Jacobian notation


cubic

1 Objective

The objective of this post is to derive the following mathematical expression:

\[\begin{equation} \left(\frac{\partial x}{\partial y}\right)_z = \frac{\displaystyle\left(\frac{\partial z}{\partial v}\right)_T \left(\frac{\partial x}{\partial T}\right)_v - \left(\frac{\partial z}{\partial T}\right)_v \left(\frac{\partial x}{\partial v}\right)_T}{\displaystyle\left(\frac{\partial z}{\partial v}\right)_T \left(\frac{\partial y}{\partial T}\right)_v - \left(\frac{\partial z}{\partial T}\right)_v \left(\frac{\partial y}{\partial v}\right)_T} \tag{1.1} \end{equation}\]

Equation (1.1) permits deriving an analytical expression for any thermodynamic property from a pressure explicit equation of state such as cubic equations of state.

2 Derivation

First, lets define a function \(z\) which depends on variables \(x\) and \(y\). Therefore, its differential form writes:

\[\begin{equation} dz\left(x,y\right) = \left(\frac{\partial z}{\partial x}\right)_y dx + \left(\frac{\partial z}{\partial y}\right)_x dy \tag{2.1} \end{equation}\]

Next step is to assume that both \(x\) and \(y\) are functions of variables \(T\) and \(v\). Therefore, differentials of \(T\) and \(v\) write:

\[\begin{equation} \left\{ \begin{array}[l] &dx\left(T,v\right) & = \displaystyle\left(\frac{\partial x}{\partial T}\right)_v dT + \left(\frac{\partial x}{\partial v}\right)_T dv \\ dy\left(T,v\right) & = \displaystyle\left(\frac{\partial y}{\partial T}\right)_v dT + \left(\frac{\partial y}{\partial v}\right)_T dv \end{array} \right. \tag{2.2} \end{equation}\]

Combining equations (2.1) and (2.2), one obtains:

\[\begin{equation} dz\left(x,y\right) = \left(\frac{\partial z}{\partial x}\right)_y \times \left[\left(\frac{\partial x}{\partial T}\right)_v dT + \left(\frac{\partial x}{\partial v}\right)_T dv\right] + \left(\frac{\partial z}{\partial y}\right)_x \times \left[\displaystyle\left(\frac{\partial y}{\partial T}\right)_v dT + \left(\frac{\partial y}{\partial v}\right)_T dv\right] \tag{2.3} \end{equation}\]

Terms in this equations can be rearranged with respect to \(dT\) and \(dv\) to get:

\[\begin{equation} dz\left(x,y\right) = \left[\left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial T}\right)_v + \left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial T}\right)_v\right]dT + \left[\left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial v}\right)_T + \left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial v}\right)_T\right]dv \tag{2.4} \end{equation}\]

Moreover, without loss of generality, it is possible to consider that function \(z\) is depends on variables \(T\) and \(v\). Thus, its differential writes:

\[\begin{equation} dz\left(T,v\right) = \left(\frac{\partial z}{\partial T}\right)_v dT + \left(\frac{\partial z}{\partial v}\right)_T dv \tag{2.5} \end{equation}\]

Because \(z\left(x,y\right) = z\left(T,v\right)\), one gets \(dz\left(x,y\right) = dz\left(T,v\right)\). After equating equations (2.4) and (2.5) and reorganizing the terms one can write:

\[\begin{equation} \left[\left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial T}\right)_v + \left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial T}\right)_v - \left(\frac{\partial z}{\partial T}\right)_v\right]dT + \left[\left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial v}\right)_T + \left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial v}\right)_T- \left(\frac{\partial z}{\partial v}\right)_T\right]dv = 0 \tag{2.6} \end{equation}\]

In order to satisfy (2.6), factors in front of \(dT\) and \(dv\) must both the zero, which is equivalent to:

\[\begin{equation} \left\{ \begin{array}[l] & \displaystyle \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial T}\right)_v + \left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial T}\right)_v - \left(\frac{\partial z}{\partial T}\right)_v = & 0 \\ \displaystyle \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial v}\right)_T + \left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial v}\right)_T- \left(\frac{\partial z}{\partial v}\right)_T =& 0 \end{array} \right. \tag{2.7} \end{equation}\]

