The Relativity of Velocities — Question 9

Question 9

A frame can remove the longitudinal component without removing all motion

A particle has laboratory components $u_x=0.60c$ and $u_y=0.30c$. Choose a boost along x so that $u_x\prime=0$. Use the inverse forms of the velocity transformations.

1. Find the required boost and the residual transverse speed.

2. Explain why this is not the full particle rest frame and find its transverse Lorentz factor in the chosen frame.

Question 9 - Solution

A frame can remove the longitudinal component without removing all motion

(a) Method and calculation

$$0=\frac{u_x-v}{1-vu_x/c^2}\quad\Rightarrow\quad\boxed{v=0.60c}.$$$$u_y\prime=\frac{u_y}{\gamma_v(1-vu_x/c^2)}=\frac{.30c}{1.25(.64)}=\boxed{.375c}.$$

(b) Interpretation and further calculation

$$\boxed{u_x\prime=0,\quad u_y\prime\ne0}.$$$$\boxed{\gamma_{u\prime}=\frac1{\sqrt{1-(3/8)^2}}=\frac8{\sqrt{55}}}.$$

Physical check

One x-directed boost removes only the longitudinal component. A further transverse boost would be needed to reach rest.