Thermal Expansion

The interaction of electrically neutral atoms can be modeled using the Lennard-Jones potential,

$$u(x) = u_0\left[\left(\frac{x_0}{x}\right)^{12}-2\,\left(\frac{x_0}{x}\right)^6\right],$$ (5.352)

where $u(x)$ is the potential energy of a pair of atoms when they are a distance $x$ apart, and $u_0$ and $x_0$ are positive constants. See Figure 5.3. We can think of a given atom in a solid made up of neutral atoms as a microscopic system interacting with a heat reservoir that consists of all of the other atoms. Let $T$ be the temperature of the reservoir. Let us treat the problem classically, which is equivalent to assuming that the temperature is sufficiently high that an atom moving in the previous potential is distributed over a large number of different quantum states. According to a straightforward generalization of the Boltzmann distribution, (5.329), the mean value of $x$ is

$$\langle x\rangle = \frac{\int_0^\infty \exp[-u(x)/(k_B\,T)]\,x\,dx}{\int_0^\infty \exp[-u(x)/(k_B\,T)]\,dx}.$$ (5.353)

Figure 5.3: The Lennard-Jones potential.

Let us assume that the temperature is sufficiently low that an atom is only likely to be found relatively close to the bottom of the potential well, $x=x_0$. We can Taylor expand the potential about $x=x_0$ to give

$$u(x) = u_0 + u_0'\,(x-x_0) + \frac{1}{2}\,u_0''\,(x-x_0)^2 + \frac{1}{6}\,u_0'''\,(x-x_0)^3+\cdots,$$ (5.354)

where

$$u_0 = u(x_0),$$ (5.355)

$$u_0' = \frac{du(x_0)}{dx}=0,$$ (5.356)

$$u_0'' = \frac{d^2u(x_0)}{dx^2}= \frac{72\,u_0}{x_0^{\,2}},$$ (5.357)

$$u_0''' = \frac{d^3u(x_0)}{dx^3}= -\frac{1512\,u_0}{x_0^{\,3}}.$$ (5.358)

Thus, Equation (5.353) gives

$$\langle x\rangle = \frac{\int_{-\infty}^\infty \exp[-u_0''\,y^2/(... ...fty}^\infty \exp[-u_0''\,y^2/(2\,k_B\,T)]\,\exp[-u_0'''\,y^3/(6\,k_B\,T)]\,dy},$$ (5.359)

where $y=x-x_0$, and we can safely replace the lower limits of integration by $-\infty$, in the integrals on the right-hand side of the previous expression, because we are assuming that the atom is very unlikely to be found a large distance from the bottom of the potential. Let us further assume that $\vert u_0'''\,y^3/(6\,k_B\,T)\vert\ll 1$. In this case, we can write

$$\langle x\rangle \simeq x_0+\frac{\int_{-\infty}^\infty \exp[-u_0... ...\infty}^\infty \exp[-u_0''\,y^2/(2\,k_B\,T)]\,[1-u_0'''\,y^3/(6\,k_B\,T)]\,dy}.$$ (5.360)

Now, $\exp[-u_0''\,y^2/(2\,k_B\,T)]$ and $y^4$ are even functions of $y$, whereas $y$ and $y^3$ are odd functions. In general, an integral over all $y$ of the product of an even and an odd function averages to zero, whereas an integral of the product of two even functions does not. Hence, the previous equation simplifies to give

$$\langle x\rangle = x_0 - \frac{u_0'''}{6\,k_B\,T}\,\frac{\int_{-\infty}^\infty y^4... ...y^2/(2\,k_B\,T)]\,dy} {\int_{-\infty}^\infty \exp[-u_0''\,y^2/(2\,k_B\,T)]\,dy}$$

$$= x_0 + \left(\frac{-u_0'''}{6\,k_B\,T}\right)\left(\frac{2\,k_B\... ...nt_{-\infty}^\infty z^4\,\exp(-z^2)\,dz}{\int_{-\infty}^\infty \exp(-z^2)\,dz}.$$ (5.361)

However, $\int_{-\infty}^\infty z^4\,\exp(-z^2)\,dz= (3/4)\,\pi^{1/2}$, and $\int_{-\infty}^\infty \exp(-z^2)\,dz= \pi^{1/2}$, so we get

$$\langle x\rangle = x_0 + \frac{(-u_0''')\,k_B\,T}{2\,(u_0'')^2}.$$ (5.362)

Note that $\langle x\rangle$ increases linearly with increasing temperature.

The coefficient of linear thermal expansion of a solid is defined

$$\alpha = \frac{1}{\langle x\rangle}\,\frac{d\langle x\rangle}{dT},$$ (5.363)

where $\langle x\rangle$ is the mean distance between nearest neighbor atoms. The previous two equations yield

$$\alpha = \frac{(-u_0''')\,k_B}{2\,x_0\,(u_0'')^2}= \frac{7\,k_B}{48\,u_0},$$ (5.364)

where use has been made of Equations (5.357) and (5.358).

For solid argon at 80 K, $x_0=3.9\times 10^{-10}\,{\rm m}$, and $u_0= 0.010$ eV. Hence, we deduce that

$$\alpha=1.3\times 10^{-3}\,{\rm K}^{-1}.$$ (5.365)

The measured value of $\alpha$ is about $2\times 10^{-3}\,{\rm K}^{-1}$.