discuss the addition higher velocities and Prove the second postulate of special theory of relativity
Answers
In relativistic physics, a velocity-addition formula is a three-dimensional equation that relates the velocities of objects in different reference frames. Such formulas apply to successive Lorentz transformations, so they also relate different frames. Accompanying velocity addition is a kinematic effect known as Thomas precession, whereby successive non-collinear Lorentz boosts become equivalent to the composition of a rotation of the coordinate system and a boost.
Standard applications of velocity-addition formulas include the Doppler shift, Doppler navigation, the aberration of light, and the dragging of light in moving water observed in the 1851 Fizeau experiment.
{\boldsymbol {\mathbf {P} }}=m_{0}{\boldsymbol {\mathbf {U} }}=(E/c,\mathbf {p} )
The energy and momentum of an object with invariant mass m0 (also called rest mass), moving with velocity v with respect to a given frame of reference, are respectively given by
{\displaystyle {\begin{aligned}E&=\gamma (\mathbf {v} )m_{0}c^{2}\\\mathbf {p} &=\gamma (\mathbf {v} )m_{0}\mathbf {v} \end{aligned}}}{\begin{aligned}E&=\gamma (\mathbf {v} )m_{0}c^{2}\\\mathbf {p} &=\gamma (\mathbf {v} )m_{0}\mathbf {v} \end{aligned}}
The factor of γ(v) comes from the definition of the four-velocity described above. The appearance of the γ factor has an alternative way of being stated, explained in the next section.
The kinetic energy, K, is defined as
{\displaystyle K=(\gamma -1)m_{0}c^{2}=E-m_{0}c^{2}\,,}{\displaystyle K=(\gamma -1)m_{0}c^{2}=E-m_{0}c^{2}\,,}
And the speed as a function of kinetic energy is given by
{\displaystyle v={\frac {c{\sqrt {K(K+2m_{0}c^{2})}}}{K+m_{0}c^{2}}}={\frac {c{\sqrt {(E-m_{0}c^{2})(E+m_{0}c^{2})}}}{E}}={\frac {pc^{2}}{E}}\,.}{\displaystyle v={\frac {c{\sqrt {K(K+2m_{0}c^{2})}}}{K+m_{0}c^{2}}}={\frac {c{\sqrt {(E-m_{0}c^{2})(E+m_{0}c^{2})}}}{E}}={\frac {pc^{2}}{E}}\,.}