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Derivation

Mandelstam Invariants and Relativistic Two-Body Kinematics

D-275 Home PU-304 Threads energy · symmetry Depends on four-vector-lorentz-invariants
Statement

For a relativistic two-body reaction \(1+2 \to 3+4\) with on-shell four-momenta \(p_i\) (metric signature \(+---\), natural units \(c=1\)), we construct the three Lorentz-invariant Mandelstam variables \(s=(p_1+p_2)^2\), \(t=(p_1-p_3)^2\), \(u=(p_1-p_4)^2\); prove the constraint \(s+t+u=\sum_i m_i^2\); identify \(\sqrt{s}\) as the total centre-of-mass (CM) energy; and derive from \(s\) alone the fixed-target threshold energy \(E_1^{\text{thr}}\) for producing a final state of invariant mass \(M\), and the CM momentum \(p^\*=\dfrac{\sqrt{\lambda(s,m_a^2,m_b^2)}}{2\sqrt{s}}\) of a two-body state built from masses \(m_a,m_b\).

Why it matters

The Mandelstam invariants are the natural coordinates of every \(2\to2\) amplitude: because they are Lorentz scalars, a cross section written as \(\frac{d\sigma}{dt}(s,t)\) is frame-independent, so the same expression describes a collider and a fixed-target experiment without re-transforming momenta. The single number \(s\) fixes what final states are energetically accessible, which is why accelerator reach is quoted as \(\sqrt{s}\).

The CM momentum \(p^\*\) and the threshold energy are the two quantities you actually compute before an experiment: \(p^\*\) sets phase space and the boost of the products, while \(E_1^{\text{thr}}\) tells you the minimum beam energy needed to open a channel — the calculation that decided, for instance, the energy of the Bevatron that discovered the antiproton.

