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Lorentz Transformations — Examples

1 · Prerequisites

2 · Summary

These examples and counterexamples evaluate the Lorentz transformations of sr-lorentz-transformations on explicit numbers. The 1+1 boost at β=3/5 is checked for its metric and determinant conditions, its inverse, the images of the events (0,0) and (13,11) and the preservation of the null lines (Worked 1+1 boost matrix: metric, determinant, inverse and interval invariance); two collinear boosts compose to the parameter 15/17 with rapidities adding as log⁡2+log⁡2=log⁡4 (Worked collinear composition and rapidity addition at beta = 3/5); and the motion of a primed chart's origin under a passive boost is computed in Worked motion of the primed frame's origin under a passive boost.

The noncommutativity of noncollinear boosts is witnessed by two exact rational matrices at β=3/5 that differ in listed entries (Counterexample: noncollinear boosts do not commute), and the composition B(35e1)B(35e2) is factored as B(3/5,12/25,0)diag⁡(1,R): the rotation acts first, followed by the boost. The rotation has cosine 40/41 and sine 9/41 in the clockwise coordinate convention of the worked example (Worked boost-rotation factorization of a composition of two perpendicular boosts). The discrete parity and time-reversal matrices show that Lorentz maps need be neither proper nor orthochronous and that det⁡L=1 alone does not force orthochrony (The parity and time-reversal maps are Lorentz maps that are not proper, and det = 1 does not imply orthochrony), while a spatial rotation is a proper orthochronous Lorentz map that is not a boost (Counterexample: a spatial rotation is a proper orthochronous Lorentz map that is not a boost). Finally the component laws are computed for a covector and a vector under the β=3/5 boost with invariant contraction (Worked covector and vector component transformation under a beta = 3/5 boost), and a constant vector field J=ρ0U, with U future timelike and unit and ρ0 an arbitrary real constant, has components (ρ0,0,0,0) in the rest chart of U, exhibiting the difference between the scalar multiplier and the coordinate density (Worked current boost law and coordinate versus rest density).

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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Worked 1+1 boost matrix: metric, determinant, inverse and interval invariance

Example

Worked computation in an inertial frame whose spatial direction is e1, with coordinates (x0,x1) and η0=diag⁡(−1,1). Take β=3/5, so γ=(1−9/25)−1/2=5/4 and the passive boost of The passive Lorentz boost with parameter beta is the linear map with matrix B=(5/4−3/4−3/45/4) on the (e0,e1) plane (the identity on the other spatial directions of the frame). Then:

(i) BTη0B=η0: the (0,0) entry is (5/4)2−(3/4)2=25/16−9/16=1 times −1, the spatial diagonal entry is 25/16−9/16=1, and the off-diagonal entry (5/4)(3/4)+(−3/4)(5/4)=0, giving η0;

(ii) det⁡B=(5/4)2−(3/4)2=1, so B is proper;

(iii) Be0 has coordinates (5/4,−3/4) and η0(Be0,Be0)=−25/16+9/16=−1, so the future timelike vector e0 is mapped to a future timelike vector;

(iv) B(−3/5)B(3/5)=(5/43/43/45/4)(5/4−3/4−3/45/4)=(1001), so the inverse is the boost with the opposite parameter;

(v) the events with coordinates (0,0) and (13,11) have images (0,0) and (8,4): x′0=(5/4)(13−(3/5)11)=(5/4)(32/5)=8 and x′1=(5/4)(11−(3/5)13)=(5/4)(16/5)=4, and the interval computed in the two frames is the same: −(8)2+42=−48=−(13)2+112 in the convention η0;

(vi) the null lines x1=±x0 are preserved: B(1,1)=(1/2,1/2) and B(1,−1)=(2,−2) are null, and B(1,−1)=4 (1/2,−1/2), so the images are exactly the same pair of lines.

Facts & Assumptions

Given: An inertial frame (o;e0,e1,e2,e3), the parameter β=3/5, and the boost B:=B(3/5 e1).

[L1]

For ∣β∣<1 the passive boost in the 1+1 case β=βe1 acts on the coordinates of the (e0,e1) plane by x′0=γ(x0−βx1), x′1=γ(x1−βx0), with γ=(1−β2)−1/2>0 and γ the rational power of the positive base 1−β2; the matrix in the ordered basis (e0,e1) is (γ−γβ−γβγ), and the map fixes every vector with x0=x1=0, namely every vector in span⁡(e2,e3) (The passive Lorentz boost with parameter beta).

[L2]

B(β) is a proper orthochronous Lorentz map with B(β)−1=B(−β), and B(β)e0=γ(e0−βe1); the map β↦B(β) is injective (The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L3]

In an inertial frame η(ea,eb)=diag⁡(−1,1,1,1)ab and e0∈C+; one has η(x0e0+x1e1,x0e0+x1e1)=−(x0)2+(x1)2 in the plane coordinates (Inertial frames and inertial charts of the affine Minkowski model).

[L4]

C+ is one of the two components of the timelike set and is the set of future timelike vectors; a Lorentz map is orthochronous exactly when it maps C+ onto C+ (Time orientation, future and past timelike/null/causal vectors, and the closed cones, The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L5]

A nonzero vector v is null when η(v,v)=0 and timelike when η(v,v)<0 (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

Proof

technique · direct
1.1L1algebra

Evaluate the 2×2 matrix B: γ=(1−9/25)−1/2=(16/25)−1/2=5/4, so B=(5/4−3/4−3/45/4), and B acts as the identity on the coordinate directions e2,e3 because β=3/5 e1 has vanishing e2,e3 components in [L1]. For the product BTη0B, the four entries are (5/4)(−5/4)+(−3/4)(−3/4)=−25/16+9/16=−1 in the (0,0) slot, (5/4)(3/4)+(−3/4)(5/4)=15/16−15/16=0 and (−3/4)(−5/4)+(5/4)(−3/4)=15/16−15/16=0 in the two off-diagonal slots, and (−3/4)(3/4)+(5/4)(5/4)=−9/16+25/16=1 in the (1,1) slot; hence BTη0B=diag⁡(−1,1)=η0.

1.2L2algebra

The determinant is det⁡B=(5/4)(5/4)−(−3/4)(−3/4)=25/16−9/16=1, so B is proper in the sense of the properness condition of [L2].

1.3L2L3L4L5algebra

By [L2], Be0=γ(e0−βe1)=(5/4)(e0−(3/5)e1), with coordinates (5/4,−3/4); [L3] gives η(Be0,Be0)=−(5/4)2+(3/4)2=−1, so Be0 is timelike by [L5]. Since B is orthochronous by [L2] and e0∈C+ by [L3], [L4] gives Be0∈B(C+)=C+, so Be0 is future timelike: the future timelike vector e0 is mapped to a future timelike vector.

1.4L1L2algebra

The product of clause (iv) is (5/43/43/45/4)(5/4−3/4−3/45/4)=(25/16−9/16−15/16+15/16−15/16+15/16−9/16+25/16)=(1001), and the first factor is the matrix of B(−3/5) by [L1]; this is the instance at β=3/5 of B(−β)B(β)=id⁡ from [L2], so the inverse is the boost with the opposite parameter.

