Lorentz-Invariant Phase Space and the Decay/Cross-Section Master Formulae

Derivation

Lorentz-Invariant Phase Space and the Decay/Cross-Section Master Formulae

D-276 Home PU-304 Threads energy · chance · symmetry Depends on Mandelstam Invariants and Relativistic Two-Body Kinematics, Fermi's Golden Rule

Statement

We construct the Lorentz-invariant \(n\)-body phase-space measure \(d\Pi_n = (2\pi)^4\,\delta^4\!\left(P-\sum_i p_i\right)\prod_i \dfrac{d^3p_i}{(2\pi)^3\,2E_i}\) from the on-shell delta-function, and use it to derive the two master formulae that convert a Lorentz-invariant squared matrix element \(|\mathcal{M}|^2\) into an observable: the differential decay width \(d\Gamma = \dfrac{1}{2M}\,|\mathcal{M}|^2\,d\Pi_n\) and the differential cross-section \(d\sigma = \dfrac{1}{F}\,|\mathcal{M}|^2\,d\Pi_n\) with invariant flux \(F = 4\sqrt{(p_1\!\cdot p_2)^2 - m_1^2 m_2^2}\). We then reduce these to the two-body results \(\Gamma = \dfrac{|\vec{p}_f|}{8\pi M^2}\,|\mathcal{M}|^2\) and \(\dfrac{d\sigma}{d\Omega} = \dfrac{1}{64\pi^2 s}\dfrac{|\vec{p}_f|}{|\vec{p}_i|}\,|\mathcal{M}|^2\).

Why it matters

Every prediction of a relativistic quantum field theory that can be measured in a detector — a decay rate, a lifetime, a scattering cross-section — is a number obtained by folding the frame-independent dynamical content \(|\mathcal{M}|^2\) against a purely kinematic weight, the phase-space measure. Getting the measure right, and Lorentz invariant, is what lets the same Feynman-diagram calculation be quoted in the lab frame of a fixed-target experiment and the centre-of-mass frame of a collider without recomputation.

These are the formulae that sit at the boundary between theory and experiment: the left-hand sides are what the Particle Data Group tabulates, the right-hand sides are what a theorist computes diagram by diagram. Their structure — a flux or lifetime prefactor, times \(|\mathcal{M}|^2\), times \(d\Pi_n\) — is universal across the Standard Model and beyond.

Assumptions

Relativistic normalisation of one-particle states, \(\langle \mathbf{p}|\mathbf{p}'\rangle = (2\pi)^3\,2E_{\mathbf p}\,\delta^3(\mathbf p - \mathbf p')\).With the non-relativistic normalisation \(2E\to 1\) the measure \(d^3p/2E\) loses its invariance and every formula below acquires frame-dependent factors of energy that spoil the covariance. External particles are on their mass shell, \(p_i^2 = m_i^2\) with \(E_i>0\).Off-shell legs (internal lines) do not carry a phase-space factor; treating a virtual particle as a real final state double-counts and violates the \(S\)-matrix reduction. Overall energy-momentum conservation, enforced by \((2\pi)^4\delta^4(P-\sum p_i)\).Drop it and the integral is not restricted to the physically accessible kinematic region; the total rate diverges and no meaningful width results. The initial state is a single particle (decay) or exactly two particles (\(2\to n\) scattering) so that \(1/2M\) or \(1/F\) is the correct flux normalisation.For three or more incoming particles the incident-flux concept is ill-defined; the naive \(1/F\) prefactor is wrong and one must return to the reduction formula with the appropriate initial-state normalisation. \(|\mathcal{M}|^2\) is a Lorentz scalar (spins summed/averaged so no free indices remain).If spin sums are left undone, \(|\mathcal{M}|^2\) carries uncontracted indices, is frame-dependent, and cannot be pulled through the invariant measure.

