# 1-form of a central force field?

1. Jul 2, 2010

### TopCat

The problem

Given Newton's law of gravitational attraction, that the force a body exerts on a particle in space is directed towards the body and has a magnitude proportional to the inverse square of the distance to the body, show that the force field is described by the 1-form

$$\frac{kx}{r^{3}}dx + \frac{ky}{r^{3}}dy + \frac{kz}{r^{3}}dz$$

where k is a postive constant and r(x,y,z) is the distance from a point to the body.

The attempt

The 1-form for a constant force field is A dx + B dy + C dz, where A, B, and C are constant and represent the work required for a unit displacement in the relevant direction. So the force is in a direction (-A, -B, -C). Then the magnitude of a radial force is the total work required for unit displacement opposed to the force, which is just the unit vector in the direction -$$F$$ or $$(\frac{A}{\sqrt{A^{2}+B^{2}+C^{2}}},\frac{B}{\sqrt{A^{2}+B^{2}+C^{2}}},\frac{C}{\sqrt{A^{2}+B^{2}+C^{2}}})$$. Thus, $$F = \sqrt{A^{2}+B^{2}+C^{2}}$$.

The problem I'm having is showing that the general 1-form of a radial force is cx dx + cy dy + cz dz, where c > 0.

If I could show that, then I know by Newton and the first part of the solution that $$\sqrt{(cx)^{2}+(cy)^{2}+(cz)^{2}} = \frac{k}{r^{2}}$$, where k is some constant. Solving for c yields $$c = \frac{k}{r^{3}}$$ and the problem is done.

Can anyone help me understand why to say that the force at (x,y,z) is radially outward means that it is of the form mentioned above?

Last edited: Jul 2, 2010