April 14th, 2016, 09:01 AM  #1 
Senior Member Joined: Sep 2012 Posts: 201 Thanks: 1  negative potential energy
Hello I am have a problem with the Leonard Jones energy curve. I just cant seem to get my head around how atoms when at equilibrium can have a negative p.e. I have try to look thought the internet for an explanation but it dose not seem to be sinking in. Is there anyone that can explain this to me. Big thanks in advance 
April 14th, 2016, 09:50 AM  #2 
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271 
Potential energies are measured as the energy required or evolved to bring a test charge, or mass from infinity to a specific point ie at a specific distance from the charge or mass generating the potential field. The potential energy of the charge is zero at infinity. This is conventional, it can be set anywhere by suitable choice of additive constant. So why is the minimum negative? Well the Lennard Jones potential is the energy of two atoms at various separations. It is a combination of short range repulsive forces and long range attractive forces, that balance at this minimum making it an equilibrium position. This is, of course, the bonding position where the two atoms are bonded. In order to pull the atoms apart you have to input energy, eventually debonding them. But the zero is set at infinity so the minimum must be negative so that adding energy will increase the PE from the minimum to zero. Do I need to post the curves? 
April 15th, 2016, 09:28 AM  #3 
Senior Member Joined: Sep 2012 Posts: 201 Thanks: 1 
Thank for the reply. I see where you coming from, so how dose do the particles reach 0 p.e when the are being pushed together? this means that there are two points where it has 0 p.e how can this be?

April 15th, 2016, 11:32 AM  #4 
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271 
I have no idea where that diagram came from but that is not the Lennard Jones diagram I recognise. Remember that I said the position of the horizontal axis is arbitrary to an additive constant so crossing this axis isn't particularly significant. The LJ diagram should be a plot of a/r where is separation distance and a is the horizontal position of the minimum. This obviously then occurs at a/r = 1. a is then a constant in the Lennard Jones equation, which has zero additive constant. What this means physically is that the equilibrium a/r=1 position occurs when the electron clouds of the two molecules just begin to overlap. If they approach more closely than this, from the bottom of the well, input energy rises rapidly towards infinity and you no longer have the original molecules approaching, but two increasingly distorted objects. Last edited by studiot; April 15th, 2016 at 11:52 AM. 
April 17th, 2016, 01:39 PM  #5 
Member Joined: Mar 2016 From: US Posts: 30 Thanks: 1 
'To save computational time and satisfy the minimum image convention, the LennardJones potential is often truncated at a cutoff distance of rc = 2.5σ, where: i.e., at rc = 2.5σ, the LennardJones potential, VLJ, is about 1/60th of its minimum value, ε (the depth of the potential well). Beyond rc, the truncated potential is set to zero.' From Wikipedia So the only way I am finding that the potential energy is equaling out to be a negative is one scenario: If the r is not truncated: 1.) r is the distance between the molecules. 2.) rm is the distance at which the potential reaches its minimum (however I am reading that as not 0 though) It still has some value. This also makes sense with the function of rm equaling a (epsilon) because 0 cannot be negative. 3.) rm = (epsilon). However when we examine this equation: We see that r cannot be negative, it is a distance. And so the rm cannot be a negative either. And this equation will always be a larger number minus a smaller number. Anything to the twelfth power will always be much larger number than anything to the sixth power times 2. Therefore when the epsilon is multiplied by this positive number, we are only going to get a negative lennardjones potential (VLJ) if the epsilon value is a negative. If the epsilon value cannot be a negative then I cannot see anyway that it is having a negative potential energy. It is only on its way to 0 potential energy as the particles move away from each other, or equaling 0 past a certain distance away from each other. This function is only about a pair of neutral atoms or molecules. It is not all about potential energy in other applications. See: Potential Energy  Something to note: If the r is truncated: VLJ is 1/60 of its minimum value epsilon. Beyond that potential is 0. One weird thing that can happen is if r is less than 2.5(theta). Then this happens: We see that past the rc the potential is 0. However, if the particles are closer than or equal to the rc, then vlj(r)vlj(rc) and we know that vlj(rc) is a negative epsilon. Therefore, vlj(r)(epsilon), or vlj(r)+epsilon. So if the distance is shorter than vlj(rc), how is it the entire function vlj(r) and then plus more in the epsilon? Can anyone explain? Last edited by GreenBeast; April 17th, 2016 at 01:44 PM. 
April 18th, 2016, 12:58 AM  #6  
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271  Quote:
Have you calculated the value of your second expression at this point? Hint 1 to the twelfth power = 1 to the sixth power = 1 Here is a spreadsheet showing the values when Rm/R is just greater than 1 (That is when the atoms are closer than the minimum) You can clearly see how quickly the negative 1 at the minimum becomes positive.  