Derivation

\[ \int \frac{d^4p}{(2\pi)^4}\,(2\pi)\,\delta(p^2-m^2)\,\theta(p^0)\,f(p) = \int \frac{d^3p}{(2\pi)^3}\,\frac{1}{2E_{\mathbf p}}\,f(E_{\mathbf p},\mathbf p),\qquad E_{\mathbf p}=\sqrt{\mathbf p^2+m^2} \]

Write \(\delta(p^2-m^2)=\delta\!\big((p^0)^2-E_{\mathbf p}^2\big)=\dfrac{1}{2E_{\mathbf p}}\big[\delta(p^0-E_{\mathbf p})+\delta(p^0+E_{\mathbf p})\big]\); the step function \(\theta(p^0)\) keeps only the positive-energy root, and doing the \(p^0\) integral collapses \(d^4p\to d^3p/2E\). B

\[ d^4p,\ \ \delta(p^2-m^2),\ \ \theta(p^0)\ \ \text{are each Lorentz invariant}\ \Rightarrow\ \frac{d^3p}{2E_{\mathbf p}}\ \text{is Lorentz invariant.} \]

\(d^4p\) is invariant (proper Lorentz transformations have unit Jacobian, \(|\det\Lambda|=1\)); \(p^2-m^2\) is a scalar; the sign of \(p^0\) is preserved by orthochronous transformations. Since the left side of Step 1 is manifestly invariant and equals \(\int d^3p/2E\,f\) for arbitrary invariant \(f\), the measure \(d^3p/2E\) is itself invariant. C

\[ \text{Direct check under a boost }\beta\text{ along }z:\quad \frac{dp_z'}{dp_z}=\gamma\!\left(1-\beta\frac{p_z}{E}\right)=\gamma\,\frac{E-\beta p_z}{E}=\frac{E'}{E},\qquad \frac{d^3p'}{E'}=\frac{d^3p}{E}. \]

With \(p_z'=\gamma(p_z-\beta E)\) and \(E'=\gamma(E-\beta p_z)\), and using \(dE/dp_z=p_z/E\) on shell; \(p_x,p_y\) are unchanged by a \(z\)-boost, so the transverse measure is inert and \(d^3p/E\) is preserved element-by-element. This confirms Step 2 explicitly. C

\[ d\Pi_n \equiv (2\pi)^4\,\delta^4\!\Big(P-\sum_{i=1}^{n} p_i\Big)\,\prod_{i=1}^{n}\frac{d^3p_i}{(2\pi)^3\,2E_i} \]

Assemble one invariant factor from Step 1 for each of the \(n\) final-state particles and impose total four-momentum conservation \(P=\sum p_i\) with the invariant \(\delta^4\). Every ingredient is a Lorentz scalar, so \(d\Pi_n\) is frame-independent. A

\[ d\Gamma = \frac{1}{2M}\,\overline{|\mathcal{M}|^2}\,d\Pi_n \]

Fermi's golden rule for a decaying particle at rest gives a transition rate \(\propto |\mathcal{M}|^2\times(\text{density of final states})\); the relativistic state normalisation contributes \(1/(2E_{\text{initial}})=1/2M\) in the rest frame, and the density of final states is precisely \(d\Pi_n\). The bar denotes averaging over initial and summing over final spins. B

\[ d\sigma = \frac{1}{F}\,\overline{|\mathcal{M}|^2}\,d\Pi_n,\qquad F = 4\,E_1 E_2\,|\vec v_1-\vec v_2| = 4\sqrt{(p_1\!\cdot p_2)^2 - m_1^2 m_2^2} \]

A cross-section is a rate per unit incident flux. The two incoming states carry normalisation \(1/(2E_1)(2E_2)\); the Møller flux factor \(|\vec v_1-\vec v_2|\) counts incident particles per unit area per unit time. The product \(4E_1E_2|\vec v_1-\vec v_2|\) is shown (Step 7) to equal the manifestly invariant form. B

\[ 4E_1E_2|\vec v_1-\vec v_2| \;\xrightarrow{\text{collinear}}\; 4\big|E_2\vec p_1-E_1\vec p_2\big| \;=\; 4\sqrt{(p_1\!\cdot p_2)^2-m_1^2m_2^2} \]

With collinear beams \(\vec v_i=\vec p_i/E_i\), so \(E_1E_2|\vec v_1-\vec v_2|=|E_2\vec p_1-E_1\vec p_2|\). Squaring and using \(E_i^2=|\vec p_i|^2+m_i^2\) together with \(p_1\!\cdot p_2=E_1E_2-\vec p_1\!\cdot\vec p_2\) reproduces \((p_1\!\cdot p_2)^2-m_1^2m_2^2\); the last form is a Lorentz scalar, so \(F\) is invariant. C

\[ F_{\text{CM}} = 4\,|\vec p_i|\,\sqrt{s},\qquad |\vec p_i|=\frac{\lambda^{1/2}(s,m_1^2,m_2^2)}{2\sqrt{s}} \]

