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The fixed-target production threshold concerns the incident accelerator beam, whereas antiproton mass from momentum and velocity concerns an outgoing secondary beam. To see the distinction, consider a free proton hitting a stationary free proton, producingThe final electric charge is $+2e$ and final baryon number is $3-1=2$, as initially. Approximate all four final rest masses by the same proton rest mass $m$. With metric signature $(+---)$, define the squared centre-of-mass energy by $W^2=c^2(P_1+P_2{)}^2$. The incident proton has total energy $E=m{c}^2+K$, and the target four-momentum is $(mc,0)$. Four-momentum conservation and the mass shell relation giveAt threshold the four final particles have zero relative momentum in their centre-of-momentum frame, so $W_{\mathrm{thr}}=4m{c}^2$. HenceOnly $2m{c}^2$ of new rest energy is created. The larger incident kinetic energy is required because the final system must also carry the initial laboratory momentum. The incident total energy at threshold is $7m{c}^2$; neither this nor $K_{\mathrm{thr}}$ is the selected secondary momentum.

The 1955 authors explicitly qualified the stationary-nucleon threshold: target nucleon motion inside an atomic nucleus can lower the required incident energy.^[ref]

The kinematic reason is visible for two on-shell incoming nucleons:A target momentum directed against the beam increases $W^2$. Applying this illustration to a bound nucleon also requires binding energy and residual-nucleus recoil; the free stationary-target $6m{c}^2$ result is therefore not a universal sharp threshold for a copper target. A low observed secondary momentum is entirely compatible with a much larger accelerator kinetic energy.

Fix the phi-fourth theory convention ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4/4!$, with mostly-minus Minkowski spacetime and $\hslash =c=1$. Define the scattering amplitude by stripping the overall momentum-conserving delta function from $S-1=i\mathcal{M}$. If instead the coupling is defined by ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4$, every vertex coupling below must be replaced by $24\lambda$; the question does not specify this factorial convention.

The momentum-space Feynman rules are a Feynman propagator $i/(k^2-m^2+i0)$ for each internal scalar line, a Feynman vertex $-i\lambda$ with four scalar legs, and four-momentum conservation at each vertex. Integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. For an amputated scattering amplitude there are no external propagators; external states have the usual relativistic normalization. Sum the connected diagrams and all inequivalent assignments of the labelled external momenta. The overall delta function is $(2\pi {)}^4{\delta }^4(\sum p_{\mathrm{in}}-\sum p_{\mathrm{out}})$.

These Feynman rules follow by expanding the Dyson series $T\exp (i\int d^{4}x{\mathcal{L}}_{\mathrm{int}})$ and applying the Wick theorem to the free-field correlation functions. A Wick contraction gives the free Feynman propagator; the $4!$ ways of contracting a vertex cancel its factorial denominator. The expansion's $1/V!$ cancels permutations of identical interaction insertions, while any remaining automorphisms give the Feynman-diagram symmetry factor. Fourier transformation produces momentum conservation, and the LSZ reduction formula amputates external propagators to obtain the scattering amplitude.

For six-point amplitudes in phi-fourth theory, an original pair of diagrams is:

In the tree-level Feynman diagram, let the internal four-momentum be $r=p_1+p_2-q_1=q_2+q_3+q_4$. The labelled diagram has symmetry factor one, so its contribution isThis is one channel. The full tree scattering amplitude sums the ten unordered partitions of the six labelled external legs into two groups of three, since $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$.

For the one-loop triangle Feynman diagram, set $P=p_1+p_2$, $Q_{12}=q_1+q_2$, $Q_{34}=q_3+q_4$, so $P=Q_{12}+Q_{34}$. A consistent routing has denominatorsWith the external labels held fixed, no nontrivial graph automorphism remains, so again the symmetry factor is one. The contribution isThere are other labelled assignments and other one-loop topologies in the full scattering amplitude; the displayed expression belongs to the particular drawn diagram.

