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 travells 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.
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.
$$
\scriptsize
\begin{split}
p_z' &= \left[ m_1 {v_1}_x' + m_2 {v_2}_x'\, \biggl| \, 0 \right] = \left[ m_1 0 + m_2 \left( \frac{{v_2}_x – u}{1-{v_2}_x\frac{u}{c^2}} \right)\, \biggl| \, 0 \right]= \left[ m \left( \frac{-v – v}{1+ v \frac{v}{c^2}} \right) \, \biggl| \, 0 \right] \\
p_z' &= \left[ – 2mv \left( \frac{1}{1+ \frac{v^2}{c^2}}\right) \, \biggl| \, 0 \right]
\end{split}
$$
$$
\scriptsize
\begin{split}
p_k' &= \left[-2mv \, \biggl| \,m_1 {v_1}_y' + m_2 {v_2}_y'\right]=\left[ -2mv \, \biggl| \, m_1 \left( \frac{{v_1}_y}{\gamma \left(1 – {v_1}_y \frac{u}{c^2}\right)} \right) + m_2 \left( \frac{{v_2}_y}{\gamma \left(1 – {v_2}_y \frac{u}{c^2}\right)} \right) \right]\\
p_k' &= \left[ -2mv \, \biggl| \, m \left( \frac{v}{\gamma \left(1 – v \frac{v}{c^2}\right)} \right) – m \left( \frac{v}{\gamma \left(1 – v \frac{v}{c^2}\right)} \right)\right]\\
p_k' &= \left[ -2mv \, \biggl| \, 0 \right]
\end{split}
$$
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?
Best Answer
Assume that the relativistic momentum is the same as the nonrelativistic momentum you used, but multiplied by some unknown function of velocity $\alpha(v)$.
$$\mathbf{p} = \alpha(v)\,\, m \mathbf{v}$$
Then in the primed frame, the total momentum before the collision is just what you had, but multiplied by $\alpha(v_i)$, with $v_i$ the speed before collision. The momentum after the collision is again what you had, but multiplied by $\alpha(v_f)$, with $v_f$ the speed after the collision.
In order to conserve momentum we must have
$$ \alpha(v_i) \frac{-2mv}{1+v^2} = -2mv \,\alpha(v_f)$$
For simplicity, I'm suppressing factors of $c$.
After the collision, you have a mistake in your velocity transformations. The vertical speed is just $v/\gamma$. That makes the speed of each ball $v_f = (v^2 + (v/\gamma)^2)^{1/2} = v \left(2-v^2\right)^{1/2}$
Plugging in $v_i$ and $v_f$ into the previous equation and canceling some like terms we have
$$ \alpha\left(\frac{2v}{1+v^2}\right) \frac{1}{1 + v^2} = \alpha\left(v[2-v^2]^{1/2}\right)$$
If you let $\alpha(v) = \gamma(v)$ and crunch some algebra you'll see that the identity above is satisfied.
As for your original point, a desire to understand why momentum has a factor $\gamma$ in it, analyzing situations like this one is helpful, but ultimately it is probably best to understand momentum as the spatial component of the energy-momentum four-vector. Since it is a four-vector, it must transform like any other four-vector, $\gamma$'s and all.