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For an incident particle of rest mass $m_1$ and total energy $E_1$ hitting a stationary target of mass $m_2>0$, four-momentum conservation gives $W^2=m_1^2{c}^4+m_2^2{c}^4+2m_2{c}^2{E}_1$. Applying the relativistic collision mass threshold gives the displayed necessary condition. The incoming energy must also satisfy $E_1\ge m_1{c}^2$; other conserved quantum numbers or reaction dynamics can impose further restrictions.

The two incoming particles have equal energies and opposite momenta, so their total four-momentum in this frame isIt is already a centre-of-momentum frame. By four-momentum conservation, the final spatial momenta sum to zero and their energies sum to $2E$. Each product has energySumming gives the necessary relativistic collision mass thresholdFor massive products the equality threshold has all of them at rest in this frame. Above it, multiple products can carry balancing momenta and relative kinetic energy. The displayed condition is necessary, as requested; a single final particle must instead have exactly the incoming invariant mass, and additional physical conservation laws could further restrict a real reaction.

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.

Choose the $M$ ball's initial velocity as $+v$; the $3M$ ball then has velocity $-v$. Both have Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$. Their total relativistic energy and signed relativistic momentum areThe laboratory frame is not the centre-of-momentum frame, despite the equal and opposite speeds. Four-momentum conservation and the relativistic energy-momentum relation give the invariant mass $M_f$ of the combined incoming system:The relativistic collision mass threshold for a final object of rest mass $5M$ is $M_f\ge 5M$. At the threshold,At equality, the outgoing object has laboratory velocity $V=c^2{p}_i/E_i=-v/2$, directed towards the initially heavier ball's motion. Its nonzero final kinetic energy is why setting $4\gamma M{c}^2=5M{c}^2$ would give the wrong threshold.

For the literal rest mass of a coalescing relativistic collision, if there is exactly one outgoing object of exactly fixed rest mass $5M$ and no radiation or other products, four-momentum conservation selects the equality above, rather than allowing every larger $v$. Speeds above this threshold require a larger final invariant mass or additional outgoing energy carriers.

Use the $+---$ metric signature, with the time component of four-momentum equal to $E/c$. The incident particle has energy $\gamma m{c}^2$ and momentum of magnitude $mc\sqrt{{\gamma }^2-1}$. Adding the stationary target's four-momentum gives total laboratory quantitiesThe centre-of-momentum energy $W$ follows from the invariant Minkowski norm of the total four-momentum:At the relativistic collision mass threshold, the two final particles have zero relative velocity in their centre-of-momentum frame. Their total energy there is therefore $2\alpha m{c}^2$. Four-momentum conservation requires $W\ge 2\alpha m{c}^2$, soThe condition $\alpha >3/2$ ensures ${\gamma }_{\min }>1$. This is the minimum required Lorentz factor; larger values allow final kinetic energy in the centre-of-momentum frame. Equivalently, the incident laboratory kinetic energy at threshold is $({\alpha }^2-9/4)m{c}^2$.

The laboratory remains the center-of-momentum frame. At the relativistic collision mass threshold, the two final W bosons can have arbitrarily small opposite momenta. The available relativistic energy is $2m_e{c}^2+eEd$, soEach incoming particle must acquire about $80.378489\mathrm{GeV}$ of kinetic energy, travelling about $16.08\mathrm{km}$. Equality is the ideal kinematic threshold; a nonzero production rate needs energy above it. This estimate neglects acceleration losses and treats the W bosons with the supplied nominal rest mass.

A linear particle accelerator avoids repeatedly bending high-energy electrons and Positrons. This greatly reduces the bending synchrotron radiation loss, which at fixed bending radius grows approximately as the fourth power of the relativistic energy for a given particle species. Its useful collision energy can therefore be increased without the severe circular electron collider radiation penalty. The price is long structures or high accelerating gradients: each bunch normally passes once, requiring substantial beam power, fresh bunches, and very tight focusing for high luminosity. A circular particle accelerator reuses its accelerating structures and circulates bunches through many collisions, but needs a large ring and substantial radiofrequency power to replace synchrotron radiation losses. Heavier particles such as protons radiate far less at a given energy and radius, although a proton's constituent structure makes the hard-collision energy less sharply specified than in an electron-positron annihilation experiment.