Equation (2.7) can be recasted in the matrix form:

\[\begin{equation} \underbrace{ \begin{bmatrix} \displaystyle\left(\frac{\partial x}{\partial T}\right)_v & \displaystyle\left(\frac{\partial y}{\partial T}\right)_v \\ \displaystyle\left(\frac{\partial x}{\partial v}\right)_T & \displaystyle\left(\frac{\partial y}{\partial v}\right)_T \end{bmatrix} }_A \times \underbrace{ \begin{bmatrix} \displaystyle\left(\frac{\partial z}{\partial x}\right)_y \\ \displaystyle\left(\frac{\partial z}{\partial y}\right)_x \end{bmatrix} }_X - \underbrace{ \begin{bmatrix} \displaystyle\left(\frac{\partial z}{\partial T}\right)_v \\ \displaystyle\left(\frac{\partial z}{\partial v}\right)_T \end{bmatrix} }_B =0 \tag{2.8} \end{equation}\]

In equation (2.8) elements of matrices \(A\) and \(B\) can be estimated from pressure explicit equation of state because they are derivatives of functions expressed with respect to variables \(T\) and \(v\). Thus, only elements of vector \(X\) are unknown and remain to be expressed with the partial derivatives expressed in matrices \(A\) and \(B\). To do so, one can rewrite (2.8) this way:

\[\begin{equation} \underbrace{ \begin{bmatrix} \displaystyle\left(\frac{\partial z}{\partial x}\right)_y \\ \displaystyle\left(\frac{\partial z}{\partial y}\right)_x \end{bmatrix} }_X = \underbrace{ \begin{bmatrix} \displaystyle\left(\frac{\partial x}{\partial T}\right)_v & \displaystyle\left(\frac{\partial y}{\partial T}\right)_v \\ \displaystyle\left(\frac{\partial x}{\partial v}\right)_T & \displaystyle\left(\frac{\partial y}{\partial v}\right)_T \end{bmatrix}^{-1} }_{A^{-1}} \times \underbrace{ \begin{bmatrix} \displaystyle\left(\frac{\partial z}{\partial T}\right)_v \\ \displaystyle\left(\frac{\partial z}{\partial v}\right)_T \end{bmatrix} }_B \tag{2.8} \end{equation}\]

From basic calculus, matrix \(A^{-1}\) has the following expression:

\[\begin{equation} A^{-1} = \frac{1}{\left(\frac{\partial x}{\partial T}\right)_v \left(\frac{\partial y}{\partial v}\right)_T - \left(\frac{\partial y}{\partial T}\right)_v \left(\frac{\partial x}{\partial v}\right)_T} \times \begin{bmatrix} \displaystyle\left(\frac{\partial y}{\partial v}\right)_T & \displaystyle -\left(\frac{\partial y}{\partial T}\right)_v \\ \displaystyle -\left(\frac{\partial x}{\partial v}\right)_T & \displaystyle\left(\frac{\partial x}{\partial T}\right)_v \end{bmatrix} \tag{2.9} \end{equation}\]

Therefore, equation (2.8) rewrites:

\[\begin{equation} \underbrace{ \begin{bmatrix} \displaystyle\left(\frac{\partial z}{\partial x}\right)_y \\ \displaystyle\left(\frac{\partial z}{\partial y}\right)_x \end{bmatrix} }_X = \underbrace{ \frac{1}{\left(\frac{\partial x}{\partial T}\right)_v \left(\frac{\partial y}{\partial v}\right)_T - \left(\frac{\partial y}{\partial T}\right)_v \left(\frac{\partial x}{\partial v}\right)_T} \times \begin{bmatrix} \displaystyle\left(\frac{\partial y}{\partial v}\right)_T & \displaystyle -\left(\frac{\partial y}{\partial T}\right)_v \\ \displaystyle -\left(\frac{\partial x}{\partial v}\right)_T & \displaystyle\left(\frac{\partial x}{\partial T}\right)_v \end{bmatrix} }_{A^{-1}} \times \underbrace{ \begin{bmatrix} \displaystyle\left(\frac{\partial z}{\partial T}\right)_v \\ \displaystyle\left(\frac{\partial z}{\partial v}\right)_T \end{bmatrix} }_B \tag{2.10} \end{equation}\]

