A proper derivation of relativistic momentum/mass?

Question

The "two balls bouncing" derivation of relativistic momentum has a flaw, I think. When the momentum for the ball going at a vertical velocity $u_y$ and moving in the $x$-direction with velocity $u_x=v$, with $v$ being the speed relative to the frame of reference in which the ball has $0$ $x$-velocity, the momentum is calculated to be $$p_y=\frac{mu_y}{\sqrt{1-\frac{v^2}{c^2}}}$$ Those derivations then move on to conclude that the momentum of any body moving at velocity $\vec{v}$ to be $$\vec{p}=\frac{m\vec{v}}{\sqrt{1-\frac{v^2}{c^2}}}$$ Now, I have been able to prove that this is the relativistic formula for momentum using the "two ball experiment" by assuming the second formula to be true and then showing it works, but I've yet to see a derivation that can properly derive the relativistic momentum formula (that is, without using four-vectors). Would anyone care to enlighten me with such a derivation?

Answer 1

A proper derivation of the formula will require knowledge of Noether's theorem which states that for every symmetry of an action, there exists a quantity that is conserved. In particular, momentum is the conserved quantity corresponding to spatial translations.

The dynamics of particle in special relativity is described by the action $$ L = - m \sqrt{1 - {\dot x}^2 }~. $$ This action is invariant under $x(t) \to x(t) + \epsilon$. The corresponding conserved quantity is then $$ p = \frac{1}{ \epsilon } \frac{ \partial L }{ \partial{\dot x } } \delta x = \frac{ m {\dot x} }{ \sqrt{ 1 - {\dot x}^2 } }~. $$

A similar derivation maybe done in Newtonian mechanics where the action is $$ L = \frac{1}{2} m {\dot x}^2 ~. $$ Then, the momentum is $$ p = \frac{1}{ \epsilon } \frac{ \partial L }{ \partial{\dot x } } \delta x = m {\dot x} ~. $$ which is the usual definition of momentum.

Answer 2

I think you're right; in particular the methods I've seen that consider a glancing collision and purport to derive the relativistic momentum formula algebraically by some limit procedure have some cleverly concealed (or inadvertent) assumption.

But one can conclude, by studying a collisions between two bodies seen from different inertial frames, that transverse momentum is Lorentz-invariant. Since we know that (rest) mass, $m$, and transverse displacement, $\Delta y$, are Lorentz invariants, the 'obvious' formula for y-momentum is obtained by dividing $m \Delta y$ by a third invariant, the proper time $\Delta \tau$, giving us $$p_y=m \frac {\Delta y}{\Delta \tau} \ = m \frac {\Delta t}{\Delta \tau} \frac{\Delta y}{\Delta t} =\ m \gamma v_y.$$ Since space is isotropic, we must have similar expressions for the $x$ and $z$ components, namely$$p_x=m \frac {\Delta x}{\Delta \tau} \ =\ \gamma m v_x \ \text{and} \ p_z=m \frac {\Delta z}{\Delta \tau} \ =\ \gamma m v_z.$$

The Lorentz factor, $\gamma$ in these expressions is $$\gamma = \sqrt{\frac{1}{1-\frac{v^2}{c^2}}}$$ in which $v$ is the body's speed in our chosen reference frame.