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Lennard-Jones-Potential: Equation of motion

  1. Jun 9, 2014 #1
    1. The problem statement, all variables and given/known data

    Hello everybody,

    i have got the following task to solve.

    Given the following potential: [itex]U_{LJ}(\vec{r}) = D \Big[ \big( \frac{R}{r}\big)^{12} - 2* \big(\frac{R}{r} \big)^{6} \Big] [/itex]
    where r means the length of [itex] \vec{r} [/itex]

    Give an equation of motion explicitly of the form [itex] m * \frac{\partial^{2} \vec{x}}{\partial t^{2}} = -\nabla U_{LJ}(\vec{r})[/itex]

    2. Relevant equations

    I normally have no problem to find the conservative force for a given potential. I would simply use this equation: [itex] \vec{F}(\vec{r}) = -\nabla U(\vec{r})[/itex]

    3. The attempt at a solution

    I finally got this equation of motion with respect to x:

    [itex]\vec{F} = - \frac{\partial}{\partial x} U_{LJ}(\vec{r})\\
    \\
    \vec{F} = - \Bigg[ \frac{\partial}{\partial x}D \Big[ \big(\frac{R}{x}\big)^{12} - 2* \big(\frac{R}{x}\big)^{6} \Big] \Bigg] \\
    \\
    \vec{F} = - \Bigg[ D * \Big[ 12 * \big(\frac{R}{x}\big)^{11} * \big(-\frac{R}{x^2}\big) - 12* \big(\frac{R}{x}\big)^{5} * \big(-\frac{R}{x^2}\big) \Big] \Bigg] \\
    \\
    \vec{F} = 12 * D * \Big[ \big(\frac{R^6}{x^7}\big) - \big(\frac{R^{12}}{x^{13}}\big) \Big] \\[/itex]

    I am not sure if this is really the solution to the task above, because it was explicitly asked for an equation of motion with respect to time. On the other hand the acceleration is a function with respect to time and part of the equation of force (Newtons second law). I am a little bit confused now, so can anybody tell me if i have the right solution? If not, please tell me what i did wrong. I thank you all in advance for every kind of help for this solution.
     
    Last edited: Jun 9, 2014
  2. jcsd
  3. Jun 15, 2014 #2

    BruceW

    User Avatar
    Homework Helper

    It looks good to me. Do they really ask for the equation of motion as a time-dependent equation? If they just said "Give an equation of motion explicitly of the form..." then I'm pretty sure you've done what they wanted you to do.
     
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