1.5L1L5algebra

For the two null directions, [L1] gives B(1,1)=((5/4)(1−3/5),(5/4)(1−3/5))=(1/2,1/2) and B(1,−1)=((5/4)(1+3/5),(5/4)(−1−3/5))=(2,−2); both images satisfy η0=0 on their squares: −(1/2)2+(1/2)2=0 and −4+4=0, so they are null by [L5]. Since B(1,1)=12(1,1) and B(1,−1)=2(1,−1), the image of the line x1=x0 is the same line and the image of the line x1=−x0 is the same line; hence the pair of null lines {x1=x0}∪{x1=−x0} is mapped onto itself, and the images are exactly that pair.

2.1L1step 1.1algebra

For the event with coordinates (13,11), [L1] gives x′0=(5/4)(13−(3/5)11)=(5/4)(13−33/5)=(5/4)(32/5)=8 and x′1=(5/4)(11−(3/5)13)=(5/4)(11−39/5)=(5/4)(16/5)=4, while (0,0) is fixed because the boost is linear. The interval in η0 is −(8)2+(4)2=−64+16=−48 and −(13)2+(11)2=−169+121=−48; this numerical equality is the instance of BTη0B=η0 from step 1.1 applied to the coordinate difference (13,11).

3.1step 1.1step 1.2step 1.3step 1.4step 2.1step 1.5∎

Steps 1.1-1.4 verify clauses (i)-(iv), step 2.1 verifies clause (v), and step 1.5 verifies clause (vi); every clause of the Example is a checked finite computation with the displayed rational entries.

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Worked collinear composition and rapidity addition at beta = 3/5

Example

Worked computation. Let e be a unit spatial direction of an inertial frame and let B(35e) be the passive boost with parameter 35 (The passive Lorentz boost with parameter beta), so γ=5/4 and, in the ordered pair of directions (e0,e), the boost has the 2×2 matrix (5/4−3/4−3/45/4) and is the identity on {y∈S:δ(e,y)=0}, the η-orthogonal complement of span⁡(e0,e) (Rapidity parametrization of collinear boosts). Then:

(i) B(35e)2=(5/4−3/4−3/45/4)2=(17/8−15/8−15/817/8)=B(1517e), because γ(1517)=(1−225289)−1/2=(64289)−1/2=178 and γ(1517)⋅1517=178⋅1517=158. This is the velocity-addition law 3/5+3/51+(3/5)(3/5)=6/534/25=1517, with ∣15/17∣<1.

(ii) The rapidities are r(35)=12log⁡(1+3/51−3/5)=12log⁡4=log⁡2 and r(1517)=12log⁡(1+15/171−15/17)=12log⁡16=log⁡4=2log⁡2, so the rapidities add under composition: r(1517)=r(35)+r(35).

(iii) The two boosts commute: the reversed product B(35e)B(35e) equals the square computed in (i), and this is the instance β1=β2=3/5 of the commutativity of collinear boosts.

Facts & Assumptions

Given: A unit spatial direction e of an inertial frame, the passive boost B(35e), and the collinear family B(βe), β∈(−1,1).

[L1]

For ∣β∣<1 the boost B(βe) acts in the frame coordinates by x′0=γ(x0−βx1), x′1=γ(x1−βx0) on the (e0,e) plane, with γ=(1−β2)−1/2>0 and β the coefficient of the unit direction; the matrix on the ordered pair (e0,e) is (γ−γβ−γβγ) and the map is the identity on {y∈S:δ(e,y)=0} (The passive Lorentz boost with parameter beta).

[L2]

For a unit spatial direction e and r∈R, the boost B(tanh⁡(r)e) has on (e0,e) the matrix (cosh⁡r−sinh⁡r−sinh⁡rcosh⁡r); in particular, with β=tanh⁡(r), γ(β)=cosh⁡(r) and γ(β)β=sinh⁡(r) (Rapidity parametrization of collinear boosts).

[L3]

For unit e and β1,β2∈(−1,1) one has B(β1e)B(β2e)=B(βe) with β=β1+β21+β1β2 and ∣β∣<1, and collinear boosts commute (Collinear boost composition and the relativistic velocity-addition law).

[L4]

For β∈(−1,1) the rapidity is r(β)=12log⁡1+β1−β, a strictly increasing bijection of (−1,1) onto R with β=tanh⁡(r(β)); for a fixed unit direction, rapidities add: B(tanh⁡(r)e)B(tanh⁡(s)e)=B(tanh⁡(r+s)e) (Rapidity parametrization of collinear boosts, The natural logarithm as the inverse of the exponential function).

[L5]

For x,y>0 one has log⁡(xy)=log⁡x+log⁡y and log⁡(x/y)=log⁡x−log⁡y (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm); the exponential satisfies er=exp⁡(r) and the hyperbolic functions are sinh⁡(r)=(er−e−r)/2, cosh⁡(r)=(er+e−r)/2 (The real exponential function and the number e by a power series).

[L6]

The Euclidean norm of a vector of R3 satisfies ∣βe∣=∣β∣ for a unit direction (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn).

Proof

technique · direct
1.1L1L2L6algebra

With β=3/5 and ∣βe∣=3/5 by [L6], [L1] gives γ=(1−9/25)−1/2=(16/25)−1/2=5/4, so the matrix of B(35e) on the ordered pair (e0,e) is (5/4−3/4−3/45/4), and the map is the identity on the η-orthogonal complement of span⁡(e0,e). This is the matrix form recorded in [L2], with the rapidity fixed by tanh⁡r=3/5; the numerical value of that rapidity is computed below.

1.2L3

By the commutativity clause of [L3], B(35e)B(35e)=B(35e)B(35e), i.e. the reversed order gives back the same product; clause (iii) is this instance, and no separate computation is needed.

2.1L3step 1.1algebra

Squaring the matrix of step 1.1 in the 2×2 block, (5/4)2+(3/4)2=25/16+9/16=34/16=17/8 on the diagonal and −(5/4)(3/4)−(3/4)(5/4)=−15/8 off the diagonal, while the identity block on the complement squares to itself; hence B(35e)2 has matrix (17/8−15/8−15/817/8) on (e0,e) and is the identity on the complement. By [L3], B(35e)2=B(βe) with β=3/5+3/51+(3/5)(3/5)=6/534/25=1517, and ∣15/17∣<1. For this β, γ(β)=(1−225/289)−1/2=(64/289)−1/2=17/8 and γ(β)β=178⋅1517=158; by [L1] the matrix of B(1517e) is therefore (17/8−15/8−15/817/8) on (e0,e), matching the square. This proves clause (i) and the velocity-addition law.

3.1L4L5step 2.1algebra

By [L4] and [L5], r(35)=12log⁡8/52/5=12log⁡4, and log⁡4=log⁡2+log⁡2 by the product law, so r(35)=12(2log⁡2)=log⁡2; similarly r(1517)=12log⁡32/172/17=12log⁡16 and log⁡16=log⁡4+log⁡4=4log⁡2, so r(1517)=12(4log⁡2)=2log⁡2=log⁡4. With log⁡2=log⁡2 this gives r(1517)=r(35)+r(35): the rapidity of the composed boost in step 2.1 is the sum of the two equal rapidities, in agreement with the addition statement of [L4].

4.1step 1.1step 1.2step 2.1step 3.1∎

Steps 1.1-2.1 prove clauses (i) and (iii) and step 3.1 proves clause (ii); every entry is an exact rational or logarithmic identity justified by the displayed suppliers.

