Problem Statement
Problem Statement
An electron and positron annihilate at rest: $e^+ + e^- \to 2\gamma$. Find the energy and wavelength of each photon. Why must there be at least 2 photons?
Given Information
- $m_e c^2 = 0.511\,\text{MeV}$; both at rest ($T = 0$)
Physical Concepts & Formulas
By energy conservation, total energy $= 2m_e c^2 = 1.022\,\text{MeV}$ shared equally between two photons (by momentum conservation — they must be back-to-back for zero total momentum). Single-photon annihilation is forbidden by energy-momentum conservation. This 511 keV gamma line is the signature used in PET scanning.
- $E_\gamma = m_e c^2 = 0.511\,\text{MeV}$ each; $\lambda = hc/E_\gamma$
Step-by-Step Solution
Step 1 — Two photons, back-to-back: By momentum conservation, $\mathbf{p}_1 + \mathbf{p}_2 = 0 \Rightarrow E_1 = E_2$.
Step 2 — Energy: $2E_\gamma = 2m_e c^2 \Rightarrow E_\gamma = 0.511\,\text{MeV}$
Step 3 — Wavelength:
$$\lambda = \frac{hc}{E_\gamma} = \frac{1240\,\text{eV·nm}}{511000\,\text{eV}} = 2.43\times10^{-3}\,\text{nm} = 2.43\,\text{pm}$$
This equals the Compton wavelength $\lambda_C = h/(m_e c)$ — a fundamental length!
Worked Calculation
$$E_\gamma = 0.511\,\text{MeV},\quad \lambda = \frac{hc}{m_e c^2} = \lambda_C = 2.43\,\text{pm}$$
Answer
$$\boxed{E_\gamma = 0.511\,\text{MeV},\quad \lambda = 2.43\,\text{pm} = \lambda_C}$$
Physical Interpretation
The 511 keV photon pair is the basis of PET imaging. Two detectors in coincidence ($\sim10$ ns window) record the photon pair; the annihilation point lies on the line joining the two detectors. With millions of such coincidences, tomographic reconstruction gives a 3D map of metabolic activity (using $^{18}$F-FDG, a glucose analog). Near-body positronium (bound $e^+e^-$) formation shifts the photon energy slightly, allowing in vivo chemistry studies.
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.
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