Mandelstam Variables

Three numbers describe a two-body collision, one equation ties them together, and so only two are free. `aq-sr-mandelstam-invariants` builds those numbers, proves the constraint $s + t + u = \sum_i m_i^2c^4$ , and shows that in elastic scattering $t$ is never positive. This article picks up where that one deliberately stops. Two questions are left open there, and both have sharp answers. First: given $s$ , exactly which values of $t$ can occur? Not a vague "some range" — a closed interval whose width is

t_{\max} - t_{\min} \;=\; \frac{\sqrt{\lambda\big(s, m_1^2c^4, m_2^2c^4\big)}\ \sqrt{\lambda\big(s, m_3^2c^4, m_4^2c^4\big)}}{s},

a Källén function for the incoming pair times one for the outgoing pair. Second: what is the shape of the physical region? The usual picture is three straight-sided wedges. That picture is exactly right for equal masses and wrong in general, and the correct boundary is a cubic curve — the vanishing of the $3\times3$ Gram determinant of the momenta, the same determinant that turned out to be the Källén function in `aq-sr-relativistic-collision-kinematics`, now one dimension up. And then the reason the apparatus exists. The same three invariants, read in three different regions of the same plane, describe three different reactions. That is crossing symmetry, and this article owns the kinematic half of it: the substitution rule, the three channels, and a proof that their physical regions never overlap.

Key takeaways

Convention lock. $(+,-,-,-)$ ; $P^\mu = (E,\ \vec p\,c)$ ; $P\cdot P = +m^2c^4$ . $s$ , $t$ , $u$ are in ENERGY², so $\sqrt{s}$ is an ENERGY and the sum rule carries $c^4$ : $s+t+u = \sum_i m_i^2 c^4$ . `aq-sr-mandelstam-invariants` boxes it; this article recalls it and does not re-derive it.
Owned result 1 — the $t$ -range at fixed $s$ . $t$ is an affine function of $\cos\theta^{*}$ , so it sweeps a closed interval as $\cos\theta^{*}$ runs over $[-1,+1]$ , of width $\sqrt{\lambda_{12}}\sqrt{\lambda_{34}}/s$ with $\lambda_{12} \equiv \lambda(s, m_1^2c^4, m_2^2c^4)$ .
Owned result 2 — the physical region is a Gram determinant (Kibble, 1960). With $G_{ij} = P_i\cdot P_j$ for $i,j \in \{1,2,3\}$ , the boundary is $\det G = 0$ and the interior is $\det G > 0$ . In invariants, $\det G$ is a cubic in $(s,t,u)$ .
And for equal masses it collapses. $\det G = \tfrac14\,s\,t\,u$ exactly. The boundary is $stu = 0$ , the $s$ -channel region is $\{t \leq 0,\ u \leq 0\}$ — and $s \geq 4m^2c^4$ is then a consequence, not an extra condition.
The wedges are not straight in general. Machine-checked counterexample: $\pi^- p \to K^0\Lambda$ at $\sqrt{s} = 1.700$ GeV has $u = +0.189$ GeV² at backward angles. $u \leq 0$ is an equal-mass accident.
Crossing symmetry, kinematically. One amplitude, three reactions: $1+2\to3+4$ , $1+\bar 3 \to \bar 2 + 4$ , $1+\bar 4 \to \bar 2 + 3$ . The rule is $P_i \to -P_i$ per crossed leg, and the three physical regions are provably disjoint.
Key idea: the sum rule makes $(s,t,u)$ a plane, the Gram determinant carves the physical region out of it, and crossing says the pieces it carves are three different experiments described by one function. The whole kinematics of two-body scattering is one cubic curve.

The core insight

The sum rule $s+t+u = \sum_i m_i^2c^4$ says the three invariants are a point on a plane. That is one line, and it is where most treatments stop. But not every point on that plane is a collision: momenta must be on-shell, angles must be real, and the four four-vectors of a $2 \to 2$ process must fit inside a three-dimensional subspace of spacetime, because conservation leaves only three independent and the scattering is planar. The condition for three four-vectors to span a subspace of the right causal type is exactly the sign of their Gram determinant — and expanded in $s$ , $t$ and $u$ , that determinant is a cubic. The physical region is where the cubic is positive; its boundary is where the momenta go collinear. Everything else — the $t$ -range, the three channels, their disjointness — reads off the same curve.

Where this lives today

Every measured cross-section at a hadron collider is reported as a function of $s$ and $t$ , for the reason this article proves: those are the invariants, and the physical region is a definite, computable set. When an experiment quotes $\mathrm{d}\sigma/\mathrm{d}t$ it is quoting something that means the same thing in every frame. The Gram-determinant boundary is not a curiosity either. Kinematic fitting codes test candidate events against $\det G \geq 0$ to reject combinations no real collision could produce, and the same determinant — one row larger — draws the boundary of a Dalitz plot for three-body final states, which is how resonances in three-body decays are found at all. Crossing survives most robustly of the three. The dynamical statement — that a single analytic function $\mathcal{M}(s,t)$ serves all three channels — is a theorem of quantum field theory, proved from locality and the spectral conditions, and it is why a calculation of Compton scattering doubles as a calculation of pair annihilation. The kinematics stops here. We have shown that the three reactions live in three disjoint regions of one plane, that the plane's physical part is bounded by a determinant, and that nothing in the invariants remembers which experiment you were doing. Why one function covers all three is a question for the field theory, and `aq-sr-mandelstam-invariants` is right to send it there.

