# Checking my work on a Lennard-Jones problem

1. Sep 29, 2013

### Emspak

EDIT: Found a silly algebra mistake. But let me know if I got this right.

1. The problem statement, all variables and given/known data

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

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 $U(r)$. I get:

$$\frac{dU(r)}{dr} = -12 \rho^{12} r^{-13} + 6 \rho^6 r^{-7}$$

and making that equal to zero we end up with $r^6 = -2 \rho^6$

So far so good, that means $r_{min} = \rho 2^{\frac{1}{6}}$

Now to show that it is a simple harmonic oscillator.

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

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

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: Sep 29, 2013
2. Sep 30, 2013

### Staff: Mentor

This equation implies that $r$ is imaginary, which is incorrect.

That is correct.

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. Sep 30, 2013

### Emspak

So would I approach it like this:

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

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:

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

which gets me

$$\frac{12 \rho^{12}}{(\rho 2^{\frac{1}{6}})^{-13}} - \frac{6 \rho^{6}}{(\rho 2^{\frac{1}{6}})^{7}} = \frac{6}{r}$$ for the first term, and $$\frac{(-156) \rho^{12}}{(\rho 2^{\frac{1}{6}})^{-14}} - \frac{(-42) \rho^{6}}{(\rho 2^{\frac{1}{6}})^{8}} = \frac{-156}{\rho^2 r^2} - \frac{42}{\rho ^2 r^2}$$ for the second term and so on.

4. Sep 30, 2013

### Staff: Mentor

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. Sep 30, 2013

### Emspak

Would it not go to zero? (if $r_{min} = r$ that would seem logical). And in that case,

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

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

I'm missing something here, obviously.

6. Sep 30, 2013

### Staff: Mentor

Exactly. So we can take
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. Sep 30, 2013

### Emspak

but wouldn't the other derivatives go to zero as well? $(r_{min}-r)^2 = 0$ if $r_{min}=r$ 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. Sep 30, 2013

### Staff: Mentor

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 $x = 0$?

It is certainly not 0. Try to derive it again.

9. Sep 30, 2013

### Emspak

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 $r_{min} = r$ 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:

$$\frac{1}{2!}(r_{min} - r)^2\frac{d^2U(r_{min})}{dr^2}$$

the part of it that is $(r_{min} - r)^2$ has to make the whole thing zero, doesn't it? What am I missing from this derivative?

10. Sep 30, 2013

### Staff: Mentor

But it is zero only at $r = r_{min}$, not for a displacement from equilibrium. Let me quote you again the OP:

11. Sep 30, 2013

### Staff: Mentor

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. Sep 30, 2013

### Emspak

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. Sep 30, 2013

### Staff: Mentor

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.