In the centre-of-momentum frame \(\vec p_1=-\vec p_2=\vec p_i\) and \((p_1\!\cdot p_2)^2-m_1^2m_2^2=\tfrac14\lambda(s,m_1^2,m_2^2)\), with the Källén function \(\lambda(a,b,c)=a^2+b^2+c^2-2ab-2bc-2ca\); this is the two-body kinematics result carried over from the prior derivation. B

\[ d\Pi_2 = (2\pi)^4\,\delta^4(P-p_1-p_2)\,\frac{d^3p_1}{(2\pi)^3 2E_1}\frac{d^3p_2}{(2\pi)^3 2E_2} \;\xrightarrow{\ \int d^3p_2\ }\; \frac{1}{(2\pi)^2}\int \frac{d^3p_1}{4E_1E_2}\,\delta\!\big(\sqrt{s}-E_1-E_2\big) \]

Work in the CM frame, \(P=(\sqrt{s},\vec 0)\). The three-momentum delta sets \(\vec p_2=-\vec p_1\), leaving the energy delta; both \(E_1,E_2\) are now functions of \(|\vec p_1|\). B

\[ d^3p_1 = |\vec p_1|^2\,d|\vec p_1|\,d\Omega,\qquad \frac{d(E_1+E_2)}{d|\vec p_1|}=|\vec p_1|\!\left(\frac{1}{E_1}+\frac{1}{E_2}\right)=\frac{|\vec p_1|\,\sqrt{s}}{E_1E_2} \]

Spherical coordinates for the surviving integral; the derivative is needed to convert \(\delta(\sqrt s-E_1-E_2)\) into a factor via \(\delta(g(x))=\delta(x-x_0)/|g'(x_0)|\), with the root at physical \(|\vec p_1|=|\vec p_f|\). A

\[ d\Pi_2 = \frac{1}{(2\pi)^2}\,\frac{|\vec p_f|^2}{4E_1E_2}\cdot\frac{E_1E_2}{|\vec p_f|\,\sqrt{s}}\,d\Omega = \frac{1}{16\pi^2}\,\frac{|\vec p_f|}{\sqrt{s}}\,d\Omega \]

Substitute the Jacobian from Step 10 into Step 9; the factors \(E_1E_2\) cancel and one power of \(|\vec p_f|\) survives, giving the compact invariant two-body measure. A

\[ \Gamma = \frac{1}{2M}\int \overline{|\mathcal{M}|^2}\,\frac{|\vec p_f|}{16\pi^2\sqrt{s}}\,d\Omega \;\xrightarrow{\ \sqrt s=M,\ \overline{|\mathcal{M}|^2}\ \text{isotropic}\ }\; \frac{|\vec p_f|}{8\pi M^2}\,\overline{|\mathcal{M}|^2} \]

Insert Step 11 into the decay master formula (Step 5) with \(\sqrt s=M\) in the rest frame; if \(\overline{|\mathcal{M}|^2}\) is angle-independent, \(\int d\Omega=4\pi\) and \(2M\sqrt s=2M^2\). A

\[ \frac{d\sigma}{d\Omega} = \frac{1}{4|\vec p_i|\sqrt s}\cdot\overline{|\mathcal{M}|^2}\cdot\frac{|\vec p_f|}{16\pi^2\sqrt s} = \frac{1}{64\pi^2 s}\,\frac{|\vec p_f|}{|\vec p_i|}\,\overline{|\mathcal{M}|^2} \]

Insert Step 11 and the CM flux \(F=4|\vec p_i|\sqrt s\) (Step 8) into the scattering master formula (Step 6); \(4\cdot16\pi^2=64\pi^2\) and \(\sqrt s\cdot\sqrt s=s\). A

