Explicit form for the most general Lorentz transformation revisited

Explicit formulae for the $4\times 4$ Lorentz transformation matrices corresponding to a pure boost and a pure three-dimensional rotation are very well-known. Significantly less well-known is the explicit formula for a general Lorentz transformation with arbitrary nonzero boost and rotation parameters. We revisit this more general formula by presenting two different derivations. The first derivation (which is somewhat simpler than previous ones appearing in the literature) evaluates the exponential of a $4\times 4$ real matrix $A$, where $A$ is a product of the diagonal matrix ${\rm diag}(+1, -1, -1, -1)$ and an arbitrary $4\times 4$ real antisymmetric matrix. The formula for $\exp A$ depends only on the eigenvalues of $A$ and makes use of the Lagrange interpolating polynomial. The second derivation exploits the observation that the spinor product $\eta^\dagger\overline{\sigma}^{\lower3pt\hbox{$\scriptstyle \mu$}}\chi$ transforms as a Lorentz four-vector, where $\chi$ and $\eta$ are two-component spinors. The advantage of the latter derivation is that the corresponding formula for a general Lorentz transformation $\Lambda$ reduces to the computation of the trace of a product of $2\times 2$ matrices. Both computations are shown to yield equivalent expressions for $\Lambda$.

Consider two inertial reference frames with coinciding origins, where one reference frame is moving with respect to the other with three-vector velocity v.The corresponding Lorentz transformation is called a Lorentz boost.The boost parameters are defined by the components of the three-vector ζ ≡ ( v/v) tanh −1 (v/c), where v ≡ | v| and c is the speed of light.However, this is not the most general Lorentz transformation.For example, let R be an arbitrary 3 × 3 orthogonal matrix of unit determinant, i.e., a proper rotation matrix parametrized by the components of the three-vector θ ≡ θn (such that θ is the angle of rotation, counterclockwise, about a fixed axis that lies along the unit vector n).Then, the transformation x ′ 0 = x 0 and x ′ = R x is also a Lorentz transformation as it leaves the Minkowski spacetime metric invariant.The corresponding matrix representations of the general Lorentz boost and three-dimensional rotation are quite well known [see eqs.(22) and (26), respectively] and are reviewed in Section 2.
A more general Lorentz transformation matrix, which shall henceforth be denoted by Λ( ζ , θ), corresponds to a simultaneous boost and rotation.As shown in Section 3, Λ( ζ , θ) can be expressed as the exponential of a 4 × 4 matrix, In contrast to Λ( ζ , 0) and Λ( 0 , θ), which correspond to a Lorentz boost matrix and a threedimensional rotation, respectively, an explicit form for Λ( ζ , θ) is much less well known.The first published formula for Λ( ζ , θ) appeared in Ref. [3].Subsequent derivations have also been given in Refs.[4][5][6].These derivations are based on the Cayley-Hamilton theorem of linear algebra (e.g., see Section 8.4 of Ref. [7]), which asserts that any n × n matrix A satisfies its own characteristic equation, p(x) = det(A − xI n ) = 0, where I n is the n × n identity matrix and p(x) is an nth-order polynomial whose roots are the eigenvalues of A. That is, p(A) is equal to the zero matrix.It follows that for any integer k ≥ n, the matrix A k can be expressed as a linear combination of I n , A, A 2 , . . .A k−1 .In particular, where each of the coefficients c k is an infinite series whose terms depend on the eigenvalues of A. Note that by setting either θ = 0 or ζ = 0 in eq. ( 1), one can easily compute the resulting matrix exponential to derive the well-known expressions given in eqs.( 22) and (26), respectively.In contrast, if both the boost vector and the rotation vector are nonzero, then the corresponding computation of the matrix exponential, which is carried out in Refs.[3,4], is significantly more difficult.In Ref. [5], this computation is performed by showing that a Lorentz transformation matrix g exists such that the 4 × 4 matrix A ≡ gAg −1 in block matrix form is made up of very simple 2 × 2 matrix blocks.The exponential exp A is then easy to evaluate directly via its Taylor series to obtain the coefficients c k , and Finally, Ref. [6] derives a system of four linear equations for the coefficients c k in eq. ( 2), whose solution provides the desired expression for exp A. Similar techniques have also been employed in Ref. [8] to obtain an explicit expression for the matrix exponential function of SO(4).
