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 \(\boldsymbol{F} = \partial U(r)/\partial \boldsymbol{r} = - (\mathrm{d}^{}U/\mathrm{d}^{}r)\boldsymbol{r}/r\); 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 §9. Angular momentum, the angular momentum of any system relative to the centre of such a field is conserved. The angular momentum of a single particle is \(\boldsymbol{M} = \boldsymbol{r}\times\boldsymbol{p}\). Since \(\boldsymbol{M}\) is perpendicular to \(\boldsymbol{r}\), the constancy of \(\boldsymbol{M}\) shows that, throughout the motion, the radius vector of the particle lies in the plane perpendicular to \(\boldsymbol{M}\).
Thus the path of a particle in a central field lies in one plane. Using polar co-ordinates \(\boldsymbol{r}\), in that plane, we can write the Lagrangian as
\[\label{equations/14.1}\tag*{(14.1)}L=\textstyle\frac{1}{2}\displaystyle m(\dot{r}^2+r^2\dot{\phi^2})-U(r)\]
see (4.5) \(L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\phi}^2+\dot{z}^2)\) . 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, \((\mathrm{d}^{}/\mathrm{d}^{}t) \partial L/\partial \dot{q}_i = \partial L/\partial q_i = 0\), 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 (9.6) \(\mathbf{M}=\mathbf{M}'+\mathbf{R}\times\mathbf{P}\) ), and we return to the known law of conservation of angular momentum:
\[\label{equations/14.2}\tag*{(14.2)}\frac{\partial L}{\partial \dot{\phi}}=mr^2\dot{\phi}=\text{constant} (=M_z)\]
This law has a simple geometrical interpretation in the plane motion
of a single particle in a central field. The expression \(\textstyle\frac{1}{2}\displaystyle r\cdot
r\mathrm{d}^{}\phi\) is the area of the sector bounded by two
neighbouring radius vectors and an element of the path
1/fig8. Calling this area \(\mathrm{d}^{}f\), we can write the angular
momentum of the particle as
\[\label{equations/14.3}\tag*{(14.3)}M=2m\dot{f}\]
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)1
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 (14.2) \(\frac{\partial L}{\partial \dot{\phi}}=mr^2\dot{\phi}=\text{constant} (=M_z)\) and substituting in the expression for the energy, we obtain
\[\label{equations/14.4}\tag*{(14.4)}E=\textstyle\frac{1}{2}\displaystyle m(\dot{r}^2+r^2\dot{\phi^2})+U(r)=\textstyle\frac{1}{2}\displaystyle m\dot{r}^2+\textstyle\frac{1}{2}\displaystyle M^2/mr^2+U(r)\]
Hence
\[\label{equations/14.5}\tag*{(14.5)}\dot{r}\equiv\frac{\mathrm{d}^{}r}{\mathrm{d}^{}t}=\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}\]
or, integrating,
\[\label{equations/14.6}\tag*{(14.6)}t=\int\mathrm{d}^{}r\bigg/\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}+\text{constant}\]
Writing (14.2) \(\frac{\partial L}{\partial \dot{\phi}}=mr^2\dot{\phi}=\text{constant} (=M_z)\) as \(\mathrm{d}^{}\phi = M\mathrm{d}^{}t/mr^2\), substituting \(\mathrm{d}^{}t\) from (14.5) \(\dot{r}\equiv\frac{\mathrm{d}^{}r}{\mathrm{d}^{}t}=\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}\) and integrating, we find
\[\label{equations/14.7}\tag*{(14.7)}\phi=\int\frac{M\mathrm{d}^{}r/r^2}{\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}}+\text{constant}\]
Formulae (14.6) \(t=\int\mathrm{d}^{}r\bigg/\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}+\text{constant}\) and (14.7) \(\phi=\int\frac{M\mathrm{d}^{}r/r^2}{\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}}+\text{constant}\) 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 (14.6) \(t=\int\mathrm{d}^{}r\bigg/\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}+\text{constant}\) 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 (14.2) \(\frac{\partial L}{\partial \dot{\phi}}=mr^2\dot{\phi}=\text{constant} (=M_z)\) shows that \(\phi\)h can never change sign.
The expression (14.4) \(E=\textstyle\frac{1}{2}\displaystyle m(\dot{r}^2+r^2\dot{\phi^2})+U(r)=\textstyle\frac{1}{2}\displaystyle m\dot{r}^2+\textstyle\frac{1}{2}\displaystyle M^2/mr^2+U(r)\) 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
\[\label{equations/14.8}\tag*{(14.8)}U_{\text{eff}}=U(r)+M^2/2mr^2\]
The quantity \(M^2/2mr^2\) is called the centrifugal energy. The values of \(r\) for which
\[\label{equations/14.9}\tag*{(14.9)}U(r)+M^2/2mr^2=E\]
determine the limits of the motion as regards distance from the centre. When equation (14.9) \(U(r)+M^2/2mr^2=E\) is satisfied, the radial velocity \(\dot{\boldsymbol{r}}\) 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 \(\dot{\boldsymbol{r}} = 0\) 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 \(r > r_{\text{min}}\), the motion is infinite: the particle comes from, and returns to, infinity.
If the range of \(r\) has two limits \(r_{\text{min}}\) and \(r_{\text{max}}\), the motion is finite and the path lies entirely within the annulus bounded by the circles \(r = r_{\text{max}}\) and \(r = r_{\text{min}}\). This does not mean, however, that the path must be a closed curve. During the time in which \(r\) varies from \(r_\text{max}\) to \(r_\text{min}\) and back, the radius vector turns through an angle \(\Delta\phi\) which, according to (14.7) \(\phi=\int\frac{M\mathrm{d}^{}r/r^2}{\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}}+\text{constant}\) , is given by
\[\label{equations/14.10}\tag*{(14.10)}\delta\phi=2\int_{r_\text{min}}^{r_\text{max}} \frac{M\mathrm{d}^{}r/r^2}{\sqrt{2m(E-U)-M^2/r^2}}\]
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 §15 Kepler's problem; the
latter is that of the space oscillator (see 1/23p3).
At a turning point the square root in (14.5) \(\dot{r}\equiv\frac{\mathrm{d}^{}r}{\mathrm{d}^{}t}=\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}\) , and therefore the integrands in (14.6) \(t=\int\mathrm{d}^{}r\bigg/\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}+\text{constant}\) and (14.7) \(\phi=\int\frac{M\mathrm{d}^{}r/r^2}{\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}}+\text{constant}\) , 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 \(\boldsymbol{r}\), i.e. the path is symmetrical about the line \(\phi = 0\). Starting, say, from a point where \(r = r_\text{max}\) the particle traverses a segment of the path as far as a point with \(r=r_\text{min}\), then follows a symmetrically placed segment to the next point where \(r = r_\text{max}\), 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 (\(r = r_\text{min}\)) to infinity.
9The 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 $-as \(r\to 0\). From the inequality
\[ \frac{1}{2}m\dot{\boldsymbol{r}}^2=E-U(r)-M^2/2mr^2 > 0, \]
or \(r^2U(r)+M^2/2m < Er^2\), it follows that \(r\) can take values tending to zero only if
\[\label{equations/14.11}\tag*{(14.11)}\left[r^2U(r)\right]_{r\to 0}<-M^2/2m\]
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 \(n > 2\).