In our pursue to study the two-body problem we face now the problem of determining the motion of one particle under the influence of a central field.
A central field is an external field with a potential energy U that depends only on the distance from the particle to the centre of the field, which is fixed.
The force produced by the field on the particle is
|F> = -∂U/∂|r> = -(dU/dR)|^r^> = -(dU/dr)|r>/r.
The magnitude of the force is a function of r only: dU(r)/dr. The direction of the force is always parallel to the radius vector (relative to the centre of the field).
We already know that for such system, |M> will be fully conserved.
Since |M> = |r> ⨯ |p>, |M> is ⟂ to |r> and |p>. Since |M> is constant, the position must always be ⟂ to it, which means that motion occurs on a plane and that this plane is ⟂ to |M>.
Since motion occurs on a plane and since we have a special point, the centre of the field, it is convenient to use polar coordinates r,ϕ, and the Lagrangian becomes
L = (m/2)·(ṙ²+r²[ϕ·]²) - U(r) .
This Lagrangian does not contain any explicit ϕ.
A generalised coordinate qᵢ that does not appear in the Lagrangian is called *cyclic*.
For a cyclic coordinate qᵢ, E-L equations are (d/dt)(∂L/∂q̇ᵢ)=∂L/∂qᵢ=0. This means that the corresponding generalised momentum pᵢ=∂L/∂q̇ᵢ is an integral of motion.
In our case, since ϕ is cyclic, the corresponding p_ϕ=∂L/∂[ϕ·] = mr²[ϕ·] is our integral of motion. See how this is just Mz, so what we obtained is just the conservation of angular momentum again. Since we are considering motion in the xy plane here, |M> = (0,0,Mz) where Mz=M is constant.
The law M = mr²[ϕ·] = constant has a geometrical explanation for motion on a plane under a central field. Consider the following triangle:
next radius vector ___
____,,....,-----''''' ↑
_____.....------'''''|dϕ area df | r·dϕ
O._···_____________________|______________________| (path)
r
previous radius vector
Its area is r·(r·dϕ)/2, and since dϕ is very small, the triangle describes the area of the sector between two consecutive radius vectors and the element of the path. We call df to this area, and then
M = mr²[ϕ·] = (2m)·(r²[ϕ·]/2) = 2mḟ,
where ḟ (=df/dt) is called the *sectorial velocity*.
Then, since (2m) is constant, the constancy of M implies the constancy of ḟ. In other words: motion is such that in equal times, the radius vector sweeps equal areas. This is Kepler's second law.
For example, when moving along an elliptical path, if the centre of the field is at one focus, motion must be faster when close to the focus, since the magnitude of the radius vector is smaller, and then it needs a faster motion. On the other hand, when far from the focus, motion becomes slow because the magnitude of the radius vector is big, and it needs little motion to produce the required equal area.
Now we pursue the complete solution of the problem of a particle under a central field.
We begin by writing the energy E, which is constant,
E = (m/2)·(ṙ²+r²[ϕ·]²) + U(r).
Now, since, we know M=mr²[ϕ·], we can eliminate [ϕ·] as
E = (1/2)·( mṙ² + M²/(mr²) ) + U(r).
From this expression, we isolate ṙ as
ṙ = dr/dt = ( 2[E-U(r)]/m - M²/(m²r²) )¹⸍².
We can now isolate dt
dt = dr / ( 2[E-U(r)]/m - M²/(m²r²) )¹⸍²
and integrate
t = ∫dr / ( 2[E-U(r)]/m - M²/(m²r²) )¹⸍² + constant.
But what we want is not t(r) but ϕ(r), so we use M=mr²[ϕ·]=mr²dϕ/dt to write
dϕ = Mdt / (mr²)
and we see that we can obtain dϕ and ϕ as
ϕ = ∫ (M/r²) · dr / ( 2m[E-U(r)] - M²/r² )¹⸍² + constant.
These two integrals, t(r) and ϕ(r) are very important since they provide a complete solution of the problem.
Since M=mr²[ϕ·] and M and m are constants and r² is always positive, [ϕ·] can never change sign. This implies that ϕ always changes monotonically with time.
Let's write again the energy E as
E = (1/2)·( mṙ² + M²/(mr²) ) + U(r).
Here we could "transfer" the term M²/(2mr²) to the potential energy, because it only depends on the magnitude of the radius, r.
E = (m/2)·ṙ² + M²/(2mr²) + U(r)
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Ueff
E = (m/2)·ṙ² + Ueff
Ueff = M²/(2mr²) + U(r).
This way, we can consider a particle moving in one dimension with an "effective" potential energy Ueff.
The quantity M²/(2mr²) is called *centrifugal energy*.
The quantity (m/2)·ṙ² can be interpreted as a "radial energy", since it tells the part of the kinetic energy due to radial motion. When using Ueff, it is the only kinetic energy, since the centrifugal term is absorbed by U.
