Finite strain equation of states

Finite strain vs. Birch-Murnaghan

Explicit expression for Birch-Murnaghan equation of states

The explicit expressions for third-order Birch-Murnaghan (BM3) equation of states (EoS) are

$E(V)=E^{0}+\tfrac{9}{2} V^{0} \kappa^{0} f^{2}\left[1+f\left(\kappa^{\prime 0}-4\right)\right]$

$P(V)=-\left(\frac{\partial E}{\partial V}\right)_{S}=3 \kappa^{0} f(1+2 f)^{\frac{5}{2}}\left[1+\tfrac{3}{2}\left(\kappa^{\prime 0}-4\right)f\right]$

where the Eulerian strain $f$ is given by

$f \equiv \tfrac{1}{2}\left[\left(V^{0} / V\right)^{\frac{2}{3}}-1\right] \,.$

The effect of using of an arbitrary $V_\mathrm{ref}$ instead of $V_0$

Since we have $f=\frac{1}{2}[(V_0/V)^{2/3}-1]$ and we assume $f' \equiv\frac{1}{2}[(V_\mathrm{ref}/V)^{2/3}-1]$ , we have

$\frac{2f + 1}{2f'+1} = \frac{(V_0/V)^{2/3}}{(V_\mathrm{ref}/V)^{2/3}} = \left(\frac{V_0}{V_\mathrm{ref}}\right)^{2/3}$

So we have

$f = \tfrac{1}{2}\left[(2f'+1)\left(\frac{V_0}{V_\mathrm{ref}}\right)^{2/3} - 1\right] = pf' + q$

Since $E(V) = a f^3 + bf^2 + d$ , we can fit with $f'$

$E(V) = a(pf' + q)^3 + b(pf'+q)^2 + d = a'f'^3 + b'f'^2 + c'f' + d'$

So with an arbitrary $V_\mathrm{ref}$ , the fitting result is equivalent.

Recovering $V_0$ , $K_0$ , and $K'_0$ from the polynomial coefficients

Changing $V_\mathrm{ref}$ reshuffles the polynomial coefficients and the strain coordinate while preserving $E(V)$ and the EoS parameters. Recover $V_0$ from the minimum of $E(x)$ . Read $K_0$ from the curvature at the minimum and $K'_0$ from how the curvature changes under compression.

Write $x=f'$ and fit

$E(x)=ax^3+bx^2+cx+d \,,$

with

$V=V_\mathrm{ref}(1+2x)^{-3/2} \,.$

Let $x_0$ mark the minimum of $E(x)$ . Convert $x_0$ back to volume with

$V_0=V_\mathrm{ref}(1+2x_0)^{-3/2} \,.$

Substitute $x_0$ into the coefficient expressions for the moduli

$K_0=\frac{2(1+2x_0)^2(3ax_0+b)}{9V_0}, \qquad K'_0=4+\frac{a(1+2x_0)}{3ax_0+b} \,.$

The constant coefficient $d$ sets the energy zero and drops out of $V_0$ , $K_0$ , and $K'_0$ .

Analytical derivatives to obtain $P(V)$

Not easy to do, because

$P = -\frac{\partial E}{\partial V} = -\frac{\partial E}{\partial f'} \frac{\partial f'}{\partial V} = -(3af'^2 + 2bf' + c)\frac{\partial f'}{\partial V}$

where

$\frac{\partial f'}{\partial V} = -\frac{1}{3}\left(\frac{V_\mathrm{ref}}{V}\right)^{2/3}V^{-1} = -\tfrac{1}{3}(2f'+1)V^{-1} \,.$

But the expression for $PV$ is a 3^rd-order polynomial vs. $f'$ ,

$PV = \tfrac{1}{3} (2f' + 1) (3af'^2 + 2bf' + c) \,.$

Gibbs free energy or enthalpy

Usually, these equation of states are for internal energy ( $U$ ), Helmholtz free energy ( $F$ ) as a function of volume $V$ . It is also possible to think about the form of enthalpy ( $H$ ), Gibbs free energy ( $G$ ) as a function of strain.

Because

$\begin{split} G &= F + PV \,,\\ H &= U + PV \,, \end{split}$

assuming $T = 0$ , from the above statement, $PV$ is a 3^rd-order polynomial of $f'$ , it is obvious that $G$ and $H$ will be 3^rd order polynomials of $f'$ too.

References

Birch–Murnaghan equation of state - Wikipedia