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Considering the Lennard-Jones potential,

$$V(r)=\frac{A}{r^{12}} − \frac{B}{r^{6}}$$

How does this relate to force or energy? I suspect that there is some physics equation I'm forgetting. I know that $F=−\nabla(V)$, but that's a partial derivative with respect to time, and I don't see how time comes into effect here. Also, the derivative of energy (work) with respect to displacement is force, but I'm not sure how to connect that to potential.

An example application that stimulated this question: given the interaction potential above, how would you find the interaction energy between two systems?

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    $\begingroup$ Tricky naming schemes come back to bite us once again. In my intro physics class, the "potential" was the amount of energy per unit something (e.g. energy per coulomb, or energy per kilogram). In this case though, "potential" simply refers to potential energy. $\endgroup$
    – chipbuster
    Oct 2, 2015 at 20:23
  • $\begingroup$ This helps, thanks so much! So for two systems, I guess it follows that one would just have to integrate to "add up" the interaction potential energy from all of system1 to all of system2. Is there I can rate this comment as best answer? $\endgroup$
    – halcyon
    Oct 3, 2015 at 16:03
  • $\begingroup$ No, but I can convert it to an answer if you'd like. Quick question though (I didn't really think too much about this on my first reading): what do you mean by the interaction energy between two systems? Could you give an example system? When I think of a "system," I think of two atoms separated by $r$, in which case the LJ potential completely describes the energy of the entire system. I don't fully understand what you mean by "two systems."If we had two of those systems, this potential becomes meaningless (because we have 4 atoms, and 10 different distances ($r$) to consider) $\endgroup$
    – chipbuster
    Oct 4, 2015 at 0:30

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We have some nomenclature issues here.

In physics, the electric potential and gravitational potential refer to energy per unit something. In the gravitational potential, it's usually energy per unit mass, e.g. kJ/kg. In the electric potential, it's usually J/C (more commonly known as a Volt).

Here, though, we're not talking about a potential in that sense. Rather, I suspect (though I may be wrong) that potential was once a shorthand for potential energy surface, which is a way to think about the potential energy of a system with respect to various parameters. In the case of an L-J potential, there is only one parameter, the distance between the two particles of the system.

As mentioned in my comment though, I don't understand what you mean by two systems. The Lennard-Jones potential describes the energy of one system. If you put two of these systems together, the potential becomes ill-defined.

Added later: I found today that the Wiki page on thermodynamic potential says the same thing, namely that the "thermodynamic potential" is actually an energy instead of a potential. It quotes an ISO/IEC document that I don't have access to. I'm still not sure of the origin of this nomenclature.

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  • $\begingroup$ About two systems, I was thinking about two slabs of some specified number densities that faced attraction to each other. Referring to your example of four atoms, why would the potential become meaningless? Couldn't you just add up the relationships, i.e. (AB+AC+AD+BC+BD+CD), for each of the six distances? What are you referring to when you say 10 different r distances? $\endgroup$
    – halcyon
    Oct 4, 2015 at 3:54
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    $\begingroup$ Oh, I see what you mean now. Yes, I suppose that's possible, but you'd be assuming that there's no other effects that come into play when you introduce multiple particles, and I don't know if that holds or not. $\endgroup$
    – chipbuster
    Oct 5, 2015 at 18:30
  • $\begingroup$ Also, for the 10 distances--I miscounted and included self-distances by mistake (AA,BB,etc.) . You're correct, it should be six. $\endgroup$
    – chipbuster
    Oct 5, 2015 at 18:30
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Potential is the same as energy. As to the force, you got the formula right, it's $F=−\nabla(V)$. This is not a partial derivative with respect to time, but one with respect to coordinates, see the definition.

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  • $\begingroup$ My mistake on the del operator, thanks for pointing that out. In general though, I remember learning that potential isn't the same thing as potential energy. Unless this case is, as @chipbuster noted, where potential refers to potential energy. Is that correct? $\endgroup$
    – halcyon
    Oct 3, 2015 at 15:57
  • $\begingroup$ Well, if you'd remember what reasons were called to justify this point of view (that potential isn't the same as potential energy), I would be able to either argue or agree with those. As it stands now, I can do neither. $\endgroup$
    – Ivan Neretin
    Oct 3, 2015 at 21:51

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