Result

\[ d\Gamma=\frac{1}{2M}\,\overline{|\mathcal{M}|^2}\,d\Pi_n,\qquad d\sigma=\frac{1}{4\sqrt{(p_1\!\cdot p_2)^2-m_1^2m_2^2}}\,\overline{|\mathcal{M}|^2}\,d\Pi_n \]

\[ \boxed{\ \Gamma=\frac{|\vec p_f|}{8\pi M^2}\,\overline{|\mathcal{M}|^2}\ }\qquad\qquad \boxed{\ \frac{d\sigma}{d\Omega}=\frac{1}{64\pi^2 s}\,\frac{|\vec p_f|}{|\vec p_i|}\,\overline{|\mathcal{M}|^2}\ } \]

Reading. The dynamics live entirely in \(|\mathcal{M}|^2\); the phase-space measure is a purely kinematic weight, positive-definite and Lorentz invariant, that counts how many ways the final-state momenta can be arranged consistent with energy-momentum conservation. A decay is that weight divided by \(2M\) (the rest-frame state normalisation); a cross-section is the same weight divided by the incident invariant flux \(F\). The boxed two-body forms are the special cases most experiments actually use, with \(|\vec p_f|\) the final-state CM momentum and \(|\vec p_i|\) the initial one.

Units check. In natural units \(d\Pi_2\) is dimensionless (each \(d^3p/2E\sim E^2\), two of them give \(E^4\), \(\delta^4\sim E^{-4}\)). For a \(1\to2\) decay \([\,\overline{|\mathcal{M}|^2}\,]=E^2\) and \(\Gamma\sim E^2/E = E\) — an inverse time, correct. For \(2\to2\), \(\overline{|\mathcal{M}|^2}\) is dimensionless, \(F\sim E^2\), so \(\sigma\sim E^{-2}\) — an area (\(1\,\text{GeV}^{-2}=0.3894\,\text{mb}\)), correct.

Limiting cases

Threshold, \(M\to m_1+m_2\): \(\lambda\to0\), so \(|\vec p_f|\to0\) and both \(\Gamma\) and \(\sigma\) vanish — no phase space at rest for the products.
Massless products, \(m_1=m_2=0\): \(|\vec p_f|=\sqrt s/2\), and \(d\Pi_2=\dfrac{1}{32\pi^2}\,d\Omega\), a constant times solid angle.
Elastic scattering, \(m_i^{\text{out}}=m_i^{\text{in}}\): \(|\vec p_f|=|\vec p_i|\), so \(\dfrac{d\sigma}{d\Omega}=\dfrac{\overline{|\mathcal{M}|^2}}{64\pi^2 s}\).
High energy, \(s\gg m_i^2\): the momentum ratio \(|\vec p_f|/|\vec p_i|\to1\) and all mass corrections to the measure fade as \(m^2/s\).
Isotropic amplitude: integrating the angular factor gives \(\sigma=\dfrac{|\vec p_f|}{16\pi s\,|\vec p_i|}\,\overline{|\mathcal{M}|^2}\).

Breaks when

Identical particles in the final state. The formulae above count each momentum configuration once; for \(k\) identical final particles one must divide the total rate or cross-section by the symmetry factor \(S=\prod_j k_j!\), else phase space is over-counted (e.g. \(\Phi\to\pi\pi\) needs a factor \(1/2\)).
Unstable / broad intermediate states. When a final particle has a width comparable to the available energy, the sharp on-shell \(\delta(p^2-m^2)\) is a poor approximation; it must be smeared into a Breit-Wigner spectral function \(\propto \dfrac{1}{(p^2-m^2)^2+m^2\Gamma^2}\), and the naive two-body result fails.
More than two incoming particles. The flux factor \(1/F=1/(4E_1E_2|\vec v_1-\vec v_2|)\) is defined only for a two-body initial state; three-body "collisions" have no single incident flux and the master cross-section formula does not apply.
Collinear/soft massless emission. For processes with massless radiated quanta, phase-space integrals develop infrared and collinear divergences; the bare formula gives a divergent rate until real and virtual contributions are combined (KLN / dressed observables).

