In Kleppner and Kolenkow chapter 13, they derive the expression of relativistic mass by considering a symmetric glancing elastic collision.
It was analyzed from two reference frames. One in which the velocity of A in the x direction was zero and another one in which the velocity of B in the x direction was zero.
Here is how the derivation goes in the book:
Our task is to find a conserved quantity analogous to classical momentum. We suppose that the momentum of a particle moving with velocity $\mathbf{w}$ is $$\mathbf{p} = m(w) \mathbf{w}$$ where $m(w)$ is a scalar quantity yet to be determined, analogous to Newtonian mass but which could depend on the speed $w$.
The x momentum in A’s frame is due entirely to particle B. Before the collision B’s speed is $w = \sqrt{V^2 + u_0^2/\gamma^2}$ and after the collision it is $w' = \sqrt{V^2 + u'^2/\gamma^2}$. Imposing conservation of momentum in the x direction yields $$m(w)V = m(w')V$$ It follows that $w=w'$, so that $$u' = u_0$$ In other words, y motion is reversed in the A frame.
Next we write the statement of the conservation of momentum in the y direction as evaluated in A’s frame. Equating the y momentum before and after the collision gives $$-m_0 u_0 + m(w) \frac{u_0}{\gamma} = m_0 u_0 - m(w) \frac{u_0}{\gamma}$$ which gives $$m(w) = \gamma m_0$$ In the limit $u_0 \rightarrow 0$, $m(u_0) \rightarrow m(0)$, which we take to be the Newtonian mass, or "rest mass" $m_0$, of the particle. In this limit, $w = V$. Hence $$m(V) = \gamma m_0 = \frac{m_0}{\sqrt{1 - V^2/c^2}}$$ Consequently, momentum is preserved in the collision provided we define the momentum of a particle moving with velocity $\mathbf{v}$ to be $$\mathbf{p} = m \mathbf{v}$$ where
$$m = \frac{m_0}{\sqrt{1 - v^2/c^2}} = \gamma m_0$$
Now I have a few problems with this derivation. They are:
- They assumed that both A and B has the same mass. Unless I'm mistaken, the momentum equation in the $x$ direction should remain unchanged because during the collision the impulse is in the $y$ direction. So suppose the masses were different, namely $m_A$ and $m_B(w)$. Then since $$m_B(w)V = m_B(w')V$$ it follows that $$u' = u_0$$ But then the y equation becomes $$-m_A u_0 + m_B(w) \frac{u_0}{\gamma} = m_0 u_0 - m_B(w) \frac{u_0}{\gamma}$$ or $$m_B(w) = \gamma m_A$$ which is weird because B's mass shouldn't depend on A's. Personally I think that their argument for $u' = u_0$ is flawed. Because it doesn't matter whether A and B's masses are different but intuitively I think it should. I don't see how the collision could be elastic and symmetrical without the two particles having the same mass. Because I could always take extreme cases when one is much more massive than the other and following their argument we would still have $u' = u_0$. Or maybe I misunderstood their argument and the masses really do matter. This is all very confusing to me.
- It seems that when writing the momentum equation in the $y$ direction, the author represented $m(u_0)$ as $m_0$ while in the final equation they meant $m_0$ to be the rest mass which makes sense because A was also moving in the y direction in A's frame so it's mass can't be just the rest mass $m_0$. However before taking the limit $u_0 \rightarrow 0$, $m(u_0) \rightarrow m(0)$, the equation for $m(w)$ was $$m(w) = \frac{m(u_0)}{\sqrt{1 - V^2/c^2}}$$ and after taking the limit it became $$m(V) = \frac{m_0}{\sqrt{1 - V^2/c^2}}$$ However both equations are supposed to be true and using the final result we should have $$m(w) = \frac{m_0}{\sqrt{1 - w^2/c^2}}$$ and similarly for A in A's frame, $$m(u_0) = \frac{m_0}{\sqrt{1 - u_0^2/c^2}}$$ Substituting this in the first equation $$m(w) = \frac{m(u_0)}{\sqrt{1 - V^2/c^2}}$$ $$ = \frac{m_0}{\sqrt{(1 - V^2/c^2)(1 - u_0^2/c^2)}}$$ $$ = \frac{m_0}{\sqrt{1 - (u_0^2/c^2 + V^2/c^2) + (Vu_0)^2/c^4}}$$ $$ = \frac{m_0}{\sqrt{1 - w^2/c^2 + (Vu_0)^2/c^4}}$$ $$ \neq \frac{m_0}{\sqrt{1 - w^2/c^2}} $$ I'm probably missing something but I can't figure out what.