Introduction
The two-body problem describes the motion of two objects interacting through mutual forces. It reduces to an equivalent one-body problem using the concept of reduced mass. This reduction separates CM motion from relative motion and is fundamental to understanding atomic systems, binary stars, and planetary motion.
Equations of Motion
For masses m1 and m2 at positions r1 and r2, interacting through central force F(r) along line joining them: m1 \(\ddot{r}\)1 = F(r)\(\hat{r}\) and m2 \(\ddot{r}\)2 = -F(r)\(\hat{r}\), where r = r2 - r1 is relative position vector. These coupled equations describe the complete motion. The system's CM position R = (m1r1 + m2r2)/(m1+m2) moves uniformly if no external forces.
Reduced Mass
Reduced mass μ = m1m2/(m1+m2) = (1/m1 + 1/m2). Properties: μ < min(m1,m2); if m2 >> m1, then μ ≈ m1; if m1 = m2 = m, then μ = m/2. The relative motion obeys: μ \(\ddot{r}\) = F(r)\(\hat{r}\), exactly like a single particle of mass μ in central potential. This reduces the two-body problem to one-body form.
\nIntroduction
The two-body problem describes the motion of two objects interacting through mutual forces. It reduces to an equivalent one-body problem using the concept of reduced mass. This reduction separates CM motion from relative motion and is fundamental to understanding atomic systems, binary stars, and planetary motion.
Equations of Motion
For masses m1 and m2 at positions r1 and r2, interacting through central force F(r) along line joining them: m1 \(\ddot{r}\)1 = F(r)\(\hat{r}\) and m2 \(\ddot{r}\)2 = -F(r)\(\hat{r}\), where r = r2 - r1 is relative position vector. These coupled equations describe the complete motion. The system's CM position R = (m1r1 + m2r2)/(m1+m2) moves uniformly if no external forces.
Reduced Mass
Reduced mass μ = m1m2/(m1+m2) = (1/m1 + 1/m2). Properties: μ < min(m1,m2); if m2 >> m1, then μ ≈ m1; if m1 = m2 = m, then μ = m/2. The relative motion obeys: μ \(\ddot{r}\) = F(r)\(\hat{r}\), exactly like a single particle of mass μ in central potential. This reduces the two-body problem to one-body form.
\nSeparation of Motion
Complete separation: (1) CM motion: MR̈ = F_ext (external force on CM); (2) Relative motion: μ \(\ddot{r}\) = F(r)\(\hat{r}\) (internal interaction). For isolated system (F_ext = 0), CM moves with constant velocity. The relative motion is equivalent to one body in central force field. This separation is exact and applies to any central force.
Energy Formulation
Total kinetic energy separates: K = 1/2MV_CM2 + 1/2μv2, where v = \(\dot{r}\) is relative velocity. Total energy: E = 1/2MV_CM2 + [1/2μv2 + U(r)]. The bracketed term is relative energy. For bound isolated systems, CM kinetic energy is constant, and relative energy determines orbit type. Binding energy calculations use reduced mass: E_n = -μZ2e4/(2(4πε0)2Ⅎn2) for hydrogen-like atoms.
\nApplications
Applications: Hydrogen atom (electron-proton system, reduced mass gives small correction to Bohr model); Binary stars (reduced mass modifies orbital period); Planetary motion (Sun moves slightly about CM, affecting observed planetary positions); Molecular vibrations (two atoms vibrate as if single mass μ on spring); Scattering problems (Rutherford scattering cross-section uses reduced mass).
\nSolved Example: Reduced Mass Correction
Calculate reduced mass for hydrogen atom and percentage correction to electron mass. Compare with deuterium (heavy hydrogen). Solution: Hydrogen: m_e = 9.11×10⻳¹ kg, m_p = 1.67×10⻲7 kg. μ_H = m_e × m_p / (m_e + m_p) = (9.11×10⻳¹ × 1.67×10⻲7) / (9.11×10⻳¹ + 1.67×10⻲7) = 1.52×10^-57 / 1.67×10⻲7 = 9.10×10⻳¹ kg. Compare to electron mass: ratio = μ/m_e = 9.10/9.11 = 0.99945. Correction is about 0.055% (or about 1 part in 1800). This affects spectral lines - hydrogen and deuterium have slightly different wavelengths. Deuterium: m_d = 2×m_p = 3.34×10⻲7 kg. μ_D = m_e × m_d / (m_e + m_d) = 9.107×10^-31 kg. Difference in reduced mass causes isotope shift in hydrogen spectrum. This small correction allows astronomers to detect deuterium in distant galaxies.
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