thermodynamics - Linear thermal expansion from statistical mechanics?

I came across a question recently regarding work done by an expanding metal and the origin of the energy used for the work, and most of the responses pointed the person to look more at the enthalpy associated with the process than the energy.

My thought was that there was some better way to describe the work as some combination of a force $J$ and a displacement $x$, but I had a hard time thinking of a force expression for this particular object. I did some searching on possibly modeling the metal as a linear spring with a temperature-dependent Young's modulus, but all the resources I found only mentioned the modulus decreasing with temperature, which would not explain the thermal expansion.

I'm wondering if anyone can point me towards or provide me with a derivation of the linear thermal expansion (and hopefully a force from which it arises) from principles of statistical mechanics.

Edit: With a little extra searching I found one site using the equation $\alpha_V = \frac{1}{V} \left(\frac{\partial V}{\partial T}\right)_P$, which is easy enough to derive for an ideal gas, but they mention the constant pressure constraint on $\frac{\partial V}{\partial T}$ being unnecessary for a solid, but this doesn't lead any closer to a force expression.

Answer

I am not entirely sure that statistical mechanics is necessary, as the resulting effect is a macroscopic one, thus applicable to thermodynamics. However, I will touch on the statistical mechanics aspects a bit.

Much of my answer derives from notes from the thermal expansion Wikipedia page (https://en.wikipedia.org/wiki/Thermal_expansion) and the book by Zel'dovich and Raizer [2002].

Let us first start with some definitions of parameters, in no particular order.

• Constants
  
  
  


• Variables
  
  
  
  • $T$ = the scalar temperature
  
  
  • $V$ = the scalar volume
  
  
  • $\epsilon$ = the specific internal energy
  
  
  • $P$ = the scalar pressure
  
  
  • Subscripts
    
    
    
    • $c$ = for components relating to elastic effects (i.e., exclusively due to forces of interaction between atoms/molecules)
    
    
    
    • $T$ = due to thermal motion of atoms/molecules
    
    
    • $e$ = due to thermal excitation of electrons
    
    
    • $o$ = initial or standard condition/value
    
    
    
    
  
  
  
  


• Coefficients/Parameters
  
  
  
  • $\kappa_{o} = \tfrac{ 1 }{ V_{o} } \left( \tfrac{ \partial V }{ \partial P } \right)_{T}$ = the isothermal compressibility
  
  
  
  • $\alpha = \tfrac{ 1 }{ V_{o} } \left( \tfrac{ \partial V }{ \partial T } \right)_{P}$ = the isobaric thermal expansion coefficient
  
  
  • $C_{v}$ = specific heat at constant volume = $3 \ k_{B} \ N$, where $N$ is the number of atoms/molecules
  
  
  • $\Gamma \left( V \right)$ = Grüneisen coefficient
  
  
  • $C_{so}$ = isothermal speed of sound = $\left( \tfrac{ V_{o} }{ \kappa_{o} } \right)^{1/2} = \left( - V^{2} \tfrac{ d P_{c} }{ d V } \right)^{1/2}$, where $N$ is the number of atoms/molecules
  
  
  
  


• Thermodynamic Identities
  
  
  
  • $\left( \tfrac{ \partial \epsilon }{ \partial V } \right)_{T} = T \left( \tfrac{ \partial P }{ \partial T } \right)_{V} - P$
  
  
  • $-1 = \left( \tfrac{ \partial P }{ \partial T } \right)_{V} \left( \tfrac{ \partial T }{ \partial V } \right)_{P} \left( \tfrac{ \partial V }{ \partial P } \right)_{T}$
  
  
  
  
  

We can separate $P$ and $\epsilon$ in three parts: elastic, thermal motion/vibration, and thermal electron excitation. The last part is generally quite small unless the material is at very high temperatures (e.g., $T \gtrsim 10^{4} \ K$). Then we can define: $$ \begin{align} \epsilon & = \epsilon_{c}\left( V \right) + 3 \ N \ k_{B} \ T \tag{1a} \\ P & = P_{c}\left( V \right) + P_{T}\left( V, T \right) \tag{1b} \end{align} $$

I discussed the physical interpretation of the elastic terms in another answer at https://physics.stackexchange.com/a/216180/59023:

Let us define $P$ as the pressure and $\epsilon$ as the internal energy of a solid material. These can be divided into two parts: an elastic (subscript $c$) and thermal part. $P_{c}$ and $\epsilon_{c}$ depend only upon the density of the material, $\rho$, or the specific volume, $V$ = $1/\rho$. These are equal to the total pressure and specific internal energy at absolute zero or $T = 0 \ K$. Let us assume that the specific volume at $T = 0$ and $P = 0$ is given by $V_{oc}$, which is only ~1-2% smaller than the specific volume at STP, $V_{o}$, for most metals.

