Decays, Thresholds, and Invariant Phase Space

A reaction is kinematically allowed when conserved four-momentum can be partitioned among future on-shell final momenta. Invariant mass determines thresholds; invariant phase space measures the allowed configurations. Neither tells us the transition probability until the dynamics is supplied.

We use $c=\hbar=1$ and the site signature $(-,+,+,+)$. For total momentum $P^\mu$, the available center-of-momentum energy is $\sqrt{s}$ with $s=-P^2$.

Required background. Events, Intervals, and Minkowski Geometry begins the Special Relativity sequence and supplies causal geometry; Relativistic Energy, Momentum, and Mass Shells supplies on-shell four-momentum; Collision Kinematics and Center-of-Momentum Frames supplies $s$, center-of-momentum existence, and two-body kinematics. Helpful background. Calculus and Linear Algebra supplies delta functions, measures, and quadratic invariants; Classical and Variational Mechanics supplies conservation laws and fixed-target intuition.

Thresholds from invariant mass

For final particles of masses $m_i$, their total momentum is future causal. In the center-of-momentum frame,

$$\sqrt{s}=\sum_i E_i^* \ge\sum_i m_i.$$

Therefore the threshold condition is

$$s_{\rm thr}=\left(\sum_i m_i\right)^2,$$

provided the final particles can all be at rest relative to one another at threshold and all additional conserved quantum numbers can be satisfied.

That proviso matters for massless particles. A genuine massless particle cannot be at rest, and zero four-momentum is not a photon state. If equality would require a massless daughter to carry zero four-momentum, $\sum_i m_i$ is a kinematic infimum approached by physical states, not an attained final configuration.

For projectile $a$ incident on a massive target $b$ at rest, $m_b\gt0$, and

$$s=m_a^2+m_b^2+2m_bE_a.$$

For an endothermic production channel whose minimum total final rest mass satisfies $M_f\gt m_a+m_b$, the projectile’s threshold total energy is

$$E_{a,\rm thr} =\frac{M_f^2-m_a^2-m_b^2}{2m_b}.$$

The threshold kinetic energy is $K_{a,\rm thr}=E_{a,\rm thr}-m_a$. It exceeds the increase in total rest mass because the final state must retain center-of-momentum motion in the fixed-target laboratory. If $M_f\le m_a+m_b$, the equality above does not define a positive beam-energy threshold. For two massive incoming particles, their mutual rest configuration already supplies invariant mass $m_a+m_b$; for a massless projectile, which has no rest frame, the kinematic infimum is instead $E_a\to0^+$ whenever the channel’s other conservation laws permit it.

Two-body decays

For a parent of mass $M$ decaying at rest into masses $m_1,m_2$, momentum conservation fixes equal and opposite three-momenta. Their magnitude is

$$p^* =\frac{\sqrt{\lambda(M^2,m_1^2,m_2^2)}}{2M},$$

with

$$\lambda(x,y,z)=x^2+y^2+z^2-2xy-2xz-2yz.$$

The daughter energies are

$$E_1^*=\frac{M^2+m_1^2-m_2^2}{2M}, \qquad E_2^*=\frac{M^2+m_2^2-m_1^2}{2M}.$$

The factorization

$$\lambda(M^2,m_1^2,m_2^2) =\bigl[M^2-(m_1+m_2)^2\bigr] \bigl[M^2-(m_1-m_2)^2\bigr]$$

separates the algebraic branches. For positive daughter masses, $M=m_1+m_2$ is an attained threshold and $p^*=0$. In the interval $|m_1-m_2|\lt M\lt m_1+m_2$, the square root is imaginary. For $M\le|m_1-m_2|$ it becomes real again, but at least one displayed daughter energy is nonpositive. Thus the physical future-directed branch is precisely $M\ge m_1+m_2$; reality of the square root alone is not enough. With a massless daughter, the equality boundary may again be only an infimum because zero four-momentum is not a physical massless-particle state.

Lorentz-invariant phase space

For one future on-shell particle, the measure

$$\frac{d^3\mathbf p}{(2\pi)^3\,2E_{\mathbf p}}$$

is invariant under proper orthochronous Lorentz transformations. The four-dimensional expression $d^4p\,\delta(p^2+m^2)\Theta(p^0)$ imposes the mass shell and selects its future sheet. Integrating out the redundant energy coordinate gives

$$\frac{d^3\mathbf p}{(2\pi)^3} \int_{-\infty}^{\infty}dp^0\, \delta\!\left(-(p^0)^2+E_{\mathbf p}^2\right)\Theta(p^0) =\frac{d^3\mathbf p}{(2\pi)^3\,2E_{\mathbf p}},$$

because the positive root of $p^2+m^2=0$ is $p^0=E_{\mathbf p}$ and the delta-function Jacobian is $2E_{\mathbf p}$. This measure matches the relativistic one-particle state normalization

$$\langle\mathbf p',\sigma'\mid\mathbf p,\sigma\rangle =(2\pi)^3\,2E_{\mathbf p} \,\delta_{\sigma'\sigma} \,\delta^{(3)}(\mathbf p'-\mathbf p).$$

The $n$-body phase-space element is

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

With this page’s convention, $d\Phi_n$ includes the displayed global $(2\pi)^4$. For a sharp-mass parent decaying in its rest frame,

$$d\Gamma =\frac{1}{2M}\,\overline{|\mathcal M|^2}\,d\Phi_n.$$

The flux factor $1/(2M)$ is the parent-rest-frame decay factor. The bar means: sum over unobserved final spins and, for an unpolarized parent ensemble, average over its initial spin states. Do not average over a parent polarization that has been prepared and specified. The amplitude $\mathcal M$ contains the interaction dynamics; $d\Phi_n$ contains kinematics.

