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\begin{document}
\begin{center}
{\large The Kepler Problem Revisited: \\
The Laplace--Runge--Lenz Vector\\ }
{\large Classical Mechanics Homework\\}
{\small March 17, 2$\infty$8 \\}
{\small John Baez}
{\small homework by C.Pro}
\end{center}
Whenever we have two particles interacting by a central force in 3d
Euclidean space, we have conservation of energy, momentum, and angular
momentum. However, when the force is gravity --- or more precisely,
whenever the force goes like $1/r^2$ --- there is an extra conserved
quantity. This is often called the {\bf Runge--Lenz vector}, but it
was originally discovered by Laplace. Its existence can be seeen in
the fact that in the gravitational 2-body problem, each particle
orbits the center of mass in an ellipse (or parabola, or hyperbola)
whose {\it perihelion does not change with time}. The Runge--Lenz
vector points in the direction of the perihelion! If the force went
like $1/r^{2.1}$, or something like that, the orbit could still be
roughly elliptical, but the perihelion would `precess' --- that is,
move around in circles.
\vskip1em
\noindent
Indeed, the first piece of experimental evidence that Newtonian gravity
was not quite correct was the precession of the perihelion of Mercury. Most
of this precession is due to the pull of other planets and other
effects, but about 43 arcseconds per century remained unexplained until
Einstein invented general relativity.
\vskip1em
\noindent
In fact, we can use the Runge--Lenz vector to simplify the proof that
gravitational 2-body problem gives motion in ellipses,
hyperbolas or parabolas. Here's how it goes. As before, let's
work with the relative position vector
\[ \q(t) = \q_1(t) - \q_2(t) \]
where $\q_1, \q_2 \maps \R \to \R^3$ are the positions of the
two bodies as a function of time. In what follows we will use
$q$ to stand for the magnitude of the vector $\q$, and $\hat \q$
to stand for a normalized vector that points in the direction of
$\q$:
\[ q = |\q|, \qquad \hat \q = {\q \over q} .\]
In a previous homework we saw that
$\q(t)$ satisfies
\[ m \ddot \q= f(q) \hat \q \]
where $f$ is the force as a function of distance, and
$m$ is the so-called `reduced mass'. Since the force of
gravity goes like $1/r^2$, we have
\[ f(q) = -k/q^2 \]
where $k$ is the same constant as in the previous homework.
We thus have
\be m \ddot \q = -k \hat \q/ q^2 \label{F=ma} \ee
and this will be our starting-point for all that follows.
\vskip1em
\noindent
1. Define the angular momentum vector $\J$ in
the usual way:
\[ \J = m \q \times \dot \q \]
We know from our previous work that angular momentum is
conserved:
\[ \dot \J = 0 . \]
Take the cross product of both sides of equation (\ref{F=ma})
with the vector $\J$. Simplify the right-hand side and show that
\be \ddot \q \times \J = k \dot {\hat \q} \label{qJ} \ee
\textit{Solution: } First we compute the time derivative of $\hat{\q}$.
Let $\q_i/q$ be the $i^{\mathrm{th}}$ component of $\hat\q$ and write
$q=\sqrt{\q\cdot\q}$. Then use the quotient rule to obtain
\ban
\frac{d}{dt}{\q_i\over q} &=&{ \dot{\q_i}q- \q_i\left({\q\cdot\dot\q\over q}\right)\over q^2}\\
&=&{\dot \q_i(\q \cdot \q) - (\q \cdot \dot \q) \cdot \q_i \over q^3}
\ean
and so, $$\dot{\hat{\q}} = {\dot \q(\q \cdot \q) - (\q \cdot \dot \q) \cdot \q \over q^3}$$
Now, using the identity
\[ \a \times (\b \times \c) = (\a \cdot \c) \b - (\a \cdot \b) \c, \]
we have
\ban
m\ddot\q\times J &=& -mk \hat \q/ q^2 \times(m\q \times \dot \q)\\
&=& {mk\over q^3} \left((\q \cdot \q)\dot\q - (\q\cdot\dot\q)\q\right)\\
&=& mk\dot{\hat\q},
\ean
and dividing both sides by $m$ we obtain the result.