Assumptions
All external particles are on their mass shell, \(p_i^2=m_i^2\).If a leg is virtual (off-shell, as inside a Feynman diagram) then \(p_i^2\neq m_i^2\) and the sum rule \(s+t+u=\sum m_i^2\) no longer holds; the invariants are still defined but are not constrained.
Four-momentum is exactly conserved, \(p_1+p_2=p_3+p_4\).Drop it and the two definitions of each invariant (e.g. \(s=(p_1+p_2)^2=(p_3+p_4)^2\)) diverge, and the CM frame of the initial state differs from that of the final state.
Special relativity with a flat Minkowski metric; no external fields.In curved spacetime or strong background fields there is no global CM frame and \(p^2\) is not globally conserved, so the invariants lose their frame-independent meaning.
Exactly four external legs (a genuine \(2\to2\) or \(1\to2\) process).For \(2\to n\) with \(n>2\) the three variables \(s,t,u\) no longer form a complete kinematic set: \(3n-4\) independent invariants are needed, and the closed form for \(p^\*\) applies only to the two-body sub-system of interest.
Derivation
1
\[ s \equiv (p_1+p_2)^2 = p_1^2 + p_2^2 + 2\,p_1\!\cdot\!p_2 \]
Definition of \(s\); expand the square using the bilinearity of the Minkowski inner product. A
2
\[ s = m_1^2 + m_2^2 + 2\,p_1\!\cdot\!p_2 \]
On-shell condition \(p_i^2=m_i^2\) (prior result: four-vector Lorentz invariants). The cross term is itself an invariant. A
3
\[ P^\mu \equiv p_1^\mu+p_2^\mu = (E_{\text{cm}},\,\mathbf{0}) \quad\text{in the CM frame} \]
The CM frame is defined by vanishing total three-momentum; such a frame exists whenever \(P^\mu\) is timelike, which holds for massive or physically produced states. B
4
\[ s = P^\mu P_\mu = E_{\text{cm}}^2 - |\mathbf{0}|^2 = E_{\text{cm}}^2 \quad\Longrightarrow\quad \sqrt{s}=E_{\text{cm}} \]
\(s\) is a scalar, so its value computed in the CM frame equals its value in any frame. This identifies \(\sqrt{s}\) with the total CM energy. A
5
\[ t \equiv (p_1-p_3)^2,\qquad u \equiv (p_1-p_4)^2 \]
Definitions of the momentum-transfer invariants; \(t\) compares the incoming particle 1 with outgoing particle 3, \(u\) with particle 4. A
6
\[ s+t+u = 3p_1^2+p_2^2+p_3^2+p_4^2 + 2p_1\!\cdot\!(p_2-p_3-p_4) \]
Add the three definitions after expanding each square and collecting the terms proportional to \(p_1\). B
7
\[ p_2-p_3-p_4 = -p_1 \;\Longrightarrow\; 2p_1\!\cdot\!(p_2-p_3-p_4) = -2p_1^2 \]
Conservation \(p_1+p_2=p_3+p_4\) rearranged. Substituting cancels the extra \(p_1^2\) terms. B
8
\[ \boxed{\,s+t+u = p_1^2+p_2^2+p_3^2+p_4^2 = m_1^2+m_2^2+m_3^2+m_4^2\,} \]
Apply on-shell conditions to each square. Only two of \(s,t,u\) are independent. A
9
\[ \text{Fixed target: } p_2=(m_2,\mathbf{0}) \;\Longrightarrow\; s = m_1^2+m_2^2+2m_2 E_1^{\text{lab}} \]
Evaluate the invariant \(s\) from Step 2 in the lab frame where target 2 is at rest, so \(p_1\!\cdot\!p_2=E_1^{\text{lab}}m_2\). Because \(s\) is invariant, this equals \(E_{\text{cm}}^2\). B
10
\[ \text{Threshold: } s_{\min}=\Big(\textstyle\sum_f m_f\Big)^2 \equiv M^2 \]
The lightest configuration of the final state has all products mutually at rest in the CM frame, so \(E_{\text{cm}}\) equals the sum of final masses. B
11
\[ \boxed{\,E_1^{\text{thr}} = \frac{M^2-m_1^2-m_2^2}{2m_2}\,} ,\qquad T_1^{\text{thr}}=E_1^{\text{thr}}-m_1 \]
Set \(s=M^2\) in Step 9 and solve for \(E_1^{\text{lab}}\). The kinetic threshold subtracts the rest energy. A
12
\[ \text{Two-body state (masses }m_a,m_b): \quad \sqrt{s}=E_a^\*+E_b^\*,\qquad \mathbf{p}_a^\*=-\mathbf{p}_b^\*\equiv\mathbf{p}^\* \]
In the CM frame the two momenta are equal and opposite; each energy satisfies \(E^\*=\sqrt{p^{\*2}+m^2}\). This applies to the initial pair \(1,2\) or to any two-body final/decay state. A
13
\[ E_a^{\*2}-E_b^{\*2}=m_a^2-m_b^2 = (E_a^\*-E_b^\*)\sqrt{s} \;\Longrightarrow\; E_a^\*-E_b^\*=\frac{m_a^2-m_b^2}{\sqrt{s}} \]
Subtract the on-shell relations (\(|\mathbf p^\*|\) common) and factor using \(E_a^\*+E_b^\*=\sqrt s\) from Step 12. B
14
\[ E_a^\* = \frac{s+m_a^2-m_b^2}{2\sqrt{s}},\qquad E_b^\* = \frac{s+m_b^2-m_a^2}{2\sqrt{s}} \]
Add and subtract the two linear relations for \(E_a^\*\pm E_b^\*\). A
15
\[ p^{\*2}=E_a^{\*2}-m_a^2 = \frac{(s+m_a^2-m_b^2)^2-4s\,m_a^2}{4s} = \frac{\lambda(s,m_a^2,m_b^2)}{4s} \]
Insert \(E_a^\*\) and simplify; the numerator is the Källén triangle function \(\lambda(x,y,z)=x^2+y^2+z^2-2xy-2yz-2zx\). C
16
\[ \boxed{\,p^\* = \frac{\sqrt{\lambda(s,m_a^2,m_b^2)}}{2\sqrt{s}} = \frac{\sqrt{[s-(m_a+m_b)^2]\,[s-(m_a-m_b)^2]}}{2\sqrt{s}}\,} \]
Take the positive root and use the factorisation \(\lambda(s,m_a^2,m_b^2)=[s-(m_a+m_b)^2][s-(m_a-m_b)^2]\). A
Result
\[ s=E_{\text{cm}}^2,\quad s+t+u=\sum_i m_i^2,\quad E_1^{\text{thr}}=\frac{M^2-m_1^2-m_2^2}{2m_2},\quad p^\*=\frac{\sqrt{\lambda(s,m_a^2,m_b^2)}}{2\sqrt{s}} \]

