advanced37 min studyLesson 605 in Physics

Expand near a physical threshold

Derive the square-root opening of a massive two-body channel, compare fixed-target and center-of-momentum thresholds, and state the dynamical assumptions behind threshold powers.

On this page

What you’ll understand

  • Expand the Källén momentum for small excess energy with positive daughter masses.
  • Translate a final-state mass threshold into a fixed-target beam requirement.
  • Separate endpoint location from phase-space and partial-wave population.

A threshold is the boundary between a real final-state momentum and a forbidden configuration. Near it, a small increase of invariant energy opens phase space in a characteristic way. The location follows from masses and conservation, while the observed rate also depends on angular momentum, interactions, and the incoming flux. An endpoint is similarly a limit of support rather than a promise that the event histogram peaks there or even populates it densely.

Expand the two-body momentum carefully

Let s=m1+m2+ϵ\sqrt s=m_1+m_2+\epsilon with positive daughter masses and excess energy much smaller than either mass. At threshold both daughters are at rest in their common center-of-momentum frame. Just above it, their nonrelativistic kinetic energies add to p∗2/(2μ)p^{*2}/(2\mu), where μ=m1m2/(m1+m2)\mu=m_1m_2/(m_1+m_2) is the reduced mass. This gives the leading square-root momentum opening. The same result follows by expanding both factors of the exact Källén function instead of assuming either daughter remains stationary.

The positivity of both masses matters. If one daughter is massless and the other has mass m, the exact momentum is (s−m2)/(2s)(s-m^2)/(2\sqrt s). With s=m+ϵ\sqrt s=m+\epsilon, its leading term is ϵ\epsilon, not the square root of excess energy. Applying the reduced-mass formula with zero reduced mass would therefore lose the leading behavior. A useful expansion states what is small, which quantities are held fixed, and which mass limits are excluded before discarding terms.

p∗=[s−(m1+m2)2][s−(m1−m2)2]2s≃2μϵ,μ=m1m2m1+m2p^*=\frac{\sqrt{[s-(m_1+m_2)^2][s-(m_1-m_2)^2]}}{2\sqrt s}\simeq\sqrt{2\mu\epsilon},\qquad \mu=\frac{m_1m_2}{m_1+m_2}
The exact expression uses +−−− and c = 1. The approximation requires 0 < ε ≪ m1,m2 with both masses positive and fixed. Momentum, masses, and ε have energy units; relative corrections vanish as excess energy becomes small compared with the masses.↔ Scroll wide equations horizontally.

Translate between beam thresholds and invariant thresholds

For a fixed target of mass mbm_b, incoming total energy gives s=ma2+mb2+2mbEas=m_a^2+m_b^2+2m_bE_a. For a final state with an accessible continuum above mass sum MfM_f, the minimum collision energy must satisfy both s≥Mf\sqrt s\ge M_f and the incoming on-shell requirement s≥ma+mb\sqrt s\ge m_a+m_b. In an endothermic channel, Mf≥ma+mbM_f\ge m_a+m_b, solving s=Mf2s=M_f^2 gives the beam requirement. For a lighter final mass sum, the kinematic kinetic-energy threshold is zero, rather than the negative value a naive mass-sum substitution could suggest. At a strictly endothermic threshold, the final particles can all be at rest relative to each other in their center-of-momentum frame, while moving together in the laboratory to carry the initial momentum.

This distinction explains why simply paying the new particle's rest energy is insufficient in a fixed-target reaction. Some laboratory energy must remain in overall recoil. The mass-sum condition is necessary only for channels allowed by conserved charges and other exact selection rules; kinematics alone cannot make a forbidden process occur. For hadronic beams, thresholds for a chosen constituent subprocess also depend on available momentum fractions and remnants. State whether a calculation refers to an exclusive complete reaction or an idealized elementary collision.

Ea,thr=max⁡ ⁣[ma,Mf2−ma2−mb22mb],Ka,thr=Ea,thr−ma,σL∝(pf∗)2L+1E_{a,\mathrm{thr}}=\max\!\left[m_a,\frac{M_f^2-m_a^2-m_b^2}{2m_b}\right],\qquad K_{a,\mathrm{thr}}=E_{a,\mathrm{thr}}-m_a,\qquad \sigma_L\propto(p_f^*)^{2L+1}
The threshold energies assume a stationary target and an allowed final-state continuum above minimum invariant mass Mf. The maximum enforces incoming total energy at least ma; a lighter final mass sum has zero kinematic kinetic threshold. The partial-wave power is a separate leading law: a short-range regular amplitude behaves as pf*^L, phase space supplies pf*, and the incoming flux must remain finite and smooth.↔ Scroll wide equations horizontally.

