Central Force: Two-Body Problem, Kepler's Laws & Scattering

1. Decoupling Center-of-Mass and Relative Coordinates

Consider two isolated particles of masses $m_1$ and $m_2$ interacting solely via an internal central force directed along the line joining them: The equations of motion are $m_1 \ddot{\vec{r}}_1 = f(r) \hat{r}$ and $m_2 \ddot{\vec{r}}_2 = -f(r) \hat{r}$. We introduce the center of mass coordinate $\vec{R}$ and relative separation vector $\vec{r}$: The individual position vectors in terms of $\vec{R}$ and $\vec{r}$ are:

2. Separation of the Lagrangian & The Reduced Mass

The total kinetic energy decomposes cleanly: where $\mu$ is the reduced mass of the system: Since the interaction potential $V(r)$ depends only on relative distance $r$, the Lagrangian separates: Because $\vec{R}$ is completely cyclic, the center of mass moves with constant linear momentum $\vec{P} = M \dot{\vec{R}} = \text{const}$. In the center-of-mass frame ($\vec{R} = 0$), the problem reduces strictly to a single particle of mass $\mu$ moving in a static central potential $V(r)$.

1. Confinement to a Fixed Plane of Motion

The torque exerted by any central force $\vec{F} = f(r)\hat{r}$ relative to the force center vanishes identically: Consequently, the orbital angular momentum $\vec{L}$ is a strict constant of motion: Since $\vec{r} \cdot \vec{L} = \vec{r} \cdot (\vec{r} \times \vec{p}) = 0$, the position vector $\vec{r}$ remains perpendicular to the fixed direction of $\vec{L}$ for all time. Thus, all central force motion is strictly confined to a two-dimensional plane.

2. Polar Coordinate Representation & Kepler's Second Law

Choosing plane polar coordinates $(r, \theta)$ in the orbital plane: The magnitude of the angular momentum is: The area swept out by the radius vector in time $dt$ is $dA = \frac{1}{2} r (r d\theta) = \frac{1}{2} r^2 \dot{\theta} dt$. The areal velocity is therefore:

Kepler's Second Law: The radius vector from the force center to the body sweeps out equal areas in equal intervals of time. This law holds universally for any central force, regardless of whether it follows an inverse-square law.

1. Total Energy in Radial Coordinates

The total mechanical energy in polar coordinates is: Eliminating $\dot{\theta}$ using the conserved angular momentum $\dot{\theta} = \frac{l}{\mu r^2}$:

2. The Effective Potential $V_{\text{eff}}(r)$

The dynamics of the radial coordinate $r(t)$ behaves identically to a 1D particle of mass $\mu$ in an effective potential: The term $\frac{l^2}{2\mu r^2}$ is the centrifugal potential barrier, representing the kinetic energy associated with angular rotation. As $r \to 0$, this barrier diverges as $+1/r^2$, preventing particles with non-zero angular momentum ($l \neq 0$) from falling into the center.

3. Classification of Orbits for Inverse-Square Gravity ($V(r) = -k/r$)

Setting $\frac{dV_{\text{eff}}}{dr} = 0$: The minimum value of the effective potential is: Orbit classification based on total energy $E$:

• $E = V_{\text{min}} = -\frac{\mu k^2}{2 l^2}$ (Circular Orbit): $\dot{r} = 0$ constantly; orbit is a circle of radius $r = r_0$ ($e = 0$).
• $V_{\text{min}} < E < 0$ (Elliptical Orbit): Bound orbit oscillating between periapsis $r_{\text{min}}$ and apoapsis $r_{\text{max}}$ ($0 < e < 1$).
• $E = 0$ (Parabolic Orbit): Unbound orbit with escape velocity; particle reaches infinity with zero residual kinetic energy ($e = 1$).
• $E > 0$ (Hyperbolic Orbit): Unbound scattering orbit; particle approaches from infinity and deflects off with non-zero residual velocity ($e > 1$).

