The Relativity of Velocities — Question 10

Question 10

Velocity uncertainty is compressed nonlinearly by a known boost

For fixed $v=0.60c$, a collinear launch has $u\prime=0.50c$ with small standard uncertainty $\sigma_{u\prime}=0.001c$. Treat the boost as exactly known. For $S\prime$ moving at +v along x, use $u_x=(u_x\prime+v)/(1+vu_x\prime/c^2)$ and $u_y=u_y\prime/[\gamma_v(1+vu_x\prime/c^2)]$, with $\gamma_v=(1-v^2/c^2)^{-1/2}$.

1. Find the laboratory velocity and derivative $du/du\prime$.

2. Propagate the uncertainty to first order and explain why simple addition would give a different uncertainty.

Question 10 - Solution

Velocity uncertainty is compressed nonlinearly by a known boost

(a) Method and calculation

$$\boxed{u/c=\frac{.50+.60}{1+.60(.50)}=11/13}.$$$$\boxed{\frac{du}{du\prime}=\frac{1-v^2/c^2}{(1+vu\prime/c^2)^2}=\frac{64}{169}}.$$

(b) Interpretation and further calculation

$$\sigma_u\simeq\left|du/du\prime\right|\sigma_{u\prime}=\boxed{0.000378698c}.$$$$u_{\mathrm{classical}}=u\prime+v\quad\Rightarrow\quad\sigma_{u,\mathrm{classical}}=0.001c.$$

Physical check

This is a local linear propagation of a small statistical uncertainty, not an exact bound for a broad velocity distribution.