Review of Force and Potential Energy

In the previous chapter, we developed a foundation for how we expect atoms to interact as they near each other. In this chapter, we will explore a mathematical model of atomic bonding and see how it can be used in computer simulation. Broadly, the learning goals are to:

  1. Construct and utilize mathematical models of bonding to make predictions
  2. Understand the fundamentals of a computational modeling technique known as molecular dynamics
  3. Explore cross-cutting physical principles using mental, mathematical and computational models.

Before getting there, we must review the physical definitions and concepts of force and potential energy to better understand the strength of the interactions between atoms.

Let's work first with a familiar system: an idealized spring, shown in Figure 4.3.1. The mechanical behavior of this spring is governed by Hooke's Law. This spring is oriented in the (x)-direction, and one end is positioned at $x = 0$ when in equilibrium. When displaced in the positive $x$-direction, there is a restoring force $F$ acting to pull the spring's end back to $x=0$ that is linearly proportional to the distance, $x$, the end of the spring was displaced. Similarly, when compressed along the $-x$-direction, there is a restorative force acting in the $+x$ direction.

Illustration of a spring and its restoring force when displaced from equilibrium.

Figure 4.3.1 Illustration of a spring and its restoring force when displaced from equilibrium.

This behavior can be written mathematically as the (perhaps familiar?) Hooke's Law:

$$F(x) = -k x \tag{4.3.1}$$

where $k$ is the spring constant - a measure of how stiff the spring is, $F$ is the force, and $x$ is the amount the spring has been displaced from equilibrium at $x = 0$. A higher $k$ means a larger restoring force when the spring is displaced. We demonstrate Hooke's law (Eq. 4.3.1) graphically in Figure 4.3.2. There, a red (solid) line showing how the force $F$ changes as a function of $x$ with a constant and arbitrary value of $k$.

Note that $F$ is positive when the spring is compressed $x< 0$, indicating (in our convention) a restorative force in the $+x$ direction. One might think of this as a "repulsion" as the spring would push your hand away. $F$ is negative when the spring is in tension $x > 0$, indicating a restorative force in the $-x$ direction. One might think of this as an "attraction" as the spring would act to pull on your hand. We'll use this idea of positive force as repulsion and negative force as attraction in the following sections.

Figure 4.3.2 The force (Eq. 4.3.1) and potential energy of an ideal spring as a function of displacement from equilibrium.

As we extend or compress the spring, we are storing mechanical energy in the system. We relate the force $F(x)$ and the potential energy $U(x)$ through the following relationship:

$$F(x) = - \frac{dU(x)}{dx} \tag{4.3.2}$$

In Exercise 4.3.1, the $U(x)$ is plotted as the purple dashed line. Eq. 4.3.2 says that the restoring force is equal to the negative derivative of the potential energy function. Intuitively, you can think of the potential function as a landscape that a ball is rolling around in. The ball feels a force that causes it to roll downhill. The magnitude of that force is proportional to the slope at the the position it is located. When potential energy is at a local minimum the force is zero (this defines equilibrium, all forces are balanced). The further the ball is displaced from this local minimum, the higher its energy.

Let's explore the spring model in Exercise 4.3.1.

Exercise 4.3.1: Spring Model Exploration
Not Currently Assigned

Note that it can be useful to communicate mathematics using $\LaTeX$. You don't need to know much about $\LaTeX$, but if you want to nicely show mathematical expressions, you can use a pallette-based editor like this one.

This exercise should take around 15 minutes.

  1. As the spring stretches, does it get harder or easier to stretch it further?

  2. Does it take more energy to pull the spring from $x=0$ to $x=1$ or from $x=1$ to $x=2$? Why?

    Note: Stretching the spring (increasing its potential energy) requires energy in the form of mechanical work. Mechanical work equals force times distance $W=Fd$.

  3. Explain why the potential energy function for the spring increases superlinearly (faster than linear) with distance based on the mathematical model we've introduced (Eq. 4.3.1).

  4. Now that you know why the potential energy function for the ideal spring increases superlinearly, show some math to show why it is specifically parabolic. In the answer box, you don’t need to type out the math if you don’t want to (although you can, of course, using this editor and encolsing your syntax in dollar signs: i.e., $F(x) = -kx$ will show $F(x) = -kx$.

    If you don't want to do that, just describe the steps you took.

  5. Assume that the spring constant $k = 1\,\mathrm{N/m}$. What is the potential energy of the spring when stretched to $1 \,\mathrm{m}$? How about $2 \,\mathrm{m}$? Make sure your answers agree with the qualitative answers from the previous parts. (i.e., define the potential energy to be zero when the spring is at equilibrium).