Central forces are forces between two particles that depend only on the distance between the particles and point from one particle to another. Central forces are conservative (why?). Furthermore, if we look at a central force between two particles we can separate the motion of the center of mass from the relative motion of the particles reducing the complexity of the problem.
We have two particles of masses and located at positions and . Furthermore, let's define . Let's assume that the potential depends only on the relative position of the two particles. We can write out the Lagrangian as
where is the direction of the acceleration due to a uniform gravitational field.
Let's define to be the position of the center of mass
so
and
Let's write out the Lagrangian using the new coordinates
Notice that the middle terms in the first two sets of parentheses and the final terms in the final set
cancel each other out and we are left with
where is the total mass and the reduced mass is
So we find that for two bodies the dynamics reduces to the motion of the center of mass (possibly in a uniform gravitational field) and the motion of single particle about the center of mass.
Let's calculate the total angular momentum of the system. We know from the results on
many-particle systems that the angular momentum of the system will separate into the angular momentum about the center of mass and that of the center of mass. Furthermore, if , the internal forces do not contribute to the change in the angular momentum.
We have
where and .
We know from the results on many-particle systems in a uniform gravitational field that the internal angular momentum () of the two-body system is conserved. We could have seen this from
separation of the internal and external degrees of freedom in the Lagrangian.
We can treat the central force between two particles as a force between a fixed center that we take to be the origin and a single particle. We have
Taking the particular case of the results from the first section where the potential does not depend on direction, we know that the angular momentum of the particle is conserved. This means that the motion will be restricted to a plane that we will take to be the equatorial plane () without a loss of generality. Let's write the Lagrangian in spherical coordinates. First let's write the position of the particle in terms of the spherical coordinates. We have
and
so
Let's write out the Lagrangian in terms of the spherical coordinates
First let's notice that so is conserved. Also
so is conserved as well.
Let's calculate the conserved momentum,
which has units of angular momentum. It is simply the angular momentum about the axis.
We can also use Lagrange's equation to get the equations of motion
If and or initial, then and stays constant. We take and initially, so the particle remains in the equatorial plane as we expected from the conservation of angular momentum.
Now for the radial motion,
Using the initial conditions we have
where the dependence of the acceleration on the angle has been included through the conserved quantity
, separating the radial equation from the angular equation.
The first term is the force between the two bodies and the second term is a 'fictitous' force because of the spherical coordinates. We will solve this differential equation for two special cases, but we can get a good idea of the nature of the motion by looking at the conserved quantities.
First, we have
so the path subtends equal areas in equal times. Given a solution for the radial motion we can determine the angular motion
Second we have the Hamiltonian,
Because the Hamiltonian is conserved, let's take its initial value to be
We can solve this for to get
which is very similar to the one-dimensional case but with an added term for the centrifugal force. Unless
as with and is negative, motion of the the particle is bounded if . This point of closest approach is called the pericenter. Depending on the potential and the initial conditions, the motion may be bounded at large as well. In particular if the potential is attractive
and positive there is always a point of maximum distance. The point of maximum radius is called the apocenter.
If the motion is bounded at both large and small radius, we can calculate the time for the particle to go from the minimum to maximum radius -- this is one half of the period of the complete radial motion as we did for the one-dimensional pendulum. The angular motion as we shall find can have a different (and not usually comeasureable) period. We have
and rearranging we have
For a spherical harmonic oscillator,
Let's first calculate the bounds of the motion we have
which yields a quadratic equation for ,
which yields the solutions
We will first use the conservation of energy to calculate the radial motion, we have
To perform this integral let's make the substitution
to give
so
with
What remains is to solve for the motion in the direction. Using the conservation of angular momentum we have
We can integrate this directly to yield
Notice that the angular frequency is , one-half of the radial frequency.
If we substitute the values of and we get
It is possible to write as a function of and find that it is the equation for an ellipse centered at the origin. This is left as an exercise for the reader.
Let's step back and try to determine instead of . First we define so we have
and using the chain rule we get
We can substitute this into the energy equation to get
and rearranging
We add a constant to both sides of the equation to complete the square
We find that the derivative of a function squared plus four times the function itself yields a constant, so the function must be a sine or cosine. We have
We can write this in terms of the radius
This is the equation for an ellipse centered on the origin. The constant in front of the cosine is related to the shape of the ellipse
where is the square of the eccentricity of the ellipse ( and are the semimajor and semiminor axes). If and only if the spring constant is negative (repulsive spring), then the eccentricity can be greater than one (up to and the curve is a hyperbola centered on the origin. An interesting case is so . This gives a straight line passing a distance from the origin.
We can solve the spherical harmonic oscillator in Cartesian coordinates as well. We have for the Lagrangian
which yields the following equation of motion for the coordinate
and similarly for the other directions. We find the following solution for the motion in the direction
with and similarly for the other coordinates. Since the value of is the same for all three directions we find that the figure is an ellipse. We can also calculate the value of
where for simplicity and without loss of generality we have restricted the motion to the plane. We have
We can rewrite these two cosines as a single cosine,
where
which has the same form as the result for the spherical analysis.
For gravity we have
yielding the following Lagrangian,
Let's first calculate the bounds of the motion we have
which yields a quadratic equation for
which yields the solutions
For the case of an attractive force we find that
if , only is greater than zero and this represents the point of closest approach. On the other hand if
we have . For a repulsive force and again only is greater than zero if . If both and are negative, we cannot find a positive value of with .
We can also solve for the radial velocity
Yielding the following equation for the time evolution
and let's make the substitution
to give the following integral
where we have assumed that . This gives
and for we have
Unfortunately we cannot invert this to find the radius as a function of time, but we can determine, the time to go from the minimum to the maximum radius in the case
Let's step back and try to determine instead of . First we define so we have
and using the chain rule we get
We can substitute this into the energy equation to get
and rearranging
Let's define the length and simplify the equation
Let's add to both sides of the equation and get
We have a function squared plus its first derivative squared being a constant. The function is clearly a trigonometric function we have
where
Rearranging the equation gives
which is the polar equation for an ellipse with its focus at the origin for . For the curve is a hyperbola. is a parabola and is a circle. (Contrast this with the
spherical harmonic oscillator where the center of the ellipse was at the origin).
If we have a repulsive force (e.g. the force between two positive charges)
because . The resulting curve is a hyperbola.
The figure shows the various conic sections as a function of the value , the eccentricity. The blue parabola has , the aqua ellipse has and the green circle has . The red hyperbola has . All of the curves have a focus at the origin and . The semi-latus rectum is the official name for the quantity which sets the size of the curve. In all cases the curve intersects the axis at . In a coordinate free way, the semi-latus rectum is the distance between the curve and the focus in the direction perpendicular to the point of closest approach.
What is the meaning of the two branches of the hyperbola? The left branch is for an attactrive force and the right branch is for a repulsive force.