San José State University

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

Classical Analysis of a 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

mr = MR
and
r + R = s

This means that

r + (m/M)r = s
so
r = (M/(m+M))s
and likewise
R = (m/(m+M))s

Note that

mr = (mM/(m+M))s

The expression mM/(m+M) is equal to the reduced mass μ of the system. The reduced mass is defined by

1/μ = 1/m + 1/M

Therefore

mr = μs
and likewise
MR = μs

Force, Energy and Angular Momentum

Let the force between the particles be given as

−Q/s²
and the potential
energy as
− Q/s

where Q is the product of the force constant and the charges of the two particles.

The angular momentum of the system is:

L = mr²ω + MR²ω
L = [mM²/(m+M)²]s²ω + [m²M/(m+M)²]s²ω
L = [mM/(m+M)][(m+M)/(m+M)]s²ω = [mM/(m+M)]s²ω

Thus

L = μs²ω
and hence
ω = (L/μ)/s²

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

½m(dr/dt)² + ½M(dR/dt)²

Since

mr = MR = μs
it follows that
m(dr/dt) = M(dR/dt) = μ(ds/dt)

and therefore the above expression for radial kinetic energy may rewritten as

½(mdr/dt)²/m + ½(MdR/dt)²/M
and hence as
½(μds/dt)²/m + ½(μds/dt)²/M
or, equivalently
½(μds/dt)²(1//m + 1/M)
and finally
½μ(ds/dt)²

Now consider the tangential kinetic energy ½m(rω)² + ½M(Rω)². This can be rewritten as

½(mrω)²/m + ½(MRω)²/M
and further to
½(μsω)²(1/m + 1/M)
and hence to
½μ(sω)²

Note that since s²ω=λ, sω is equal to λ/s. Thus the tangential kinetic energy can be expressed as

½μλ²/s²

Thus the total energy E can be expressed as

E = ½μ(ds/dt)² + ½μλ²/s² −Q/s

The above equation can be solved for (ds/dt)²; i.e.,

(ds/dt)² = 2(E+Q/s)/μ − λ²/s²

Making θ the Independent Variable

Note that

(ds/dθ) = (ds/dt)/(dθ/dt)
but (dθ/dt)=ω=λ/s² so
(ds/dθ) = (ds/dt)(s²/λ)
and hence
(ds/dθ)² = (ds/dt)²(s4/λ²)

Therefore

(ds/dθ)² = (s4/λ²)[2(E+Q/s)/μ] − s²
which reduces to
(ds/dθ)² = s4(E+Q/s)/(μλ²) − s²

This latter formula will be used later.

Development of a Simple Differential Equation

Now let u=1/s so

(du/dθ) = −(1/s²)(ds/dθ)
and since (ds/dθ)=(ds/dt)(s²/λ)
(du/dθ) = −(1/s²)(ds/dt)/λ

Then furthermore

(d²u/dθ²) = d/dt(du/dθ)/(dθ/dt) = (d/dt(−(ds/dt)/λ)/ω
(d²u/dθ²) = −[(d²s/dt²)/λ]/(λ/s²)
(d²u/dθ²) = −(d²s/dt²)/(u²λ²)

The dynamic equations for the two particles are

m(d²r/dt²) + mrω² = −Q/s²
M(d²R/dt²) + MRω² = −Q/s²

Since mr=MR=μs the above two equations are the same when expressed in terms of s; i.e.,

μ((d²s/dt²) + μsω² = −Q/s²

When ω is replaced with λ/s² and the result divided through by μ the above equation becomes

(d²s/dt²) + λ²/s³ = −(Q/μ)/s²

The ratio Q/μ occurs often enough to justify it having its own symbol, say q.

Thus

(d²s/dt²) = −qu² + λ²u³

Therefore

(d²u/dθ²) = − (d²s/dt²)/(u²λ²) = q/λ² − u

By defining w equal to (u−q/λ²) the above equation reduces to

(d²w/dθ²) + w = 0

which has the solution

w = W·cos(θ−θ0)
where W is
a constant
and hence
u = W·cos(θ−θ0) + q/λ²
and thus
s = 1/(W·cos(θ−θ0) + q/λ²)

This latter solution can be put into the form

s = (λ²/q)/(1+ε·cos(θ−θ0))

where ε turns out to be the eccentricity of the elliptical orbit;.

By a suitable choice of the polar coordinate system; θ0 can be made equal to zero. Thus the solution for the system is

s(θ) = (λ²/q)/(1+ε·cos(θ))

Note for later use that

s(1+ε·cos(θ)) = (λ²/q)

Particle Velocities

The next step is to derive the velocities of each particle. vm and vM. where

vm² = (dr/dt)² + (rω)²
vM² = (dR/dt)² + (Rω)²

From the solution for s

(ds/dt) = [(λ²/q)ε·sin(θ)/(1+ε·cos(θ))²](dθ/dt)
(ds/dt) = [(λ²/q)ε·sin(θ)/(1+ε·cos(θ))²]ω
(ds/dt) =[(λ²/q)ε·sin(θ)/(1+ε·cos(θ))²](λ/s²)

Since (1+ε·cos(θ))²s² is equal to (λ²/q)².

(ds/dt) =[(λ²/q)ε·sin(θ)/(λ²/q)²
which reduces to
(ds/dt) = (q/λ)ε·sin(θ)
and
(ds/dt)² = (q/λ)²ε²sin²(θ)

The tangential velocity is rω, but rω is equal to (M/(m+M))sω and sω is equal to λ/s. Thus

(sω)² = λ²(q/λ²)²(1+ε·cos(θ))²
(sω)² = (q/λ)²(1+ε·cos(θ))²
(sω)² = (q/λ)²(1+2ε·cos(θ)+ε²cos²(θ))

Let v²=μ(ds/dt)²+(sω)².

Combining the two terms for v² gives

v² = (q/λ)²(1+2ε·cos(θ)+ε²)

Then

vm = (M/(m+M))v
and
vM = (m/(m+M))v

The probability density per unit path length for both particles are proportional to

1/|v| = (λ/q)/(1+2ε·cos(θ)+ε²)½

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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