San José State University


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The Classical Analysis of a More General Case of the Two Body Problem 

Consider two particles, one of mass m and the other of a mass M. Let the distance of the first particle from the center of mass of the system be denoted by r and that of the second particle by R. Let s designate the separation distance of the two particles. The center of the mass of the system is such that
This means that
Note that
The expression mM/(m+M) is equal to the reduced mass μ of the system. The reduced mass is defined by
Therefore
Let the force between the particles be given as
where Q is the product of the force constant and the charges of the two particles.
The angular momentum of the system is:
Thus
The expression (L/μ) plays a crucial role in the further analysis and will be denoted as λ. Thus ω=λ/s².
The radial kinetic energy E of the system is
Since
and therefore the above expression for radial kinetic energy may rewritten as
Now consider the tangential kinetic energy ½m(rω)² + ½M(Rω)². This can be rewritten as
Note that since s²ω=λ, sω is equal to λ/s. Thus the tangential kinetic energy can be expressed as
Thus the total energy E can be expressed as
The above equation can be solved for (ds/dt)²; i.e.,
Note that
Therefore
This latter formula will be used later.
Now let u=1/s so
Then furthermore
The dynamic equations for the two particles are
Since mr=MR=μs the above two equations are the same when expressed in terms of s; i.e.,
When ω is replaced with λ/s² and the result divided through by μ the above equation becomes
The ratio Q/μ occurs often enough to justify it having its own symbol, say q.
Thus
Therefore
By defining w equal to (u−q/λ²) the above equation reduces to
which has the solution
This latter solution can be put into the form
where ε turns out to be the eccentricity of the elliptical orbit;.
By a suitable choice of the polar coordinate system; θ_{0} can be made equal to zero. Thus the solution for the system is
Note for later use that
The next step is to derive the velocities of each particle. v_{m} and v_{M}. where
From the solution for s
Since (1+ε·cos(θ))²s² is equal to (λ²/q)².
The tangential velocity is rω, but rω is equal to (M/(m+M))sω and sω is equal to λ/s. Thus
Let v²=μ(ds/dt)²+(sω)².
Combining the two terms for v² gives
Then
The probability density per unit path length for both particles are proportional to
This gives a probability density function with one maximum and one minimum. The constants of proportionality disappear in the process of normalization. The probability density has one point at which it is a maximum and another point at which it is a minimum. Here is an example for the case of eccentricity equal to 0.25. The horizontal axis is in terms of θ measured in radians.
(To be continued.)
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