Exercise: The Relativistic Kinematics of a Two-Body Decay

Introduction

This is one of the exercises in the relativity series. It is a set of worked problems in the relativistic kinematics of a two-body decay, using the framework and the notation of the companion article The Relativistic Two-Body Problem in Biquaternionic Form. That article is the parent of this exercise: it sets up the kinematics, and what follows applies it. Nothing new is introduced, and every result below is obtained from the tools already defined there.

What is assumed. The biquaternion algebra $\mathbb{B} = \mathbb{C}\otimes_\mathbb{R}\mathbb{H}$, with quaternion basis $e_0 = 1, e_1, e_2, e_3$ satisfying $e_k^2 = -e_0$, and the scalar imaginary $i$ with $i^2 = -1$. The anti-Hermitian subspace $\mathbb{M}_-$ (imaginary scalar part, real vector part) and the Hermitian subspace $\mathbb{M}_+$. The biquaternion norm $N(\tilde{Q}) = \tilde{Q}\tilde{Q}^{\natural}$ and the invariant pairing $\langle \tilde{A}, \tilde{B}\rangle = \mathrm{Sc}(\tilde{A}\tilde{B}^{\natural})$ on $\mathbb{M}_-$. The four-momentum $\tilde{P} = m\tilde{U} = iE/c\,e_0 + \mathbf{p}$ with $N(\tilde{P}) = -m^2c^2$, the unit four-velocity $\tilde{u} = \tilde{U}/c$ with $N(\tilde{u}) = -1$, the rotor conjugation $\tilde{Q} \mapsto \tilde{\Lambda}\tilde{Q}\tilde{\Lambda}^{*}$ with $\tilde{\Lambda}\tilde{\Lambda}^{\natural} = e_0$, the boost biquaternion $\tilde{\Lambda} = \cosh\frac{\psi}{2} + i\sinh\frac{\psi}{2}\hat{\mathbf{u}}$ with $\tanh\psi = u/c$, and the relative rapidity $\psi_{\rm rel}$ with $\cosh\psi_{\rm rel} = -\langle \tilde{u}_1, \tilde{u}_2\rangle$. Throughout, $c$ is the speed of light in the medium, $c = 1/\sqrt{\epsilon\mu}$, and $c_0$ is the vacuum value.

What is to be shown. The problems are: (1) the daughter energies and momentum in the centre-of-momentum frame of a parent at rest; (2) the general case of a moving parent, through the boost biquaternion the parent constructs; (3) the invariant mass of each pair in a three-body decay; (4) the threshold and the opening-angle conditions; (5) a numerical instance, checked to several digits; (6) the non-relativistic limit. Each problem is stated and then solved in full; the value of an exercise article is in the solutions.

Notation for the decay. A parent of mass $M$ decays into two daughters of masses $m_1, m_2$, $$ A \;\longrightarrow\; 1 + 2, \qquad \tilde{P}_A = \tilde{P}_1 + \tilde{P}_2 . $$ In the rest frame of the parent the total four-momentum is $\tilde{P}_A = iMc\,e_0$, so the parent rest frame is the centre-of-momentum (COM) frame of the daughters. Starred quantities refer to that frame: $\mathbf{p}_1^* = -\mathbf{p}_2^* = \mathbf{p}^*$, $p^* = |\mathbf{p}^*|$, and $E_1^* + E_2^* = Mc^2$. Unstarred quantities refer to the laboratory frame. Because the two-body article already derived the kinematics, several intermediate steps below reproduce its equations; they are included so that the exercise is self-contained.

This exercise also serves to test the parent. The direction of the boost biquaternion that carries the COM frame to the laboratory — a point that repays care — is worked out explicitly in Problem 2.

Problem 1: Daughter Energies and Momentum in the COM Frame

Statement. For $A \to 1+2$ with the parent at rest, derive the daughter energies $E_1^*, E_2^*$, the common momentum magnitude $p^*$, the speeds $v_a^*$, and the released energy $Q$. Show that the decay is allowed exactly when $M \ge m_1 + m_2$.

