Wigner Rotation and Its SO(3) Model: an Active-Frame Approach Leehwa Yeh

Wigner rotation and its SO(3) model: an active-frame approach Leehwa Yeh

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Leehwa Yeh. Wigner rotation and its SO(3) model: an active-frame approach. 2021. hal-03196008v2

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HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Copyright Wigner rotation and its SO(3) model: an active-frame approach

Leehwa Yeh*

Shing-Tung Yau Center, National Yang Ming Chiao Tung University, Hsinchu 30010, Taiwan

As an important issue in special relativity, Wigner rotation is notoriously dicult for begin- ners for two major reasons: this physical phenomenon is highly unintuitive, and the mathematics behind it can be extremely challenging. To remove the rst obstacle, we introduce a clear and easy toy model under the guidance of group theory. To overcome the second, a concise mathematical method is developed by the integration of geometric algebra and the active-frame formalism.

I. INTRODUCTION

First discovered by L. Silberstein then rediscovered by L. Thomas [1,2], the phenomenon that two successive non-parallel boosts (i.e. pure Lorentz transformations) lead to a boost and a rotation is generally called Wigner rotation [3]. It has been studied by many authors for almost a century [4-7], the mysterious aura still persists nevertheless. The spatial rotation resulting from the successive application of two non-parallel Lorentz transformations have been declared every bit as paradoxical as the more frequently discussed apparent violations of common sense, such as the so-called `twin paradox.' But the present apparent paradox has important applications... said H. Goldstein in his classic work Classical Mechanics [8]. Intuitively, Wigner rotation seems to violate the transitive relation of parallelism, which is certainly unreasonable. As a matter of fact, as long as there is a relative velocity, the apparent parallelism between the spatial parts of two reference frames is just a three-dimensional illusion. The situation will be quite dierent if we switch to the four-dimensional perspective. Consider two reference frames with no relative velocity at rst, we may assume their individual spatial axes are parallel to each other. However, when one of the frames begins to move with a constant velocity, the parallelism vanishes at least partially. This is because boost is essentially a pseudo- rotation in four-dimensional spacetime.

When the problem involves two non-parallel boosts and three reference frames, say K0, K1 and K2, it makes sense that some of the spatial axes of K0 and K2 might be non-parallel since not all of the spatial axes of K0 (K1) are parallel to that of K1 (K2). What contradicts intuition is that although another frame K3 can be constructed from K2 by a boost transformation so that K3 and K0 are at rest with respect to each other, some spatial axes of K3 and K0 still dier by a rotationthe Wigner rotationeven though there is no temporal dimension involved. To comprehend this fact, it is better to use a geometric picture to replace the physical one, i.e., consider a series of frame pseudo-rotations (K0→K1→K2 →K3) instead of a snap shot of these four frames. Even then, since the pseudo-rotation is quite dierent to the ordinary one,

*Electronic mail: [email protected]

1 it is hard to build a clear picture in one's mind. The best policy is to nd a toy model for this process which contains only ordinary rotations. Although Wigner rotation emerges whenever the two boost velocities are not parallel, people usually let these velocities be perpendicular to each other to simplify the calculations. It will be called the simple Wigner rotation in this paper. Interpreting this simple case as a series of pseudo-rotations, we are able to build an SO(3) toy model to mimic this process (Section III). Being a model, it contains the essence of the original problem nonetheless. Thus we can use what we learn to study the simple Wigner rotation (Section IV) then generalize it to the general case (Section V). This model may be considered as the rst achievement of this paper. As for the second achievement, by working directly on the active frame we show there is no need to consider the passive coordinate transformation at all (Figure 1). Allying with geometric algebra, the active-frame formalism enables us to derive all the important results of Wigner rotation and condense them into three neat geometric theorems.

Figure 1: Passive transformation of coordinates is generated by active transformation of frame, using the former to describe the latter is indirect and unintuitive. When we need to study the transformation of a frame, it is better to use the active-frame formalism.

II. PRELIMINARIES A. Active frame

To make the mathematics amiable, the two-dimensional Euclidean space R2 is served as our rst example. For a point with coordinates (x, y), the position vector of this point is xxˆ + yyˆ = (ˆx yˆ)(x y)>, where > is the notation for matrix transpose and (ˆx, yˆ) is the frame which may be taken as a set of bases at the origin. From now on, we will always use these bases instead of coordinate axes, the only dierence is the former have unit length while the length of the latter is innite. Now consider a passive linear transformation in this space, if the transformation law for

2 the coordinates is (x0 y0)> = [T ](x y)>, where [T ] is the matrix representation of the transformation T , then the frame must obey (ˆx0 yˆ0) = (ˆx yˆ)[T ]−1 to balance the change made by [T ] and render the position vector intact. Conversely, in order to generate the passive coordinate transformation, the frame actively transforms from its original orientation to the new one. When the transformation is a two-dimensional rotation, i.e., an element of the orthogonal group SO(2), we have [T ]−1 = [T ]>so that the transformations for the coordinates and frame are formally the same.

x0 x cos ω sin ω x = [T ] = , y0 y − sin ω cos ω y

xˆ0 xˆ cos ω sin ω xˆ = [T ] = . (1) yˆ0 yˆ − sin ω cos ω yˆ For a positive ω, (1) represents a counterclockwise (i.e. positive-sense) rotation of the frame (ˆx, yˆ). We shall think of this rotation as taking place in the xy-plane instead of around some axis, the reason will be clear soon. If the coordinates (x, y) are replaced by (x, ct), where c is light speed, the space corresponding to the new coordinates is called the two-dimensional Minkowski space R1,1. It diers from R2 in the following aspects. 1. Although the second coordinate is ct with the dimension of length, the corresponding basis is tˆ which is a dimensionless quantity like xˆ and yˆ. 2. There is no ordinary rotation in R1,1, instead we have the pseudo-rotation (or more precisely the hyperbolic rotation) dened by

x0 cosh Ω − sinh Ω x = , (2) ct0 − sinh Ω cosh Ω ct which leaves x2 − c2t2 unchanged. We may think of it as taking place in the xt-plane just like ordinary rotations in an Euclidean plane. 3. The transformation matrix in (2) belongs to the group SO(1, 1) instead of SO(2). They are symmetric but not orthogonal unless Ω = 0. 4. With regard to the coordinate transformation (2), the transformation law for the Minkowski frame is