It is now possible to get the expressions for \(\left(\frac{\partial z}{\partial x}\right)_y\) and \(\left(\frac{\partial z}{\partial y}\right)_x\):

\[\begin{equation} \left\{ \begin{array}[l] &\displaystyle\left(\frac{\partial z}{\partial x}\right)_y = & \frac{\displaystyle\left(\frac{\partial y}{\partial v}\right)_T \left(\frac{\partial z}{\partial T}\right)_v - \left(\frac{\partial y}{\partial T}\right)_v \left(\frac{\partial z}{\partial v}\right)_T}{\displaystyle\left(\frac{\partial y}{\partial v}\right)_T \left(\frac{\partial x}{\partial T}\right)_v - \left(\frac{\partial y}{\partial T}\right)_v \left(\frac{\partial x}{\partial v}\right)_T}\\ \displaystyle\left(\frac{\partial z}{\partial y}\right)_x = & \frac{\displaystyle\left(\frac{\partial x}{\partial T}\right)_v \left(\frac{\partial z}{\partial v}\right)_T - \left(\frac{\partial x}{\partial v}\right)_T \left(\frac{\partial z}{\partial T}\right)_v}{\displaystyle\left(\frac{\partial x}{\partial T}\right)_v \left(\frac{\partial y}{\partial v}\right)_T - \left(\frac{\partial x}{\partial v}\right)_T \left(\frac{\partial y}{\partial T}\right)_v} \end{array} \tag{2.11} \right. \end{equation}\]

Because variables \(x\), \(y\) and \(z\) are dummy variables, both equations in system (2.11) are equivalent and only one can be considered. This allows to write:

\[\begin{equation} \left(\frac{\partial x}{\partial y}\right)_z = \frac{\displaystyle\left(\frac{\partial z}{\partial v}\right)_T \left(\frac{\partial x}{\partial T}\right)_v - \left(\frac{\partial z}{\partial T}\right)_v \left(\frac{\partial x}{\partial v}\right)_T}{\displaystyle\left(\frac{\partial z}{\partial v}\right)_T \left(\frac{\partial y}{\partial T}\right)_v - \left(\frac{\partial z}{\partial T}\right)_v \left(\frac{\partial y}{\partial v}\right)_T} \tag{2.12} \end{equation}\]

which is equation (1.1).

3 Jacobian notation

One can observe that the numerator and the denominator of equation (2.12) can be expressed as determinants:

\[\begin{equation} \left(\frac{\partial x}{\partial y}\right)_z = \frac{ \begin{vmatrix} \displaystyle\left(\frac{\partial x}{\partial T}\right)_v & \displaystyle\left(\frac{\partial x}{\partial v}\right)_T \\ \displaystyle\left(\frac{\partial z}{\partial T}\right)_v & \displaystyle\left(\frac{\partial z}{\partial v}\right)_T \end{vmatrix} } { \begin{vmatrix} \displaystyle\left(\frac{\partial y}{\partial T}\right)_v & \displaystyle\left(\frac{\partial y}{\partial v}\right)_T \\ \displaystyle\left(\frac{\partial z}{\partial T}\right)_v & \displaystyle\left(\frac{\partial z}{\partial v}\right)_T \end{vmatrix} } \tag{3.1} \end{equation}\]

