---
title: 14. Motion in a central field
---

On reducing the two-body problem to one of the motion of a single body, we arrive at the problem of determining the motion of a single particle in an external field such that its potential energy depends only on the distance r from some fixed point. This is called a central field. The force acting on the particle is $\v{F} = \partial U(r)/\partial \v{r} = - (\dd{U}/\dd{r})\v{r}/r$; its magnitude is likewise a function of $r$ only, and its direction is everywhere that of the radius vector.

As has already been shown in `1/9`, the angular momentum of any system relative to the centre of such a field is conserved. The angular momentum of a single particle is $\v{M} = \v{r}\times\v{p}$. Since $\v{M}$ is perpendicular to $\v{r}$, the constancy of $\v{M}$ shows that, throughout the motion, the radius vector of the particle lies in the plane perpendicular to $\v{M}$.

Thus the path of a particle in a central field lies in one plane. Using polar co-ordinates $\v{r}$, in that plane, we can write the Lagrangian as

```load
1/14.1
```

see `1/4.5`. This function does not involve the co-ordinate $\phi$ explicitly. Any generalised co-ordinate $q_i$ which does not appear explicitly in the Lagrangian is said to be cyclic. For such a co-ordinate we have, by Lagrange's equation, $(\dd{}/\dd{t}) \partial L/\partial \dot{q}_i = \partial L/\partial q_i = 0$, so that the corresponding generalised momentum $p_i = \partial L/\partial \dot{q}_i$ is an integral of the motion. This leads to a considerable simplification of the problem of integrating the equations of motion when there are cyclic co-ordinates.

In the present case, the generalised momentum $p_\phi=mr^2\dot{\phi}$ is the same as the angular momentum $M_z=M$ (see `1/9.6`), and we return to the known law of conservation of angular momentum:

```load
1/14.2
```

This law has a simple geometrical interpretation in the plane motion of a single particle in a central field. The expression $\mfrac{1}{2}r\cdot r\dd{\phi}$ is the area of the sector bounded by two neighbouring radius vectors and an element of the path `1/fig8`. Calling this area $\dd{f}$, we can write the angular momentum of the particle as

```load
1/14.3
```

where the derivative $\dot{f}$ is called the *sectorial velocity*. Hence the conservation of angular momentum implies the constancy of the sectorial velocity: in equal times the radius vector of the particle sweeps out equal areas (*Kepler's second law*)[^\dagger]

[^\dagger]: The law of conservation of angular momentum for a particle moving in a central field
is sometimes called the area integral.

`1/fig8`

The complete solution of the problem of the motion of a particle in a central field is most simply obtained by starting from the laws of conservation of energy and angular momentum, without writing out the equations of motion themselves. Expressing in terms of $M$ from `1/14.2` and substituting in the expression for the energy, we obtain

```load
1/14.4
```
Hence

```load
1/14.5
```

or, integrating,

```load
1/14.6
```

Writing `1/14.2` as $\dd{\phi} = M\dd{t}/mr^2$, substituting $\dd{t}$ from `1/14.5` and integrating, we find

```load
1/14.7
```

Formulae `1/14.6` and `1/14.7` give the general solution of the problem. The latter formula gives the relation between $r$ and $\phi$, i.e. the equation of the path.  Formula `1/14.6` gives the distance $r$ from the centre as an implicit function of time. The angle $\phi$, it should be noted, always varies monotonically with time, since `1/14.2` shows that $\phi$h can never change sign.

The expression `1/14.4` shows that the radial part of the motion can be regarded as taking place in one dimension in a field where the "effective potential energy" is

```load
1/14.8
```
The quantity $M^2/2mr^2$ is called the *centrifugal energy*. The values of $r$ for which

```load
1/14.9
```

determine the limits of the motion as regards distance from the centre.  When equation `1/14.9` is satisfied, the radial velocity $\dot{\v{r}}$ is zero. This does not mean that the particle comes to rest as in true one-dimensional motion, since the angular velocity $\phi$ is not zero. The value $\dot{\v{r}} = 0$ indicates a *turning point* of the path, where $r(t)$ begins to decrease instead of increasing, or *vice versa*.

If the range in which $r$ may vary is limited only by the condition $r > r_{\min}$,
the motion is infinite: the particle comes from, and returns to, infinity.

If the range of $r$ has two limits $r_{\min}$ and $r_{\max}$, the motion is finite and the path lies entirely within the annulus bounded by the circles $r = r_{\max}$ and $r = r_{\min}$. This does not mean, however, that the path must be a closed curve.  During the time in which $r$ varies from $r_\max$ to $r_\min$ and back, the radius vector turns through an angle $\Delta\phi$ which, according to `1/14.7`, is given by

```load
1/14.10
```

The condition for the path to be closed is that this angle should be a rational fraction of $2\pi$, i.e. that $\Delta\phi=2\pi m/n$, where $m$ and $n$ are integers. In that case, after $n$ periods, the radius vector of the particle will have made $m$ complete revolutions and will occupy its original position, so that the path is closed.