xˆ0 cosh Ω sinh Ω xˆ = . (3) tˆ0 sinh Ω cosh Ω tˆ It is obvious that the transformed frame is not orthonormal as long as Ω 6= 0, i.e., the new bases will no longer have unit length and will no longer be perpendicular to each other (Figure 2). From the physical perspective, there are several more noteworthy points. 5. If the frame (ˆx, tˆ) is interpreted as a physical reference frame, then the spatial basis xˆ is usually thought of as a rigid rule and the temporal basis tˆ a set of clocks xed on the rule. When talking about a moving frame, we mean the rule carries those clocks (not the basis tˆ!) moving along the x-direction. 6. If we interpret the hyperbolic angle Ω as the rapidity of the relative speed u between the two frames, i.e., u = c tanh Ω, then (2) and (3) become the boost transformations of the spacetime coordinates (x, ct) and the frame (ˆx, tˆ) respectively. Note that the rigid rules corresponding to the spatial bases xˆ and xˆ0 are still parallel to each other from the one-dimensional point of view. 7. We can see from (3) the length of each of the new bases is larger than the old one, meaning the scale of the frame increases after the transformation. This change of scale is, from geometric

3 point of view, the origin of length contraction and time dilation in special relativity. Hence these two phenomena can be easily demonstrated and understood when we use active frame as the mathematical tool.

Figure 2: Hyperbolic rotation of the Minkowski frame (ˆx, tˆ).

B. SO(2, 1) and SO(3) groups

Although physical spacetime is the four-dimensional Minkowski space R3,1, we work on its subspace in many cases without losing generality. For example, when discussing a boost transformation between two frames (ˆx, y,ˆ z,ˆ tˆ) and (ˆx0, yˆ0, zˆ0, tˆ0), we may assume the relative velocity is along the x-direction and consider just the transformation of coordinates (x, ct) which is depicted by (2). Similarly, since the problem of Wigner rotation involves only two relative velocities, it is legitimate to put them in the xy-plane so that none of the z-components shows up in the calculations. Therefore the space R2,1 and the transformation group SO(2, 1) are sucient for us to derive all of the related results. From the point of view of group theory, there are many similarities between SO(2, 1) and SO(3), the rotation group of three-dimensional Euclidean space R3. For example, the invariants of these two groups are x2 + y2 − c2t2 and x2 + y2 + z2 respectively. This allows us to model the Wigner rotation problem with a series of rotations in R3 which provides a concrete and clear picture. Our strategy is to study this model in detail and then apply what we learn to the original problem.

C. Geometric algebra Wigner rotation, like many other problems in special relativity, is usually studied by using vectors and matrices as the mathematical tools. However, there is an alternative choice named geometric algebra (also known as Cliord algebra) which might be more suitable. Putting it simply, geometric algebra is nothing but the traditional vector algebra plus a new operation, the so called geometric product. Since none of the original structure is aected after the new operation is added, we may say geometric algebra is a straightforward extension of vector algebra. For the four-dimensional Minkowski space R3,1, the geometric product is dened as below.

4 1. xˆxˆ =y ˆyˆ =z ˆzˆ = 1 and tˆtˆ = −1. The rst three correspond to the unit length of those bases and the fourth one reects the Minkowskiness of R3,1. 2. The geometric products of dierent bases obey the anti-commutative relation, e.g., xˆyˆ = −yˆx,ˆ yˆtˆ= −tˆyˆ, etc. 3. The dagger conjugation changes the ordering in the products, e.g., (ˆxyˆ)† =y ˆx,ˆ (ˆxyˆzˆ)† = zˆyˆxˆ, etc. With the aid of geometric algebra, we can express (1) and its three-dimensional version in the following neat forms.

xˆ0 cos ω sin ω xˆ xˆ = = R R†, yˆ0 − sin ω cos ω yˆ yˆ xˆ0  cos ω sin ω 0 xˆ xˆ † yˆ0 = − sin ω cos ω 0 yˆ = R yˆ R , (4) zˆ0 0 0 1 zˆ zˆ where ω is called rotor. Note that behaves like a constant because R = exp(− 2 xˆyˆ) zˆ (ˆxyˆ)ˆz =z ˆ(ˆxyˆ) and ω ω . Using the identity , can be expanded as exp(− 2 xˆyˆ)ˆz =z ˆexp(− 2 xˆyˆ) (ˆxyˆ)(ˆxyˆ) = −1 R ω ω and its action to the bases can be calculated easily. For example, cos 2 − xˆyˆsin 2 ω ω ω ω xˆ0 = RxRˆ † = cos − xˆyˆsin xˆ cos +x ˆyˆsin =x ˆ cos ω +y ˆsin ω. (5) 2 2 2 2 If we replace with , ω becomes but the rotation transformation ω ω + 2π R = exp(− 2 xˆyˆ) −R (4) is not aected at all. Therefore we adopt the identication R ≡ −R. The rotor R introduced above can be derived rigorously from the so called Cartan-Dieudonné theorem [6,9]. Here we start with the semi-nished result R = C(ˆx +x ˆ0)ˆx = C(ˆy +y ˆ0)ˆy, where xˆ0 and yˆ0 are given by (1) and C is the real normalization constant in order that RR† = 1. The term xˆ in R = C(ˆx +x ˆ0)ˆx may be interpreted metaphorically as initial position and (ˆx +x ˆ0) on its left as halfway between xˆ and xˆ0. This combination may be taken as a general rule. Since R = C(1 +x ˆ0xˆ) and xˆ0xˆ0 =x ˆxˆ = 1, RxRˆ † = C2(1 +x ˆ0xˆ)ˆx(1 +x ˆxˆ0) = C2xˆ0(1 +x ˆ0xˆ)(1 +x ˆxˆ0) =x ˆ0, which is consistent with (5). By using the explicit expression of xˆ0 and some trigonometric identities, it is straightforward to derive ω ω . Obviously we can use 0 to obtain the same result. R = cos 2 − xˆyˆsin 2 R = C(ˆy +y ˆ )ˆy The rotors for the hyperbolic rotations in R2,1 are similar to those in R3. For example, when the rotation is in the -plane, the corresponding rotor takes the form Ω ˆ and the xt B exp(− 2 xˆt) basis xˆ is transformed as Ω Ω Ω Ω xˆ∗ = BxBˆ † = cosh − xˆtˆsinh xˆ cosh +x ˆtˆsinh =x ˆ cosh Ω + tˆsinh Ω, 2 2 2 2 where the hyperbolic functions come from the identity (ˆxtˆ)(ˆxtˆ) = 1. Analogous to (4), the transformation of the frame (ˆx, y,ˆ tˆ) is

