [8, equ.(3.9.11) on p.303 and fact 3.17.2 on p.334]:det (LT?) = det = det(L) — VTL == det(L) (u° - -u,LLxu\ Wd,et, ( / TLA)eQu-(S)= det(L) I u° — „o “1(126)Annex 31: proof of equ.(27)

[0149] In order to prove equ.(27) we need the following identity:u2(1 — cos'd) (£TU) + cos'd<=> uTv — cos'd = (1 — cos'd) (127)

[0150] Now we can prove equ.(27):equ.(58)Q =(F — 2)2+ 8 (1 — cos'd) (M° — 1)equ.(127)Q - (F - 2)2= uTv — cos'd8(u° - 1)Q — (P — 2)2<=> cos'd = u7v8(u° - 1)This is equ.(27).Annex 32: proof of equ.(33)

[0151] We start from equ.(32):equ.(32) sgn sgn [sW] x / 2x / l—cosdy / u0+ IF equ.(26),(27) = sgn [sm$] • sgn ^8(1 / ° — 1) (uTv — cos'd)'1 V+ (cos'd + 1) (u° + l)x x / 2\ / l—cosdy / u0+ IF = 2 sgn [smi?] • y (u° — 1) (uTv — cos'd)U+V+ (u° + 1) (1 + cos"d)x Pyj (u° + 1) (1 — cos'd) equ.(55) = sgn [sm$] • sgn [xTu] \ / (u0 —1) (uTv — cosd) (u + v) + 2^ / (u° + 1) (1 + cosd)x Pyj (iz° + 1) (1 — cosd) sgn V uTv — v^cosd (u + v) + 2(iz° + 1W(1 + cosd)x = sgn [smi?]P (u° + 1) Vl — cos-d This proves equ.(33).Annex 33: proof of equ.(28)

[0152] We start from equ.(45):equ.(45)(1 — COS'1 2 cos'd 7^ —12sz •n 2 vo + cosidu2— - — I- cos'dvi21 + COsd 1 + cos'd 1 2 X\ 2 1 + cos'd cos'd 7^ —1 1 2uTv — u2cos'd + uTvcosd — v? cost'd = x 2 1 2 u2(128)

[0153] Although this equation was derived under the assumption cos'd 7^ —1, it is also valid for cos'd = — 1, becauseequ.(55)n T / u + v\ 1cos'd = — 1 => x ( - j = 0

[0154] In order to resolve equ.(128) for cos'd, we need to calculate the discriminant:

[0155] Solving equ.(128) for cos'd yields:12This proves equ.(28).Annex 34

[0156] In this annex we construct the general three dimensional rotation matrix R imaging a vector a into a vector b with a2= b20 and a — b (if a2= b2= 0, every rotation images a into b. If a = — b, it is easy to see that the rotation angle must be d = and the rotation axis GJ can be any axis fulfilling the condition GJ1a = jTb = 0).

[0157] As stated in equ.(16) the matrix R of a general rotation about the axis GJ by the angle d has the formR = 1 + (smi?)[w]x+ (1 - cosd)([d)]x)2We are looking for the most general rotation fulfilling the conditionRa = bO Ra = a + (sinG)Gj x a + (1 — cos'd} [d) x (GJ X a)] = = a + [(d)Ta)d> — a] + (smd)d) x a — cosG [d) x (d) x a)] == (d)Ta)d) + x a + cosd [(a) x a) x w] = b

[0158] By forming the scalar product of the last line with GJ we getGj'a = GJT\J (129) and by forming the scalar product of the same line with (d) x a) x GJ we get[(d) x a) x d?]Tb [(d) x a) x d)]TbC0 > =[d> x a) x GJ}2 =(GJ X a)2

[0159] From equation (129), one can derive that the axis GJ must he in the plane spanned by vectors a + b and a x b, i.e.GJ = i (a + b) + 2(a x b) (131)

