Talk:Lennard-Jones potential/Archive 1

This is an archive of past discussions about Lennard-Jones potential. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.

Archive 1

Old Posts

Doesn't the Lennard-Jones potential describe the potential energy between two atoms at a distance r and not just the strength of the two forces when they balance as stated by the first paragraph? Zapateria 17:52, May 25, 2005 (UTC)

Good point. The article assumes the reader knows that the force is given by

f(r)=-\nabla V(r)

. Perhaps it should state it?

I would be interested in seeing analytical solutions for properties of the Lennard-Jones potential, assuming that there are any. I've not been very successful in finding any myself or in the literature. 66.21.139.18 12:42, 27 March 2006 (UTC)Gabriel Hanna

Probably the most interesting analytical property is the position of the minimum, which can be obtained using differential calculus (I think it was something like sigma*2^(1/12) but I'm too lazy to check right now...). Itub 23:53, 8 April 2006 (UTC)

The minima of the potential is at 2^1/6 sigma. The first zero point is at sigma. I'm not sure what exactly you mean by analytical solutions for the properties since it's unclear what properties you are talking about. Simpleliquid 17:40, 18 December 2006 (UTC)

Or if you are thinking about rare gases have a look here: http://link.aps.org/abstract/PRB/v73/e064112; doi: 10.1103/PhysRevB.73.064112.

Hard sphere diameter

I think that "hard sphere diameter" should be defined, or at least there should be a link to a page that defines it. Zeroparallax 21:23, 1 September 2006 (UTC)

Oh! And shouldn't it be the "hard sphere radius" rather than "diameter"? Zeroparallax 21:26, 1 September 2006 (UTC)

The hard sphere diameter is not a well defined quantity for a Lennard-Jones system. I removed the sentence that said that sigma was the hard-sphere radius because it was wrong. The Stokes-Einstein relationship with slip conditions, the viscosity, temperature and Diffusion constant of a LJ fluid can be used to obtain an effective hard-sphere diameter but I think this is beyond the scope of this stub. Simpleliquid 17:30, 18 December 2006 (UTC)

Mistake in formula for FORCE

F = -dV/dr = 4*epsilon*( .. ) and not 8*epsilon*( .. ) as displayed!

MISTAKE in bottom page L-J potential

... the 2 seems odd, especially since it's only on the right hand term. It should be 4 epsilon on both terms.

Definition of r

I readded a line explaining what r is in the equations as it was not defined anywhere in the article after the previous edit. Zapateria (talk) 22:13, 18 February 2008 (UTC)

Parameter

What are the numerical parameters used in the plot?

Van der Waals forces

Aren't van der Waals forces occasionally repulsive? The opening paragraph suggests they are always attractive.128.243.220.21 11:46, 21 March 2007 (UTC)

Yes, I know they can be attractive (for high r) or repulsive (for low r), as the other type of primary and secondary bonds. In fact the diagram is related to any type of bond (intermolecular or intramolecular). --Aushulz (talk) 18:00, 2 September 2010 (UTC)

I don't see how the opening sentence implies this I think this is an old post and will remove to archive soon.

Phancy Physicist (talk) 11:46, 4 September 2010 (UTC)

Origin of the L-J potential

I looked up the article which is cited as the origin for this potential, but Lennard-Jones did not, in fact, use a 12-6 potential in that paper. He used 5 for the power of the attractive term, and tried values of 14 1/3, 21, and 25 for the repulsive term. Where was the 12-6 potential first introduced? —Preceding unsigned comment added by Pkeastman (talk • contribs) 00:08, 15 April 2010 (UTC)

I don't know the origin of 12-6 exponent, but I read that for ionic bond and dipole-dipole bond the correct exponent are 9-1, while 12-6 are used for covalent bond, metallic bond and Van der Waals forces. --Aushulz (talk) 17:42, 2 September 2010 (UTC)

