# Figure Description of Equilibrium in a Crystal Lattice

I am reading an Introductory Semiconductor Physics by Parker. The textbook states the following:

The ionic crystal itself is formed by the ions coming together in an equilibrium configuration. Figure 1.2 shows how such an equilibrium configuration can come about. There is an atomic separation, $$r_0$$, at which there is balance between the repulsive and attractive electrostatic forces.

Figure 1.2:

1. Why is it that the energy ($$E_0$$) at "balance" ($$r_0$$) is still negative? Naively, it seems like it should be $$0$$?

2. After a certain distance, "repulsion" goes to $$0$$ and, it seems, remains there, whilst "attraction" remains negative and, it seems, continues to remain negative, only equalling $$0$$ at a distance of infinity?

3. What is "total energy" supposed to be, and why does it equal "attraction" at some point slightly beyond $$r_0$$ and remain equal as far as can be seen?

4. What does negative "Energy" on the $$y$$-axis even mean? There exists the concept of "negative energy" in physics, but, according to the Wikipedia article, it is only used to describe a phenomena in certain (other) areas of physics.

I would greatly appreciate it if people could please take the time to clarify this.

1. In physics, when we talk about equilibrium we invoke the local extrema of the energy. The dynamics of a system are completely determined by the forces that applies on it (Newton's second law). The gradient of an energy is the "force" on the system. The energy's "sign" is not necessarily relevant, only the sign of it's gradient. Where the gradient of the energy is null, the force is null and thus the system is in "equilibrium" (stable or unstable equilibrium depends of the sign of the curvature of the energy at that point), whatever the sign of the energy is.

As an example, the total energy of a system can always be changed by a constant because this constant doesn't affect the gradient. Thus, the force on the system is unaffected by such a constant and the dynamics of the system are untouched. In that way, we could just add an arbitrary constant to the energy above to make it positive and it wouldn't change the system at all.

2. In your example, when $$r\rightarrow 0$$, the energy goes to infinity really fast and thus, the gradient of the energy blows up as well. Thus if you approach $$r=0$$, the "force" to go back to equilibrium will be very strong. On the contrary, if $$r\rightarrow\infty$$, we see that the gradient of the energy is slowly going to 0. Thus, the force on the system vanishes at very far distances.

3. The potential you are looking at looks a lot like the Lennard-Jones potential which is made of 2 parts:

$$E(r)=\epsilon\left[\left(\frac{\sigma}{r}\right)^{12}-\left(\frac{\sigma}{r}\right)^6\right]$$

In this particular case, the total energy is considered to be the sum of an attractive interaction (the power to the 6th part) and a repulsive interaction. Taken separately, you see that the attractive part is dominant at intermediate distance (near $$r_0$$), that the repulsive part is dominant when $$r\rightarrow 0$$ whilst all interactions vanishes at large distance. So the sum of both terms gives the graph you see.

4. As explained above, the sign of the energy does not really matter in physics because you can always add a constant to it and it would not change the dynamics of the system. The "negative energy" you pointed in your question are related to negative mass which is a totally different topic.