Solution. The conservation law is the single biquaternion equation $\tilde{P}_A = \tilde{P}_1 + \tilde{P}_2$. In the parent rest frame $\tilde{P}_A = iMc\,e_0$, and writing $\tilde{P}_a^* = iE_a^*/c\,e_0 + \mathbf{p}_a^*$, the conservation law splits, under the real-linear projections onto the scalar and vector parts of $\mathbb{M}_-$, into $$ E_1^* + E_2^* = Mc^2, \qquad \mathbf{p}_1^* + \mathbf{p}_2^* = 0 . $$ The second equation says the daughters are emitted back to back: $\mathbf{p}_1^* = -\mathbf{p}_2^* = \mathbf{p}^*$, so $p_1^{*2} = p_2^{*2} = p^{*2}$. The mass-shell relations are $$ E_1^{*2} = m_1^2c^4 + p^{*2}c^2, \qquad E_2^{*2} = m_2^2c^4 + p^{*2}c^2 . $$ Subtracting eliminates $p^{*2}$, $$ E_1^{*2} - E_2^{*2} = (m_1^2 - m_2^2)c^4 \quad\Longrightarrow\quad (E_1^* - E_2^*)(E_1^* + E_2^*) = (m_1^2 - m_2^2)c^4, $$ and with $E_1^* + E_2^* = Mc^2$ this gives $E_1^* - E_2^* = (m_1^2 - m_2^2)c^2/M$. Combining the sum and the difference, $$ E_1^* = \frac{\left(M^2 + m_1^2 - m_2^2\right)c^2}{2M}, \qquad E_2^* = \frac{\left(M^2 + m_2^2 - m_1^2\right)c^2}{2M}. $$ Substituting $E_1^*$ into the mass shell and factoring, $$ p^{*2}c^2 = E_1^{*2} - m_1^2c^4 = \frac{c^4}{4M^2}\left[\left(M^2 + m_1^2 - m_2^2\right)^2 - 4M^2m_1^2\right]. $$ The bracket factors as a difference of squares, first as $\left[(M-m_1)^2 - m_2^2\right]\left[(M+m_1)^2 - m_2^2\right]$ and then as the product of two differences of squares, so that $$ p^* = \frac{c}{2M}\sqrt{\left[M^2 - (m_1+m_2)^2\right]\left[M^2 - (m_1-m_2)^2\right]} . $$ The speeds follow from $\mathbf{p}_a^* = \gamma_a^* m_a \mathbf{v}_a^*$ and $E_a^* = \gamma_a^* m_a c^2$, giving $$ v_a^* = \frac{p^*c^2}{E_a^*}, \qquad \tanh\psi_a^* = \frac{p^*c}{E_a^*} . $$

Threshold. A physical parent always has $M > |m_1 - m_2|$, so the second factor under the square root is positive. The radicand is therefore non-negative exactly when $M \ge m_1 + m_2$. At $M = m_1 + m_2$ one has $p^* = 0$ and $E_a^* = m_ac^2$: the daughters emerge at rest, and the decay has no kinetic energy to distribute. Hence the decay is kinematically allowed if and only if $M \ge m_1 + m_2$, i.e. if and only if the released energy $$ Q := (M - m_1 - m_2)c^2 $$ is non-negative.

Released energy. The kinetic energies $K_a := E_a^* - m_ac^2$ satisfy $$ K_1 + K_2 = (E_1^* + E_2^*) - (m_1 + m_2)c^2 = Q , $$ so $Q$ is exactly the total kinetic energy of the daughters in the parent rest frame. An exact rearrangement used in Problem 6 is $$ K_1 = \frac{Q\left(2m_2 + Q/c^2\right)}{2M}, \qquad K_2 = \frac{Q\left(2m_1 + Q/c^2\right)}{2M}, $$ obtained from $K_1 = E_1^* - m_1c^2 = \left[(M-m_1)^2 - m_2^2\right]c^2/(2M)$ by writing $M - m_1 - m_2 = Q/c^2$ and $M - m_1 + m_2 = 2m_2 + Q/c^2$.

Problem 2: The Moving Parent and the Boost Biquaternion

Statement. Let the parent have four-momentum $$ \tilde{P}_A = i\frac{E}{c}\,e_0 + \mathbf{P}, \qquad E = \Gamma Mc^2, \qquad \mathbf{P} = \Gamma M\mathbf{V}, \qquad \Gamma = \frac{1}{\sqrt{1 - V^2/c^2}}, $$ in the laboratory, so that $\mathbf{V}$ is the parent's velocity in the lab. (a) Construct the boost biquaternion $\tilde{\Lambda}_{\rm CM}$ that carries the lab to the COM frame, and verify $\tilde{\Lambda}_{\rm CM}\tilde{P}_A\tilde{\Lambda}_{\rm CM}^{*} = iMc\,e_0$. (b) Identify the rotor that carries the COM daughter four-momenta to the lab. (c) Obtain the lab energies and momenta and the range of a daughter's lab energy.

