I was reading a recent paper,

Capture into rydberg states and momentum distributions of ionized electrons. N. I. Shvetsov-Shilovski et al. Laser Physics 19, 1550 (2009).

which studies the ionization of an electron from a hydrogen atom by a strong low-frequency laser field, treating the electrons as classical particles under the influence of the laser field and the Coulomb potential of the nucleus.

Among other things, the paper presents an expression for the asymptotic momentum $\mathbf{p}$ of the electron, as a function of the position $\mathbf{r}$ and the momentum at the end of the laser pulse $\mathbf{q}$, which they state as

$\mathbf{p} = p \frac{p(\mathbf{A}\times\mathbf{M})-\mathbf{A}}{1+p^2M^2}$

in terms of the angular momentum $\mathbf{M} = \mathbf{r}\times\mathbf{q}$ and the Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{q}\times\mathbf{M}-\mathbf{r}/r$ of the electron.

Is this expression correct? How can it be proved, and is it possible to derive it in a way that explains its physical meaning?


1 Answer 1



The short answer is that the expression in the paper contains a misprint, and that it is indeed possible to prove it and to provide a construction that shows why it should have this form.

In the following, I will present an expanded derivation and geometric justification for the correct expression of the asymptotic momentum. This is based on an outline of the proof provided as a personal communication by one of the authors of the paper [1]. In this answer, I will first provide all the classical dynamics and geometry required to understand the derivation. If you are confident with this knowledge, the derivation can be found at the end of the answer.

Angular momentum

Defintion of angular momentum

Angular momentum is the rotational equivalent of linear momentum. It is defined as the cross product of the position and momentum vectors: $$\vec{M} = \vec{r}\times\vec{p}$$

What makes angular momentum relevant to this problem, is its conservation. In a closed system, the net angular momentum remains constant if the net torque on a rotating body is 0 (pg 18). A closed system is one which prevents the transfer of matter, energy or momentum in or out of the system.

Conservation of angular momentum

Let's now derive this conservation law by differentiating the angular momentum with respect to time.

$$\frac{d\vec{M}}{dt} = \frac{d}{dt}(\vec{r}\times\vec{p}) = \frac{d\vec{r}}{dt}\times p + \vec{r}\times\frac{dp}{dt} = \vec{v}\times m\vec{v} + \vec{r}\times \vec{F}$$

$\vec{v}$ and $\vec{mv}$ point in the same direction, because m is a scalar. Therefore, their cross product is 0. Hence,

$$\frac{d\vec{M}}{dt} = \vec{r}\times \vec{F} = \tau$$

The torque, $\tau$, in a closed system is always 0, because no external force acts upon the system. Therefore, angular momentum must be a conserved quantity since its time derivative is 0.

(A derivation for the conservation of angular momentum using the Lagrangian can be found in the appendix)

Laplace-Runge-Lenz vector

Definition of Laplace-Runge-Lenz vector

The Laplace-Runge-Lenz vector is an integral of motion (function of position and velocity which remains constant along orbits) and is independent of time. The vector remains constant only for fields with potential $U = -\frac{\alpha}{r}$. It is defined as: $\vec{A}$= $\vec{v}\times\vec{M} + \frac{\alpha\vec{r}}{r}$.

Geometric interpretation

The direction of the Laplace-Runge-Lenz vector is along the major axis from the focus to the perihelion. The easiest way to prove this, is by examining the case in which r is equal to the distance from the origin to the perihelion.

Consider the motion of a particle along a hyperbola.

Since $\vec{r}$ is defined as the vector from the focus to the point along a path, $\vec{r}$ will point along the major axis from the focus to the perihelion. Hence the term $\frac{\alpha\vec{r}}{r}$ will point in the same direction since $\alpha$ and $r$ are scalars.

The velocity always acts tangentially to the path, therefore it will point in the y axis. The angular momentum will always point in the z direction. Since, $\vec{v}\times\vec{M}$ will point in the direction perpendicular to both v and M, its direction will be along the x axis. Hence, the Laplace-Runge-Lenz vector points along the major axis from the focus to the perihelion.