Reading. The Mandelstam variables package all frame-independent kinematics of a \(2\to2\) process into two independent numbers (a third is fixed by the mass sum rule). \(\sqrt{s}\) is the energy available in the CM to make new particles; the threshold formula converts that requirement into a minimum beam energy for a stationary target; and \(p^\*\) gives the common magnitude of the back-to-back CM momenta, vanishing exactly as \(\sqrt{s}\to(m_a+m_b)\).

Units check. With \(c=1\), masses, energies and momenta are all measured in the same unit (GeV in HEP practice). Then \([s]=[t]=[u]=\text{GeV}^2\), and the sum rule balances GeV\(^2\) on both sides. \(\lambda\) has units GeV\(^4\), so \(\sqrt{\lambda}/(2\sqrt{s})\) is GeV\(^2\)/GeV\(=\)GeV, a momentum — correct. Restoring factors, \(s\) carries units of (energy)\(^2\)/\(c^4\) and \(p^\*\) of energy/\(c\).

Limiting cases
  • Non-relativistic threshold. When \(M-m_1-m_2\ll m_2\), \(T_1^{\text{thr}}\approx\dfrac{M^2-(m_1+m_2)^2}{2m_2}\) reduces to the familiar Galilean requirement that CM kinetic energy equal the reaction \(Q\)-value.
  • Equal masses \(m_a=m_b=m\). \(p^\*=\tfrac12\sqrt{s-4m^2}\) and \(E_a^\*=E_b^\*=\tfrac12\sqrt s\): the CM energy is shared equally.
  • Massless product \(m_b=0\). \(p^\*=E_b^\*=\dfrac{s-m_a^2}{2\sqrt s}\); the photon (or neutrino) carries momentum equal to its energy.
  • Ultra-relativistic collider \(\sqrt s\gg m_i\). \(s\approx2\,p_1\!\cdot\!p_2\); for head-on beams of energy \(E\), \(\sqrt s\approx2E\), whereas fixed target gives only \(\sqrt s\approx\sqrt{2m_2E_1}\).
  • Exact threshold \(s=(m_a+m_b)^2\). \(\lambda\to0\Rightarrow p^\*\to0\): products are produced at rest in the CM, the physical origin of Step 10.
Breaks when
  • Off-shell / internal legs. Inside a diagram a propagator has \(p^2\neq m^2\); the sum rule \(s+t+u=\sum m_i^2\) fails and \(t,u\) can wander outside their physical (on-shell) ranges — the basis of analytic continuation but not of on-shell kinematics.
  • More than two particles in the final state. For \(2\to n\ (n\geq3)\) the invariant mass of the recoiling system is not fixed, \(s,t,u\) no longer close, and \(p^\*\) applies only after you select a two-body subsystem; a Dalitz-type description with additional invariants is required.
  • Spacelike total momentum. If one tries to define a "CM frame" for a spacelike \(P^\mu\) (e.g. treating a \(t\)-channel exchange as if it had a rest frame) then \(s<0\) there and \(\sqrt s\) is imaginary — no physical CM frame exists.
  • Massless target at fixed-target threshold. The formula \(E_1^{\text{thr}}=(M^2-m_1^2-m_2^2)/2m_2\) diverges as \(m_2\to0\): you cannot use a massless particle as a stationary target, because it has no rest frame.
Failure modes
  • Threshold as sum of masses in the lab. Setting \(E_1^{\text{thr}}=\sum_f m_f - m_1\) ignores that the products must recoil; the correct condition is \(s=M^2\), giving the extra \(M^2/2m_2\) penalty for fixed targets.
  • Confusing \(\sqrt s\) with beam energy. Writing \(\sqrt s=E_1^{\text{lab}}\) for a fixed target; in fact \(\sqrt s=\sqrt{m_1^2+m_2^2+2m_2E_1^{\text{lab}}}\), growing only as \(\sqrt{E_1}\).
  • Metric-sign slip. Using \(p^2=E^2+|\mathbf p|^2\) (Euclidean) instead of \(E^2-|\mathbf p|^2\); every invariant comes out wrong, and \(s\) loses its interpretation as an energy squared.
  • Dropping \(m_a\ne m_b\) terms in \(p^\*\). Using \(p^\*=\tfrac12\sqrt{s-4m^2}\) for unequal masses instead of the full triangle function \(\lambda\).
  • Sign of \(t\). Reporting \(t\) as positive; for physical elastic scattering \(t\leq0\) (spacelike momentum transfer).
  • Forgetting the second on-shell branch. Taking \(\lambda\) without the factorised form and missing that \(p^{\*2}<0\) (imaginary \(p^\*\)) signals a kinematically forbidden state, not an arithmetic slip.
Discussion

