Invariant Phase Space Preview

Relativistic phase space counts allowed final momenta with an invariant measure and energy–momentum conservation. It becomes a cross section only after weighting by an amplitude and dividing by incident flux. This preview derives the two-body measure and its use in elementary rates; general multiparticle integration and field-theory amplitude calculations lie beyond its scope. The one-particle mass-shell derivation belongs to Relativistic Phase Space.

Required background. Relativistic Phase Space provides $d^3p/(2E)$; Relativistic Normalization fixes the state and amplitude factors. Helpful background. QFT Bridge: S-Matrix explains asymptotic multiparticle states.

Covariant amplitudes and final-state counting

Use $\hbar=c=1$ and future-directed on-shell momenta throughout. Normalize each external particle by $\langle p',r|p,s\rangle=2E_p(2\pi)^3\delta^3(\mathbf p'-\mathbf p)\delta_{rs}$. Define the connected scattering coefficient by

$$\langle f|S-I|i\rangle =i(2\pi)^4\delta^4(P_f-P_i)\mathcal M_{fi}.$$

Take the one-particle measure as input:

$$d\Pi_k=\frac{d^3k}{(2\pi)^3\,2E_k}.$$

For total momentum $P$, define

$$d\Phi_n(P)= (2\pi)^4\delta^4\!\left(P-\sum_{a=1}^n k_a\right) \prod_{a=1}^n d\Pi_{k_a}.$$

The $(2\pi)^4$ is inside this definition. The PDG kinematics review places it outside its phase-space symbol; translating that convention gives the same rates. All integrations here use positive-energy particles, not an integral over both mass-shell sheets.

The displayed product labels the outgoing momenta. If integrating over all ordered momenta of $n_j$ identical final particles of each species $j$, include

$$\frac1{\mathcal S_f},\qquad \mathcal S_f=\prod_j n_j!.$$

Alternatively integrate each physical final state once. Do not apply both restrictions. Sum final polarizations; for an unpolarized initial beam, average over initial spin states only after specifying their population weights. Symmetrization of the amplitude and the counting factor in the integral have different roles.

Incident flux and elementary rates

For two incoming particles define

$$\mathcal F= 4\sqrt{(p_1\cdot p_2)^2-m_1^2m_2^2}.$$

Then, for fixed initial spin labels,

$$d\sigma_{2\to n} =\frac1{\mathcal F\,\mathcal S_f} \sum_{\text{final spins}}|\mathcal M_{fi}|^2\,d\Phi_n.$$

In a chosen beam frame this flux is $4E_1E_2v_{\rm M}$, where

$$v_{\rm M} =\frac{\sqrt{(p_1\cdot p_2)^2-m_1^2m_2^2}}{E_1E_2}.$$

This relative-flux factor is sometimes called the Møller velocity. It is not itself a Lorentz scalar or a signal speed. For noncollinear beams it need not equal $|\mathbf v_1-\mathbf v_2|$. The invariant combination $\mathcal F$ is the useful general statement.

In the center-of-momentum frame let $s=(p_1+p_2)^2$ and let $p_i$ be the magnitude of either incoming momentum. Since the incoming momenta are opposite,

$$\mathcal F=4p_i\sqrt s.$$

A parent of mass $M$ decaying at rest instead has

$$d\Gamma_{\rm rest} =\frac1{2M\mathcal S_f} \sum_{\text{final spins}}|\mathcal M|^2\,d\Phi_n.$$

Initial parent spin is fixed or averaged according to the preparation. The same decay in a frame where the parent energy is $E_P$ has a coordinate-time rate smaller by $M/E_P$. Decay and two-beam scattering use different initial flux factors.

Reducing the two-body delta function

For two final masses $m_3,m_4$, work in the frame $P=(\sqrt s,\mathbf0)$. The spatial delta sets $\mathbf k_4=-\mathbf k_3$. Calling their common magnitude $p$ and writing $E_j=\sqrt{p^2+m_j^2}$ gives

$$\begin{aligned} d\Phi_2&=\frac{p^2dp\,d\Omega}{16\pi^2E_3E_4}\\ &\quad\times\delta\!\left(\sqrt s-E_3-E_4\right). \end{aligned}$$

The positive root is

$$p_f=\frac{\sqrt{\lambda(s,m_3^2,m_4^2)}}{2\sqrt s},$$

where

$$\begin{aligned} \lambda(x,y,z)&=x^2+y^2+z^2\\ &\quad-2xy-2xz-2yz,\\ \lambda(s,m_3^2,m_4^2) &=[s-(m_3+m_4)^2]\\ &\quad\times[s-(m_3-m_4)^2]. \end{aligned}$$

The physical condition is $\sqrt s\geq m_3+m_4$ with positive individual energies. A real square root on the other algebraic branch is not an additional allowed two-particle channel.