The potential energy curve, or curve defining $\epsilon_{c}$, is qualitatively similar to the potential energy curve describing the interaction between two atoms as a function of the intranuclear distance, $\Delta x_{n}$. When $V > V_{oc}$, the attractive forces dominate but fall off rapidly as the intranuclear distances increase (e.g., as $T$ increases). In other words, when the atoms move further apart $\epsilon_{c}$ will asymptotically increase to some value $U$, which is roughly the binding energy of the atoms in the body. Thus, $U$ represents the energy required to remove all atoms from the object to infinity, which is roughly equal to the heat of vaporization for the material ... Thus, $\epsilon_{c} (V) \rightarrow U$ as $\Delta x_{n} \rightarrow \sim 2$.

Conversely, the repulsive forces dominate if $V < V_{oc}$. We can define this quantitatively by considering that the work done by compressing the material will be equal to the increase in internal energy. In other words: $$ P_{c} = - \left( \frac{ d \epsilon_{c} }{ d V } \right)_{T = 0} $$ which is equivalent to saying it is the isothermic/isentropic equation for cold compression. The negative sign shows that if a tensile force were applied to the body, the binding forces between atoms would act as a restoring force.

If we use the first thermodynamic identity above we find two results, one for the elastic parameters and one for the thermal: $$ \begin{align} \left( \frac{ \partial \epsilon_{c} }{ \partial V } \right)_{T} & = \frac{ d \epsilon_{c} }{ d V } = - P_{c}\left( V \right) \tag{2a} \\ \left( \frac{ \partial \epsilon_{T} }{ \partial V } \right)_{T} & = 0 = T \left( \frac{ \partial P_{T} }{ \partial T } \right)_{V} - P_{T} \tag{2b} \end{align} $$ In the linear limit, Equation 2b suggests we can express $P_{T}$ as a function dependent upon volume times the temperature. This allows use to write $P_{T}$ as: $$ \begin{align} P_{T} & = T \left( \frac{ \partial P_{T} }{ \partial T } \right)_{V} = \phi\left( V \right) \ T \tag{3a} \\ & = \Gamma \left( V \right) \frac{ C_{v} \ T }{ V } \tag{3b} \\ & = \Gamma \left( V \right) \frac{ \epsilon_{T} }{ V } \tag{3c} \end{align} $$

Note that Equation 3 shows that $\Gamma \left( V \right)$ is a parameter that characterizes the ratio of $P_{T}$ to $\epsilon_{T}$, called the Grüneisen coefficient/parameter.

If we use the second thermodynamic identity above and Equation 3a combined with the definitions for $\kappa_{o}$ and $\alpha$, we can show: $$ \begin{align} -1 & = \left( \frac{ \partial P }{ \partial T } \right)_{V} \left( \frac{ \partial T }{ \partial V } \right)_{P} \left( \frac{ \partial V }{ \partial P } \right)_{T} \tag{4a} \\ & = - \kappa_{o} \ V_{o} \ \left( \frac{ \partial P }{ \partial T } \right)_{V} \left( \frac{ \partial T }{ \partial V } \right)_{P} \tag{4b} \\ & = - \frac{ \kappa_{o} \ V_{o} }{ \alpha \ V_{o} } \left( \frac{ \partial P }{ \partial T } \right)_{V} \tag{4c} \\ & = - \frac{ \kappa_{o} \ P_{T} }{ \alpha \ T } \tag{4c} \\ & = - \frac{ \kappa_{o} \ C_{v} }{ \alpha \ V_{o} } \ \Gamma \left( V_{o} \right) \tag{4d} \end{align} $$