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Worked boost-rotation factorization of a composition of two perpendicular boosts

Example

Worked computation in a fixed inertial frame (o;e0,e1,e2,e3) with coordinates (x0,x1,x2,x3) and η0=diag⁡(−1,1,1,1). Take the two passive boosts B1:=B((3/5,0,0)) and B2:=B((0,3/5,0)) (The passive Lorentz boost with parameter beta), each acting as the identity on the orthogonal complement of its own coordinate plane, and let L:=B1B2 be their composition (matrix product in the frame's ordered basis). Then:

(i) L=(25/16−3/4−15/160−15/165/49/160−3/405/400001), and direct multiplication gives LTη0L=η0 and det⁡L=1; since B1 and B2 are proper orthochronous Lorentz maps, L is a proper orthochronous Lorentz map.

(ii) u:=Le0 has coordinates (25/16,−15/16,−3/4,0), with η0(u,u)=−625/256+225/256+9/16=−256/256=−1 and u0=25/16>0, so u is a future timelike unit vector. The parameter β:=uvec/u0 of Unique boost-rotation factorization of proper orthochronous Lorentz maps is β=(−3/5,−12/25,0), with ∣β∣2=369/625 and γ(β)=(1−369/625)−1/2=25/16, consistent with u0=γ(β).

(iii) B(β)L=diag⁡(1,R) with R=(40/419/410−9/4140/410001), whose extension diag⁡(1,R) is a spatial rotation of the frame: RTR=I3 and det⁡R=1 (using 402+92=412). Hence L=B(−β)diag⁡(1,R)=B(35,1225,0)diag⁡(1,R) is the unique factorization of Unique boost-rotation factorization of proper orthochronous Lorentz maps; since R≠I3, L is not a passive boost.

(iv) The map R fixes e3 and acts on the e1-e2 plane as the rotation with cosine parameter 40/41 and sine parameter 9/41 in the clockwise coordinate convention C(c∗,s∗):=(c∗s∗−s∗c∗), c∗2+s∗2=1 (Carroll, equation (1.17)); this rotation is nontrivial because 9/41≠0. The two boosts therefore compose to a nonzero rotation followed by a boost of the e1-e2 plane; this composition of the explicit pair B1,B2 has a rotation factor R, called its Thomas-Wigner rotation. No gyroscope, precession or physical observation claim is attached here.

Facts & Assumptions

Given: A fixed inertial frame (o;e0,e1,e2,e3) with coordinates x=(x0,x1,x2,x3), the boosts B1:=B((3/5,0,0)), B2:=B((0,3/5,0)) and their composition L:=B1B2.

[L1]

For 0<∣β∣<1 the passive boost has, in the frame's ordered basis, the matrix (γ−γβT−γβI3+γ−1∣β∣2ββT) with γ=(1−∣β∣2)−1/2; for β=0 one defines B(0)=id⁡V, whose matrix is I4 (The passive Lorentz boost with parameter beta).

[L2]

For every ∣β∣<1 the boost B(β) is a proper orthochronous Lorentz map with B(β)−1=B(−β); B(β)e0=γ(e0−β1e1−β2e2−β3e3); and if u is a future timelike unit vector with coordinates (u0,uvec), u0>0, then B(uvec/u0)u=e0 (The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L3]

For a proper orthochronous Lorentz map L with u:=Le0, coordinates (u0,uvec), the parameter β:=uvec/u0 satisfies ∣β∣<1; the map R^:=B(β)L satisfies R^e0=e0, R^(S)=S and R^∣S:S→S is a determinant-one δ-isometry and R^ is a spatial rotation of the frame, so that L=B(−β)R^; the factorization with a spatial rotation R^ and a parameter in the open unit ball is unique; and L is a passive boost if and only if R^ is the identity (Unique boost-rotation factorization of proper orthochronous Lorentz maps).

[L4]

A spatial rotation of the frame is an extension of a linear map R with Re0=e0, R(S)⊆S, δ(Rw,Rz)=δ(w,z) on S and det⁡(R∣S)=+1; via the δ-orthonormal basis (e1,e2,e3) these are exactly the maps whose spatial matrix A satisfies ATA=I3 and det⁡A=+1 (Spatial rotations of an inertial frame).

[L5]

In the frame, η(ea,eb)=diag⁡(−1,1,1,1)ab, e0∈C+, and the ordered basis is positively oriented; η(x0e0+x1e1+x2e2+x3e3,x0e0+x1e1+x2e2+x3e3)=−(x0)2+(x1)2+(x2)2+(x3)2 (Inertial frames and inertial charts of the affine Minkowski model).

[L6]

The Euclidean norm on R3 satisfies ∣β∣2=β12+β22+β32 (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn).

[L7]

Composition of proper orthochronous Lorentz maps is a proper orthochronous Lorentz map, and every Lorentz map satisfies η(Lv,Lw)=η(v,w) and det⁡L=±1 (Lorentz maps are isomorphisms of determinant ±1 that preserve causal character and permute the cones; the Lorentz sets are groups).

Proof

technique · direct
1.1L1algebra

By [L1] with β=(3/5,0,0) and ∣β∣2=9/25, one has γ=(16/25)−1/2=5/4 and B1=(5/4−3/400−3/45/40000100001); with β=(0,3/5,0) one has B2=(5/40−3/400100−3/405/400001). Multiplying, Lac=∑bB1abB2bc; row by row: the (0,⋅) entries are (5/4)(5/4)=25/16, (5/4)⋅0+(−3/4)⋅1=−3/4, (5/4)(−3/4)+0=−15/16 and 0; the (1,⋅) entries are (−3/4)(5/4)=−15/16, (−3/4)⋅0+(5/4)⋅1=5/4, (−3/4)(−3/4)=(9/16) and 0; the (2,⋅) entries are 1⋅(−3/4)=−3/4, 0, 1⋅(5/4)=5/4 and 0; the (3,⋅) entries are 0,0,0,1. This proves the displayed matrix of clause (i).

1.2L5algebra

To verify LTη0L=η0, the matrix LTη0L is symmetric, and its upper triangle is computed from the columns of L: the (0,0) entry is −(25/16)2+(15/16)2+(3/4)2=(−625+225+144)/256=−1; the (1,1) entry is −(3/4)2+(5/4)2+02=(−9+25)/16=1; the (2,2) entry is −(15/16)2+(9/16)2+(5/4)2=(−225+81+400)/256=1; the (3,3) entry is −0+0+0+1=1; the (0,1) entry is −(25/16)(−3/4)+(−15/16)(5/4)+0=75/64−75/64=0; the (0,2) entry is −(25/16)(−15/16)+(−15/16)(9/16)+(−3/4)(5/4)=(375−135−240)/256=0; the (1,2) entry is −(−3/4)(−15/16)+(5/4)(9/16)+0=−45/64+45/64=0; the remaining upper-triangle entries involving the fourth row or column are 0 because the fourth row and column of L are those of the identity. Hence LTη0L=η0. Expanding det⁡L along the fourth row gives det⁡L=det⁡(25/16−3/4−15/16−15/165/49/16−3/405/4)=2516⋅2516+34(−7564+2764)−1516⋅1516=625−225256−3664=400−144256=1.