Solution (a). From $\tilde{P}^{\natural}_A = iE/c\,e_0 - \mathbf{P}$, $$ -\frac{i}{Mc}\tilde{P}^{\natural}_A = \frac{E}{Mc^2}e_0 + i\frac{\mathbf{P}}{Mc} = \cosh\Psi\,e_0 + i\sinh\Psi\,\hat{\mathbf{V}}, $$ with $\cosh\Psi = E/(Mc^2) = \Gamma$, $\sinh\Psi = |\mathbf{P}|/(Mc) = \Gamma V/c$, hence $\tanh\Psi = V/c$. This is an element of $\mathbb{M}_+$ of unit norm, since $\cosh^2\Psi - \sinh^2\Psi = 1$. Its principal square root, with $\mathrm{Sc} > 0$, is the boost biquaternion $$ \tilde{\Lambda}_{\rm CM} = \sqrt{-\frac{i}{Mc}\tilde{P}^{\natural}_A} = \cosh\frac{\Psi}{2} + i\sinh\frac{\Psi}{2}\,\hat{\mathbf{V}}, \qquad \tilde{\Lambda}_{\rm CM}\tilde{\Lambda}^{\natural}_{\rm CM} = e_0 . $$ To verify that it rotates $\tilde{P}_A$ to $iMc\,e_0$, use the component action of the rotor with parameter $\hat{\mathbf{V}}$, which is the boost to the frame moving with velocity $\mathbf{V}$. On a four-momentum $(E', \mathbf{P}')$ it acts as $$ E'' = \Gamma\left(E' - \mathbf{V}\cdot\mathbf{P}'\right), \qquad \mathbf{P}'' = \mathbf{P}' + \frac{\Gamma-1}{V^2}\left(\mathbf{V}\cdot\mathbf{P}'\right)\mathbf{V} - \Gamma\frac{E'}{c^2}\mathbf{V}. $$ Inserting $E' = E = \Gamma Mc^2$ and $\mathbf{P}' = \mathbf{P} = \Gamma M\mathbf{V}$ gives $$ E'' = \Gamma\left(\Gamma Mc^2 - \Gamma MV^2\right) = \Gamma^2 Mc^2\left(1 - V^2/c^2\right) = Mc^2, $$ and, using $\mathbf{V}\cdot\mathbf{P} = \Gamma MV^2$, $$ \mathbf{P}'' = \Gamma M\mathbf{V} + (\Gamma-1)\Gamma M\mathbf{V} - \Gamma^2 M\mathbf{V} = 0 . $$ Hence $\tilde{\Lambda}_{\rm CM}\tilde{P}_A\tilde{\Lambda}_{\rm CM}^{*} = iMc\,e_0$, which is precisely the statement that $\tilde{\Lambda}_{\rm CM}$ is the boost to the COM frame.

Solution (b). The rotation generated by $\tilde{\Lambda}_{\rm CM}$ carries the lab to the COM frame. The inverse rotation is generated by the quaternion conjugate, $$ \tilde{\Lambda}^{\natural}_{\rm CM} = \cosh\frac{\Psi}{2} - i\sinh\frac{\Psi}{2}\,\hat{\mathbf{V}} = \cosh\frac{\Psi}{2} + i\sinh\frac{\Psi}{2}\,(-\hat{\mathbf{V}}), $$ which is the boost biquaternion of the same rapidity in the opposite direction. The COM daughter four-momenta are therefore carried to the laboratory by $$ \tilde{P}_a = \tilde{\Lambda}^{\natural}_{\rm CM}\,\tilde{P}_a^*\,\tilde{\Lambda}^{\natural}_{\rm CM}^\dagger . $$

Remark on the parent's boost convention. The parent writes the star-to-lab rotation as $\tilde{P}_a = \tilde{\Lambda}\tilde{P}_a^*\tilde{\Lambda}^{*}$ with $\tilde{\Lambda} = \cosh\frac{\Psi}{2} - i\sinh\frac{\Psi}{2}\hat{\mathbf{V}}$, the quaternion conjugate of the rotor that carries the laboratory frame to the parent rest frame - the sign of the vector part being opposite to that of the lab-to-rest rotor, as the parent states. That rotor produces $E_a = \gamma(E_a^* + \mathbf{V}\cdot\mathbf{p}_a^*)$, in agreement with the component formula quoted immediately below it in the parent; the biquaternion and component forms therefore agree. We work with the physical convention, in which a forward-emitted daughter ($\mathbf{V}\cdot\mathbf{p}_a^* > 0$) gains energy, and which corresponds to the inverse rotor above.

