In Kleppner and Kolenkow chapter 13, they derive the expression of relativistic mass by considering a symmetric glancing elastic collision.

enter image description here

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.

enter image description here

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:

  1. 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.
  2. 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.
  • 1
    $\begingroup$ @ Brain Stroke Patient This is a comment rather than an answer. I once examined several versions of the glancing collision used to 'derive' the formula for relativistic momentum. Despite first appearances, none of the arguments was totally watertight. I concluded that the best argument claimed plausibility only, but had the merit of transparency and very simple algebra. It goes like this... (1) Transverse momentum must be a Lorentz invariant (for a collision not to appear different in the y-direction in different frames) (2) We know that $mv_y=m\frac{\Delta y}{\Delta t}$ in which m is $\endgroup$ Commented Jul 31, 2020 at 15:49
  • $\begingroup$ I'm following modern practice (well justified, imo) of not using the concept of 'relativistic mass'. Rest mass can then simply be called 'mass' and designated simply by $m$, $\endgroup$ Commented Jul 31, 2020 at 16:12
  • $\begingroup$ I'm familiar with that argument. It's a very popular one. I'd still like to know though whether the argument provided in Kleppner is flawed or whether I'm missing some points. A lot of older textbooks and even some new ones still use glancing collision derivations so I'd like to understand them. $\endgroup$ Commented Jul 31, 2020 at 16:41
  • $\begingroup$ a Lorentz invariant constant for the body is not a Lorentz invariant, because Δ𝑡 is not Lorentz invariant. (3) But if we replace Δ𝑡 by the proper time interval, Δ𝜏 (for a body to traverse Δ𝑦) we have a Lorentz invariant quantity, $$p_y=m \frac{\Delta y}{\Delta \tau}=m\gamma \frac{\Delta y}{\Delta t}=m\gamma v_y \ \ \ \ \ \text{in which} \ \ \ \ \ \gamma=\left(1-\frac{v^2}{c^2}\right)^{-1/2}$$ (4) From the assumed isotropy of space, we know that the momenta in the x and z directions must be given by similar formulae. These collapse to the Newtonian formula when 𝑣<<𝑐. v << c . $\endgroup$ Commented Jul 31, 2020 at 16:45
  • $\begingroup$ You are clearly not entirely happy with the K and K argument, and my point is that I doubt if pursuing it will give you a better justification for the relativistic momentum formula. But no doubt you have your own reasons. $\endgroup$ Commented Jul 31, 2020 at 17:00

1 Answer 1


I don't see how the collision could be elastic and symmetrical without the two particles having the same mass.

You're right, but this isn't a mistake in the argument. The authors are assuming a specific situation and using that to derive general constraints. If you change the assumptions, you would get a different, more complicated setup, which would not be useful.

What you are saying is analogous to this:

Klepper: Let $x$ be the number of cows. Since you can't have a negative number of cows, $x \geq 0$.

You: But what if $x$ isn't the number of cows? Then it could be negative, so your argument is flawed.

In response to comments: there's indeed another step that Kleppner has let go implicit. Kleppner has assumed that it is possible for such a collision to happen. And as you point out, it wouldn't be possible if the masses weren't equal, either in relativistic or nonrelativistic physics.

So here's an argument for why it's possible when the masses are equal. The initial state in the lab frame has equal mass particles traveling with opposite velocities. As long as "momentum" flips sign when the sign of the velocity is flipped, the initial momentum must be zero. By the same logic, the final momentum is also zero. So the setup is consistent with momentum conservation.

Then you might ask, how do you know that momentum flips sign when the velocity flips sign? But this is similar to asking "how do you know that $x$ means the number of cows?" We are searching for conserved quantities in a new context, and a conserved quantity would only deserve the name "momentum" if it satisfied that basic requirement.

It seems that when writing the momentum equation in the $y$ direction, the author represented $m(u_0)$ as $m_0$

The argument is correct, but their notation is very confusing, because it's not explicit enough. By making all mass-dependence in the $m$'s and $\gamma$'s explicit, their $y$-momentum equation rearranges to $$m(u_0) u_0 = m(w) u_0 / \gamma(V).$$ Upon cancelling a $u_0$ and taking $u_0 \to 0$ on both sides we have $$m(0) = m(V) / \gamma(V)$$ which is precisely the desired conclusion. Now, your question is whether this is self-consistent, if we substitute it back into the original equation. If we do that, we get $$\gamma(u_0) m(0) u_0 = \gamma(w) m(0) u_0 / \gamma(V)$$ and cancelling factors gives $$\gamma(u_0) \gamma(V) = \gamma(w).$$ Taking the inverse square of both sides gives $$(1 - u_0^2)(1-V^2) = (1-w^2)$$ where I set $c = 1$. Simplifying a bit gives $$u_0^2 + V^2 - u_0^2 V^2 = w^2.$$ Since the velocity $w$ has components $V$ and $u_0 / \gamma(V)$, we have $$w^2 = V^2 + (u_0 / \gamma(V))^2 = V^2 + u_0^2 (1 - V^2)$$ which precisely matches the desired left-hand side. So it is self-consistent.

  • $\begingroup$ Ah I see what mistake I did in the second part now. I assumed $w^2 = V^2 + u_0^2$ which was wrong. But I still don't understand your first point. In your example, we use the fact that the number of cows can't be negative to claim that $x \ge 0$ but in Kleppner's proof of $u' = u$, he never uses the fact that the two masses are equal $\endgroup$ Commented Aug 2, 2020 at 8:36
  • $\begingroup$ I could always choose the initial velocities to be such that at least they approach each other in a symmetric way and that shouldn't depend on their mass. Could it be that if their masses are different the collision has to be inelastic so that their rest masses before and after collision is different? Because that would explain the x direction momentum equation and not imply that $u' = u$ $\endgroup$ Commented Aug 2, 2020 at 8:45
  • $\begingroup$ @BrainStrokePatient I added some more detail, does it answer your question? $\endgroup$
    – knzhou
    Commented Aug 2, 2020 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.