# COM frame in Relativistic particle physics

I have a home work:

Consider a head-on collision of a 'bullet' of rest mass $$M$$ with a stationary 'target' of rest mass $$m$$. Prove that the post-collision γ-factor of the bullet cannot exceed:$$(m^2 + M^2)/(2mM).$$ [Hint: if $$P, P'$$ are the pre- and post- collision four-momenta of the bullet, and $$Q, Q'$$ those of the target,show, by going to the CM frame, that $$(P' − Q)^2 ≥ 0$$; in fact, in the CM frame $$P' − Q$$ has no spatial components.]

I have solved problem from the hint that $$(P' − Q )^2 ≥ 0$$ and $$P' − Q$$ should have non-spatial component. The first hint is easy, because four-momentum is time-like hence it should always be non-negative. What i don't understand is that why $$P' − Q$$ should have non-spatial component in COM?

• In the CM frame you have $E=P+Q=P'+Q'$ (where $E$ is total rest mass), so you can express $P'-Q$ in terms of $Q$ and $Q'$. Commented Oct 6, 2018 at 17:38
• @CuspyCode and then? I still don’t get it. Commented Oct 6, 2018 at 17:51
• $P'-Q=E-(Q+Q')$. The total rest four-momentum $E$ obviously has no spatial components, and $Q+Q'$ also has no spatial components in a 100% elastic collision (by symmetry, in the CM frame). Energy considerations require that the maximal $\gamma$-factor involves an elastic collision, I think. Commented Oct 6, 2018 at 18:16
• @CuspyCode Q and Q’ represents the 4-momentum of target. You mean the target can’t be deflected? Commented Oct 6, 2018 at 18:21
• The bullet should be $K$, the target $P$, and $Q=K-K'$. Calling something that isn't momentum transfer $Q$ is a problem. I mean if someone asked you," a mass, $g$, falls a distance, $t$, in $x$ seconds, what is the gravitational acceleration, $m$?", would you answer them?
– JEB
Commented Oct 6, 2018 at 22:36

(The problem in the OP appears to be from Rindler, Relativity: Special, General, and Cosmological, 2ed, Exercise 6.7.)

The statement
"in the CM [zero-momentum, center of momentum] frame $$(P'-Q)$$ has no spatial components" means that $$(P'-Q)$$ is parallel to the total-4-momentum $$(P+Q)$$, also $$(P'+Q')$$ [by conservation]. This appears to hold true when the collision is elastic.

As I was writing this, I eventually arrived at a simpler solution in the end.
But I've kept what I started with because it may be useful to others (as it was to me).

Inspired by the Mandelstam variables ( https://en.wikipedia.org/wiki/Mandelstam_variables )
(which I am learning more about) I can prove it using dot-products... but can't easily explain it in those terms... but I suspect there is a simpler method using dot-products using the 4-momenta directly... when things get sorted out.

In these terms, let me define the following 4-vectors:
$$S=P+Q=P'+Q'$$ (the total-4-momentum) ,
$$T=P-P'=Q'-Q$$, (the 4-momentum transfer) and
$$U=P-Q'=P'-Q$$ (some kind of cross term),
where I have used momentum conservation ($$P+Q=P'+Q'$$) to write the second equality in each.
It's easy to show that $$S\cdot T=0$$ (the momentum-transfer is orthogonal to the total-momentum).

What you want to show is that,
for elastic collisions, $$U$$ is parallel to $$S$$ by showing that $$U\cdot T=0$$.
(By elastic, we have, in addition to $$P+Q=P'+Q'$$,
the conditions $$P^2=P'^2$$ and $$Q^2=Q'^2$$.)

First note that $$S^2=(P+Q)^2=P^2+2P\cdot Q+Q^2.$$ Next, note that $$S+T+U=(3P+Q-P'-Q')=2P,$$ where I invoked momentum conservation.
It can be shown that $$S^2+T^2+U^2=P^2+Q^2+P'^2+Q'^2$$, which equals $$2(P^2+Q^2)$$ in the elastic case.