2.1L2L7step 1.2algebra

B1 and B2 are proper orthochronous Lorentz maps by [L2], so their composition L is a proper orthochronous Lorentz map by [L7]; this, with the direct metric and determinant checks of step 1.2, is clause (i). The coordinates of Le0 are the entries of the first column of L, namely u=Le0=(25/16,−15/16,−3/4,0).

3.1L3L5L6step 2.1algebra

With u as in step 2.1, [L5] gives η0(u,u)=−(25/16)2+(15/16)2+(3/4)2=(−625+225+144)/256=−1, and u0=25/16>0; thus u is timelike, and because L is orthochronous by step 2.1 and e0∈C+ by [L5], u=Le0 lies in C+, so u is a future timelike unit vector. Its parameter as in [L3] is β=uvec/u0=((−15/16)⋅(16/25),(−3/4)⋅(16/25),0)=(−3/5,−12/25,0); by [L6], ∣β∣2=(9/25)+(144/625)=(225+144)/625=369/625<1, and γ(β)=(1−369/625)−1/2=(256/625)−1/2=25/16=u0.

4.1L1step 3.1algebra

By [L1] with β=(−3/5,−12/25,0): γ=25/16, the coefficient is γ−1∣β∣2=9/16369/625=9⋅62516⋅369=625656, and the matrix of B(β) is (25/1615/163/4015/16881/65645/16403/445/16450/4100001), because B01=−γβ1=15/16, B02=−γβ2=3/4, B11=1+625656⋅925=1+225656=881656, B12=625656⋅35⋅1225=45164 and B22=1+625656⋅144625=1+144656=5041.

5.1L1L2step 1.1step 4.1algebra

Multiply M:=B(β)L using steps 1.1 and 4.1. The first column is B(β)Le0=B(β)u=e0 by [L2] applied to the future timelike unit vector u of step 3.1, so it is (1,0,0,0)T. The first row entries are M0j=(25/16)L0j+(15/16)L1j+(3/4)L2j: for j=0, (625−225−144)/256=1; for j=1, −75/64+75/64=0; for j=2, (−375+135+240)/256=0; and for j=3, 0. The fourth row and column are those of the identity because L and B(β) fix e3 and preserve the splitting. The remaining 2×2 block is Mij=Bi0L0j+Bi1L1j+Bi2L2j for i,j∈{1,2}: M11=−45/64+881656⋅54=(−1845+4405)/2624=25602624=4041, M12=−225/256+881656⋅916+45164⋅54=(−9225+7929+3600)/10496=230410496=941, M21=−9/16+45164⋅54=(−369+225)/656=−144656=−941, and M22=−45/64+45164⋅916+5041⋅54=(−1845+405+4000)/2624=25602624=4041. Hence B(β)L=diag⁡(1,R) with R as displayed.

6.1L4step 5.1algebra

For the displayed R one has RTR=I3: the (1,1) and (2,2) entries are (40/41)2+(9/41)2=(1600+81)/1681=1 and the (1,2) entry is (40/41)(9/41)+(−9/41)(40/41)=0, with the third row and column those of the identity. Moreover det⁡R=4041⋅4041−941⋅(−941)=16811681=1 (using 402+92=1681=412). Since R acts on S=span⁡(e1,e2,e3) in the δ-orthonormal basis (e1,e2,e3) and satisfies RTR=I3, det⁡R=1, the extension diag⁡(1,R) is a spatial rotation of the frame by [L4].

6.2L3step 3.1step 5.1algebra

By [L3] applied to the proper orthochronous Lorentz map L with u=Le0 and β=uvec/u0 from step 3.1, the factorization L=B(−β)R^ with R^ the extension of a spatial rotation is the unique such factorization, and by step 5.1 the map R^ equals diag⁡(1,R); hence L=B(35,1225,0)diag⁡(1,R), since −β=(3/5,12/25,0). Since R≠I3 (the entry 9/41≠0), [L3] clause (v) gives that L is not a passive boost.

7.1step 5.1step 6.1algebra

Finally, R fixes e3 and multiplies the coordinate pair (e1,e2) by the matrix (40/419/41−9/4140/41), whose entries have squares summing to (1600+81)/1681=1; it is therefore C(40/41,9/41) in the clockwise coordinate convention defined in (iv), and it is nontrivial because 9/41≠0. This is clause (iv): L is this nonzero rotation followed by the boost (the rightmost factor acts first) of the e1-e2 plane.

8.1step 1.1step 1.2step 2.1step 3.1step 4.1step 5.1step 6.1step 6.2step 7.1∎

Steps 1.1, 1.2 and 2.1 prove clause (i), steps 3.1, 4.1 and 5.1 prove clause (ii), steps 5.1, 6.1 and 6.2 prove clause (iii), and step 7.1 proves clause (iv); every assertion is a finite rational computation from the displayed matrices.

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Counterexample: noncollinear boosts do not commute

Statement refuted

The statement "for all β1,β2∈R3 with ∣βi∣<1 the passive boosts satisfy B(β1)B(β2)=B(β2)B(β1)" is false.

Facts & Assumptions

Given: An inertial frame with ordered basis (e0,e1,e2,e3) and η0=diag⁡(−1,1,1,1), and the boost parameters β1=(3/5,0,0) and β2=(0,3/5,0).

[L1]

For 0<∣β∣<1 the passive boost has matrix entries B00=γ, B0i=Bi0=−γβi and Bij=δij+γ−1∣β∣2βiβj in the frame's basis, with γ=(1−∣β∣2)−1/2; for β=0 one defines B(0)=id⁡V, whose matrix is I4 (The passive Lorentz boost with parameter beta).

[L2]

For boosts in a common unit direction e one has B(β1e)B(β2e)=B(β1+β21+β1β2e) and the two collinear boosts commute (Collinear boost composition and the relativistic velocity-addition law).

[L3]

Every B(β) is a proper orthochronous Lorentz map, and products of Lorentz maps are Lorentz maps; the proper orthochronous Lorentz maps form a group (The passive boost is a proper orthochronous Lorentz map with inverse B(-beta), Lorentz maps are isomorphisms of determinant ±1 that preserve causal character and permute the cones; the Lorentz sets are groups).

[L4]

The decomposition L=B(−β)R^ of a proper orthochronous Lorentz map into a boost and a spatial rotation, with β=uvec/u0 for u=Le0, is unique, and its rotation factor is the identity if and only if L is a passive boost (Unique boost-rotation factorization of proper orthochronous Lorentz maps).

[L5]

Coordinates of the frame are read in the ordered basis (e0,e1,e2,e3) (Inertial frames and inertial charts of the affine Minkowski model).

Counterexample

1.1L1L5algebra

For β1=(3/5,0,0) and β2=(0,3/5,0) one has ∣β1∣=∣β2∣=3/5 and γ=(1−9/25)−1/2=5/4, while (γ−1)/∣βi∣2=(1/4)/(9/25)=25/36; substituting into [L1] gives B(β1)=(5/4−3/400−3/45/40000100001) and B(β2)=(5/40−3/400100−3/405/400001): the contributions (25/36)βiβj place the entries 25/36⋅9/25=1/4 on the diagonal blocks, giving 5/4 on each boosted spatial diagonal and leaving the transverse directions at 1.