xˆ∗ cosh Ω 0 sinh Ω xˆ xˆ † yˆ∗ =  0 1 0  yˆ = B yˆ B . tˆ∗ sinh Ω 0 cosh Ω tˆ tˆ To derive the hyperbolic rotors, we need to use the Minkowski version of Cartan-Dieudonné theorem. The procedure is similar to that of the rotors in R3. First we have B = C0(ˆx +x ˆ∗)ˆx = 0 ˆ ˆ∗ ˆ with the condition † . Then it can be proved Ω ˆ Ω as long −C (t + t )t BB = 1 B = cosh 2 − xˆt sinh 2 as the identication R ≡ −R is generalized to include hyperbolic rotors.

5 D. Composition of velocities In special relativity, a boost transformation takes place between two inertial frames, hence each boost is dened by a constant velocity which is the relative velocity between the frames. Since Wigner rotation involves two successive boosts, it is inherently related to the problem of adding two velocities relativistically. We give a brief review of this problem below. In the four-dimensional Minkowski space R3,1, the transformation law for a general four- vector Q is Q0 = [L]Q, where [L] is an element of the matrix group SO(3, 1) which contains all the possible passive boosts and rotations as well as their products. When a four-velocity undergoes a boost B(V~ ) with V~ being the relative velocity between the two frames, the transformation is W 0 = [B(V~ )]W , where W and W 0 are the four-velocities of the old and the new frames respectively, and [B(V~ )] is the matrix representation of B(V~ ) which can be proved to be always symmetric. Conversely, the inverse boost transformation W = [B(V~ )]−1W 0 = [B(−V~ )]W 0 allows us to calculate the four-velocity of the old frame from that of the new one. Now consider an object resting in the new frame, since its three-velocity relative to the old frame equals the relative velocity between the two frames, transforming its four-velocity from the new frame back to the old one reveals the information of the boost velocity.     0 Vx −1 0 Vy W = [B(V~ )]   = γ(V~ )   , (6) 0 Vz  c c

1 V where γ(V~ ) = = cosh(tanh−1 ). q V 2 c 1 − c2 The above result has a remarkable geometric representation in the active-frame formalism. Given a boost transformation B(V~ ), the temporal basis tˆ0 of the new frame is proportional to the four-velocity of the boost velocity,

0 0 ~ 0 0 0 0 0 0 −1 0 γ(V ) tˆ = (ˆx yˆ zˆ tˆ)   = (ˆx yˆzˆtˆ)[B(V~ )]   = (V~ + ctˆ). (7) 0 0 c 1 1

When the problem involves three inertial frames and two successive boosts, say rst B(V~1) then B(V~2), the four-velocity of a rest object in the third frame can be transformed to that of the rst frame by

0 0 ~ ~ −1 0 ~ −1 ~ −1 0 W = ([B(V2)][B(V1)])   = [B(V1)] [B(V2)]   . 0 0 c c The three-velocity contained in this four-velocity is the composition of the two boost velocities in that order and is usually denoted by V~1 ⊕ V~2. Therefore the above formula is equivalent to       0 0 (V~1 ⊕ V~2)x ~ ~ ~ −1 ~ −1 0 ~ ~ −1 0 ~ ~ (V1 ⊕ V2)y [B(V1)] [B(V2)]   = [B(V1 ⊕ V2)]   = γ(V1 ⊕ V2)   , (8) 0 0 (V~1 ⊕ V~2)z  c c c

6 and (7) is accordingly generalized to

γ(V~ ⊕ V~ ) tˆ00 = 1 2 [V~ ⊕ V~ + ctˆ] or ctˆ00 = γ(V~ ⊕ V~ )[V~ ⊕ V~ + ctˆ]. (9) c 1 2 1 2 1 2

00 In short, ctˆ equals the four-velocity which corresponds to the three-velocity V~1 ⊕V~2 that denes the composite boost. If the order of the two boosts is exchanged, then (8) becomes

      0 0 (V~2 ⊕ V~1)x ~ ~ ~ −1 ~ −1 0 ~ ~ −1 0 ~ ~ (V2 ⊕ V1)y [B(V2)] [B(V1)]   = [B(V2 ⊕ V1)]   = γ(V2 ⊕ V1)   . 0 0 (V~2 ⊕ V~1)z  c c c

An important identity γ(V~1 ⊕ V~2) = γ(V~2 ⊕ V~1) can be proved by using the following two facts: −1 −1 −1 −1 (i) [B(V~1)] [B(V~2)] and [B(V~2)] [B(V~1)] share the same diagonal elements since boost −1 −1 matrices are all symmetric, (ii) γ(V~1 ⊕ V~2) equals the (4, 4) element of [B(V~1)] [B(V~2)] and −1 −1 γ(V~2 ⊕V~1) equals that of [B(V~2)] [B(V~1)] . This identity together with the marvelous relation (9) will prove useful in later discussion. It is worthwhile to elaborate (8) a little further before closing this section. First notice (8) leads to

0 0 0 ~ −1 0 ~ −1 ~ ~ −1 0 ~ ~ ~ −1 0 [B(V2)]   = [B(−V1)] [B(V1 ⊕ V2)]   = [B(−V1 ⊕ (V1 ⊕ V2))]   , 0 0 0 c c c

~ ~ ~ ~ ~ ~ ~ which implies V2 = −V1 ⊕ (V1 ⊕ V2). Comparing with VA|B = VA − VB (the non-relativistic formula for the relative velocity of object A with respect to object B), we get the formula for relativistic relative velocity, ~ ~ ~ VA|B = −VB ⊕ VA.

