A relativistic particle with momentum p and total energy E moves along the x axis of the frame K. Demonstrate that in the frame `K^'` moving with a constant velocity `V` relative to the frame K in the positive direction of its axis x the momentum and the total energy of the given particle are defined by the formula:
`p_x^'=(p_x-EV//c^2)/(sqrt(1-beta^2))`, `E^'=(E-p_xV)/(sqrt(1-beta^2))`
where `beta=V//c`

Explanation

By definition,
`E=m_0(c^2)/(sqrt(1-(v_x^2)/(c^2)))=(m_0c^3dt)/(ds), p_x=m_0(v_x)/(sqrt(1-v^2/c^2))=(cm_0dx)/(ds)`
where `ds^2=c^2dt^2-dx^2` is the invariant interval `(dy=dz=0)`
Thus, `p_x^'=cm_0(dx^')/(ds)=cm_0gamma((dx-Vdt))/(ds)=(p_x-VE//c^2)/(sqrt(1-V^2//c^2))`
`E^'=m_0c^3(dt^')/(ds)-c^3m_0gamma((dt-(Vd_x)))/(ds)=(E-Vp_x)/(sqrt(1-V^2/c^2))`