Such cases are exceptional, however, and when the form of $U(r)$ is arbitrary the angle is not a rational fraction of $2pi$. In general, therefore, the path of a particle executing a finite motion is not closed. It passes through the minimum and maximum distances an infinity of times, and after infinite time it covers the entire annulus between the two bounding circles. The path shown in `1/fig9` is an example.

There are only two types of central field in which all finite motions take place in closed paths. They are those in which the potential energy of the particle varies as $1/r$ or as $r^2$. The former case is discussed in `1/15`; the latter is that of the space oscillator (see `1/23p3`).

At a turning point the square root in `1/14.5`, and therefore the integrands in `14.6` and `14.7`, change sign. If the angle $\phi$ is measured from the direction of the radius vector to the turning point, the parts of the path on each side of that point differ only in the sign of $\phi$ for each value of $\v{r}$, i.e. the path is symmetrical about the line $\phi = 0$. Starting, say, from a point where $r = r_\max$ the particle traverses a segment of the path as far as a point with $r=r_\min$, then follows a symmetrically placed segment to the next point where $r = r_\max$, and so on. Thus the entire path is obtained by repeating identical segments forwards and backwards. This applies also to infinite paths, which consist of two symmetrical branches extending from the turning point ($r = r_\min$) to infinity.

```fig
9
```

The presence of the centrifugal energy when $M \neq 0$, which becomes infinite as $1/r^2$ when $r\to 0$, generally renders it impossible for the particle to reach the centre of the field, even if the field is an attractive one. A "fall" of the particle to the centre is possible only if the potential energy tends suffi-
ciently rapidly to $-\infty as $r\to 0$. From the inequality

$$
\frac{1}{2}m\dot{\v{r}}^2=E-U(r)-M^2/2mr^2 > 0,
$$

or $r^2U(r)+M^2/2m < Er^2$, it follows that $r$ can take values tending to zero only if

```load
14.11
```

i.e. $U(r)$ must tend to $-\infty$ either as $-\alpha/r^2$ with $\alpha > M^2/2m$, or proportionally
to $-1/r^n$ with $n > 2$.

<!-- PROBLEMS -->
<!-- PROBLEM 1. Integrate the equations of motion for a spherical pendulum (a particle of mass -->
<!-- m moving on the surface of a sphere of radius l in a gravitational field). -->
<!-- SOLUTION. In spherical co-ordinates, with the origin at the centre of the sphere and the -->
<!-- polar axis vertically downwards, the Lagrangian of the pendulum is -->
<!-- 1ml2(02 + 62 sin20) +mgl cos 0. -->
<!-- 2* -->
<!-- 34 -->
<!-- Integration of the Equations of Motion -->
<!-- §14 -->
<!-- The co-ordinate is cyclic, and hence the generalised momentum Po, which is the same as the -->
<!-- z-component of angular momentum, is conserved: -->
<!-- (1) -->
<!-- The energy is -->
<!-- E = cos 0 -->
<!-- (2) -->
<!-- = 0. -->
<!-- Hence -->
<!-- (3) -->
<!-- where the "effective potential energy" is -->
<!-- Ueff(0) = COS 0. -->
<!-- For the angle o we find, using (1), -->
<!-- do -->
<!-- (4) -->
<!-- The integrals (3) and (4) lead to elliptic integrals of the first and third kinds respectively. -->
<!-- The range of 0 in which the motion takes place is that where E > Ueff, and its limits -->
<!-- are given by the equation E = Uell. This is a cubic equation for cos 0, having two roots -->
<!-- between -1 and +1; these define two circles of latitude on the sphere, between which the -->
<!-- path lies. -->
<!-- PROBLEM 2. Integrate the equations of motion for a particle moving on the surface of a -->
<!-- cone (of vertical angle 2x) placed vertically and with vertex downwards in a gravitational -->
<!-- field. -->
<!-- SOLUTION. In spherical co-ordinates, with the origin at the vertex of the cone and the -->
<!-- polar axis vertically upwards, the Lagrangian is sin2x) -mgr cos a. The co- -->
<!-- ordinate is cyclic, and Mz = mr2 sin²a is again conserved. The energy is -->
<!-- = a. -->
<!-- By the same method as in Problem 1, we find -->
<!-- == -->
<!-- The condition E = Ueff(r) is (if M + 0) a cubic equation for r, having two positive roots; -->
<!-- these define two horizontal circles on the cone, between which the path lies. -->
<!-- PROBLEM 3. Integrate the equations of motion for 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 (Fig. 2, §5). -->
<!-- SOLUTION. In the Lagrangian derived in §5, Problem 2, the co-ordinate x is cyclic. The -->
<!-- generalised momentum Px, which is the horizontal component of the total momentum of the -->
<!-- system, is therefore conserved -->
<!-- Px = cos = constant. -->
<!-- (1) -->
<!-- The system may always be taken to be at rest as a whole. Then the constant in (1) is zero -->
<!-- and integration gives -->
<!-- (m1+m2)x+m2) sin = constant, -->
<!-- (2) -->
<!-- which expresses the fact that the centre of mass of the system does not move horizontally. -->
