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In the centre-of-mass frame, the two equal incoming rest masses have opposite momenta and total energy $2E$. The squared total four-momentum fixes the composite invariant mass $\mathcal{M}$:A change of inertial frame, including going to either particle's rest frame, cannot change this invariant mass. It includes the internal kinetic energy and is generally larger than $2M_1$.

There is a terminology convention worth making explicit. The older “relativistic mass” means frame-dependent total energy divided by $c^2$. To stop particle 1 requires a boost with Lorentz factor $\gamma =E/(M_1{c}^2)$, so the system has total energy $2\gamma E=2E^2/(M_1{c}^2)$ in that frame. Under that alternative convention, its relativistic mass would be $2E^2/(M_1{c}^4)$. This is not its invariant mass; both quantities coincide only when $E=M_1{c}^2$ and the incoming particles already have zero relative motion.

Four-momentum conservation fixes the outgoing system's invariant mass to the same $2E/c^2$. In the rest frame of particle 3, let particle 4 have energy $E_4^{\prime }$ and momentum ${\mathbf{p}}_4^{\prime }$. Each outgoing particle has rest mass $M_2$, soConsequentlyHere $E_4^{\prime }$ is the total energy, including rest energy. At the production threshold $E=M_2{c}^2$, it reduces to $M_2{c}^2$, as it should. Production is kinematically possible only for $E\ge M_2{c}^2$. The initial rest mass $M_1$ drops out because the conserved total centre-of-mass energy, rather than the identity of the incoming particles, determines this outgoing relative motion.

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.

In the parent's rest frame, the identical daughters have equal and opposite momenta, and each has total energy $110\mathrm{GeV}$. With daughter rest energy $Q=90\mathrm{GeV}$, each speed in this frame isTransform to the rest frame of one daughter using the collinear relativistic velocity addition law. The other daughter then has speedAn independent invariant mass check is useful. In the rest frame of one daughter the other's total energy satisfies $W^2=2Q^2+2Q{E}^{\prime }$, so $E^{\prime }=178.8889\mathrm{GeV}$ and the relative Lorentz factor is ${\gamma }_{\mathrm{rel}}=E^{\prime }/Q=161/81$. Hence $w/c=\sqrt{1-{\gamma }_{\mathrm{rel}}^{-2}}$, the same result.

The answer is unchanged if the parent is produced by a fixed-target collision. Its laboratory velocity then differs, but its rest mass and two-body decay kinematics do not. The scalar product of the daughters' four-momenta, equivalently their relative Lorentz factor, depends only on $W$ and $Q$. A common boost cannot change the speed of one measured in the rest frame of the other.^[ref]

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.

Choose particle 1's incoming velocity positive and particle 2's negative. The final particle is stationary, so four-momentum conservation requires zero total incoming momentum:It follows that ${\gamma }_1{\beta }_1=1/\sqrt{2}$. Squaring gives ${\beta }_1^2/(1-{\beta }_1^2)=1/2$, henceThe initial total energy becomes the final rest energy. ThereforeThe extra invariant mass is the incoming kinetic energy retained as internal energy of the merged particle: $(M-3m)c^2$. It does not violate conservation of energy or four-momentum conservation; the sum of the separate rest masses is not a conserved quantity in an inelastic collision.

The threshold must conserve both energy and momentum. For head-on photons, the total energy is $E_1+E_2$, while the total momentum has magnitude $|E_2-E_1|/c$. Their combined invariant mass $M$ satisfiesAt the threshold for photon-photon pair production, the electron and Positron have no relative momentum in their centre-of-momentum frame, so $M=2m_e$. ThereforeSimply equating the laboratory energy to $2m_e{c}^2$ would miss the very large laboratory momentum: the newly produced pair must carry it away.

Use energy units $m=m_p{c}^2$, $M=m_A{c}^2$, $e=E_{\gamma }$ and $P=pc=\sqrt{E^2-m^2}$. The soft-photon resonant capture threshold follows from four-momentum conservation:For fixed proton energy the left side is largest at head-on incidence. Its threshold is $E+P=2E_0$. Combining that with $(E-P)(E+P)=m^2$ givesEquality is allowed for a head-on photon. Here $e=k_{B}T=2.3482\times 10^{-4}\mathrm{eV}$, soA very soft photon needs an enormous proton Lorentz factor to reach the specified invariant mass. A true thermal photon distribution and a finite-width resonance would smear this monoenergetic threshold.

