1-form of a central force field?

[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

[tex]\frac{kx}{r^{3}}dx + \frac{ky}{r^{3}}dy + \frac{kz}{r^{3}}dz[/tex]

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 -[tex]F[/tex] or [tex](\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}}})[/tex]. Thus, [tex]F = \sqrt{A^{2}+B^{2}+C^{2}}[/tex].

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 [tex]\sqrt{(cx)^{2}+(cy)^{2}+(cz)^{2}} = \frac{k}{r^{2}}[/tex], where k is some constant. Solving for c yields [tex]c = \frac{k}{r^{3}}[/tex] 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?

 

[mighty2000]

Sure, let's break this down step by step.

First, let's define what we mean by a radial force. A radial force is a force that is directed towards or away from a central point. In the case of gravitational attraction, this central point would be the body exerting the force.

Now, let's look at the 1-form \frac{kx}{r^{3}}dx + \frac{ky}{r^{3}}dy + \frac{kz}{r^{3}}dz. This 1-form represents a vector field, where each component (kx/r^3, ky/r^3, kz/r^3) is a vector pointing in the x, y, and z directions respectively. We can see that each component is directly proportional to the position coordinates (x, y, z) and inversely proportional to the cube of the distance (r^3) from the central point. This is exactly what we would expect for a radial force - as the distance increases, the force decreases, and it is directed towards the central point.

To further understand this, we can look at a specific example. Let's say we have a body at the origin (0,0,0) exerting a force on a particle at (1,1,1). The distance between the two points would be r = \sqrt{3}. Plugging this into the 1-form, we get \frac{k}{3\sqrt{3}}(dx + dy + dz). This represents a vector pointing in the direction of (1,1,1) with a magnitude of k/(3\sqrt{3}), which is exactly what we would expect for a radial force.

Now, let's look at the general form of a 1-form for a radial force, cx dx + cy dy + cz dz. We can see that this is similar to the previous 1-form, with the only difference being the constant c. As we mentioned before, this constant represents the work required for a unit displacement in the relevant direction. In the case of a radial force, this work would be directly proportional to the distance from the central point. So, c would be equal to k/r^3, where k is a positive constant and r is the distance from the central point. This shows that the general form of a 1-form for a radial force is indeed cx dx + cy dy + cz dz, where c > 0.

In