Distinguish a support boundary from its population

For a regular short-range two-body channel, a higher outgoing orbital angular momentum suppresses the amplitude with additional momentum powers. Combining that amplitude behavior with the opening phase space produces the threshold power in the equation. Long-range Coulomb effects, a nearby pole, singular interactions, or an initial channel also at threshold can modify the law. It is an approximation with a dynamical domain, not a consequence of energy conservation alone. A fitted exponent should therefore be interpreted with the interaction and flux model stated.

Invariant-mass endpoints in multibody decays can constrain mass combinations even when one product is invisible. Their ideal positions follow from a constrained maximization, but detector smearing and selection can move an observed edge or remove events near it. A cascade introduces additional on-shell constraints and can produce an endpoint below the naive parent-minus-daughter mass difference. Before extracting a mass, identify the topology, invisible masses, and which intermediate states are actually on shell. A measured edge is evidence about that whole model rather than a mass measurement independent of it.

The recoil cost of producing a new particle

Consider an allowed idealized reaction a+b→a+b+Xa+b\to a+b+X with ma=mb=1 GeVm_a=m_b=1\,\mathrm{GeV} and mX=2 GeVm_X=2\,\mathrm{GeV}. Particle b is stationary. Find the minimum projectile total and kinetic energies.

  1. At threshold the final mass sum is Mf=1+1+2=4 GeVM_f=1+1+2=4\,\mathrm{GeV}, so the minimum invariant is sthr=16 GeV2s_{\mathrm{thr}}=16\,\mathrm{GeV^2}. The products have no relative kinetic energy in their common rest frame.

  2. The laboratory invariant is s=1+1+2Eas=1+1+2E_a. Solving 16=2+2Ea16=2+2E_a gives total beam energy Ea,thr=7 GeVE_{a,\mathrm{thr}}=7\,\mathrm{GeV}.

  3. Subtract the projectile rest energy to find Ka,thr=6 GeVK_{a,\mathrm{thr}}=6\,\mathrm{GeV}. The total laboratory energy is eight GeV; the final system must retain overall motion rather than ending at rest in the laboratory.

The result

The threshold is seven GeV of projectile total energy, or six GeV of projectile kinetic energy. The new particle's two-GeV rest mass alone does not determine the fixed-target energy cost.

Explore the model

Interactive lab

Interactive lab · function

Inspect the quadratic relation behind a square-root opening

Reset and set coefficient a = 1 and position x = 0.5, predicting y = 0.25. Compare after a fresh Reset with a = 1 and x = 1, predicting y = 1. Read the y values. The toy curve y = ax² illustrates that a fourfold energy-like quadratic value corresponds to doubling a momentum-like x. It is not a reaction threshold solver and has no masses, Källén function, or partial-wave cross sections.

Inspect the quadratic relation behind a square-root openingThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe leading massive threshold relation is quadratic in momentum and linear in excess energy.

Check your understanding

When does the leading two-body relation p* ≈ √(2με) require a different expansion?

Try it yourself

Two equal daughters each have mass 1 GeV1\,\mathrm{GeV}. In the leading small-excess approximation compare ϵ=0.002 GeV\epsilon=0.002\,\mathrm{GeV} with 0.008 GeV0.008\,\mathrm{GeV}. Find the momenta and the ratios of ideal S-wave and P-wave production rates, assuming a smooth nonzero incoming flux and regular short-range amplitudes.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The reduced mass is one half GeV. Momentum doubles when excess energy quadruples; then apply the stated threshold powers.

Compare with the solution

Here 2μ=1 GeV2\mu=1\,\mathrm{GeV}, so the approximate momenta are 0.002≈0.04472 GeV\sqrt{0.002}\approx0.04472\,\mathrm{GeV} and 0.008≈0.08944 GeV\sqrt{0.008}\approx0.08944\,\mathrm{GeV}. Their ratio is two. The leading S-wave rate, proportional to p*, increases by two; the P-wave rate, proportional to p* cubed, increases by eight. These are leading ratios within the declared regime. The exact Källén momentum adds small corrections, and a singular amplitude or vanishing incoming momentum could change the rate law.

Take it with you

  • Massive two-body momentum opens as the square root of small excess energy.
  • A fixed-target beam must supply overall recoil as well as final rest masses.
  • Threshold powers and observed endpoint shapes require dynamical and detector assumptions.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.