2.1step 1.1algebra

By matrix multiplication, B(β1)B(β2)=(25/16−3/4−15/160−15/165/49/160−3/405/400001): the first row is 5/4 times the first row of B(β2) plus −3/4 times its second row, giving (25/16,0,−15/16,0)+(0,−3/4,0,0), and the remaining rows are computed the same way from the displayed matrices. Reversing the order gives B(β2)B(β1)=(25/16−15/16−3/40−3/45/400−15/169/165/400001).

3.1L2L3step 1.1step 2.1algebra

The two products differ in the (0,1) entry (−3/4 versus −15/16), the (0,2) entry (−15/16 versus −3/4) and the (2,1) entry (0 versus 9/16); a single differing entry already refutes the asserted commutation, so B(β1)B(β2)≠B(β2)B(β1) and the group of proper orthochronous Lorentz maps is not abelian. This is consistent with [L2], which asserts commutation only for boosts in a common direction, and the counterexample makes no claim that the two products fail to be Lorentz maps: both are proper orthochronous Lorentz maps as products of such maps by [L3].

4.1L2L3L4step 3.1algebra∎

The result is consistent with [L2], which asserts commutation only for boosts along one common direction, and with [L4], which factors any proper orthochronous Lorentz map as a boost times a spatial rotation: the product L:=B(β1)B(β2) is a proper orthochronous Lorentz map by [L3], so [L4] represents it as such a product, and no contradiction with the collinear law arises. The counterexample itself proves only the noncommutation asserted in the refuted statement, namely that the group of proper orthochronous Lorentz maps is not abelian.

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The parity and time-reversal maps are Lorentz maps that are not proper, and det = 1 does not imply orthochrony

Statement refuted

The statements "every Lorentz map is proper and orthochronous" and, separately, "every Lorentz map with det⁡L=1 is orthochronous" are false.

Facts & Assumptions

Given: An inertial frame (o;e0,e1,e2,e3) with η0=diag⁡(−1,1,1,1), and the coordinate columns x=(x0,x1,x2,x3) of a vector in that frame.

[L1]

A linear map L is a Lorentz map exactly when its matrix M satisfies MTη0M=η0 in the frame's ordered basis; L is proper when det⁡L=1 and orthochronous when L(C+)=C+ (Linear and affine Lorentz maps, and the Lorentz group).

[L2]

A Lorentz map satisfies det⁡L=±1; L(C+) is one of the two components C+,C−; and L(C+)=C+ if and only if L(C+)⊆C+ (Lorentz maps are isomorphisms of determinant ±1 that preserve causal character and permute the cones; the Lorentz sets are groups).

[L3]

Since e0∈C+ in an inertial frame, the components are described in the frame's coordinates by C+={x:x0>∣xvec∣} and C−={x:x0<−∣xvec∣} (Inertial frames and inertial charts of the affine Minkowski model, Time orientation, future and past timelike/null/causal vectors, and the closed cones).

[L4]

A vector is timelike when η(x,x)<0, and η(x,x)=−(x0)2+∣xvec∣2 in an η-orthonormal basis (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L5]

The determinant of a diagonal matrix is the product of its diagonal entries (The determinant of a triangular matrix is the product of its diagonal entries), and the matrix product of two diagonal matrices is the diagonal matrix of the products of the corresponding entries (Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes).

Counterexample

1.1L1L5algebra

Define Π and T by their matrices in the frame basis, Π=diag⁡(1,−1,−1,−1) and T=diag⁡(−1,1,1,1). For any diagonal M=diag⁡(m0,m1,m2,m3), MTη0M=diag⁡(−m02,m12,m22,m32) by [L5]; for Π and for T each ma2=1, so both matrices equal η0=diag⁡(−1,1,1,1). Hence Π and T are Lorentz maps by the matrix criterion [L1].

1.2L1L5algebra

By the diagonal product rule [L5], det⁡Π=1⋅(−1)3=−1 and det⁡T=−1, so by [L1] neither Π nor T is proper: they are Lorentz maps failing properness.

1.3L2L3algebra

For every x, (Πx)0=x0 and (Πx)vec=−xvec, so ∣(Πx)vec∣=∣xvec∣; if x∈C+ then x0>∣xvec∣ by [L3], hence (Πx)0=x0>∣(Πx)vec∣ and Πx∈C+ by [L3] again. Thus Π(C+)⊆C+, and [L2] gives Π(C+)=C+: the improper map Π is orthochronous, so Π refutes "orthochronous implies proper".

1.4L2L3L4algebra

For every future timelike x, (Tx)0=−x0<0 and ∣(Tx)vec∣=∣xvec∣<x0=−(Tx)0, so Tx∈C− by [L3]; since T is a bijection with inverse T, T(C+)=C− and T is not orthochronous.

1.5L1L2L3L5algebra

By the diagonal product rule [L5], ΠT=diag⁡(−1,−1,−1,−1)=−I4, whose determinant is (−1)4=+1; and for every x∈C+ one has x0>∣xvec∣≥0, so (ΠT)x=−x has ((ΠT)x)0=−x0<0 and ∣((ΠT)x)vec∣=∣xvec∣<x0=−((ΠT)x)0, whence (ΠT)x∈C− by [L3] and (ΠT)(C+)⊆C−; since ΠT is a bijection, ΠT(C+)=C− by [L2]. Hence det⁡(ΠT)=+1 and ΠT is not orthochronous: det⁡L=1 does not imply orthochrony, and this proper non-orthochronous map refutes "proper implies orthochronous".

2.1step 1.1step 1.2step 1.3step 1.4step 1.5L1L2∎

Steps 1.1 and 1.2 exhibit Lorentz maps that are not proper (Π and T), step 1.3 exhibits a Lorentz map that is not proper yet is orthochronous, and steps 1.4 and 1.5 exhibit Lorentz maps that are not orthochronous, including one, ΠT, with det⁡=+1. The identity supplies the proper orthochronous sign class, so together with Π,T,ΠT all four classes are realized. Therefore O(V,η) strictly contains SO+(V,η), the four combinations of the sign of the determinant and orthochrony are genuinely distinct, and the convention of Linear and affine Lorentz maps, and the Lorentz group(e), which reserves "Lorentz transformation" for the proper orthochronous maps, is not automatic. Both refuted statements are false.

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Worked covector and vector component transformation under a beta = 3/5 boost

Example

Worked computation. In an inertial frame with coordinates (x0,x1,x2,x3) consider the passive boost Λ:=B(35e1) (The passive Lorentz boost with parameter beta), so that on the (x0,x1) plane Λ=(5/4−3/4−3/45/4) and Λ−1=B(−35e1)=(5/43/43/45/4), and let x′ be an inertial chart with x′=Λx+b (Inertial charts are related by proper orthochronous affine Lorentz maps, and every such map is a chart transition). Take the covector k with components (k0,k1)=(−1,1) and the vector w with components (w0,w1)=(1,1), with transverse components zero. A covector k≠0 is called null here when η−1(k,k)=0, where the inverse form is defined by η−1(k,l):=η(k♯,l♯) and η(k♯,v)=k(v) for every v∈V. In this orthonormal basis the unique vector k♯ has coordinates (−k0,k1,k2,k3), verifying both existence and uniqueness directly; hence η−1(k,k)=−k02+k12+k22+k32. This uses the nondegenerate Lorentz form of Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space. Then:

(i) ka′=(Λ−1)bakb gives k0′=(5/4)(−1)+(3/4)(1)=−1/2 and k1′=(3/4)(−1)+(5/4)(1)=1/2, so k′=(−1/2,1/2).