If the full integration treats identical final particles as labeled, divide by

$$\prod_\alpha n_\alpha!,$$

where $n_\alpha$ is the multiplicity of each identical species, unless the chosen integration domain or amplitude convention already removes those permutations.

The Particle Data Group uses a nearby convention without the global $(2\pi)^4$ in its phase-space definition. Explicitly,

$$d\Phi_n^{\rm page}=(2\pi)^4d\Phi_n^{\rm PDG}.$$

Thus the formula above is identical to PDG Eq. (49.11), which writes the global $(2\pi)^4$ in $d\Gamma$ instead. Keeping the full formula with the state normalization prevents an invisible $2\pi$ mismatch.

Worked example: two-photon pair threshold

Two photons with energies $E_1,E_2$ meet at angle $\theta$ between their propagation directions. Since each is null,

$$s=-(k_1+k_2)^2 =2E_1E_2(1-\cos\theta).$$

Producing a pair of particles of mass $m$ requires

$$2E_1E_2(1-\cos\theta)\ge4m^2.$$

Parallel photons have $\theta=0$ and $s=0$, however large their individual energies; they have no center-of-momentum frame and cannot produce the pair in isolation. For a head-on collision with $E_1=E_2=E$, $s=4E^2$, so the threshold is $E=m$.

Phase-space interpretation and limits

Near a nondegenerate two-body threshold, the available momentum $p^*$ tends to zero, and the integrated phase-space volume is suppressed. This often suppresses the rate, but the power law and angular pattern also depend on $\mathcal M$, spin, symmetry, and partial waves.

For an unstable resonance, a sharp on-shell mass may be inadequate; its spectral line shape and interference with nonresonant amplitudes matter. In external fields, finite volumes, media, or curved spacetime, translation symmetry and the usual global four-momentum delta function may require modification.

The measure is not a probability distribution by itself. A statement such as “decays uniformly in phase space” is a dynamical model that takes $|\mathcal M|^2$ to be constant over the allowed region.

The normalization and recursive structure of Lorentz-invariant phase space are given in Miller, Tovey, and Jackson 2025, §49.4, p. 2, Eqs. (49.11)–(49.13), PDF, and the two-body formulas in Miller, Tovey, and Jackson 2025, §49.4.2, p. 3, Eqs. (49.16)–(49.18), PDF. The phase-space translation above makes the different placement of $(2\pi)^4$ explicit.

Exercises

The real but unphysical lower branch

Take $M=1$, $m_1=3$, and $m_2=1$ in a common energy unit. Evaluate $\lambda$, $p^*$, and both daughter energies. Why does a real $p^*$ not make this a physical decay?

Solution

The Källén function is

$$\lambda(1,9,1)=45, \qquad p^*=\frac{\sqrt{45}}{2}.$$

However,

$$E_1^*=\frac{1+9-1}{2}=\frac92, \qquad E_2^*=\frac{1+1-9}{2}=-\frac72.$$

The second daughter is past-directed. The lower algebraic branch makes the momentum magnitude real but violates the requirement that every physical final four-momentum be future-directed.

Massless projectile threshold

A massless projectile strikes a target of mass $m_b\gt0$ at rest and produces a final state of minimum mass $M_f\gt m_b$. Find the threshold projectile energy.

Solution

Set $m_a=0$ in the fixed-target invariant:

$$s=m_b^2+2m_bE_a.$$

At threshold $s=M_f^2$, hence

$$E_{a,\rm thr} =\frac{M_f^2-m_b^2}{2m_b}.$$

The positivity assumption $M_f\gt m_b$ makes this a genuine production threshold. If $M_f\le m_b$, the invariant-mass condition alone supplies only the infimum $E_a\to0^+$ rather than a negative threshold; other conserved charges may still forbid the channel. In every case the target-plus-projectile system must be isolated and the proposed final state must respect all conserved charges.

The next chapter, Accelerated Observers and Reference Frames, extends local observer measurements along accelerated and rotating families of worldlines.

References

• Miller, D., D. R. Tovey, and J. D. Jackson. “Kinematics.” In S. Navas et al. (Particle Data Group), Review of Particle Physics, Physical Review D 110, 030001 (2024), 2025 update. Official PDF.

Further reading

• Tong, David. Lectures on Dynamics and Relativity, §7.5, printed pp. 145–149, especially Eqs. (7.34)–(7.35), University of Cambridge, 2013.