In this paper, we shall provide a somewhat simpler and more straightforward evaluation of Λ( ζ , θ) as compared to the derivations given in Refs.[3][4][5][6].In Section 2, we first exhibit the explicit forms for the general Lorentz boost and the three-dimensional rotation matrices of Minkowski spacetime, which correspond to special cases of the more general 4 × 4 Lorentz transformation matrix, as noted above.In Section 3, an expression for the most general Lorentz transformation is then derived.Indeed, it is sufficient to consider the set of all Lorentz transformations that are continuously connected to the identity, known as the proper orthochronous Lorentz transformations (e.g., see Ref. [1]).The matrix representation of any element of this latter set can be expressed in the form given by eq. ( 1), as discussed below eq.(40).In Section 4, we explicitly evaluate eq. ( 1) for arbitrary boost and rotation parameters.We then demonstrate that an alternative derivation of Λ( ζ , θ) can be given that only involves the manipulation of 2 × 2 matrices, by making use of two-component spinors.In particular, we show in Section 5 that the most general proper orthochronous Lorentz transformation matrix can be expressed as a trace of the product of four 2 × 2 matrices, which is then explicitly evaluated.Both methods for computing Λ( ζ , θ) are carried out in pedagogical detail.In Section 6, we check that both computations yield the same expression for Λ( ζ , θ).Final remarks are presented in Section 7, and some related discussions are relegated to the appendices.

Lorentz transformations-special cases
In a first encounter with special relativity, a student learns how the spacetime coordinates change between two inertial reference frames K and K ′ .If the spacetime coordinates with respect to K are (ct; x, y, z) and the spacetime coordinates with respect to K ′ are (ct ′ ; x ′ , y ′ , z ′ ), where K ′ is moving relative to K with velocity v = v x in the x direction, then where c is the speed of light and It is straightforward to generalize the above results for an arbitrary velocity v by writing where x is the projection of x along the direction of v ≡ c β, and x ⊥ is perpendicular to v (so that x • x ⊥ = 0).The definition of x implies that where β ≡ | β|.Note that 0 ≤ β < 1 for any particle of non-zero mass.
In light of eq. ( 10), eqs.( 4)-( 7) are equivalent to x where γ ≡ (1 − | β| 2 ) −1/2 .Note that 1 ≤ γ < ∞ for any particle of non-zero mass.More explicitly, which yield β• x = β• x and β• x ⊥ = 0 as required.Inserting the expressions given in eq. ( 14) back into eqs.( 11)-( 13), we end up with the well-known result (e.g., see eq. (11.19) of Ref. [9]): Following eq.(11.20) of Ref. [9], it is convenient to introduce the boost parameter ζ (also called the rapidity), since the definitions of β and γ are consistent with the relation cosh 2 ζ − sinh 2 ζ = 1.In particular, note that 0 ≤ ζ < ∞.We then define the boost vector ζ to be the vector of magnitude ζ that points in the direction of β.Since eq. ( 17) yields β = tanh ζ, it follows that In terms of the boost vector ζ and its magnitude ζ ≡ | ζ|, eqs.( 15) and ( 16) yield Before proceeding, it is instructive to distinguish between active and passive Lorentz transformations (e.g., see Ref. [1]).The Lorentz transformation discussed above is a passive transformation, since the reference frame K (specified by the coordinate axes) is transformed into K ′ , while leaving the observer fixed.Equivalently, one can consider an active transformation, in which the coordinate axes are held fixed while the location of the observer in spacetime is boosted using the inverse of the transformation specified by eqs.(19) and (20).That is, a spacetime point of the observer located at (ct ; x) is transformed by the boost to (ct ′ ; x ′ ) using eqs.(19) and (20) with ζ replaced by − ζ.Henceforth, all Lorentz transformations treated in this paper will correspond to active transformations.
