Adittional ways of writing down the gamma in relation to speen direction

I have managed to proove below that equation $\gamma(v')=\gamma \gamma(v)\left(1-v_x \, u\!/\!c^2\right)$ holds if whole speed $v$ is in the direction of axis $x,x'$. So $v$ equals $v_x$ and therefore $v_x'=\frac{v_x-u}{\left(1-v_x\, u\!/\!c^2\right)}$.

$u$ is a relative speed between primed and unprimed coordinate systems.

$v$ whole speed.

$\gamma=\frac{1}{\sqrt{1-\,u^2\!/\!c^2}}$

$\gamma(v)=\frac{1}{\sqrt{1-\,v^2\!/\!c^2}}$.

The proof:

$$\scriptsize \begin{split} \gamma(v_x') &= \gamma \, \gamma(v_x) \! \left( 1 - v_x \frac{u}{c^2} \right)\\ \frac{1}{\sqrt{1-\frac{{v_x'}^2}{c^2}}} &= \frac{1}{\sqrt{1-\frac{u^2}{c^2}}} \frac{1}{\sqrt{1- \frac{{v_x}^2}{c^2}}} \! \left( 1 - v_x \frac{u}{c^2} \right)\\ \frac{1}{\sqrt{1-\frac{ \left[ \bigl( v_x -u \bigl) / \left(1 - v_x\frac{u}{c^2}\right)\right]^2}{c^2}}} &= \frac{1}{\sqrt{1-\frac{u^2}{c^2}}} \frac{1}{\sqrt{1- \frac{{v_x}^2}{c^2}}} \! \left( 1 - v_x \frac{u}{c^2} \right)\\ \frac{1}{1-\frac{\bigl( v_x -u \bigl)^2 / \left(1 - v_x\frac{u}{c^2}\right)^2}{c^2}} &= \frac{1}{1-\frac{u^2}{c^2}} \frac{1}{1- \frac{{v_x}^2}{c^2}} \! \left( 1 - v_x \frac{u}{c^2} \right)^2\\ \frac{1}{\frac{ c^2 - \bigl( v_x -u \bigl)^2 / \left(1 - v_x\frac{u}{c^2}\right)^2}{c^2}} &= \frac{1}{1-\frac{u^2}{c^2} - \frac{{v_x}^2}{c^2} + \frac{u^2 {v_x}^2}{c^4}} \left( 1 - v_x \frac{u}{c^2} \right)^2\\ \frac{c^2}{c^2 - \frac{\bigl( v_x -u \bigl)^2}{\left(1 - v_x\frac{u}{c^2}\right)^2}} &= \frac{c^2}{c^2-u^2 - {v_x}^2 + \frac{u^2 {v_x}^2}{c^2}} \left( 1 - v_x \frac{u}{c^2} \right)^2\\ \frac{c^2}{\frac{c^2 \left( 1 - v_x \frac{u}{c^2}\right)^2 - \bigl( v_x -u \bigl)^2}{\left(1 - v_x\frac{u}{c^2}\right)^2}} &= \frac{c^2}{c^2-u^2 - {v_x}^2 + \frac{u^2 {v_x}^2}{c^2}} \left( 1 - v_x \frac{u}{c^2} \right)^2\\ \frac{c^2}{c^2 \left( 1 - v_x \frac{u}{c^2}\right)^2 - \bigl( v_x -u \bigl)^2} \left(1 - v_x\frac{u}{c^2}\right)^2 &= \frac{c^2}{c^2-u^2 - {v_x}^2 + \frac{u^2 {v_x}^2}{c^2}} \left( 1 - v_x \frac{u}{c^2} \right)^2\\ \frac{1}{c^2 \left( 1 - 2v_x \frac{u}{c^2} + {v_x}^2\frac{u^2}{c^4}\right) - \bigl( {v_x}^2 - 2 v_x u + u^2 \bigl)} &= \frac{1}{c^2-u^2 - {v_x}^2 + \frac{u^2 {v_x}^2}{c^2}} \\ \frac{1}{c^2 - 2v_x u + \frac{u^2 {v_x}^2}{c^2} - {v_x}^2 + 2 v_x u - u^2} &= \frac{1}{c^2-u^2 - {v_x}^2 + \frac{u^2 {v_x}^2}{c^2}} \\ \frac{1}{c^2 - u^2 - {v_x}^2 + \frac{u^2 {v_x}^2}{c^2}} &= \frac{1}{c^2-u^2 - {v_x}^2 + \frac{u^2 {v_x}^2}{c^2}} \\ \end{split}$$

Question 1: Does this equation hold also if $v$ equals $v_y$ and $v_y'=\frac{v_y}{\gamma\left(1-v_x\, u\!/\!c^2 \right)}$ ?

Question 2: Does this equation holds only if $u$ is paralel to axix $x,x'$? How does it change if $u$ is parallel to axis $y,y'$?

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 I'm not sure from where you get the very first relation you started from, but such problems usually solved by simply writing down Lorenz transformation between primed and unprinted systems and differentiating them straight ahead to get speeds, and obviously you should get different results for $v_{x},v_{y}$ – TMS Dec 2 '12 at 16:41