The deep reason the three invariants are the "right" variables is symmetry. A \(2\to2\) amplitude is a Lorentz scalar, so it can depend only on scalar combinations of the momenta; with four on-shell legs and one conservation law there are exactly two independent such scalars. Choosing them to be \(s\), \(t\), \(u\) (with the linear constraint) exposes the crossing symmetry of quantum field theory: the same analytic function \(\mathcal M(s,t,u)\) describes \(1+2\to3+4\) in the region \(s>0,\,t,u<0\), and its antiparticle "crossed" reactions in other regions of the same \((s,t,u)\) plane. This is why \(s\), \(t\), \(u\) are named for the \(s\)-, \(t\)- and \(u\)-channels.

The threshold and CM-momentum formulas are the practical face of the same invariance. Because \(s\) is computed identically in every frame, an experimenter designs a beam by demanding \(s\geq M^2\) once, then reads off the lab energy from the frame-specific expression \(s=m_1^2+m_2^2+2m_2E_1^{\text{lab}}\). The quadratic growth of \(s\) with lab energy — versus linear growth for colliders — is precisely why particle physics moved from fixed-target machines to colliding beams: to reach a given \(\sqrt s\) a fixed target demands a beam energy that scales as \(s/2m_2\), which becomes prohibitive at the TeV scale.

The Källén function \(\lambda\) recurs far beyond this derivation: it is the two-body phase-space factor, it sets the velocity \(\beta^\*=p^\*/E^\*\) of decay products, and its vanishing at \(s=(m_a\pm m_b)^2\) marks the two square-root branch points of every two-body amplitude in the complex-\(s\) plane. The lower branch point \((m_a+m_b)^2\) is the physical threshold; the "pseudo-threshold" \((m_a-m_b)^2\) governs analytic structure. Reading \(p^\*(s)\) as a function on the complex plane is the entry point to dispersion relations and unitarity.

At a still deeper level, \(s\), \(t\), \(u\) are the Casimir-like invariants attached to the different ways the four-particle state can be grouped into two subsystems, and the mass-shell constraints plus conservation define a two-dimensional physical region (the Mandelstam or Dalitz region) bounded by the curve \(\lambda(s,t,u\text{-boundaries})=0\). The boundary is exactly where some \(p^\*\) vanishes, so the geometry of the allowed \((s,t)\) region is nothing but the simultaneous positivity of the triangle functions of every two-body subsystem — kinematics and analyticity meeting in one algebraic curve.