[0160] Evaluating the square of equ.(131) yields:d)2= A2(a + b)2+ 2AiA2(a + b)T(a x b) + A^a x b)2== A2(a + b)2+ A^a x b)2= A2(a2+ 2aTb + b2) + A^a x b)2== 2A2(1 + aTb) + A^a x b)2= 2A2(1 + aTb) + A^l — (aTb)2] = 121 — 2A2(l + aTb) 1 — A2(a + b)2=> A 9 = - 7 - = - 7 - (132)1 — (aTb)2(a x b)2

[0161] Evaluating the scalar product of equ.(131) with a yields:equ.(131)d)7a i [Ai (a + b) + A2(a x b)]Ta = Ai(l + aTb) (133)

[0162] With these relations equ.(130) can be transformed. The denominator of equ.(130) can be transformed as followsequ.(133)(d> x a)2= 1 — (d;Ta)2= 1 — A2(l + aTb)2and the numerator of equ.(130) can be transformed as follows[(d) x a) x d)]Tb = [a — (d)Ta)d)]Tb = aTb — (d)Ta)(d)Tb) = equ.(129) equ.(133)= a b — (u? a) = a b — Ax(l + a b)

[0163] Inserting this in equ.(130) yieldsaTb — A2(l + aTb)2cos'd =(d> x a)21 — A2(l + aTb)2=> [1 — A2(l + aTb)2]cosd —Tb + A2(l +Tb)2= 0cosd — aTb + [1 — cosd]A2(l + aTb)2= 0 aTb — cos'd(134)— cosd)(l + aTb)2We have thus Ai in equ.(131) expressed in terms of a, b and d.

[0164] With this result we can also express A2in equ.(131) in terms of a, b and d:x x „ „ aTb — cos'dequ.(132) equ.(134) 1 — 2 - — j_ 1 - 2Aj(l + aTb) j_ (1 - cosd)(l + aTb)1 — (aTb)21 — (aTb)21 (1 — cosd)(l + aTb) — 2aTb + 2cosd 1 — (aTb)2(1 — cosd)(l + aTb) 1 1 + aTb — cos'd — cos'0aT\) — 2aTb + 2cosd 1 — (aTb)2(1 — cosd)(l + aTb)1 1 — cosdaTb — aTb + cos'd 1 — (aTb)2(1 — cosd)(l + aTb) 1 (1 + cosd)(l — aTb) 1 + cos'd 1 — (aTb)2(1 — cosd)(l + aTb) (1 — cosd)(l + aTb)21 + cos'dO A2= [±] (135)(1 — cosd)(l + aTb)2

[0165] From equ.(131), (134) and (135) thus followsaTb — cos'd 1 + cos'd.(a + b) [±] - — (a x b) (136)\ (1 — cosd)(l + aTb)2(1 — cosd)(l + aTb)2

[0166] The radicand of the second square root is always positive and the radicand of the first square root is positive as long ascos'd < cos[<(a, b)]if 0 < <(a, b) < %=> <(a, b) < d < %which defines the possible range for the angle d, which is consistent with simple geometrical considerations.

[0167] Thus for a given cos'd, i.e. for the angles ±d, there are due to the two independent sign ambiguities ± and [±] four possible axis as can also be seen from simple geometrical considerations, which also show that for each angle 'd there are two possible different axis. In order to determine the two possible axis for a given angle 'd (and to check our derivation of equ.(136)), we insert (136) into equ.(16) and demand Ra = b. Note that equ.(16) contains not only cos'd as equ.(136), but also sind and thus fully specifies the angle 'd.

[0168] In order to determine Ra using equ.(16)R= [sind) [d>]x+ (1 — cos,d)<jj<jjT+ cos'd!we have to determineequ.(136)[d>]xa = Cd x a =1 + cos'd (1 — cosd)(l + aTb)2aTb — cos'd 1 + cos'd _- x — a x b x a) = (1 — cosd)(l + aTb)21 + cos'd [b — (aTb)a](1 — cosd)(l + aTb)2andequ.(136)aTb — cos'd(1 + bTa)2(1 — cosd)(l + aTb)