April 18th, 2016, 04:58 AM  #7 
Member Joined: Mar 2016 From: US Posts: 30 Thanks: 1 
Allright, the Wikipedia page is not explaining that as well as it could. So when the pair of atoms are bonded, the potential energy is just 1 units. And as soon as they separate even by a small amount, the potential energy goes to positive number which can be calculated given this VLJ, and then as they move farther apart, the potential energy goes to 0. It seems arbitrary that the potential energy is 1 unit when they are bonded. The positive value going out to 0, could be calculated with this function, however I do not understand how you can apply this function and get a negative number such as 1, when the epsilon and RM are calculated in terms of each other. Even if you multiplied 1 to both sides of RM=(1) epsilon you will have a negative RM distance which does not make sense because they are either bonded or not, and when they are not that distance will always be a positive measurable number. For RM=epsilon, even if r and RM are both 1, in that second equation, this also seems to cause a controversy, because the potential will be epsilon times 0. So a potential of 0, at 1 unit distance. If the minimum and the maximum are the same, it should be a measurable number I think, not 0, and not1 where they are bonded. For that particular case, it is almost like it goes from 1 at bonded, skips a measurable number that we need for maximum and minimum, and goes right to 0. So it is almost like we are describing a pair that has no potential energy whether it is bonded or not. Thereby proving it cannot be 1 bonded. Last edited by GreenBeast; April 18th, 2016 at 05:40 AM. 
April 18th, 2016, 07:36 AM  #8  
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271  Quote:
I would agree that often to understand Wikipedia you have to pretty much already know the subject. Wikipedia often only provides what is known as an executive summary. So when you quote impressive formula from there it is a good idea to indicate whether you understand the formula and where it comes from. The original diagram, posted by taylor, appears to come from a different Wiki LennardJones Potential  Chemwiki This does not make it clear what is being plotted on the axes. The energy on the vertical potential axis is measured in terms of epsilon units The distance along the distance axis in terms of sigma (or Rm in our notation) and so is R divided by Rm or r/sigma The Wiki calls this r(sigma), that is r as a function of sigma. This means that the LennardJones Curve can be applied to any pair of molecules using suitable observed values of epsilon and sigma (or V and Rm). Now you seem to be questioning the following arithmetic. $\displaystyle \varepsilon \left\{ {{{\left( {\frac{{{r_m}}}{r}} \right)}^{12}}  2{{\left( {\frac{{{r_m}}}{r}} \right)}^6}} \right\}$ Note that $\displaystyle {\rm{\varepsilon ,}}\;{{\rm{r}}_{\rm{m}}}\;{\rm{are}}\;{\rm{constan ts, r}}\;{\rm{is}}\;{\rm{a}}\;{\rm{variable}}{\rm{.}}$ and when $\displaystyle {\rm{r = }}{{\rm{r}}_m}$ So $\displaystyle \frac{{{r_m}}}{r} = 1$ Substituting $\displaystyle \varepsilon \left\{ {{{\left( 1 \right)}^{12}}  2{{\left( 1 \right)}^6}} \right\} =  1(\varepsilon )$ and is the most negative the potential gets. Last edited by studiot; April 18th, 2016 at 07:39 AM.  
April 18th, 2016, 09:50 AM  #9 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
I think it's a historical artifact. Basically, when experiments were done on potential energy curves or nuclear radii in the lab, the curves would look messy with data points all over the shop. Therefore, in order to make the curves as neat as possible, you can plot them so that the smooth tail at the latter end of the curve, which is a fairly common feature for all nuclei, tends to zero. Translating the yaxis up so that the minimum of the curve is zero is fine for one curve, but looks horrible for many curves overlaid on top of each other, hence the definition. Therefore, it makes sense to define theoretical curves using similar axes, consistent with the definitions. Remember, energy transfers don't work any differently just because there's a minus sign in front. Me giving you 10 J is exactly the same as you giving me 10 J, it just looks odd at first glance. 
April 18th, 2016, 10:16 AM  #10 
Member Joined: Mar 2016 From: US Posts: 30 Thanks: 1 
studio, Your spreadsheet rightmost column should be to the twelfth power minus to the sixth power. Yep, you are right about my math being wrong. It was not 0 times epsilon for r=1 and rm=1. That ends up being 1 times epsilon. So VLJ=1 for r=1 and rm=1. If rm is a constant at 1 (or 1) epsilon since it does not seem to make a difference, and the VLJ=1 or VLJ=1 at r=1(epsilon energy unit), then as r increases it seems that this expression is just being used to say that there is one potential energy unit, either positive or negative, and since that is not clearly defined, then noone knows whether it the potential energy should increase or decrease as the particle moves farther away. Certainly past a certain point it is 0. Agreed. However infinitely close together without bonding and the potential is high. And when they do bond, it is either 0 or negative depending on this equation only seemingly. I say seemingly, because only gravitational potential energy can be negative because: 'The potential energy U is equal to the work you must do against that force to move an object from the U=0 reference point to the position r.'  link So an object fighting against gravity gets a negative PE, not an electrical charge attraction as I am guessing this LennardJones energy curve is defining. In the Joules example, do you think that is based on some kind of unknown logic, without scientific basis. You giving me 10KJ makes sense. You giving me 10KJ does not make sense. These numbers have to be used in formulas, outside of their original formulas to solve for them. (for example at the power plant to find out how much power to route to a persons house. How are they supposed to work with KJ. They need the final number to be positive only for simple arithmetic. You cannot charge someone money for a negative product.) Maybe people long ago were trying to balance the conservation of energy equations, and without success, just began with to many constants that make it seem like a negative energy unit is the same as a positive energy unit. Last edited by GreenBeast; April 18th, 2016 at 11:13 AM. 

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