The transformation that boosts the spacetime point (ct ; x) to (ct ′ ; x ′ ) is given by where the 4 × 4 matrix Λ( ζ , 0) can be written in block matrix form as after converting eqs.(19) and (20) to an active transformation via ζ → − ζ.In eq. ( 22), where the Latin indices i, j ∈ {1, 2, 3} refer to the x, y, and z components of the three-vector ζ, and there is an implicit sum over the repeated index j on the right hand side of eq. ( 21).
The matrix Λ( ζ, 0) is sometimes inaccurately called the Lorentz transformation matrix.In fact, this matrix represents a special type of Lorentz transformation consisting of a boost without rotation [the latter is indicated by the second argument of Λ( ζ , 0)].Furthermore, note that Λ( 0, 0) = I 4 is the 4 × 4 identity matrix.Any Lorentz transformation of the form Λ( ζ , 0) can be continuously deformed into the identity matrix by continuously shrinking the vector ζ to the zero vector.
Another example of a Lorentz transformation is a three-dimensional proper rotation of the vector x into the vector x ′ = R x by an angle θ, counterclockwise, about a fixed axis n, where R is a 3 × 3 orthogonal matrix of unit determinant, and the time coordinate is not transformed.In this notation, n = (n 1 , n 2 , n 3 ) is a unit vector (i.e., n• n = 1).It is then convenient to define a three-vector quantity called the rotation vector, where 0 ≤ θ ≤ π.In the case of a proper three-dimensional rotation, the transformation of the spacetime point (ct ; x) to (ct ′ ; x ′ ) is given by where the 4 × 4 matrix Λ( 0 , θ) can be written in block matrix form as where 0 j [0 i ] are the components of the zero row [column] vector (with i, j ∈ {1, 2, 3}), and In eq. ( 27), the Levi-Civita symbol is defined by ǫ ijk = +1 [−1] when ijk is an even [odd] permutation of 123, and ǫ ijk = 0 if any two of the indices coincide.Eq. ( 27) is known as Rodrigues' rotation formula (e.g., see Refs.[10,11]).A clever proof of this formula is provided in Appendix A.

General Lorentz transformations
Consider a four-vector v µ = (v 0 ; v).Under an active Lorentz transformation, the spacetime components of the four-vector v µ transform as where the Greek indices such as µ, α ∈ {0, 1, 2, 3}, and there is an implied sum over any repeated upper/lower index pair.The quantities Λ µ α can be viewed as the elements of a 4 × 4 real matrix, where µ labels the row and α labels the column.In special relativity, the metric tensor (in a rectangular coordinate system) is given by the diagonal matrix.
where the so-called mostly minus convention for the metric tensor has been chosen.
To construct a Lorentz-invariant scalar quantity that is unchanged under a Lorentz transformation, one only needs to combine tensors in such a way that all upper/lower index pairs are summed over and no unsummed indices remain.For example, Using eqs.( 28) and (30), it follows that Since the four-vector v is arbitrary, it follows that Eq. ( 32) defines the most general Lorentz transformation matrix Λ.The set of all such 4 × 4 Lorentz transformation matrices is a group (under matrix multiplication) and is denoted by O(1, 3).Here, the notation (1, 3) refers to the number of plus and minus signs in the metric tensor η µν [cf.eq. ( 29)].In particular, O(1, 3) is a Lie group, appropriately called the Lorentz group (e.g., see Refs.[1,2,11]).
After taking the determinant of both sides of eq. ( 32), one obtains (det Λ) 2 = 1.Hence, Moreover, by setting α = β = 0 in eq. ( 32) and summing over µ and ν, one obtains The Lie group SO(1, 3) is the group of proper Lorentz transformation matrices that satisfy det Λ = +1.The elements of the subgroup of SO(1, 3) that also satisfy Λ 0 0 ≥ +1 are continuously connected to the identity element [the 4×4 identity matrix, denoted by I 4 ] and constitute the set of proper orthochronous Lorentz transformations, which is often denoted by SO 0 (1,3).Three examples of Lorentz transformations that are not continuously connected to the identity are as follows In particular, there is no way to continuously change the parameters of a proper orthochronous Lorentz transformation to yield a Lorentz transformation with det Λ = −1 and/or Λ 0 0 ≤ −1 in light of eqs.(33) and (34).
