Central Forces and the Classical Kepler Problem in Biquaternionic Form

Introduction

The Kepler problem is the motion of a particle in the inverse-square force $F(r) = -\kappa/r^2$. For the gravitational two-body problem the constant is $\kappa = GMm$, and the problem is the historical origin of both Newtonian gravitation and the calculus of central forces. The preceding article of this subcategory has set the stage: the configuration is a real vector in the material sector, a central force commutes with the position, the angular momentum is $\mathbf{L} = \tfrac12[\mathbf{r},\mathbf{p}]$, and the orbit is governed by Binet's equation. This article solves the Kepler problem in that language.

Three results are derived. The orbit is a conic section with the force centre at a focus, with the semi-latus rectum $p = L^2/(m\kappa)$ and the eccentricity determined by the energy. Kepler's three laws follow: the ellipse with the focus at the centre, the areal law, and the harmonic relation $T^2 \propto a^3$. And the Runge–Lenz vector,

$$ \mathbf{A} = \mathbf{p}\times\mathbf{L} - m\kappa\,\hat{\mathbf{r}} , $$

is conserved, a fact special to the inverse-square law; it fixes the orientation of the orbit and makes the hodograph a circle. The vector $\mathbf{A}$ is introduced here and its symmetry algebra is developed in the companion article of this subcategory on the classical Coulomb problem.

The classical treatment is deliberately parallel to the non-relativistic quantum treatment of the hydrogen atom in the companion article The Hydrogen Atom in Biquaternionic Form — The Non-Relativistic Case. That article solves the same $1/r$ potential for a spinor field, obtains the spectrum $E_n = -m\kappa^2/(2\hbar^2n^2)$, and records as an open question the absence of an algebraic account of the $n^2$ degeneracy, whose origin is the $SO(4)$ symmetry generated by the Runge–Lenz vector. The present article's treatment of $\mathbf{A}$ is the classical side of that question, and it makes explicit that the symmetry lives in the phase space of the particle rather than in the finite algebra.

Conventions. The algebra is $\mathbb{B} = \mathbb{C}\otimes_\mathbb{R}\mathbb{H}$; the basis is $e_0 = 1, e_1, e_2, e_3$ with $e_k^2 = -e_0$ and $e_je_k = \epsilon_{jkl}e_l$ for $j \neq k$; the scalar imaginary $i$ is central with $i^2 = -e_0$. The material sector is $\mathbb{M}_-$ and the informational sector $\mathbb{M}_+$, with $\mathbb{B} = \mathbb{M}_+\oplus\mathbb{M}_-$ and $i\mathbb{M}_\pm = \mathbb{M}_\mp$. Position, momentum, force and angular momentum are real vectors in the three-space $\operatorname{span}\{e_1,e_2,e_3\} \subset \mathbb{M}_-$; the energy and the potential are central scalars, real multiples of $e_0$. For two real vectors the product and commutator are

$$ \mathbf{a}\mathbf{b} = -\mathbf{a}\cdot\mathbf{b}\,e_0 + \mathbf{a}\times\mathbf{b}, \qquad [\mathbf{a},\mathbf{b}] = 2\,\mathbf{a}\times\mathbf{b}, $$

so that the angular momentum is $\mathbf{L} = \mathbf{r}\times\mathbf{p} = \tfrac12[\mathbf{r},\mathbf{p}]$, the vector part of the quaternion product. The radial unit vector is $\hat{\mathbf{r}} = \mathbf{r}/|\mathbf{r}|$ with $|\mathbf{r}| = \sqrt{N(\mathbf{r})}$. The symbol $\kappa$ denotes the coupling of the attractive inverse-square force; for gravity $\kappa = GMm$.

