Mandelstam Invariants

At CERN, 220 metres down the beamline from the point where the LHC's protons meet, there are detectors that live in movable steel pots. During a physics run they are driven to within millimetres of a beam carrying enough energy to melt a ton of copper. They are there to catch protons scattered by about fifty microradians — three thousandths of a degree, a deflection so small that after 220 metres of flight the proton has wandered about a centimetre off course. Why such trouble for such a small angle? Because the quantity the theorists want is not the angle. It is a number called $t$ , and at the LHC an angle of 49 microradians is $t = -0.1\ \mathrm{GeV}^2$ . Every experiment on Earth reports its scattering data in $t$ , not in degrees, because degrees are a frame-dependent accident and $t$ is not. Boost to another frame and the angle changes. $t$ does not. There are three such numbers for a two-body collision — $s$ , $t$ and $u$ — and between them they hold everything Lorentz invariance permits you to know about the kinematics. This article proves it, in the form of a single identity:

s + t + u = \big(m_1^2 + m_2^2 + m_3^2 + m_4^2\big)c^4

Three invariants; one constraint; two independent numbers. A $2\to2$ collision has exactly two kinematic degrees of freedom — how hard, and how sideways — and $s$ and $t$ are they. Everything else is bookkeeping. And the last of the three carries a punchline that outlives the kinematics: in the physical region, $t$ is never positive. We prove that in one line — and then watch it forbid the particle exchanged in a $t$ -channel diagram from ever being a real particle at all.

Key takeaways

Signature $(+,-,-,-)$ , $p\cdot p = +m^2c^2$ . All three invariants are in ENERGY² units (GeV², J²), matching Centre-of-Momentum Frame and Threshold Energy: $\sqrt{s} = E_{\mathrm{cm}}$ is an energy.
For $1 + 2 \to 3 + 4$ :

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

with $(A)^2 \equiv A^0A^0 - |\vec A|^2$ evaluated on four-vectors scaled to energy units. Each is a Lorentz scalar: the same number in every frame.

The sum rule — the owned result:

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

Three invariants, one constraint, two independent. That matches the two real degrees of freedom of a $2\to2$ collision exactly: $\sqrt{s}$ (how hard) and $\theta$ (how sideways).

$s$ is energy. $\sqrt{s} = E_{\mathrm{cm}}$ . Fixed target: $s = m_1^2c^4 + m_2^2c^4 + 2m_2c^2E_1^{\mathrm{lab}}$ . Collider: $\sqrt{s} = 2E_{\mathrm{beam}}$ .
$t$ is momentum transfer, and in the CM frame (elastic, equal masses) $t = -2p^2c^2(1-\cos\theta)$ — zero at $\theta=0$ , most negative at $\theta = 180°$ .
$t \leq 0$ in the physical region, always, and the cleanest proof is an identity: for elastic scattering off a target at rest, $t = -2m_2c^2 T_{\mathrm{recoil}}$ . The recoil kinetic energy cannot be negative, so $t$ cannot be positive.
Consequence: a $t$ -channel exchanged quantum has a spacelike four-momentum, $t < 0$ , and no real particle can. It is off shell — virtual — by construction, not by approximation.
Key idea: the angle is an accident of your frame; $t$ is not. Kinematics written in $s$ and $t$ is kinematics written in a language that every observer in the universe agrees on — and the price of that agreement is exactly one constraint, the sum rule.

The core insight

A $2 \to 2$ collision has four four-momenta — sixteen numbers. Mass shells and conservation knock that down; Lorentz invariance, which says the answer cannot depend on how you orient or boost your laboratory, knocks it down again. What survives is two numbers. The Mandelstam invariants are the natural coordinates on that two-dimensional space: you can build three manifestly invariant scalars from the momenta, but they are chained by $s+t+u = \sum_i m_i^2c^4$ , so only two are free. The count is the content. Once a collision is a two-parameter object, every cross-section, every angular distribution and every resonance in the subject is a function of $s$ and $t$ alone — and looks the same to every observer in the universe.

Where this lives today

Open any experimental paper on elastic or diffractive scattering and the $x$ -axis is $|t|$ . Open any theory paper on a $2\to2$ process and the amplitude is $\mathcal{M}(s,t)$ . This is not a convention that could have gone the other way: an angle is not a property of a collision, and a cross-section that depended on the observer would not be physics. The invariants also draw the map of where the interesting things are. A resonance — a short-lived particle formed and then decaying — appears as a bump at a particular value of $s$ , because $\sqrt{s}$ is the energy available to make it and it can only be made at its own mass. The $Z$ boson is a bump at $\sqrt{s} = 91.2\ \mathrm{GeV}$ . Meanwhile the forward peak that dominates every hadronic cross-section lives at small $|t|$ , because most collisions are glancing. And the differential cross-section is written $d\sigma/dt$ , not $d\sigma/d\theta$ , precisely so that the answer means the same thing in every frame. Computing it requires an amplitude, and amplitudes are quantum field theory: $\overline{|\mathcal{M}|^2}$ , a function of $s$ and $t$ , is where the dynamics lives, and a future QFT track owns it. The kinematics stops here — having already fixed the shape of the physical region, the sign of $t$ , and the virtuality of every exchanged quantum in the subject.