The physical picture

A two-body reaction $1+2 \to 3+4$ has four four-momenta, and conservation kills one: $P_4 = P_1 + P_2 - P_3$ . Three are left, each on its mass shell. What remains free is the relative geometry of three four-vectors, and after the sum rule that is exactly two numbers — take them to be $s$ and $t$ . So a collision is a point in a plane. The question this article is really about is: which points? Not all of them. Pick $s$ and $t$ at random and you will usually be describing something impossible — a scattering angle whose cosine is $1.4$ . The set of $(s,t)$ that can occur is the physical region, and its boundary is where the geometry degenerates: where the outgoing particle goes exactly forward or exactly backward, and the three independent four-momenta stop spanning a three-dimensional space. That is a Gram determinant condition, and it is the same idea `aq-sr-relativistic-collision-kinematics` used to explain the Källén function — there, two momenta and a $2\times2$ determinant; here, three momenta and a $3\times3$ one. Kinematic boundaries are always determinants going to zero, because a boundary is always a degeneracy. The last piece is what Mandelstam was actually after. The invariants do not know which reaction produced them. Feed them a point in a different region of the same plane and they describe $1 + \bar 3 \to \bar 2 + 4$ — a different experiment, with different particles in the beam. Crossing symmetry is the statement that one function of $(s,t,u)$ serves all three, and the kinematics of it — which region is which, and why they never overlap — we can prove here with no quantum field theory at all.

The core equations

The invariants, recalled from the owner (`aq-sr-mandelstam-invariants`), with the metric re-derived from $\eta$ , not transcribed. For $1+2\to3+4$ , with $P^\mu = (E,\ \vec p\,c)$ so that $P\cdot P = \eta_{\mu\nu}P^\mu P^\nu = E^2 - |\vec p\,|^2c^2 = m^2c^4$ :

s = (P_1+P_2)^2, \qquad t = (P_1-P_3)^2, \qquad u = (P_1-P_4)^2 \qquad (1)

where: all three are in energy². Physical reading: $\sqrt{s}$ is the total CM energy, and $t$ is the invariant square of the momentum transfer. The constraint, recalled and not re-derived — `aq-sr-mandelstam-invariants` proves it:

\boxed{\;s + t + u \;=\; \sum_{i=1}^{4} m_i^2c^4\;} \qquad (2)

where: note the $c^4$ . Physical reading: three invariants, one equation, two free numbers — geometrically, any collision is a point on a plane. The general $t$ , at fixed $s$ , as a function of the CM scattering angle:

\boxed{\;t \;=\; m_1^2c^4 + m_3^2c^4 - 2E_1^{*}E_3^{*} + 2\,p_1^{*}p_3^{*}c^2\cos\theta^{*}\;} \qquad (3)

where: $E_i^{*}$ and $p_i^{*}$ come from `aq-sr-centre-of-momentum-frame`: $E_1^{*} = (s + m_1^2c^4 - m_2^2c^4)/(2\sqrt{s})$ , $p_1^{*} = \lambda^{1/2}_{12}/(2c\sqrt{s})$ , with $\lambda_{12} \equiv \lambda(s, m_1^2c^4, m_2^2c^4)$ , and $3\to4$ for the outgoing pair. Physical reading: $t$ is affine in $\cos\theta^{*}$ — the general, unequal-mass statement, of which the elastic $t = -2p^2c^2(1-\cos\theta)$ boxed by `aq-sr-mandelstam-invariants` is the case $m_1 = m_3$ , $m_2 = m_4$ . Hence the first owned result — the width of the physical $t$ -interval:

\boxed{\;t_{\max} - t_{\min} \;=\; 4\,p_1^{*}p_3^{*}c^2 \;=\; \frac{\sqrt{\lambda_{12}}\ \sqrt{\lambda_{34}}}{s}\;} \qquad (4)

where: $t_{\max}$ is at $\cos\theta^{*} = +1$ (forward), $t_{\min}$ at $-1$ (backward). Physical reading: the entire allowed range of momentum transfer, from the four masses and $s$ alone. It vanishes when either pair is at its threshold — no phase space, nothing to transfer. And the second — the boundary of the physical region (Kibble, 1960). With $G_{ij} = P_i\cdot P_j$ for the three independent momenta:

\boxed{\;\det G \;\geq\; 0 \quad\text{in the physical region}, \qquad \det G = 0 \ \text{ on its boundary}\;} \qquad (5)

where: $\det G$ , in invariants, is a cubic in $(s,t,u)$ . Physical reading: the boundary is where the three momenta stop spanning three dimensions — exactly forward or exactly backward scattering. For equal masses the cubic factorises completely:

\boxed{\;\det G \;=\; \tfrac{1}{4}\,s\,t\,u \qquad (\text{all } m_i = m)\;} \qquad (6)

where: the sum rule has been used. Physical reading: the three straight-sided wedges of the textbook picture are the three components of $stu \geq 0$ that run to infinity. In the $s$ -channel wedge $s>0$ , $t\leq0$ , $u\leq0$ , so $stu \geq 0$ ✓ — and $s \geq 4m^2c^4$ follows from $t,u \leq 0$ and the sum rule rather than being imposed.

