Advanced Classical Mechanics/Central Forces

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.

Separation of MotionEdit

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  

LagrangianEdit

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.

Angular MomentumEdit

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.

General ResultsEdit

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

 

LagrangianEdit

Let's write out the Lagrangian in terms of the spherical coordinates

 

Conserved QuantitiesEdit

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.


Equations of MotionEdit

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.

Bounded MotionEdit

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

 

Specific ResultsEdit

Spherical Harmonic OscillatorEdit

Central Force SolutionEdit

For a spherical harmonic oscillator,  

Bounds of MotionEdit

Let's first calculate the bounds of the motion we have

 

which yields a quadratic equation for  ,

 

which yields the solutions

 

Radial MotionEdit

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

 

Angular MotionEdit

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.

The Shape of the OrbitEdit

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.

Cartesian SolutionEdit

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.

GravityEdit

For gravity we have

 

yielding the following Lagrangian,

 

Bounds of the MotionEdit

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  .

Radial MotionEdit

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

 

The Shape of the OrbitEdit

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.