Lorentz-Invariant Phase Space

The $n$-particle Lorentz-invariant phase-space measure is the product of positive-energy on-shell measures, constrained by one four-momentum delta distribution. It can be reduced recursively by inserting the invariant mass of an intermediate cluster. The decisive normalization checks are $\Phi_2=\sqrt{\lambda}/(8\pi s)$ and, for three labeled massless particles, $\Phi_3=s/(256\pi^3)$.

Required background. Relativistic Scattering Kinematics supplies invariant masses, thresholds, and the Källén function.

Helpful background. Delta Distributions, Weak Derivatives, Pullbacks, and Pushforwards supplies the change-of-variables rule used to restrict momenta to their mass shells.

Restricting a four-momentum to its positive-energy shell

For a particle of mass $m$,

$$\begin{aligned} \mathrm d\Pi_m(p) &\equiv \frac{\mathrm d^4p}{(2\pi)^3} \theta(p^0)\delta(p^2-m^2)\\ &=\frac{\mathrm d^3\mathbf p}{(2\pi)^3 2E_{\mathbf p}}, \qquad E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}. \end{aligned}$$

The first line is manifestly invariant under proper orthochronous Lorentz transformations. To obtain the second, use

$$\delta\!\left((p^0)^2-E_{\mathbf p}^2\right) =\frac{\delta(p^0-E_{\mathbf p})+ \delta(p^0+E_{\mathbf p})}{2E_{\mathbf p}},$$

then let $\theta(p^0)$ retain the future root. This measure is reciprocal to the covariant state norm; changing the state normalization changes the measure in completeness at the same time.

The n-body measure

For total incoming momentum $P$ and $n$ labeled final momenta,

$$\boxed{ \mathrm d\Phi_n(P;p_1,\ldots,p_n) =(2\pi)^4\delta^{(4)}\!\left(P-\sum_{i=1}^n p_i\right) \prod_{i=1}^n \frac{\mathrm d^3\mathbf p_i}{(2\pi)^3 2E_i} }.$$

Every factor is invariant, so $\mathrm d\Phi_n$ is invariant. In four dimensions,

$$[\mathrm d\Phi_n]=2n-4.$$

The measure is supported only where $P$ can be written as a sum of future-directed on-shell momenta. It therefore vanishes below threshold. Schwartz derives the same measure from covariantly normalized S-matrix states in Schwartz 2014, § 5.1, pp. 59–62.

Two-body reduction

Let $P^2=s>0$ and work temporarily in the $P=(\sqrt{s},\mathbf0)$ frame. Use the spatial delta function to set $\mathbf p_2=-\mathbf p_1\equiv-\mathbf p$. Then

$$\mathrm d\Phi_2 =\frac{1}{16\pi^2} \frac{|\mathbf p_*|}{\sqrt{s}}\,\mathrm d\Omega,$$

where

$$|\mathbf p_*| =\frac{\sqrt{\lambda(s,m_1^2,m_2^2)}}{2\sqrt{s}}.$$

Equivalently,

$$\boxed{ \mathrm d\Phi_2 =\frac{\sqrt{\lambda(s,m_1^2,m_2^2)}}{32\pi^2s}\,\mathrm d\Omega, \qquad \Phi_2 =\frac{\sqrt{\lambda(s,m_1^2,m_2^2)}}{8\pi s}. }$$

The angular integral gives $4\pi$. At threshold $\lambda\to0$, so the available two-body phase space closes linearly in $|\mathbf p_*|$. The reduction and its use in rates are checked in Schwartz 2014, § 5.1.2, pp. 62–63 and Srednicki 2007, § 11, pp. 93–101.

Recursive factorization

Group $p_2,\ldots,p_n$ into $q=\sum_{i=2}^n p_i$. Insert

$$1=\int\mathrm d^4q\, \delta^{(4)}\!\left(q-\sum_{i=2}^n p_i\right), \qquad 1=\int\mathrm dQ^2\,\delta(Q^2-q^2).$$

Regrouping the factors of $2\pi$ and renaming $Q^2$ as $q^2$ yields the compact identity

$$\boxed{ \mathrm d\Phi_n(P;p_1,\ldots,p_n) =\frac{\mathrm dq^2}{2\pi} \mathrm d\Phi_2(P;p_1,q) \mathrm d\Phi_{n-1}(q;p_2,\ldots,p_n). }$$

For $P^2=s$ and future-directed momenta, the allowed range is

$$\left(\sum_{i=2}^n m_i\right)^2 \le q^2\le (\sqrt{s}-m_1)^2.$$

The lower limit is the threshold of the $q\to p_2+\cdots+p_n$ subspace; the upper limit is the threshold of the complementary $P\to p_1+q$ step. Repeating the factorization produces sequential invariant-mass and angular variables. This is a measure identity, not a dynamical factorization of $\mathcal M$.