The physical picture

Take the simplest violent thing that can happen: two particles go in, two come out, $1 + 2 \to 3 + 4$ . You want to describe what happened. The obvious description is the one your apparatus gives you: particle 3 came off at 37 degrees with 4.2 GeV. It is a perfectly good description with one fatal defect — it describes your laboratory, not the collision. An observer drifting past at half the speed of light sees a different angle and a different energy, and no experiment can say which of you is right, because there is nothing to be right about. Angles and energies are coordinates, not facts. So build the facts. Take the four-momenta $P_1, P_2, P_3, P_4$ and contract them into scalars — objects with no free index, and therefore no dependence on your frame. The mass shells $P_i \cdot P_i = m_i^2c^4$ give you nothing new: they are the masses you already knew. The cross terms are the new information, and of the possible combinations three natural ones stand out, one for each way of pairing the four particles:

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

Read them physically. $s$ pairs the two incoming particles: the squared total energy available in the CM frame, and $\sqrt{s}$ is the number a collider is built to. $t$ pairs particle 1 with the outgoing particle it most resembles — "the same particle, deflected" — so $t$ measures how much four-momentum was transferred. $u$ pairs 1 with the other outgoing particle, and it matters when the two outgoing particles are identical and "deflected" cannot be told from "knocked out". Now the surprise. These three are not independent; they obey an identity, true for every event:

s + t + u = \big(m_1^2 + m_2^2 + m_3^2 + m_4^2\big)c^4

The right-hand side is pure mass. It does not know what the collision did. So the three invariants live on a plane, and a collision is a point on a two-dimensional surface — which is exactly right, because a $2\to2$ collision has two degrees of freedom and never had more. Fix the energy and the angle and there is nothing left to fix. That is the first payoff. The second is what happens to $t$ .

The core equations

The three invariants, defined on four-momenta written in energy units, $P^\mu = (E,\ \vec p\,c)$ , so that $P \cdot P = E^2 - |\vec p\,|^2c^2 = m^2c^4$ :

\boxed{\;s \equiv (P_1+P_2)\cdot(P_1+P_2), \qquad t \equiv (P_1-P_3)\cdot(P_1-P_3), \qquad u \equiv (P_1-P_4)\cdot(P_1-P_4)\;} \qquad (1)

where: each is a contraction of four-vectors, hence a Lorentz scalar — one number, agreed on by every observer. Physical reading: all three carry units of energy², so $\sqrt{s}$ is an energy. This is the convention Centre-of-Momentum Frame and Threshold Energy already lock: $\sqrt{s} = 13\ \mathrm{TeV}$ at the LHC. The competing convention writes $P\cdot P$ in momentum² units and converts with $s = (P \cdot P)c^2$ ; both are consistent, mixing them is a factor-of- $c^2$ catastrophe, and this article never mixes them. The identity, and the reason there are only two numbers here and not three:

\boxed{\;s + t + u \;=\; \big(m_1^2 + m_2^2 + m_3^2 + m_4^2\big)\,c^4\;} \qquad (2)

where: the sum runs over the four external masses — two in, two out. Physical reading: the right-hand side is pure mass. It carries no information about the collision at all. So knowing $s$ and $t$ determines $u$ , and the kinematic content of a two-body collision is exactly two real numbers. Note the $c^4$ : with $s$ in energy² units the masses must appear as $m^2c^4$ , i.e. rest energies squared. (The AllQFT base for this article writes $c^2$ here, having defined $s$ in momentum² units two cards earlier and then used energy² units two cards later. It is a units error, and it is corrected throughout this article.) $s$ evaluated where you actually have data — a beam of total energy $E_1^{\mathrm{lab}}$ on a target of mass $m_2$ at rest:

s = m_1^2c^4 + m_2^2c^4 + 2\,m_2c^2\,E_1^{\mathrm{lab}} \qquad\Longrightarrow\qquad \sqrt{s}\;\propto\;\sqrt{E_1} \qquad (3)

where: $E_1^{\mathrm{lab}}$ is the beam's total energy, rest plus kinetic. Physical reading: $s$ is linear in the beam energy, so the useful energy $\sqrt{s}$ grows only as its square root. Threshold Energy owns that comparison and plots it; Centre-of-Momentum Frame derives this formula. Here we only use it. $t$ , in the CM frame, for elastic scattering of equal masses through an angle $\theta$ :