Conservation of Laplace-Runge-Lenz vector

Firstly, to demonstrate the conservation of the Laplace-Runge-Lenz vector, we must show that the force due to a field with potential $U = -\frac{\alpha}{r}$ is $\frac{\alpha\vec{r}}{r^3}$.

Since $\vec{r} = (x, y, z)$, its magnitude is $\sqrt{x^2 + y^2 + z^2}$. To find the force, we must take the negative derivative of the potential, however, because $\vec{r}$ is a function of x, y and z, we must differentiate partially with respect to each component separately.

$$\frac{\partial U}{\partial x} = \alpha \frac{\partial}{\partial x}\left(\frac{1}{\sqrt{x^2 + y^2 + z^2}}\right) $$

To solve this, let's define some function m = $$x^2 + y^2 + z^2$$ and $n = m^{\frac{1}{2}}$ .$\frac{\partial m}{\partial x} = 2x$ and $\frac{\partial n}{\partial m} = \frac{m^{-\frac{1}{2}}}{2}$. Using the chain rule, $$ \frac{\partial m}{\partial x}\times\frac{\partial n}{\partial m} = \frac{\partial n}{\partial x} = \frac{\partial }{\partial x}\left(\sqrt{x^2 + y^2 + z^2}\right) = 2x\times\frac{1}{2}\left(\frac{1}{\sqrt{x^2 + y^2 + z^2}}\right) = \frac{x}{{\sqrt{x^2 + y^2 + z^2}}}. $$

Using the quotient rule, we get $$-\frac{\partial U}{\partial x} = \frac{\alpha x}{r^3}$$ By symmetry, it is the same for the y and z components.

Therefore, the force is equal to $\frac{\alpha \vec{r}}{r^3}$.

To show that the Laplace-Runge-Lenz vector is conserved, we must take its time derivative. (This part of the derivation is here (pg 39) , however all the steps are included below for complete understanding). $$\frac{d}{dt}(\vec{v}\times\vec{M}) = \dot{\vec{v}}\times\vec{M} $$ $$ \frac{d}{dt}(\frac{\alpha\vec{r}}{r}) = \dot{\vec{r}}\frac{\alpha}{r} + \vec{r}\frac{d}{dt}(\frac{\alpha}{r}) = \frac{\alpha\vec{v}}{r} - \frac{\alpha\vec{r}(\vec{v}\cdot\vec{r})}{r^3} $$ Hence, $$ \frac{dA}{dt} = \dot{\vec{v}}\times\vec{M} + \frac{\alpha\vec{v}}{r} - \frac{\alpha\vec{r}(\vec{v}\cdot\vec{r})}{r^3}$$

Substituting M = $m\vec{r}\times\vec{v}$ and using the vector triple product expansion rule : $$ (\vec{b} \times \vec{c})\times \vec{a} = (\vec{a} \cdot \vec{b})\vec{c} - (\vec{a} \cdot \vec{c})\vec{b}$$

$$ \frac{dA}{dt} = m\vec{r}(\vec{v}\cdot\dot{\vec{v}}) - m\vec{v}(\vec{r}\cdot\dot{\vec{v}}) + \frac{\alpha\vec{v}}{r} - \frac{\alpha\vec{r}(\vec{v}\cdot\vec{r})}{r^3} $$

From above, we know that $m\dot{\vec{v}} = \frac{\alpha \vec{r}}{r^3}$. Substituting $\dot{\vec{v}}$ for $\frac{\alpha \vec{r}}{r^3 m}$ we get:

\begin{align} \frac{dA}{dt} & = m\vec{r}\left(\vec{v}\frac{\alpha \vec{r}}{r^3 m}\right) - m\vec{v} \left(\vec{r}\frac{\alpha \vec{r}}{r^3 m}\right) + \frac{\alpha\vec{v}}{r} - \frac{\alpha\vec{r}(\vec{v}\cdot\vec{r})}{r^3} \\ & = \vec{r}\left(\vec{v}\frac{\alpha \vec{r}}{r^3 }\right) - \vec{v} \left(\vec{r}\frac{\alpha \vec{r}}{r^3 }\right) + \frac{\alpha\vec{v}}{r} - \frac{\alpha\vec{r}(\vec{v}\cdot\vec{r})}{r^3} \\ & = \frac{\alpha\vec{r}(\vec{v}\cdot\vec{r})}{r^3} - \vec{v} \left(\vec{r}\frac{\alpha \vec{r}}{r^3 }\right) + \frac{\alpha\vec{v}}{r} - \frac{\alpha\vec{r}(\vec{v}\cdot\vec{r})}{r^3} \\ & = - \frac{\alpha\vec{v}}{r^3}(\vec{r}\cdot\vec{r}) + \frac{\alpha\vec{v}}{r} \\ & = 0 \end{align}

