central force

Central fields have the following properties:

1. The force happens at a distance between two particles.
2. The force gives both the particles an equal amount of force but in opposite directions (Newton's Third Law).
3. The force magnitude only depends on the distance between the two particles in question, and maybe some other property intrinsic to both the particles.
4. If you draw a straight line between the two points, the force vector has to be parallel to that line.
5. The force works in the same no matter where in the universe you are.

Where these five properties are all common attributes of everyday forces. What these properties are basically saying is that we want a function only dependent on the vector between two particles \(P_{1}\) and \(P_{2}\) that are experiencing the force, and it is also parallel to this vector, as well as some symmetry constraints. Therefore, we can define a central force:

A central field is defined as follows:

\begin{align*} \vec{f}(\vec{r}) = f(\vec{r})\hat{r}. \end{align*}

Using the , we can find the curl of the central force. We will use the Cartesian del operator first because it requires less math to understand, despite the calculation being longer. Therefore, we must switch coordinates:

\begin{align*} f(\vec{r})\hat{r} = f(x, y, z)\frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} \\ \vec{\nabla} \times \vec{f} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f(x, y, z)\frac{x}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} & f(x, y, z)\frac{y}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} & f(x, y, z)\frac{z}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} \end{vmatrix} \end{align*}

If we can figure out one of these derivatives, then by symmetry we can figure out all these other derivatives.

\begin{align*} \frac{\partial}{\partial y}f(x, y, z)\frac{z}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} = f'(x, y, z)\frac{z}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} + f(x, y, z)(\frac{z}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}})' \\ \frac{\partial}{\partial y}z(x^{2} + y^{2} + z^{2})^{-\frac{1}{2}} = -yz(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} \\ \frac{\partial}{\partial y}f(x, y, z)\frac{z}{(x^{2} + y^{2} + z^{2})^{-\frac{1}{2}}} = f'(x, y, z)\frac{z}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} -yz f(x, y, z)(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} \\ \frac{\partial}{\partial z}f(x, y, z)\frac{y}{(x^{2} + y^{2} + z^{2})^{-\frac{1}{2}}} = f'(x, y, z)\frac{y}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} -zy f(x, y, z)(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} \\ (\frac{\partial}{\partial y}f(x, y, z)\frac{z}{(x^{2} + y^{2} + z^{2})^{-\frac{1}{2}}} - \frac{\partial}{\partial z}f(x, y, z)\frac{y}{(x^{2} + y^{2} + z^{2})^{-\frac{1}{2}}})\hat{i} = \vec{0} \end{align*}

And finally, by symmetry,

\begin{align*} \vec{\nabla} \times \vec{f} = \vec{0} \end{align*}

Because the computation for \(\hat{j}\) and \(\hat{k}\) are the same. Also, this implies that all central forces are conservative forces, so by :

\begin{align*} \vec{f} = \vec{\nabla}V \end{align*}