A massless three-body benchmark

For three labeled massless particles, set $q=p_2+p_3$. The invariant mass lies in $0\le q^2\le s$. Using the integrated two-body formula twice,

$$\begin{aligned} \Phi_3(s;0,0,0) &=\int_0^s\frac{\mathrm dq^2}{2\pi} \Phi_2(s;0,\sqrt{q^2})\Phi_2(q^2;0,0)\\ &=\int_0^s\frac{\mathrm dq^2}{2\pi} \frac{s-q^2}{8\pi s}\frac{1}{8\pi}\\ &=\boxed{\frac{s}{256\pi^3}}. \end{aligned}$$

This result has dimension two, matching $2n-4$ for $n=3$. It is the volume for labeled particles before any identical-final-state rate factor.

The same benchmark has a useful Dalitz form. With $s_{12}=(p_1+p_2)^2$ and $s_{23}=(p_2+p_3)^2$, the massless physical region is

$$s_{12}\ge0, \qquad s_{23}\ge0, \qquad s_{12}+s_{23}\le s,$$

and, after integrating the overall orientation,

$$\mathrm d\Phi_3 =\frac{\mathrm ds_{12}\,\mathrm ds_{23}}{128\pi^3s}.$$

The triangle has area $s^2/2$, immediately reproducing $\Phi_3=s/(256\pi^3)$. This form is for three labeled particles and a four-dimensional parent with $s>0$.

Labels, identical particles, and cuts

$\mathrm d\Phi_n$ as defined above integrates each labeled momentum independently. If $k$ final particles are identical and the integration domain counts all $k!$ label permutations, the rate contains $1/k!$. One may instead integrate over a permutation-ordered region with no factorial, but never do both. This quantum-statistical counting is required even when a detector can order the measured momenta by energy or angle.

Restrictions such as detector cuts or a measurement function multiply the integrand; they do not change the definition of the underlying phase-space measure. Hadronic convolutions, finite-volume spectra, and maintained Monte Carlo generators require additional structures and are not derived here.

Common pitfalls

“$\mathrm d^3\mathbf p$ is Lorentz invariant.” It is not under boosts. The shell Jacobian $1/(2E_{\mathbf p})$ is essential.

“The symmetry factor belongs inside $\mathrm d\Phi_n$.” The displayed measure is labeled. A factorial belongs to the rate only when the chosen domain overcounts indistinguishable configurations.

“Recursive phase space means the amplitude factorizes.” The identity only reorganizes integration variables. Dynamical factorization requires a pole, approximation, or theorem of its own.

“A negative Källén function gives an imaginary phase-space volume.” Physical phase space is empty there. Analytic square roots used in amplitudes are a different construction.

Check your understanding

1. Show that $\mathrm d\Phi_2$ is dimensionless in four dimensions.

   Answer

   Each on-shell measure $\mathrm d^3\mathbf p/(2E)$ has dimension two, so their product has dimension four. The four-dimensional delta distribution has dimension minus four. The total is dimension zero.

2. For two massless particles, evaluate the integrated phase space.

   Answer

   Since $\lambda(s,0,0)=s^2$ for $s>0$, the formula gives $\Phi_2=s/(8\pi s)=1/(8\pi)$.

3. Integrate the massless Dalitz measure above and check both its normalization and dimension.

   Answer

   The triangular integral is

   $$\int_0^s\mathrm ds_{12} \int_0^{s-s_{12}}\mathrm ds_{23} =\frac{s^2}{2}.$$

   Multiplication by $1/(128\pi^3s)$ gives $s/(256\pi^3)$. Two invariant-mass differentials have dimension four and division by $s$ leaves dimension two, as required for $\mathrm d\Phi_3$.

Continue

Cross Sections and Decay Rates supplies flux, spin sums, and identical-particle factors. Phase-Space Integration and Monte Carlo Estimators develops numerical integration and validation; it should reproduce the two benchmarks above before tackling a nontrivial integrand.

References

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge: Cambridge University Press, 2014. DOI.
• Srednicki, Mark. Quantum Field Theory. Cambridge: Cambridge University Press, 2007. DOI.