\boxed{\;t = -2p^2c^2\,(1-\cos\theta)\;}\qquad\text{and}\qquad u = -2p^2c^2\,(1+\cos\theta) \qquad (4)

where: $p$ is the common CM three-momentum magnitude, the same for all four particles in the equal-mass elastic case. Physical reading: $t$ runs from $0$ at $\theta = 0$ (nothing happened) to $-4p^2c^2$ at $\theta = 180°$ (turned right around). It is a measure of how violent the deflection was, and it is never positive. Add the two and check: $t + u = -4p^2c^2$ , and since $s = 4(p^2c^2 + m^2c^4)$ for equal masses, $s + t + u = 4m^2c^4$ — the sum rule, verified. And the identity that makes the sign of $t$ inevitable. For elastic scattering (particle 4 is the same species as particle 2) off a target at rest:

\boxed{\;t \;=\; -\,2\,m_2c^2\;T_{\mathrm{recoil}}\;} \qquad (5)

where: $T_{\mathrm{recoil}} = E_4 - m_2c^2$ is the kinetic energy the struck target walks away with. Physical reading: this is the whole story of $t$ in one line. $t$ is (minus twice the target's rest energy times) the energy you managed to knock into it. A target cannot recoil with negative kinetic energy, so $t \leq 0$ , identically, in every elastic process, in every frame. It is also the most useful practical form of $t$ there is: measure the recoil energy of the target and you have measured $t$ , with no angles and no boosts.

Deriving the invariants

Derivation AThe sum rule: three invariants, one constraint, and therefore exactly two independent numbers

We will build $s$ , $t$ and $u$ , confirm that each really is frame-independent, and then prove that they are chained together by an identity whose right-hand side contains nothing but masses. The point is not the algebra — the algebra is four lines — but the count that comes out of it: a two-body collision, stripped of every frame-dependent accident, is a two-parameter object, and the sum rule is the statement that the three obvious invariants overdetermine it by exactly one.

Step 1. Establish that each is invariant. $s$ , $t$ and $u$ are each of the form $A \cdot A$ where $A$ is a sum or difference of four-vectors. Four-vectors form a vector space under the Lorentz transformation — if $P_1$ and $P_3$ each transform as $P \to \Lambda P$ , then so does $P_1 - P_3$ , by linearity. And the contraction of any four-vector with itself is a scalar. Why: this is the entire justification, and it is worth being slow about. We are not claiming $t$ is "approximately frame-independent" or "independent in the CM frame". We are claiming it is a scalar, in the technical sense — Four-Vectors owns that machinery, and this article is simply cashing it.

Step 2. Use conservation to eliminate one momentum. Conservation of Four-Momentum gives $P_1 + P_2 = P_3 + P_4$ , so

P_2 = P_3 + P_4 - P_1

and this single substitution is the whole trick. Everything below is expansion. Why: with $P_2$ expressed through the others, the four momenta are no longer independent, and the "three" invariants get a chance to collapse.

Step 3. Add the three definitions and expand:

s+t+u = (P_1{+}P_2)^2 + (P_1{-}P_3)^2 + (P_1{-}P_4)^2 = 3P_1^2 + P_2^2 + P_3^2 + P_4^2 + 2P_1\!\cdot\!P_2 - 2P_1\!\cdot\!P_3 - 2P_1\!\cdot\!P_4

Why: nothing has happened yet. Every cross term involving $P_1$ has been collected, because $P_1$ is the particle that appears in all three definitions — that is why it is the one that will do the cancelling.

Step 4. Collect the cross terms and kill them. They are all proportional to $P_1$ contracted with something:

2P_1\cdot\big(P_2 - P_3 - P_4\big) \;=\; 2P_1\cdot\big(-P_1\big) \;=\; -2P_1^2

using Step 2. Why: this is the cancellation, and it is why the identity exists at all. The bracket $P_2 - P_3 - P_4$ is exactly $-P_1$ by conservation — the cross terms do not merely simplify, they turn into a mass-shell term. There is nothing left that knows about the dynamics.

Step 5. Substitute back and impose the mass shells $P_i \cdot P_i = m_i^2c^4$ :

s+t+u = 3P_1^2 - 2P_1^2 + P_2^2 + P_3^2 + P_4^2 = P_1^2 + P_2^2 + P_3^2 + P_4^2 = \boxed{\big(m_1^2+m_2^2+m_3^2+m_4^2\big)c^4}

Why: the three $P_1^2$ from the expansion and the $-2P_1^2$ from the cross terms leave exactly one, and the four squares reassemble into the four masses. The right-hand side is a constant of the particles, not of the collision.

