The Lorentz Transformation — Question 4

Question 4

A shifted origin must be subtracted before applying the boost

The common zero event of two frames is at $(ct_0,x_0)=(600,300)\,\mathrm m$ in a laboratory chart. An event E has $(ct,x)=(1200,900)\,\mathrm m$. The new frame moves at $0.60c$. 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. Write the affine coordinate transform relative to the common zero event.

2. Find E in the new chart and classify its interval from the zero event.

Question 4 - Solution

A shifted origin must be subtracted before applying the boost

(a) Method and calculation

$$ct\prime=\gamma[(ct-ct_0)-\beta(x-x_0)].$$$$x\prime=\gamma[(x-x_0)-\beta(ct-ct_0)].$$

(b) Interpretation and further calculation

$$\boxed{ct\prime=x\prime=300\,\mathrm m}.$$$$s^2=(600)^2-(600)^2=\boxed{0\quad\text{(null)}}.$$

Physical check

Translations change individual coordinates but leave event differences and their intervals unchanged.