Special Relativity - Non-conservation of Newtonian momentum I'm studying special relativity and I don't quite understand why Newtonian momentum isn't conserved, because everytime I think of the math involved or imagine myself observing some system from various reference frames, it never makes sense to me...
I'll be very grateful to anyone who can explain this non-conservation of Newtonian momentum to me.
 A: Newtonian linear momentum is not conserved because your idea of how to calculate how different reference frames see the same thing is flawed. Usually when people get stuck on this point it's because they have trouble grasping the concept of how the passage of time is reference frame dependent in special relativity (SR). The Minutephysics channel did a great pair of videos with visualizations of how relative time works when applying the Lorentz transformations between reference frames:
$$\begin{align} t' & = \frac{t - \frac{vx}{c^2}}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \\
x' & = \frac{x - vt}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \\
y' &= y \\
z' & = z,\end{align}$$ and how to apply them to resolve the "twins paradox".
If I'm right about what's confusing you, you're stuck on what is known as Galilean addition of velocities:$$\mathbf{v}' = \mathbf{v} - \mathbf{u}_o,$$ where if an object is moving with velocity $\mathbf{v}$ and an observer is moving with velocity $\mathbf{u}_o$ then that observer will measure the objects velocity to be $\mathbf{v}'$. The derivation of that formula assumed an absolute rate of time's passage. The relativistic addition of velocities formula is somewhat more complicated, and can be derived from the Lorentz transformations (note: Wikipedia uses a different sign convention). If we break the vectors down into components parallel to (eg $\mathbf{v}_{||}$) and perbendicular to ($\mathbf{v}_\perp$) $\mathbf{u}_o$ then we get the following transformations: $$\begin{align}\mathbf{v}'_{||} & = \frac{\mathbf{v}_{||} - \mathbf{u}_o}{1 - \frac{\mathbf{v}\cdot \mathbf{u}_o}{c^2}} \\ 
\mathbf{v}'_\perp &= \left(\frac{\sqrt{1 - \frac{u_o^2}{c^2}}}{1 - \frac{\mathbf{v}\cdot \mathbf{u}_o}{c^2}}\right)\mathbf{v}_\perp.\end{align}$$
If you apply the rules correctly, you'll find that what is conserved is called the 4-momentum. It has four components, 0 through 3, that correspond to timelike (component zero) and space-like (components 1 through 3, often denoted as the ordinary vector part) parts. It is: $$\begin{align}p_0 &= \frac{m c}{\sqrt{1 - \frac{v^2}{c^2}}} \\
\mathbf{p} & = \frac{m \mathbf{v}}{\sqrt{1 - \frac{v^2}{c^2}}}.
\end{align}$$ You should recognize the space-like part of the 4-momentum as the Newtonian momentum times $\gamma = 1 \left/ \sqrt{1 - \frac{v^2}{c^2}}\right.$, and the time like part as the relativistic energy, $E = \gamma mc^2$, divided by $c$.
