0
$\begingroup$

I have the two following graphs representing a particles potential energy in function of its distance to another particle and another one which represents the attraction and repulsion forces between two particles with respect to their intermolecular distance.

enter image description here

As you can see $E_p$ is minimal when $F=0$. However potentially energy (as I know it) is given by $E_p = -\int \vec{F}.d\vec{r}$. So if F=0 this multiplication should clearly result in 0, but it doesn't. Why?

PS: follow-up question: knowing that $\frac{-dU}{dx} = F(x)$. I cannot see this relationship between the two graphs,i.e. completely on the left of the $E_p$,r - graph, you can clearly see the curve has a negative slope. Meaning the derivative should be negative, however this is not what is being shown on the lower graph. The only "correct" thing I see is that when the slope of the upper graph is 0 (at $r_0$)the derivative indicates that correctly.

Do I read the graphs incorrectly?

$\endgroup$
  • $\begingroup$ $E_p=-\int F\cdot dr$ is an integral, and it will only equal the potential energy if you integrate it from $r=\infty$ up to whatever $r$ value you're interested in... $\endgroup$ – lemon Aug 22 '16 at 12:05
  • $\begingroup$ Not quite. It's just indefinite integral, and it's defined up to an additive constant, as energy is. $\endgroup$ – Ruslan Aug 22 '16 at 12:05
0
$\begingroup$

I believe this is a Lennard-Jones potential or something like that (Buckingham maybe?).

I personnaly prefer to define the force acting on the particle as the derivative of the potential it sees. In this case: $$\overrightarrow{F}(\overrightarrow{r})=-\overrightarrow{\nabla}(E_p(\overrightarrow{r}))$$

Just like you said, in this form you can define $E_p$ adding any constant you want without changing the value of the force. Be careful as these graphs are plotted with regards to the radius $\overrightarrow{r}$ between two particles. It may be confusing to use $x$ as you have to understand that in this case, you're in the reference frame of one of the particles.

In simple word, the energy value of a single point is of no real interest and it's more interesting to see the difference between states. In this regards it is common to set the 0 (i.e. the reference state) as the radius at which two particles do not see each other (obviously infinity) and LJ potentials are nicely converging to that value as $r$ goes bigger. The depth of the well $E(r_0)$ is then the difference between a state where two particles do not see one another and the equilibrium state where there is compensation of attractive and repulsive forces. Usually the potential are set so that the energy curve is set to zero at the radius $\sigma$ where particles start to penetrate each other.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.