Question: In the frame of reference shown, a stationary particle of mass m₀ explodes into two identical particles of mass m moving in opposite directions at 0.6c . Momentum is obviously conserved in this frame. Verify explicitly that it is conserved in a frame of reference moving to the right at 0.6c .

A stationary particle of mass in a lab's frame explodes into two identical particles moving in opposite directions assumed on the x-axis at . In a moving frame, initially, a particle of mass is moving along the negative x-axis at a velocity of . After the explosion, the particle at the right is stationary and the particle at the left is moving along a relative velocity of

$$\begin{gathered} v_r=-\left(\frac{0.6c+0.6c}{1+{\left(0.6\right)}^2}\right) \\ =-0.882c \end{gathered}$$

In the frame moving at to the right, the particle that was stationary in the lab frame is now moving at to the left. Hence the initial momentum of the system will be

$$\begin{gathered} P_{initial}={\gamma }_{0.6c}{m}_o\left(-0.6c\right) \\ =-\frac{5}{4}{m}_o\left(0.6c\right) \\ =-0.75m_{o}c \end{gathered}$$

After the explosion, the particle of mass is moving to the left at a relative speed of. The momentum after the explosion is

$$\begin{gathered} P_{final}={\gamma }_{0.88c}m\left(-0.882c\right) \\ =\left(\frac{1}{\sqrt{1-{\left(0.882\right)}^2}}\right)\left(-0.882c\right)m \\ =-1.872mc \end{gathered}$$

Total Energy will be conserved in the lab as well as moving frame. For simplicity, we are using it in the lab frame.

$$\begin{gathered} m_o{c}^2={\gamma }_{0.6c}m{c}^2+{\gamma }_{0.6c}m{c}^2 \\ {m}_o=2{\gamma }_{0.6c}m \\ {m}_o=2\left(\frac{5}{4}\right)m \\ m=0.4m_o \end{gathered}$$

Putting relation of in expression for final momentum.

$$\begin{gathered} P_{final}=-0.75m_{0}c \\ {P}_{initial}=P_{final} \end{gathered}$$

Hence, momentum is conserved in the frame moving at 0.6c to right.