(ii) w′a=Λabwb gives w′0=(5/4)(1−(3/5)⋅1)=1/2 and w′1=(5/4)(1−(3/5)⋅1)=1/2, so w′=(1/2,1/2).

(iii) k is null for the inverse form and w is null for η0: −1+1=0, and so are k′ and w′: −1/4+1/4=0.

(iv) The contraction is invariant: ka′w′a=(−1/2)(1/2)+(1/2)(1/2)=0=(−1)(1)+(1)(1)=kawa.

The example shows all three features of the component law of Transformation of tensor, covector, derivative and current components under affine Lorentz maps: the covariant slot transforms with Λ−1, the contravariant slot with Λ, and the contraction is unchanged. The common factor 1/2 on the null covector is the numerical form of the transformation law; no frequency or Doppler interpretation is attached to it here (that belongs to the light-and-doppler page).

Facts & Assumptions

Given: An inertial frame with coordinates (x0,x1,x2,x3) and η0=diag⁡(−1,1,1,1), the passive boost Λ=B(35e1) on the (x0,x1) plane, an inertial chart x′ with x′=Λx+b, the covector k with components (−1,1,0,0) and the vector w with components (1,1,0,0).

[L1]

For ∣β∣<1 the boost B(βe1) acts on the (x0,x1) coordinates by x′0=γ(x0−βx1), x′1=γ(x1−βx0), with γ=(1−β2)−1/2>0, and its matrix on (e0,e1) is (γ−γβ−γβγ); B(βe1)−1=B(−βe1) (The passive Lorentz boost with parameter beta).

[L2]

For every proper orthochronous linear Lorentz map L and every event c there is an inertial frame whose chart xL satisfies xL(p)=ML−1(x(p)−x(c)), where ML=E−1∘L∘E; the transition x′∘x−1 of inertial charts x,x′ is affine with linear part a proper orthochronous Lorentz matrix (Inertial charts are related by proper orthochronous affine Lorentz maps, and every such map is a chart transition).

[L3]

Let x,x′ be inertial charts with transition x′=Λx+b, Ω open, and let J, k, T be Cℓ vector, covector and type-(r,s) tensor fields with components Ja, ka, Ta1…arb1…bs in x and primed components in x′. Then (i) J′a=ΛabJb; (ii) ka′=(Λ−1)bakb; (iii) every contraction of one contravariant with one covariant index is invariant (Transformation of tensor, covector, derivative and current components under affine Lorentz maps, A type (r,s) tensor on a finite-dimensional vector space).

[L4]

A nonzero vector is null when η(v,v)=0, and in the frame η(v,v)=−(v0)2+(v1)2+(v2)2+(v3)2 (Timelike, null, spacelike and causal vectors of a Lorentz vector space, Inertial frames and inertial charts of the affine Minkowski model).

Proof

technique · direct
1.1L1L2algebra

By [L1] with β=3/5, γ=(1−9/25)−1/2=5/4, so the matrix of Λ=B(35e1) on the (x0,x1) plane is (5/4−3/4−3/45/4) and its inverse is the matrix of B(−35e1), namely (5/43/43/45/4); the existence of the chart x′ with this linear part is [L2].

2.1L3step 1.1algebra

Clause (i): by L3, ka′=(Λ−1)bakb, where (Λ−1)ba is the matrix entry in row b and column a of Λ−1 from step 1.1. Hence k0′=(5/4)(−1)+(3/4)(1)=−5/4+3/4=−1/2 and k1′=(3/4)(−1)+(5/4)(1)=−3/4+5/4=1/2.

2.2L3step 1.1algebra

Clause (ii): by L3, w′a=Λabwb, so w′0=(5/4)(1)+(−3/4)(1)=1/2 and w′1=(−3/4)(1)+(5/4)(1)=1/2, which agrees with the displayed forms (5/4)(1−3/5)=1/2; the components in the transverse directions are unaffected.

3.1L3L4step 2.1step 2.2algebra

Clause (iii): the inverse form defined in the Example gives η−1(k,k)=−(−1)2+12=0, while [L4] gives η(w,w)=−(1)2+12=0. Both objects are nonzero, so the covector and vector are null in their respective spaces. In the primed orthonormal frame the inverse metric matrix is again diag⁡(−1,1,1,1), the inverse of its metric matrix; the computed components give η−1(k,k)=−(−1/2)2+(1/2)2=0 and η(w,w)=−(1/2)2+(1/2)2=0. Thus the primed components describe the same null covector and null vector; no covector is inserted into the form on V×V.

3.2L3step 2.1step 2.2algebra

Clause (iv): by L3 the contraction kawa is invariant under the chart change, so ka′w′a=kawa; directly, ka′w′a=(−1/2)(1/2)+(1/2)(1/2)=−1/4+1/4=0 and kawa=(−1)(1)+(1)(1)=0, consistent with the invariant statement.

4.1step 1.1step 2.1step 2.2step 3.1step 3.2∎

Steps 1.1-2.2 prove clauses (i)-(ii), step 3.1 proves clause (iii) and step 3.2 proves clause (iv); every entry is an exact rational matrix computation from the displayed law, and no frequency, wave or Doppler interpretation is attached to the null covector.

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Worked current boost law and coordinate versus rest density

Example

Worked computation. In an inertial frame with coordinates (x0,x1,x2,x3) consider the constant vector field U with components (U0,U1,U2,U3)=(5/4,3/4,0,0) on an open set Ω (C^k tensor fields on affine Minkowski space and their inertial components), so that η0(U,U)=−25/16+9/16=−1 with U0>∣Uvec∣=3/4; U is a future timelike unit vector field. Let ρ0 be a constant scalar field and put J:=ρ0U. Choose the primed inertial chart x′=Bxx, where Bx is the matrix of B(35e1); it exists by Inertial charts are related by proper orthochronous affine Lorentz maps, and every such map is a chart transition. Thus the primed frame basis has e′a=B(−35e1)ea and the primed component columns are [U]x′=Bx[U]x, [J]x′=Bx[J]x; these are changes of components of the same fields. Then:

(i) J has components (J0,J1)=(5ρ0/4,3ρ0/4), and ∂aJa=0 because the components are constant; the continuity equation of Transformation of tensor, covector, derivative and current components under affine Lorentz maps reads ∂0J0+∂1J1=0 identically.

(ii) Under the passive boost B=B(35e1) one computes BU=e0: the first component of BU is (5/4)(5/4−(3/5)(3/4))=(5/4)(16/20)=1, the second is (5/4)(3/4−(3/5)(5/4))=(5/4)⋅0=0, and the transverse components vanish. Correspondingly [J]x′=(ρ0,0,0,0): in the primed chart the field J has those components. Hence the coordinate density J0 is 5ρ0/4 in the original chart and ρ0 in the chart in which U=e′0, while the scalar multiplier ρ0 is the same in both; this is the worked instance of clause (vii) of Transformation of tensor, covector, derivative and current components under affine Lorentz maps.

(iii) The general boost law of clause (vi) of Transformation of tensor, covector, derivative and current components under affine Lorentz maps reproduces (ii): J′0=γ(J0−βJ1)=(5/4)(5ρ0/4−(3/5)(3ρ0/4))=(5/4)(16ρ0/20)=ρ0 and J′1=γ(J1−βJ0)=(5/4)(3ρ0/4−(3/5)(5ρ0/4))=0.