You are correct to observe that in J. E. Jones, “On the determination of molecular fields. II. From the equation of state of a gas,” Proc. R. Soc. A 106(738), pp. 463–477, 1924. doi:10.1098/rspa.1924.0082, he uses an exponent of 5 for the attractive term and a variety of exponents (including 14 1/3) for the repulsive term (and furthermore, these are the exponents for the interaction force, so the exponents for the potential are really 4 and 13 1/3). In a subsequent paper (J. E. Lennard-Jones, “Cohesion,” Proc. Phys. Soc. 43(5), pp. 461–482, 1931. doi:10.1088/0959-5309/43/5/301), he uses an exponent of 7 for the attractive force and an exponent of 13 for the repulsive force, which correspond to the exponents of 6 and 12 in the potential that are now in common use. --Eyliu (talk) 06:00, 6 October 2011 (UTC)

Condon-Morse potential

"Condon-Morse potential" and "Lennard-Jones potential" are the same or not? --Aushulz (talk) 17:42, 2 September 2010 (UTC)

The Morse potential is certainly different from the LJ potential. -- 140.142.20.229 (talk) 20:52, 28 September 2010 (UTC)

Lennard-Jones potential

Hello. You removed the formula with A and B coefficients. Why?

-unsigned

First, thank you for checking up on a page edit. This is the section your are referring to.

==Alternative expressions==
The simplest formulation, often used internally by simulation software, is
: $V(r)={\frac {A}{r^{12}}}-{\frac {B}{r^{6}}},$ 
where A = 4εσ¹² and B = 4εσ⁶; conversely, σ = ⁶√A/B and ε = B²/(4A).

The equation with the A and B coefficients did nothing but replace one set of constants with another set of constants. Also, it is a dubious claim since,

V(r)=4\epsilon \left[\left\{\left({\frac {\sigma }{r}}\right)^{6}\right\}^{2}-\left({\frac {\sigma }{r}}\right)^{6}\right],

is more computationally efficient. This section didn't add anything to the article and its factually questionable.

Phancy Physicist (talk) 13:21, 28 December 2010 (UTC)

1. AB form is numerically more efficient and is used in almost all programs.

2. AB form traditionally used by computer chemists, in contrast to the sigma-epsilon form.

3. The coefficient B has the physical meaning (the magnitude of the dispersion interaction), in contrast to the sigma-epsilon form.

P99am (talk) 15:11, 28 December 2010 (UTC)

Responce to P99am:

1. If you pre-calculate A,B and 4ε then both methods take 5 operations. If you don't, the AB form takes 10 operations and the εσ form takes 6 operations. However, when pre-calculating the AB form has 2 divisions, 2 exponents and 1 subtraction whereas the εσ form has 1 division, 1 multiplications, 2 exponents and 1 subtraction. Multiplication is faster then division for a processor, allowing the εσ form to be called faster. If it truly is used by most programs, then supporting information needs to be given to make this claim.

2. If this form is traditionally used by computer chemists, then this information should be added to the section in that context to denote where its use comes from.

3. Again if this section is added back it should be done with more description as to why computational chemists use it and why A and B are useful to them. This added information would make the section a valuable contribution to the article.

If there is fault with my analysis in 1. please point it out to me.

Phancy Physicist (talk) 17:25, 28 December 2010 (UTC)

OK

1. Simplified code from a real program:

           r_2     = x*x + y*y + z*z;
           lIr     = 1.0 / sqrt (r_2);
           lIr_6   = lIr * lIr;    
           lIr_6 *= lIr_6 * lIr_6;
           VdW = (A * lIr_6 - B ) * lIr_6;

We never use two divisions. It's too long operation. (- 1, + 2, * 8, Sqrt 1, / 1)

Sigma-epsilon form is really prevalent in the theoretical literature. But in real programs and real calculations common form is AB. I believe that both forms should be presented in the article.

-unsigned

Responce to code

Sorry about the extra division. I have gotten the code that I use now for comparison.

Count the operations in your code: 6 variable writes, 1 division, 1 square root, 9 multiplication, 2 additions and 1 subtraction.

Code in εσ form:

           r_2     = x*x + y*y + z*z;
           lIr     = sig2/r_2;
           lIr_6   = lIr * lIr *lIr;    
           VdW = epsilon*lIr_6*(lIr_6 - 1);

Operation count: 4 variable writes, 1 division, 0 square root, 7 multiplication, 2 additions and 1 subtraction. You could possibly trade 2 multiplications for 1 exponent but the multiplication is faster.