The complete list of Lorentz transformations is then given by Consequently, to determine the explicit form of the most general Lorentz transformation, it suffices to consider the explicit form of the most general proper orthochronous Lorentz transformation.
The Lie algebra of the Lorentz group is obtained by considering an infinitesimal Lorentz transformation, where A is a 4×4 matrix that depends on infinitesimal Lorentz group parameters.In particular, terms that are quadratic or of higher order in the infinitesimal group parameters are neglected.Inserting eq.(37) into eq.(32), and denoting G = diag(+1, −1, −1, −1) to be the 4 × 4 matrix whose matrix elements are η µν , it follows that Keeping only terms up to linear order in the infinitesimal group parameters, we conclude that That is, GA is a 4 × 4 real antisymmetric matrix.Hence, the Lie algebra of the Lorentz group, henceforth denoted by so(1, 3), consists of all 4 × 4 real matrices A such that GA is an antisymmetric matrix.
To construct a proper orthochronous Lorentz transformation, one can choose any 4 × 4 real matrix A that satisfies eq. ( 39), and consider a large positive integer n such that A/n is an infinitesimal quantity.Then, a proper orthochronous Lorentz transformation is obtained by applying a sequence of n infinitesimal Lorentz transformations in the limit as n → ∞, Note that Λ is continuously connected to the identity matrix since one can continuously deform A into the zero matrix.Hence, it follows that Λ ∈ SO 0 (1, 3).However, one can make a stronger statement: the exponential map, exp : so(1, 3) → SO 0 (1, 3), is surjective.A proof of this result can be found in Section 6.3 of Ref. [11].That is, the set of all proper orthochronous Lorentz transformations consists of matrices of the form exp A, where GA is a 4 × 4 real antisymmetric matrix.
Let us first reconsider the two special cases examined in Section 2. A matrix representation of an infinitesimal boost is obtained by evaluating eq. ( 22) to leading order in ζ, where the three matrices k = (k 1 , k 2 , k 3 ) are defined by (42) Similarly, a matrix representation of an infinitesimal rotation is obtained by evaluating eqs.( 26) and ( 27) to leading order in θ (with where the three matrices s = (s 1 , s 2 , s 3 ) are defined by The six matrices k = (k 1 , k 2 , k 3 ) and s = (s 1 , s 2 , s 3 ) satisfy the following commutation relations: where i, j, ℓ ∈ {1, 2, 3} and there is an implicit sum over the repeated index ℓ.
Using eqs.( 41) and ( 43), it follows that the matrix representation of a general infinitesimal Lorentz transformation, to linear order in the boost and rotation parameters, is given by Note that we also could have written Λ( ζ , θ) ≃ Λ( ζ , 0)Λ( 0 , θ) in eq. ( 46), since the infinitesimal Lorentz transformations commute at linear order.
In light of the remarks below eq.( 40), one can conclude that the most general proper orthochronous Lorentz transformation matrix Λ( ζ, θ) is a 4 × 4 matrix given by Here, we follow the conventions of Refs.[12,13].Note that in the notation of Ref. [9], k = i K and s = i S, where the 4 × 4 matrix representations of K and S are given in eq. ( 11.91) of Ref. [9] and yield Λ = exp( θ• S + ζ • K).The argument of exp differs by an overall sign with eq. ( 11.93) of Ref. [9], where a passive Lorentz transformation is employed, which amounts to replacing { ζ, θ} with {− ζ, − θ}.