Therefore, the Laplace-Runge-Lenz vector is a conserved quantity and is independent of time.

Scaling used in this paper

The scaling for this paper will be used according to this (pg215) document. The value of $\kappa$ is set to 1. Note, that there is a typo in the scaling of velocity; it should read $\kappa$.



In mathematics, an ellipse is a plane curve surrounding two focal points, such that for all points on the curve, the sum of the two distances to the focal points is a constant. As such, it generalizes a circle, which is the special type of ellipse in which the two focal points are the same.

An ellipse is obtained by cutting a piece of a cone using a plane which intersects the cone's vertical axis, but not perpendicularly.

In Cartesian coordinates, the equation for an ellipse is:

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 $$

In an ellipse, the ratio between the distance to the focus and the distance to the directrix (a line outside the focal point) is constant. This constant is the eccentricity, $e$, which is given by the equation: $e = \frac{c}{a}$.

$e$ is always between 0 and 1 in an ellipse.

The semi-latus rectum, $l$, is half the length of the perpendicular chord to one of the focal points. It is defined by: $l = \frac{b^2}{a} = a(1-e^2)$.

In polar coordinates, the equation for an ellipse is:

$$ r(\theta) = \frac{b}{\sqrt{1-(e \cos{\theta})^2}} $$

See here for the derivation.

If the origin is at one focus, where $\theta$ is measured from the major axis, the equation becomes:

$$ r(\theta) = \frac{a(1-e^2)}{1 + e\cos{\theta}} = \frac{l}{1 + e\cos{\theta}} $$


A Hyperbola is defined as a plane curve surrounding two focal points, such that for all points on the curve, the difference between the two distances to the focal points is a constant.

In Cartesian coordinates, the equation for a hyperbola is:

$$ \frac{x^2}{a^2}-\frac{y^2}{b^2} = 1 $$

As a conic section, a hyperbola is the intersection of a plane with both halves of a double cone.

The eccentricity is given by: $e = \sqrt{1+\frac{b^2}{a^2}}$.

Following the same pattern as with the Cartesian equations, the equation for a hyperbola in polar coordinates only differs from that of an ellipse due to a change in sign:

$$ r(\theta) = \frac{b}{\sqrt{(e \cos{\theta})^2 - 1}} $$

See here for the derivation. If the origin is at one focus:

$$ r(\theta) = \frac{a(1-e^2)}{1 - e\cos{\theta}} = \frac{l}{1 - e\cos{\theta}} $$

Derivation for path of Kepler's problem in polar coordinates

Kinetic energy in polar coordinates

Consider the motion of a point mass around a circle with radius r where the angle of rotation is $\theta$

The x and y components of $\vec{r}$ are r$\cos{\theta}$ and r$\sin{\theta}$ respectively. Differentiating x and y to find the velocities in each direction we get:

\begin{align} \dot{x} & = \dot{r}\cos{\theta} - r\dot{\theta}\sin{\theta} \\ \dot{y} & = \dot{r}\sin{\theta} + r\dot{\theta}\cos{\theta} \end{align}

Squaring both sides:

\begin{align} \dot{x}^{2} & = \dot{r}^2\cos{\theta}^2 + r^2\dot{\theta}^2\sin{\theta}^2 - 2r\dot{r}\dot{\theta}\sin{\theta}\cos{\theta} \\ \dot{y}^{2} & = \dot{r}^2\sin{\theta}^2 + r^2\dot{\theta}^2\cos{\theta}^2 + 2r\dot{r}\dot{\theta}\sin{\theta}\cos{\theta} \end{align}

