Irodov Problem 6.280 — Threshold Energy for Pion Production

Problem Statement

Problem Statement

Find the threshold kinetic energy of a proton hitting a stationary proton for the reaction $p + p \to p + p + \pi^0$ (neutral pion production). $m_\pi^0 c^2 = 135\,\text{MeV}$, $m_p c^2 = 938.3\,\text{MeV}$.

Given Information

• $m_p c^2 = 938.3\,\text{MeV}$, $m_{\pi^0} c^2 = 135\,\text{MeV}$

Physical Concepts & Formulas

At threshold, all final particles are at rest in the CM frame. The invariant mass squared $s = (\sum E_i)^2/c^2 – (\sum \mathbf{p}_i)^2$ is Lorentz invariant. In the lab (target at rest): $s = 2m_p^2 c^4 + 2m_p c^2 T_{th}$. At threshold: $s = (2m_p + m_\pi)^2 c^4$. Equating gives $T_{th}$.

• $s_{lab} = 2m_p^2 c^4 + 2m_p c^2 T$ (lab invariant)
• $s_{th} = (2m_p + m_\pi)^2 c^4$ (threshold condition)
• $T_{th} = \dfrac{(2m_p + m_\pi)^2 c^4 – 2m_p^2 c^4}{2m_p c^2}$

Step-by-Step Solution

$$s_{th} = (2\times938.3 + 135)^2 = (2011.6)^2 = 4{,}046,539\,\text{MeV}^2$$

$$2m_p^2 c^4 = 2\times(938.3)^2 = 2\times880,407 = 1{,}760,814\,\text{MeV}^2$$

$$T_{th} = \frac{s_{th} – 2m_p^2 c^4}{2m_p c^2} = \frac{4{,}046,539 – 1{,}760,814}{2\times938.3} = \frac{2{,}285,725}{1876.6} = 1217.8\,\text{MeV}$$

Worked Calculation

$$T_{th} = \frac{(2m_p + m_\pi)^2 – 2m_p^2}{2m_p}c^2 = \frac{4m_p^2 + 4m_p m_\pi + m_\pi^2 – 2m_p^2}{2m_p}c^2 = 2m_p c^2 + 2m_\pi c^2 + \frac{m_\pi^2 c^2}{2m_p}$$

$$= 2\times135 + 2\times135 + \frac{135^2}{2\times938.3} \approx 280 + 9.7 \approx 290\,\text{MeV}… $$

Correct formula: $T_{th} = \frac{4m_pm_\pi + m_\pi^2}{2m_p}c^2 = 2m_\pi c^2 + m_\pi^2c^2/(2m_p) = 270 + 9.7 = 279.7\,\text{MeV}$

Wait: $T_{th} = (s_{th} – 2m_p^2c^4)/(2m_pc^2)$: $s_{th} = (2m_p+m_\pi)^2c^4 = (4m_p^2+4m_pm_\pi+m_\pi^2)c^4$

$T_{th} = (4m_pm_\pi+m_\pi^2)c^2/(2m_p) = (2m_\pi + m_\pi^2/(2m_p))c^2 = (270 + 9.7)\,\text{MeV} = 279.7\,\text{MeV}$

Answer

$$\boxed{T_{th} = 2m_\pi c^2\!\left(1 + \frac{m_\pi}{4m_p}\right)\approx 280\,\text{MeV}}$$

Physical Interpretation

The threshold of 280 MeV is much larger than $m_\pi c^2 = 135\,\text{MeV}$ because a significant fraction of the kinetic energy goes into CM motion of the final state. This “kinematic inefficiency” is why collider experiments (where both beams move) are far more efficient than fixed-target experiments for creating massive particles — at a collider, all the energy is available for mass creation, while in a fixed-target experiment, most energy is “wasted” on CM motion.

Given Information

• $\hbar c = 197.3\,\text{MeV·fm}$; $\hbar = 6.582\times10^{-25}\,\text{GeV·s}$
• $\alpha_{EM} = 1/137.036$; $G_F/(\hbar c)^3 = 1.166\times10^{-5}\,\text{GeV}^{-2}$
• $m_p c^2 = 938.3\,\text{MeV}$; $m_\pi c^2 = 135\,\text{MeV}$ (neutral), $140\,\text{MeV}$ (charged)
• $m_\mu c^2 = 105.7\,\text{MeV}$; $m_W = 80.4\,\text{GeV}/c^2$; $m_Z = 91.2\,\text{GeV}/c^2$; $m_H = 125.1\,\text{GeV}/c^2$

Physical Concepts & Formulas

Elementary particle physics is governed by the Standard Model — a quantum field theory based on gauge symmetry $SU(3)_C \times SU(2)_L \times U(1)_Y$. Conservation laws (baryon number, lepton number, strangeness in strong interactions, energy-momentum, charge) determine which processes are allowed. Kinematics of relativistic reactions uses the Lorentz-invariant Mandelstam variables $s, t, u$; the threshold condition requires $\sqrt{s} = \sum m_{final} c^2$ at minimum.

• $s = (p_1 + p_2)^2 c^2 = E_{CM}^2$ — Mandelstam $s$ (squared CM energy)
• $T_{th} = [(\sum m_f)^2 – (\sum m_i)^2]c^4/(2m_{target}c^2)$ — threshold kinetic energy
• $\tau = \hbar/\Gamma$ — particle lifetime from decay width
• $P(\nu_\alpha \to \nu_\beta) = \sin^2(2\theta)\sin^2(1.27\Delta m^2 L/E)$ — neutrino oscillation probability

Step-by-Step Solution

Step 1 — Apply 4-momentum conservation: In any particle reaction, total 4-momentum is conserved. Compute the invariant mass $\sqrt{s}$ of the initial state and match to the final state mass sum at threshold.

$$\sqrt{s} = \sqrt{(E_1+E_2)^2/c^2 – (\mathbf{p}_1+\mathbf{p}_2)^2} = \sum_{final} m_i c^2 \quad (\text{at threshold})$$

Step 2 — Check all conservation laws: Baryon number $B$, lepton numbers $L_e, L_\mu, L_\tau$, charge $Q$, strangeness $S$ (strong/EM only), isospin $I$ (strong only).

Step 3 — Calculate observable quantities: Masses, lifetimes ($\tau = \hbar/\Gamma$), branching ratios, oscillation lengths, cross-sections.

Worked Calculation

Applying the threshold formula or Breit-Wigner resonance formula to the given particle masses and widths, using the conversion $\hbar = 6.582\times10^{-25}\,\text{GeV·s}$ for lifetime calculations and $\hbar c = 197.3\,\text{MeV·fm}$ for cross-section estimates.

Answer

$$\boxed{\text{See derivation above for specific numerical results}}$$

Physical Interpretation

The Standard Model has been tested to extraordinary precision across 17 orders of magnitude in energy, from atomic physics to LHC collisions at 13 TeV. Despite this success, it cannot be the final theory: it does not include gravity, dark matter, dark energy, or a mechanism for the observed matter-antimatter asymmetry. Neutrino masses (discovered via oscillations) require SM extensions. The hierarchy problem (why $m_H \ll m_{Planck}$) motivates supersymmetry, extra dimensions, or compositeness. The next generation of experiments (HL-LHC, DUNE, CMB-S4, gravitational wave observatories) aims to discover the physics that lies beyond.