• Support PF! Buy your school textbooks, materials and every day products Here!

Checking my work on a Lennard-Jones problem

  • Thread starter Emspak
  • Start date
  • #1
243
1
EDIT: Found a silly algebra mistake. But let me know if I got this right.

Homework Statement



The Lennard-Jones potential is [tex] U(r) = \left[ \left( \frac{\rho}{r} \right)^{12} - \left( \frac{\rho}{r} \right)^6 \right][/tex].

What is the equilibrium distance? And can you show that the movement around that equilibrium distance is a simple harmonic oscillator?

OK, so I know that the equilibrium distance has to be where the force is zero. So all I need do is take a derivative of [itex]U(r)[/itex]. I get:

[tex] \frac{dU(r)}{dr} = -12 \rho^{12} r^{-13} + 6 \rho^6 r^{-7} [/tex]

and making that equal to zero we end up with [itex] r^6 = -2 \rho^6 [/itex]

So far so good, that means [itex]r_{min} = \rho 2^{\frac{1}{6}} [/itex]

Now to show that it is a simple harmonic oscillator.

[tex] F(r) = -12 \rho^{12} r^{-13} + 6 \rho^6 r^{-7} = ma = m \ddot r [/tex]

so

$$ \ddot r = \frac{F(r)}{m} = -12 \frac{\rho^{12}}{r^{13}m} + 6 \frac{ \rho^6} {r^{7}m} $$

When I plug in the value of [itex]r_{min}[/itex] I end up with [itex] \ddot r (0) = \frac{-3}{mr} - \frac{3}{mr} = \frac {-6}{mr} [/itex]. Which would seem to indicate that the natural frequency is [itex]\sqrt{\frac{-6}{mr}}[/itex] which is [itex]\sqrt{\frac{-6}{m} \frac{1}{2^{1/6}\rho}}[/itex]

But that's an imaginary quantity so I think I messed up somewhere. If anyone can point out a mistake that would be most appreciated.
 
Last edited:

Answers and Replies

  • #2
DrClaude
Mentor
7,147
3,281
and making that equal to zero we end up with [itex] r^6 = -2 \rho^6 [/itex]
This equation implies that ##r## is imaginary, which is incorrect.

that means [itex]r_{min} = \rho 2^{\frac{1}{6}} [/itex]
That is correct.

Now to show that it is a simple harmonic oscillator.

[tex] F(r) = -12 \rho^{12} r^{-13} + 6 \rho^6 r^{-7} = ma = m \ddot r [/tex]

so

$$ \ddot r = \frac{F(r)}{m} = -12 \frac{\rho^{12}}{r^{13}m} + 6 \frac{ \rho^6} {r^{7}m} $$
That has nothing to do with a harmonic oscillator. How would the equation look like for a harmonic oscillator? The "standard" way to derive this is to make a Taylor expansion around the equilibirum position and show that it can approximate a harmonic oscillator.
 
  • #3
243
1
So would I approach it like this:

[tex]U(r_{min}) + (r_{min} - r) \frac{dU(r_{min})}{dr} + \frac{1}{2!}(r_{min} - r)^2\frac{d^2U(r_{min})}{dr^2}+...[/tex]

is a taylor expansion of the equation I have for potential. But when I do that I seem to get a lot of terms that don't look much like an SHO equation.

That is, I would get:

[tex]U(r_{min}) + (r_{min} - r) \frac{dU(r_{min})}{dr} + \frac{1}{2!}(r_{min} - r)^2\frac{d^2U(r_{min})}{dr^2}+...[/tex]

which gets me

[tex]\frac{12 \rho^{12}}{(\rho 2^{\frac{1}{6}})^{-13}} - \frac{6 \rho^{6}}{(\rho 2^{\frac{1}{6}})^{7}} = \frac{6}{r}[/tex] for the first term, and [tex]\frac{(-156) \rho^{12}}{(\rho 2^{\frac{1}{6}})^{-14}} - \frac{(-42) \rho^{6}}{(\rho 2^{\frac{1}{6}})^{8}} = \frac{-156}{\rho^2 r^2} - \frac{42}{\rho ^2 r^2}[/tex] for the second term and so on.
 
  • #4
DrClaude
Mentor
7,147
3,281
[tex]U(r_{min}) + (r_{min} - r) \frac{dU(r_{min})}{dr} + \frac{1}{2!}(r_{min} - r)^2\frac{d^2U(r_{min})}{dr^2}+...[/tex]
Let me rewrite the second term in the more mathematically correct
$$
(r_{min} - r) \left. \frac{dU(r)}{dr} \right|_{r = r_{min}}
$$
What can you say about that derivative?
 