Adding the squared velocities in each direction to obtain $v^2$:

\begin{align} \dot{x}^{2} + \dot{y}^{2} & = \dot{r}^2\cos{\theta}^2 + \dot{r}^2\sin{\theta}^2 + r^2\dot{\theta}^2\sin{\theta}^2 + r^2\dot{\theta}^2\cos{\theta}^2 \\ & =\dot{r}^2(\cos{\theta}^2 + \sin{\theta}^2) + r^2\dot{\theta}^2(\cos{\theta}^2 + \sin{\theta}^2) \\ & = \dot{r}^2 + r^2\dot{\theta}^2 \end{align}

Therefore, the kinetic energy $\frac{1}{2}mv^2$ becomes $\frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2)$.

Let's analyse this expression.$\frac{1}{2}m\dot{r}^2$ is the kinetic energy due to the radial velocity. However, the meaning behind the term $\frac{1}{2}mr^2\dot{\theta}^2$ is not so simple to understand. $\frac{1}{2}mr^2\dot{\theta}^2$ is the kinetic energy due to rotational motion. It is more commonly written as $\frac{1}{2}I\omega^2$ (I is the moment of intertia and $\omega$ is the angular velocity). This "kinetic energy" is actually named centrifugal potential energy. This is because, in order to move closer to the centre of the field, work must be done against the centrifugal force. Therefore potential energy is gained as it moves inwards at the expense of its rotational kinetic energy.

Derivation for path of motion in central field

Now that we can express kinetic energy in polar coordinates, the total mechanical energy can be written as:

$$E = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) + U(r)$$

Since M = $mr^2\dot{\theta} (mr^2 = I, \dot{\theta} = \omega), r^2\dot{\theta}^2$ can be written as $\frac{M^2}{mr^2}$ . Hence,

$$E = \frac{1}{2}m(\dot{r}^2 + \frac{M^2}{mr^2}) + U(r)$$

In order to find the equation of the path, we must find a function for $\theta$. To do this, let's isolate $\dot{r}$.

\begin{align} E - U(r) & = \frac{1}{2}m(\dot{r}^2 + \frac{M^2}{mr^2}) \\ \frac{2}{m}(E - U(r)) & = \dot{r}^2 + \frac{M^2}{mr^2} \\ \dot{r}^2 & = \frac{2}{m}(E - U(r)) - \frac{M^2}{mr^2} \\ \dot{r} & = \sqrt{\frac{2}{m}(E - U(r)) - \frac{M^2}{mr^2}} \end{align}

$\dot{r}$ can be written as $\frac{dr}{dt}$. Therefore,

$$ \int_{}{}dt = t = \int_{}{}\frac{dr}{\sqrt{\frac{2}{m}(E - U(r)) - \frac{M^2}{mr^2}}}$$

From the definition of angular momentum: $d\theta = M\frac{dt}{mr^2}$ (M= $I\omega$), we can integrate both sides to get $\theta = M\frac{t}{mr^2}$ and $t = \frac{\theta mr^2}{M}$. By substitution,

\begin{align} \frac{\theta mr^2}{M} & = \int_{}{}\frac{dr}{\sqrt{\frac{2}{m}(E - U(r)) - \frac{M^2}{mr^2}}} \\ \theta & = \int_{}{}\frac{M\frac{dr}{r^2}}{\sqrt{2m(E - U(r)) - \frac{M^2}{r^2}}} \end{align}

Now, let's solve this integral. Dividing both the numerator and denominator by M and transferring the $\frac{1}{r^2}$ to the denominator,

$$ \theta = \int_{}{}\frac{dr}{r^2\sqrt{\frac{2mE}{M^2}-\frac{2mU}{M^2}-\frac{1}{r^2}}}$$

This integral can be solved by substitution. Let $u = \frac{1}{r}$. Hence, $\frac{du}{dr} = -\frac{1}{r^2}$ and $dr = -du r^2$. Apllying the substitution,

\begin{align} \theta & = \int_{}{}\frac{-du r^2}{r^2\sqrt{\frac{2mE}{M^2}-\frac{2mU}{M^2}-u^2}} \\ & = -\int_{}{}\frac{du}{\sqrt{\frac{2mE}{M^2}-\frac{2mU}{M^2}-u^2}} \end{align}

