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.


2 Answers 2


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.)


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.

  • 1
    $\begingroup$ Would a $\cdot$ (\cdot) to represent the scalar product make the notation here more explicit and readable? $\endgroup$ Commented Apr 24, 2018 at 17:49

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.