Editing Laplace–Runge–Lenz vector (section)

== Definition ==
An inverse-square central force acting on a single particle is described by the equation 
<math display="block">
\mathbf{F}(r)=-\frac{k}{r^{2}}\mathbf{\hat{r}};
</math>
The corresponding [[potential energy]] is given by <math>V(r) = - k / r</math>. The constant parameter {{mvar|k}} describes the strength of the central force; it is equal to {{math|[[Gravitational constant|''G'']]''Mm''}} for gravitational and {{math|−{{sfrac|1|4''π''[[Vacuum electric permittivity|''ε''<sub>0</sub>]]}}''Qq''}} for electrostatic forces. The force is attractive if {{math|''k'' > 0}} and repulsive if {{math|''k'' < 0}}.

[[File:Laplace Runge Lenz vector.svg|thumb|right|400px|Figure 1: The LRL vector {{math|'''A'''}} (shown in red) at four points (labeled 1, 2, 3 and 4) on the elliptical orbit of a bound point particle moving under an inverse-square central force. The center of attraction is shown as a small black circle from which the position vectors (likewise black) emanate. The angular momentum vector {{math|'''L'''}} is perpendicular to the orbit. The coplanar vectors {{math|'''p''' × '''L'''}} and {{math|(''mk''/''r'')'''r'''}} are shown in blue and green, respectively; these variables are defined [[#Mathematical definition|below]]. The vector {{math|'''A'''}} is constant in direction and magnitude.]]

The LRL vector {{math|'''A'''}} is defined mathematically by the formula<ref name="goldstein_1980" />
{{Equation box 1
|indent =:
|equation = <math> \mathbf{A} = \mathbf{p} \times \mathbf{L} - m k \mathbf{\hat{r}},</math>
|cellpadding= 6
|border
|border colour = #0073CF
|background colour=#F9FFF7}}
where
* {{mvar|m}} is the [[mass]] of the point particle moving under the central force,
* {{math|'''p'''}} is its momentum vector,
* {{math|1='''L''' = '''r''' × '''p'''}} is its angular momentum vector,
* {{math|'''r'''}} is the position vector of the particle (Figure 1),
* <math>\mathbf{\hat{r}}</math> is the corresponding [[unit vector]], i.e., <math>\mathbf{\hat{r}} = \frac{\mathbf{r}}{r}</math>, and
* {{mvar|r}} is the magnitude of {{math|'''r'''}}, the distance of the mass from the center of force.

The [[International System of Units|SI units]] of the LRL vector are joule-kilogram-meter (J⋅kg⋅m). This follows because the units of {{math|'''p'''}} and {{math|'''L'''}} are kg⋅m/s and J⋅s, respectively. This agrees with the units of {{mvar|m}} (kg) and of {{mvar|k}} (N⋅m<sup>2</sup>).

This definition of the LRL vector {{math|'''A'''}} pertains to a single point particle of mass {{mvar|m}} moving under the action of a fixed force. However, the same definition may be extended to [[two-body problem]]s such as the Kepler problem, by taking {{mvar|m}} as the [[reduced mass]] of the two bodies and {{math|'''r'''}} as the vector between the two bodies.

Since the assumed force is conservative, the total energy {{mvar|E}} is a constant of motion,
<math display="block">
E = \frac{p^{2}}{2m} - \frac{k}{r} = \frac{1}{2} mv^{2} - \frac{k}{r}.
</math>

The assumed force is also a central force. Hence, the angular momentum vector {{math|'''L'''}} is also conserved and defines the plane in which the particle travels. The LRL vector {{math|'''A'''}} is perpendicular to the angular momentum vector {{math|'''L'''}} because both {{math|'''p''' × '''L'''}} and {{math|'''r'''}} are perpendicular to {{math|'''L'''}}. It follows that {{math|'''A'''}} lies in the plane of motion.

Alternative formulations for the same constant of motion may be defined, typically by scaling the vector with constants, such as the mass {{math|''m''}}, the force parameter {{math|''k''}} or the angular momentum {{math|''L''}}.<ref name="goldstein_1975_1976" /> The most common variant is to divide {{math|'''A'''}} by {{math|''mk''}}, which yields the eccentricity vector,<ref name="taff_1985" /><ref name="hamilton_1847_quaternions" /> a [[dimensionless quantity|dimensionless]] vector along the semi-major axis whose modulus equals the eccentricity of the conic:
<math display="block">
\mathbf{e} = \frac{\mathbf{A}}{m k} = \frac{1}{m k}(\mathbf{p} \times \mathbf{L}) - \mathbf{\hat{r}}.
</math>
An equivalent formulation<ref name="arnold_1989b" /> multiplies this eccentricity vector by the major semiaxis {{mvar|a}}, giving the resulting vector the units of length. Yet another formulation<ref name="symon_1971b">{{cite book | last=Symon | first=K. R. | author-link=Keith Symon | date=1971 | title=Mechanics | edition=3rd | publisher=Addison Wesley | pages=130–131}}</ref> divides {{math|'''A'''}} by <math>L^2</math>, yielding an equivalent conserved quantity with units of inverse length, a quantity that appears in the solution of the Kepler problem
<math display="block">
u \equiv \frac{1}{r} = \frac{km}{L^2} + \frac{A}{L^2} \cos\theta
</math>
where <math>\theta</math> is the angle between {{math|'''A'''}} and the position vector {{math|'''r'''}}. Further alternative formulations are given [[#Alternative scalings, symbols and formulations|below]].