Eqs. (42), ( 44) and (47) imply that As anticipated in eq.(39), GA is the most general 4 × 4 real antisymmetric matrix, which depends on six real independent parameters ζ i and θ i (i ∈ {1, 2, 3}).The {s i , k i } satisfy the commutation relations [eq.( 45)] of the real Lie algebra so(1, 3).As indicated in eq. ( 48), A is a real linear combination of the six Lie algebra generators {−is i , −ik i } and thus constitutes a general element of so(1, 3).In Section 4, we provide an explicit computation of exp A.
Before moving on, we shall introduce a useful notation that assembles the matrices {s i , k i } into six independent non-zero matrices, s ρλ = −s λρ (with λ, ρ ∈ {0, 1, 2, 3}) such that Note that eq.(49) implies that s ij = ǫ ijℓ s ℓ , so that the six independent matrices can be taken to be s ij (i < j) and s 0i (i, j ∈ {1, 2, 3}).The matrix elements of the s ρλ are given by where µ indicates the row and ν indicates the column of the corresponding matrix.

An explicit evaluation of Λ( ζ , θ) = exp A
We now proceed to evaluate exp A, where A is given by eq. ( 48).First, we compute the characteristic polynomial of A, where Solving eq. ( 55) for a 2 and b 2 yields Note that a 2 ≥ 0 and b 2 ≥ 0 so that a, b ∈ R. The individual signs of a and b are not determined, but none of the results that follow depend on these signs.The eigenvalues of A, denoted by λ i (i = 1, 2, 3, 4), are the solutions of p(x) = 0, which are given by If ab = 0, then the four eigenvalues of A [eq. (58)] are distinct, which implies that A is a diagonalizable matrix.
To evaluate exp A for a diagonalizable matrix A, we shall make use of a formula [eq.( 60) below] that is based on the Lagrange interpolating polynomial.Consider an n × n matrix A with n eigenvalues of which m are distinct and denoted by λ i (i = 1, 2, . . ., m).The matrix A is diagonalizable if and only if where I n is the n × n identify matrix. 1 Note that if m = n (i.e., all n eigenvalues are distinct), then A is diagonalizable, since in this case eq. ( 59) is automatically satisfied due to the Cayley-Hamilton theorem.Any function of a diagonalizable matrix A is given by the following formula:2 Applying eq.(60) to f (A) = exp A, where A is given by eq. ( 48), under the assumption that ab = 0, it follows that Simplifying the above expression yields, (62) Combining terms, we end up with exp where a and b are defined in eq. ( 55) and the f k (a, b) are given by in agreement with the results previously obtained in Refs.[3][4][5][6].
The matrix A and its powers can be conveniently written in block matrix form, and The ij element of A 3 can be simplified by noting that the ij element of any 3 × 3 antisymmetric matrix must be of the form ǫ ijk C k (after summing over the repeated index k).Thus, Multiplying the above equation by ǫ ijm and summing over i and j yields That is, we have derived the identity Thus, the matrix A 3 [eq.( 67)] can be rewritten in a more convenient form, Consider separately the case of ab = 0.The eigenvalues given in eq. ( 58) are no longer distinct.If a = 0 and b = 0, then the matrix A is diagonalizable since A satisfies eq. ( 59), i.e., A(A 2 − b 2 I 4 ) = 0.In particular, if a = 0 then eq. ( 55 One can check that eq. ( 72) coincides with the a → 0 limit of eqs.( 63)-( 65) after making use of Likewise, if b = 0 and a = 0, then the matrix A is diagonalizable since A satisfies eq. ( 59), i.e., A(A 2 +a 2 I 4 ) = 0.In particular, if b = 0, then eq. ( 55) implies that θ• ζ = 0 and a 2 = | θ| 2 −| ζ| 2 .Plugging these results into eqs.(66) and (71) yields A 3 + a 2 A = 0. Consequently, one can make use of eq. ( 60) with m = 3 to obtain One can check that eq. ( 73) coincides with the b → 0 limit of eqs.( 63)-( 65) after making use of Finally, in the case of a = b = 0, eq. ( 55) yields θ• ζ = 0 and | ζ| 2 = | θ| 2 .Using eq. ( 71), it then follows that A 3 = 0. Thus, the Taylor series of the exponential terminates and one obtains Although one cannot directly employ eq. ( 60) in this final case (since A is no longer diagonalizable), one can still recover eq. ( 74) either by taking the b → 0 limit of eq. ( 72) or the a → 0 limit of eq. ( 73).