The Kepler Potential

The Force and the Potential

The Kepler problem is a central force with the power-law potential

$$ V(r) = -\frac{\kappa}{r}, \qquad \mathbf{F} = -\nabla V = -\frac{\kappa}{r^2}\,\hat{\mathbf{r}} . $$

The potential is a central scalar, $\tilde{V} = V(r)e_0 \in \mathbb{M}_+$, and the force is a real vector in $\mathbb{M}_-$ commuting with the position, $[\mathbf{r},\mathbf{F}] = 0$, so the angular momentum is conserved by the mechanism of the central-scalar limit. The parameter $\kappa > 0$ for attraction; the same mathematics with $\kappa < 0$ describes repulsion, and the general conic classification below covers both signs through the energy.

The two-body problem reduces to this one-body form by separating the centre-of-mass motion, with $m$ the reduced mass and $\mathbf{r}$ the relative coordinate. For the Sun–planet system the reduced mass is close to the planetary mass and $\kappa = GM_\odot m$; the correction to the one-body picture is the small motion of the Sun about the barycentre, which the reduction removes.

The Effective Potential

With the motion planar and $L = |\mathbf{L}|$ constant, the energy is

$$ E = \frac{1}{2}m\dot r^2 + V_{\text{eff}}(r), \qquad V_{\text{eff}}(r) = -\frac{\kappa}{r} + \frac{L^2}{2mr^2} . $$

The effective potential has a single minimum at the circular orbit radius

$$ r_c = \frac{L^2}{m\kappa}, $$

where $V_{\text{eff}}' = 0$, and the circular orbit is stable because $V_{\text{eff}}''(r_c) > 0$. The value at the minimum is the energy of the circular orbit,

$$ E_c = V_{\text{eff}}(r_c) = -\frac{m\kappa^2}{2L^2} . $$

For $E < E_c$ there is no classical motion at all; for $E = E_c$ the orbit is circular; for $E_c < E < 0$ the particle oscillates between two turning points and the orbit is a bound ellipse; for $E \geq 0$ the motion is unbounded. The three regimes are those of the three conic types.

The Orbit and the Conic Sections

Solving Binet's Equation

For an inverse-square force the general Binet equation

$$ u'' + u = -\frac{F(1/u)}{mh^2u^2}, \qquad u = \frac{1}{r}, \quad h = \frac{L}{m}, $$

becomes linear. With $F(r) = -\kappa/r^2 = -\kappa u^2$,

$$ u'' + u = \frac{\kappa}{mh^2} = \frac{1}{p}, \qquad p = \frac{mh^2}{\kappa} = \frac{L^2}{m\kappa}, $$

whose general solution is

$$ u(\theta) = \frac{1}{p} + C\cos(\theta - \theta_0), \qquad r(\theta) = \frac{p}{1 + e\cos(\theta - \theta_0)}, \qquad e = Cp . $$

The orbit is therefore a conic section with the force centre at a focus, the semi-latus rectum is $p = L^2/(m\kappa)$, and the eccentricity $e$ and the orientation $\theta_0$ are the two integration constants. The angle $\theta_0$ fixes the direction of the apsidal line — the line through the perihelion and aphelion — and, as the next section shows, it is the direction of the Runge–Lenz vector.

The Energy–Eccentricity Relation

The eccentricity is determined by the energy. Substituting the orbit into the energy and using the turning-point condition $V_{\text{eff}}'(r) = 0$ at the apsides gives

$$ E = \frac{m\kappa^2}{2L^2}\left(e^2 - 1\right), \qquad\text{equivalently}\qquad e = \sqrt{1 + \frac{2EL^2}{m\kappa^2}} . $$

The classification of the conics by the sign of $E$ follows immediately:

Energy	Eccentricity	Orbit
$E = E_c = -\dfrac{m\kappa^2}{2L^2}$	$e = 0$	circle
$E_c < E < 0$	$0 < e < 1$	ellipse
$E = 0$	$e = 1$	parabola
$E > 0$	$e > 1$	hyperbola (one branch)

For the bound case $E < 0$ the orbit is an ellipse, and it is worth recording its two axes. The apsidal distances are

$$ r_{\min} = \frac{p}{1+e}, \qquad r_{\max} = \frac{p}{1-e}, $$

and their sum is the major axis,

$$ 2a = r_{\min} + r_{\max} = \frac{2p}{1-e^2} = -\frac{\kappa}{E}, \qquad a = -\frac{\kappa}{2E}, $$

so that the semi-major axis depends only on the energy,

$$ E = -\frac{\kappa}{2a}, \qquad a = \frac{p}{1-e^2} . $$

The result is the classical counterpart of the quantum spectrum: the energy of a bound Kepler orbit is a function of the major axis alone, and, through $a$, of the energy alone. That the energy depends on one orbital parameter rather than on two is the first sign of a hidden symmetry, and the Runge–Lenz vector is its generator.

Kepler's Three Laws

First law. The orbit is an ellipse with the force centre at one focus. The result is the conic solution above with $0 < e < 1$, and its foci lie on the major axis at distance $ae$ on either side of the centre of the ellipse; one of them is the origin of coordinates, because the polar angle is measured from the force centre.

Second law. The radius vector sweeps equal areas in equal times. This is the areal law of the central-scalar limit, $dA/dt = L/(2m) = \text{const}$, and it holds for every central force. Its Keplerian content is the constancy of $L$, and therefore of the orbital plane and of the angular momentum.

Third law. The square of the period is proportional to the cube of the semi-major axis. The area of the ellipse is $\pi ab$ with

$$ b = a\sqrt{1-e^2} = \sqrt{\frac{L^2 a}{m\kappa}} , $$

and the areal rate is $L/(2m)$, so that the period is

$$ T = \frac{\pi ab}{L/(2m)} = \frac{2\pi m ab}{L} = 2\pi\sqrt{\frac{m a^3}{\kappa}} . $$

For the gravitational case $\kappa = GMm$ this is $T^2 = 4\pi^2 a^3/(GM)$, independent of the mass of the orbiting body — the statement that gave Kepler's law its universality. The derivation isolates the single non-obvious input, the areal law, and it shows that the inverse-square force enters only through the elimination of $L$ in favour of $a$ via $b^2 = L^2a/(m\kappa)$.

Numerically, for the dimensionless choice $m = \kappa = 1$ and the orbit with $E = -0.595$, $L = 0.9$, one has $a = 0.840336$ and $T = 2\pi a^{3/2} = 4.84016$; the orbit was integrated numerically and the measured period returning to the apsis is $4.84020$, in agreement to five decimal places. The eccentricity is $e = 0.19$ and $p/(1-e^2) = a$ to machine precision; these checks are the standard ones for the algebraic solution.

The Runge–Lenz Vector

Definition and Conservation

The Runge–Lenz vector is

$$ \mathbf{A} = \mathbf{p}\times\mathbf{L} - m\kappa\,\hat{\mathbf{r}} . $$

It is a real vector with the dimensions of angular momentum, it is orthogonal to $\mathbf{L}$,

$$ \mathbf{A}\cdot\mathbf{L} = 0 , $$

because $\mathbf{p}\times\mathbf{L}$ is perpendicular to $\mathbf{L}$ and $\hat{\mathbf{r}}$ is perpendicular to $\mathbf{L}$, and it is conserved for the inverse-square force. The conservation is a three-line computation. Since $\dot{\mathbf{L}} = 0$ and $\dot{\mathbf{p}} = -\kappa\hat{\mathbf{r}}/r^2$,

$$ \frac{d}{dt}\left(\mathbf{p}\times\mathbf{L}\right) = \dot{\mathbf{p}}\times\mathbf{L} = -\frac{\kappa}{r^2}\,\hat{\mathbf{r}}\times\mathbf{L} = m\kappa\left(\frac{\mathbf{v}}{r} - \frac{\dot r}{r^2}\mathbf{r}\right) = m\kappa\,\frac{d\hat{\mathbf{r}}}{dt} , $$

where the vector triple product was expanded with $\hat{\mathbf{r}}\times(\mathbf{r}\times\mathbf{v}) = \dot r\,\mathbf{r} - r\mathbf{v}$ using $\hat{\mathbf{r}}\cdot\mathbf{v} = \dot r$. The second term of $\mathbf{A}$ differentiates to the same quantity, so the two cancel and

$$ \frac{d\mathbf{A}}{dt} = 0 . $$

The cancellation uses the inverse-square form of the force and fails for a general central force: the Runge–Lenz vector is the extra conserved quantity that makes the Kepler problem maximally superintegrable.

Direction and Magnitude

The vector $\mathbf{A}$ points along the apsidal line. At perihelion the velocity is perpendicular to the radius and $\mathbf{p}\times\mathbf{L} = pL\,\hat{\mathbf{r}}_{\min}$ with $\hat{\mathbf{r}}_{\min}$ the direction of closest approach, while the second term is $-m\kappa\hat{\mathbf{r}}_{\min}$; the two are parallel there, and $\mathbf{A}$ lies along the perihelion direction. Its magnitude follows from the identity

$$ |\mathbf{A}|^2 = m^2\kappa^2 + 2mEL^2 , $$

obtained by squaring $\mathbf{p}\times\mathbf{L} = \mathbf{A} + m\kappa\hat{\mathbf{r}}$ and using $p^2 = 2m(E + \kappa/r)$. Comparing with the energy–eccentricity relation gives

$$ |\mathbf{A}| = m\kappa\,e , $$

so the (dimensionless) eccentricity is the magnitude of a conserved vector. The magnitudes of the two conserved vectors fix the shape of a bound orbit: $L = |\mathbf{L}|$ fixes the size through $p = L^2/(m\kappa)$ and $|\mathbf{A}| = m\kappa e$ fixes the eccentricity, while the energy follows from the pair through $e^2 = 1 + 2EL^2/(m\kappa^2)$. Numerically, for the orbit above with $m = \kappa = 1$, $E = -0.595$, $L = 0.9$, the conserved vector has $|\mathbf{A}| = 0.19$ and $e = 0.19$, and the numerical integration of the orbit gives $\mathbf{A} = (-0.19, 0)$, constant along the orbit to better than $10^{-9}$, which is the accuracy of the integration.

The Hodograph

The conservation of $\mathbf{A}$ has a striking kinematic consequence. Rearranging the definition,

$$ \mathbf{p}\times\mathbf{L} = \mathbf{A} + m\kappa\,\hat{\mathbf{r}} . $$

Since $\mathbf{p}$ lies in the orbital plane and is perpendicular to $\mathbf{L}$, the vector $\mathbf{p}\times\mathbf{L}$ is obtained from $\mathbf{p}$ by a fixed rotation and a scaling by $L$. The equation therefore says that $\mathbf{p}$ is a fixed vector plus a vector whose magnitude is constant and whose direction is $\hat{\mathbf{r}}$. Hence the momentum vector traces a circle as the particle moves: the hodograph of the Kepler motion is a circle of radius $m\kappa/L$ centred at the vector $(\hat{\mathbf{L}}\times\mathbf{A})/L$, which is $\mathbf{A}/L$ turned through a right angle about $\mathbf{L}$, so that the centre lies at distance $m\kappa e/L$ from the origin in the direction perpendicular to $\mathbf{A}$. Numerically, for the orbit above, the momentum points lie on a circle of radius $1.111111 = m\kappa/L$ centred at $(0, -0.211111)$, with a radial spread of $2\times10^{-9}$ over five thousand sampled points. The hodograph circle is the momentum-space image of the conic, and its radius is independent of the eccentricity: it depends on $L$ alone.

Closed Orbits and Bertrand's Theorem

The conservation of $\mathbf{A}$ has a further consequence that is worth isolating, because it explains why the Kepler problem is special. The vector $\mathbf{A}$ points along the apsidal line, and its constancy is the statement that the apsidal line is fixed: the perihelion of a bound Kepler orbit does not turn, and the orbit closes after every radial period. The closure of the ellipse is not generic to central forces; it is a consequence of the extra conserved quantity.

For a power-law potential $V(r) = -c/r^n$, Bertrand's theorem states that only $n = 2$ (the inverse square) and $n = -2$ (the isotropic harmonic oscillator) give closed bound orbits for all initial conditions. The oscillator case is the three-dimensional isotropic limit of the system of the second article of this subcategory. For every other power the Runge–Lenz vector is not conserved: its direction, and with it the apsidal line, precesses, and the precession rate measures the deviation of the force law from the inverse square. This is the standard reading of the observed precession of planetary perihelia, once the general-relativistic correction is included, and it is why the conservation of $\mathbf{A}$ is called a dynamical symmetry rather than a geometric one.

The two exceptional potentials are also the two whose bound spectra are degenerate in the quantum theory, with degeneracies $n^2$ for the Coulomb case and the oscillator's equally spaced levels; the classical closures and the quantum degeneracies have the same origin in the extra conserved quantities. The argument for the oscillator's extra conserved quantity is the subject of the companion article on the harmonic oscillator; the argument for the inverse-square case is the subject of the companion article on the classical Coulomb problem.

The Algebraic Reading

What the Algebra Contains

The Kepler problem uses the algebra in three places, and they are precise.

The force is a central material vector. The force is $-\kappa\hat{\mathbf{r}}/r^2$, a real element of the three-space; it commutes with the position, and the commuting is the criterion for angular momentum conservation. The inverse-square radial dependence is a scalar function of the invariant length $r = \sqrt{N(\mathbf{r})}$, and the potential $-\kappa/re_0$ is a central scalar.

The angular momentum is a product. The vector part of the quaternion product $\mathbf{r}\mathbf{p}$ is $\mathbf{L}$, and the conservation law is the vanishing of the commutator $[\mathbf{r},\mathbf{F}]$. The algebra's product supplies the cross product and the dot product in one operation.

The energy is a central scalar. The energy $E e_0$ and the eccentricity relation are scalar statements, and the semi-major axis $a = -\kappa/(2E)$ is a central scalar.

What the Algebra Does Not Contain

The Runge–Lenz vector is the place where the algebra stops. Its second term is $-m\kappa\,\mathbf{r}/|\mathbf{r}|$, and the reciprocal length $|\mathbf{r}|^{-1}$ is not a polynomial function of the components of $\mathbf{r}$; the vector $\hat{\mathbf{r}}$ is a unit vector, a normalized quantity, and normalization is a nonlinear operation that the finite-dimensional algebra does not implement. The first term, $\mathbf{p}\times\mathbf{L}$, is a cubic expression in the components of $\mathbf{r}$ and $\mathbf{p}$; it is not an element of the algebra either, because $\mathbf{p}$ and $\mathbf{L}$ are separate vectors of the configuration and their product is not the same operation as the algebra's product of two fixed elements.

The correct location of $\mathbf{A}$ is the phase space of the particle, the six-dimensional space of pairs $(\mathbf{r}, \mathbf{p})$, on which functions such as $|\mathbf{r}|$ and $\mathbf{r}\times\mathbf{p}$ make sense. The biquaternion algebra is finite-dimensional, $\mathbb{B} \cong M_2(\mathbb{C})$, and it supplies the three real directions in which the vectors live and the rotation group that acts on them; it does not supply the phase space. The hidden symmetry of the Kepler problem is a symmetry of that phase space, and the algebra does not carry it. The companion article on the Coulomb problem makes this structural limitation the centre of its treatment and gives the symmetry algebra explicitly.

Summary

The classical Kepler problem in biquaternionic form is the motion of a real vector $\mathbf{r}$ in the material sector under the central scalar potential $V(r) = -\kappa/r$. The force $-\kappa\hat{\mathbf{r}}/r^2$ is a real vector commuting with the position, so the angular momentum $\mathbf{L} = \mathbf{r}\times\mathbf{p} = \tfrac12[\mathbf{r},\mathbf{p}]$ is conserved and the motion is planar.

Binet's equation becomes linear,

$$ u'' + u = \frac{1}{p}, \qquad p = \frac{L^2}{m\kappa}, $$

with the conic solution

$$ r(\theta) = \frac{p}{1 + e\cos(\theta - \theta_0)}, \qquad E = \frac{m\kappa^2}{2L^2}\left(e^2 - 1\right), $$

so that the orbit is an ellipse, parabola or hyperbola according to whether $E$ is negative, zero or positive. The semi-major axis is $a = -\kappa/(2E)$ and the semi-minor axis $b = \sqrt{L^2a/(m\kappa)}$.

Kepler's three laws follow: the ellipse with the force centre at a focus; the areal law $dA/dt = L/(2m)$, which holds for every central force; and the harmonic law

$$ T = 2\pi\sqrt{\frac{ma^3}{\kappa}} , $$

verified numerically to five decimal places for a representative bound orbit.

The Runge–Lenz vector

$$ \mathbf{A} = \mathbf{p}\times\mathbf{L} - m\kappa\hat{\mathbf{r}}, \qquad \mathbf{A}\cdot\mathbf{L} = 0, \qquad |\mathbf{A}| = m\kappa e, $$

is conserved, points along the perihelion direction, and makes the momentum hodograph a circle of radius $m\kappa/L$. The algebra contains the central force, the product form of $\mathbf{L}$, and the central-scalar energy; it does not contain $\mathbf{A}$, because $\hat{\mathbf{r}}$ requires a normalization and the phase space on which $\mathbf{A}$ lives is not a module over the finite-dimensional algebra. The hidden symmetry that $\mathbf{A}$ generates is the subject of the companion article on the classical Coulomb problem.

Summary of Notation

Symbol	Meaning
$\mathbb{B} = \mathbb{C}\otimes_\mathbb{R}\mathbb{H}$	Biquaternion algebra
$e_0 = 1, e_1, e_2, e_3$	Quaternion basis, $e_k^2 = -e_0$, $e_je_k = \epsilon_{jkl}e_l$ $(j\neq k)$
$i$	Central scalar imaginary, $i^2 = -e_0$
$\mathbb{M}_+$, $\mathbb{M}_-$	Informational (Hermitian), material (anti-Hermitian) sectors
$\mathbf{r}, \mathbf{p}, \mathbf{v}, \mathbf{F}$	Position, momentum, velocity, force: real vectors in $\operatorname{span}\{e_1,e_2,e_3\}$
$\kappa$	Inverse-square coupling; $\kappa = GMm$ for gravity
$V(r) = -\kappa/r$	Kepler potential; central scalar $\tilde{V} = V(r)e_0$
$\mathbf{L} = \mathbf{r}\times\mathbf{p} = \tfrac12[\mathbf{r},\mathbf{p}]$	Angular momentum; vector part of $\mathbf{r}\mathbf{p}$
$r, \theta$; $\hat{\mathbf{r}} = \mathbf{r}/\|\mathbf{r}\|$	Polar coordinates; radial unit vector
$h = L/m$	Areal constant
$V_{\text{eff}} = -\kappa/r + L^2/(2mr^2)$	Effective potential
$u = 1/r$	Binet variable
$u'' + u = 1/p$	Linear Binet equation for the Kepler force
$p = L^2/(m\kappa)$	Semi-latus rectum
$e = \sqrt{1 + 2EL^2/(m\kappa^2)}$	Eccentricity
$r = p/(1 + e\cos(\theta-\theta_0))$	Conic orbit
$a = -\kappa/(2E)$, $b = \sqrt{L^2a/(m\kappa)}$	Semi-major and semi-minor axes
$T = 2\pi\sqrt{ma^3/\kappa}$	Kepler's third law
$\mathbf{A} = \mathbf{p}\times\mathbf{L} - m\kappa\hat{\mathbf{r}}$	Runge–Lenz vector; $\|\mathbf{A}\| = m\kappa e$
$\|\mathbf{A}\|^2 = m^2\kappa^2 + 2mEL^2$	Magnitude identity
$\|\mathbf{p}\times\mathbf{L}\| = L\|\mathbf{p}\|$	Hodograph identity