[0169] Thus Ra assumes the formRa = (swi$)w x a + (1 — cos,d)'<jj \6JT+ cosda == (sin'd)-aTb — cos'd \ (1 — cos$)(l + aTb)2+cosda = aTb — cosd » - x — smut a x b) + \ (1 — cos$)(l + aTb)21 + cos'd sinidb+ (1 — cos$)(l + aTb) 1 + cos'd., - x — sift'd a b a+ (1 — cos$)(l + aTb)2(a x b) + aTb — cos'd +(1 — cos'd) (1 + bTa)(1 — cos$)(l + aTb)

[0170] For this expression of Ra to be equal to b the factor in front of a x b must be equal to 0:aTb — cos'd.. - <71?? / — I— (1 — cos$)(l + aTb)2aTb — cos'd\ (1 — cos$)(l + aTb)2= ±√(â^T b̂ − cosϑ / (1−cosϑ)(1+â^T b̂)²) · [−sinϑ [±] =\ (1 - cos<l + aTb)2[— 1 [±] sgn[sm$]] = 0 Va,b,i? => [±] = sgn[sm$]

[0171] The factor in front of b must be equal to 1:r, 1 + cos'd.n±. - x — Sffl»+ \ (1 — cos$)(l + aTb)2aTb — cos'd aTb — cos'd +(1 — cos'd)(1 — cos$)(l + aTb) (1 — cos$)(l + aTb)21[±] A / 1 + cos'd sin'd+ (1 — cos$)(l + aTb)2aTb — cos'd +(1 — cos'd)1[±] A / 1 + cos'd sgn [sm$] |sm$| + (1 — cos$)(l + aTb)2+\ / l — cos$(aTb — cos'd)1[±] A / 1 + cos'd sgn [sm$] v / (l — cos'd) (1 + cos'd) + (1 — cos$)(l + aTb)2[+A / 1 — cos'd(aTb — cos'd) > =1[±] (1 + cos'd) sgn [sW] + aTb — cos'd > = 1 V a, b, id1 + aTb[±] (1 + cos'd) sgn [sin??] + aTb — cos'd = 1 + aTb[±] sgn [sm ] (1 + cos'd) — cos'd = 1 [±] = sgn [sm^]

[0172] The factor in front of a must be equal to 0:1 + cos'd sin'd(aTb) + (1 — cos??)(l + aTb)21[=p] A / 1 + cos'd sinid)aTb) + (1 — cos$)(l + aTb)2[=p] A / 1 + cos'd sgn [sm??] |s?n??| (aTb) ++\ / l — cos'd (aTb — cos'd ) + cos'd =1.. [T] Vl + cos?? sgn [sinid] ^ / (l - cos'd) (1 + cos??)(aTb) + \ (1 — cos'd))! + a1b)2I> + cos'd =— — ~ S [T] (1 + cos'd) sgn [sin??] (aTb) ++aTb — cos'd + cos'd = 0 V a, b, id=^- [=p] sgn [sm-z?] (1 + cos-z?)(aTb) + aTb — cos'd = — (1 + aTb)cosid+> [+] sgn [smd] (1 + cosd)(aTb) = — (aTb) {cos'd + 1)[=p] sgn = -1 <4 [±] = sgn [sm$]

[0173] Thus for Ra = b to be true, [±] = sgn [smd] must apply and for a given angle 'd the rotation axis can assume the following two forms due to one sign ambiguity:aTb — cos'd 1 + cos'dr(a + b) + sgn [smd] - — a x b)\ (1 — cosd)(l + aTb)2(1 — cosd)(l + aTb)2±y aTb — cosd(a + b) + sgn [smd] y / 1 + cosd(a x b)(1 + aTb)y / l — cos'dThis is equ.(39), which defines the two possible axis for a given angle d, which is consistent with elementary geometrical considerations.

[0174] If the angle 'd is chosen to bethe rotation axis is given byy aTb + l(a + b), a + b a + b — —~- OJ = ± -X— — — = ±—, = ±—, = ±a + b(l + aTb)y / 2 ^2 + 2aTb i / fa + bVwhich is equ.(63), and the rotation matrix assumes according to equ.(16) the form- 1which is equ.(64).

