Consider the product of the four-momentum of an incoming particle $P_{\pi}$, and that of a target particle $P_{p}$. Making use of the Lorentz invariance of the scalar product of 2 four-momenta, the product in the lab-frame $(L)$ of the target particle $p$ should be the same as the product in the center of mass - frame $(\text{com})$. (for which holds that $\vec{p}^{com}_{p}+ \vec{p}^{com}_{\pi}=0$) $$P^{L}_{\pi}P^{L}_{p}=P^{\text{com}}_{\pi}P^{\text{com}}_{p} $$

Now the solution of an exercise in my book makes use of the (handy) expression

$|\vec{p}^{\text{com}}_{\pi}|=\frac{M_{p}\sqrt{T^{L}_{\pi}(T^{L}_{\pi}+2M_{\pi})}}{\sqrt{(M_{p}+M_{\pi})^{2}+2T^{L}_{\pi} M_{p}}}$

for the momentum in the center of mass-frame ($T$ stands for the kinetic energy)

Question : I can't seem to find the derivation for this (handy) formula and also wonder whether it's valid for all cases or whether there's a hidden assumption in it.


I think your formula

$$\displaystyle |\vec{p}^{\text{com}}_{\pi}|=\frac{M_{p}\sqrt{T^{L}_{\pi}(T^{L}_{\pi}+2M_{\pi})}}{\sqrt{(M_{p}+M_{\pi})^{2}+2T^{L}_{\pi} M_{p}}}$$

assumes that the target particle $P_p$ is initially at rest in the lab frame.
This condition was used in @JEB 's answer.

Here's a geometric interpretation of the formula, followed by the more general formula.
(I'll admit that it took me a while to realize this. I recall your question and the quick answer from when these were first posted. Upon its recent bump by Community, I thought about it again.)

The strategy below can be used to geometrically interpret other similar formulas.

Your formula gives the altitude (segment BC) of a triangle [representing conservation of total 4-momenta] in energy-momentum space ,
where the base (segment OZ) is the system's total momentum 4-vector (which I will write as $\tilde p + \tilde \pi$ [to declutter the formulas])
and one side (segment OB) is the target's 4-momentum $\tilde p$, which is (by assumption) initially at rest in the lab-frame $\hat L$.
(The rapidities are shown. In addition, I marked that segment BC (spatial momentum in the COM-frame) is Minkowski-perpendicular to segment OZ (the worldline of the COM-frame).


Let's write your formula using rapidities (the Minkowski-angle $\theta$ between future timelike vectors, where $\beta=\tanh\theta$ and $\gamma=\cosh\theta$). This will reveal what the numerator and denominator represent.
With rapidities, the 4-momentum $\tilde \pi$ can be expressed in terms of

  • rest mass $m_{\pi}=\pi =\sqrt{\tilde \pi \cdot \tilde \pi}$,
  • energy $E_{\pi}=\pi\cosh\theta=\tilde\pi\cdot\hat t$ (in a general frame $\hat t$),
  • momentum $p_{\pi}=\pi\sinh\theta = \tilde\pi\cdot\hat t_{\bot}$ , and
  • kinetic energy $T_{\pi}=\pi (\cosh\theta -1)$

So, your formula

$$\displaystyle |\vec{p}^{\text{com}}_{\pi}|=\frac{M_{p}\sqrt{T^{L}_{\pi}(T^{L}_{\pi}+2M_{\pi})}}{\sqrt{(M_{p}+M_{\pi})^{2}+2T^{L}_{\pi} M_{p}}}$$