In the limit of low temperature, we can see from Equation 4d that: $$ \Gamma \left( V_{o} \right) \equiv \Gamma_{o} = \frac{ \alpha \ V_{o} }{ \kappa_{o} \ C_{v} } \tag{5} $$ which shows that $\Gamma_{o}$ is independent of temperature. This is only true for cases when $P_{e}$ and $\epsilon_{e}$ are small. We can also relate $\Gamma_{o}$ to the isothermal sound speed by: $$ \Gamma_{o} = \frac{ \alpha \ C_{so}^{2} }{ C_{v} } \tag{6} $$

In the limit of high temperature (i.e., when $k_{B} \ T \gg h \ \nu$ or when thermal pressures/energies dominate over vibrational [for lattice configurations]), the specific free energy is given by: $$ F = \epsilon_{c}\left( V \right) + 3 \ N \ k_{B} \ T \ \ln \frac{ h \ \bar{\nu} }{ k_{B} \ T} \tag{7} $$ where $\bar{\nu} = e^{-1/3} \ k_{B} \ \theta$ is the average vibrational frequency and $\theta$ is the Debye temperature. We can relate $F$ to $P$ and $\epsilon$ through the following: $$ \begin{align} \epsilon & = F - T \frac{ \partial F }{ \partial T } \tag{8a} \\ & = \epsilon_{c}\left( V \right) + 3 \ N \ k_{B} \ T \tag{8b} \\ & = \epsilon_{c} + \epsilon_{T} \tag{8c} \\ P & = - \frac{ \partial F }{ \partial V } \tag{8d} \\ & = - \frac{ \partial \epsilon_{c} }{ \partial V } - 3 \ N \ k_{B} \ T \frac{ \partial \ln h \ \bar{\nu} }{ \partial V } \tag{8e} \\ & = P_{c} + \epsilon_{T} \frac{ \partial \ln h \ \bar{\nu} }{ \partial V } \tag{8f} \\ & = P_{c} + P_{T} \tag{8g} \end{align} $$ We can use Equation 8 and 3c to show that: $$ \begin{align} P_{T} = \epsilon_{T} \frac{ \partial \ln h \ \bar{\nu} }{ \partial V } & = \Gamma \left( V \right) \frac{ \epsilon_{T} }{ V } \tag{9a} \\ & = \frac{ V \ P_{T} }{ \Gamma \left( V \right) } \frac{ \partial \ln h \ \bar{\nu} }{ \partial V } \tag{9b} \\ & = \frac{ P_{T} }{ \Gamma \left( V \right) } \frac{ \partial V }{ \partial \ln V } \frac{ \partial \ln h \ \bar{\nu} }{ \partial V } \tag{9c} \\ & = \frac{ P_{T} }{ \Gamma \left( V \right) } \frac{ \partial \ln h \ \bar{\nu} }{ \partial \ln V } \tag{9d} \end{align} $$ which gives us the expression for $\Gamma \left( V \right)$ in terms of $\bar{\nu}$, given by: $$ \Gamma \left( V \right) = \frac{ \partial \ln h \ \bar{\nu} }{ \partial \ln V } \tag{10} $$

We can approximate $\bar{\nu}$ as ~$C_{so}/r_{o}$, where $r_{o}$ is the interatomic spacing, but this means that $$ \bar{\nu} \sim V^{2/3} \left( - \frac{ d P_{c} }{ d V } \right) \tag{11} $$

We can now see why solids expand when heated, if we examine Equations 1b and 11. If one heats a solid, then $P_{T}$ increases. Thus, upon heating the elastic pressure, $P_{c}$, becomes negative in order to maintain pressure balance, but the body of the material expands [quote on page 700 of Zel'dovich and Raizer, 2002]:

...up to that point when the binding forces holding the atoms in the lattice, or the negative pressure, will no longer counterbalance the repulsive effect of the positive thermal pressure...

1. Zel'dovich, Ya.B., and Yu.P. Raizer (2002) Physics of Shock Waves and High-Temperature Hydrodynamic Phenomena, Ed. by W.D. Hayes and R.F. Probstein, Mineola, NY, Dover Publications, inc., The Dover Edition; ISBN-13: 978-0486420028.