Failure modes

Non-relativistic normalisation leak. Using \(\int d^3p/(2\pi)^3\) without the \(1/2E\) (Schrödinger convention) while keeping the relativistic \(|\mathcal{M}|^2\); the two conventions must match or spurious factors of \(2E\) appear.
Forgetting the \(1/2M\) or \(1/F\). Writing \(\Gamma=\int|\mathcal{M}|^2 d\Pi_n\) with no initial-state normalisation — the single most common dimensional error, off by a factor of energy.
Wrong momentum in the two-body factor. Using \(|\vec p_i|\) where \(|\vec p_f|\) belongs (or vice versa) in \(\Gamma\); the decay uses the final momentum only, but scattering carries the ratio \(|\vec p_f|/|\vec p_i|\).
Omitting the identical-particle symmetry factor when integrating over the full \(4\pi\) for two identical products.
Averaging vs summing spins. Forgetting to average over initial spins (dividing by \((2s_1+1)(2s_2+1)\)) while summing over final ones — the unpolarised cross-section needs both.
Using \(s=M^2\) off resonance. Setting \(\sqrt s=M\) in the scattering formula (that substitution belongs only to the rest-frame decay).

Discussion

The deep point is factorisation: an observable rate splits cleanly into a dynamical scalar \(|\mathcal{M}|^2\) and a kinematic measure \(d\Pi_n\). This is not an accident of perturbation theory — it follows from the LSZ reduction of the \(S\)-matrix, where each external leg contributes exactly one invariant momentum-space factor \(d^3p/(2\pi)^3 2E\), and the connected amputated Green's function supplies \(\mathcal{M}\). Because the measure is manifestly Lorentz invariant, the whole apparatus is frame-agnostic: the same \(|\mathcal{M}|^2\) computed once yields the collider cross-section and the fixed-target cross-section merely by evaluating the flux and phase space in the relevant frame.

The two prefactors have complementary physical readings. The \(1/2M\) in the decay is time dilation in disguise: a particle boosted to energy \(E=\gamma M\) has its rest-frame width \(\Gamma\) reduced to a lab decay rate \(\Gamma/\gamma\), and the invariant \(1/2E\) normalisation is what encodes that dilation covariantly. The invariant flux \(F=4\sqrt{(p_1\!\cdot p_2)^2-m_1^2m_2^2}\) is the covariant statement of "beams sweeping through each other": it reduces to \(4|\vec p_i|\sqrt s\) in the CM frame and to \(4 m_2 |\vec p_1^{\,\text{lab}}|\) against a fixed target, so a single computed cross-section transfers between the two setups without touching \(|\mathcal{M}|^2\).

The measure also organises multi-body final states through recursive structure: \(d\Pi_n\) factorises into \(d\Pi_2\) for a subsystem times \(d\Pi_{n-1}\) times an invariant-mass integral \(dm^2/2\pi\). This is the engine behind the Dalitz plot for three-body decays, where the density of events across the \((m_{12}^2,m_{23}^2)\) plane is flat for constant \(|\mathcal{M}|^2\) — so any structure a Dalitz plot shows is dynamics, resonances and interference, cleanly separated from kinematics.

At the most rigorous level, the invariance in Step 2 rests on the mass-shell hyperboloid \(p^2=m^2,\ p^0>0\) being an orbit of the orthochronous Lorentz group, and \(d^3p/2E\) being (up to normalisation) the unique group-invariant measure on that orbit — a special case of the Haar measure on a homogeneous space \(SO^+(3,1)/SO(3)\). This is why no other combination of \(d^3p\) and energy could serve: invariance fixes the measure uniquely, and the relativistic state normalisation \(\langle p|p'\rangle=2E(2\pi)^3\delta^3\) is precisely the choice that makes the resolution of the identity \(\int \frac{d^3p}{(2\pi)^3 2E}|p\rangle\langle p|=\mathbb{1}\) covariant.

Common misconceptions. Phase space is not a probability and not dimensionless "number of states" in any absolute sense — it is a measure, and only ratios or products with a normalised \(|\mathcal{M}|^2\) are physical. And \(|\vec p_f|\) appearing in the two-body width does not mean "faster products decay faster"; it is the density-of-states Jacobian, reflecting how much momentum room the kinematics allow.