Solution (c). Applying the component form of the rotation with the inverse rotor — equivalently, the standard Lorentz transformation with velocity $\mathbf{V}$ — gives $$ E_a = \gamma\left(E_a^* + \mathbf{V}\cdot\mathbf{p}_a^*\right), \qquad \mathbf{p}_a = \mathbf{p}_a^* + \frac{\gamma-1}{V^2}\left(\mathbf{V}\cdot\mathbf{p}_a^*\right)\mathbf{V} + \gamma\frac{E_a^*}{c^2}\mathbf{V}, $$ with $\gamma = \cosh\Psi = \Gamma$. Writing $\cos\theta_a^*$ for the angle between $\mathbf{V}$ and $\mathbf{p}_a^*$, and using $v_a^* = p^*c^2/E_a^*$, $$ E_a = \gamma E_a^*\left(1 + \frac{Vv_a^*}{c^2}\cos\theta_a^*\right). $$ Hence for a fixed daughter the lab energy lies in the range $$ \gamma E_a^*\left(1 - \frac{Vv_a^*}{c^2}\right) \;\le\; E_a \;\le\; \gamma E_a^*\left(1 + \frac{Vv_a^*}{c^2}\right), $$ the upper end being forward emission and the lower end backward emission. The two lab four-momenta still sum to $\tilde{P}_A$, because the rotation is linear: $$ \tilde{\Lambda}^{\natural}_{\rm CM}\left(\tilde{P}_1^* + \tilde{P}_2^*\right)\tilde{\Lambda}^{\natural}_{\rm CM}^\dagger = \tilde{\Lambda}^{\natural}_{\rm CM}\left(iMc\,e_0\right)\tilde{\Lambda}^{\natural}_{\rm CM}^\dagger = \tilde{P}_A . $$

Problem 3: The Invariant Mass of Each Pair in a Three-Body Decay

Statement. Let $A \to 1 + 2 + 3$, with the parent at rest. For each pair define $M_{ij}$ by $$ N\!\left(\tilde{P}_i + \tilde{P}_j\right) = -M_{ij}^2c^2, \qquad M_{ij} \ge 0 . $$ (a) Express $M_{ij}$ through the invariant pairing. (b) Show that $M_{12}^2c^4 = (Mc^2 - E_3^*)^2 - p_3^{*2}c^2$. (c) Find the range of $M_{12}$. (d) Establish the sum rule $\sum_{i

Solution (a). Because $N$ is a quadratic form with polarization $\langle \cdot, \cdot\rangle$, $$ N\!\left(\tilde{P}_i + \tilde{P}_j\right) = N(\tilde{P}_i) + N(\tilde{P}_j) + 2\langle \tilde{P}_i, \tilde{P}_j\rangle, $$ and with $N(\tilde{P}_i) = -m_i^2c^2$ this gives $$ M_{ij}^2c^2 = m_i^2c^2 + m_j^2c^2 - 2\langle \tilde{P}_i, \tilde{P}_j\rangle . $$ This is the parent's definition of the invariant mass of a pair, applied to a partial sum of three four-momenta. It requires only that $\mathbb{M}_-$ is a real vector space closed under addition and that $N$ be a quadratic form; no new primitive is needed. (The parent defines the pair mass for a two-body total; the same definition applied to a subset of a three-body system is the natural extension used in Dalitz-plot kinematics.)

Solution (b). In the parent rest frame the three momenta sum to zero, $\mathbf{p}_1^* + \mathbf{p}_2^* + \mathbf{p}_3^* = 0$, so with $\tilde{P}_{12} = \tilde{P}_1 + \tilde{P}_2$, $$ N\!\left(\tilde{P}_{12}\right) = \left(i\frac{E_1^* + E_2^*}{c}\right)^2 + \left(\mathbf{p}_1^* + \mathbf{p}_2^*\right)^2 = -\frac{\left(Mc^2 - E_3^*\right)^2}{c^2} + p_3^{*2}, $$ where we used $\mathbf{p}_1^* + \mathbf{p}_2^* = -\mathbf{p}_3^*$ and $E_1^* + E_2^* = Mc^2 - E_3^*$. Hence $$ M_{12}^2c^4 = \left(Mc^2 - E_3^*\right)^2 - p_3^{*2}c^2 = M^2c^4 + m_3^2c^4 - 2Mc^2E_3^*, $$ the second equality using $E_3^{*2} = m_3^2c^4 + p_3^{*2}c^2$. The pair mass is thus fixed by the energy of the third particle.