Next, if there are three successive boosts, B(V~1), B(V~2) and B(V~3) in this order, then (8) is generalized to

0 0 ~ −1 ~ −1 ~ −1 0 ~ ~ ~ −1 0 [B(V1)] [B(V2)] [B(V3)]   = [B(V1 ⊕ (V2 ⊕ V3))]   , (10) 0 0 c c which means the composite boost velocity is V~1 ⊕ (V~2 ⊕ V~3). Accordingly, if more boosts are added, the composite boost velocity becomes

~ V~1 ⊕ (V2 ⊕ (V~3 ⊕ (V~4 ⊕ · · · (V~n−1 ⊕ V~n) ··· )). The parentheses are indispensable since the velocity composition rule is not associative, which will be proved in the Appendix. So much for the general review. In fact, since the problem we are dealing with may be restricted to R2+1, there is no chance for us to use the term three-velocity or four-velocity hereafter. In order to avoid confusion, the counterpart of four-velocity in R2,1 will be called (2+1)-velocity.

7 III. THE TOY MODEL A. process a and Theorem 1

To construct the SO(3) toy model of the simple Wigner rotation, rst we dene a series of rotations of an Euclidean frame (ˆx, y,ˆ zˆ) as below and name it process a. Step 1. Rotate the frame in the zx-plane (equivalent to rotating it around the y-axis) by an angle , and call the new frame 0 0 0 , where 0 θ (ˆxa, yâ, zâ) yâ =y. ˆ Step 2. Rotate the new frame in the new yz-plane (equivalent to rotating it around the new -axis) by an angle , and call the newer frame 00 00 00 , where 00 0 x φ (ˆxa, yâ , zâ ) xâ =x â. Step 3. Rotate the newer frame in the plane spanned by and 00 to the extent that the zˆ zâ nal 000 coincides with the original . zâ zˆ Using the formulation of geometric algebra, these three rotations can be expressed as follows:

xˆ0  xˆ a θ θ 1. 0 † yˆa = exp(− zˆxˆ) yˆ exp(− zˆxˆ) , 0 2 2 zˆa zˆ

xˆ00 xˆ0  a φ a φ 2. 00 0 0 0 0 0 † yâ  = exp(− yâzâ) yâ exp(− yâzâ) 00 2 0 2 zâ zâ xˆ θ φ φ θ = exp(− zˆxˆ) exp(− yˆzˆ) yˆ exp(− yˆzˆ)† exp(− zˆxˆ)† 2 2   2 2 zˆ xˆ † † = XY yˆ Y X , zˆ

 000  00   xâ xâ xˆ 3. 000 00 † † † † yâ  = M yâ  M = MXY yˆ Y X M , 000 00 zâ zâ zˆ where θ and φ , and is constructed by employing the Cartan- X = exp(− 2 zˆxˆ) Y = exp(− 2 yˆzˆ) M Dieudonné theorem,

00 00 00 M = C(ˆza +z ˆ)ˆza = C(1 +z ˆzˆa ) = C(1 + X†Y †2X†) (11) with the condition MM † = 1. Note that zXYˆ = X†Y †zˆ has been used to obtain (11). In the light of (4), the expressions of the rst two steps can be easily transformed to matrix forms. In contrast, the elegance of (11) will be lost if we use matrix formulation to replace geometric algebra. Now we are ready to prove that the result of process a is generated by a rotor in the original xy-plane. Theorem 1. generates the toy Wigner rotation, where the toy MXY = exp(ˆxyˆ2 ) =: Rw Wigner angle is dened by

θ φ tan = tan tan with ∈ (−π, π]. 2 2 2

8 The proof is provided by some simple calculations with three notes as follows:

MXY = C(1 + X†Y †2X†)XY = C(XY + X†Y †) θ φ θ φ = C exp(− zˆxˆ) exp(− yˆzˆ) + exp( zˆxˆ) exp( yˆzˆ) 2 2 2 2 θ φ θ φ = 2C cos cos +x ˆyˆsin sin 2 2 2 2 θ φ = 2C cos cos sec cos +x ˆyˆsin 2 2 2 2 2 = exp(ˆxyˆ ). 2 Note 1. A normalized M implies the product MXY is also normalized. Hence we may assert the coecient of equals unity without doing practical calculation. exp(ˆxyˆ2 ) Note 2. Since − , a positive corresponds to a clockwise (i.e. negative- Rw = exp(−xˆyˆ 2 ) sense) rotation according to (4). If the range of θ and φ is taken to be (−π, π], then > 0 if and only if θφ > 0. The maximum of the toy Wigner angle is π which corresponds to, for example, π and θ = 2 φ = π. Note 3. This theorem tells us although the basis zˆ comes back to its original orientation at the end of the process, the other two bases deviate from the original (ˆx, yˆ) by a toy Wigner rotation (Figure 3).

     †  xˆ xˆ RwxRˆ w † † † † † MXY yˆ Y X M = Rw yˆ Rw = RwyRˆ w . zˆ zˆ zˆ

Figure 3: The frame rotates clockwise in the xy-plane by a toy Wigner angle at the end of process a. The result of process b is the same except the rotation is counterclockwise.

9 B. process b and Theorem 2 & 3 To fully model the problem of simple Wigner rotation, we have to construct another process which will be named process b. The main dierence between these two processes is the order of the rst two rotations. In process b, we let the frame rotate in the yz-plane by an angle φ rst and then let the new frame rotate in the new zx-plane by an angle θ. The result of this process can be expressed as

 000  00   xˆb xˆb xˆ 000 00 † † † † yˆb  = N yˆb  N = NYX yˆ X Y N , 000 00 zˆb zˆb zˆ where X and Y are the same as those in process a, while N as the counterpart of M can be constructed in a similar way,

0 00 00 0 00 N = C (ˆzb +z ˆ)ˆzb = C (1 +z ˆzˆb ) = C0(1 + Y †X†2Y †) (12) with the condition NN † = 1. It is easy to prove NYX is also a rotor in the xy-plane. Moreover, it is the inverse of the rotor MXY . Theorem 2. † † −1. NYX = (MXY ) = Rw = Rw The proof is also made of a few simple calculations:

NYX = C0(1 + Y †X†2Y †)YX = C0(YX + Y †X†) C0 = C0Y †X†(1 + XY 2X) = Y †X†M † = (MXY )†, C where C = C0 comes from both MXY and NYX are normalized. This theorem implies that, except for the sense of the rotation, the result of process b is the same as that of process a.

     †  xˆ xˆ RwxRˆ w † † † † † NYX yˆ X Y N = Rw yˆ Rw = RwyRˆ w . zˆ zˆ zˆ

In addition to Theorem 2, there is another interesting relation between the rotors N and M. Theorem 3. † RwNRw = M. The trick of the proof is substituting † † † for and for † . X Y N Rw NYX Rw An important relation 00 † 00 can be derived from Theorem 3 when is expressed Rwzˆb Rw =z ˆa M by 00 and by 00 . It implies the two rotations generated by and C(1 +z ˆzˆa ) N C(1 +z ˆzˆb ) M N performing in two dierent planes and the N-plane can be transformed to the M-plane by the toy Wigner rotation given by Theorem 1 (Figure 4). In summary, each theorem of this toy model has a precise geometric meaning. Theorem 1. process a brings about a toy Wigner rotation of the frame in the xy-plane. Theorem 2. The only dierence between the results of process a and process b is the sense of the toy Wigner rotation (clockwise vs. counterclockwise). Theorem 3. The third rotation planes of the two processes dier from each other by a toy Wigner rotation, hence the angle between them equals the toy Wigner angle. These statements may be illustrated in one schematic diagram as shown in Figure 5.

10 Figure 4: The third rotation plane of process b (N-plane) can be transformed to that of process a (M-plane) by the toy Wigner rotation.

Figure 5: Schematic diagram for Theorem 1, 2 & 3 of the toy Wigner rotation.

11 IV. SIMPLE WIGNER ROTATION A. process c and Theorem I

After familiarizing ourselves with the SO(3) toy model, we are ready to study the processes that yield the simple Wigner rotations. First we dene process c which contains three hyperbolic rotations of a Minkowski frame (ˆx, y,ˆ tˆ). Step 1. Rotate the frame in the xt-plane by a hyperbolic angle Θ, and call the new frame 0 0 ˆ0 , where 0 (ˆxc, yˆc, tc) yˆc =y. ˆ Step 2. Rotate the new frame in the new yt-plane by a hyperbolic angle Φ, and call the newer frame 00 00 ˆ00 , where 00 0 (ˆxc , yˆc , tc ) xˆc =x ˆc. Step 3. Rotate the newer frame hyperbolically in the plane spanned by ˆand ˆ00 to the extent t tc that the nal ˆ000 coincides with the original ˆ. tc t The expressions of these three hyperbolic rotations are similar to those of process a. xˆ0  xˆ c Θ Θ 1. 0 ˆ ˆ † yˆc = exp(− xˆt) yˆ exp(− xˆt) , ˆ0 2 ˆ 2 tc t

xˆ00 xˆ0  c Φ c Φ 2. 00 0 ˆ0 0 0 ˆ0 † yˆc  = exp(− yˆctc) yˆc exp(− yˆctc) ˆ00 2 ˆ0 2 tc tc xˆ Θ Φ Φ Θ = exp(− xˆtˆ) exp(− yˆtˆ) yˆ exp(− yˆtˆ)† exp(− xˆtˆ)† 2 2   2 2 tˆ xˆ † † = XY yˆ Y X , tˆ

 000  00   xˆc xˆc xˆ 3. 000 00 † † † † (13) yˆc  = M yˆc  M = MX Y yˆ Y X M , ˆ000 ˆ00 ˆ tc tc t where Θ ˆ Φ ˆ and † †2 † are the analogues of X = exp(− 2 xˆt), Y = exp(− 2 yˆt) M = C(1 + X Y X ) X,Y and M of process a respectively. The construction of M is analogous to (11), ˆ00 ˆ ˆ00 ˆˆ00 M = −C(tc + t)tc = C(1 − ttc ) = C(1 + X †Y†2X †) (14) with the condition MM† = 1. With regard to this process, we can deduce a theorem which is analogous to Theorem 1 of the toy model. Theorem I. ε is the rotor of the simple Wigner rotation, where the MX Y = exp(ˆxyˆ2 ) =: Rw simple Wigner angle ε is dened by ε Θ Φ π π tan = tanh tanh with ε ∈ (− , ). 2 2 2 2 2 The proof is omitted owing to its resemblance to that of Theorem 1. Similar to the toy Wigner angle , the simple Wigner angle ε is positive if and only if ΘΦ > 0. However, the range of is half of that of because the range of Ω is while that of ε tanh 2 (−1, 1) ω is . tan 2 (−∞, ∞)

12 B. process d and Theorem II & III Imitating the procedure for constructing the toy model, we now exchange the rst two rotations in process c, i.e., let the frame rotate in the yt-plane by a hyperbolic angle Φ rst and then let the new frame rotate in the new xt-plane by a hyperbolic angle Θ. This new process will be named process d and its result can be expressed as

 000  00   xˆd xˆd xˆ 000 00 † † † † yˆd  = N yˆd  N = NYX yˆ X Y N , ˆ000 ˆ00 ˆ td td t where N is the counterpart of M, 0 ˆ00 ˆ ˆ00 0 ˆˆ00 N = −C (td + t)td = C (1 − ttd ) = C0(1 + Y†X †2Y†) (15) with the condition NN † = 1. Thanks to the isomorphisms between (11) and (14) and between (12) and (15), we acquire the following two theorems by change of notations. Theorem II. † −1. NYX = (MX Y) = Rw † Theorem III. RwNR w = M.