Worked examples

Example 1 — Two-body decay width and lifetime.

\[ M=1.00\ \text{GeV},\quad m_1=0.140\ \text{GeV},\quad m_2=0.500\ \text{GeV},\quad \overline{|\mathcal{M}|^2}=0.0100\ \text{GeV}^2\ (\text{constant}). \]

Set up a distinct-mass two-body decay (distinct masses avoid an identical-particle symmetry factor). A

\[ \lambda=M^4+m_1^4+m_2^4-2M^2m_1^2-2M^2m_2^2-2m_1^2m_2^2 = 1+3.84\!\times\!10^{-4}+0.0625-0.0392-0.500-0.0098=0.5139\ \text{GeV}^4 \]

Källén function with \(M^2=1,\ m_1^2=0.0196,\ m_2^2=0.250\) (all GeV\(^2\)). A

\[ |\vec p_f|=\frac{\lambda^{1/2}}{2M}=\frac{0.7168}{2.00}=0.3584\ \text{GeV} \]

Final-state CM momentum from Step 8 of the derivation. A

\[ \Gamma=\frac{|\vec p_f|}{8\pi M^2}\,\overline{|\mathcal{M}|^2}=\frac{0.3584}{8\pi(1.00)}\times0.0100=\frac{0.3584}{25.13}\times0.0100=1.43\times10^{-4}\ \text{GeV} \]

Insert into the boxed two-body width. A

\[ \tau=\frac{\hbar}{\Gamma}=\frac{6.582\times10^{-25}\ \text{GeV·s}}{1.43\times10^{-4}\ \text{GeV}}=4.6\times10^{-21}\ \text{s} \]

Convert width to lifetime with \(\hbar=6.582\times10^{-25}\) GeV·s. A

\[ \Gamma\approx1.43\times10^{-4}\ \text{GeV}=0.143\ \text{MeV},\qquad \tau\approx4.6\times10^{-21}\ \text{s} \]

Reading. A weak-strength constant amplitude produces a sub-MeV width and a strong-interaction-scale lifetime; note \(\Gamma\) scales linearly with \(|\vec p_f|\), so this width would collapse to zero as \(M\to m_1+m_2=0.640\) GeV.

Units check. \(\text{GeV}^2/(\text{GeV}^2)\times\text{GeV}^0\)… explicitly \([|\vec p_f|/M^2]\times[|\mathcal{M}|^2]=\text{GeV}^{-1}\times\text{GeV}^2=\text{GeV}\), an inverse time. Good.

Example 2 — Total cross-section for a constant amplitude.

\[ \sqrt{s}=10.0\ \text{GeV},\quad m_i\approx0\ \Rightarrow\ |\vec p_f|=|\vec p_i|=\tfrac{\sqrt s}{2},\quad \overline{|\mathcal{M}|^2}=0.300\ (\text{dimensionless}). \]

High-energy, massless \(2\to2\) with an isotropic constant amplitude. A

\[ \frac{d\sigma}{d\Omega}=\frac{1}{64\pi^2 s}\,\frac{|\vec p_f|}{|\vec p_i|}\,\overline{|\mathcal{M}|^2}=\frac{\overline{|\mathcal{M}|^2}}{64\pi^2 s} \]

Elastic/massless ratio \(|\vec p_f|/|\vec p_i|=1\). A

\[ \sigma=\int\frac{d\sigma}{d\Omega}\,d\Omega=\frac{\overline{|\mathcal{M}|^2}}{64\pi^2 s}\times4\pi=\frac{\overline{|\mathcal{M}|^2}}{16\pi s} \]

Isotropic integrand, \(\int d\Omega=4\pi\). A

\[ \sigma=\frac{0.300}{16\pi(100\ \text{GeV}^2)}=\frac{0.300}{5027\ \text{GeV}^2}=5.97\times10^{-5}\ \text{GeV}^{-2} \]

Numerics with \(s=100\) GeV\(^2\), \(16\pi=50.27\). A

\[ \sigma=5.97\times10^{-5}\ \text{GeV}^{-2}\times0.3894\ \frac{\text{mb}}{\text{GeV}^{-2}}=2.32\times10^{-5}\ \text{mb}=23.2\ \text{nb} \]

Convert with \(1\ \text{GeV}^{-2}=0.3894\) mb; \(1\) mb\(=10^{6}\) nb. A

\[ \sigma\approx6.0\times10^{-5}\ \text{GeV}^{-2}\approx23\ \text{nb} \]

Reading. A constant amplitude gives \(\sigma\propto1/s\), the hallmark \(1/s\) fall-off of a pointlike cross-section; doubling \(\sqrt s\) quarters the rate.

