# Lorentz-Transformation of Velocity

In a reference system $\Sigma$ a particle has velocity $\vec{v} = v \hat{z}$. What is the velocity of the particle in a system $\Sigma'$ moving with velovity $\vec{u} = u \hat{z}$ relative to $\Sigma$?

One takes the 4-vector $v^\mu = \gamma(1,\vec{v}/c) = \gamma(1,0,0,v/c)$ and Lorentz-transforms it according to:

$$v^{\mu '} = \begin{pmatrix} \gamma & 0 & 0 & - \beta \gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - \beta \gamma & 0 & 0 & \gamma \end{pmatrix} \begin{pmatrix} \gamma \\ 0 \\ 0 \\ \gamma v /c \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ \gamma^2 (-\beta + v / c ) \end{pmatrix} \cong \begin{pmatrix} \gamma \\ 0 \\ 0 \\ \gamma v' /c \end{pmatrix}$$

and therefore $v' = \gamma (-u + v).$

However the known formula for relativistic velocity transformations states: $$v' = \frac{v + u}{1 + \frac{v u }{c^2}},$$

which is not the same as above. Where did I do a mistake?

• Thus: $$v^{\mu '} = \begin{pmatrix} \gamma & 0 & 0 & - \beta \gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - \beta \gamma & 0 & 0 & \gamma \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ v /c \end{pmatrix} = \begin{pmatrix} \gamma - \beta \gamma v/c \\ 0 \\ 0 \\ \gamma (-\beta + v / c ) \end{pmatrix} \cong \begin{pmatrix} \gamma' \\ 0 \\ 0 \\ \gamma' v' /c \end{pmatrix}$$ – Breaking Mad Aug 28 '16 at 12:01
• and this gives: $\gamma' = \gamma (1 - u v /c^2)$ and thus, $$v' = \frac{- u + v}{1- \frac{uv}{c^2}}.$$ The minus sign of $u$ arises probably due to another sign convention, or not? – Breaking Mad Aug 28 '16 at 12:03
• @Breaking M_a_t There are three different gamma's: in the 4x4 matrix you have $\gamma_u = (1 - u^2/c^2)^{-1/2}$, in the first 4x1 matrix there's a $\gamma_v = (1 - v^2/c^2)^{-1/2}$, and in the last 4x1 matrix there's a $\gamma' = (1 - v'^2/c^2)^{-1/2}$. – Pulsar Aug 28 '16 at 13:00