One can also express this scalar triangle Feynman integral using Feynman parameters. With $x+y+z=1$, completing the square in $x{D}_0+y{D}_1+z{D}_2$ givesThe identity $1/(D_0{D}_1{D}_2)=2{\int }_{x,y,z\ge 0}dxdydz\delta (1-x-y-z)/(x{D}_0+y{D}_1+z{D}_2{)}^3$ and the shifted momentum integral yieldThus ${\mathcal{M}}_{△}=-{\lambda }^3/(16{\pi }^2)$ times the displayed parameter integral. It is ultraviolet finite in four dimensions, although physical threshold singularities must retain the Feynman i-epsilon prescription.

For the fixed-target production threshold, the incoming four-momenta are $(E,\mathbf{p})$ and $(m,0)$, with $E=\sqrt{m^2+|\mathbf{p}{|}^2}$. Their invariant mass obeysIn the centre-of-momentum frame, four final particles have total energy at least $4m$, attained when all four are at rest in that frame. Therefore $s\ge 16m^2$, equivalently $E\ge 7m$, andThis is the kinematic threshold for $m>0$. Exactly at equality the final phase space has zero volume; a nonzero production rate requires a strict inequality.

Let the incoming projectile momentum be ${\mathbf{p}}_1$. The stationary target has four-momentum $(m_{2}c,0)$, so the total incoming norm isHere the relativistic energy-momentum relation eliminated $|{\mathbf{p}}_1{|}^2$. Write $W$ for the total energy in the centre-of-momentum frame; then $W^2=c^2{P}_{\mathrm{in}}^2$. If $M_f={\sum }_i{\mu }_i$, the relativistic collision mass threshold gives $W\ge M_f{c}^2$. Squaring and rearranging with $m_2>0$ yields the fixed-target production thresholdThe actual incident total energy also obeys $E_1\ge m_1{c}^2$. Thus a negative algebraic reaction bound would not allow a negative-energy projectile.

For a photon, $m_1=0$ and $E_1=h\nu$, where $h$ is the Planck constant. The necessary bound becomesIn the usual massive multi-product channel with $M_f>m_2$, conservation alone permits the threshold configuration in which all products are at rest relative to one another in the centre-of-momentum frame. Boosting that configuration supplies the required laboratory momentum. Hence the photon fixed-target production threshold isFor a multi-product channel with $M_f\le m_2$, there is no positive mass threshold: the lower limit of allowed positive frequencies is zero. The compact expression for that lower limit is $\boxed{{\nu }_{\inf }=\max \{0,M_f^2-m_2^2\}{c}^2/(2m_{2}h)}$; it need not correspond to a zero-frequency incident photon.

For completeness, if the only product is one particle of mass ${\mu }_1$, its invariant mass must match exactly. A positive incident frequency is possible only when ${\mu }_1>m_2$, and is then fixed at $\nu =({\mu }_1^2-m_2^2)c^2/(2m_{2}h)$; frequencies above that value do not produce the same single particle while conserving four-momentum. If a multi-product channel includes massless particles, an all-products-at-rest configuration is impossible for those particles with nonzero energy, so the nominal massive threshold may be only an infimum. These qualifications distinguish the requested necessary inequalities from unrestricted claims of sufficiency.

Put $M_f=m_p+m_{{\pi }^{+}}+m_{\Lambda }+m_{K^0}$. In natural units, a stationary target proton and a beam proton of total energy $E_b$ have the Mandelstam invariantAt fixed-target production threshold the final particles have zero relative momentum in their centre-of-mass frame, so $s=M_f^2$. Restoring $c$, the total beam energy and its kinetic energy areUsing rounded masses in $\mathrm{MeV}/c^2$, $(938.27,139.57,1115.68,497.61)$, gives $E_{b,\mathrm{thr}}\simeq 2.921\mathrm{GeV}$ and $T_{b,\mathrm{thr}}\simeq 1.983\mathrm{GeV}$.