By squaring $$T+U=2P-S=P-Q$$ and isolating $$2T\cdot U$$,
I get \begin{align} 2 T\cdot U &=P^2-2P\cdot Q+Q^2 -[T^2+U^2]\\ &=P^2-2P\cdot Q+Q^2 -[2(P^2+Q^2)-S^2]\\ &=-P^2-2P\cdot Q-Q^2 +S^2\\ &=0 \end{align} There's likely a simpler calculation with dot-products [without invoking Mandelstam] which might be suggested by the following energy-momentum diagram. (update: there is)

Now here's an energy-momentum diagram, which suggests a simpler geometric interpretation.

The 4-vector $$U=P'-Q=P'-Q_{copy}$$ is shown, connecting the tip of $$Q_{copy}$$ to $$P'$$.

The momentum-transfer segment $$T$$ (connecting A and C), which is orthogonal to the total-momentum $$S$$, is actually bisected by $$S$$ when the collision is elastic... that is when $$C$$ (the tip of $$P'$$ and the tail of $$Q'$$) is at the intersection of the two mass-shells determined by $$P$$ and $$Q$$.

So, this means that in the zero-momentum COM frame, the spatial part of $$Q$$ (which is projected onto half of $$T$$) is equal to the spatial part of $$P'$$ (which is projected onto the other half of $$T$$). Thus, the difference in their spatial components is zero.

Thus, $$U\cdot T=(P'-Q)\cdot T=P'\cdot T - Q\cdot T=P'_{x,COM}-Q_{x,COM}=0$$.

Okay, after writing this up and thinking about it some more...
Here is a simpler calculation. (I'll leave the earlier stuff up there because it may be useful. It was to me.)

\begin{align} 0&\stackrel{?}{=}U\cdot T\\ &=(P'-Q)\cdot T\\ &=(P'-[P'+Q'-P])\cdot T \qquad \mbox{invoking conservation}\\ &=(P'+P-[P'+Q'])\cdot T\\ &=(P'+P)\cdot T-[P'+Q']\cdot T\\ &=(P'+P)\cdot T- \qquad 0\qquad\quad \mbox{since S\cdot T=0}\\ &=(P'+P)\cdot (P-P')\\ &=P'^2-P^2\\ &=\qquad 0 \qquad \qquad \qquad\qquad \mbox{since P'^2=P^2 for an elastic collision}\\ \end{align}

One more update.
I think the following variant of the last calculation constructs $$U$$ as a multiple of $$S$$:

\begin{align} U&=(P'-Q)\\ &=(P'-[P'+Q'-P])\qquad \mbox{invoking conservation}\\ &=(P'+P-[P'+Q'])\\ &=(P'+P)-[P'+Q']\\ &=(P'+P) -S\qquad\qquad\quad \mbox{since S=P'+Q'}\\ \end{align} Since $$P'$$ and $$P$$ are future-pointing timelike, then so is $$(P'+P)$$.
For an elastic collision $$(P'+P)$$ is a multiple of $$S$$, because of symmetry [recall $$P^2=P'^2$$ for an elastic collision]... and it seems we need that $$P$$, $$P'$$, and $$S$$ are coplanar.
From the diagram, $$(P'+P)$$ has magnitude $$2P\cosh\theta_{PS}$$, where $$\theta$$ is the rapidity between $$P$$ and $$S$$.
So, it seems (assuming no errors in the calculation and the diagram)
that since $$S$$ has magnitude $$P\cosh\theta_{PS}+Q\cosh\theta_{QS}$$,
then, for an elastic collision [with $$P$$, $$P'$$,and $$S$$ coplanar],
$$U$$ is a multiple of $$S$$ and
$$U$$ has magnitude $$| P\cosh\theta_{PS}-Q\cosh\theta_{QS}|$$,
the difference of the relativistic-energies-in-the-COM-frame.

• I still have one question:if $U\cdot T$ and $S\cdot T$ is $0$, how do you conclude that U and S are parallel? What do you mean by parallel? Commented Oct 7, 2018 at 18:42
• I'm presuming that all of these 4-vectors are in the same plane... like in a typical spacetime diagram with one-spatial dimension. By parallel in the plane, I mean that S and U are proportional, multiples of each other: $\tilde U=k \tilde S$. Commented Oct 7, 2018 at 18:50
• Is this a proved property of 4-vectors? Commented Oct 7, 2018 at 18:54
• What if they are not in the same plane? Commented Oct 7, 2018 at 18:57
• If they are not coplanar, then you probably have more work to do. Commented Oct 7, 2018 at 18:58