It is interesting to set the radial energy equal to 0, since this condition will give us the points for which ṙ=0. In a one-dimensional motion it means that the particle comes to rest, although we know that in a plane, it means that the velocity of the particle is purely angular. The points for which this condition holds are called *turning points*. In them, r turns from increasing to decreasing or vice versa.
If we obtain a single turning point, it means that motion is infinite, since it is not bounded on one side. But if we obtain two values, motion is bounded between them. In such case we can call them rmin and rmax, and we know that motion is bounded between the annulus formed by the two rings r=rmin and r=rmax.
This does not mean that the path is closed. The path can be bounded but never repeating itself!
Whether the path is closed or not depends on the angle turned during the time taken by the particle to go from rmin to rmax and back (or from rmax to rmin and back), in other words, the period T. We call Δθ to this angle.
The path is closed if after a whole number n of periods the total angle increment is a whole number m of full turns. Since Δθ is the angle turned per one period T, we obtain the condition
n·Δθ = 2πm.
The expression for Δθ is
Δθ = 2·∫{r;rmin;rmax}dr·(M/r²) / ( 2m[E-U] - M²/r² )¹⸍².
In general, the condition for a closed path will not be satisfied, so the path will not be closed even if the motion is bounded by two circles. If this occurs, it will oscillate between rmin and rmax and back forever, and after an infinite time, the path will cover the entire annulus between the two circles. This property of visiting all available points is called *ergodicity*.
There are only *two* types of central field in which *all* bounded motions take place in *closed* paths. They are those for which the potential energy varies as 1/r or as r². We will eventually study both cases. The case 1/r is what we have in electrostatics and in classical gravitation. The case r² is called the space oscillator.
At a turning point, the quantity
ṙ = dr/dt = ( 2[E-U(r)]/m - M²/(m²r²) )¹⸍².
changes sign, and *only the sign*. As this square root appears in our integral solutions for t(r) and ϕ(r), if we define ϕ=0 for a turning point, then the path must be symmetric about the line ϕ=0.
If we define ϕ=0 for r=rmin, then the paths for positive or negative dϕ are symmetrical about the line ϕ=0 and the total path covered until reaching r=rmin will be the same in both cases.
Then, the entire path is obtained by repeating the same path increment back and forth. This also applies to infinite paths, where from r=rmin we obtain two identical branches up to infinity.
The centrifugal energy M²/(2mr²), if M ≠ 0, explodes to infinity when r → 0. This implies that the particle can never reach the centre of the field. Even if the field is attractive!
Since the total energy is
E = (m/2)·ṙ² + M²/(2mr²) + U(r),
the kinetic term (the radial energy) is positive,
(m/2)·ṙ² = E - M²/(2mr²) - U(r) > 0.
From this, we deduce
E > M²/(2mr²) + U(r) => Er² > M²/(2m) + r²·U(r).
By observing the latter expression, we can ask whether is possible to reach r=0. The term Er² tends to 0 if r goes to 0, so we need that for r tending to 0, the sum r²U(r) + M²/(2m) tends to something < 0. This is only possible if r²U(r) tends to a negative quantity that can be finite for U(r) ∝ -1/r² or infinite for U(r) ∝ -1/rⁿ for n>2. Otherwise, the fall to the centre is impossible.
Integrate the equations of motion for a *spherical pendulum* (a particle of mass m moving on the surface of radius l in a gravitational field). It is like a pendulum with a rigid and massless "string" that moves in 3d.
We write the Lagrangian in spherical coordinates,
L = (m/2)·l²·( [θ·]^2 + [ϕ·]²sin²θ ) + mglcosθ.
The angle θ is not cyclic because it makes an explicit appearance in U. However, the angle ϕ is cyclic, which means that the associated momentum is ∂L/∂[ϕ·] = p_ϕ = constant = (ml²[ϕ·])·sin²θ = M·sin²θ =Mz (recall section 9 for the last step).
The energy is
E = (m/2)l²·([θ·]² + [ϕ·]²sin²θ) - mglcosθ
and we get rid of [ϕ·]² by using Mz,
E = (m/2)l²·[θ·]² + ml²Mz²/2·sin²θ - mglcosθ.
We define Ueff as
Ueff = ml²Mz²/2·sin²θ - mglcosθ.
We can write, then
E = (m/2)l²·[θ·]² + Ueff => => E-Ueff = (m/2)l²·dθ²/dt² => => dt² = (m/2)l²dθ² / (E-Ueff) => => dt = (ml/2)¹⸍²ldθ / (E-Ueff)¹⸍² => => t = (ml/2)¹⸍²l∫dθ / (E-Ueff)¹⸍².
This integral leads to an elliptic integral of the first kind.
But if instead of t(θ) we prefer ϕ(θ), we use
Mz=ml²[ϕ·]sin²θ => => Mz·dt = ml²sin²θdϕ => => dϕ = Mz·dt / (ml²sin²θ).
and then
dϕ = Mz/(ml²sin²θ)·(ml/2)¹⸍²ldθ / (E-Ueff)¹⸍² => => ϕ = Mz/(2ml²)¹⸍²·∫dθ / sin²θ / (E-Ueff)¹⸍².