For $\theta =0$, $E-P=m^2/(E+P)\le m$, whereas $2E_0\gg m$, so production is impossible. For each nonzero angle, sufficiently large energy is possible, and $P\simeq E$ givesTo find the maximum deflection in soft-photon capture, note that the final particle's transverse momentum comes entirely from the photon:Writing $x=\cos \theta$, maximization gives $(1-x)(2x+1)=0$ and the interior maximum $x=-1/2$. Therefore, to the same extraordinarily accurate ultrarelativistic order,The neglected relative corrections are of order $m^2/E_0^2$ and $e/E_0$, both around $10^{-24}$. The final particle is effectively collinear with the proton despite a large photon incidence angle.

Let the inertial frame $S^{\prime }$ move at speed $v$ in the positive $x$ direction relative to $S$, with parallel axes and coincident origins at zero time. The inverse Lorentz transformation givesHere ${\gamma }_v=(1-v^2/c^2{)}^{-1/2}$ and ${\mathbf{u}}^{\prime }=d{\mathbf{x}}^{\prime }/d{t}^{\prime }$ is the particle's velocity in $S^{\prime }$. Dividing spatial differentials by $dt$ gives the relativistic velocity addition lawFor the collision, set $c=1$ during the algebra. Let the incident projectile have energy $E>m_1$ and momentum $p=\sqrt{E^2-m_1^2}$. The target is initially at rest. The squared invariant mass, boost speed and Lorentz factor of the center-of-momentum frame areIn that frame the projectile has $p_{\ast }=m_{2}p/\sqrt{s}$ and $E_{\ast }=(m_1^2+m_{2}E)/\sqrt{s}$. An elastic collision changes the direction of its momentum, through an angle $\phi$, while preserving these magnitudes. Transforming the outgoing four-momentum back to the laboratory givesThe maximum occurs at $\cos \phi =-1/B$. Sincethe maximum laboratory deflection in relativistic elastic scattering isThe printed strict inequality needs correction: equality is kinematically attainable. For example, in $c=1$ units take $m_1=2$, $m_2=1$, and incident $E=3$. A projectile scattered through $30^{\circ }$ with final energy $16/7$ has momentum $(3\sqrt{5}/7,\sqrt{15}/7)$. The target then has energy $12/7$ and momentum $(4\sqrt{5}/7,-\sqrt{15}/7)$; both outgoing particles satisfy their mass shells and conserve total four-momentum. This gives $\sin \theta =1/2=m_2/m_1$ exactly. The inequality is strict only away from the maximizing direction.

The covariance of a selected particle momentum sum follows because, for a fixed set of particle four-momenta, their sum $P_A^{\mu }={\sum }_{i\in A}{p}_i^{\mu }$ is a four-vector by linearity of the Lorentz transformation. Thereforeis invariant. The subset need not comprise the entire plasma, and its particles need not move together. This expression includes cross terms between different particles; it is not the sum of their individual rest-mass squares in a.

The comparison must use the same subset and the transformed four-momenta at the same chosen events. Selecting “the lowest-energy half” anew in each frame need not do that. For example, take four photons along $\pm x$, with energies $E_0,2E_0$ in each direction. The two lowest in the original frame are an opposite pair and have invariant energy squared $4E_0^2$. After a boost at $c/2$ along $+x$, both forward photons are redshifted enough to become the lowest pair; that new collinear pair has zero invariant mass. This does not contradict covariance, because the subset changed.

If energies are resampled on a different simultaneity slice in an interacting plasma, they can also refer to different events. A physical conserved subsystem total must then include exchanged field energy and any boundary flux as appropriate. Such operational conservation requirements are distinct from the exact algebraic covariance of a sum of fixed four-vectors.

Each photon has energy $\epsilon =\hslash \omega$. Their total four-momentum isIts squared invariant is $P^2=2{\epsilon }^2(1-\cos \theta )/c^2$. Therefore the system's rest energy isTwo individually massless particles can thus have positive total invariant mass. Back-to-back photons give rest energy $2\hslash \omega$, whereas exactly parallel photons have zero invariant mass and no massive-system rest frame.

For a massive particle on a timelike world line, the four-velocity is $U^{\mu }=d{X}^{\mu }/d\tau =\gamma (c,\mathbf{v})$, with $U^2=c^2$. Its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$ and its four-force is $F^{\mu }=d{P}^{\mu }/d\tau$. Here $\tau$ is proper time. A massless trajectory does not admit this massive-particle proper-time parametrization.

The asserted classification of arbitrary orthogonal vectors is false. With the chosen Minkowski metric, $D=(0,1,0,0)$ and $G=(0,0,1,0)$ are nonzero, orthogonal and both spacelike vectors. Also $D=(1,1,0,0)$ and $G=2D$ are orthogonal null vectors. The valid statement is that a nonzero vector orthogonal to a timelike vector is spacelike. To prove it, choose the rest frame of the timelike vector, $D=(d,0,0,0)$; then $D\cdot G=d{G}^0=0$ forces $G^0=0$, so $G^2=-|\mathbf{G}{|}^2<0$.

