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My question is fairly simple, but I do need clarification on how to get the inverse of the Lennard-Jones potential V(x). I am working with the following expression: $$ V(x) = e\times[(R/x)^{12} -2\times(R/x)^6] $$ So given a value $V$, how can I find $x(V)$ ? |
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What you are asking is to take the inverse of the function $V(x)$. Using the $z=(R/x)^6$ as suggested, you will get the quadratic equation: $$z^2-2z-V/e=0$$ and the solution are $$ z_{\pm} = 1 \pm \sqrt{1+V/e}$$ and so $$x_{\pm}(V) = \frac{R}{(1\pm\sqrt{1+V/e})^{1/6}} \tag{1}$$ Note that there are at most two real solutions (not twelve) as shown in the figure. The equation (1) needs the condition $z_{\pm}>0$ hold which means $ x_{+} $ exists for $V>-e$ and $x_{-}$ exists for $-e>V>0$
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Substitute $u=\left(\frac{R}{x}\right)^6$: $V=e(u^2-2u)$ $u^2-2u+\frac V e=0$ Solve the quadratic equation: $u_{1,2}=1\pm \sqrt{1+\frac V e}$ Insert into the substitution: $x_{1,2}=R u^{-\frac 1 6}=R \left( 1\pm \sqrt{1+\frac V e}\right)^{-\frac 1 6}$ |
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