Common misconceptions. \(s\) is not the beam energy and \(\sqrt s\) is not additive across collisions; \(t\) is negative for physical scattering, not a "transfer" you add on; and the threshold is set by the invariant mass of the final state, not by summing lab kinetic energies. Finally, \(p^\*\) is the CM momentum of a chosen pair — for a \(1\to2\) decay the "collision" \(s\) is simply the parent's \(M^2\).

Worked examples

Example 1 — Antiproton production threshold \(p+p\to p+p+p+\bar p\) on a hydrogen (proton) target. Take \(m_p=0.938\ \text{GeV}\).

1
\[ M=\sum_f m_f = 4m_p \quad\Longrightarrow\quad M^2=16m_p^2 \]
Final state is four nucleons (three \(p\) plus one \(\bar p\)); baryon number conservation forbids fewer. A
2
\[ E_1^{\text{thr}}=\frac{M^2-m_1^2-m_2^2}{2m_2}=\frac{16m_p^2-2m_p^2}{2m_p}=7m_p \]
Symbolic threshold with \(m_1=m_2=m_p\). Numbers deferred to the last step. A
3
\[ E_1^{\text{thr}}=7(0.938)=6.57\ \text{GeV},\qquad T_1^{\text{thr}}=E_1^{\text{thr}}-m_p=6m_p=5.63\ \text{GeV} \]
Insert numbers; kinetic threshold subtracts the beam proton's rest energy. A
\[ T_1^{\text{thr}} = 6m_p \approx 5.63\ \text{GeV} \]

Reading. A stationary-target machine needs a \(5.63\ \text{GeV}\) kinetic beam — six proton rest energies, five of them "wasted" giving the CM the recoil momentum it must carry. This is the number the Bevatron was built to exceed.

Units check. \(M^2/m_p\) is GeV\(^2\)/GeV = GeV, an energy. Correct.

Example 2 — Pion momentum in \(\rho^0\to\pi^+\pi^-\), a two-body decay. Take \(m_\rho=0.775\ \text{GeV}\), \(m_\pi=0.140\ \text{GeV}\), equal-mass products.

1
\[ s=m_\rho^2 \quad(\text{parent at rest defines the CM of the decay}) \]
For a decay the "collision energy" is the parent invariant mass. A
2
\[ p^\* = \frac{\sqrt{[s-(2m_\pi)^2]\,[s-0]}}{2\sqrt s} = \frac12\sqrt{m_\rho^2-4m_\pi^2} \]
Equal masses: \((m_a-m_b)^2=0\), so \(\lambda=(s-4m_\pi^2)\,s\) and \(\sqrt s\) cancels. Symbols first. A
3
\[ p^\*=\tfrac12\sqrt{(0.775)^2-4(0.140)^2}=\tfrac12\sqrt{0.6006-0.0784}=\tfrac12\sqrt{0.5222} \]
Substitute numbers (units GeV\(^2\) inside the root). A
4
\[ p^\* = \tfrac12(0.7227)=0.361\ \text{GeV},\qquad E_\pi^\*=\tfrac12 m_\rho = 0.388\ \text{GeV} \]
Each pion energy is half the parent mass by symmetry (Step 14 with \(m_a=m_b\)). A
\[ p^\*\approx 0.361\ \text{GeV}/c,\qquad \beta^\*=p^\*/E_\pi^\*\approx0.93 \]

Reading. Each pion emerges with \(361\ \text{MeV}\) momentum, back-to-back, at \(93\%\) of \(c\). This fixes the opening kinematics used to reconstruct \(\rho\) decays in detectors.

Units check. \(\sqrt{\text{GeV}^2}=\text{GeV}\), a momentum with \(c=1\); \(\beta^\*\) is dimensionless. Correct.

