Question

In a particle collider experiment, particle I is moving to the right at 0.99c and particle 2 to the left at 0.99c, both relative to the laboratory. What is the relative velocity of the two panicles according to (an observer moving with) particle 2?

Short answer

The velocity of the particle 1 and 2 to the observer is 0 and 0.9999c respectively.

Step-by-step solution

Write the given data from the question.

The velocity of the particle 1 is 0.99c.

The velocity of the particle 2 is 0.99c .

Determine the formulas to calculate the velocity of the two-particle to the overseer.

The expression for the Lorentz velocity transformation is given as follows.

$\mathbf{u}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{u}\mathbf{-}\mathbf{v}}{\mathbf{1}\mathbf{-}{\displaystyle \frac{\mathbf{uv}}{{\mathbf{c}}^{\mathbf{2}}}}}$ …… (i)

Here, u' is the velocity of the object in frame s' , u is the velocity of object in frame S , v is the velocity of frame S' relative to S and c is the speed of the light.

Calculate the velocity of the two-particle to observer.

Calculate the velocity of the particle which to observer moving with particle 2.

Substitute 0.99c for u and 0.99c for v into equation (i).

$$\begin{gathered} u\text{'}=\frac{0.99c-0.99c}{1-{\displaystyle \frac{\left(0.99c\right)\left(0.99c\right)}{c^2}}} \\ u\text{'}=\frac{0}{1-0.99\times 0.99} \\ u\text{'}=0 \end{gathered}$$

Therefore, the velocity of the particle 2 to the observer is 0 .

Calculate the velocity of the particle 1 to the observer.

Substitute -0.99c for u and 0.99c for v into equation (i).

$$\begin{gathered} u\text{'}=\frac{-0.99c-0.99c}{1-{\displaystyle \frac{\left(-0.99c\right)\left(0.99c\right)}{c^2}}} \\ u\text{'}=\frac{-1.98c}{1+0.9801} \\ u\text{'}=\frac{-1.98c}{1.9801} \\ u\text{'}=0.9999c \end{gathered}$$

Therefore, the velocity of the particle 1 to the observer is 0.9999c.

Hence the velocity of the particle 1 and 2 to the observer is 0 and 0.9999c respectively.