Since $U = -\frac{\alpha}{r} = -\alpha u$,

$$ \theta = -\int_{}{}\frac{du}{\sqrt{\frac{2mE}{M^2}+\frac{2m\alpha }{M^2}u-u^2}} $$

To simplify this integral, let's complete the square of the quadratic $\frac{2mE}{M^2}+\frac{2m\alpha u}{M^2}-u^2$.

\begin{align} -u^2 + \frac{2m\alpha }{M^2}u + \frac{2mE}{M^2} & = -\left(u^2 - \frac{2m\alpha u}{M^2} - \frac{2mE}{M^2}\right) \\ & = -\left(\left(u- \frac{m\alpha }{M^2}\right)^2 - \frac{m^2\alpha^2 }{M^4} - \frac{2mE}{M^2}\right) \\ & = \frac{m^2\alpha^2+2mEM^2}{M^4} - \left(u- \frac{m\alpha }{M^2}\right)^2 \\ & = \left(\frac{\sqrt{m^2\alpha^2+2mEM^2}}{M^2}\right)^2 - \left(u- \frac{m\alpha }{M^2}\right)^2 \end{align}


$$ \theta = -\int_{}{}\frac{du}{\sqrt{\left(\frac{\sqrt{m^2\alpha^2+2mEM^2}}{M^2}\right)^2 - \left(u- \frac{m\alpha }{M^2}\right)^2}} $$

Inspecting the derivative of inverse cosine $\frac{d}{dx} \left( \arccos\left(\frac{x}{a}\right)\right) = -\frac{1} {\sqrt{a^2-x^2}}$, we can see that our integral takes the same form as $-\frac{1} {\sqrt{a^2-x^2}}$ where $a = \frac{\sqrt{m^2\alpha^2+2mEM^2}}{M^2}$ and $x = u- \frac{m\alpha }{M^2}$. Hence,

\begin{align} \theta & = \arccos\left(\frac{u- \frac{m\alpha }{M^2}}{\frac{\sqrt{m^2\alpha^2+2mEM^2}}{M^2}}\right) + C \\ & = \arccos\left(\frac{(u- \frac{m\alpha}{M^2})M^2}{\sqrt{m^2\alpha^2+2mEM^2}}\right)+ C \\ &= \arccos\left(\frac{uM^2- m\alpha}{\sqrt{m^2\alpha^2+2mEM^2}}\right)+ C \end{align}

Dividing both the numerator and denominator by M and replacing u with $\frac{1}{r}$,

$$ \theta = \arccos\left(\frac{\frac{M}{r}- \frac{m\alpha}{M}}{\sqrt{\frac{m^2\alpha^2}{M^2}+2mE}}\right)+ C $$

Formula for asymptotic momentum


For the duration of a laser pulse, the electron travels in both the laser field and Coulomb field. However, at the end of the laser pulse the electron is affected only by the Coulomb field. From the initial condition, $\vec{r}, \vec{q}$, which is the displacement vector and the momentum vector at the end of the laser pulse respectively, the asymptotic momentum (where $r \xrightarrow{}\infty$) can be calculated.

Value of semi-latus rectum and eccentricity from solution of path in Kepler's problem

Recall the general formula for the motion of a particle in a central field in polar coordinates:

$$\theta = \arccos\left({\frac{\frac{M}{r}- \frac{m\alpha}{M}}{\sqrt{\frac{m^2\alpha^2}{M^2}+2mE}}}\right)+ C$$

As the electron is liberated, the attractive Coulomb force from the proton results in the electron following the path of a hyperbola. Since the equation for the path along a hyperbola is $\frac{l}{r} = 1 + e\cos{\theta}$ we can rearrange the general equation to fit the one of a hyperbola. By making the origin a focus, C = 0. Taking cosine of both sides and transferring the denominator,

$$\sqrt{\frac{m^2\alpha^2}{M^2}+2mE}\cos{\theta} = \frac{M}{r}- \frac{m\alpha}{M}$$

