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There is a derivation of a deformation formula for rocks in one of my textbooks which I don't quite follow. As the problem is mathematical, I've decided to post it here The derivation goes as follows:

Denote the internal energy per unit volume of rock as $u(\epsilon_{ij},v,s)$ and the free energy as $f(\epsilon_{ij}, p, T) = u - Ts - pv$. For isothermal deformations, infinitesimal variations $d\epsilon_{ij}$ and $dp$ result in a variation $df$:

$$df = \sigma_{ij} d\epsilon_{ij} - vdp$$

Because $df$ is a total differential one finds:

$$\left( \frac{\partial \sigma_{ij}}{\partial p} \right)_{\epsilon_{ij}} = - \left(\frac{\partial v}{\partial \epsilon_{ij}} \right)_p$$

I follow everything until the last step. I know the definition of a differential, but I am a bit confused since the last expression contains partial differentials of $\sigma$ and $v$. If someone can illustrate to me how we get from the $df = \sigma_{ij} d\epsilon_{ij} - vdp$ to the final expression, I would be extremely grateful.

Community
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Kristian
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2 Answers2

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Observe that the total differential $df$ may be written from calculus as

$$df=\left(\frac{\partial f}{\partial \epsilon_{ij}}\right)_p d\epsilon_{ij} +\left(\frac{\partial f}{\partial p}\right)_{\epsilon_{ij}} dp$$ From this we identify $\sigma_{ij}=\left(\dfrac{\partial f}{\partial \epsilon_{ij}}\right)_p$ and $v=-\left(\dfrac{\partial f}{\partial p}\right)_{\epsilon_{ij}}.$ Assuming the symmetry of second derivatives (i.e. Schwarz's theorem) we conclude that

$$ \frac{\partial^2 f}{\partial p \partial\,\epsilon_{ij}} =\left(\frac{\partial \sigma_{ij}}{\partial p}\right)_{\epsilon_{ij}} =\frac{\partial^2 f}{\partial \epsilon_{ij}\,\partial p } =-\left(\frac{\partial p}{\partial \epsilon_{ij}}\right)_{p} $$

Semiclassical
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This comes from Schwartz Theorem.

Indeed, from the expression $ df = \sigma_{ij} d \epsilon_{ij} - \nu dp $, you have

$$ \begin{cases} \displaystyle \left( \frac{\partial f}{\partial \epsilon_{ij}} \right)_{p} = \sigma_{ij} \\ \displaystyle \left( \frac{\partial f}{\partial p} \right)_{\epsilon_{ij}} = - \nu \end{cases} $$

Schwartz Theorem gives you that

$$ \frac{\partial^{2} f}{\partial p \partial \epsilon_{ij}} = \frac{\partial^{2} f}{\partial \epsilon_{ij} \partial p} \, ,$$

from which you immediately obtain the desired expression $\displaystyle \left( \frac{\partial \sigma_{ij}}{\partial p} \right)_{\epsilon_{ij}} = - \left( \frac{\partial v}{\partial \epsilon_{ij}} \right)_{p}$

jibe
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