There is a passage in a physics textbook I don't quite follow. Since my question is mathematical, I've decided to post it here. The book says:

Let $V$ be the volume of a molecule and assume $V = nr^3$. Then, because the incompressibility $K$ is defined as $K = -V \frac{\partial P}{\partial V}$, and pressure $P$ is defined as $P = - \frac{\partial U}{\partial V}$, one finds $K =V \frac{\partial ^2 U}{\partial V^2}$.

Using $V = nr^3$, $P = -\frac{dU}{dr} \frac{dr}{dV} = - \frac{1}{3nr^2} \frac{dU}{dr}$. Thus the incompressibility becomes

$$K = -nr^3 \frac{dP}{dr} \frac{dr}{dV} = \frac{r}{9n} \frac{d}{dr} \left[\frac{1}{r^2} \frac{dU}{dr} \right]$$

Question 1

I don't quite see how this last expression is obtained. I know that since $V = nr^3$ we have $\frac{dV}{dr} = 3nr^2$, so $\frac{dr}{dV} = \frac{1}{3nr^2}$. Thus, I see how the expression for $P$ is obtained. Also I see that for $K$ we have $K = -nr^3 \frac{d}{dr} \left[ -\frac{1}{3nr^2} \frac{dU}{dr} \right] \frac{dr}{dV}$, but I have a hard time working this out algebraically to get the expression above. If anyone can show me the intermediate steps here, then I would be very grateful.

The book further states:

At the equilibrium position $r = r_0$, $\frac{dU}{dr} = 0$. Thus

$$K_0 = \frac{1}{9nr_0} \left(\frac{d^2 U}{dr^2} \right)_0$$

Question 2

This confuses me. If $\frac{dU}{dr} = 0$ and $K = \frac{r}{9n} \frac{d}{dr} \left[\frac{1}{r^2} \frac{dU}{dr} \right]$, the shouldn't we get simply $K = 0$?

Any help on any of these questions will really be appreciated!


1. $$ K=-V\frac{\partial P}{\partial V}=-V\frac{\partial P}{\partial r}\frac{\partial r}{\partial V} =-nr^3\frac{\partial P}{\partial r}\frac{\partial r}{\partial V} =-nr^3\frac{\partial P}{\partial r}\frac{1}{3nr^2}\\ =-nr^3\frac{1}{3nr^2}\frac{\partial}{\partial r}\left[-\frac1{3nr^2}\frac{\partial U}{\partial r}\right] =\frac{r}{9n}\frac{\partial}{\partial r}\left[\frac1{r^2}\frac{\partial U}{\partial r}\right] $$ 2. $$ \frac{\partial}{\partial r}\left[\frac1{r^2}\frac{\partial U}{\partial r}\right]=-\frac2{r^3}\frac{\partial U}{\partial r}+\frac1{r^2}\frac{\partial^2 U}{\partial r^2} $$

  • $\begingroup$ Great! Thanks a lot. Really appreciate it! $\endgroup$ – Kristian Jul 1 '14 at 18:22

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\large\bf\mbox{Question 1}:}$ \begin{align} \partiald{}{r}&=\partiald{V}{r}\,\partiald{}{V} = 3nr^{2}\,\partiald{}{V} \ \imp\ \partiald{}{V} = {1 \over 3n}\,{1 \over r^{2}}\,\partiald{}{r} \ \imp\ \partiald[2]{}{V}={1 \over 9n^{2}r^{2}}\,\partiald{}{r}\pars{{1 \over r^{2}}\,\partiald{}{r}} \end{align}

\begin{align} K&= V\,\partiald[2]{U}{V}\ =\ \overbrace{nr^{3}}^{\ds{=\ V\quad}}\ \overbrace{% {1 \over 9n^{2}r^{2}}\,\partiald{}{r}\pars{{1 \over r^{2}}\,\partiald{U}{r}}} ^{\ds{=\ \partiald[2]{U}{V}}}\ =\ {r \over 9n}\,\,\partiald{}{r}\pars{{1 \over r^{2}}\,\partiald{U}{r}} \\[3mm]&={r \over 9n}\braces{% \partiald{\pars{1/r^{2}}}{r}\,\partiald{U}{r} +{1 \over r^{2}}\,\partiald{}{r}\bracks{\partiald{U}{r}}} =-\,{2 \over 9nr^{2}}\,\partiald{U}{r} + {1 \over 9nr}\,\partiald[2]{U}{r} \end{align}

$\ds{\large\bf\mbox{Question 2}}$:

$\ds{\pars{\partiald[2]{U}{r}}_{0}}$ is a 'short notation': The second derivative is evaluated at the point where the first derivative is zero. It doesn't need to be zero. For example: $$ \fermi\pars{x}\equiv x^{2} - 2x\,,\qquad \fermi'\pars{1} = 0\,,\qquad\fermi''\pars{1} = 2 \not= 0 $$

  • $\begingroup$ Thanks a lot! Much appreciated! Wish I could mark both answers as accepted! $\endgroup$ – Kristian Jul 1 '14 at 18:23

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