The Lorentz Transformation — Question 2

Question 2

Recovering an event uses the inverse boost sign

In $S\prime$, an event has $(ct\prime,x\prime)=(500,100)\,\mathrm m$. The primed frame moves at $+0.60c$ relative to S. Use $ct\prime=\gamma(ct-\beta x)$, $x\prime=\gamma(x-\beta ct)$, $y\prime=y$, $z\prime=z$, where $\beta=v/c$ and $\gamma=(1-\beta^2)^{-1/2}$.

1. Derive the inverse equations by replacing the boost velocity with its negative.

2. Find the original coordinates and verify one forward transformation.

Question 2 - Solution

Recovering an event uses the inverse boost sign

(a) Method and calculation

$$\boxed{ct=\gamma(ct\prime+\beta x\prime)},\qquad\boxed{x=\gamma(x\prime+\beta ct\prime)}.$$$$\gamma=1.25,\qquad\beta=0.60.$$

(b) Interpretation and further calculation

$$\boxed{ct=700\,\mathrm m},\qquad\boxed{x=500\,\mathrm m}.$$$$x\prime=1.25(500-0.60\cdot700)=100\,\mathrm m.$$

Physical check

The inverse changes the boost sign; it does not simply interchange coordinate labels while retaining the same minus signs.