Step 6. Read the count. Three invariants, one identity, so two are independent — and a $2\to2$ collision with fixed masses genuinely has two kinematic degrees of freedom, the total energy and the scattering angle. (Azimuth is a third, but nothing unoriented can depend on it.) Why: the count is the content. Were there no sum rule, the invariants would be claiming a collision has three degrees of freedom, and they would be lying. *Verified in verify_aq-sr-mandelstam-invariants.py: the sum rule is proved symbolically in sympy on generic four-momenta with $P_2 = P_3+P_4-P_1$ substituted — the identity $s+t+u-\sum_i P_i^2 = 0$ reduces to exactly zero with no assumption about the masses, the frame, or the process; and the equal-mass elastic forms of $s$ , $t$ and $u$ are shown to satisfy it as $s+t+u = 4m^2c^4$ .*

Go deeper · Derivation B · 3 minWhy t is never positive, and why that makes the exchanged particle a fiction

Derivation BWhy t is never positive, and why that makes the exchanged particle a fiction

The sign of $t$ looks like a technicality and is not. It is the reason the phrase " $t$ -channel exchange" must be read with care, and the reason a Feynman diagram's internal line is not a particle in any sense an experiment could check. We prove $t \leq 0$ twice — once by brute force in the CM frame, once by an identity that makes it obvious — and then read off what it costs. The second proof is the one worth remembering: it holds in any frame, for any elastic process, and it takes one line.

Step 1. Set up the CM frame for elastic, equal-mass scattering. The incoming pair have equal and opposite momenta of magnitude $p$ ; so, by conservation and the mass shell, do the outgoing pair, and all four particles have the same energy $E = \sqrt{p^2c^2 + m^2c^4}$ . Put particle 1 along $+z$ and let particle 3 emerge at angle $\theta$ :

P_1 = (E,\ 0,\ 0,\ pc), \qquad P_3 = (E,\ pc\sin\theta,\ 0,\ pc\cos\theta)

Why: the CM frame is chosen because it makes the energies equal, so the time component of $P_1 - P_3$ vanishes and $t$ becomes a pure spatial norm — which is what forces its sign.

Step 2. Compute $t$ directly:

t = (P_1 - P_3)^2 = \underbrace{(E-E)^2}_{=\;0} - \big|\vec p_1 - \vec p_3\big|^2c^2 = -\big(p^2 + p^2 - 2p^2\cos\theta\big)c^2 = \boxed{-2p^2c^2(1-\cos\theta)}

Why: the energy difference is zero, so $t$ is minus a squared length. That is the proof. $1-\cos\theta \in [0,2]$ , so $t$ runs from $0$ (forward, nothing happened) to $-4p^2c^2$ (backscatter), and it is negative everywhere in between.

Step 3. Object to the proof, because it deserves it. The vanishing of $E - E$ used the CM frame and elastic equal masses. Why: a proof that leans on a special frame still proves the number — $t$ is a scalar, so negative in one frame is negative in all. But a proof that leans on a special process proves less than we want.

Step 4. Prove it again, generally, with an identity. Take any elastic process — particle 4 is the same species as particle 2, mass $m_2$ — and evaluate $t$ using the other pairing that defines it, $t = (P_2-P_4)^2$ , in the frame where the target 2 is at rest:

t = P_2^2 + P_4^2 - 2P_2\cdot P_4 = 2m_2^2c^4 - 2\,(m_2c^2)(E_4) = -2m_2c^2\big(E_4 - m_2c^2\big) = \boxed{-2\,m_2c^2\,T_{\mathrm{recoil}}}

because with $P_2 = (m_2c^2,\vec 0\,)$ at rest only the time components contract. Why: and now the sign is not a calculation, it is a physical fact. $T_{\mathrm{recoil}} \geq 0$ — a struck target cannot recoil with negative kinetic energy — so $t \leq 0$ , with equality only if nothing was transferred. No frame was privileged: $t$ is a scalar, and we merely computed it somewhere convenient.

Step 5. Now spend it. In a $t$ -channel exchange diagram, an unobserved quantum carries the four-momentum $q = P_1 - P_3$ from particle 1 to particle 2. Its invariant norm is, by definition, $q \cdot q = t$ . If that quantum were a real particle of mass $\mu$ , it would be on shell: $q \cdot q = \mu^2c^4 \geq 0$ . But $t \leq 0$ . Why: the two statements are incompatible, and the physical region wins. The exchanged quantum has a spacelike four-momentum. It is not on any mass shell. It is not a particle that could be caught in a detector, ever, even in principle — the phrase "the electrons exchanged a photon" describes a term in a calculation, not an event.

