Calculate the coefficient of compressibility and the bulk modulus for MgO

The coefficient of compressibility ($K$ or $\beta$) is determined by measuring the change of volume ($V$) of the material as a function of the pressure applied ($P$), while the temperature ($T$) is held constant:


\begin{displaymath} K = \beta = - \frac{1}{V} \left( \frac{ \partial V}{ \partial P} \right) _T \end{displaymath}

The definition includes the normalisation with respect to the volume in order to take into the account the size of the system; $K$ is therefore an intrinsic quantity.

When $P$ increases, $V$ decreases: the derivative in formula above is inherently negative. Thus, inclusion of the minus sign guarantees that tabulated values of K are positive numbers.
Note: The compressibility is defined analogously to to the thermal expansion coefficient:

\begin{displaymath} \alpha = - \frac{1}{V} \left( \frac{ \partial V}{ \partial T} \right) _P \end{displaymath}


The bulk modulus $B$ is given by


\begin{displaymath} B = \frac{1}{K} = - V \left( \frac{ \partial P }{ \partial V } \right) _T \end{displaymath}


The pressure can be expressed as follows by means of the first principle of thermodynamics: the internal energy ($U$) of the system under investigation can change when an amount of work ($ -P dV $ ) is performed or when an amount of heat ($ T dS $) is transferred to the system.


\begin{displaymath} d U = -P dV + T dS \end{displaymath}

In our calculations, $T$=0 K and the internal energy corresponds to the total energy of the system obtained at the end of the self-consistent field procedure.


\begin{displaymath} P = - \frac{ d U }{ d V} = - \frac{ d E }{ d V } \end{displaymath}

The expression above can be inserted in the definition of $B$:


\begin{displaymath} B = - V \left( \frac{ \partial ( - \frac{ d E }{ d V } ) }{ \partial V } \right) _T \end{displaymath}

and by simplifying the notation, the bulk modulus can be obtained at the athermal limit by applying the formula


\begin{displaymath} B = V \left( \frac{ d^2 E } { d V^2 } \right) \end{displaymath}



At this point, it is necessary to tabulate the energy as a function of the volume, to fit the data and to calculate $\frac{ d^2 E } { d V^2 }$.

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