Units check. Dimensionless \(|\mathcal{M}|^2\) over \(s\ (=\text{GeV}^2)\) gives \(\text{GeV}^{-2}\), an area. Good.

Problems

A particle of mass \(M=0.500\) GeV decays to two products of equal mass \(m=0.100\) GeV. Compute the final-state CM momentum \(|\vec p_f|\).
Solution
For equal masses \(|\vec p_f|=\dfrac{M}{2}\sqrt{1-\dfrac{4m^2}{M^2}}\). Here \(4m^2/M^2=4(0.0100)/0.250=0.160\), so \(\sqrt{1-0.160}=\sqrt{0.840}=0.9165\), giving \(|\vec p_f|=0.250\times0.9165=0.229\) GeV.
Using the result of Problem 1 (\(M=0.500\) GeV, \(m=0.100\) GeV, \(|\vec p_f|=0.229\) GeV), and a constant \(\overline{|\mathcal{M}|^2}=0.0200\ \text{GeV}^2\), find the decay width \(\Gamma\). The two products are distinct, so no symmetry factor.
Solution
\(\Gamma=\dfrac{|\vec p_f|}{8\pi M^2}\overline{|\mathcal{M}|^2}=\dfrac{0.229}{8\pi(0.250)}\times0.0200=\dfrac{0.229}{6.283}\times0.0200=0.03645\times0.0200=7.29\times10^{-4}\) GeV \(=0.729\) MeV.
Show that the two-body decay width vanishes at threshold, and quantify it: for \(m_1=m_2=0.140\) GeV, compute \(|\vec p_f|\) at (a) \(M=0.280\) GeV and (b) \(M=0.300\) GeV.
Solution
\(|\vec p_f|=\tfrac{M}{2}\sqrt{1-4m^2/M^2}\). (a) At \(M=0.280=2m\): \(4m^2/M^2=4(0.0196)/0.0784=1\), so \(\sqrt{1-1}=0\), \(|\vec p_f|=0\) — exact threshold, \(\Gamma=0\). (b) At \(M=0.300\): \(4m^2/M^2=0.0784/0.0900=0.8711\), \(\sqrt{1-0.8711}=\sqrt{0.1289}=0.359\), \(|\vec p_f|=0.150\times0.359=0.0539\) GeV. The width turns on continuously from zero as the mass rises above \(m_1+m_2\).
Two massless particles scatter elastically at \(\sqrt s=5.00\) GeV with a constant \(\overline{|\mathcal{M}|^2}=0.500\). Compute the total cross-section in GeV\(^{-2}\) and in nb.
Solution
Massless/elastic gives \(|\vec p_f|/|\vec p_i|=1\), so \(\sigma=\dfrac{\overline{|\mathcal{M}|^2}}{16\pi s}=\dfrac{0.500}{16\pi(25.0)}=\dfrac{0.500}{1257}=3.98\times10^{-4}\ \text{GeV}^{-2}\). Convert: \(3.98\times10^{-4}\times0.3894\ \text{mb}=1.55\times10^{-4}\) mb \(=155\) nb.
(Harder.) The invariant flux is \(F=4\sqrt{(p_1\!\cdot p_2)^2-m_1^2m_2^2}\). Show that in the fixed-target (lab) frame, where particle 2 is at rest, \(F=4\,m_2\,|\vec p_1^{\,\text{lab}}|\). Then evaluate it for a \(p_1^{\text{lab}}\)-momentum beam of \(|\vec p_1|=3.00\) GeV striking a target of mass \(m_2=0.938\) GeV.
Solution
In the lab, \(p_2=(m_2,\vec 0)\) and \(p_1=(E_1,\vec p_1)\), so \(p_1\!\cdot p_2=E_1 m_2\). Then \((p_1\!\cdot p_2)^2-m_1^2m_2^2=m_2^2 E_1^2-m_1^2m_2^2=m_2^2(E_1^2-m_1^2)=m_2^2|\vec p_1|^2\). Taking the square root, \(F=4\sqrt{m_2^2|\vec p_1|^2}=4m_2|\vec p_1^{\,\text{lab}}|\). Numerically, \(F=4(0.938)(3.00)=11.3\ \text{GeV}^2\). (Note the beam energy \(E_1\) cancels entirely — the lab flux depends only on the target mass and the beam three-momentum, as expected for a stationary target.)