Relativistic Momentum: Deriving Why p=mγv

[71GA]

I have been trying to derive why relativistic momentum is defined as $p=\gamma mv$.

I set up a collision between 2 same balls ($m_1 = m_2 = m$). Before the collision these two balls travel one towards another in $x$ direction with velocities ${v_1}_x = (-{v_2}_x) = v$. After the collision these two balls travel away from each other with velocity ${v_1}_y = (-{v_2}_y) = v$. Coordinate system travels from left to right with velocity $u=v$ at all times (after and before collision).

Please see the pictures below where picture (a) shows situation before collision and picture (b) after collision.

http://shrani.si/f/2A/m3/4kDXDQo1/momentum.png

Below is a proof that Newtonian momentum $mv$ is not preserved in coordinate system $x'y'$. I used $[\, | \,]$ to split $x$ and $y$ components. $p_z'$ is momentum before collision where $p_k'$ is momentum after collision.

$\scriptsize<br /> \begin{split}<br /> p_z' &= \left[ m_1 {v_1}_x' + m_2 {v_2}_x'\, \biggl| \, 0 \right] = \left[ m_1 0 + m_2 \left( \frac{{v_2}_x - u}{1-{v_2}_x\frac{u}{c^2}} \right)\, \biggl| \, 0 \right]= \left[ m \left( \frac{-v - v}{1+ v \frac{v}{c^2}} \right) \, \biggl| \, 0 \right] \\<br /> p_z' &= \left[ - 2mv \left( \frac{1}{1+ \frac{v^2}{c^2}}\right) \, \biggl| \, 0 \right]<br /> \end{split}$

$\scriptsize<br /> \begin{split}<br /> p_k' &= \left[-2mv \, \biggl| \,m_1 {v_1}_y' + m_2 {v_2}_y'\right]=\left[ -2mv \, \biggl| \, m_1 \left( \frac{{v_1}_y}{\gamma \left(1 - {v_1}_y \frac{u}{c^2}\right)} \right) + m_2 \left( \frac{{v_2}_y}{\gamma \left(1 - {v_2}_y \frac{u}{c^2}\right)} \right) \right]\\<br /> p_k' &= \left[ -2mv \, \biggl| \, m \left( \frac{v}{\gamma \left(1 - v \frac{v}{c^2}\right)} \right) - m \left( \frac{v}{\gamma \left(1 - v \frac{v}{c^2}\right)} \right)\right]\\<br /> p_k' &= \left[ -2mv \, \biggl| \, 0 \right]<br /> \end{split}$

It is clear that $x$ components differ by factor $1/\left(1+\frac{v^2}{c^2}\right)$.

QUESTION: I want to know why do we multiply Newtonian momentum $p=mv$ by factor $\gamma = 1/ \sqrt{1 - \frac{v^2}{c^2}}$ and where is the connection between $\gamma$ and factor $1/\left(1+\frac{v^2}{c^2}\right)$ which i got?

FURTHER EXPLAINATION:
In the proof above I used velocity transformations derived below (derivation is taken from here):

$v_x' = \frac{dx'}{dt'}=\frac{\gamma (d x - u d t)}{\gamma \left(d t - d x \frac{u}{c^2} \right)} = \frac{d x - u d t}{d t - d x \frac{u}{c^2}} = \frac{\frac{d x}{d t} - u \frac{d t}{d t}}{\frac{d t}{d t} - \frac{d x}{d t} \frac{u}{c^2}} \Longrightarrow \boxed{v_x' = \frac{v_x - u}{1- v_x \frac{u}{c^2}}}$


$v_y' = \frac{dy'}{dt'}=\frac{d y}{\gamma \left(d t - d x \frac{u}{c^2} \right)} = \frac{\frac{dy}{dt}}{\gamma \left(\frac{dt}{dt} - \frac{dx}{dt} \frac{u}{c^2} \right)} \Longrightarrow \boxed{v_y' = \frac{v_y}{\gamma \left(1 - v_x \frac{u}{c^2} \right)}}$

 

[Nugatory]

Look carefully at the x component of the momentum in the first expression in the first line of your derivation of $p_k^{'}$

 

[71GA]

Look carefully at the x component of the momentum in the first expression in the first line of your derivation of $p_k^{'}$

You mean this expression?

$p_k' = \left[ -2mv\, \biggl| \, m_1 {v_1}_y + m_2 {v_2}_y \right]$

Coordinate system and observer in its origin move with $u=v$ from left to right at all times. This is why after collision observer in $xy$ would say that balls don't move in $x$ direction, while observer in $x'y'$ would state, that they both move in opposite direction. This is where i got my $x$ component $m_1 {v_1}_x' + m_2 {v_2}_x' = m (-v) + m(-v) = -2mv$.

 

[greswd]

Using the equations in your first post, you can already prove it.

full proof here:

http://en.wikibooks.org/wiki/Special_Relativity/Dynamics#Mass_and_energy

 

[71GA]

Using the equations in your first post, you can already prove it.
full proof here:
http://en.wikibooks.org/wiki/Special_Relativity/Dynamics#Mass_and_energy

I will take a look into it and report back if there are unclarities...