Solution (c). Because $M_{12}$ is invariant, it may be evaluated in the parent rest frame, where it is a function of $E_3^*$ alone. The energy $E_3^*$ has a minimum $m_3c^2$, attained when $\mathbf{p}_3^* = 0$, and a maximum $$ E_3^{*\max} = \frac{\left[M^2 + m_3^2 - (m_1+m_2)^2\right]c^2}{2M}, $$ attained when $\mathbf{p}_1^* = -\mathbf{p}_2^*$, that is, when the pair $(1,2)$ is at relative rest and the decay is effectively the two-body decay $A \to (12) + 3$ with the composite mass $m_1 + m_2$. Since $M_{12}^2c^4 = M^2c^4 + m_3^2c^4 - 2Mc^2E_3^*$ decreases with $E_3^*$, the endpoints are $$ M_{12}^{\max} = M - m_3 \quad \left(\text{at } E_3^* = m_3c^2\right), \qquad M_{12}^{\min} = m_1 + m_2 \quad \left(\text{at } E_3^{*\max}\right), $$ so that $(m_1+m_2) \le M_{12} \le M - m_3$, with the cyclic permutations for the other two pairs. These are the three boundary curves of the Dalitz plot.

Solution (d). Summing the squared pair masses and using part (a), $$ \sum_{i

Problem 4: Threshold and the Opening Angle

Statement. (a) State the threshold condition and show that it follows from $p^{*2} \ge 0$. (b) Show that the opening angle between the daughters in the COM frame is $\pi$. (c) Derive the laboratory angle of a daughter. (d) For two massless daughters, derive the minimum opening angle $\Theta_{\min} = \arccos(1 - 2/\gamma^2)$ and its beamed limit.

Solution (a). Problem 1 gives $$ p^{*2}c^2 = \frac{c^4}{4M^2}\left[M^2 - (m_1+m_2)^2\right]\left[M^2 - (m_1-m_2)^2\right]. $$ Since $M > |m_1 - m_2|$ for any physical parent, the second factor is positive, and the condition $p^{*2} \ge 0$ is equivalent to $$ M \ge m_1 + m_2 \iff Q \ge 0 . $$ At threshold $Q = 0$, $p^* = 0$, and the daughters emerge with no relative motion. Below threshold there is no solution of $\tilde{P}_A = \tilde{P}_1 + \tilde{P}_2$ with all four-momenta on the future mass shell and real momenta.

Solution (b). In the COM frame the momenta are back to back, $\mathbf{p}_1^* = -\mathbf{p}_2^*$, so the opening angle $\Theta^*$ between them satisfies $$ \cos\Theta^* = \frac{\mathbf{p}_1^*\cdot\mathbf{p}_2^*}{p^{*2}} = -1, \qquad \Theta^* = \pi . $$ The daughters are collinear and oppositely directed.

Solution (c). Boost to the lab with the parent's velocity $\mathbf{V}$ along $\hat{\mathbf{z}}$, and let $\theta^*$ be the angle between $\mathbf{p}^*$ and $\mathbf{V}$ for daughter 1. From Problem 2 its lab momentum components perpendicular to and along $\mathbf{V}$ are $$ p_{1\perp} = p^*\sin\theta^*, \qquad p_{1\parallel} = \gamma\left(p^*\cos\theta^* + \frac{VE_1^*}{c^2}\right), $$ so the lab angle $\theta_1$ measured from $\mathbf{V}$ satisfies $$ \tan\theta_1 = \frac{p^*\sin\theta^*}{\gamma\left(p^*\cos\theta^* + VE_1^*/c^2\right)} = \frac{\sin\theta^*}{\gamma\left(\cos\theta^* + \beta/\beta_1^*\right)}, $$ where $\beta = V/c$ and $\beta_a^* = p^*c/E_a^* = v_a^*/c$ is the daughter's speed in the COM frame. For a massless daughter $\beta_a^* = 1$, and the angle is the classic result $$ \tan\theta_a = \frac{\sin\theta^*}{\gamma\left(\cos\theta^* + \beta\right)} . $$