Problems
  1. (Sum rule.) For elastic proton–proton scattering \(p+p\to p+p\), evaluate \(s+t+u\) numerically with \(m_p=0.938\ \text{GeV}\).
    Solution All four masses equal \(m_p\), so \(s+t+u=4m_p^2=4(0.938)^2=4(0.8798)=3.52\ \text{GeV}^2\). Note this is independent of beam energy — raising \(s\) forces \(t+u\) down by the same amount.
  2. (Photoproduction threshold.) Find the lab photon energy needed for \(\gamma+p\to p+\pi^0\) on a proton at rest. Use \(m_p=0.938\), \(m_{\pi^0}=0.135\ \text{GeV}\), \(m_\gamma=0\).
    Solution \(M=m_p+m_{\pi^0}\), \(m_1=m_\gamma=0\), \(m_2=m_p\). \(E_\gamma^{\text{thr}}=\dfrac{(m_p+m_\pi)^2-m_p^2}{2m_p}=\dfrac{m_\pi^2+2m_pm_\pi}{2m_p}=m_\pi+\dfrac{m_\pi^2}{2m_p}=0.135+\dfrac{0.0182}{1.876}=0.135+0.0097=0.145\ \text{GeV}\approx145\ \text{MeV}\), matching the measured pion-photoproduction threshold.
  3. (Collider vs fixed target.) Two \(7\ \text{TeV}\) protons collide head-on. (a) Give \(\sqrt s\). (b) What beam energy would a fixed-target machine need to reach the same \(\sqrt s\)? Use \(m_p=0.938\ \text{GeV}\).
    Solution (a) Head-on equal beams: \(\sqrt s\approx2E=14\ \text{TeV}=1.4\times10^4\ \text{GeV}\). (b) Fixed target: \(s=2m_pE_1^{\text{lab}}\Rightarrow E_1^{\text{lab}}=\dfrac{s}{2m_p}=\dfrac{(1.4\times10^4)^2}{2(0.938)}=\dfrac{1.96\times10^8}{1.876}\approx1.0\times10^8\ \text{GeV}=10^5\ \text{TeV}\). Seven orders of magnitude larger — the case for colliders.
  4. (CM momentum.) In \(e^+e^-\to\mu^+\mu^-\) at \(\sqrt s=10.0\ \text{GeV}\), find the muon CM momentum. \(m_\mu=0.106\ \text{GeV}\).
    Solution Equal final masses: \(p^\*=\tfrac12\sqrt{s-4m_\mu^2}=\tfrac12\sqrt{100-4(0.0112)}=\tfrac12\sqrt{100-0.0449}=\tfrac12\sqrt{99.955}=\tfrac12(9.9977)=4.999\ \text{GeV}\). The muons are ultra-relativistic; \(p^\*\approx\tfrac12\sqrt s\) up to a \(2\times10^{-4}\) correction.
  5. (Massless legs and the sum rule.) For Compton scattering \(\gamma+e^-\to\gamma+e^-\), show \(s+t+u=2m_e^2\), and using \(s=(p_\gamma+p_e)^2\) with the electron at rest, express \(s\) in terms of the incident photon energy \(E_\gamma\).
    Solution Masses are \(m_\gamma=m_\gamma=0\) and \(m_e=m_e\), so \(\sum_i m_i^2=0+m_e^2+0+m_e^2=2m_e^2\); hence \(s+t+u=2m_e^2\). With the electron at rest \(p_e=(m_e,\mathbf0)\) and \(p_\gamma=(E_\gamma,E_\gamma\hat{\mathbf n})\): \(s=m_e^2+2p_\gamma\!\cdot\!p_e=m_e^2+2E_\gamma m_e=m_e(m_e+2E_\gamma)\). As \(E_\gamma\to0\), \(s\to m_e^2\) (threshold), consistent with elastic scattering having no production threshold.