Question

An observer in frame$S\text{'}$is moving to the right (+x-direction) at speed u= 0.600caway from a stationary observer in$S\text{'}$ frame S. The observer inmeasures the speed $v\text{'}$of a particle moving to the right away from her . What speed v does the observer in Smeasure for the particle$v\text{'}=0.990c$?

Short answer

The speed of the observer in Smeasure for the particle $v^{\text{'}}=0.990c$ is 0.9997 c .

Step-by-step solution

Velocity transformation in relativity

When two particle or a body moves with a velocity which is approaching to speed of light then normal method of finding relative velocity does not work as it exceeds the velocity of body greater than velocity of light. Therefore relative velocity for two particles moving with velocity approaching to speed of light is given by velocity addition method.

Velocity addition method

According the velocity addition method the velocity of particle in lab frame observed by observer in moving frame is given by

$u_x^{\text{'}}=\frac{u_x-v}{1-{\displaystyle \frac{v{u}_x}{c^2}}}----\left(i\right)$

For the velocity of particle moving in moving frame or relativistic frame observed by observer in rest frame is given by inverse Lorentz transformation of first(i) formula

That is

$u_x=\frac{u_x^{\text{'}}+v}{1+{\displaystyle \frac{v{u}_x^{\text{'}}}{c^2}}}$

Where, ${u^{\text{'}}}_x$is the velocity of body in $S\text{'}$frame (moving frame),$u_x$ is velocity of body in frame S (lab frame) and v is velocity of the moving frame.

The calculation of speed of observer in S measure for the particle

$v\text{'}=0.990c$

Using

$u_x=\frac{u_c^{\text{'}}+v}{1+{\displaystyle \frac{v{u}_x^{\text{'}}}{c^2}}}$

Now put the value of constants in above equation

$$\begin{gathered} u_x=\frac{0.99c+0.6c}{1+{\displaystyle \frac{0.6c\times 0.99c}{c^2}}} \\ {u}_x=\frac{795}{797}c=0.997c \end{gathered}$$

Thus, the speed of the observer in Smeasure for the particle $v^{\text{'}}=0.990c$is 0.997 c .