Step 6. Note what this does not say. $t > 0$ is perfectly possible — in a different process. Crossing takes the reaction $1 + \bar 3 \to \bar 2 + 4$ , in which the quantity we called $t$ plays the role of the total CM energy squared and is positive, and can sit on a resonance. That is the germ of crossing symmetry, and a future QFT track owns it. Why: the same three numbers describe three different physical processes, each in its own region of the $(s,t,u)$ plane, and the regions do not overlap. The invariants are more general than the process that motivated them. *Verified in verify_aq-sr-mandelstam-invariants.py: $t = -2p^2c^2(1-\cos\theta)$ and $u = -2p^2c^2(1+\cos\theta)$ are derived symbolically from explicit CM four-momenta; $t \leq 0$ is asserted over the full angular range; and the identity $t = -2m_2c^2T_{\mathrm{recoil}}$ is proved symbolically from $t = (P_2-P_4)^2$ with the target at rest.*

Figure

THE MANDELSTAM PLANE — the sum rule s + t + u = Σmᵢ²c⁴ is the equation of a PLANE, so the three invariants of any collision are a single point on it, and only TWO of the three are free. The shaded wedge is the physical region of the s-channel process 1+2 → 3+4: it requires s ≥ (m₁+m₂)²c⁴ AND t ≤ 0 AND u ≤ 0. The dashed axis t = 0 is a WALL the physical region can touch but never cross — that is the geometric statement of t ≤ 0, and it is why a t-channel exchanged quantum is spacelike and can never be a real particle. Crossing carries the SAME three numbers into the t-channel and u-channel wedges, which are different physical processes

Limiting cases

Case	Result	Meaning
$\theta = 0$ (forward)	$t = 0$ , $u = -4p^2c^2$	nothing was transferred — the beam sailed past
$\theta = 180°$ (backward)	$t = -4p^2c^2$ , $u = 0$	maximal transfer; $t$ hits its floor
$\theta = 90°$ (CM)	$t = u = -2p^2c^2$	the symmetric point; $t$ and $u$ swap under $\theta \to \pi - \theta$
all four masses zero	$s + t + u = 0$	the massless sum rule — the only case where it is homogeneous
$p \to 0$ (threshold)	$t, u \to 0$ , $s \to (m_1+m_2)^2c^4$	no momentum, nothing to transfer
elastic, target at rest	$t = -2m_2c^2T_{\mathrm{recoil}}$	$t \leq 0$ made obvious — recoil energy cannot be negative
fixed target, $E_1 \gg mc^2$	$s \to 2m_2c^2E_1$	$\sqrt{s} \propto \sqrt{E_1}$ — Threshold Energy owns this
collider, equal beams	$\sqrt{s} = 2E_{\mathrm{beam}}$	the lab frame is the CM frame

Common trapThe units of s are ENERGY squared — and the base for this article gets it wrong in print

There are two self-consistent conventions in the literature and they differ by $c^2$ . Either

energy² units (used here, and by Centre-of-Momentum Frame and Threshold Energy, and by every experiment that says " $\sqrt{s} = 13\ \mathrm{TeV}$ "): $s = (\sum E)^2 - |\sum \vec p\,|^2c^2$ , so $\sqrt{s}$ is an energy and the sum rule reads $s+t+u = \sum_i m_i^2 \mathbf{c^4}$ ; or
momentum² units: $s = P\cdot P$ with $P^\mu = (E/c, \vec p\,)$ , so $s$ is a momentum² and the sum rule reads $\sum_i m_i^2c^2$ .

Both are fine. Mixing them is not, and the mixing is not hypothetical: the AllQFT base for this article writes the sum rule as $s+t+u = \sum_i m_i^2c^2$ (momentum² units) and then, one card later, writes $E_{\mathrm{cm}} = \sqrt{s}\,c$ with an energy²-unit radicand inside the root. Those two cards cannot both be right, and a reader who trusts them will be wrong by $c^2$ — a factor of $9\times10^{16}$ in SI.

The cheapest possible defence is dimensional. Ask what $\sqrt{s}$ is. If your answer is "13 TeV", then $s$ is an energy², the masses must enter as $m^2c^4$ , and $\sqrt{s}\,c$ is not an energy — it is nonsense. This article uses energy² units everywhere.

Worked examples

1 · Elastic proton–proton scattering at the LHC: from 13 TeV to 49 microradians

The LHC collides protons head-on at $\sqrt{s} = 13\ \mathrm{TeV}$ . TOTEM measures the elastic differential cross-section $d\sigma/dt$ . What scattering angle corresponds to $t = -0.1\ \mathrm{GeV}^2$ , and how much energy does the struck proton walk away with?

