# Forces from potential energy

#### Prerequisites

From our analysis of the work-energy theorem and conservative forces, we learned that we could express the work done by certain forces (gravity, electricity, and springs, for example) in terms of a simple function of position (separations of the interacting objects) -- the potential energy. In many situations, this is a much more convenient quantity to work with than forces since it is a scalar -- it has no direction. Forces always have to be added vectorially with all the complexity that entails. To add potential energies, you just add them, of course being careful of the signs!

Despite the fact that the potential energy is a scalar without a direction associated with it, nonetheless, it still contains the vector information about the direction of forces. The way it does this is by how it changes in space -- its derivatives. Recall that the basic definition of the potential is the negative of the work done by a conservative force along the direction of motion:

$$F^{\parallel}_{B \rightarrow A} \Delta r_A = - \Delta U = U(r_i) - U(r_f)$$

If we reverse this, solving for the force, we get:

$$F^{\parallel}_{B \rightarrow A} = -\frac{\Delta U}{\Delta r}.$$

If we have a force of a particular conservative type (gravity, spring, or electric), and we only look along one dimension, when we take our $\Delta r$ small enough, we can write it with a derivative:

$$F^{type} = -\frac{dU^{type}}{dr}$$

The force in the direction of the displacement is the derivative of the potential energy in that direction. We use the marker "type" to indicate that the kind of force we get comes from the kind of PE we start with. Electric PE gives the electric force, gravitational PE gives the gravitational force, etc.

If our displacement was in the x direction, we would get the more familiar derivative:

$$F^{type}_x = -\frac{dU^{type}}{dx}$$

and we would get the x-component of the corresponding force.

We would get analogous results if we took our displacements in the y or z directions. We can combine these to get the full vector of the force corresponding to the potential energy:

$$\overrightarrow{F}^{type} = -\bigg( \frac{\partial U^{type}}{dx} \hat{i} + \frac{\partial U^{type}}{dy} \hat{j} + \frac{\partial U^{type}}{dz} \hat{k} \bigg) = -\overrightarrow{\nabla}U^{type}$$

This particular combination of the spatial derivatives gives the full vector of the force. The upside-down vector delta ($\nabla$, called "nabla") on the right just is a shorthand for the stuff in parentheses. It is called the gradient and is useful in talking about how a function changes in space. (We use the curly or "partial" dees, "$\partial$", to remind ourselves that $U$ is a function of all three variables, $(x, y, z)$, and when we take a derivative with respect to one of those variables, we hold the other variables fixed.)

Although this is using somewhat advanced math (vector calculus), all it is really saying is:

The force associated with a potential energy function points in the direction that the potential energy is falling the fastest.

To see how this works in 1, 2, and 3D, check out the follow-on examples.

Joe Redish and Wolfgang Losert 11/26/12

Article 454