[0175] If the angle 'd is chosen to be equal to the angle between a and b, the rotation axis is given by. \ / l + aTb(a x b)= sgn [ sift'd ] -. =(1 + aTb)y 1 — aTbr.Q1(a x b) (a x b) — "y- = sgn [ sift'd ], = sgn [ sift'd ]. = sgn [ sift'd ] a x b ^l - (aTb)2V ta x b)2which is equ.(65).

[0176] The rotation matrix assumes in this case according to equ.(16) the formR(d, u>) = (smd) sgn [smd]-I 2|smd|x / 1 — cos2d221 + aTb[8, fact 4.12.1 xi) on p.384]= 1 + b̂âT− âb̂T+ 1 / (1+âTb̂) (b̂âT− âb̂T)1 + aTb ' which is equ.(66).Annex 35: proof of equ. (68)

[0177] In this annex equ. (44), (46), (47) and (50) are inserted into equ. (67) and it is demanded that the result is equal to Co.

[0178] The denominator of equ.(67) is transformed as follows:equ. (47)(u² + vTu)√1 − cosϑ = u²√1 − cosϑ [(1 − cosϑ)(ω̂Tû)² + 1 + cosϑ] (137)

[0179] The numerator of equ.(67) is transformed as follows:±y VTU — u2cos'd (v + u) + sgn [smd] A / 1 + cos'd (v x u) =equ. (44), (46), (47), (50) -. - = ±y (1 — cos'd) + cosidu2— u2cosd-• (sired)co x u + (1 — cos'd) (d / uj & + cosdu + u] ++ sgn [smd] A / 1 + cos'd (s / iid) (co x u) x u + (1 — cos'd) (c<)Tu) x u| = r \ f x = ±v (1 — cos'd) u)2•• [-(swi$)w x u + (1 — cos'd) (c<)Tu) & + (cos'd + 1) u| ++ sgn [smd] A / 1 + cos'd [(smd)u x (w x u) + (1 - cos'd) (ci)Tu) w x u] == ±^(1 — cos'd) (c<)Tu) [-(sm^)w x u + (1 — cos'd) (c<)Tu) £ + (cos'd + 1) u| ++ sgn [smd] x / 1 + cos'd [(smd) [u2a> — (c<)Tu)u] + (1 — cos'd) (c<)Tu) x u| ={ / 7, 2 _ / \ |=F / (1 — cos'd) (u)Tu (sired) + sgn [smd] x / l + cosd(l — cos'd) (u)Tu > co x u+ { I ~. 2 / \ _ | ±1 / (1 — cos'd) (u)Tu (1 — cos'd) (a)Tu + sired) sgn [smd] x / 1 + cosdu2> u>+ | / .. 2 _ / \ | + < ±v (1 — cos'd}) (^Tu) (cos'd + 1) — (sired}) sgn [smd] A / 1 + cos'd (^Tu) > u

[0180] For equ.(67) to be equal to co the factor in front of u in the numerator of equ.(67) must be equal to 0:±1 / (1 — cosd) (u / u) (cosd 1) — (sind) sgn [smi?] 1 cos'd (u)Tu= x / 1 + cosd < ±y (1 — cosd) (cos'd ± 1) a>Tu — |smd| u / u sgn u / u= |smd| p' ul x / 1 + cos'd / ±1 — sgn u / u } = 0 V d, co, u=> ±1 — sgn u / u = 0 => ± = sgn u / u = sgn u / u

[0181] For equ.(67) to be equal to co the factor in front of co in the numerator of equ.(67) must be equal to the denominator:±y (l — cosd) (£TU) (1 — cosd) (CL / U') ± (sind) sgn [smd] x / 1 ± cosdv? == ±i / (l — cos'd) / u / u') (1 — cos'd) fd / u) ± |smd| x / 1 ± cosdu2== ±y (1 — cos'd) (k>Tu) (1 — cos'd) (k>Tu) ± y (1 ± cos'd) (1 — cos'd) x / 1 ± cosdu2== u2-\ / l — cos'd j ± sgn [u)Tu] (c<)Tu') (1 — cos'd) ± 1 ± cos'd equ.(137)(1 — cos'd) ± 1 ± cos'd x / 1 — cos'd V d, co, u=s ±sgn u / u = 1 <=> ± = sgn ±7u