Figure

THE PHYSICAL REGION IS A GRAM DETERMINANT. The Mandelstam plane, drawn in the symmetric 120-degree coordinates in which s, t and u are the three signed distances to the three axes (the sum rule s + t + u = sum of m-squared c-to-the-fourth makes that possible). For EQUAL MASSES the Gram determinant collapses to det G = s t u / 4, so its zero set is the three straight lines s = 0, t = 0, u = 0, and the physical regions are the three shaded wedges running to infinity — one per channel, each with one invariant above threshold and the other two negative. The small central triangle also has stu greater than zero, but NO channel is above threshold there: that is the MANDELSTAM TRIANGLE, unphysical, and the domain in which the amplitude is real. For UNEQUAL masses the straight lines bend: the boundary is the full cubic curve det G = 0, and the region can poke into u greater than zero, as pi-minus p to K-zero Lambda actually does

The boundary, and the three channels

Derivation AThe t-range at fixed s, for arbitrary masses, and why its width is a product of two Kaellen functions

`aq-sr-mandelstam-invariants` boxes $t = -2p^2c^2(1-\cos\theta)$ , which is exact and which is elastic: it assumes $m_1 = m_3$ and $m_2 = m_4$ , so $p_1^{*} = p_3^{*} \equiv p$ . Real experiments are usually not elastic. This derivation does the general case, and the payoff is a closed formula for how much momentum transfer a given $\sqrt{s}$ makes available.

Step 1. Work in the CM frame, where $\vec p_1^{\,*} = -\vec p_2^{\,*}$ , $\vec p_3^{\,*} = -\vec p_4^{\,*}$ , and $\theta^{*}$ is the angle between $\vec p_1^{\,*}$ and $\vec p_3^{\,*}$ . Expand $t = (P_1 - P_3)\cdot(P_1-P_3)$ :

t = m_1^2c^4 + m_3^2c^4 - 2\,P_1\!\cdot\! P_3 = m_1^2c^4 + m_3^2c^4 - 2E_1^{*}E_3^{*} + 2p_1^{*}p_3^{*}c^2\cos\theta^{*} .

Why: only the last term contains the angle, and it enters linearly. $t$ is an affine function of $\cos\theta^{*}$ , so as $\cos\theta^{*}$ sweeps its physical range $[-1,+1]$ , $t$ sweeps a closed interval and nothing else.

Step 2. Read off the endpoints. Forward scattering ( $\cos\theta^{*} = +1$ ) gives the largest $t$ ; backward ( $\cos\theta^{*} = -1$ ) the smallest:

t_{\max,\min} = m_1^2c^4 + m_3^2c^4 - 2E_1^{*}E_3^{*} \pm 2p_1^{*}p_3^{*}c^2 .

Why: no dynamics has entered. The amplitude decides how often each $t$ occurs; kinematics decides which are possible at all, and that is settled by the masses and $s$ .

Step 3. Take the difference and substitute the CM momenta from `aq-sr-centre-of-momentum-frame`, $p_i^{*} = \lambda^{1/2}/(2c\sqrt{s})$ :

t_{\max} - t_{\min} = 4\,p_1^{*}p_3^{*}c^2 = 4c^2\cdot\frac{\sqrt{\lambda_{12}}}{2c\sqrt{s}}\cdot\frac{\sqrt{\lambda_{34}}}{2c\sqrt{s}} = \boxed{\ \frac{\sqrt{\lambda_{12}}\,\sqrt{\lambda_{34}}}{s}\ }

with $\lambda_{12} = \lambda(s, m_1^2c^4, m_2^2c^4)$ and $\lambda_{34} = \lambda(s, m_3^2c^4, m_4^2c^4)$ . Why: the incoming pair and the outgoing pair each contribute one Källén function, and the width is their geometric product over $s$ . Every factor of $c$ cancels, as it must. And by the Heron reading of `aq-sr-relativistic-collision-kinematics`, each $\lambda$ is minus sixteen times a squared triangle area — so the width of the $t$ -range is set by two failed triangles.

Step 4. Sanity-check against the owner. For elastic scattering, $\lambda_{12} = \lambda_{34}$ , so the width is $\lambda_{12}/s = 4p^{*2}c^2$ ; and $t_{\max} = 0$ because $E_1^{*} = E_3^{*}$ and $p_1^{*} = p_3^{*}$ make the constant term $2(m_1^2c^4 - E_1^{*2} + p_1^{*2}c^2) = 0$ . So $t$ runs over $[-4p^{*2}c^2,\ 0]$ — exactly `aq-sr-mandelstam-invariants`' $t = -2p^2c^2(1-\cos\theta)$ . Why: the general result must contain the special one, and it does. Equation (3) is the elastic formula with the assumption removed. *Verified in verify_aq-sr-mandelstam-variables.py: equation (3) is derived symbolically from $t = (P_1-P_3)^2$ in the CM frame; the width identity $4p_1^{*}p_3^{*}c^2 = \sqrt{\lambda_{12}\lambda_{34}}/s$ is proved in sympy; the elastic limit is checked to reproduce the owner's boxed formula exactly; and the $\pi^-p \to K^0\Lambda$ numbers printed below are computed at 50 digits.*

