# Potential energy graphs of chemical systems

I really wish someone could dissect this for me in the language of physics.

The system is that of two atoms that approximate each other along 'r' (or distance themselves apart).

My concerns are these:

• What does negative potential energy mean here?
• Where is the "zero" reference defined?
• Why is there reference to repulsion and attraction in the PE graph?
• If $F=-dU/dr$, shouldn't the graph of the attractive force be flipped upside down? (Specifically they say the Energy of attraction is $-1.436/r$ and the energy of repulsion is $7.32x10^-6/r^8$.)
• My textbook makes reference to $U=\int F\,dr$. Why isn't it defined as the negative work in this system?
• Are these ions?
– BMS
Aug 21 '14 at 5:19
• @BMS Not necessarily. It could be 2 hydrogen atoms approaching each other.
– LDC3
Aug 21 '14 at 5:47

What does negative potential energy mean here?

Not much. The particular value of potential energy isn't important at all in classical physics. But changes in potential energy are. You could shift everything up so that $U>0$ everywhere, and you'd still get the same physics. (Why, you ask? Well, would the force change?)

So why would people choose to have negative potential energy? Read on.

Where is the "zero" reference defined?

The typical choice one makes in systems where there is no interaction at infinite distance is to set $U=0$ at infinity. That's the zero reference here; $U(r=\infty)=0$ (in non-careful physicist speak). The fact that there is a negative potential energy is a consequence of this (and of having a long-range attractive force).

Why is there reference to repulsion and attraction in the PE graph?

From far away the molecules are attracted to each other, while at close range they repel each other. You will have to read your textbook more closely to see why your problem is choosing to model the interaction like this. But the attractive term that goes as $-1/r$ looks like Coulomb attraction to me. Others might be able to identify the repulsive $1/r^8$ term.

By the way, take a look at the Lennard-Jones potential if you're unfamiliar with it. It's a way to model the attraction and repulsion of two neutral molecules.

(These next two point are related, so don't read just one of them.)

[...] shouldn't the graph of the attractive force be flipped upside down?

They've done something a bit odd and defined an attractive force to be positive, which seems to go against the coordinate system in the potential energy diagram. An attractive force (meaning toward the origin) should have a negative sign (if $r>0$).

So it seems they're using $F=dU/dr$ for the relation between force and potential energy. This is related to the next question...

Why isn't it defined as the negative work in this system?

It depends what is meant by $F$. Most physicists would write $U=-\int F\,dr$ and know that, loosely speaking, $F$ is the conservative force between the two particles. But you could choose to have $F$ be an external force that overcomes the internal one, which would introduce another negative sign.

So perhaps they mean that $F$ is the force you would have to exert in order to keep the two atoms stationary. In the "attractive" region, they want to move toward each other, so the force on whatever particle whose position is labeled by $r$ should be negative, meaning you need to exert a positive force.

If you look at the 2 graphs, the force varies directly as the slope of the potential energy curve ($F\propto$ slope of $PE$)(or is it the other way around?). There is a reference to attraction in the PE graph because the atoms will stay together if the distance keeps the PE negative. If the distance were to decrease so that the PE is positive, the atoms would fly apart (and never come together again).

When the atoms are far apart and approaching each other the force is increasing, so the PE is decreasing. When the PE is at a minimum, there is no force acting on the 2 atoms.

If the distance becomes shorter, there is a force to repel the atoms away from each other.

In this diagram F is the force each of the atoms feels towards the other atom. If these were ions (such as sodium and chloride) then you would know that the attractive force is from unlike charges. For 2 hydrogen atoms, the attractive force is a combination of the protons and electrons to form a chemical bond.

What does negative potential energy mean here?

It's similar to negative temperature. You need to pick a zero point somewhere.

My textbook makes reference to $U=∫Fdr$. Why isn't it defined as the negative work in this system?

It's because the limits are not what you expect. The integral is actually $U=\int ^{0}_{\infty }{Fdr}$ instead of $U=\int ^{\infty }_{0}{Fdr}$.

If you take 2 rubber balls as a model, you can see how the repulsive force is responsible for keeping the centers apart. Unfortunately, our model lacks the strong attractive force that the graphs represents. A similar force on the same scale would be 2 very strong magnets placed inside the balls. If the balls are touching and you move the balls apart, work is required. If the balls are touching and you move the centers closer together, again work is required. The minimum PE could not be used as a zero point since it is different for each set of atoms.

• What made them define the "attractive energy" as a function of the form -1/r ? I guess that's what bugs me the most: why a negative function?
– DLV
Aug 23 '14 at 14:50

In general, the absolute energy of a physical system does not matter for these potential problems. However, for convenience sake, one often choses to make the potential energy term at infinite separation equal to zero.

If you were to hold the two atoms infinitely far away from each other, they will first experience an attractive force, because the potential energy decreases as they get closer. The most potential energy that one can extract from this attraction is E_0. Below the radial distance at which the system has its minimal energy, the force becomes repulsive, and one would have to expend energy to push the two atoms closer together. As for the sign, if I see this correctly, integrating from plus infinity towards zero with a positive definition of F will make dr negative, which will lead to a negative U, so the sign seems to be chosen consistently. Another way to see this is to calculate the force as the derivative of the potential energy, which, again, has the correct sign for the way it's shown in the diagrams. Note that the force curve is negative where the potential is falling, goes trough zero where the potential has its minimum and then becomes positive between the potential minimum and infinity.