Velocity Addition

We have derived the velocity addition formula using two Lorentz transformations (in the same direction) and rapidity.

\bgroup\color{black}$ \displaystyle \beta=\tanh(\phi_1+\phi_2)=\frac{\tanh\phi_1...
...}{1 + \tanh\phi_1 \tanh\phi_2}
={\beta_1+\beta_2\over 1+\beta_1\beta_2} $\egroup

We can also compute this in terms of the usual \bgroup\color{black}$ {dx_i\over dt}$\egroup going from one frame in which an object is moving to a frame boosted along the \bgroup\color{black}$ x$\egroup direction. (This should give the same result as above if the particle is moving in the \bgroup\color{black}$ -x$\egroup direction.)

Let the velocity in the original system be \bgroup\color{black}$ \vec{u}={d\vec{x}\over dt}$\egroup.

  $\displaystyle u'_i={dx'_i\over dt'}$    
  $\displaystyle u'_1={\gamma\left(dx_1-vdt\right)\over \gamma\left(dt-{v\over c^2...
...\over dt}-v\over 1-{v\over c^2}{dx_1\over dt}} = {u_1-v\over 1-{vu_1\over c^2}}$    
  $\displaystyle u'_2={dx_2\over \gamma\left(dt-{v\over c^2}dx_1\right)}={{dx_2\ov... c^2}{dx_1\over dt}\right)} = {u_2\over \gamma\left(1-{vu_1\over c^2}\right)}$    
  $\displaystyle u'_3= {u_3\over \gamma\left(1-{vu_1\over c^2}\right)}$    

This formula gives the same result as above but is more general since it allows us to add velocities that are in different directions.

\bgroup\color{black}$ \displaystyle u'_i={dx'_i\over dt'} $\egroup
\bgroup\color{black}$ \displaystyle u'_1= {u_1-v\over 1-{vu_1\over c^2}} $\egroup
\bgroup\color{black}$ \displaystyle u'_2= {u_2\over \gamma\left(1-{vu_1\over c^2}\right)} $\egroup
\bgroup\color{black}$ \displaystyle u'_3= {u_3\over \gamma\left(1-{vu_1\over c^2}\right)} $\egroup

Jim Branson 2012-10-21