The relativistic energy-momentum relation givesTherefore $\boxed{P^2=0⟹m=0}$. A photon is an example: $E=c|\mathbf{p}|$. A collection of photons need not itself have zero invariant mass.

Put $\epsilon =\hslash \omega$. The total four-momentum of the perpendicular photons isAlthough each constituent is massless, the sum has $P_{\mathrm{tot}}^2=2{\epsilon }^2/c^2$. Its invariant mass is thereforeThe rest energy is the total energy in the frame where the two photon momenta cancel, not the sum of individual photon rest energies.

With $c=1$, the total four-momentum is $P_{\mathrm{tot}}={\sum }_i{P}_i$. A sum of Lorentz four-vectors transforms as a four-vector, sois a Lorentz invariant. Its value is the squared invariant mass of the complete system, irrespective of whether its constituents interact. Isolation makes the total four-momentum constant in time.

For a photon gas, write ${\mathbf{p}}_i=W_i{\mathbf{n}}_i$ with $|{\mathbf{n}}_i|=1$. Its individual null norms do not make the total norm vanish:For an isotropic gas in its momentum rest frame, ${\sum }_i{\mathbf{p}}_i=0$ and $s^2=W_{\mathrm{tot}}^2$. Zero is possible only when all nonzero photons propagate in the same direction. Two equal-energy photons at right angles, for example, give $s^2=2W^2$, rather than zero. Without an angular distribution, the pairwise expression is the general answer; isotropy is an additional assumption.

An invariant has the same value when the same physical event or state is described in different inertial frames. A conserved quantity has the same value as the physical system evolves, subject to its equations and isolation conditions. These are different comparisons. The invariant mass of a heated object can change with time, while an isolated system's conserved energy has different numerical values in boosted frames.

Under a Lorentz transformation $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$, a Lorentz four-vector transforms asA covector transforms by the inverse matrix, $W_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{W}_{\nu }$. A general tensor has one corresponding factor for each index; for example,Contraction with the Minkowski metric produces Lorentz scalars; it does not by itself make those scalars conserved.

Use $c=1$ and define the total four-momentum $P^{\mu }=(E,\mathbf{P})={\sum }_i(E_i,{\mathbf{p}}_i)$. For an inertial frame moving with velocity $\mathbf{V}$ relative to the laboratory, the Lorentz boost givesChoose $\mathbf{V}$ parallel to $\mathbf{P}$ and require zero total three-momentum. ThenThis requires $E>|\mathbf{P}|$, so that the total four-momentum is timelike. Its invariant mass is $\sqrt{E^2-|\mathbf{P}{|}^2}$, the total relativistic energy in the center-of-momentum frame. If all particles are exactly parallel photons, the total four-momentum is null and no finite-velocity center-of-momentum frame exists.

Sign qualification. The minus sign printed in the source describes the laboratory's velocity in the center-of-momentum frame, or the parameter of an active Lorentz boost with a plus-sign spatial rule. It is not the velocity of the center-of-momentum frame relative to the laboratory under question 1's passive convention. For example, a single particle with $(E,P_x)=(5,3)$ has its rest frame moving at $V_x=3/5$. The printed $V_x=-3/5$ would give $P_x^{\prime }=(5/4)(3+3)=15/2$, which is not zero. This explicitly resolves the sign rather than assigning a rest frame moving away from the particle.

Two photons meeting at angle $\alpha$ can produce an electron and a Positron when their combined four-momentum has sufficient invariant mass. For a head-on collision the threshold is $E_1{E}_2=m_e^2{c}^4$. Threshold means zero relative momentum in the centre-of-momentum frame, not necessarily in the laboratory.

A particle decay into three daughter particles subject to four-momentum conservation and the relativistic energy-momentum relation. The daughter energies generally vary continuously even in the parent's rest frame. Grouping two daughters into a recoil system of variable invariant mass turns the remaining kinematics into a relativistic two-body decay.

In 1974, Samuel Chao Chung Ting's Brookhaven team and Burton Richter's SPEAR team independently found the narrow state now called the J/psi meson. Each discovery paper notes learning of the other result during manuscript preparation.^[ref][ref]

The common question was whether the observed excess represented a new particle resonance, rather than a peculiarity of particle identification, acceptance or calibration. The important difference was the horizontal coordinate of the search: Brookhaven reconstructed a final-state invariant mass; SPEAR scanned an initial-state center-of-mass energy. The invariant-mass selection of bubble-chamber events uses the first kind of reasoning, despite a very different recording medium.

A scalar particle has zero intrinsic spin. A scalar-field excitation provides its relativistic description. Spin zero alone does not force its decay matrix element to be independent of internal invariant masses; resonance dynamics can still produce a nonuniform three-body distribution.