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1/14-motion-in-a-central-field.md

@@ -2,32 +2,22 @@
 title: 14. Motion in a central field
 ---
 
-On reducing the two-body problem to one of the motion of a single body,
+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.
-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
+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 $\v{M}$.
-from some fixed point. This is called a central field. The force acting on the
+
-particle is F = du(r)/dr = - (dU/dr)r/r; its magnitude is likewise a func-
+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
-tion of r only, and its direction is everywhere that of the radius vector.
+
-As has already been shown in §9, the angular momentum of any system
+```load
-relative to the centre of such a field is conserved. The angular momentum of a
+1/14.1
-single particle is M = rxp. Since M is perpendicular to r, the constancy of
+```
-M shows that, throughout the motion, the radius vector of the particle lies
+
-in the plane perpendicular to M.
+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.
-Thus the path of a particle in a central field lies in one plane. Using polar
+
-co-ordinates r, in that plane, we can write the Lagrangian as
+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:
-(14.1)
+
-see (4.5). This function does not involve the co-ordinate explicitly. Any
+```load
-generalised co-ordinate qi which does not appear explicitly in the Lagrangian
+1/14.2
-is said to be cyclic. For such a co-ordinate we have, by Lagrange's equation,
+```
-(d/dt) aL/dqi = aL/dqi = 0, so that the corresponding generalised momen-
+
-tum Pi = aL/dqi is an integral of the motion. This leads to a considerable
+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`
-simplification of the problem of integrating the equations of motion when
-there are cyclic co-ordinates.
-In the present case, the generalised momentum is the same as
-the angular momentum M z = M (see (9.6)), and we return to the known law
-of conservation of angular momentum:
-M = mr2o = constant. =
-(14.2)
-This law has a simple geometrical interpretation in the plane motion of a single
-particle in a central field. The expression 1/2 . rdo is the area of the sector
-bounded by two neighbouring radius vectors and an element of the path