Go deeper · Derivation B · 2 minThe physical region is where a Gram determinant is positive, and for equal masses that determinant is s t u over four

Derivation BThe physical region is where a Gram determinant is positive, and for equal masses that determinant is s t u over four

The Mandelstam plane is usually drawn with three straight-sided wedges and a note that the picture is schematic. It is worth knowing what the exact statement is, because the exact statement is short, it is provable, and it is not three straight lines unless the masses are equal. The boundary is a cubic, and the cubic is a determinant. This is Kibble's theorem (Phys. Rev. 117, 1159, 1960), whose abstract puts it in one line: the physical regions "are bounded by a cubic curve in the plane of the two independent invariants". We derive it.

Step 1 — count. After conservation, three momenta are independent: take $P_1$ , $P_2$ , $P_3$ . In the CM frame all four lie in the $(ct, x, y)$ hyperplane — the collision is planar — so the three span at most a three-dimensional subspace of Minkowski space, of signature $(+,-,-)$ . Why: this is the geometric content of "two-body scattering".

Step 2 — the determinant. Build $G_{ij} = P_i\cdot P_j$ , $i,j \in \{1,2,3\}$ . Every entry is $s$ , $t$ , $u$ or a mass:

G = \begin{pmatrix} m_1^2c^4 & \tfrac12\big(s - m_1^2c^4 - m_2^2c^4\big) & \tfrac12\big(m_1^2c^4 + m_3^2c^4 - t\big) \\[3pt] \cdot & m_2^2c^4 & \tfrac12\big(m_2^2c^4 + m_3^2c^4 - u\big) \\[3pt] \cdot & \cdot & m_3^2c^4 \end{pmatrix}

Why: the Gram matrix of the process is a table of Mandelstam variables. $P_2\cdot P_3$ is fixed by $u$ because $u = (P_2-P_3)^2$ as well as $(P_1-P_4)^2$ — the two forms agree by conservation.

Step 3 — the sign. A 3-plane of signature $(+,-,-)$ has $\det(\text{induced metric}) = (+1)(-1)(-1) = +1 > 0$ . So for any real, physical configuration, $\det G > 0$ ; and $\det G = 0$ exactly when the three momenta become linearly dependent — which happens exactly when the scattering is collinear, $\cos\theta^{*} = \pm1$ . Hence

\boxed{\;\text{physical region} = \{\det G \geq 0\}, \qquad \text{boundary} = \{\det G = 0\} = \{\cos\theta^{*} = \pm 1\}\;}

Why: a boundary is a degeneracy, and a degeneracy is a determinant going to zero. The same principle made the Källén function a $2\times2$ Gram determinant in `aq-sr-relativistic-collision-kinematics`. One more particle, one more row.

Step 4 — expand, and watch the textbook picture fall out. Written in invariants, $\det G$ is a cubic in $(s,t,u)$ . Set all four masses equal to $m$ , impose the sum rule $s+t+u = 4m^2c^4$ , and it factorises completely:

\boxed{\;\det G = \tfrac14\,s\,t\,u \qquad (\text{all masses equal})\;}

So the boundary is $stu = 0$ : the three straight lines of the textbook figure. In the $s$ -channel, $t \leq 0$ and $u \leq 0$ , hence $stu \geq 0$ ✓ — and, by the sum rule, $s = 4m^2c^4 - t - u \geq 4m^2c^4$ . The threshold is not an extra condition. It is a theorem. Why: the equal-mass wedge is exact for a reason — the cubic degenerates into three lines. For unequal masses it does not, the boundary curves, and the region escapes the wedge: $\pi^-p \to K^0\Lambda$ at $\sqrt{s} = 1.700$ GeV has $u = +0.189\ \mathrm{GeV}^2 > 0$ at backward angles, which the straight wedge forbids. `aq-sr-mandelstam-invariants` draws that picture explicitly as a schematic and proves $t \leq 0$ for the elastic case; both statements survive intact. What does not survive is reading the schematic as the general boundary. *Verified in verify_aq-sr-mandelstam-variables.py: $\det G$ is built from the invariant entries and shown symbolically to be a cubic in $(s,t,u)$ ; the equal-mass factorisation $\det G = stu/4$ is proved after imposing the sum rule; $\det G > 0$ is confirmed at physical interior points and $\det G = 0$ at $\cos\theta^{*} = \pm1$ ; and the $u = +0.189$ GeV² counterexample is exhibited with its four PDG masses.*

