Question

By expanding an arbitrary$\mathit{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ in a power series about a local minimum assumed to be at$\mathit{x}\mathbf{=}\mathit{a}$ , prove that the effective spring constant is given by equation$\left(10-3\right)$ .

Short answer

The effective spring constant is given by the expansion ${\overline{)k=\frac{d^{2}U\left(x\right)}{d{x}^2}}}_a$.

Step-by-step solution

Given data

a power series expansion of $U\left(x\right)$about local minimum at $x=a$is given by the expansion ${\overline{)k=\frac{d^{2}U\left(x\right)}{d{x}^2}}}_a$

Concept of power series expansion

the expansion about $\mathit{x}\mathbf{=}\mathit{a}$is given by,

$\mathit{f}\left(a\right)\mathbf{+}\frac{\mathbf{f}\mathbf{\text{'}}\left(a\right)}{\mathbf{1}\mathbf{!}}\left(x-a\right)\mathbf{+}\frac{\mathbf{f}\mathbf{\text{'}}\left(a\right)}{\mathbf{2}\mathbf{!}}{\left(x-a\right)}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

Determine the potential energy between two protons

We need to prove that a power series expansion of $U\left(x\right)$about local minimum at $x=a$is given by the expansion

${\overline{)k=\frac{d^{2}U\left(x\right)}{d{x}^2}}}_a$

Applying $\left(1\right)$on $U\left(x\right)$about $x=a$, we get

$U\left(x-a\right)=U\left(a\right)+\frac{U\text{'}\left(a\right)}{1!}\left(x-a\right)+\frac{U\text{'}\left(a\right)}{2!}{\left(x-a\right)}^2+.....$

The potential energy of diatomic molecule for small oscillation with an equilibrium separation and spring constant ' $k$ ' is given by

$U\left(x-a\right)\approx U_0+\frac{1}{2}k{\left(x-a\right)}^2$

Comparing the respective coefficients of (2) and (3), we get

$$\begin{gathered} U_0=U\left(a\right)\frac{U\text{'}\left(a\right)}{2!}{\left(x-a\right)}^2 \\ =\frac{1}{2}k{\left(x-a\right)}^{2}k \\ =U\text{'}\text{'}\left(a\right) \\ {\overline{)k=\frac{d^{2}U\left(x\right)}{d{x}^2}}}_a \end{gathered}$$