Question

A laboratory experiment shoots an electron to the left at $0.90c$. What is the electron’s speed, as a fraction of c, relative to a proton moving to the right at $0.90c$?

Short answer

The electron's speed is $0.994c$.

Step-by-step solution

Given Information

We have given that a laboratory experiment shoots an electron to the left at $0.90c$.

We have to find the electron’s speed, as a fraction of $c$, relative to a proton moving to the right at $0.90c$.

Simplify

Frame of Earth : $S$

Frame of Proton :$S\text{'}$

$S$moves at$v=0.9c$. In the laboratory frame, the electron's velocity is $-0.9c$.

Using, the Lorentz velocity transformation equation,

$$\begin{gathered} u\text{'}=\frac{u-v}{1-{\displaystyle \frac{uv}{c^2}}} \\ u\text{'}=\frac{-0.9c-0.9c}{1-{\displaystyle \frac{(-0.9c)(0.9c)}{c^2}}} \\ u\text{'}=-0.994c. \end{gathered}$$

Therefore the electron's speed is $0.994c$.