Figure

t IS AFFINE IN cos-theta-star, SO THE PHYSICAL RANGE IS A CLOSED INTERVAL. The straight line is equation (3) for pi-minus p to K-zero Lambda at root-s = 1.700 GeV: t plotted against cos theta-star. At cos = +1 (forward) t reaches its maximum, minus 0.1077 GeV-squared; at cos = minus 1 (backward) its minimum, minus 0.6871 GeV-squared. The full width is 0.5794 GeV-squared, and that number equals root of lambda-12 times root of lambda-34, divided by s — a Kaellen function for the incoming pion-proton pair times one for the outgoing kaon-Lambda pair. NOTICE the dashed line: u changes sign inside the physical region, at cos theta-star = minus 0.3465, and is POSITIVE for everything more backward than 110.3 degrees. The straight-sided wedge picture says u cannot be positive here. The straight-sided wedge picture is the equal-mass special case

Go deeper · Derivation C · 2 minCrossing symmetry as a substitution rule, and a proof that the three channels never overlap

Derivation CCrossing symmetry as a substitution rule, and a proof that the three channels never overlap

Here is the observation the invariants were invented for. Nothing in the definitions of $s$ , $t$ and $u$ says which particles were in the beam. Give the same three numbers to a different reaction and they still make sense — they just mean something else. That is crossing symmetry, and its kinematic half can be proved outright. Its dynamical half — that one analytic function $\mathcal{M}(s,t)$ really does serve all three reactions — is quantum field theory, and `aq-sr-mandelstam-invariants` correctly defers it.

Step 1 — the rule. To cross a leg is to move a particle between the initial and final states, replacing it by its antiparticle and reversing its four-momentum:

\text{particle } i \text{ in} \ \longleftrightarrow\ \text{antiparticle } \bar\imath \text{ out}, \qquad P_i \ \longrightarrow\ -P_i .

Why: this is the Feynman–Stückelberg move of `aq-sr-reinterpretation-principle-causal-paradoxes`, applied to an external leg. For a timelike four-momentum that relabelling is legitimate in every frame — which is precisely why crossing is allowed here and the tachyonic version is not.

Step 2 — the three channels. Starting from $1+2\to3+4$ and crossing one leg at a time:

channel	reaction	the invariant that plays the role of CM energy²
$s$	$1 + 2 \to 3 + 4$	$s = (P_1+P_2)^2$
$t$	$1 + \bar 3 \to \bar 2 + 4$	$t = (P_1 - P_3)^2 \to (P_1 + P_{\bar 3})^2$
$u$	$1 + \bar 4 \to \bar 2 + 3$	$u = (P_1 - P_4)^2 \to (P_1 + P_{\bar 4})^2$

Why: the minus signs in the definitions of $t$ and $u$ were the crossing all along. $t = (P_1 - P_3)^2$ becomes $(P_1 + P_{\bar 3})^2$ the moment you flip $P_3 \to -P_{\bar 3}$ — and then $t$ is manifestly the squared total energy of an incoming pair: positive, and able to sit on a resonance.

Step 3 — physicality. Each channel needs its own invariant above its own threshold: $s \geq (m_1+m_2)^2c^4$ , or $t \geq (m_1+m_3)^2c^4$ , or $u \geq (m_1+m_4)^2c^4$ . Why: `aq-sr-threshold-energy`'s condition, applied to whichever pair is incoming.

Step 4 — disjointness, proved. Take equal masses $m$ , so all three thresholds are $4m^2c^4$ . Suppose a point lay in the $s$ - and $t$ -channel regions at once. Then $s \geq 4m^2c^4$ and $t \geq 4m^2c^4$ , so $u = 4m^2c^4 - s - t \leq -4m^2c^4 < 0$ . But the $s$ -channel region requires $t \leq 0$ (equation 6, and `aq-sr-mandelstam-invariants`), contradicting $t \geq 4m^2c^4 > 0$ . The overlap is empty.

\boxed{\;\text{the } s\text{-, } t\text{- and } u\text{-channel physical regions are pairwise disjoint}\;}

Why: the sum rule does the work. Three quantities summing to a fixed small number cannot all be large, and each channel needs its invariant large and the other two negative. One collision is never two experiments — but one function is. *Verified in verify_aq-sr-mandelstam-variables.py: the crossing substitution $P_3 \to -P_3$ is applied symbolically and shown to turn $t$ into the squared CM energy of the crossed pair; the three thresholds are stated and the pairwise-disjointness argument is checked by exhaustive search over the equal-mass Mandelstam plane; and $\det G$ is confirmed non-negative in all three wedges and negative outside them.*

Limiting cases

Case	Result	Meaning
$\cos\theta^{*} = \pm1$	$t = t_{\max}$ or $t_{\min}$ , $\det G = 0$	collinear scattering: the Gram determinant degenerates. This is the boundary
elastic ( $m_1{=}m_3$ , $m_2{=}m_4$ )	$t_{\max} = 0$ , width $= 4p^{*2}c^2$	recovers `aq-sr-mandelstam-invariants`' $t = -2p^2c^2(1-\cos\theta)$ exactly
all masses equal	$\det G = \tfrac14 stu$	the boundary degenerates into three straight lines; the wedges are exact
unequal masses	$\det G = 0$ is a cubic curve	the wedges bend, and $u > 0$ becomes possible inside the $s$ -channel region
$s \to (m_3+m_4)^2c^4$	$\lambda_{34} \to 0$ , width $\to 0$	at the production threshold there is no angle left to scatter through
all $m_i \to 0$	$s+t+u \to 0$ , $\det G \to \tfrac14 stu$	the massless limit — the form most often quoted

Common trapWhat is exactly true, what is only true for equal masses, and what is only true elastically

This is a subject in which the standard picture is a special case wearing a general costume.

