Deriving relativistic momentum (wikibooks)

Question

I asked a question about a derivation of relativistic momentum here, but I didn't really get an answer that helped me. So I looked up a different but similar derivation on Wikibooks (see here), and I have a different question now about this other proof.

Somewhere near the end they say:

I don't get this. What exactly is the principle of relativity here? I thought it was that the laws of physics are the same in each reference frame, so how did they come up with this equation? I see they're equating the change in classical momentum for R to the change in classical momentum for B, but wasn't the whole point of this thought experiment that we're deriving a relativistic momentum?

If someone could help me with this, that'd be great, because it's the only step I don't get!

Answer 1

First a little clarification: this derivation is, although effective, a bit old-fashioned since it uses a concept like the relativistic increase of mass, which is now considered an outdated and confusing interpretation of processes in special relativity.

Nevertheless, the answer to your question comes from a few lines above the ones you cited: at a certain point, after having found that $u_{yR} \neq u_{yB}$, the derivation states:

If the mass were constant between collisions and between frames then although $2mu'_{yR} = 2mu'_{yB}$ it is found that $2mu_{yR} \neq 2mu_{yB}$

After that, the (deprecated) relativistic mass $m_A$ and $m_B$ are introduced in order for the inequality above to become an equality. The principle of relativity here is to be understood as "the equality which holds in a frame must hold in every other frame too"; in this case the equality is the conservation of linear momentum: if the equality didn't hold, you would have a frame in which it is conserved and a frame in which is not. But conservation of momentum is a basic law, therefore a new definition of momentum has to be introduced in order to restore conservation.

Answer 2

You can't really derive the equation for relativistic momentum', though sometimes 'plausibility arguments' are dressed up to look like derivations. [As for the Wikibooks treatment; I think it loses the incredible simplicity of the essential Physics in the algebra.]

Here's what I find a compelling plausibility argument. I apologise for the scanned page that follows, but I did write the stuff:

The only way, surely, to satisfy both these equations, whatever the relative velocity between frames S and S', is if, for both particles, $p_y' = p_y.$ In other words transverse momentum must be a relativistic invariant.

The Newtonian formula, $p_y=mu_y=m\frac{\Delta y}{\Delta t}$, in which $\Delta t$ is the time taken to traverse the transverse distance $\Delta y,$ clearly won't work, because although $m$ and $\Delta y$ are invariant between the frames, $\Delta t$ is not. But all we have to do is to put the invariant proper time, $\Delta \tau,$ in place of $\Delta t$, and we have a Lorentz-invariant candidate expression for $p_y.$ Crucially, the expression collapses to the Newtonian expression at low velocities, for it is easy to show that $\Delta t=\gamma \Delta \tau,$ in which $\gamma=\left(1-\frac{u^2}{c^2}\right)^{-1},$ $u$ being the body's speed.

Thus we have $$p_y=m\frac{\Delta y}{\Delta \tau}=m \gamma\frac{\Delta y}{\Delta t}=m \gamma u_y$$

Similar expressions must hold for $p_x$ and $p_z$, since there is nothing special about the $y$ direction. Therefore we have $$\vec p=m \gamma \vec u.$$

Of course our confidence in this equation is really built on the equation, together with $E^2-c^2p^2=c^4m^2$ and other equations of relativistic dynamics and electromagnetism, forming a self-consistent system, and one that has been confirmed by a myriad of experiments.