C. The physical meanings To discuss the physical meanings of the processes and theorems introduced in this section, we begin with identifying the rotors X and Y with the boosts dened by the velocities ~u = c tanh Θˆx and ~v = c tanh Φˆy respectively. Then the (2+1)-dimensional version of (9) gives us the following results,

ˆ00 ˆ; ˆ00 ˆ (16) ctc = γ(~u ⊕ ~v)[~u ⊕ ~v + ct] ctd = γ(~v ⊕ ~u)[~v ⊕ ~u + ct], where γ(~u ⊕ ~v) = γ(~v ⊕ ~u) as proved at the end of Section II. Since ˆ00 †ˆ000 †ˆ according to (13), it implies that † −1 as a boost is tc = M tc M = M tM M = M dened by the velocity ~u ⊕ ~v . Therefore the boost M is dened by −(~u ⊕ ~v), and for the same reason N † is dened by ~v ⊕ ~u and N by −(~v ⊕ ~u). Now we are ready to use physical language to rephrase the two processes in this section (omitting the suxes c and d). process c: Step 1. A frame (ˆx0, yˆ0, tˆ0) is found which moves with velocity ~u relative to the original frame (ˆx, y,ˆ tˆ). Step 2. Another frame (ˆx00, yˆ00, tˆ00) is found which moves with velocity ~v relative to (ˆx0, yˆ0, tˆ0). Step 3. The third frame (ˆx000, yˆ000, tˆ000) is found which moves with velocity −(~u ⊕ ~v) relative to (ˆx00, yˆ00, tˆ00). Since tˆ000 = tˆ, we know from (7) there is no relative velocity between the third and the original unprimed frames. The spatial bases of these two frames dier by a simple Wigner rotation according to Theorem I.

 000      †  xˆ xˆ xˆ RwxˆRw 000 † † † † † yˆ  = MX Y yˆ Y X M = Rw yˆ Rw = RwyˆRw . tˆ000 tˆ tˆ tˆ With the aid of Theorem III, we can transform the above relation to the following form,

13 xˆ xˆ xˆ † † † † † † (17) XY yˆ Y X = M Rw yˆ RwM = RwN yˆ NRw, tˆ tˆ tˆ which means the action of the rst two boosts to the original frame is equivalent to that of a boost of velocity ~u ⊕ ~v preceded by a simple Wigner rotation, or a boost of velocity ~v ⊕ ~u followed by the same rotation. However, if the term Wigner rotation connotes no temporal dimension being involved, the latter should be ruled out as a Wigner rotation because N †tˆN will be rotated by the pair † . (Rw, Rw) process d: Step 1. A frame (ˆx0, yˆ0, tˆ0) is found which moves with velocity vˆ relative to the original frame (ˆx, y,ˆ tˆ). Step 2. Another frame (ˆx00, yˆ00, tˆ00) is found which moves with velocity ~u relative to (ˆx0, yˆ0, tˆ0). Step 3. The third frame (ˆx000, yˆ000, tˆ000) is found which moves with velocity −(~v ⊕ ~u) relative to (ˆx00, yˆ00, tˆ00). Similarly we can conclude that (i) Theorem II asserts the spatial bases of the third and the original frames dier by an inverse simple Wigner rotation,

 000      †  xˆ xˆ xˆ RwxˆRw 000 † † † † † yˆ  = NYX yˆ X Y N = Rw yˆ Rw = RwyˆRw , tˆ000 tˆ tˆ tˆ (ii) the action of the rst two boosts to the original frame is equivalent to a boost of velocity ~v ⊕ ~u preceded by an inverse simple Wigner rotation, or a boost of velocity ~u ⊕ ~v followed by the same rotation which may be ruled out as a Wigner rotation,

xˆ xˆ xˆ † † † † † † (18) YX yˆ X Y = N Rw yˆ RwN = RwM yˆ MRw. tˆ tˆ tˆ In addition to the above results, Theorem III leads to a relation ˆ00 † ˆ00 which is Rwtd Rw = tc the analogue of 00 † 00 of the toy model. We can use (16) to convert this relation to Rwzˆb Rw =z ˆa † which implies the angle between these two composite velocities equals Rw(~v ⊕ ~u)Rw = ~u ⊕ ~v the simple Wigner angle. In summary, each of the three theorems in this section has a precise physical meaning. Theorem I. process c brings about a simple Wigner rotation of the frame in the xy-plane. Theorem II. The only dierence between the results of process c and process d is the sense of the simple Wigner rotation. Theorem III. The composite velocities of the two processes dier from each other by a simple Wigner rotation, implying the angle between them equals the simple Wigner angle. These statements are illustrated in one schematic diagram as shown in Figure 6. Lastly, although the explicit expressions of ~u ⊕ ~v and ~v ⊕ ~u are not needed here, their derivations are provided as a demonstration. Comparing (16) with the expression of ˆ00 which can be calculated easily, tc ˆ00 ˆ tc = sinh Θ cosh Φˆx + sinh Φˆy + cosh Θ cosh Φt tanh Φ = cosh Θ cosh Φ tanh Θˆx + yˆ + tˆ cosh Θ γ(~u)γ(~v) v = uxˆ + yˆ + ctˆ , c γ(~u)

14 h i we obtain and v For the same reason the expression of ˆ00 γ(~u ⊕ ~v) = γ(~u)γ(~v) ~u ⊕ ~v = u, γ(~u) . td h i leads to u . ~v ⊕ ~u = γ(~v) , v

Figure 6: Schematic diagram for Theorem I, II & III of the simple Wigner rotation.

V. GENERAL WIGNER ROTATION

A. process e and Theorem I0 Now we release the 90◦ constraint on the two boost velocities, allowing the angle between them to be arbitrary. Without loss of generality, we still put the velocity ~u along the x-direction, while the velocity deviates from the -direction clockwise by an angle π π . ~v y η ∈ (− 2 , 2 ) The basis yˆ can be rotated to the direction of ~v by a rotor and the new basis will be called wˆ, i.e., η η (Figure 7). Using this new basis, we generalize wˆ = exp( 2 xˆyˆ)ˆy exp(− 2 xˆyˆ) =x ˆ sin η+y ˆcos η process c to the following one which will be named process e. Step 1. Rotate the frame in the xt-plane by a hyperbolic angle Θ, and call the new frame 0 0 ˆ0 , where 0 Accordingly is transformed to 0 . (ˆxe, yê, te) yê =y. ˆ wˆ wê Step 2. Rotate the new frame in the new wt-plane by a hyperbolic angle Φ, and call the newer frame 00 00 ˆ00 . Note that 00 0 . (ˆxe , yê , te ) xê 6=x ˆ e Step 3. Rotate the newer frame hyperbolically in the plane spanned by ând ˆ00 to the extent t te that the nal ˆ000 coincides with the original ˆ. te t

15 Figure 7: The orientations of velocities ~u and ~v in the problem of general Wigner rotation. The range π π is adequate for all possible congurations since can be any real number. η ∈ (− 2 , 2 ) Φ