The example carries no physical charge assignment: it illustrates the transformation of a specified vector field, and the identification of J with a physical current is a model premise of the electromagnetic framework.

Facts & Assumptions

Given: An inertial frame with coordinates (x0,x1,x2,x3) and η0=diag⁡(−1,1,1,1), an open set Ω, the constant future unit timelike vector field U with components (5/4,3/4,0,0), a constant scalar field ρ0, and J:=ρ0U.

[L1]

A Ck vector field on Ω is a map assigning to each point a vector, with components in the inertial chart; for a constant field all partial derivatives of the components vanish, and for a C1 vector field J in an inertial chart the continuity equation ∂aJa=0 is equivalent to ∂0J0+∑i∂iJi=0; under the passive boost with parameter β, Λ=B(β), the components transform as J′0=γ(J0−β⋅Jcov) and Jcov′=Jcov+γ−1∣β∣2(β⋅Jcov)β−γβJ0 (Transformation of tensor, covector, derivative and current components under affine Lorentz maps).

[L2]

For a C1 future timelike vector field U and a C1 scalar field ρ0, J=ρ0U is a C1 vector field with first component J0=ρ0U0; the normalized chart direction u:=U/U0 satisfies u0=1 and ρ:=J0=ρ0U0 is the coordinate density, while ρ0 is the scalar multiplier; ρ=ρ0 holds exactly when ρ0(U0−1)=0, hence at points with ρ0≠0 exactly when U0=1 (Transformation of tensor, covector, derivative and current components under affine Lorentz maps).

[L3]

For β=βe1 with ∣β∣<1 the boost acts by x′0=γ(x0−βx1), x′1=γ(x1−βx0) on the (e0,e1) plane with γ=(1−β2)−1/2, and B(β) is a proper orthochronous Lorentz map (The passive Lorentz boost with parameter beta, The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L4]

In the frame, e0∈C+ and the coordinates identify the timelike sets: T+={v:v0>∣vvec∣}, T−={v:v0<−∣vvec∣}, with C+=T+; a nonzero vector with η(v,v)<0 is timelike, and η(v,v)=−(v0)2+(v1)2+(v2)2+(v3)2 (Inertial frames and inertial charts of the affine Minkowski model, The timelike set has exactly two convex components, described in orthonormal coordinates, Time orientation, future and past timelike/null/causal vectors, and the closed cones, Timelike, null, spacelike and causal vectors of a Lorentz vector space).

Proof

technique · direct
1.1L1L2L4algebra

By [L4], η0(U,U)=−(5/4)2+(3/4)2=(−25+9)/16=−1, so U is timelike, and U0=5/4>3/4=∣Uvec∣ places U in T+=C+: U is future timelike unit. Since ρ0 is constant, [L2] gives J=ρ0U with components (J0,J1,J2,J3)=ρ0(5/4,3/4,0,0)=(5ρ0/4,3ρ0/4,0,0), and all partial derivatives of the components vanish; the sum ∂0J0+∑i∂iJi is then a sum of vanishing terms, so the continuity equation ∂aJa=0 holds by [L1].

1.2L2L3algebra

By [L3] with β=(3/5,0,0), γ=5/4: the first component of BU is (5/4)(5/4−(3/5)(3/4))=(5/4)(5/4−9/20)=(5/4)(16/20)=1, the second is (5/4)((3/4)−(3/5)(5/4))=(5/4)(3/4−3/4)=0, and the transverse components of U vanish, so BU=e0. By linearity BJ=ρ0e0 as an active image in V, and by the primed-chart component law [L1], [J]x′=Bx[J]x=(ρ0,0,0,0). Likewise [U]x′=(1,0,0,0), so U=e′0 and J=ρ0e′0 intrinsically. This is clause (ii).

2.1L1step 1.1step 1.2algebra

For β=35e1 the general law of [L1] reduces, since J2=J3=0 and β⋅Jcov=35J1, to J′0=γ(J0−35J1) and J′1=γ(J1−35J0) (the correction term (γ−1)J1e1 combining with J1 to give γJ1). Substituting γ=5/4, J0=5ρ0/4 and J1=3ρ0/4: J′0=(5/4)(5ρ0/4−(3/5)(3ρ0/4))=(5/4)(25ρ0/20−9ρ0/20)=(5/4)(16ρ0/20)=ρ0 and J′1=(5/4)(3ρ0/4−(3/5)(5ρ0/4))=(5/4)(3ρ0/4−3ρ0/4)=0, reproducing the components of step 1.2; this is clause (iii).

3.1step 1.1step 1.2step 2.1∎

Clause (i) is step 1.1, clause (ii) is step 1.2, and clause (iii) is step 2.1. Every entry is an exact rational evaluation of the boost matrix on the displayed components, and no physical identification of J with a charge-current is made.

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Worked motion of the primed frame's origin under a passive boost

Example

Worked computation. Let B=B(βe1) be a passive boost with parameter β∈(−1,1) along the first spatial direction of an inertial frame with coordinates (x0,x1) (The passive Lorentz boost with parameter beta); in matrix expressions write Bx:=E−1∘B∘E and x′=Bxx for the transformed coordinates. Then:

(i) the locus x′1=0 of the primed coordinates (with x′2,x′3 fixed) is, in the unprimed chart, x1=βx0 with x2,x3 fixed constants. The primed spatial origin is the case x2=x3=0; arbitrary fixed transverse coordinates give parallel loci: indeed x′1=γ(x1−βx0)=0 forces x1=βx0, and the transverse coordinates are unchanged because B fixes them. With x0=ct this is uniform motion with coordinate velocity βc along e1;

(ii) conversely the unprimed origin x1=0 satisfies x′1=γ(0−βx0)=−βγx0=−βx′0, since x′0=γx0 when x1=0; so the unprimed frame moves with parameter −β in the primed chart, consistent with B−1=B(−β);

(iii) for β=3/5 the concrete values are x1=(3/5)x0 in the first chart and x′1=−(3/5)x′0 in the second; the locus x′1=0 passes through the event (x0,x1)=(0,0) and through (5,3) (choosing x0=5), which the boost maps to (x′0,x′1)=((5/4)(5−(3/5)3),(5/4)(3−(3/5)5))=(4,0), confirming x′1=0.

Facts & Assumptions

Given: An inertial frame (o;e0,e1,e2,e3) with coordinates x=(x0,x1,x2,x3), a parameter β∈(−1,1), and the boost B:=B(βe1); write x′=Bxx.

[L1]

For β=βe1 the boost acts by x′0=γ(x0−βx1) and x′1=γ(x1−βx0), with γ=(1−∣β∣2)−1/2=(1−β2)−1/2>0, and it fixes the transverse coordinates: x′2=x2, x′3=x3; the Euclidean norm ∣β∣ is that of the frame's spatial coordinates (The passive Lorentz boost with parameter beta, The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn).

[L2]

B(βe1) is a proper orthochronous Lorentz map and B(βe1)−1=B(−βe1); the maps B(βe1) are injective in β (The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L3]

An inertial frame has η-orthonormal basis with η(ea,eb)=diag⁡(−1,1,1,1)ab, and its chart x(p)=E−1(p−o) is affine (Inertial frames and inertial charts of the affine Minkowski model).

