# Deriving relativistic momentum (wikibooks)

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!

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a bit old-fashioned since it uses a concept like the relativistic increase of mass, which is now considered an outdated and confusing interpretation of processes in special relativity.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.

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.