\vskip1em
\noindent
2. Use parts 1 to show that
\[ {d\over dt}(\dot \q \times \J) = k \dot{ \hat \q} .\]
\textit{Solution :}
\ban
{d\over dt}(\dot \q \times \J) &=& \ddot \q \times \J + \q\times\dot \J\\
&=& k\dot{\hat\q} + \q\times 0\\
&=& k\dot{\hat\q}.
\ean
\vskip1em
\noindent
3.
Use part 2 to show that
\[ \dot \q \times \J = k \hat \q + \x \]
for some vector $\x \in \R^3$ that is {\it independent of time.}
\textit{Solution: } Note 2 says
$${d\over dt}((\dot \q \times \J) - k\hat \q)=0$$
and so the two functions must only differ by some constant vector, say $x$.
\vskip 1em
\noindent
It will be handy to divide this vector by $k$, obtaining the
{\bf Runge--Lenz vector}:
\[ \A = {\x \over k} \]
which clearly is also independent of time.
In other words, the Runge--Lenz vector
\be \A = {\dot \q \times \J \over k} - \hat \q \label{RL} \ee
is a conserved quantity for the Kepler problem:
\[ \dot \A = 0 .\]
\vskip1em
\noindent
4. Use equation (\ref{RL}) to show that
\be \A \cdot \q = {\J \cdot \J \over km} - q . \label{Aq} \ee
\textit{Solution: } Using the identity $(a\times b)\cdot c =(c\times a)\cdot b$
we have
\ban
\A\cdot\q &=& {\dot \q \times \J \over k}\cdot\left({m\q\over m}\right) - \hat \q\cdot \q\\
&=& {(m\q\times \dot\q)\cdot J\over mk} - {q^2\over q}\\
&=& {J\cdot J\over mk} -q
\ean
as desired.
\vskip1em
\noindent
5. Now write
\[ \A \cdot \q = Aq \cos \theta \]
where $A = |\A|$ is the magnitude of the Runge--Lenz vector
and $\theta$ is the angle between $\q$ and $\A$.
Using equation (\ref{Aq}), show that
\[ q = {\J \cdot \J \over km} {1 \over 1+ A \cos \theta } .\]
\textit{Solution: } From equation (\ref{Aq})
$$ \A \cdot \q +q = {\J \cdot \J \over km} $$
so
$$q( A\cos\theta +1) = {\J \cdot \J \over km}$$
and therefore
\[ q = {\J \cdot \J \over km} {1 \over 1+ A \cos \theta } .\]
\vskip 2em
\noindent
{\textit{\textbf{Now, let me explain
what you have achieved!}}}
\vskip 2em
\noindent
The above equation looks almost like this equation in our previous
homework about the Kepler problem:
\[ r = {p \over 1 + e \cos (\theta - \theta_0)} . \]
And indeed, they are really just different ways of writing
the same equation, except that now we
have rotated our polar coordinate system so that $\theta_0 = 0$.
Another way of saying this is that in these coordinates,
$\theta$ is zero at the perihelion of the orbit.
So, you've just given a new proof
that the orbit in the Kepler problem
must be an ellipse, parabola or hyperbola! Moreover,
comparing the two equations above we see that
\[ e = A , \]
so {\it the magnitude of the Runge--Lenz vector is the
eccentricity of the orbit}. We also see that
{\it the Runge--Lenz vector points in the direction of the orbit's
perihelion}. Thus the conservation of the Runge--Lenz vector
is just a way of saying the eccentricity and perihelion don't
change with time! Finally, we see that
\[ p = {\J \cdot \J \over km} \]
which we already saw last time.
\vskip1em
\noindent
The story of the Runge--Lenz vector goes much
deeper than this, and I encourage you to look at my webpage about it,
and also the excellent Wikipedia article:
\begin{itemize}
\item
John Baez, The mysteries of the gravitational 2-body problem, \hfill \break
\href{http://math.ucr.edu/home/baez/gravitational.html}{http://math.ucr.edu/home/baez/gravitational.html}
\item
Wikipedia, Laplace--Runge--Lenz vector, \hfill \break
\href{http://en.wikipedia.org/wiki/Laplace-Runge-Lenz_vector}{http://en.wikipedia.org/wiki/Laplace-Runge-Lenz$\underline{\;}$vector}
\end{itemize}
\noindent
Here's the story in a nutshell. First,
the Runge--Lenz vector is conserved not only
in the classical, but also in the {\it quantum-mechanical}
theory of two particles attracted by an inverse square force law.