Also what is the line,

           lIr_eps = lIr * lIepsilon;

for? I assume you mean:

           r_2     = x*x + y*y + z*z;
           lIr     = 1.0 / sqrt (r_2);
           lIr_6   = lIr * lIr;    
           lIr_6 *= lIr_6 * lIr_6;
           VdW = (A * lIr_6 - B ) * lIr_6;

Optimize your code a bit and get:

           r_2     = x*x + y*y + z*z;
           lIr     = 1 / r_2;
           lIr_6   = lIr * lIr * lIr;
           VdW = lIr_6*(A * lIr_6 - B );

Which would leave you with the exact same number and type of operations in my code.

You get almost identical speeds but the difference will end up being the time difference to divide 1 by a number and to divide a number by a number and the time it takes to subtract a number by 1 and a number by a number. I believe that the divisions take the same amount of time if so the εσ form would be slightly faster.

However this was not my point. My point was that without citations to back it up we can not claim that any algorithm is the most widely used. I agree that the AB form is worth putting in if you say why chemists prefer it and why A and B are useful. But not the way it is written now.

Phancy Physicist (talk) 19:36, 28 December 2010 (UTC)

I brought you a piece of real code. Real code takes into account not only what is written in Wikipedia, but also many other things. No doubts that this little piece can be rewritten no less effectively (but not more) into the sigma-epsilon form. But this changes nothing. AB form is more commonly used. There is no need to prove it. I simply confirm that the author of this paragraph was right. If you think that he was not, then you have to give the references. P99am (talk) 10:15, 29 December 2010 (UTC)

I also brought a real piece of code that I personally use and along with others in my research group. I simply changed the code variables to follow your example. I have shown that the two codes can be done almost identically. Again, you can not state that most of anything commonly uses something without proof. I not claiming that most use εσ form just that you need to proof it if you claim they mostly use AB. This discussion has gone longer than I thought it would and since It seems we will never agree, I am moving this to the talk page of the article.

Phancy Physicist (talk) 15:20, 29 December 2010 (UTC)

Sorry, unfortunately I had to insist what you have to give the evidence because you have made changes to the article. This section has been in existence for 3 years. It edited by many people. Therefore, I believe that there are at least a few people who consider it valuable. In order to remove it we need more essential reasons. In this circumstances I am returning it back. If you strongly disagree with this, please, resorting to the standard public discussion to resolve the conflict.P99am (talk) 16:51, 29 December 2010 (UTC)

As I said above after you told me that the code was indeed used in software, I agreed that it is a valuable section.

I don't know enough about the reasons for using the AB form to add the information needed to make it clear why it is used. I am unsure why having dispersion strength explicit rather than calculated is preferred. My reason for moving this discussion to the talk page is that I was hoping someone would rewrite the section that has knowledge of this method of simulation. I'll try to research it but unless I find references my changing would be mostly guess work so I don't want to change it. I shouldn't have reverted P99am's adding of the section back into the article but I thought we would work on the section in talk before it was reintroduced. P99am is right that we should leave it and work on it while it is in the article so that the information in some form is available.

So would someone please rewrite the section describing the use and benefits of this form.

Phancy Physicist (talk) 17:51, 29 December 2010 (UTC)

This is an archive of past discussions about Lennard-Jones potential. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.

Archive 1

Assessment comment

The comment(s) below were originally left at Talk:Lennard-Jones potential/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

In the article it says:

" $\sigma$ is the (finite) distance at which the interparticle force is zero."

This is not quite true. Like someone else mentioned, $\sigma$ is the distance at which the potential is zero. The interparticle force will be zero when the first derivative of the potential (i.e. the force, up to a sign) is zero. That happens at $2^{\frac {1}{6}}\sigma$ .

Jasonrunguitar 16:49, 12 July 2007 (UTC)

Last edited at 16:49, 12 July 2007 (UTC). Substituted at 15:15, 1 May 2016 (UTC)