[0182] For equ.(67) to be equal to co the factor in front of co x u in the numerator of equ.(67) must be equal to 0:I. 2 _ / \ =FA / (1 — cosi?) + sgn [smi?] A / 1 + — cosi?) larui == x / 1 — cos'd |=p |c<)Tu| + sgn [sm'd] y^(l ± cos'd) (1 — cos'd) | == \ / 1 — cos'd I =p u)Tu (sind) ± sgn [smd] |smd| sgn u / u u / u f=(smd)TU x / 1 — cos'd =pl + sgn ci)Tu = O V i?,w, u= =fl + sgn uzTu = 0 => ± = sgn u)Tu = sgn u)Tu

[0183] Thus for all three factors we get the conditionequ.(55)± = sgn [ω̂Tû] = sgn [ω̂Tu] ≜ sgn [χTu] sgn[sinϑ]This is equ.(68).Annex 36: proof of equ.(92)

[0184] We start from equ.(36):Z / 'i'z / 'J + ufu2+ 1U2U1Z / 'l'z / 'J +u'|u2 + 1U2U'

[0185] Before we continue the transformation we calculate the numerator of the fac-[(«? + 1) (u® + 1) + ufu2] + ufu2+ 1) ++ (u? + (u® + ufu2+ (ufu2) +T- 1 I- f I7,0 “I |- 1 ] \ I I f„ 1 / . O2—I |- J ]- \ I.. T. U.2_ L. O f I?1 / / 02—r | 1 -L \ I ~ r | „ ^.o2 uxU2== + 1) (u® + 1) + 2ufu2

[0186] Now we can continue the transformation:1 TU2U1U? U2+ ufu2+ 1 This is equ.(92).Annex 37: proof of equ.(75)

[0187] We start with the left hand side of equ.(75):^2u2+ 1 ^2u2+ 1 2U°UT2w°u 2w°u 1 + 2uu7^This is the right hand side of equ.(75).

[0188] The following paragraph lists all cited documents:Bibliography[1] Donald E. Fahnline, A covariant four- dimensional expression for Lorentz transformations American Journal of Physics, Vol.50, Issue No.9, pp.818-821, September 1982, https: / / doi.org / 10.1119 / 1.12748[2] Donald E. Fahnline, Manifestly covariant, coordinate-free dyadic expression for planar homogenous Lorentz transformations, Journal of Mathematical Physics, Vol.24, No.5, p.1080-1086, May 1983, https: / / doi.org / 10.1063 / 1.525833[3] Jose Angel Cid and F. Adrian F. Tojo, A Lipschitz condition along a transversal foliation implies local uniqueness for ODEs, Electronic Journal of Qualitative Theory of Differential Equations, 2018, No.13, p.1-14, published 16 February 2018, https: / / doi.org / 10.14232 / ejqtde.2018.1.13[4] Abraham A. Ungar, THOMAS ROTATION AND THE PARAMETRIZATION OF THE LORENTZ TRANSFORMATION GROUP, Foundations of Physics Letters, Vol.1, No.1, pp.57-89, 1988, https: / / link. springer. com / article / 10.1007 / BF00661317[5] Howard Percy Robertson, Thomas W. Noonan, Relativity and Cosmology, 1968, W. B. Saunders company, Philadelpia, London, Toronto Library of Congress catalog card number 68-23690[6] Eric Gourgoulhon, Special Relativity in General Frames - From Particles to Astrophysics, 2013, ISBN 978-3-642-37276-6 (eBook), Springer Berlin Heidelberg[7] Oliver Davis Johns, Analytical Mechanics for Relativity and Quantum Mechanics, 2nd edition published 2011, first published in paperback 2016, Oxford University press, ISBN 978-0-19-876680-3 (PBK)[8] Dennis S. Bernstein, Scalar, Vector, and Matrix Mathematics: Theory, Facts and Formulas, Revised and Expanded Edition, 2018, Princeton University Press, ISBN 9780691151205 (hardcover), ISBN 9780691176536 (paperback)[9] John David Jackson, Classical Electrodynamics, Third Edition, John Wiley & Sons, 1998, ISBN 978-0-471-30932-1