The main example is the {\it hydrogen atom}. The surprising
degeneracy of energy levels for the hydrogen atom, also important
for the periodic table of elements, is special to an inverse square
force law and due to the conservation of the Runge--Lenz vector.
\vskip 1em
\noindent
Mathematically, conservation of angular momentum is due to rotation
symmetry: symmetry under the group $\SO(3)$. Any central force
has this symmetry. But the inverse square force law actually has
symmetry under a bigger group, $\SO(4)$ or $\SO(3,1)$.
$\SO(4)$ is the rotational group in 4 dimensions, while $\SO(3,1)$
is the Lorentz group. These are both 6-dimensional groups, so
they give 6 conserved quantities: the 3 components of $\J$ and
the three components of $\K$, which is a rescaled version of the
Runge--Lenz vector:
\[
\K = k \sqrt{\frac{m}{2|H|}} \A
\]
where $H$ is the Hamiltonian. This new vector $\K$ is conserved
because $\A$ and $H$ are.
\vskip 1em
\noindent
To see this in detail, one can calculate that
\[ \{ J_1, J_2 \} = J_3 \qquad \textrm{and cyclic permutations} \]
\[ \{ J_1, K_2 \} = K_3 \qquad \textrm{and cyclic permutations} \]
and, oddly,
\[ \{ K_1, K_2 \} = \pm K_3 \qquad \textrm{and cyclic permutations} \]
where the sign is always $+$ if the energy is negative, and
$-$ if the energy is positive. (Here by `and cyclic permutations',
I mean we can cyclically permute the indices $1,2,3$ and get other true
equations.)
\vskip 1em
\noindent
This means that if we restrict to
$$ X_- = \{ (q,p) \in \R^3 \times \R^3 \colon H(q,p) < 0 \} $$
we get the above equations with the plus sign, and these are precisely
the formulas for the Lie bracket in $\so(4)$. If we restrict to
$$ X_+ = \{ (q,p) \in \R^3 \times \R^3 \colon H(q,p) < 0 \} $$
we get the above equations with the minus sign, which are the formulas
for the Lie bracket in $\so(3,1)$.
\vskip 1em
\noindent
The spaces $X_-$ and $X_+$ are
Poisson manifolds in their own right. $X_-$ is the phase space for
particles in elliptical orbits, while $X_+$ is the phase space for
particles in hyperbolic orbits. With some work, one can show that
$\SO(4)$ acts on $X_-$ and $\SO(3,1)$ acts on $X_+$. Both actions
are Hamiltonian, and the corresponding observables are conserved
quantities: just $\J$ and $\K$.
\vskip 1em
\noindent
In short: {\it the problem of a particle
in an elliptical orbit in gravity has $\SO(4)$ symmetry!}
But {\it why?} What does gravity have to do with 4d rotations?
An explanation was given by the Russian physicist Fock in 1935.
\vskip 1em
\noindent
Fock showed that $X_-$ is isomorphic, as a Poisson manifold,
to $T^* S^3$ --- the phase space for a particle on the sphere
$S^3 \subseteq \R^4$. Even better, the whole problem of a particle in an
elliptical orbit in an inverse-square force law is {\it isomorphic
to the problem of a free particle on $S^3$!} The latter problem
has an {\it obvious} symmetry under $\SO(4)$.
\vskip 1em
\noindent
Of course, all this is puzzling in its own right. It goes to show
that even the most classic of classical mechanics problems still holds
mysteries! Here is a good book on the subject:
\begin{itemize}
\item
Victor Guillemin and Shlomo Sternberg,
{\sl Variations on a Theme by Kepler}, American Mathematical Society,
Providence, Rhode Island, 1990.
\end{itemize}
\end{document}