One possible strong interaction quark-line flavour rearrangement isIt needs newly created $d\bar{d}$ and $s\bar{s}$ pairs, with no net change of any quark flavor. For example, keep the first $uud$ as the outgoing proton; allocate the second incoming $u,d$ and the new $s$ to ${\Lambda }^0$, the remaining incoming $u$ and new $\bar{d}$ to ${\pi }^{+}$, and the new $d,\bar{s}$ to $K^0$. Two gluon splittings illustrate this flow. This is a possible hadronization topology, not a unique perturbative description of the proton collision.

After several short mean lifetimes, the surviving coherent beam is approximately $|K_L⟩=(|K^0⟩-|{\bar{K}}^0⟩)/\sqrt{2}$. In ordinary matter the forward scattering amplitudes of $K^0$ and ${\bar{K}}^0$ differ. For example, the $s$ in a neutral antikaon can enter a strange baryon through strong absorption; the opposite strangeness assignment has different available channels. If their forward transmission amplitudes through a thin block are $t_0$ and ${\bar{t}}_0$, respectively, thenThus the transmitted forward beam contains a regenerated short-lived component, with prompt two-pion decays after the block. Differential absorption changes the intensities as well as the relative phase. Scattered particles outside the forward beam can provide an incoherent component. This neutral kaon regeneration does not require CP violation: the material itself distinguishes matter and antimatter.

Let $P=m_p{c}^2$ denote the proton rest energy and $W=220\mathrm{GeV}$ the desired particle's rest energy. For an antiproton of total energy $E_{\bar{p}}$ striking a stationary proton, the incoming squared centre-of-mass energy isFor production of the stated single particle, four-momentum conservation requires $s=W^2$. ThusThe numerical value uses $P=0.938272\mathrm{GeV}$. It is a total incident energy; the corresponding kinetic energy is $E_{\bar{p}}^{\mathrm{fixed}}-P\simeq 25.7902\mathrm{TeV}$. Much of the laboratory energy supplies motion of the outgoing system rather than its invariant mass.

For equal and opposite beam momenta, the laboratory is the centre-of-mass frame. Each beam has the same total energy and their sum is $W$, givingThese are the fixed-target production threshold and collider kinematic comparison for the specified final state. Extra final particles or restrictions from conserved quantum numbers would require a separately specified reaction.

Write $m_e$ for the electron rest mass and $\epsilon =m_e{c}^2$. The incident positron has total energy $E$ and satisfies $p^2{c}^2=E^2-{\epsilon }^2$. Adding the target electron's four-momentum, the squared total centre-of-momentum energy isAt the fixed-target production threshold, all three final particles are at rest relative to one another, so $W=3M_0{c}^2$. ConsequentlyThis is the positron's total energy, not its kinetic energy. It assumes $3M_0\ge 2m_e$ and addresses kinematic allowance; other selection rules could still forbid a proposed reaction. For $M_0{c}^2=135\mathrm{MeV}$ and $m_e{c}^2\simeq 0.510999\mathrm{MeV}$, $E_{\min }\simeq 160.494\mathrm{GeV}$.

At threshold the three products share the centre-of-momentum frame's laboratory velocity. Their common Lorentz factor and speed areFor the subsequent two-photon decay, each photon has rest-frame energy $E_{\gamma }^{\ast }=M_0{c}^2/2$. A Lorentz transformation of a photon emitted at rest-frame angle ${\theta }^{\ast }$ to the boost gives $E_{\gamma }={\gamma }_{\mathrm{CM}}{E}_{\gamma }^{\ast }(1+{\beta }_{\mathrm{CM}}\cos {\theta }^{\ast })$. Therefore the kinematically allowed laboratory range isor numericallyThe two photon energies in a single event are correlated: their sum is the parent's laboratory energy ${\gamma }_{\mathrm{CM}}{M}_0{c}^2$. When one attains the upper endpoint, the other attains the lower endpoint. Angular distributions and detector acceptance can affect which part of this range is observed.