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