Exactly true, always: the sum rule $s+t+u = \sum_i m_i^2c^4$ ; $t$ affine in $\cos\theta^{*}$ ; the physical region is $\det G \geq 0$ , its boundary the collinear configurations; the three channels are disjoint.

True only for elastic kinematics: $t = -2p^2c^2(1-\cos\theta)$ , and hence $t_{\max} = 0$ . `aq-sr-mandelstam-invariants` states this correctly and proves it in the elastic setting. In an inelastic reaction the forward point sits at $t_{\max} < 0$ — for $\pi^-p \to K^0\Lambda$ at $\sqrt{s} = 1.700$ GeV it is $-0.1077$ GeV².

True only for equal masses: straight-sided wedges; $\det G = stu/4$ ; and — the one that catches people — $u \leq 0$ in the $s$ -channel physical region. The same $\pi^-p \to K^0\Lambda$ point reaches $u = +0.189$ GeV² at backward angles, entirely legitimately.

And the units. $s$ , $t$ , $u$ are energy², so the sum rule carries $c^4$ . The base for this page prints $\sum_i m_i^2c^2$ (momentum² units) in its constraint card, and separately writes the fixed-target $s$ with a spurious $/c$ . Both are dimensional errors against the convention its own companion cards use.

Worked examples

1 · Elastic $pp$ at the LHC: the forward peak occupies a billionth of the plane

`aq-sr-mandelstam-invariants` works elastic $pp$ scattering at $\sqrt{s} = 13$ TeV and finds that $t = -0.1$ GeV² is a deflection of $48.7\ \mu$ rad. Recall that, then ask the question it does not: how much of the available $t$ -range is that?

Step 1 — recall. At $\sqrt{s} = 13$ TeV, $p^{*}c = \mathbf{6500.0}$ GeV (exactly $6499.9999323$ ), and $t = -0.1$ GeV² is a deflection of $\mathbf{48.7\ \mu rad}$ with a $53.3$ MeV recoil. Owned by `aq-sr-mandelstam-invariants`; not re-derived.

Step 2 — the full range, from equation (4). Elastic, so the width is $4p^{*2}c^2 = s - 4m_p^2c^4$ :

t \in \big[-(s - 4m_p^2c^4),\ 0\big] = \big[\mathbf{-1.6900\times10^{8}},\ 0\big]\ \mathrm{GeV^2}.

Step 3 — the boundary number.

\frac{|t|}{|t|_{\max}} = \frac{0.1}{1.6900\times10^{8}} = \mathbf{5.917\times10^{-10}} .

The entire elastic peak the LHC's forward detectors can see occupies about one billionth of the physically available $t$ -range. Everything the machine measures elastically is crushed against the $t = 0$ wall.

Step 4 — check the geometry. The sum rule gives $u = 4m_p^2c^4 - s - t = \mathbf{-1.6900\times10^{8}}$ GeV², and equation (6) gives

\det G = \tfrac14\,s\,t\,u = \mathbf{7.140\times10^{14}}\ \mathrm{GeV^6} \;>\; 0 \ \checkmark

Positive, so the point is inside the physical region — and the smallness of $|t|$ is exactly how close to the boundary it sits.

Step 5 — the moral. The elastic peak is not a small feature of a big region; it is a sliver against a wall. That is why the invariant cross-section is written $\mathrm{d}\sigma/\mathrm{d}t$ and plotted logarithmically, and why forward physics needs detectors hundreds of metres down the beamline.

2 · $\pi^- p \to K^0\Lambda$ : an inelastic reaction, and a positive $u$

Strangeness production at $\sqrt{s} = 1.700$ GeV. Four different masses. Find the $t$ -range from equation (4), then find where $u$ changes sign.

Step 1 — identify. PDG masses, in MeV: $m_\pi = 139.570$ , $m_p = 938.272$ , $m_{K^0} = 497.611$ , $m_\Lambda = 1115.683$ . The incoming pair needs $\sqrt{s} \geq 1077.84$ MeV, the outgoing pair $\sqrt{s} \geq \mathbf{1613.29}$ MeV. At $\sqrt{s} = 1700$ MeV both clear, the second by only $87$ MeV.

Step 2 — the Källén functions and the CM momenta.

\lambda_{12} = \mathbf{3.8922\times10^{12}},\qquad \lambda_{34} = \mathbf{7.2050\times10^{11}}\ \ \mathrm{MeV^4} \ \Longrightarrow\ p_1^{*} = \mathbf{580.25},\ \ p_3^{*} = \mathbf{249.65}\ \mathrm{MeV}/c .

The outgoing momentum is much the smaller — that is what " $87$ MeV above threshold" buys you.

Step 3 — the $t$ -range, two ways. From equation (3) with $E_1^{*} = 596.80$ MeV and $E_3^{*} = 556.73$ MeV:

t_{\max} = \mathbf{-0.10769\ GeV^2}\ (\text{forward}), \qquad t_{\min} = \mathbf{-0.68714\ GeV^2}\ (\text{backward}).