1. Derivation of Binet's Differential Orbit Equation

To determine the geometric trajectory $r(\theta)$ directly without solving for time $t$, we substitute $u = 1/r$. The radial velocity is: Differentiating once more with respect to $t$: Substituting $\ddot{r}$ into the radial equation of motion $\mu(\ddot{r} - r\dot{\theta}^2) = f(r)$: Multiplying by $-\frac{\mu}{l^2 u^2}$ yields Binet's Formula:

2. Exact Solution for the Gravitational Force ($f(r) = -k/r^2 = -k u^2$)

Substituting $f(1/u) = -k u^2$: This is an inhomogeneous linear second-order differential equation with constant coefficients. Its general solution is: Setting $\theta_0 = 0$ along the periapsis direction and defining the semi-latus rectum $p$ and eccentricity $e$: Inverting $u = 1/r$ delivers the universal polar equation of conic sections: The orbital eccentricity is related to total energy $E$ by:

3. Kepler's Three Laws Derived

1. First Law (Law of Ellipses): For bound states ($E < 0$), $0 \le e < 1$; the trajectory $r(\theta)$ is an ellipse with the gravitational center at one focus.
2. Second Law (Law of Equal Areas): Areal velocity $dA/dt = l/(2\mu)$ is constant.
3. Third Law (Harmonic Law): The area of an ellipse is $A = \pi a b$, where $a = p/(1-e^2)$ is the semi-major axis and $b = a\sqrt{1-e^2} = \sqrt{a p} = l\sqrt{a}/\sqrt{\mu k}$ is the semi-minor axis. The orbital period $\tau$ is:
   $$\tau = \frac{\text{Area}}{dA/dt} = \frac{\pi a b}{l / (2\mu)} = \frac{2\pi \mu a b}{l} = 2\pi a^{3/2} \sqrt{\frac{\mu}{k}}$$
   Squaring both sides and setting $k = G m_1 m_2$ and $\mu = m_1 m_2 / (m_1 + m_2)$:
   $$\tau^2 = \frac{4\pi^2 a^3}{G(m_1 + m_2)}$$

1. The Laplace-Runge-Lenz (LRL) Conserved Vector

In addition to energy $E$ and angular momentum $\vec{L}$, the $1/r$ Kepler potential possesses an additional conserved vector quantity, the Laplace-Runge-Lenz vector $\vec{A}$: Proof of conservation: Since $\dot{\vec{p}} = -\frac{k}{r^2}\hat{r}$ and $\vec{L} = \mu \vec{r} \times \dot{\vec{r}}$: Thus $\frac{d\vec{A}}{dt} = -\mu k \dot{\hat{r}} - (-\mu k \dot{\hat{r}}) = 0$.

Physical Significance: The vector $\vec{A}$ lies permanently in the orbital plane, pointing directly from the focus toward the periapsis with magnitude $|\vec{A}| = \mu k e$. Its constancy prevents orbital precession, ensuring that Keplerian orbits are strictly closed. In modern physics, this reflects an underlying dynamic $SO(4)$ symmetry.

2. The Virial Theorem for Central Potentials

Consider the quantity $G = \sum_{i=1}^N \vec{p}_i \cdot \vec{r}_i$. Its time derivative is: Taking the long-term time average $\langle X \rangle = \lim_{\tau \to \infty} \frac{1}{\tau} \int_0^\tau X dt$: since for bound systems $G(t)$ remains bounded. Hence: For power-law potentials $V(r) = a r^n$, $\vec{F} = -\nabla V = -n a r^{n-2} \vec{r}$, so $\vec{F} \cdot \vec{r} = -n V(r)$. This yields the Virial Theorem:

• Gravitational / Coulomb Potential ($n = -1$): $2\langle T \rangle = -\langle V \rangle \implies E = \langle T \rangle + \langle V \rangle = -\langle T \rangle = \frac{1}{2}\langle V \rangle$.
• Harmonic Oscillator ($n = 2$): $2\langle T \rangle = 2\langle V \rangle \implies \langle T \rangle = \langle V \rangle = \frac{1}{2} E$.

1. Kinematics of Elastic Scattering

Consider a projectile of mass $m$ and initial speed $v_0$ incident from infinity with impact parameter $b$ (the perpendicular distance from the scattering center to the incident velocity line). The total angular momentum and energy are:

2. The Scattering Angle $\Theta$

From the orbital equation in polar coordinates: For a repulsive Coulomb potential $V(r) = \frac{k}{r} = \frac{q_1 q_2}{4\pi \varepsilon_0 r}$, evaluating the integral delivers the relation between impact parameter $b$ and scattering angle $\theta$:

3. The Differential Scattering Cross-Section

Particles incident within an annular area $d\sigma = 2\pi b db$ scatter into a solid angle $d\Omega = 2\pi \sin\theta d\theta$. The differential cross-section $\frac{d\sigma}{d\Omega}$ is: Differentiating $b(\theta)$: Using $\sin\theta = 2 \sin(\theta/2) \cos(\theta/2)$: Substituting $k = \frac{q_1 q_2}{4\pi \varepsilon_0}$ yields the celebrated Rutherford Scattering Formula: