The Laplace–Runge–Lenz vector

1. A conserved direction inside the orbital plane

Energy determines orbital size, and angular momentum fixes the plane and controls eccentricity. Neither vector angular momentum nor scalar energy specifies which way the ellipse points within its plane. In inverse-square attraction the periapsis direction stays fixed. The Laplace–Runge–Lenz vector encodes that additional geometrical information.

We use relative position $\mathbf r$, relative momentum $\mathbf p=\mu\dot{\mathbf r}$, angular momentum $\mathbf L=\mathbf r\times\mathbf p$, and attractive force $\dot{\mathbf p}=-k\hat{\mathbf r}/r^2$, with $k>0$. The definition is

The order of the cross product matters: $\mathbf L\times\mathbf p$ has the opposite sign. Both terms have dimensions of mass times energy times length, so $\mathbf A/(\mu k)$ is dimensionless.

2. The vector identity used in the proof

Prerequisite. The vector triple-product identity is

It expresses a double cross product as a combination of the two vectors inside the brackets. Its sign can be checked with perpendicular unit vectors: $\hat{\mathbf x}\times(\hat{\mathbf x}\times\hat{\mathbf y})=-\hat{\mathbf y}$. Cross products are not associative, so brackets cannot be moved without changing the expression.

Also, differentiating $\hat{\mathbf r}=\mathbf r/r$ by the quotient rule gives

The numerator is the tangential part of momentum. Radial motion changes the length of $\mathbf r$ without changing its direction.

3. Proof of conservation

Differentiate $\mathbf A$. Central-force angular momentum is constant, so

Insert the force and expand the triple product in the first term:

Since $\mathbf r=r\hat{\mathbf r}$ and $\hat{\mathbf r}\cdot\mathbf r=r$, this is

The two derivatives cancel exactly:

The cancellation uses the precise inverse-square force. Repeating the calculation with a general $F(r)$ would not give the factor $k/r$ required here. Thus $\mathbf A$ is an extra conserved vector for this special force, not a property of every central-force orbit.

4. Direction and the orbit equation

First, $\mathbf A\cdot\mathbf L=0$: the cross-product term is perpendicular to $\mathbf L$, and $\hat{\mathbf r}\cdot\mathbf L=0$. Thus $\mathbf A$ lies in the orbital plane.

Next dot the definition with $\mathbf r$. The scalar triple-product rule allows a cyclic rearrangement:

Therefore

Let $\theta$ be the angle between $\mathbf A$ and $\mathbf r$, assuming $A=|\mathbf A|>0$. Then $Ar\cos\theta=\ell^2-\mu kr$, and

Comparison with the conic equation identifies

The eccentricity vector $\mathbf e$ points toward periapsis, where $\cos\theta=1$ and the radius is smallest. For an ellipse it lies along the major axis toward the nearer endpoint from the force center. For a circular orbit $A=0$, so a periapsis direction is undefined, as rotational symmetry requires.

5. Magnitude, energy, and independence

Square the definition:

Since $\mathbf p\cdot\mathbf L=0$, the first term is $p^2\ell^2$, where $p^2=\mathbf p\cdot\mathbf p$. The dot product in the last term is $\ell^2/r$, from the previous section. Hence

The bracket is energy, so

Dividing by $\mu^2k^2$ reproduces $e^2=1+2E\ell^2/(\mu k^2)$. Thus $E$, the three components of $\mathbf L$, and the three components of $\mathbf A$ are not seven independent constants. They satisfy $\mathbf A\cdot\mathbf L=0$ and the magnitude relation, leaving five independent orbital constants for a generic nonradial Kepler orbit. A position along that orbit at a reference time supplies the remaining initial information.

The fixed $\mathbf A$ explains why the apsidal direction does not precess in the exact inverse-square problem. It is related to the additional symmetry of the Kepler problem; a full algebraic symmetry treatment is beyond this course.

6. Applications

Example 1 · Eccentricity and orientation from one state

Use dimensionless units $\mu=k=1$. At an instant let $\mathbf r=(1,0,0)$ and $\mathbf p=(0,\sqrt{3/2},0)$. Then $\mathbf L=(0,0,\sqrt{3/2})$ and

Therefore $e=1/2$ and periapsis points along $+x$. Energy is $E=p^2/2-1=3/4-1=-1/4$, giving $a=2$. The current radius equals $a(1-e)=1$, confirming that this state is periapsis.