The expressions for this process are as follows:

 0    xê xˆ 0 Θ Θ 1. yê  ˆ yˆ ˆ †  ˆ0  = exp(− xˆt)  ˆ exp(− xˆt) ,  te  2  t  2 0 wê wˆ

xˆ00 xˆ0  e Φ e Φ 2. 00 0 ˆ0 0 0 ˆ0 † yê  = exp(− wête) yê exp(− wête) ˆ00 2 ˆ0 2 te te xˆ Θ Φ Φ Θ = exp(− xˆtˆ) exp(− wˆtˆ) yˆ exp(− wˆtˆ)† exp(− xˆtˆ)† 2 2   2 2 tˆ xˆ † † = XW yˆ W X , tˆ

 000  00   xê xê xˆ 3. 000 00 † † † † yê  = M yê  M = MXW yˆ W X M , ˆ000 ˆ00 ˆ te te t where Θ ˆ , Φ ˆ and X = exp(− 2 xˆt) W = exp(− 2 wˆt)

ˆˆ00 † †2 † with † M = C(1 − tte ) = C(1 + X W X ) MM = 1.

16 Now we can generalize Theorem I of the simple Wigner rotation to the general case. Theorem I0.

ε MXW = exp(ˆxyˆ ) =: R , 2 w ε cos η π π where tan = with ε ∈ (− − η, − η). 2 Θ Φ 2 2 coth 2 coth 2 + sin η The proof is contained in the following calculations,

MXW = C(XW + X †W†) Θ Φ Θ Φ = C exp(− xˆtˆ) exp(− wˆtˆ) + exp( xˆtˆ) exp( wˆtˆ) 2 2 2 2 Θ Φ Θ Φ = 2C cosh cosh +x ˆwˆ sinh sinh 2 2 2 2 Θ Φ Θ Φ Θ Φ = 2C cosh cosh + sin η sinh sinh +x ˆyˆcos η sinh sinh 2 2 2 2 2 2 ε ε = cos +x ˆyˆsin . 2 2 The limits of corresponds to Θ Φ . ε coth 2 coth 2 → ±1

B. process f and Theorem II0 & III0 Just like process e is the generalization of process c, process f is generalized from process d.

 000  00   xˆf xˆf xˆ 000 00 † † † † yˆf  = N yˆf  N = N WX yˆ X W N , ˆ000 ˆ00 ˆ tf tf t where 0 ˆˆ00 0 † †2 † with † . N = C (1 − ttf ) = C (1 + W X W ) NN = 1 Because the expressions of the rotors M and N are respectively isomorphic to those of M and N of the simple Wigner rotations, Theorem II and Theorem III are generalized to the following. Theorem 0. † −1. II N WX = (MXW) = Rw Theorem 0. † . III RwNRw = M In summary, except for the fact that the formula for the Wigner angle is more complicated, the related theorems and their physical meanings have no signicant change for the general Wigner rotation. For example, the analogue of (17) is

xˆ xˆ xˆ † † † † † † (19) XW yˆ W X = M Rw yˆ RwM = RwN yˆ NRw, tˆ tˆ tˆ and that of (18) is

xˆ xˆ xˆ † † † † † † (20) WX yˆ X W = N Rw yˆ RwN = RwM yˆ MRw. tˆ tˆ tˆ

17 VI. CONCLUSION

By the joint eort of group theory, geometric algebra and active-frame formalism, the ge- ometry of Wigner rotation problem is claried and the mathematics is simplied. Among other things, the SO(3) toy model provides an easy way to comprehend the essence of this problem. As a byproduct, the active-frame formulas (7) and (9) suggest an alternative approach to teaching relativistic kinematics.

APPENDIX: MATRIX-COORDINATE FORMALISM

The results in this paper are mainly obtained and expressed by geometric algebra. In order to make comparisons with those in other literature, there is a need to translate these results into matrix formulation. First of all, let's practice with an example from process c,

xˆ Θ Φ Φ Θ exp(− xˆtˆ) exp(− yˆtˆ) yˆ exp(− yˆtˆ)† exp(− xˆtˆ)† 2 2   2 2 tˆ 1 0 0  xˆ Θ Θ = exp(− xˆtˆ) 0 cosh Φ sinh Φ yˆ exp(− xˆtˆ)†, 2     2 0 sinh Φ cosh Φ tˆ 1 0 0  cosh Θ 0 sinh Θ xˆ = 0 cosh Φ sinh Φ  0 1 0  yˆ . 0 sinh Φ cosh Φ sinh Θ 0 cosh Θ tˆ

Using the aforementioned notations X and Y for the rotors, and denoting the corresponding matrices by [X ] and [Y], the above equality becomes

xˆ xˆ † † XY yˆ Y X = [Y][X ] yˆ . tˆ tˆ

The correspondence between these two formulations may be put formally as X Y ⇐⇒ [Y][X ]. However, in order to make those matrices act on the coordinates (x, y, ct) instead of the frame (ˆx, y,ˆ tˆ), we need to go further to take the transpose-inverse of [Y][X ]. Hence in this example the correspondence between the rotor-frame formalism and matrix-coordinate formalism should be X Y ⇐⇒ [Y]−>[X ]−>. Now we are ready to apply this rule to the results of general Wigner rotation discussed in Section V. Starting with Theorem I0 and bearing in mind the boost matrix is symmetric while the rotation matrix orthogonal, we have

−1 −1 −1 MXW = Rw ⇐⇒ [W] [X ] [M] = [Rw]. If we use the notation B(V~ ) in Section II to unify those hyperbolic rotors, since B(V~ ) corresponds to passive coordinate transformations, we have [W]−1 = [B(~v)], [X ]−1 = [B(~u)] and [M]−1 = [B(−(~u ⊕ ~v))], and the above matrix equality becomes

[B(~v)][B(~u)][B(−(~u ⊕ ~v))] = [Rw],

or [B(~v)][B(~u)] = [Rw][B(~u ⊕ ~v)]. (21)

18 Applying the same rule to Theorem II0 gives us a similar result,

−1 [B(~u)][B(~v)] = [Rw] [B(~v ⊕ ~u)].