Step 1 — identify. Elastic, equal masses, and (because the beams are head-on and equal) the lab frame is the CM frame. So equation (4) applies directly, with no boost anywhere in the problem.

Step 2 — get $p$ . For equal masses, $s = 4(p^2c^2 + m_p^2c^4)$ , so

pc = \sqrt{\tfrac{s}{4} - m_p^2c^4} = \sqrt{\tfrac{(13000)^2}{4} - (0.938272)^2}\ \mathrm{GeV} = \mathbf{6500.0\ GeV}

to seven figures — the proton mass is utterly negligible against 6.5 TeV, and the CM momentum is just the beam momentum.

Step 3 — apply. From $t = -2p^2c^2(1-\cos\theta)$ , and for the tiny angles in play, $1 - \cos\theta \approx \theta^2/2$ , so $|t| \approx p^2c^2\theta^2$ and

\theta \;\approx\; \frac{\sqrt{|t|}}{pc} \;=\; \frac{\sqrt{0.1}\ \mathrm{GeV}}{6500\ \mathrm{GeV}} \;=\; 4.87\times10^{-5}\ \mathrm{rad} \;=\; \mathbf{48.7\ \mu rad}

Step 4 — the recoil, from equation (5). $T_{\mathrm{recoil}} = |t|/(2m_pc^2) = 0.1/(2 \times 0.938272) = \mathbf{53.3\ MeV}$ . The struck proton is deflected by 49 microradians and picks up 53 MeV — about 5.7 % of its own rest energy — while the beam proton it came from carries 6.5 TeV. That is what "soft" means at the energy frontier.

Step 5 — the boundary number, and why the pots exist. $49\ \mu\mathrm{rad}$ over the $220\ \mathrm{m}$ flight to the Roman pots is a transverse displacement of about $1.1\ \mathrm{cm}$ — which is why the detectors must be movable, driven to within millimetres of the beam once it is stable, and why they sit 220 m away rather than in the cavern. TOTEM reaches down to $|t| \approx 8\times10^{-4}\ \mathrm{GeV}^2$ , which is $\theta = \mathbf{4.4\ \mu rad}$ and a recoil of $0.43\ \mathrm{MeV}$ . The entire apparatus is an instrument for measuring a Lorentz invariant, and its hardest engineering constraint is set by how small $t$ you want.

2 · The antiproton threshold, in the language of $s$

Rewrite the classic Bevatron threshold — $p + p \to p + p + p + \bar p$ — as a statement about $s$ , and check the sum rule is not needed.

Step 1 — the condition. Threshold Energy owns this: the reaction runs iff $\sqrt{s} \geq \sum_f m_fc^2$ . Baryon number forces four baryons out, so

\sqrt{s} \;\geq\; 4m_pc^2 = 3.753\ \mathrm{GeV} \qquad\Longleftrightarrow\qquad s \;\geq\; \mathbf{14.09\ GeV^2}

Step 2 — convert to a beam energy with equation (3), $m_1 = m_2 = m_p$ : $16m_p^2c^4 = 2m_p^2c^4 + 2m_pc^2E_1$ , giving $E_1 = 7m_pc^2$ and a kinetic energy $T_1 = 6m_pc^2 = \mathbf{5.630\ GeV}$ .

Step 3 — the point. Notice what was not used: $t$ , $u$ , the sum rule, the scattering angle. A threshold is a statement about $s$ alone, because it asks only "is there enough energy?" and not "where did it go?". $t$ and $u$ enter the moment you ask about the angular distribution — which is dynamics, and which is where $d\sigma/dt$ lives.

Figure

t IS THE DEFLECTION, MEASURED INVARIANTLY. Left: t = −2p²c²(1 − cos θ) sweeps from 0 at θ = 0 (nothing happened) down to its floor −4p²c² at θ = 180° (turned right around). The curve NEVER crosses into t > 0 — that is the whole content of t ≤ 0. Right: the same statement without any frame at all — for elastic scattering off a target at rest, t = −2m₂c² T_recoil, so t is just minus the recoil kinetic energy, scaled. A target cannot recoil with negative kinetic energy. At the LHC with pc = 6500 GeV, t = −0.1 GeV² is a 48.7 microradian deflection and a 53.3 MeV recoil

Physics of the portrait

The invariants were not invented to make undergraduate kinematics tidy. They were invented because a scattering amplitude needed somewhere to live.

