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Checking my work on a Lennard-Jones problem

  1. Sep 29, 2013 #1
    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 [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: Sep 29, 2013
  2. jcsd
  3. Sep 30, 2013 #2

    DrClaude

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    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.
     
  4. Sep 30, 2013 #3
    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.
     
  5. Sep 30, 2013 #4

    DrClaude

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    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?
     
  6. Sep 30, 2013 #5
    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.
     
  7. Sep 30, 2013 #6

    DrClaude

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    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##)
     
  8. Sep 30, 2013 #7
    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.
     
  9. Sep 30, 2013 #8

    DrClaude

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    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.
     
  10. Sep 30, 2013 #9
    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?
     
  11. Sep 30, 2013 #10

    DrClaude

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    Staff: Mentor

    But it is zero only at ##r = r_{min}##, not for a displacement from equilibrium. Let me quote you again the OP:
     
  12. Sep 30, 2013 #11

    DrClaude

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    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.
     
  13. Sep 30, 2013 #12
    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!
     
  14. Sep 30, 2013 #13

    DrClaude

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    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.
     
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