Multiplying both sides by $\frac{M}{m\alpha}$,

$$\sqrt{1+\frac{2EM^2}{m\alpha^2}}\cos{\theta} = \frac{M^2/m\alpha}{r}$$

Hence, $l = \frac{M^2}{m\alpha}, e = \sqrt{1+\frac{2EM^2}{m\alpha^2}}$ ($l = M^2, e = \sqrt{1+2EM^2}$ in atomic units). Since the electron is liberated, E $>$ 0 and therefore e $>$ 1, further proving that the electron's path is along a hyperbola.

Relation between Runge-Lenz vector and eccentricity

A vital relation in order to derive the expression for the asymptotic momentum, is the Runge-Lenz vector's magnitude to the eccentricity, $A = e\alpha\ = e (a.u)$. To derive this, let's first take the dot product of the Runge-Lenz vector with itself:

\begin{align} \vec{A}\cdot\vec{A} & = \left(\vec{v}\times\vec{M} + \alpha\frac{\vec{r}}{r}\right) \cdot \left(\vec{v}\times\vec{M} + \alpha\frac{\vec{r}}{r}\right) \\ & = (\vec{v}\times\vec{M}) \cdot (\vec{v}\times\vec{M}) + 2 \alpha (\vec{v}\times\vec{M}) \cdot \left(\frac{\vec{r}}{r}\right) + \alpha^2 \left(\frac{\vec{r}}{r}\right) \cdot \left(\frac{\vec{r}}{r}\right) \end{align}

Since $\frac{\vec{r}}{r}$ is a unit vector and therefore magnitude 1,

$$A^2 = (\vec{v}\times\vec{M}) \cdot (\vec{v}\times\vec{M}) + 2 \alpha (\vec{v}\times\vec{M}) \cdot \left(\frac{\vec{r}}{r}\right) + \alpha^2$$

Using the cross product rule $(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d}) = (\vec{c}\cdot\vec{a})(\vec{b}\cdot\vec{d})-(\vec{b}\cdot\vec{c})(\vec{a}\cdot\vec{d})$, we can simplify the term, $$ (\vec{v}\times\vec{M})\cdot(\vec{v}\times\vec{M}) = (\vec{v}\cdot\vec{v})(\vec{M}\cdot\vec{M})-(\vec{M}\cdot\vec{v})(\vec{v}\cdot\vec{M}) = v^2M^2 -(\vec{M}\cdot\vec{v})(\vec{v}\cdot\vec{M}) $$

$\vec{M}$ and $\vec{v}$ are perpendicular, therefore their dot product is 0. Hence,

$$ (\vec{v}\times\vec{M})\cdot(\vec{v}\times\vec{M}) = v^2M^2 $$

Using another cross product rule $(\vec{u}\times\vec{v})\cdot\vec{w} = (\vec{v}\cdot\vec{w})\cdot\vec{u} = (\vec{w}\cdot\vec{u})\cdot\vec{v}$, we can simplify the term,

\begin{align} 2\alpha(\vec{v}\times\vec{M}) \cdot \left(\frac{\vec{r}}{r}\right) & = 2\alpha\left(\frac{\vec{r}}{r}\times\vec{v}\right) \cdot\vec{M} = \frac{2\alpha}{r}(\vec{r}\times\vec{v})\cdot\vec{M} \\ & = \frac{2\alpha}{rm}(\vec{r}\times m\vec{v})\cdot\vec{M} = \frac{2\alpha}{rm}\vec{M}\cdot\vec{M} \\ & = \frac{2\alpha M^2}{rm} \end{align}


$$A^2 = v^2M^2 + \frac{2\alpha M^2}{rm} + \alpha^2$$

Dividing by $\alpha^2$,

$$\frac{A^2}{\alpha^2} = \frac{v^2M^2}{\alpha^2} + \frac{2 M^2}{rm\alpha} + 1$$

Factorising by $\frac{2M^2}{m\alpha^2}$ we get,

$$\frac{A^2}{\alpha^2} = \frac{2M^2}{m\alpha^2}\left(\frac{m}{2}v^2 + \frac{\alpha}{r}\right) + 1$$

$(\frac{m}{2}v^2 + \frac{\alpha}{r})$ is kinetic + potential energy = E. Hence,

$$\frac{A^2}{\alpha^2} = 1 +\frac{2EM^2}{m\alpha^2} $$

Thererfore $\frac{A^2}{\alpha^2} = e^2, A^2 = \alpha^2e^2$.

