--- title: Mandelstam Variables and Lorentz Invariants module: Relativistic Dynamics moduleNumber: 3 lessonNumber: 4 order: 304 summary: > For a two-to-two process the three Mandelstam invariants s, t, and u encode all the kinematics in frame-independent form. They obey a single linear constraint, the sum of the four squared masses, so only two are independent. s is the center-of-momentum energy squared, t and u are momentum transfers tied to the scattering angle, and crossing symmetry relates one amplitude across three channels through these variables. topics: [Relativistic Dynamics] draft: false sources: - book: PDG ref: "Review of Particle Physics — Kinematics review, §47.4.3 Invariant variables (Mandelstam s, t, u)" - book: Carroll ref: "Lecture Notes on General Relativity, §1 — Special Relativity and Flat Spacetime (invariants), arXiv:gr-qc/9712019" --- The [collision kinematics of the previous lesson](/relativity/relativistic-dynamics/collisions-thresholds-and-the-cm-frame) reduced everything to Lorentz invariants — dot products of four-momenta that every frame computes to the same number. For a two-to-two process $1 + 2 \to 3 + 4$ there are exactly three independent such invariants of dimension energy squared, and Mandelstam's choice — $s$, $t$, $u$ — makes their symmetry manifest. They are not independent: a single linear relation ties them to the four masses, so the scattering of two particles into two is a two-parameter problem no matter how many frames one writes it in. This lesson defines the three variables, proves their constraint, reads off their physical meaning, and states the crossing symmetry that lets one amplitude serve three related reactions. Signature $(-,+,+,+)$; each four-momentum satisfies $p_i\cdot p_i = -m_i^2 c^2$, and for a four-vector $A$ the shorthand $A^2 \equiv A\cdot A$ is the Minkowski square. ## The three invariants Label the incoming four-momenta $p_1, p_2$ and the outgoing $p_3, p_4$, with conservation $p_1 + p_2 = p_3 + p_4$. The Mandelstam variables are the squared sums and differences that group the particles into the three ways two-and-two can be paired, written in energy-squared units: $$ s = -\,(p_1 + p_2)^2\,c^2 = -\,(p_3 + p_4)^2\,c^2, $$ $$ t = -\,(p_1 - p_3)^2\,c^2 = -\,(p_2 - p_4)^2\,c^2, $$ $$ u = -\,(p_1 - p_4)^2\,c^2 = -\,(p_2 - p_3)^2\,c^2 . $$ The second equality in each line uses conservation and makes the pairing plain: $s$ groups the two incoming particles, $t$ pairs particle $1$ with outgoing $3$, and $u$ pairs $1$ with outgoing $4$. Each is manifestly Lorentz-invariant, being the Minkowski square of a four-vector. > **Definition (Mandelstam variables).** For $1 + 2 \to 3 + 4$, define $s = > -(p_1{+}p_2)^2 c^2$, $t = -(p_1{-}p_3)^2 c^2$, $u = -(p_1{-}p_4)^2 c^2$. Then > $\sqrt{s}$ is the total center-of-momentum energy, and $t$ and $u$ are the > squared four-momentum transfers between the incoming and the two outgoing > particles. ## The constraint The three variables are not independent. Summing them, $$ s + t + u = -c^2\Bigl[(p_1{+}p_2)^2 + (p_1{-}p_3)^2 + (p_1{-}p_4)^2\Bigr]. $$ Expanding each square and collecting the $p_1$ terms, $$ s + t + u = -c^2\Bigl[3p_1^2 + p_2^2 + p_3^2 + p_4^2 + 2p_1\!\cdot\!(p_2 - p_3 - p_4)\Bigr]. $$ Conservation gives $p_2 - p_3 - p_4 = -p_1$, so the cross term is $2p_1\cdot(-p_1) = -2p_1^2$, leaving $3p_1^2 - 2p_1^2 = p_1^2$. With every $p_i^2 = -m_i^2c^2$, $$ s + t + u = -c^2\bigl(p_1^2 + p_2^2 + p_3^2 + p_4^2\bigr) = \bigl(m_1^2 + m_2^2 + m_3^2 + m_4^2\bigr)c^4 . $$ > **Theorem (Mandelstam constraint).** For any two-to-two process, $s + t + u = > \sum_{i=1}^4 m_i^2 c^4$. Only two of the three invariants are independent; the > full kinematics of $2\to2$ scattering lives on a plane. $$ % caption: The constraint that s, t, and u sum to a fixed value is the geometry % of an equilateral triangle: the perpendicular distances from any interior point % to the three sides always sum to the triangle's height, so each point is one % allowed set of invariants. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % equilateral triangle, side 4.6 \coordinate (A) at (0,0); \coordinate (B) at (4.6,0); \coordinate (C) at (2.3,3.98); \draw[very thick] (A) -- (B) -- (C) -- cycle; \node[black!70, below] at (2.3,-0.1) {constant $s$ side}; \node[black!70, left] at (1.0,2.1) {constant $t$}; \node[black!70, right] at (3.6,2.1) {constant $u$}; % interior point \coordinate (P) at (2.05,1.35); \fill[acc] (P) circle (1.6pt); \node[acc, right] at (2.1,1.35) {one event}; % three perpendiculars \draw[black, dashed] (P) -- (2.05,0); \draw[black, dashed] (P) -- (0.83,1.97); \draw[black, dashed] (P) -- (3.24,1.72); \node[black, below] at (2.6,3.9) {heights sum to a constant}; \end{tikzpicture} $$ ## Physical meaning of s, t, and u Evaluate the invariants in the CM frame, where the incoming particles have equal and opposite momenta $\vec p_1^{\,\ast} = -\vec p_2^{\,\ast}$ with magnitude $p_i^\ast$, and the outgoing pair leaves at CM scattering angle $\theta^\ast$. - **$s$ is the total energy squared.** In the CM frame the spatial momenta cancel, so $s = (E_1^\ast + E_2^\ast)^2 = E_{\text{CM}}^2$. It is the same quantity that set the production threshold in the last lesson, and the physical region for the reaction requires $s \ge (m_1 + m_2)^2 c^4$ and $s \ge (m_3 + m_4)^2 c^4$. - **$t$ is the momentum transfer.** Expanding $t = -c^2(p_1 - p_3)^2$ in the CM frame for elastic scattering of equal masses ($E_1^\ast = E_3^\ast$, $p_1^\ast = p_3^\ast \equiv p^\ast$), $$ t = -\,2\,(p^\ast c)^2\,(1 - \cos\theta^\ast), $$ which runs from $t = 0$ at forward scattering ($\theta^\ast = 0$) to $t = -4(p^\ast c)^2$ at backscattering. A small $|t|$ is a glancing collision with little deflection; a large $|t|$ is a hard, wide-angle scatter. Because $t \le 0$ throughout the physical scattering region, it is often called spacelike momentum transfer. - **$u$ is the crossed transfer**, pairing incoming $1$ with the other outgoing particle $4$. For identical outgoing particles $t$ and $u$ exchange under $\theta^\ast \to \pi - \theta^\ast$, so $u$ carries the backward-angle information that $t$ carries in the forward direction. The angle dependence sits entirely in $t$ (and $u$): fixing $s$ fixes the total energy, and varying $t$ sweeps the scattering angle. Differential cross sections are written $\d\sigma/\d t$ precisely because $t$ is the invariant that tracks the deflection. $$ % caption: The scattering angle in the center-of-momentum frame maps directly to % the momentum transfer t: forward scattering gives t near zero, backward % scattering gives the most negative t, with s fixing the overall energy. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % incoming axis \draw[->, black] (-2.8,0) -- (2.9,0) node[right] {beam axis}; \draw[->, very thick] (-2.4,0) -- (-0.15,0); \node[above] at (-1.6,0.05) {incoming $1$}; \fill[black!80] (0,0) circle (1.6pt); % outgoing at angle \draw[->, acc, very thick] (0,0) -- (2.0,1.35) node[right] {outgoing $3$}; \draw[->, black, very thick] (0,0) -- (-2.0,-1.35) node[left] {outgoing $4$}; % angle arc \draw[black] (1.0,0) arc (0:34:1.0); \node[black!70, right] at (1.05,0.3) {CM angle}; \node[black!70, below] at (1.7,-0.6) {small angle: $t\to 0$}; \node[black!70, below] at (1.7,-1.05) {large angle: $t$ most negative}; \end{tikzpicture} $$ ## CM energies, momentum, and the massless limit Fixing $s$ fixes the CM energies and momentum of the incoming particles exactly as a parent mass fixes its decay products, because a collision is the time-reverse of a formation. In the CM frame, $$ E_1^\ast = \frac{s + m_1^2 c^4 - m_2^2 c^4}{2\sqrt{s}}, \qquad E_2^\ast = \frac{s + m_2^2 c^4 - m_1^2 c^4}{2\sqrt{s}}, $$ and the common CM momentum is set by the Källén function of $s$ and the two masses, $$ p^\ast c = \frac{\sqrt{\lambda\!\left(s,\, m_1^2 c^4,\, m_2^2 c^4\right)}}{2\sqrt{s}}, \qquad \lambda(a,b,c) = a^2 + b^2 + c^2 - 2ab - 2bc - 2ca . $$ The same $\lambda$ governs the outgoing pair with $m_3, m_4$, and its vanishing is exactly the threshold $\sqrt{s} = m_3 c^2 + m_4 c^2$ where the outgoing momentum goes to zero. The boundary of the physical region — the curve drawn above in the $s$-$t$ plane — is the locus where these Källén functions and the Gram determinant of the momenta vanish. When every energy is large compared to the masses, the rest terms drop out. The constraint collapses to $$ s + t + u \to 0 \qquad (\text{all } m_i c^2 \ll \sqrt{s}\,), $$ and the momentum transfers take the clean forms $$ t = -\frac{s}{2}\,(1 - \cos\theta^\ast), \qquad u = -\frac{s}{2}\,(1 + \cos\theta^\ast), $$ with $\theta^\ast$ the CM scattering angle. Forward scattering ($\theta^\ast = 0$) sends $t \to 0$ and $u \to -s$; backscattering swaps them. In this limit a differential cross section written as $\d\sigma/\d t$ depends only on the ratio $t/s$, so the angular distribution is scale-free — the reason high-energy scattering data are plotted against the dimensionless $t/s$. ## Physical regions and crossing symmetry A scattering amplitude for $2\to2$ is a function $\mathcal{M}(s,t)$ (with $u$ fixed by the constraint). The three channels correspond to the three ways the same four external lines can be split into "in" and "out" pairs: - **$s$-channel:** $1 + 2 \to 3 + 4$, physical for $s \ge (m_1{+}m_2)^2 c^4$, with $t, u \le 0$. - **$t$-channel:** $1 + \bar 3 \to \bar 2 + 4$, physical when $t$ plays the role of the CM energy squared, $t \ge (m_1{+}m_3)^2 c^4$. - **$u$-channel:** $1 + \bar 4 \to \bar 2 + 3$, physical when $u$ is the CM energy squared. Here $\bar n$ denotes the antiparticle, moved from the final to the initial state. The three physical regions occupy disjoint parts of the $s$-$t$ plane, separated by the mass thresholds; between them lies unphysical territory where no real process runs but the amplitude is still defined by analytic continuation. $$ % caption: The s-channel physical region in the s-t plane: energy at or above the % production threshold and momentum transfer between the forward and backward % limits, bounded by a curve set by the masses; outside it no real process runs. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,-2.7) -- (0,1.0) node[above] {transfer $t$}; \draw[->, black] (0,0) -- (6.2,0) node[right] {energy $s$}; % threshold line \draw[black, densely dotted] (1.6,-2.6) -- (1.6,0.7); \node[black!70, above] at (1.6,0.72) {threshold}; % t = 0 upper boundary \draw[black, thick] (1.6,0) -- (5.8,0); \node[black!70, above] at (3.4,0.06) {forward $t=0$}; % shade region \fill[acc!12] (1.6,0) -- (5.8,0) -- (5.8,-2.31) -- cycle; % lower boundary (backward limit) \draw[acc, very thick] (1.6,0) -- (5.8,-2.31); \node[acc, above, rotate=-29] at (4.6,-1.65) {backward limit}; \node[black!75] at (3.3,-0.6) {physical region}; \end{tikzpicture} $$ > **Definition (Crossing symmetry).** One analytic function $\mathcal{M}(s,t,u)$ > describes all three channels: the amplitude for the $s$-channel reaction, > continued to the region where $t$ (or $u$) exceeds its threshold, gives the > amplitude for the $t$-channel (or $u$-channel) reaction with the appropriate > particles crossed to antiparticles. Moving a particle across the reaction arrow > negates its four-momentum and swaps its role in $s$, $t$, $u$. Crossing is why the same calculation yields, for example, both electron-electron and electron-positron scattering: the two are the $t$- and $s$-channel readings of one amplitude. The Mandelstam variables are the coordinates in which this identity is visible, because they are frame-independent and treat the four particles symmetrically. $$ % caption: Crossing symmetry: the same amplitude serves three reactions, read as % the s-, t-, or u-channel depending on which pair of external lines is taken as % incoming, with crossed lines becoming antiparticles. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % s-channel \draw[very thick] (-0.6,0.6) -- (0,0) -- (-0.6,-0.6); \draw[very thick] (0,0) -- (0.9,0); \draw[very thick] (0.9,0) -- (1.5,0.6); \draw[very thick] (0.9,0) -- (1.5,-0.6); \node[black!70, above] at (0.45,0.75) {s-channel}; \node[black!70, below] at (0.45,-0.75) {$1,2 \to 3,4$}; % t-channel \begin{scope}[xshift=4.0cm] \draw[very thick] (-0.6,0.6) -- (0,0.3) -- (0.9,0.3) -- (1.5,0.6); \draw[very thick] (0,0.3) -- (0,-0.3); \draw[very thick] (0.9,0.3) -- (0.9,-0.3); \draw[very thick] (-0.6,-0.6) -- (0,-0.3); \draw[very thick] (0.9,-0.3) -- (1.5,-0.6); \node[black!70, above] at (0.45,0.75) {t-channel}; \node[black!70, below] at (0.45,-0.75) {crossed lines}; \end{scope} \node[black!80] at (2.55,0) {vs}; \end{tikzpicture} $$ The invariants collapse frame-dependent energies and angles into a two-parameter description that every observer shares. Later modules put the same idea to work on fields rather than particles: the electromagnetic field carries its own Lorentz-invariant combinations, and the [covariant formulation of electromagnetism](/relativity/covariant-electrodynamics/the-electromagnetic-field-tensor) classifies a field by invariants exactly as $s$, $t$, $u$ classify a collision.[^pdg][^carroll] [^pdg]: Particle Data Group, _Review of Particle Physics_, Kinematics review, invariant variables $s$, $t$, $u$ and the $s+t+u$ relation. [^carroll]: Carroll, _Lecture Notes on General Relativity_, §1 (Lorentz invariants of flat spacetime).