Origins · 2 minMandelstam variables and how this was discovered

Mandelstam variables

1958 · three scalars for a two-body collision · a plane with three wedges on it · and the coordinates in which an amplitude is an analytic function

By the mid-1950s quantum field theory was in trouble: the strong interaction had no small coupling constant, so perturbation theory — the one tool that worked for electrodynamics — was useless for the pions and nucleons filling the bubble chambers. The response was to constrain the scattering amplitude directly, from general principles alone: Lorentz invariance, unitarity, and analyticity. But a function of what? Angles and energies were useless — they change under a boost, and an amplitude must not. Stanley Mandelstam, South-African-born and Birmingham-trained under Rudolf Peierls, supplied the answer in a 1958 Physical Review paper on the pion–nucleon amplitude: use the invariants $s$ , $t$ and $u$ . Because the sum rule ties them together only two are free, so the amplitude becomes a function of two complex variables — in which Mandelstam wrote his double dispersion relation, now the Mandelstam representation. The variables outlived the programme that produced them. The bootstrap did not survive the arrival of QCD; $s$ , $t$ and $u$ are on every page of every scattering calculation done since, and every differential cross-section ever published has been plotted against $t$ . They are what is left when you take a collision and remove the observer.

History

1905

— Einstein's kinematics makes the observer's frame a choice, so any statement of physical content must be frame-independent. The search for invariants begins.

1906

— Poincare's Palermo memoir (received July 1905, published 1906) sets out the four-dimensional formalism and the invariant quadratic form.

1908

— Minkowski's "Raum und Zeit" gives spacetime its geometry; contractions of four-vectors become the natural currency of relativistic physics.

1923

— Compton scatters X-rays off electrons and finds a shift that depends only on angle. It is the first scattering experiment whose analysis is pure four-momentum bookkeeping.

1958

— Stanley Mandelstam introduces s, t and u in Physical Review 112, 1344, as coordinates for the double dispersion relation for the pion-nucleon amplitude.

2011

— TOTEM at the LHC measures elastic proton-proton scattering down to |t| below 0.001 GeV squared, using detectors 220 m from the collision point to catch microradian deflections.

Going further

Texts. Griffiths, Introduction to Elementary Particles, 2nd ed., §3.4–3.5 · Halzen & Martin, Quarks and Leptons, ch. 4 · Rindler, Introduction to Special Relativity, 2nd ed., §35 · Byckling & Kajantie, Particle Kinematics (the definitive reference on the physical regions) · S. Mandelstam, "Determination of the Pion-Nucleon Scattering Amplitude from Dispersion Relations and Unitarity. General Theory", Physical Review 112, 1344 (1958) · Particle Data Group, Review of Particle Physics, "Kinematics".

Provenance. Signature $(+,-,-,-)$ ; $p\cdot p = +m^2c^2$ . All three invariants are in ENERGY² units, so $\sqrt{s}$ is an energy — matching the live convention of Centre-of-Momentum Frame and Threshold Energy. Base defect corrected: the AllQFT base writes the sum rule as $s+t+u = \sum_i m_i^2c^2$ (momentum² units) while a neighbouring card writes $E_{\mathrm{cm}} = \sqrt{s}\,c$ with an energy²-unit radicand — two mutually inconsistent unit systems inside one card list. The sum rule is re-derived here in energy² units as $\sum_i m_i^2 c^4$ . verify_aq-sr-mandelstam-invariants.py proves the sum rule symbolically and re-derives every number printed above: $s = 1.690\times10^8\ \mathrm{GeV}^2$ , $4m_p^2c^4 = 3.5214\ \mathrm{GeV}^2$ , $pc = 6500.0$ GeV, $\theta = 48.7\ \mu$ rad and $T = 53.3$ MeV at $t = -0.1\ \mathrm{GeV}^2$ , $\theta = 4.4\ \mu$ rad at $|t| = 8\times10^{-4}\ \mathrm{GeV}^2$ , the 1.1 cm displacement at 220 m, and the antiproton threshold $s = 14.09\ \mathrm{GeV}^2$ .

Scope. This article is kinematics. It builds the invariants, proves the sum rule, and fixes the sign of $t$ — and it stops there. The dynamics — the invariant differential cross-section $d\sigma/dt = \overline{|\mathcal{M}|^2}/(64\pi s\, p_{\mathrm{cm}}^2)$ , the amplitude $\mathcal{M}(s,t)$ , crossing symmetry as a theorem rather than an observation, and the analytic structure Mandelstam actually wanted the variables for — is quantum field theory and is owned by a future track. Mandelstam Variables (S7) revisits these invariants from that side; this article owns the kinematics it recalls.

Builds on. Relativistic Four-Momentum · Invariant Mass · Energy-Momentum Relation · Relativistic Energy · Centre-of-Momentum Frame. Leads to. Massless Particles · Mandelstam Variables · Threshold Energy.

Why bother with s and t at all, when the scattering angle is what I actually measure?

Because the angle is a fact about your laboratory, not about the collision. Boost to a frame moving at $0.5c$ and the same event shows a different angle and a different outgoing energy — so an angular distribution is not, by itself, a statement about physics; it is a statement about physics plus your velocity. The invariants strip the observer out. $t$ is a number that every observer in the universe computes and agrees on, so a plot of $d\sigma/dt$ from a fixed-target experiment can be compared directly against a plot from a collider, with no transformation and no argument. You still measure angles. You just never report them.

Is s really the same thing as the invariant mass squared?

Yes — of the system, not of any particle. $\sqrt{s}$ is exactly $Mc^2$ where $M$ is the invariant mass of the two-particle system: Centre-of-Momentum Frame derives it, Invariant Mass owns the concept. The trap is that $M \neq m_1 + m_2$ — a system's invariant mass is not the sum of its parts' masses unless they are all mutually at rest. Two protons approaching each other at the LHC have $\sqrt{s} = 13\ \mathrm{TeV}$ against a combined rest mass of $1.88\ \mathrm{GeV}$ : a factor of seven thousand, all of it kinetic energy that the invariant mass counts and the rest masses do not.

Can t ever be positive?

Not in the physical region of the process $1+2 \to 3+4$ — that is what the derivation proves, and the cleanest way to see it is $t = -2m_2c^2T_{\mathrm{recoil}}$ , with a recoil energy that cannot be negative. But the symbol $t$ can certainly take positive values, because the same three invariants describe three different processes. In the reaction $1 + \bar 3 \to \bar 2 + 4$ — the " $t$ -channel process" — the quantity called $t$ plays the role that $s$ plays here, and it is positive, and it can sit right on a resonance. That is the observation crossing symmetry is built from. The three processes occupy three disjoint wedges of the $(s,t,u)$ plane, and no event is ever in two of them.

If the exchanged particle in a t-channel diagram can never be real, in what sense is it there?

In no sense that an experiment could ever confirm, and that is the honest answer. Its four-momentum $q = P_1 - P_3$ has $q\cdot q = t \leq 0$ — spacelike — while any real particle of mass $\mu$ has $q\cdot q = \mu^2c^4 \geq 0$ . The two cannot be reconciled, so the internal line is not a particle: it is a factor in an integral, a propagator, a piece of bookkeeping in a perturbation series. Language like "the electrons exchanged a photon" is a picture drawn over an equation, and the equation forbids taking the picture literally. The kinematics is what tells you this, and it tells you before any field theory is written down — which is why it belongs in a special-relativity article and not a QFT one.

Does the sum rule hold for inelastic collisions, where the outgoing particles are different?

Yes, and nothing in the derivation assumed otherwise. The four masses are simply those of whatever went in and whatever came out; they need not be equal, pairwise equal, or non-zero. The only inputs are four-momentum conservation and the four mass shells. So $\pi^- p \to K^0 \Lambda$ obeys $s+t+u = (m_\pi^2 + m_p^2 + m_K^2 + m_\Lambda^2)c^4$ exactly as $pp \to pp$ obeys $s+t+u = 4m_p^2c^4$ . What does require elasticity is the tidy identity $t = -2m_2c^2T_{\mathrm{recoil}}$ , which needs particle 4 to be the same species as particle 2.

The invariants have three names but only two are free. Isn't u just redundant clutter?

It would be, if computing were all you ever did. But $u$ earns its keep twice. When particles 3 and 4 are identical — electron–electron scattering — there is no fact of the matter about which outgoing electron is "the deflected one", so $t$ and $u$ describe the two indistinguishable possibilities and the amplitude must be built symmetrically from both; drop $u$ and you get the wrong answer. And in the crossing picture $u$ labels a genuine third physical process with its own wedge in the plane. The redundancy is the price of a symmetric description.

Mandelstam Invariants

Where this lives today

The physical picture

The core equations

Deriving the invariants

Limiting cases

Worked examples

1 · Elastic proton–proton scattering at the LHC: from 13 TeV to 49 microradians

2 · The antiproton threshold, in the language of $s$

Physics of the portrait

Going further

Before this

Where this leads

Mandelstam Invariants

Where this lives today

The physical picture

The core equations

Deriving the invariants

Limiting cases

Worked examples

1 · Elastic proton–proton scattering at the LHC: from 13 TeV to 49 microradians

2 · The antiproton threshold, in the language of sss

Physics of the portrait

Going further

Before this

Where this leads

2 · The antiproton threshold, in the language of $s$