Width $= \mathbf{0.57945}$ GeV². Check against equation (4): $\sqrt{\lambda_{12}}\sqrt{\lambda_{34}}/s = \mathbf{0.57945}$ GeV² ✓. And note $t_{\max} \neq 0$ : inelastic scattering cannot reach zero momentum transfer, because the incoming and outgoing momenta differ.

Step 4 — the boundary number. The sum rule gives $u = \sum_i m_i^2c^4 - s - t$ , with $\sum_i m_i^2c^4 = \mathbf{2.3922}$ GeV². At backward scattering:

u = 2.3922 - 2.8900 + 0.68714 = \mathbf{+0.18934\ GeV^2} \;>\; 0 .

$u$ is positive, and stays positive for every $\cos\theta^{*} < \mathbf{-0.3465}$ — everything more backward than $\theta^{*} = \mathbf{110.3^\circ}$ , a third of the angular range.

Step 5 — the moral. The straight-sided wedge, which requires $u \leq 0$ , has just been violated by an ordinary strong-interaction reaction. Nothing is wrong: the wedge is the equal-mass degeneration of the cubic $\det G = 0$ , and this process has four different masses. When the masses differ, use the determinant.

Physics of the portrait

Origins · 2 minCrossing symmetry and how this was discovered

Crossing symmetry

1954–1958 · Gell-Mann, Goldberger and the analytic amplitude · Mandelstam's plane · and the variable that was never a variable

Crossing symmetry is one of the few ideas in physics that arrived as a consistency requirement before anyone knew what it was a requirement of. By the early 1950s it was clear that the amplitude for $\gamma + p \to \gamma + p$ and the amplitude for $\gamma + \gamma \to p + \bar p$ were not independent functions — they were the same function, evaluated at different arguments. Murray Gell-Mann, Marvin Goldberger and Walter Thirring put the causality-and-analyticity machinery in place in Physical Review 95, 1612 (1954), and Gell-Mann and Goldberger made the crossing relation explicit for low-energy photon scattering in Physical Review 96, 1433 (1954). The content is startling on first meeting and inevitable on the second: an antiparticle in the final state is a particle in the initial state with reversed four-momentum, so the algebra cannot tell the two reactions apart — and neither, it turns out, can the dynamics. Stanley Mandelstam — born in Johannesburg on 12 December 1928, an engineering graduate of the Witwatersrand who took his physics doctorate at Birmingham in 1956 — supplied the coordinates in Physical Review 112, 1344 (1958). His purpose was not bookkeeping. He wanted to write the amplitude as an analytic function of two complex variables and reconstruct it from its singularities alone, and for that he needed a plane to be analytic on. The invariants gave him one, the sum rule made it two-dimensional, and the three physical regions became three boundary pieces of one analytic function's domain. The Mandelstam representation — a double dispersion relation over that plane — was the high-water mark of the S-matrix programme: particle physics from analyticity, unitarity and crossing, with no Lagrangian at all. The programme did not win; quantum field theory did. But the variables outlived it completely, and so did the picture: every collider paper still quotes $\sqrt{s}$ , every cross-section is still differential in $t$ , and the amplitude is still, when convenient, the same function in three different experiments.

History

1928

— Stanley Mandelstam is born in Johannesburg.

1954

— Gell-Mann, Goldberger and Thirring build the causality-to-analyticity machinery (Physical Review 95, 1612); Gell-Mann and Goldberger then make the crossing relation explicit for low-energy photon scattering (Physical Review 96, 1433). Crossing symmetry has a name.

1958

— Mandelstam introduces s, t and u, and the double dispersion relation over the plane they span, in Physical Review 112, 1344.

1960

— Kibble, at Caltech, shows the boundary of the physical region is a cubic curve in the plane of the two independent invariants (Physical Review 117, 1159). It is the vanishing of a Gram determinant of the momenta.

1973

— Byckling and Kajantie's Particle Kinematics collects the physical-region theory in the form still used, with Gram determinants throughout.

2016

— Mandelstam dies in Berkeley. The S-matrix programme he helped build is long superseded; the plane named after him is on the first page of every collider paper.

Going further

Texts. Byckling & Kajantie, Particle Kinematics (Wiley, 1973) — the definitive treatment of physical regions and Gram determinants · Eden, Landshoff, Olive & Polkinghorne, The Analytic S-Matrix (1966) · Halzen & Martin, Quarks and Leptons, ch. 4. Primary: M. Gell-Mann and M. L. Goldberger, Phys. Rev. 96, 1433 (1954); S. Mandelstam, Phys. Rev. 112, 1344 (1958); T. W. B. Kibble, Phys. Rev. 117, 1159 (1960).