Example 2 · Recognizing an apoapsis

Again take $\mu=k=1$, but now $\mathbf r=(1,0,0)$ and $\mathbf p=(0,1/\sqrt2,0)$. Then $\mathbf L=(0,0,1/\sqrt2)$ and $\mathbf A=(-1/2,0,0)$. The eccentricity is still $1/2$, but periapsis is along $-x$. The particle is currently at apoapsis. Its energy is $1/4-1=-3/4$, so $a=2/3$ and $r_a=a(1+e)=1$.

The same radial direction can be periapsis or apoapsis depending on speed. The sign of the eccentricity vector resolves the distinction without plotting the whole orbit.

Example 3 · Deflection during an attractive encounter

Let a particle approach from infinity with speed $v_\infty$ and impact parameter $b_{\rm imp}>0$. Then $E=\mu v_\infty^2/2$ and $\ell=\mu b_{\rm imp}v_\infty$. The eccentricity relation gives

The asymptotic position directions are at $\theta=\pm\theta_\infty$, where $\cos\theta_\infty=-1/e$. Incoming velocity points inward along the incoming asymptote, whereas outgoing velocity points outward. The deflection between those velocity directions is therefore $\chi=2\theta_\infty-\pi$, with $0<\chi<\pi$.

Since $\sin(\chi/2)=\sin(\theta_\infty-\pi/2)=-\cos\theta_\infty=1/e$,

For $\mu=k=v_\infty=1$ and $b_{\rm imp}=2$, $\chi=2\arctan(1/2)\simeq53.13^\circ$. Larger impact parameter or higher speed reduces deflection. This is an application of the orbit constants, not a requirement to study scattering cross sections.

Example 4 · What changes when the force is perturbed?

Add a small radial force $\delta F(r)\hat{\mathbf r}$ to $-k\hat{\mathbf r}/r^2$. Angular momentum stays constant. In the derivative of $\mathbf A$, the inverse-square contribution still cancels the unit-vector derivative, leaving

This generally does not vanish. The osculating eccentricity direction can rotate or change magnitude, even though the orbital plane stays fixed. For an extra attractive inverse-cube term, the precessing solution in Chapter 18 demonstrates exactly this loss of a fixed periapsis direction. A nonzero instantaneous derivative need not imply secular rotation after every cycle; the accumulated effect depends on the perturbation.

7. The constants of the Kepler problem

The angular-momentum vector gives the orbital plane, its magnitude contributes the semi-latus rectum, energy fixes semimajor size for a bound orbit, and the Runge–Lenz vector gives eccentricity and periapsis direction. Their algebraic relations prevent double-counting independent information. Together they connect symmetry, dynamics, and the geometry of conic sections.

In a written conservation proof, retain the derivative of $\hat{\mathbf r}$, the triple-product expansion, and the cancellation. In a geometrical proof, retain the dot product with $\mathbf r$. Those two calculations explain both why the vector is constant and what it means.

Exercises

Hints and solutions are collected separately.

1. 22.1 Define the Runge–Lenz vector and state the cross-product order.

2. 22.2 Derive the time derivative of the radial unit vector.

3. 22.3 Prove conservation of $\mathbf A$ for inverse-square attraction.

4. 22.4 Prove $\mathbf A\cdot\mathbf L=0$.

5. 22.5 Derive the conic equation from $\mathbf A\cdot\mathbf r$.

6. 22.6 Prove the magnitude–energy relation for $\mathbf A$.

7. 22.7 With $\mu=k=1$, $\mathbf r=(2,0,0)$ and $\mathbf p=(0,1,0)$, find $\mathbf A$ and classify the orbit.

8. 22.8 What is the Runge–Lenz vector of a circular orbit?

9. 22.9 For an encounter with $k/(\mu b_{\rm imp}v_\infty^2)=1$, find the deflection angle.

10. 22.10 Why are $E,\mathbf L,\mathbf A$ not seven independent orbit constants?

11. 22.11 For an added radial perturbation $\delta F\hat{\mathbf r}$, find $\dot{\mathbf A}$.

12. 22.12 Explain how to reconstruct a Kepler orbit from one relative position and velocity.