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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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