Proof for $p=\gamma_Pmu$ As I'm reading about Relativistic Momentum, my book states the following:
$$p=m \frac{\Delta x}{\Delta t}=m\frac{\Delta x}{\sqrt{(1-u^2/c^2)}\Delta t}=\frac{mu}{\sqrt{1-u^2/c^2}}=\gamma_Pmu$$
"Whether this is the 'correct' expression for p depends on whether the total momentum P is conserved when the velocities of a system of particles are transformed with the Lorentz velocity transformation equations. The proof is rather long and tedious..."
I'm interested in seeing the proof that they are describing as "long and tedious."
 A: The answer to this question depends on what you mean by "derive."  If you were to define relativistic momentum by the expression $\gamma m v$, then you could show, in real, physical experiments, that the relativistic momentum of an isolated system of particles is conserved.  Moreover, one can show mathematically that this conservation law is Lorentz-covariant in the sense that it holds in all inertial frames, and that it reduces to the Newtonian expression at low speeds.
The covariance property is what the quote is referring to as far as I can tell, and I think this is what Michael Brown is referring to in his comment as well.  The idea is that in writing an expression for the relativistic momentum, we want one that leads to conservation of this quantity in all inertial frames, and one that reduces to the Newtonian expression $mv$ in the limit of small velocities.  If we were to find an expression for relativistic momentum that weren't to satisfy these criteria, then we would be inclined to call it "incorrect."
Having said this, here is how you would proceed with the proof of Lorentz-covariance of relativistic momentum in the case of a single space dimension and two particles colliding (the generalization to higher dimensions and more particles is more tedious but doesn't really add to understanding).
First, one shows that relativistic momentum transforms as follows under a change of inertial frame from frame $S$ to frame $S'$:
$p' = \gamma(p - v E/c^2)$
where $p$ and $p'$ are the momenta in the respective frames, $v$ is the relative velocity of the frames, and $E$ is the energy ($\gamma m c^2$).  Then one notes that if we have conservation of energy and momentum in frame $S$
$p_{1i} + p_{2i} = p_{1f} + p_{2f}, \qquad E_{1i} + E_{2i} = E_{1f} + E_{2f}$
then in $S'$ one has
$p'_{1f} + p'_{2f} = \gamma(p_{1f} + p_{2f} - v(E_{1f}+E_{2f})/c^2)
=\gamma(p_{1i} + p_{2i} - v(E_{1i}+E_{2i})/c^2) = p'_{1i} + p'_{2i}$
so that momentum is conserved in $S'$, and similarly for energy.
Please let me know if there are typos.  Hope this helps!
Cheers!