5 An explicit evaluation of Λ µ ν = 1 2 Tr M † σ µ M σ ν In Section 3, we remarked that a general element of the Lie algebra so(1, 3) is a real linear combination of the six generators {−is i , −ik i }.In particular, the matrix A defined in eq. ( 48) provides a four-dimensional matrix representation of so(1, 3).The corresponding 4 × 4 matrix that represents a general element of the proper orthochronous Lorentz group, SO 0 (1,3), is then obtained by exponentiation, Λ( ζ, θ) = exp A. In this section, we will take advantage of the existence of a two-dimensional matrix representation of so (1,3).It is noteworthy that by exponentiating this two-dimensional representation, one obtains a two-dimensional matrix representation of the group of complex 2 × 2 matrices with unit determinant, which defines the Lie group SL(2, C).Thus, the two-dimensional matrix representation of SL(2, C) provides representation matrices M [defined in eq. ( 81) below] for the elements of SO 0 (1,3).However, in this case, the 2 × 2 matrices M and −M of SL(2, C) represent the same element of SO 0 (1,3) [cf.eq.(91)].For example, consider the general element of the two-dimensional representation of SL(2, C) that is given by where ζ and θ are the boost and rotation vectors that parametrize an element of the proper orthochronous Lorentz group and σ = (σ 1 , σ 2 , σ 3 ) are the three Pauli matrices assembled into a vector whose components are the 2 × 2 matrices, It is convenient to define a fourth Pauli matrix, σ 0 = I 2 , where I 2 is the 2 × 2 identity matrix.We can then define the four Pauli matrices in a unified notation.Following the notation of Refs.[12,13], we define σ µ = (I 2 ; σ) , σ µ = (I 2 ; − σ) , where µ ∈ {0, 1, 2, 3}.Note that these sigma matrices have been defined with an upper (contravariant) index.They are related to sigma matrices with a lower (covariant) index in the usual way: However, the use of the spacetime indices µ and ν is slightly deceptive since the sigma matrices defined above are fixed matrices that do not change under a Lorentz transformation.It is also convenient to introduce the set of 2 × 2 matrices, One can then rewrite eq.(81) in the following form that is reminiscent of eq. ( 53), That is, the six independent −iσ µν matrices are generators of the Lie algebra of SL(2, C), henceforth denoted by sl(2, C).It is straightforward to check that the 2×2 matrices σ µν possess the same commutation relations as the 4 × 4 matrices s µν [cf.eq. ( 51)], which establishes the isomorphism so(1, 3) ≃ sl(2, C).
Under an active Lorentz transformation, a two-component spinor χ α (where α ∈ {1, 2}) transforms as Suppose that χ and η are two-component spinors and consider the spinor product η † σ µ χ.Under a Lorentz transformation, We assert that the quantity η † σ µ χ transforms as a Lorentz four-vector, The standard proof of this assertion based on the analysis of infinitesimal Lorentz transformations is given in Appendix B. (See also Appendix C, where the corresponding result is obtained by employing the four-component spinor formalism.)Eqs. ( 88) and (89) imply that the following identity must be satisfied: Multiplying eq. ( 90) on the right by σ ρ and using Tr(σ ν σ ρ ) = 2δ ν ρ , it follows that It is now convenient to introduce the complex vector, z ≡ ζ + i θ, and the associated quantity, One can now evaluate the matrix exponential M = exp − 1 2 z • σ [cf.eq. ( 81)] by making use of eq. ( 60) if ∆ = 0.The corresponding eigenvalues of − 1 2 z• σ are λ = ± 1 2 ∆.Hence, Note that the limit as ∆ → 0 is continuous and yields M = I 2 − 1 2 z• σ.Since the Pauli matrices are hermitian, We shall evaluate Λ µ ν in four separate cases depending whether the spacetime index is 0 or i ∈ {1, 2, 3}.In particular, using block matrix notation, eq.(91) yields where we have used σ j = −σ j to obtain the final matrix expression above.