Magnitude of asymptotic momentum

At $t\xrightarrow{}\infty$, r$\xrightarrow{}\infty$. Therefore the potential$\xrightarrow{}0$. By conservation of total energy (written in atomic units), $$E = \frac{q^2}{2} - \frac{1}{r} = \frac{p^2}{2}$$, where p is the magnitude of the asymptotic momentum.

Asymptotic momentum equation

Now that we have the magnitude of the asymptotic momentum, all that is left is to find the direction of its vector. This can be done by examining the asymptotes of the hyperbola.

Three orthogonal vectors $\vec{e_1},\vec{e_2},\vec{e_3}$ describe the orientation of the hyperbola and are defined as:

$$ \vec{e_1} = \frac{\vec{A}}{A}, \qquad \vec{e_2} = \frac{\vec{A\times M}}{AM}, \qquad \vec{e_3} = \frac{\vec{M}}{M} $$

The vector $\vec{e_1}$ is the unit vector of the Runge-Lenz vector. Hence, it points from the origin to the perihelion. The vector $\vec{e_3}$ is the unit vector of the angular momentum, which acts in the z-axis. The vector $\vec{e_2}$ is defined as the cross product of the vector $\vec{e_1}$ and the vector $\vec{e_3}$.

In order to obtain the asymptotic momentum, it is necessary to find a vector in the direction of the asymptotes. In the diagram, this vector is $\vec{g}$. g is the hypotenuse of the right-angled triangle formed by a and b. $\vec{g}$ has a horizontal component (in the direction of -e1) of magnitude a and vertical component (in the direction of e2) of magnitude b. Hence, $\vec{g}$ can be written as,

$$\vec{g} = b\vec{e_2}-a\vec{e_1}$$

Therefore, we have an expression for the asymptotic momentum,

$$\vec{p} = p\hat{g} = p\frac{b\vec{e_2}-a\vec{e_1}}{\sqrt{a^2+b^2}}$$

Using the substitutions,

\begin{align} a & = \frac{l}{e^2-1} = \frac{M^2}{2EM^2} = \frac{1}{2E} \\& = \frac{1}{p^2}(a.u) \\ b & = \frac{l}{\sqrt{e^2-1}} = \frac{M^2}{\sqrt{2EM^2}} = \frac{M}{\sqrt{2E}} \\ & = \frac{M}{p}(a.u) \end{align}

we can transform the formula for the asymptotic momentum into,

\begin{align} \vec{p} & = p\frac{\frac{M}{p}\frac{(\vec{M}\times \vec{A})}{MA}-\frac{1}{p^2}\frac{\vec{A}}{A}}{\sqrt{\frac{1}{p^4}+\frac{M^2}{p^2}}} = p\frac{\frac{(\vec{M}\times \vec{A})}{pA}-\frac{\vec{A}}{p^2A}}{\sqrt{\frac{1}{p^4}+\frac{M^2}{p^2}}} = p\frac{\frac{p(\vec{M}\times \vec{A})-\vec{A}}{p^2A}}{\sqrt{\frac{1}{p^4}+\frac{M^2}{p^2}}} \\ & = p\frac{p(\vec{M}\times \vec{A})-\vec{A}}{A\sqrt{1+M^2p^2}} \end{align}

Since $A = e = \sqrt{1+2EM^2} = \sqrt{1+p^2M^2}$ (a.u),

$$ \vec{p} = p\frac{p(\vec{M}\times \vec{A})-\vec{A}}{1+M^2p^2}(a.u) $$


Conservation of angular momentum using the Lagrangian

First, we introduce a new function, the Lagrangian. The Lagrangian is equal to the difference between the total potential energy and kinetic energy of a system. If a system is rotated by angle $\phi$, the Lagrangian remains unchanged. After all, it's a scalar, and shouldn't change due to a vector rotation. The vector $\vec{\delta\phi}$ represents the infinitesimal rotation of the system. It's direction is that of the axis of rotation, and magnitude $\delta\phi$ is the angle of rotation of the system in radians.

