The Lorentz Transformation — Question 3

Question 3

The three-dimensional interval checks a transformed displacement

Two events differ by $(c\Delta t,\Delta x,\Delta y,\Delta z)=(1000,600,300,0)\,\mathrm m$. Apply a boost at $0.60c$ along x. 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. Find the transformed displacement and interval squared.

2. Find the elapsed proper time of an inertial clock connecting the events; take $c=3.00\times10^8\,\mathrm{m/s}$.

Question 3 - Solution

The three-dimensional interval checks a transformed displacement

(a) Method and calculation

$$(c\Delta t\prime,\Delta x\prime,\Delta y\prime,\Delta z\prime)=(800,0,300,0)\,\mathrm m.$$$$\boxed{s^2=1000^2-600^2-300^2=800^2-300^2=550000\,\mathrm m^2}.$$

(b) Interpretation and further calculation

$$c\Delta\tau=\sqrt{s^2}=100\sqrt{55}\,\mathrm m.$$$$\boxed{\Delta\tau=\frac{\sqrt{55}}3\,\mu\mathrm s\simeq2.47207\,\mu\mathrm s}.$$

Physical check

Eliminating x motion does not eliminate y motion. The chosen primed frame is not the connecting clock rest frame.