This is Example 2.3.4 from Analytical Mechanics 7th Ed by Fowles & Cassiday. It uses the Morse function $V(x)$, given as $$V(x) = V_0[1 - e^{-(x-x_0)/\delta}]^2 - V_0.$$

The question is:

"Show that for the separation distances $x$ close to $x_0$, the potential energy function is parabolic and the resultant force on each atom of the pair is linear, always directed toward the equilibrium position."

Answer given:

"All we need to do here is expand the potential energy function near the equilibrium position." $$V(x) = V_0[1 - [1 - [ \frac{x-x_0}{\delta}]]]^2 - V_0.$$

I understand the steps after this, but I'm confused as to how we got here from the potential energy function and what "expand...near the equilibrium" means. I tried to apply a Taylor expansion and didn't get anything close to the same answer.


It means a Taylor expansion about the point $x=x_0.$ Recall that the general Taylor expansion formula is:$$f(x) \approx f(x_0)+ f'(x_0)~(x - x_0) + \frac12 f''(x_0)~(x-x_0)^2 + \dots,$$where the terms on the right hand side come from ensuring that the $n^\text{th}$ derivative at the point $x_0$ is the same on both sides, hence being $f^{(n)}(x_0)~(x-x_0)^n/(n!).$

The equilibrium position can be more formally defined as the position $x_0$ where $V'(x_0) = 0$, this is because the first derivative of potential energy is closely tied to forces and "equilibrium" means roughly "there's no net force either way".

The trickery they're using

Taylor expansions have a property which allows you to also expand inside an expression, if you have some insight about it. For example if the Taylor expansion around $u=0$ for $g(u)$ is $g_0 + g_1~u + \frac12 g_2~u^2 + \dots$ and the Taylor expansion around $x=x_0$ for $f(x)$ is $f_0 + f_1~(x-x_0) + \frac12 f_2 ~(x-x_0)^2 + \dots$ then it is formally allowed to write the Taylor expansion of $g(f(x))$ as the composition, $$g\big(f(x)\big) = \sum_{m}\frac1{m!}g_m~\left(\sum_n \frac1{n!}f_n~ (x - x_0)^n\right)^m.$$ This is a very complicated expression, but one can then begin to extract out terms of their lowest orders just by reasoning "this is the only way to get an $(x - x_0)^0$ term, this is the only way to get an $(x - x_0)^1$ term, these are the only ways to get $(x - x_0)^2$ terms," and so on. In your case $g(u)=u^2$ is a really simple expression to do this with, squaring the above series gives $$\big[f(x)\big]^2 = \sum_{mn} \frac1{m!}\frac1{n!} f^{(m)}(x_0)~f^{(n)}(x_0)~(x-x_0)^{m+n}.$$To get a constant term one finds that the only contribution is $(m, n)=(0, 0)$ and the constant term is just $[f(x_0)]^2$. To get a first order term one only needs to consider $(m,n) = (0,1)$ and $(m,n) = (1, 0)$ and this leads to the first-order term $2~f(x_0)~f'(x_0)$, which you can confirm by the chain rule. To get the second-order term you need to consider $(0,2)$, $(1,1)$, and $(2,0):$ these give an expression $f(x_0)~f''(x_0) + f'(x_0)~f'(x_0).$

What they are observing is that this inner expression $f(x) = 1-e^{-(x-x_0)/\delta}$ has been carefully crafted so that $f(x_0) = 0.$ As you can see this means that the expansion is something like $\alpha~(x-x_0) + \beta~ (x - x_0)^2 + \dots$ and after squaring the only quadratic term is going to be $\alpha^2~(x-x_0)^2$ coming from this $f'(x_0)~f'(x_0)$ term; the $f(x_0)=0$ fact nukes the constant term, the first derivative term, and the part of the second derivative term that would come from the second derivative of the exponential. They just "saw" this without needing any further remark, and as a result $x_0$ is indeed the point where $V'(x_0) = 0$ and therefore it is the proper point to expand around.


This just uses a Taylor expansion of $e^{u}$ about $u=0$ truncated at first order:$$e^{u}=1+u+\frac{u}{2}+...$$ where in this case $u=\frac{x-x_0}{\delta}$


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