Figure 5 depicts the the infinitesimal rotation of a system, where O is the origin, and $\vec{r}$is the position vector of the particle. $\vec{\delta r}$ is the change in $\vec{r}$ during the system's rotation. The trajectory of the particle as it's rotated maps out a cone around $\vec{\delta\phi}$ as it rotates 2$\pi$ around O. The radius of the circular trajectory is $r\sin{\theta}$. Since, in radians, arc length is the radius multiplied by the angle in a sector, $|\vec{\delta r}| = r\sin{\theta}\vec{\delta\phi}$. The direction of $\vec{\delta r}$ is tangential to $\vec{r}$, creating a angle of $\pi/2$ with it. The plane $\vec{\delta \phi}$ , $\vec{r}$ bisects $\vec{\delta r}$. Therefore, $\vec{\delta r}$ is also perpendicular to $\vec{\delta \phi}$.

Since, by definition, the cross product of two perpendicular vectors produces a vector perpendicular to both original vectors, we can construct the equation: $\vec{\delta r} = \vec{\delta\phi} \times \vec{r}$ (1). We can show the same for velocity by differentiating both sides with respect to time. Since we take the same two snapshots in time, $\vec{\delta\phi}$ is constant. Hence, we get $\vec{\delta v} = \vec{\delta\phi} \times \vec{v}$ (2).

The equation for the Lagrangian is $L = T-V$, where $T$ is the kinetic energy and $V$ is the potential energy of the system. $L$ is a function of both ${\vec{r}}_a$ and ${\vec{v}}_a$. The infinitesimal change of a function f(x,y) is as follows: $\delta f = (\frac{\partial f}{\partial x})\delta x + (\frac{\partial f}{\partial y})\delta y$.

Using the fact that the Lagrangian remains unchanged, we can write:

$$ \delta L = \sum_{a}^{}\left(\frac{\partial L}{\partial \vec{r_a}} \cdot \delta\vec{r_a} + \frac{\partial L}{\partial \vec{v_a}} \cdot \delta\vec{v_a}\right) = 0 $$ ($a$ represents the sum over every particle in the system)

We know $L = T - V$. $T$ is a function of $p$ and $V$ is a function of $r$. Hence, $\frac{\partial L}{\partial \vec{r_a}} = {\dot{{\vec{p}}}}_a$ and $\frac{\partial L}{\partial \vec{v_a}} = {\vec{p}}_a$. Substituting using (1) and (2):

$$\sum_{a}^{}(\dot{\vec{p_a}} \cdot \vec{\delta\phi} \times \vec{r} + \vec{p_a} \cdot \vec{\delta\phi} \times \vec{v_a}) = 0 $$

Factorising $\vec{\delta\phi}$ and swapping the cross products to then divide by -1, we get:

$$\vec{\delta\phi}\sum_{a}^{}(\vec{r_a} \times \dot{\vec{p}} + \vec{v_a} \times \vec{p_a}) = 0$$

Since v is the derivative of r:

$$\vec{\delta\phi}\sum_{a}^{}(\vec{r_a} \times \dot{\vec{p}} + \dot{\vec{r_a}} \times \vec{p_a}) = 0 $$

It is now clear to see that the sum is the expansion of a time derivative using the product rule:

$$\vec{\delta\phi} \cdot \frac{d}{dt} \sum_{a}^{}(\vec{r_a} \times \vec{p_a}) = 0 $$

As $\vec{\delta\phi}$ is arbitrary, $\frac{d}{dt} \sum_{a}^{}(\vec{r_a} \times \vec{p_a}) = 0$ and therefore the angular momentum of the system is conserved as its time derivative is 0. This proof is more concisely found here (18-19)


[1]. N. Shvetsov-Shilovsky, "Asymptotic momentum". Personal communication (2022)


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