As for Theorem III0, the rotor-matrix correspondence is

† −1 −1 −1 RwNRw = M ⇐⇒ [Rw] [N ] [Rw] = [M] −1 or [Rw] [B(−(~v ⊕ ~u))][Rw] = [B(−(~u ⊕ ~v))] −1 or [Rw] [B(~v ⊕ ~u)][Rw] = [B(~u ⊕ ~v)]. (22) Substituting (22) into (21), the result is the correspondent of (19),

[B(~v)][B(~u)] = [Rw][B(~u ⊕ ~v)] = [B(~v ⊕ ~u)][Rw]. (23) Taking the transpose of (23) gives us the correspondent of (20),

−1 −1 [B(~u)][B(~v)] = [B(~u ⊕ ~v)][Rw] = [Rw] [B(~v ⊕ ~u)]. (24) Needless to say, (23) and (24) can also be obtained by applying the correspondence rule to the rotor-frame formulas (19) and (20). Next, the inverse and transpose-inverse of (23) are respectively

−1 −1 −1 −1 −1 −1 [B(~u)] [B(~v)] = [B(~u ⊕ ~v)] [Rw] = [Rw] [B(~v ⊕ ~u)] , (25) and

−1 −1 −1 −1 [B(~v)] [B(~u)] = [Rw][B(~u ⊕ ~v)] = [B(~v ⊕ ~u)] [Rw]. (26) If we apply (25) and (26) to a (2+1)-velocity 0 := (0, 0, c)>, the result is not only compatible with (8) but also provides an elucidation of the velocity composition rule. For example,

−1 −1 −1 −1 −1 −1 [B(~u)] [B(~v)] 0 = [B(~u ⊕ ~v)] [Rw] 0 = [Rw] [B(~v ⊕ ~u)] 0 shows the (2+1)-velocity thus constructed corresponds to ~u ⊕ ~v instead of ~v ⊕ ~u. This mirrors the fact that the second pair † of (17) and the second pair † of (19) may be (Rw, Rw) (Rw, Rw) ruled out as Wigner rotations. Similarly,

−1 −1 −1 −1 [B(~v)] [B(~u)] 0 = [Rw][B(~u ⊕ ~v)] 0 = [B(~v ⊕ ~u)] [Rw]0 evidently just corresponds to ~v ⊕ ~u. Moreover, we can use either (25) or (26) to prove the velocity composition rule is non- associative. Taking V~ ⊕ (~u ⊕ ~v) 6= (V~ ⊕ ~u) ⊕ ~v as an example, where V~ is another velocity in the xy-plane. V~ ⊕ (~u ⊕ ~v) corresponds to the (2+1)-velocity

[B(V~ )]−1[B(~u ⊕ ~v)]−10 −1 −1 −1 = [B(V~ )] [B(~u)] [B(~v)] [Rw]0 = [B(V~ )]−1[B(~u)]−1[B(~v)]−10, which is consistent with (10).

19 On the other hand, (V~ ⊕ ~u) ⊕ ~v corresponds to a dierent (2+1)-velocity

[B(V~ ⊕ ~u)]−1[B(~v)]−10 −1 −1 −1 = [B(V~ )] [B(~u)] [R˜ w][B(~v)] 0 −1 −1 −1 −1 = [B(V~ )] [B(~u)] [R˜ w][B(~v)] [R˜ w] 0 = [B(V~ )]−1[B(~u)]−1[B(~v0)]−10, where ˜ is the Wigner rotation induced by the boosts ~ and , and 0 ˜ † ˜ . Rw B(V ) B(~u) ~v = Rw~v Rw Lastly, the explicit forms of these boost matrices are provided below for reference.

 cosh Θ 0 − sinh Θ cosh Θ 0 sinh Θ −1 [B(~u)] =  0 1 0  , [B(~u)] =  0 1 0  . − sinh Θ 0 cosh Θ sinh Θ 0 cosh Θ

 cos η sin η 0 1 0 0  cos η − sin η 0 [B(~v)] = − sin η cos η 0 0 cosh Φ − sinh Φ sin η cos η 0 0 0 1 0 − sinh Φ cosh Φ 0 0 1  cos η2 + sin η2 cosh Φ sin η cos η(cosh Φ − 1) − sin η sinh Φ = sin η cos η(cosh Φ − 1) sin η2 + cos η2 cosh Φ − cos η sinh Φ , − sin η sinh Φ − cos η sinh Φ cosh Φ

 cos η2 + sin η2 cosh Φ sin η cos η(cosh Φ − 1) sin η sinh Φ −1 [B(~v)] = sin η cos η(cosh Φ − 1) sin η2 + cos η2 cosh Φ cos η sinh Φ . sin η sinh Φ cos η sinh Φ cosh Φ

P + Q2 QR −PQ  1 [B(~u ⊕ ~v)] = QR P + R2 −PR , P   −PQ −PRP 2 − P

P + Q2 QR P Q  1 [B(~u ⊕ ~v)]−1 = QR P + R2 PR , P   PQPRP 2 − P

where P = 1 + cosh Θ cosh Φ + sin η sinh Θ sinh Φ, Q = sinh Θ cosh Φ + sin η cosh Θ sinh Φ, R = cos η sinh Φ.

According to (6), we can read out γ(~u ⊕~v) and ~u ⊕~v from the third column of [B(~u ⊕~v)]−1, i.e., and c γ(~u ⊕ ~v) = P − 1 ~u ⊕ ~v = P −1 [Q, R]. With the following modications for Q and R, the two matrices above can also be used to represent [B(~v ⊕ ~u)] and [B(~v ⊕ ~u)]−1, and ~v ⊕ ~u can be extracted by using the same rule.

Q = cos2 η sinh Θ + sin η cosh Θ sinh Φ + sin2 η sinh Θ cosh Φ, R = cos η cosh Θ sinh Φ + sin η cos η sinh Θ(cosh Φ − 1).

20 SUPPLEMENTARY MATERIAL

An animation of the six processes discussed in this paper is available at https://www.youtube.com/watch?v=HyVouwd7X2o

ACKNOWLEDGMENTS

This paper is in memory of my thesis advisor Professor Georey F. Chew (1924-2019). I would like to express my sincere gratitude to Professor Charles Lineweaver for his valuable comments, and to Professor Kuu-Young Young for his moral support over the past two decades.

REFERENCES

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