Solution (d). For two massless daughters (for example $\pi^0 \to \gamma\gamma$) one has $E_1^* = E_2^* = Mc^2/2$ and $p^* = Mc/2$, so $\beta_a^* = 1$. Emit the two photons at COM angles $\theta^*$ and $\pi - \theta^*$ relative to $\mathbf{V}$; their lab angles $\theta_1, \theta_2$ measured from $\mathbf{V}$ obey the formula of part (c), and the opening angle is $\Theta = \theta_1 + \theta_2$. A direct calculation gives $$ \cos\Theta = \frac{\gamma^2 - 2 - \gamma^2\beta^2\cos^2\theta^*}{\gamma^2\left(1 - \beta^2\cos^2\theta^*\right)} . $$ As $\cos^2\theta^*$ runs from $0$ to $1$ this expression decreases monotonically, its derivative with respect to $\cos^2\theta^*$ being $-\,2\beta^2/\left[\gamma^2\left(1 - \beta^2\cos^2\theta^*\right)^2\right]$, which is negative for $\gamma > 1$. Hence $\cos\Theta$ is largest at $\cos\theta^* = 0$ (symmetric emission) and smallest at $\cos^2\theta^* = 1$ (collinear emission). The minimum opening angle is therefore $$ \cos\Theta_{\min} = 1 - \frac{2}{\gamma^2}, \qquad \Theta_{\min} = \arccos\!\left(1 - \frac{2}{\gamma^2}\right), \qquad \tan\frac{\Theta_{\min}}{2} = \frac{1}{\gamma\beta}, $$ and the maximum is $\Theta = \pi$. For $\gamma \gg 1$, $\Theta_{\min} \approx 2/\gamma$, the familiar collimation of the decay products into a cone of half-angle $1/\gamma$ about the parent's direction. Numerically $\Theta_{\min} = 60.00^\circ$ at $\gamma = 2$ and $23.07^\circ$ at $\gamma = 5$, the latter within $0.7\%$ of $2/\gamma$ radians. For massive daughters the single-daughter formula of part (c) still holds (with $\beta_a^* < 1$), but the opening angle is then a more complicated function of $\theta^*$.

Problem 5: A Numerical Instance

Statement. Evaluate the two-body formulas for $K^+ \to \pi^+ \pi^0$, and for the limits $\pi^0 \to \gamma\gamma$ (two massless daughters) and $\pi^+ \to \mu^+ \nu_\mu$ (one massless daughter). Compare with the non-relativistic approximation.

Solution. We use the standard masses $m_{K^+} = 493.677$, $m_{\pi^+} = 139.57039$, $m_{\pi^0} = 134.9768$, $m_\mu = 105.6583755$ MeV$/c^2$, and set $c = 1$ so that energies are in MeV.

For $K^+ \to \pi^+\pi^0$, with $M = m_{K^+}$, $m_1 = m_{\pi^+}$, $m_2 = m_{\pi^0}$: $$ Q = 219.1298\ \text{MeV}, $$ $$ E_1^* = E_{\pi^+}^* = 248.1158\ \text{MeV}, \qquad E_2^* = E_{\pi^0}^* = 245.5612\ \text{MeV}, \qquad E_1^* + E_2^* = 493.6770\ \text{MeV} = Mc^2, $$ $$ p^*c = 205.1379\ \text{MeV}, \qquad v_{\pi^+}^*/c = 0.826783, \qquad v_{\pi^0}^*/c = 0.835384 . $$ The kinetic energies are $K_{\pi^+} = 108.5454$ MeV and $K_{\pi^0} = 110.5844$ MeV, whose sum is $219.1298$ MeV $= Q$, as required; the mass-shell residual $E_{\pi^+}^{*2} - m_{\pi^+}^2c^4 - p^{*2}c^2$ vanishes to round-off.

The non-relativistic approximation of Problem 6 gives, with $\mu = m_1m_2/(m_1+m_2) = 68.6176$ MeV, $$ p^*_{\rm NR}c = \sqrt{2\mu Q} = 173.4137\ \text{MeV}, $$ which is low by $15.46\%$; and the partition $K_1/Q$ is $0.495348$ exactly, against the approximation $m_2/(m_1+m_2) = 0.491634$, an error of $0.75\%$. The momentum formula is poor here because $Q/(m_1+m_2) \approx 0.80$, far from the non-relativistic regime.

A decay with a smaller release, $\Lambda \to p\pi^-$ ($M = 1115.683$, $m_p = 938.2721$, $m_{\pi^-} = 139.5704$ MeV$/c^2$, so $Q = 37.8405$ MeV, only $3.51\%$ of the mass sum), illustrates the opposite regime: $$ E_p^* = 943.6476\ \text{MeV}, \qquad E_{\pi^-}^* = 172.0354\ \text{MeV}, \qquad p^*c = 100.5797\ \text{MeV}, $$ $$ v_p^*/c = 0.106586, \qquad v_{\pi^-}^*/c = 0.584645, \qquad \frac{K_p}{Q}\Big|_{\rm exact} = 0.142057, \qquad \frac{m_{\pi^-}}{m_p+m_{\pi^-}} = 0.129491 . $$ Here $\sqrt{2\mu Q} = 95.8908$ MeV, low by $4.66\%$, and the partition is in error by $8.9\%$.

