The Lorentz Transformation — Question 8

Section 37-6 · Physics pages 6146–6147

Question 8 Null coordinates turn a boost into reciprocal scale changes Define U = ct −x and V = ct + x. Use ct′ = γ(ct −βx), x′ = γ(x −βct), y′ = y, z′ = z, where β = v/c and γ = (1 −β2)−1/2. x ct V = 0 U = 0 Spacetime sketch; the vertical coordinate is ct. Labels specify events and worldlines. (a) Derive the transformations of U and V. (b) For β = 0.60, find both scale factors and show the interval product is unchanged.
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Question 8 - Solution Null coordinates turn a boost into reciprocal scale changes x ct V = 0 U = 0 (a) Method and calculation U′ = ct′ −x′ = γ(1 + β)U. V ′ = ct′ + x′ = γ(1 −β)V. (b) Interpretation and further calculation U′ = 2U , V ′ = V/2 . U′V ′ = UV = c2t2 −x2 . Physical check One null coordinate expands by precisely the factor that contracts the other.

Question 8

Null coordinates turn a boost into reciprocal scale changes

Define U=ct−xU=ct-x and V=ct+xV=ct+x. Use ct′=γ(ct−βx)ct\prime=\gamma(ct-\beta x), x′=γ(x−βct)x\prime=\gamma(x-\beta ct), y′=yy\prime=y, z′=zz\prime=z, where β=v/c\beta=v/c and γ=(1−β2)−1/2\gamma=(1-\beta^2)^{-1/2}.

Diagram shown in the original SVG worksheet.

  1. Derive the transformations of U and V.

  2. For β=0.60\beta=0.60, find both scale factors and show the interval product is unchanged.

Question 8 - Solution

Null coordinates turn a boost into reciprocal scale changes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

U′=ct′−x′=γ(1+β)U.U\prime=ct\prime-x\prime=\gamma(1+\beta)U.V′=ct′+x′=γ(1−β)V.V\prime=ct\prime+x\prime=\gamma(1-\beta)V.

(b) Interpretation and further calculation

U′=2U,V′=V/2.\boxed{U\prime=2U},\qquad\boxed{V\prime=V/2}.U′V′=UV=c2t2−x2.\boxed{U\prime V\prime=UV=c^2t^2-x^2}.

Physical check

One null coordinate expands by precisely the factor that contracts the other.