[L4]

Fix a real constant c>0 and define the reparametrized temporal coordinate t:=x0/c, so x0=ct. Coordinate velocity means the derivative dx1/dt with respect to this coordinate. This rescaling makes no assertion about physical clock readings or the empirical value or SI units of c.

[L5]

For every proper orthochronous linear Lorentz map L:V→V and every c0∈A the tuple (c0;Le0,Le1,Le2,Le3) is an inertial frame whose chart xL satisfies xL(p)=ML−1(x(p)−x(c0)), where ML=E−1∘L∘E (Inertial charts are related by proper orthochronous affine Lorentz maps, and every such map is a chart transition).

Proof

technique · direct
1.1L1L4algebra

By [L1], x′1=γ(x1−βx0) and γ>0, so x′1=0 holds exactly when x1=βx0; the equations x′2=x2, x′3=x3 of [L1] show that the transverse coordinates are arbitrary constants, unchanged by the boost. Under the convention x0=ct of [L4], the locus is x1=βct, that is, uniform motion along e1 with coordinate velocity dx1/dt=βc.

1.2L1L2algebra

For the unprimed origin set x1=0 (with x2=x3=0). [L1] gives x′0=γx0 and x′1=−γβx0=−βx′0, so in the primed coordinates the unprimed origin moves along the locus x′1=−βx′0. This is consistent with [L2]: applying the inverse boost B(−βe1) to a point with x′1=−βx′0 gives x1=γ(x′1+βx′0)=γ(−βx′0+βx′0)=0, so the inverse map sends that locus to the unprimed origin, as B−1=B(−β) requires.

1.3L1L2L3L5algebra

Finally, the transformed coordinates x′=Bxx are the coordinates of an inertial chart of the model: by [L2] the map L:=B(−βe1) is proper orthochronous, so [L5] applied with c0:=o provides an inertial frame whose chart is xL(p)=ML−1(x(p)−x(o))=Bx(βe1)x(p)=x′(p), where Bx denotes the boost matrix and ML=Bx(−βe1) for every event p, using x(o)=0 and L−1=B(βe1). Thus the worked motion is the motion of the primed inertial chart's spatial origin in the unprimed chart, and conversely; the computation uses only the 1+1 boost formulas and the affine chart structure.

2.1L1step 1.1step 1.2algebra

For β=3/5 one has γ=(1−9/25)−1/2=(16/25)−1/2=5/4; step 1.1 gives the locus x1=(3/5)x0 and step 1.2 gives x′1=−(3/5)x′0. The event (x0,x1)=(5,3) satisfies 3=(3/5)⋅5, so it lies on the locus, and [L1] gives x′0=(5/4)(5−(3/5)3)=(5/4)(5−9/5)=(5/4)(16/5)=4 and x′1=(5/4)(3−(3/5)5)=(5/4)(3−3)=0; the origin (0,0) is fixed because the boost is linear. Hence the displayed values of clause (iii) are correct.

3.1step 1.1step 1.2step 2.1step 1.3∎

Steps 1.1-1.2 and 2.1 verify clauses (i)-(iii) of the Example and step 1.3 identifies the primed coordinates with an inertial chart; every assertion is a finite computation with the displayed rational values.

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Counterexample: a spatial rotation is a proper orthochronous Lorentz map that is not a boost

Statement refuted

The statement "every proper orthochronous Lorentz map is a passive boost B(β) for some β with ∣β∣<1" is false.

Facts & Assumptions

Given: An inertial frame (o;e0,e1,e2,e3) with η(ea,eb)=diag⁡(−1,1,1,1)ab, and the linear maps defined below by their action on the frame basis.

[L1]

A linear Lorentz map is a linear L:V→V with η(Lv,Lw)=η(v,w) for all v,w; it is proper when det⁡L=+1 and orthochronous when L(C+)=C+ (Linear and affine Lorentz maps, and the Lorentz group).

[L2]

A spatial rotation of the frame is a linear R:V→V with R(e0)=e0, R(S)⊆S, δ(Rw,Rz)=δ(w,z) on the spatial three-space S, and det⁡(R∣S)=+1 in (e1,e2,e3) (Spatial rotations of an inertial frame).

[L3]

The extension R^=diag⁡(1,R) of a determinant-one δ-isometry R:S→S is a proper orthochronous Lorentz map (Unique boost-rotation factorization of proper orthochronous Lorentz maps).

[L4]

For every ∣β∣<1 the passive boost satisfies B(β)e0=γ(e0−β1e1−β2e2−β3e3) and B(0)=id⁡V; the map β↦B(β)e0 is injective (The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L5]

The determinant of a square matrix is computed by Laplace cofactor expansion, det⁡(A)=∑kakjCkj(A) along the column j (Laplace expansion computes the determinant along every row and every column over a commutative ring, Deleted-row-and-column minors, cofactors, the cofactor matrix and the adjugate over a commutative ring); in particular det⁡(abcd)=ad−bc by expanding along the first column, since the (0,0)-cofactor is d and the (1,0)-cofactor is −b.

Counterexample

1.1L2L5algebra

Define R^:V→V by R^e0=e0, R^e1=e2, R^e2=−e1, R^e3=e3. In the ordered basis (e0,e1,e2,e3) its matrix is (100000−1001000001), and its restriction to S=span⁡(e1,e2,e3) has the matrix A=(0−10100001) in (e1,e2,e3). The columns of A are orthonormal in the standard inner product on R3 (they are (0,1,0), (−1,0,0), (0,0,1)), so ATA=I3 and A preserves the form δ on S; and Laplace expansion along the third column gives det⁡A=1⋅det⁡(0−110)=1⋅(0⋅0−(−1)⋅1)=1 by [L5]. Hence R^(e0)=e0, R^(S)=S, δ(R^w,R^z)=δ(w,z) for all w,z∈S, and det⁡(R^∣S)=+1: all four defining conditions of [L2] hold, so R^ is a spatial rotation of the frame.

2.1L1L3step 1.1

By step 1.1 and [L3], the extension R^=diag⁡(1,R^∣S) is a proper orthochronous Lorentz map. In particular R^ is a member of SO+(V,η) in the sense of [L1], and it is not the identity because R^e1=e2≠e1.

3.1L4step 1.1step 2.1algebra

Suppose, for contradiction, that R^=B(β) for some β with ∣β∣<1. Applying both sides to e0 and using [L4] gives γ(e0−β1e1−β2e2−β3e3)=e0; comparing the e0 coordinates gives γ=1 and then comparing the spatial coordinates gives γβi=0 for i=1,2,3, hence β=0 and B(β)=B(0)=id⁡V by [L4]. But R^e1=e2≠e1=id⁡V(e1), a contradiction. Hence R^ is a proper orthochronous Lorentz map that is not a passive boost, and the refuted statement is false.

4.1L3step 2.1algebra∎

Consequently the set of passive boosts is a proper subset of SO+(V,η): the strict inclusion is witnessed by R^. The factorization theorem L=B(−β)R^β of Unique boost-rotation factorization of proper orthochronous Lorentz maps therefore cannot drop its rotation factor in general: for L=R^ the theorem produces the factorization R^=B(0)R^ with rotation factor R^∣S≠id⁡S, and by clause (v) of that theorem L is a passive boost if and only if that factor is the identity.

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