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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 |
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Answer» Solution :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))` |
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