6 Reconciling the results of Sections 4 and 5 In this section, we shall verify that the explicit expressions for Λ( ζ , θ) obtained, respectively, in Sections 4 and 5 coincide in the general case of non-zero boost and rotation parameters.First, it is convenient to rewrite eqs.( 56) and (57) as follows: where ∆ is defined in eq. ( 92).As noted below eq.( 57), a, b ∈ R but their undetermined signs have no impact on the expressions obtained for the matrix elements of Λ( ζ , θ).Using eq. ( 55), we can fix the relative sign of a and b by choosing ab = θ• ζ.It then follows that After taking the positive square root, the signs of a and b are now fixed by identifying One can check that eqs.(99)-(102) are unchanged if ∆ → −∆ and/or ∆ → ∆ * .This reflects the fact that the expressions obtained for the matrix elements of Λ( ζ , θ) do not depend on the choice of signs for a and b.Thus, eqs.( 63)-( 66) and (71) yield: after making use of eq. ( 106).We now employ the following two identities: Hence, eqs.( 108) and (109) yield in agreement with eq.(99).Next, eqs.( 63)-( 66) and (71) yield Using eq. ( 55), it follows that | ζ| 2 − | θ| 2 = b 2 − a 2 and θ• ζ = ab [the latter with the sign conventions adopted above eq.( 107)].Inserting these results into eq.( 113), we obtain We can rewrite eq. ( 114) with the help of some identities.It is straightforward to show that Collecting the results obtained above, we end up with in agreement with eq.(100).The computation of Λ i 0 is nearly identical.The only change is due to the change in the sign multiplying the term proportional to the Levi-Civita tensor.Consequently, it is convenient to replace eq. ( 117) with an equivalent form: Hence, we end up with in agreement with eq. ( 101).
Finally, we use eqs.( 63)-( 66) and (71) to obtain The following identities can be derived: Note that the terms proportional to ǫ ijk in eq.(121) combine nicely and yield after putting z k = ζ k + iθ k .

Final remarks
The main goal of this paper is to exhibit an explicit form for the 4 × One can also obtain the most general proper orthochronous Lorentz transformation in another way by invoking the following theorem (e.g., see Section 1.5 of Ref. [1], Section 6.6 of Ref. [17], or Section 4.5 of Ref. [18]): In contrast to eq. ( 130), when considering infinitesimal Lorentz transformations, the boost matrix [eq.( 41)] and the rotation matrix [eq.( 43)] commute at linear order, which results in eq. ( 46).The effects of the noncommutativity appear first at quadratic order in the boost and rotation parameters.Given the parameters { ζ ′ , θ ′ } (or { ζ ′′ , θ ′′ }), it would be quite useful to be able to obtain expressions for the corresponding parameters of Λ( ζ, θ).The formulae that determine { ζ, θ} in eq. ( 129) are quite complicated [21], although they could in principle be derived by using the explicit matrix representations given in this paper.This is left as an exercise for the reader.

1 2
4 proper orthochronous Lorentz transformation matrix as a function of general boost and rotation parameters ζ and θ.Whereas the matrices Λ( ζ, 0) and Λ( 0, θ) are well known and appear in many textbooks, the explicit form for more general Λ( ζ, θ) is much less well known.Two different derivations are provided for Λ( ζ, θ).One derivation evaluates the exponential of a real 4 × 4 matrix A that satisfies (GA) T = −GA [where G ≡ diag(1, −1, −1, −1)], and a second derivation evaluates Tr M † σ µ Mσ ν , where the 2 × 2 matrix M = exp{− 1 2 ( ζ + i θ)• σ}.Although the results obtained by the two computations look somewhat different at first, we have verified by explicit calculation that these two results are actually equivalent.