The two limiting checks confirm the formula in the regime $\beta_a^* = 1$. For the massless decay $\pi^0 \to \gamma\gamma$ ($M = m_{\pi^0} = 134.9768$ MeV$/c^2$), $$ E_\gamma^* = p^*c = \frac{Mc^2}{2} = 67.4884\ \text{MeV}, \qquad Q = Mc^2 . $$ For $\pi^+ \to \mu^+\nu_\mu$ with a massless neutrino ($M = m_{\pi^+} = 139.57039$, $m_1 = m_\mu = 105.6583755$, $m_2 = 0$ MeV$/c^2$), $$ E_\mu^* = 109.7782\ \text{MeV}, \qquad E_\nu^* = 29.7921\ \text{MeV}, \qquad p^*c = 29.7921\ \text{MeV}, \qquad v_\mu^*/c = 0.271385, $$ with $Q = 33.9120$ MeV and $K_\mu = 4.1199$ MeV. In this decay $\mu = 0$, so the non-relativistic momentum formula $\sqrt{2\mu Q}$ vanishes and fails completely: a massless, necessarily ultrarelativistic daughter lies outside the non-relativistic limit.

Problem 6: The Non-Relativistic Limit

Statement. Expand the exact formulas of Problems 1 and 2 for $Q \ll (m_1+m_2)c^2$, recovering the Newtonian partition of the released energy, the momentum $p^* \approx \sqrt{2\mu Q}$, and Galilean velocity addition for the moving parent. Quantify the accuracy of the limit.

Solution. Write $$ M = m_1 + m_2 + \frac{Q}{c^2}, \qquad \mu = \frac{m_1m_2}{m_1+m_2} . $$

Energies. From $K_1 = Q(2m_2 + Q/c^2)/(2M)$ and $2M = 2(m_1+m_2) + O(Q/c^2)$, $$ K_1 = \frac{m_2}{m_1+m_2}\,Q + O\!\left(\frac{Q^2}{(m_1+m_2)c^2}\right), \qquad K_2 = \frac{m_1}{m_1+m_2}\,Q + O\!\left(\frac{Q^2}{(m_1+m_2)c^2}\right), $$ so the light daughter carries the larger share of the released energy — the parent's non-relativistic decay result.

Momentum. The two factors under the square root are $$ M^2 - (m_1+m_2)^2 = \frac{Q}{c^2}\left(2(m_1+m_2) + \frac{Q}{c^2}\right), \qquad M^2 - (m_1-m_2)^2 = \left(2m_1 + \frac{Q}{c^2}\right)\left(2m_2 + \frac{Q}{c^2}\right). $$ Keeping the leading term in each, $$ p^* \approx \frac{c}{2M}\sqrt{\frac{Q}{c^2}\,2(m_1+m_2)\cdot 4m_1m_2} = \sqrt{\frac{2Qm_1m_2}{m_1+m_2}} = \sqrt{2\mu Q}. $$ Equivalently, since $v_a^* = p^*c^2/E_a^* \approx p^*/m_a$ and the daughters move oppositely, the relative speed is $v_{\rm rel} = v_1^* + v_2^* = p^*(1/m_1 + 1/m_2) = p^*/\mu$, so $$ p^* \approx \mu\,v_{\rm rel}, $$ in agreement with the parent's non-relativistic COM momentum. The kinetic energies follow as $K_a \approx p^{*2}/(2m_a)$; using $p^{*2} = 2\mu Q$ reproduces the partition above, and $Q \approx \tfrac12 \mu v_{\rm rel}^2$ makes contact with the parent's relative-rapidity relation $Mc^2 = (m_1+m_2)c^2 + \tfrac12\mu v_{\rm rel}^2 + O(v_{\rm rel}^4/c^2)$.

Moving parent. Expanding the lab formulas of Problem 2 for $V, v_a^* \ll c$, $$ E_a = \gamma\left(E_a^* + \mathbf{V}\cdot\mathbf{p}_a^*\right) \approx m_ac^2 + \tfrac12 m_a\left|\mathbf{V} + \mathbf{v}_a^*\right|^2, \qquad \mathbf{p}_a \approx m_a\left(\mathbf{V} + \mathbf{v}_a^*\right), $$ so the daughter velocities add Galileanly, and the total lab momentum is $$ \sum_a \mathbf{p}_a \approx (m_1+m_2)\mathbf{V} + m_1\mathbf{v}_1^* + m_2\mathbf{v}_2^* = (m_1+m_2)\mathbf{V}, $$ because $\mathbf{p}_1^* = -\mathbf{p}_2^*$ implies $m_1\mathbf{v}_1^* + m_2\mathbf{v}_2^* = 0$ at leading order. This is the Newtonian statement that the parent moves with velocity $\mathbf{V}$ while the daughters share the released energy in their relative motion.