Above threshold the magnitude of the radial delta derivative is

$$\left|\frac{d}{dp}(\sqrt s-E_3-E_4)\right|_{p_f} =\frac{p_f\sqrt s}{E_3E_4}.$$

Thus

$$d\Phi_2=\frac{p_f}{16\pi^2\sqrt s}\,d\Omega =\frac{\sqrt{\lambda(s,m_3^2,m_4^2)}}{32\pi^2s}\,d\Omega.$$

Integrating over the full solid angle gives

$$\Phi_2=\frac{p_f}{4\pi\sqrt s} =\frac{\sqrt{\lambda(s,m_3^2,m_4^2)}}{8\pi s}.$$

These are the bare kinematic measures, without spin sums or identical-particle factors. At threshold the formulas are interpreted as limits from above. For two massless particles $\Phi_2=1/(8\pi)$.

Two-body scattering and decay checks

For distinguishable final particles and fixed spin labels,

$$\frac{d\sigma_{2\to2}}{d\Omega} =\frac{|\mathcal M|^2}{64\pi^2s}\frac{p_f}{p_i}.$$

When the appropriate spin-summed or averaged squared amplitude is angle independent, the full-angle result is $\sigma=|\mathcal M|^2p_f/(16\pi s p_i)$. An identical final pair contributes an additional $1/2!$ if the full ordered solid angle is integrated.

For an isotropic two-body decay into distinguishable particles,

$$\Gamma_{\rm rest} =\frac{|\mathcal M|^2p_f}{8\pi M^2}.$$

For example, a spinless parent with a constant amplitude $g_d$ to two massless distinguishable scalars has $\Gamma=|g_d|^2/(16\pi M)$. Here $g_d$ has mass dimension one, so the rate has mass dimension one. For identical daughters, with the same specified physical amplitude, the full-angle counting factor halves this result.

At a two-massive-particle threshold, $p_f$ vanishes as the square root of the excess center-of-momentum energy. A finite nonsingular amplitude therefore acquires this kinematic suppression. Angular momentum barriers, singular amplitudes, long-range forces, or a simultaneously vanishing incoming flux can change the complete cross-section threshold law. The measure by itself does not settle it.

Exercises

1. A parent with mass $M=5$ decays into daughters with masses $3$ and $0$, in a consistent natural-unit system. Find $p_f$, both energies, and the integrated bare phase space.

Solution

$p_f=(25-9)/(2\cdot5)=8/5$, $E_3=(25+9)/(2\cdot5)=17/5$, and $E_4=8/5$. Their energies sum to $5$. $\Phi_2=p_f/(4\pi M)=2/(25\pi)$. This is not a decay rate until multiplied by the squared amplitude and $1/(2M)$.

2. Derive the relative-flux identity $v_{\rm M}^2=|\mathbf v_1-\mathbf v_2|^2- |\mathbf v_1\times\mathbf v_2|^2$.

Solution

Use $p_1\cdot p_2=E_1E_2(1-\mathbf v_1\cdot\mathbf v_2)$ and $m_a^2=E_a^2(1-\mathbf v_a^2)$. After division by $E_1^2E_2^2$, the radicand becomes $\mathbf v_1^2+\mathbf v_2^2-2\mathbf v_1\cdot\mathbf v_2 -\mathbf v_1^2\mathbf v_2^2+(\mathbf v_1\cdot\mathbf v_2)^2$. The cross-product identity gives the stated expression. For opposite nearly lightlike beams $v_{\rm M}\simeq2$; this flux factor is not a velocity of one object relative to another.

3. An elastic spinless $2\to2$ channel has constant $\mathcal M=\lambda_0$ and distinguishable particles. Find the total cross section and its dimension.

Solution

Elastic kinematics has $p_f=p_i$, so $\sigma=|\lambda_0|^2/(16\pi s)$. In four spacetime dimensions this invariant $2\to2$ amplitude is dimensionless, giving cross section of mass dimension $-2$. Restoring energy-based SI units multiplies by $(\hbar c)^2$. The constant-amplitude assumption is dynamical input, not a prediction of phase space.

References

• Miller, D., and D. R. Tovey, originally written by J. D. Jackson. “Kinematics.” In S. Navas et al. (Particle Data Group), Review of Particle Physics, Physical Review D 110, 030001 (2024), 2025 update, sections 49.3–49.5. Review. Its phase-space definition leaves $(2\pi)^4$ outside.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014, chapter 5. doi:10.1017/9781139540940. Scattering normalization and rates.
• Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge University Press, 1995, chapter 3. Asymptotic states, transition probabilities, and cross sections.