Learning about force and motion from the potential energy

We learnt above about how the force leads to the concept of potential energy. However, it is the potential energy that is easier to specify than the force. The reason is very simple: force is a vector quantity and as such in specifying it we have to give its three components as a function of position. On the other hand, potential energy is a scalar quantity and is easier to write as a function of position. For the same reason, many a times it is easier to calculate the potential energy than to calculate the force, as we will see in an example below. Thus generally we give the potential energy of a particle to tell about the force field in which the particle is moving. In this section we discuss what can we learn about the motion of a particle by looking at its potential energy.

First we discuss how do we get the force from the potential energy. Let us first look at one-dimensional case. Employing the definition of potential energy, we find that for a small displacement Δx

which means that the force is given by the formula

This is the key formula relating the force to the potential energy. On the basis of this formula, we can infer a lot about the nature of motion by looking at the potential energy curve. First if , then the force is towards the negative x-direction and if , the force is towards the positive x-direction. Thus the force is in the direction of decreasing U(x). What if ? In that case the particle in either on a maximum or a minimum of the potential and there is no force on the particle. The particle is therefore in equilibrium. The equilibrium will be stable one, that is the particle will come back to the equilibrium point when displaced slightly from that point, if it is at the potential energy minimum or equivalently where . On the other hand at the maximum of the potential energy, the particle will rush away from that point if it is disturbed. Thus at the potential energy maximum, where , the equilibrium is unstable. We see that a particle tends to move towards its potential energy minimum and move away from its potential energy maximum. All these concepts can be shown nicely with a bead moving on a smooth frictionless wire bent in the shape of a curve with many maxima and minima and held in the vertical plane (see the figure below). The potential energy of the bead is then proportional to the height of the curve and as such the wire itself represents the potential energy curve in the figure below.

Now with a bead sliding over the wire, you can easily check that all the points made above about the relationship between the force on the bead and the mathematical properties of the potential energy curve are correct. Further the minima and maxima of the curve are clearly observed to be stable and unstable equilibrium points, respectively.

In three dimensions the equivalent of the derivative is the gradient operator. Thus the force in two or three dimensions is given as

Thus the force is in the direction opposite to that of increasing U. Further, it vanishes wherever the gradient of the potential energy is zero. Individual components of the force are given as

A word of caution is needed here. does not mean that if we transform to some other co-ordinates system (say spherical) then

will be correct. This is not even dimensionally correct. To get the correct answer, one must properly transform from Cartesian to polar co-ordinates. The result then is

Thus in spherical polar coordinate system, the force components are given as

Similarly in cylindrical coordinate system the force is related to the potential energy as

With the individual force components

Having given you the prescription for obtaining force from the potential energy let us now apply it to find the field of an electric dipole using its scalar potential.

Example: As an application of finding force from the potential, let us calculate the electric field due to a dipole.

Let the dipole be situated at the origin along the x-axis. Let the charges -q and +q be separated by distance 2a (see figure below) so that the dipole moment is . Then potential and field at any point can be calculated by adding the field due to the two charges. Adding the field in this case becomes a bit difficult because we have to obtain three components of the field for each charge and add them. On the other hand, finding the potential is relatively easy because it is a scalar quantity and we obtain it by adding the potential due to two charges. Then the gradient gives the field. In the calculation we assume that a → 0 and q is correspondingly very large so that their product is finite. We will be using this by keeping term only linear in a and neglecting higher orders.

The potential (potential energy per unit charge) due the two charges is given as

Now taking the gradient we get the three components of the force, which are

Similarly

Combining these results together we get for the field of the dipole

I would like you to get the same result by adding the fields of the charges together and compare the answers.

In these lectures, we have learnt: the work-energy theorem, definition of potential and its relationship with the force field, concept of conservative forces and the principle of conservation of energy. I leave these lectures by giving you a few exercises.

Exercise 1: Consider one-dimensional motion in a potential U(x). Show that if a particle of mass m is displaced slightly from its equilibrium position at a potential energy minimum at x₀ , it will perform simple harmonic oscillations. Find the corresponding frequency.

Exercise 2: Consider two different inertial frames moving with respect to one another with a constant velocity. Starting from the work-energy theorem in one frame, prove that it is true in the other frame also.