Accuracy. The expansions carry relative errors of order $Q/((m_1+m_2)c^2)$ and $v_a^{*2}/c^2$. Problem 5 makes this concrete: for $\Lambda \to p\pi^-$ ($Q/(m_1+m_2) \approx 3.5\%$) the momentum approximation is good to $4.7\%$ and the energy partition to $8.9\%$; for $K^+ \to \pi^+\pi^0$ ($Q/(m_1+m_2) \approx 0.80$) the momentum approximation already fails at the $15\%$ level. The limit requires both daughters to be non-relativistic, so it fails outright when one daughter is massless.

Summary

We have worked the relativistic kinematics of a two-body decay as an application of the two-body article.

1. COM frame ($A$ at rest). The daughters are back to back with $E_1^* + E_2^* = Mc^2$, individual energies $E_a^* = (M^2 + m_a^2 - m_b^2)c^2/(2M)$, common momentum $p^* = \frac{c}{2M}\sqrt{[M^2-(m_1+m_2)^2][M^2-(m_1-m_2)^2]}$, speeds $v_a^* = p^*c^2/E_a^*$, and released energy $Q = (M-m_1-m_2)c^2 = K_1 + K_2$. The decay is allowed iff $M \ge m_1 + m_2$.

2. Moving parent. The parent constructs $\tilde{\Lambda}_{\rm CM} = \sqrt{-i\tilde{P}^{\natural}_A/(Mc)} = \cosh\frac{\Psi}{2} + i\sinh\frac{\Psi}{2}\hat{\mathbf{V}}$, which carries the lab to the COM frame; the inverse rotation, which carries the COM daughters to the lab, is generated by the quaternion conjugate $\tilde{\Lambda}^{\natural}_{\rm CM}$. It gives $E_a = \gamma(E_a^* + \mathbf{V}\cdot\mathbf{p}_a^*)$ and the standard momentum formula, with the lab energy of each daughter confined to $\gamma E_a^*(1 \pm Vv_a^*/c^2)$.

3. Three-body decay. The pair mass is $M_{ij}^2c^2 = m_i^2c^2 + m_j^2c^2 - 2\langle\tilde{P}_i,\tilde{P}_j\rangle$; equivalently $M_{12}^2c^4 = (Mc^2 - E_3^*)^2 - p_3^{*2}c^2$, with $(m_1+m_2) \le M_{12} \le M - m_3$ and $\sum_{i

4. Threshold and opening angle. The threshold is $M \ge m_1+m_2$; in the COM frame the opening angle is $\pi$, and for two massless daughters the lab opening angle has minimum $\Theta_{\min} = \arccos(1 - 2/\gamma^2) \approx 2/\gamma$ at symmetric emission.

5. Numbers. For $K^+ \to \pi^+\pi^0$ the exact values are $p^*c = 205.1379$ MeV and $K_{\pi^+} = 108.5454$ MeV, $K_{\pi^0} = 110.5844$ MeV; for $\Lambda \to p\pi^-$, $p^*c = 100.5797$ MeV. The massless limits $\pi^0 \to \gamma\gamma$ and $\pi^+ \to \mu^+\nu_\mu$ reproduce $E_\gamma^* = Mc^2/2$ and the standard pion-decay energies.

6. Non-relativistic limit. The exact formulas reduce to $K_a \approx (m_b/(m_1+m_2))Q$, $p^* \approx \sqrt{2\mu Q} = \mu v_{\rm rel}$, and Galilean velocity addition, with relative errors of order $Q/((m_1+m_2)c^2)$.

Summary of Notation

Further Reading

• Lev Landau and Evgeny Lifshitz, The Classical Theory of Fields (Pergamon, 1975), for the standard relativistic two-body kinematics and invariant masses.
• J. D. Jackson, Classical Electrodynamics (Wiley, 1999), for the Lorentz transformation of four-momenta and fixed-target kinematics.
• Particle Data Group, Review of Particle Physics, for the particle masses used in Problem 5.
• Chris Doran and Anthony Lasenby, Geometric Algebra for Physicists (Cambridge, 2003), for the geometric algebra treatment of relativistic rotors and multiparticle kinematics.
• David Hestenes, Space-Time Algebra (Gordon and Breach, 1966), for the original spacetime-algebra formulation of relativistic mechanics.
• The companion articles of this series: Relativistic Mechanics in Biquaternionic Form, The Lorentz Transformation as a Biquaternionic Rotation, and The Relativistic Two-Body Problem in Biquaternionic Form.