Such determinants are commonly written:

\[\begin{equation} \left\{ \begin{array}[l] &\left(\partial x\right)_z &= \begin{vmatrix} \displaystyle\left(\frac{\partial x}{\partial T}\right)_v & \displaystyle\left(\frac{\partial x}{\partial v}\right)_T \\ \displaystyle\left(\frac{\partial z}{\partial T}\right)_v & \displaystyle\left(\frac{\partial z}{\partial v}\right)_T \end{vmatrix} \\ \left(\partial y\right)_z &= \begin{vmatrix} \displaystyle\left(\frac{\partial y}{\partial T}\right)_v & \displaystyle\left(\frac{\partial y}{\partial v}\right)_T \\ \displaystyle\left(\frac{\partial z}{\partial T}\right)_v & \displaystyle\left(\frac{\partial z}{\partial v}\right)_T \end{vmatrix} \end{array} \right. \tag{3.2} \end{equation}\]

with \(\left(\partial x\right)_z\) and \(\left(\partial y\right)_z\) being the determinant of the jacobian matrices.

Finally, equation (3.1) writes in a simpler form:

\[\begin{equation} \left(\frac{\partial x}{\partial y}\right)_z = \frac{\left(\partial x\right)_z}{\left(\partial y\right)_z} \end{equation}\]

4 Properties of \(\left(\partial x\right)_z\)

4.1 Linearity of \(\left(\partial x\right)_z\)

In the following section, \(\lambda\) is a scalar value while \(a\) and \(b\) are functions of variables \(T\) and \(v\).

\[\begin{equation} \left[\partial \left(a + \lambda b\right)\right]_z = \begin{vmatrix} \displaystyle\left[\frac{\partial \left(a + \lambda b\right)}{\partial T}\right]_v & \displaystyle\left[\frac{\partial \left(a + \lambda b\right)}{\partial v}\right]_T \\ \displaystyle\left(\frac{\partial z}{\partial T}\right)_v & \displaystyle\left(\frac{\partial z}{\partial v}\right)_T \end{vmatrix} \end{equation}\]

Because of the alternated n-linearity of the determinant, one can write:

\[\begin{equation} \left[\partial \left(a + \lambda b\right)\right]_z = \begin{vmatrix} \displaystyle\left(\frac{\partial a}{\partial T}\right)_v & \displaystyle\left(\frac{\partial a}{\partial v}\right)_T \\ \displaystyle\left(\frac{\partial z}{\partial T}\right)_v & \displaystyle\left(\frac{\partial z}{\partial v}\right)_T \end{vmatrix} + \lambda\begin{vmatrix} \displaystyle\left(\frac{\partial b}{\partial T}\right)_v & \displaystyle\left(\frac{\partial b}{\partial v}\right)_T \\ \displaystyle\left(\frac{\partial z}{\partial T}\right)_v & \displaystyle\left(\frac{\partial z}{\partial v}\right)_T \end{vmatrix} = \left(\partial a \right)_z + \lambda \left(\partial b \right)_z \end{equation}\]

A similar result can be obtained with respect to variable \(z\):

\[\begin{equation} \left(\partial x\right)_{a + \lambda b} = \left(\partial x\right)_{a} + \lambda \left(\partial x\right)_{b} \end{equation}\]

4.2 Antisymetry of \(\left(\partial x\right)_z\)

Knowing that interverting two lines (or rows) in a determinant implies to multiply the determinant by \(-1\), one immediately gets:

\[\begin{equation} \left(\partial x\right)_z = \begin{vmatrix} \displaystyle\left(\frac{\partial x}{\partial T}\right)_v & \displaystyle\left(\frac{\partial x}{\partial v}\right)_T \\ \displaystyle\left(\frac{\partial z}{\partial T}\right)_v & \displaystyle\left(\frac{\partial z}{\partial v}\right)_T \end{vmatrix} = -\begin{vmatrix} \displaystyle\left(\frac{\partial z}{\partial T}\right)_v & \displaystyle\left(\frac{\partial z}{\partial v}\right)_T \\ \displaystyle\left(\frac{\partial x}{\partial T}\right)_v & \displaystyle\left(\frac{\partial x}{\partial v}\right)_T \end{vmatrix} \end{equation}\]

Therefore:

\[\begin{equation} \left(\partial x\right)_z = - \left(\partial z\right)_x \end{equation}\]