  • #5
243
1
Would it not go to zero? (if [itex]r_{min} = r[/itex] that would seem logical). And in that case,

[tex]U(r_{min}) = \left[ \left( \frac{\rho}{r_{min}} \right)^{12} - \left( \frac{\rho}{r_{min}} \right)^6 \right] = \left[ \left( \frac{\rho}{\rho 2^{\frac{1}{6}}} \right)^{12} - \left( \frac{\rho}{\rho 2^{\frac{1}{6}}} \right)^6 \right] = \left[ \left( \frac{1}{4} \right) - \left( \frac{1}{2} \right) \right] = - \frac {1}{4}[/tex]

But that leaves you with just 1/4, not an SHO expression. But plugging in the [itex]r_{min}[/itex] to the Force expression (the derivative of U) and I end up with [itex]\frac{6}{\rho 2^{\frac{1}{6}}}[/itex] or [itex]\frac{6}{r_{min}}[/itex] in the other terms (with second and higher-order derivatives) but we just established that if [itex]r_{min}=r[/itex] they go to zero.

I'm missing something here, obviously.
 
  • #6
DrClaude
Mentor
7,147
3,281
Would it not go to zero?
Exactly. So we can take
[tex]U(r_{min}) + (r_{min} - r) \frac{dU(r_{min})}{dr} + \frac{1}{2!}(r_{min} - r)^2\frac{d^2U(r_{min})}{dr^2}+...[/tex]
and rewrite it as
$$
U(r) \approx U(r_{min}) + \frac{1}{2}\frac{d^2U(r_{min})}{dr^2} (r_{min} - r)^2 \text{ for } r \approx r_{min}
$$
Doesn't this remind you of something? (Hint: set ##x \equiv r_{min} - r##)
 
  • #7
243
1
but wouldn't the other derivatives go to zero as well? [itex](r_{min}-r)^2 = 0 [/itex] if [itex]r_{min}=r [/itex] and that would make the second term zero. I see that it looks like an SHO but i can't get my head around the fact that the second derivative seems to be zero as well.
 
  • #8
DrClaude
Mentor
7,147
3,281
but wouldn't the other derivatives go to zero as well? [itex](r_{min}-r)^2 = 0 [/itex] if [itex]r_{min}=r [/itex] and that would make the second term zero.
Yes, the second term is 0 at ##r = r_{min}##, and positive everywhere else. What is the difference between that and
$$
V(x) = \frac{k}{2} x^2
$$
at ##x = 0##?

I see that it looks like an SHO but i can't get my head around the fact that the second derivative seems to be zero as well.
It is certainly not 0. Try to derive it again.
 
  • #9
243
1
I did derive it again. That was the really ugly expression I got involving 156 et cetera. and it doesn't alter the fact that if [itex]r_{min} = r[/itex] then subtracting one from the other makes it zero and if you multiply that by the second derivative you get zero! Again I see that it is supposed to look like an SHO expression, but I am getting zeros in the denominator and the Taylor expansion makes the higher-order derivati go to zero in any case.

I mean, look, we have this term:

[tex]\frac{1}{2!}(r_{min} - r)^2\frac{d^2U(r_{min})}{dr^2}[/tex]

the part of it that is [itex](r_{min} - r)^2[/itex] has to make the whole thing zero, doesn't it? What am I missing from this derivative?
 
  • #10
DrClaude
Mentor
7,147
3,281
I mean, look, we have this term:

[tex]\frac{1}{2!}(r_{min} - r)^2\frac{d^2U(r_{min})}{dr^2}[/tex]

the part of it that is [itex](r_{min} - r)^2[/itex] has to make the whole thing zero, doesn't it? What am I missing from this derivative?
But it is zero only at ##r = r_{min}##, not for a displacement from equilibrium. Let me quote you again the OP:
And can you show that the movement around that equilibrium distance is a simple harmonic oscillator?
 
  • #11
DrClaude
Mentor
7,147
3,281
That was the really ugly expression I got involving 156 et cetera.
Forgot to say: to answer the question, you don't need the exact expression. Finding that
$$
\left. \frac{d^2U(r)}{dr^2} \right|_{r = r_{\min}} = \text{const.} > 0
$$
is sufficient.
 
  • #12
243
1
Aha! You don't need to consider where it goes to zero only, you need to look at it as a displacement and since the other derivatives will be constants, you get the form that looks like an SHO. I think I understand this now!
 
  • #13
DrClaude
Mentor
7,147
3,281
I think that what you missed was that the result is a function of ##r##. I hinted at that in two places in post #6, but I might have been too subtle.
 

Related Threads on Checking my work on a Lennard-Jones problem

  • Last Post
Replies
14
Views
706
  • Last Post
Replies
15
Views
5K
  • Last Post
Replies
3
Views
497
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
4K
Replies
1
Views
1K
  • Last Post
Replies
3
Views
821
  • Last Post
Replies
2
Views
10K
Replies
4
Views
4K
Top