This integral leads to an elliptic integral of the third kind.
What are the limits of integration? In other words, what is the range of θ? Remember that E must always be ≥ Ueff, so that the limits come from E = Ueff. This produces a cubic equation for cosθ, with two roots between -1 and +1. These two roots define two circles of constant latitude in the sphere. The path must lie between them.
Although we cannot provide an analytical answer in this case, it is good to *know* that we cannot, and that we have to use elliptic integrals, for which we can jump to numerical solutions, or even approximate them through series expansions. However, approximations would accumulate errors and would eventually lead to a wrong path, but for short times of evolution they would be useful. Knowing that the path must lie between two latitudes is also very useful.
Integrate the equations of motion for a particle moving on the surface of a cone (of vertical angle 2α) placed vertically and with vertex downwards in a gravitational field. We understand that the particle is restricted to the surface and cannot leave it. Or we can see it as a particle being inside the cone and sliding on that interior surface.
We use spherical coordinates with the origin at the vertex of the cone. The polar axis is positive upwards. The Lagrangian is
L = (m/2)·(ṙ² + r²[ϕ·]²sin²α) - mgrcosα.
The angle ϕ is cyclic, and its associated momentum is mr²[ϕ·]sin²α = Mz, is conserved. From here, we can isolate dϕ as
dϕ = Mz·dt/(mr²sin²α).
The energy E is
E = (m/2)·ṙ² + Mz²/(2mr²)·sin²α + mgrcosα
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Ueff(r)
We use now the same method as in the previous problem. First, we isolate dt as
dt = dr / (2·(E-Ueff)/m)¹⸍².
And we can also write
dϕ = Mz/(r²sin²α)·dr / (2m·(E-Ueff))¹⸍².
The condition E=Ueff(r) is a cubic equation for r. There are two positive roots which define two horizontal circles on the cone. The path lies between them.
Integrate the equations of motion of a pendulum of mass m₂ with a mass m₁ at the point of support, which can move on a horizontal line lying in the plane in which m₂ moves. We have already written this Lagrangian in section 5, problem 2.
In that Lagrangian we can see that x is cyclic. The generalised momentum Px is
Px = (m₁+m₂)ẋ + m₂l[ϕ·]cosϕ = constant.
The centre of mass the system may be taken as zero. See how Px is the x-velocity of the centre of mass. So Px=0.
The energy is
E = (m₁+m₂)ẋ²/2 +(m₂/2)·(l²[ϕ·]²+2lẋ[ϕ·]cosϕ) - m₂glcosϕ = = (ẋ/2)·( (m₁+m₂)ẋ + m₂·l[ϕ·]cosϕ ) + (ẋ/2)·m₂l[ϕ·]cosϕ + (m₂/2)l²[ϕ·]² - m₂glcosϕ = = ẋ·Px/2 + (ẋ/2)·m₂l[ϕ·]cosϕ + (m₂/2)l²[ϕ·]² - m₂glcosϕ = = (ẋ/2)·m₂l[ϕ·]cosϕ + (m₂/2)l²[ϕ·]² - m₂glcosϕ = = (m₂/2)l[ϕ·]·( ẋcosϕ + l[ϕ·]) - m₂glcosϕ =
Now we use that (m₁+m₂)ẋ + m₂l[ϕ·]cosϕ = 0 to write
ẋ = -m₂/(m₁+m₂)·l·[ϕ·]·cosϕ,
which allows us to finally write
E = (m₂/2)l²[ϕ·]²·( 1 - m₂/(m₁+m₂)·cos²ϕ ) - m₂glcosϕ.
From here we can isolate dt as
E + m₂glcosϕ = (m₂/2)l²[ϕ·]²·( 1 - m₂/(m₁+m₂)·cos²ϕ ) dt² = (m₂/2)l²dϕ²·(1 - m₂/(m₁+m₂)·cos²ϕ)/(E+m₂glcosϕ) dt = dϕ·l·[m₂ / 2 · (m₁ + m₂sin²ϕ) / (m₁+m₂) / (E+m₂glcosϕ) ]¹⸍².
If we integrate Px wrt time we get
(m₁+m₂)x + m₂lsinϕ = constant = C.
We isolate x as x = (C - m₂lsinϕ)/(m₁+m₂) and substitute into the Cartesian coordinated for particle 2,
x₂ = x + lsinϕ = C/(m₁+m₂) + m₁l/(m₁+m₂)·sinϕ y₂ = lcosϕ.
This means that particle 2 follows an elliptical path with horizontal semi-axis m₁l/(m₁+m₂) and vertical semi-axis l.
For m₁ → ∞ the horizontal semi-axis becomes l and we recover the circular motion of the simple pendulum.