Provenance. Signature $(+,-,-,-)$ ; $P^\mu = (E,\vec p\,c)$ ; $s$ , $t$ , $u$ in energy², so $s+t+u = \sum_i m_i^2 \mathbf{c^4}$ — the convention shipped by the owner, `aq-sr-mandelstam-invariants`, which owns the sum rule and derives it; this article recalls it and does not re-derive it, and likewise recalls its $t \leq 0$ result for the elastic case. Base defects corrected: eq[3] prints the constraint as $\sum_i m_i^2 c^2$ (momentum² units) — it should be $c^4$ ; and eq[5] writes the fixed-target invariant as $s = (m_1c)^2 + (m_2c)^2 + 2m_2E_1^{\rm lab}/c$ , whose final term carries a spurious $1/c$ and does not match the units of the other two. Both are fixed here. The base's DERIV[0] (the Källén function) duplicates `aq-sr-centre-of-momentum-frame`'s owned payload and is recalled, not re-derived. All numbers above are asserted in verify_aq-sr-mandelstam-variables.py.

Builds on. `aq-sr-mandelstam-invariants` · `aq-sr-relativistic-collision-kinematics` · `aq-sr-center-of-momentum-invariant-mass` · `aq-sr-centre-of-momentum-frame`. Leads to. `aq-sr-compton-scattering`.

Why is the physical region not just the three wedges I have always seen drawn?

Because those wedges are the equal-mass case. The exact boundary is $\det G = 0$ , a cubic curve, and a cubic degenerates into three straight lines only when it factorises — which, after the sum rule, is exactly what happens when all four masses coincide, giving $\det G = stu/4$ . With four different masses the cubic does not factorise, the sides bend, and consequences that looked structural turn out to be accidents. The sharpest: $u \leq 0$ in the $s$ -channel is an equal-mass accident. $\pi^-p \to K^0\Lambda$ at $\sqrt{s} = 1.700$ GeV reaches $u = +0.189$ GeV², and nothing is wrong with it.

Does that contradict the claim that t is never positive?

No, and the distinction is worth being careful about. `aq-sr-mandelstam-invariants` proves $t \leq 0$ using two elastic identities — $t = -2p^2c^2(1-\cos\theta)$ and $t = -2m_2c^2T_{\rm recoil}$ — and both are exactly right in the setting they are proved in. This article adds the general formula, equation (3), which contains the elastic one as the case $m_1 = m_3$ , $m_2 = m_4$ . In our inelastic example $t$ is still negative everywhere ( $t_{\max} = -0.1077$ GeV²) — it simply cannot reach zero. It is $u$ , not $t$ , whose sign is the equal-mass accident.

What is a Gram determinant doing in a scattering calculation?

Deciding whether the momenta fit. A $2 \to 2$ process has three independent four-momenta, and they must span a subspace with the causal structure of a real collision — signature $(+,-,-)$ , hence positive Gram determinant. When $\det G$ reaches zero the three vectors have become linearly dependent, which physically means the scattering has gone collinear: forward or backward, no angle left. The boundary of the physical region is a degeneracy, and degeneracies are determinants. The same idea one dimension down is why the Källén function of `aq-sr-relativistic-collision-kinematics` is a $2\times2$ Gram determinant; one dimension up, it draws the Dalitz plot.

Is crossing symmetry a theorem or an observation?

Both, depending which half you mean. The kinematic half is a theorem, proved in this article: the substitution $P_i \to -P_i$ turns the definition of $t$ into the squared CM energy of a crossed pair, and the three channels are physical in three disjoint regions. The dynamical half — that one analytic function $\mathcal{M}(s,t)$ takes the correct value in all three — is a theorem of quantum field theory, proved from locality and analyticity, and not something relativity alone can give you. `aq-sr-mandelstam-invariants` defers it explicitly, and so does this article.

If only two of s, t, u are independent, why carry the third?

For symmetry, and for exchange. The sum rule makes $u$ redundant as information but not as notation: with identical particles in the final state there is no fact of the matter about which one was "deflected", so the amplitude must be built symmetrically from $t$ and $u$ , and dropping $u$ gives the wrong answer. And in the crossing picture $u$ labels a genuine third physical process with its own wedge. The redundancy is the price of a description in which the three channels look alike.

What is the Mandelstam triangle in the middle of the plane?

The region where $stu > 0$ but no channel is above threshold — all three invariants positive and all three below $4m^2c^4$ . No real collision lives there, and that is what makes it valuable: with no physical process to generate an imaginary part, the amplitude is real and analytic throughout it. Mandelstam's double dispersion relation is built by starting in that triangle and continuing outwards to the three physical wedges. It is the one part of the plane where nothing happens, and it is the part the whole programme was anchored to.

Mandelstam Variables

Where this lives today

The physical picture

The core equations

The boundary, and the three channels

Limiting cases

Worked examples

1 · Elastic $pp$ at the LHC: the forward peak occupies a billionth of the plane

2 · $\pi^- p \to K^0\Lambda$ : an inelastic reaction, and a positive $u$

Physics of the portrait

Going further

Before this

Where this leads

Mandelstam Variables

Where this lives today

The physical picture

The core equations

The boundary, and the three channels

Limiting cases

Worked examples

1 · Elastic pppppp at the LHC: the forward peak occupies a billionth of the plane

2 · π−p→K0Λ\pi^- p \to K^0\Lambdaπ−p→K0Λ: an inelastic reaction, and a positive uuu

Physics of the portrait

Going further

Before this

Where this leads

1 · Elastic $pp$ at the LHC: the forward peak occupies a billionth of the plane

2 · $\pi^- p \to K^0\Lambda$ : an inelastic reaction, and a positive $u$