[0010] Michael Tsamparlis, Special Relativity - An Introduction with 200 Problems and Solutions, 2019, ISBN 978-3-030-27346-0, 2nd edition, Springer Nature Switzerland AG

[0011] Goetz Trenkler, On the rotations taking one vector into another, International journal of mathematical education in science and technology, Volume 37, Issue 5, pp.614-618, 2006, ISSN: 1464-5211, published online 23.11.2006, https: / / doi.org / 10.1080 / 00207390600597591

[0012] Bernhard Strohmayer, Expression for the four- dimensional Frenet-Serret frame of a given timelike worldline in Minkowski space, which expression encompasses the cases that the third 4D-curvature (second torsion, hyper-torsion or bi-torsion) and possibly also the second 4D-curvature (first torsion) and possibly also the first 4D-curvature are vanishing, WIPO patent publication number WO 2023 / 161916, publication date 31.08.2023https: / / patentscope, wipo.int / search / en / detail.jsf?docId=W O2023161916&_cid=P 11- LSOIWA- 77100-1

[0013] Bernhard Strohmayer, USE OF MATRIX PARTITIONING TO DETERMINE THE WIGNER ROTATION MATRIX (THOMAS ROTATION MATRIX) IN RODRIGUES FORM SPECIFYING THE WIGNER ANGLE (THOMAS ANGLE) AND THE AXIS BY MULTIPLICATION OF THREE LORENTZ BOOST MATRICES, WIPO patent publication number WO 2024 / 075105, publication date 11.04.2024https: / / patentscope, wipo.int / search / de / detail.jsf?docId=W 02024075105&_cid=P20- MBQPK3-19935-1

[0014] Bernhard Strohmayer, METHOD OF DETERMINING THE TRANSPORT PROPERTIES OF LOCAL FRAMES OF TIMELIKE WORLDLINES IN FOUR DIMENSIONAL SPACE WITH MINKOWSKI METRIC H USING MATRIX PARTITIONING, WIPO patent publication number WO 2024 / 084476, publication date 25.04.2024https: / / patentscope.wipo.int / search / de / detail.jsf? docId=WO2024084476&_cid=Pll- MDD5ZN-42785-1

Claims

Claims1. Method of determining for a given restricted Lorentz transformation matrix Lη, which can always be partitioned in the formuLand which can always be factorised as the product of a boost matrix Bη and a 4-dim. rotation matrix1 00 Rthe angle d or the axis of the three dimensional matrix R contained in four dimensional rotation matrix Rη = (Bη)⁻¹(Lη), characterised in that the angle d or the axis are expressed in terms of u⁰, P, Q, u, v and χ. with the 3-dim. vector χ being defined byand with P being the trace of Lq and with Q being the trace of(LT;)2.

2. Method as defined in claim 1, characterised in that the angle d is determined from any of equ.(25)-(30).

3. Method as defined in claim 1, characterised in that the axis is determined from any of equ.(31)-(34).

4. Method of determining for a given restricted Lorentz transformation matrix Lη, which can always be partitioned in the formuu Land which can always be factorised as the product of a boost matrix Bη and a 4-dim. rotation matrixthe three dimensional matrix R contained in four dimensional rotation matrix Rη =(Bη) characterised in that equ.(36) is used.

5. Method of determining the axis of the most general 3-dim. rotation matrix R with rotation angle ϑ ∉ 2πℤ, which images a 3-dim. vector a into a 3-dim. vector b, i.e. b = Ra, with a² = b²≠ 0 ∧ â ≠ −b̂, characterised in that the axis is expressed in dependence of the rotation angle d.

6. Method as defined in claim 5, characterised in that the axis is determined by equ.(39).

7. Method as defined in any preceding claim, characterised in that the calculations are performed by means of software with a computer.

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