becomes... (where I will manipulate the numerator and denominator separately) \begin{align} |\pi\sinh\theta^C| &= \frac{ p\sqrt{\pi(\cosh\theta^L-1) (\pi(\cosh\theta^L-1)+2\pi)} } { \sqrt{ (p+\pi)^2+2\pi(\cosh\theta^L-1)p } } \\ &= \frac{ p\sqrt{\pi(\cosh\theta^L-1) \pi(\cosh\theta^L+1)} } { \sqrt{ (p^2+\pi^2+2p\pi) + 2\pi\cosh\theta^L-2\pi p } } \\ &= \frac{ p\pi \sqrt{\sinh^2\theta^L} } { \sqrt{ p^2+\pi^2+ 2\pi\cosh\theta^L } } \end{align} Since (by assumption) $\tilde p$ is at rest in the lab-frame $\hat L$,
the rapidity-with-respect-to-the-lab $\theta^L$ is equal to the rapidity between the 4-momenta $\tilde \pi$ and $\tilde p$: \begin{align} |\pi\sinh\theta^C| &= \frac{ p\pi \sqrt{\sinh^2\theta_{between\ \pi\ and\ p}} } { \sqrt{ p^2+\pi^2+ 2\pi\cosh\theta_{between\ \pi\ and\ p} } } \\ &= \frac{ p\pi \sinh\theta_{between} } { \sqrt{ (\tilde p + \tilde \pi)\cdot(\tilde p + \tilde \pi) } } \\ &= \frac{ p\pi \sinh\theta_{between} } { |\tilde p + \tilde \pi| } = \frac{| \tilde p \times \tilde \pi |}{ | \tilde p + \tilde \pi | } \\ &= \frac{ \mbox{area of parallelogram [or kite] with sides $\tilde p$ and $\tilde \pi$} } { \mbox{diagonal of parallelogram [or kite] with sides $\tilde p$ and $\tilde \pi$} }\\ &= \frac{ \mbox{2(area of triangle with sides $\tilde p$ and $\tilde \pi$)} } { \mbox{base of triangle with sides $\tilde p$ and $\tilde \pi$} }\\ &= (\mbox{altitude of triangle with sides $\tilde p$ and $\tilde \pi$ }) \end{align}

So, the more general relation is $$\displaystyle |\vec{p}^{\text{com}}_{\pi}|=\frac{M_{p}\sqrt{T^{p}_{\pi}(T^{p}_{\pi}+2M_{\pi})}}{\sqrt{(M_{p}+M_{\pi})^{2}+2T^{p}_{\pi} M_{p}}}$$ which involves " the kinetic energy of $\tilde \pi$ in the frame of $\tilde p$ ''.

If your target $\tilde p$ is not at rest in the lab-frame $\hat L$, then you can substitute $$\theta_{\pi p}=\theta_{\pi L} - \theta_{p L},$$ expand out using hyperbolic-trig-identities and translate into masses, energies, and momenta. Then your expression will become more complicated, involving components of $\tilde p$ in the lab-frame.

If you take the calculation with rapidities a little further (and I'll make the $\pi$ explicit in $\theta_{\pi}^C$ and drop the magnitude on the left), one gets The Law of Sines in Minkowski-spacetime: \begin{align} \pi \sinh\theta_\pi^C &= \frac{ p\pi \sinh\theta_{between} } { |\tilde p + \tilde \pi| }\\ \sinh\theta_\pi^C &= \frac{ p \sinh\theta_{between} } { |\tilde p + \tilde \pi| }\\ \frac{\sinh\theta_\pi^C}{p} &= \frac{ \sinh\theta_{between} } { |\tilde p + \tilde \pi| } = \frac{\sinh\theta_p^C}{\pi} \end{align} where $\theta_\pi^C$ is the Minkowski-angle opposite to $\tilde p$
and $\theta_{between}$ is the external Minkowski-angle at the vertex B opposite to $\tilde p + \tilde \pi$,
and, by symmetry, $\theta_p^C$ is the Minkowski-angle opposite to $\tilde \pi$.
(We can't use the "internal angle" at B because that would be a Minkowski-angle between a future-timelike and past-timelike vector. No arc of a hyperbola is cut by those vectors. With some work, one might be able to define such a thing.)

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You should just derive it. Start in the lab:

$$ s = p_{\pi}p_P = (m + T, {\bf p}_{\pi})(M, {\bf 0}) = (m+T)M$$

Here $s$ is the Mandelstam variable called the invariant mass squared, and as you pointed out: it is Lorentz invariant. Note the only variable here is the lab kinetic energy $T$. The two masses are fixed.

Now consider the center-of-moentum system. Here the only variable is ${\bf p}$, the beam/target momentum (of course I am making them collinear):

$$ s' = p'_{\pi}p'_P = (E_{\pi}, {\bf p})(E_p, -{\bf p})=E_{\pi}E_p+p^2$$

$$ s' = \sqrt{m^2+p^2}\sqrt{M^2+p^2}+p^2$$

Noting $s=s'$:

$$\sqrt{m^2+p^2}\sqrt{M^2+p^2}+p^2 = (m+T)M$$

Solve that for $|p|$, and you should get your formula.

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  • 1
    $\begingroup$ Would a $\cdot$ (\cdot) to represent the scalar product make the notation here more explicit and readable? $\endgroup$ – dmckee --- ex-moderator kitten Apr 24 '18 at 17:49

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