I am trying to understand the role that the work conjugate pair of stress and strain plays in the enthalpy: $H = U - V_0\mathbf{P}:\mathbf{F}$, where $\mathbf{P} = J\sigma\mathbf{F}^{-T}$ is the 1st Piola-Kirchhoff stress, $\sigma$ is the Cauchy tensor, $\mathbf{F}$ is the deformation gradient, and $J=\det{\mathbf{F}}$. It is easy to show that $V_0(J\sigma\mathbf{F}^{-T}:\mathbf{F})=JV_0\mathrm{Tr}(\mathbf{F}^{-1}\sigma^T\mathbf{F})=V\mathrm{Tr}(\sigma)$.

On the other hand, we have the usual definition of the pressure $p=-\mathrm{Tr}(\sigma)/3$, and it gives $H=U+3pV$. I didn't expect to see coefficient 3 and I am wondering what I have missed here.


The original question did not use the density of the thermodynamic potential, thus it has a unit volume in the equation. If $H$ and $U$ are density-like properties, the results I had would then become $H = U - J\mathrm{Tr}(\sigma)=U+3pJ$, whereas the enthalpy per unit volume is given by $H = U + pJ$. $J = \det{\mathbf{F}} = V/V_0$. I can still not figure out what leads to this inconsistency.

  • $\begingroup$ Do you have a reference for the first expression of $H$? I would expect the time integral $\int\boldsymbol P:\dot{\boldsymbol F}\mathrm dt$, which does not yield $\boldsymbol P:\boldsymbol F$. $\endgroup$
    – neerby
    Commented Oct 7, 2022 at 14:08
  • $\begingroup$ @neerby I am a layman in continuum mechanics. Thanks to your comment, now I see where the problem was. Thank you. $\endgroup$
    – oukore
    Commented Oct 22, 2022 at 8:35

2 Answers 2


Clarifications the proposed result

The premise is not true in general: the enthalpy is path dependent and known only after its rate $$ H = \int_0^t \dot U - \boldsymbol P:\dot{\boldsymbol F}\mathrm dV_0 \mathrm \;d\tau\\ \neq U +\boldsymbol P:{\boldsymbol F}\mathrm dV_0. $$ Here we write $\mathrm dV_0=J^{-1}\mathrm dV$ the infinitesimal volume on the initial configuration.

  • For the given expression (with $-\boldsymbol P:{\boldsymbol F}$) to make sense in principle, we would need $\boldsymbol F$ and $\dot{\boldsymbol F}$ to remain proportional. The only way is for $\dot{\boldsymbol F}$ to be proportional to the initial deformation gradient $\boldsymbol I$. This corresponds to a spheric deformation, with only volume changes, which explains why only the pressure term appears at the end of OP's development. In general there is another deviatoric term.

  • For the given expression to be applicable to a concrete case, one would need to find $(\boldsymbol \sigma(t),\boldsymbol F(t))$ such that, $\forall t$: $$ \int_0^t\boldsymbol P(\boldsymbol \sigma,\boldsymbol F):\dot{\boldsymbol F}\mathrm d\tau =\boldsymbol P:{\boldsymbol F}(t). $$ This is tedious and far from general.

So in this case the problem was in the question formulation. But we can still address some questions connected to the original post.

What do the Cauchy stress and its pressure term conjugate with?

Substituting $\boldsymbol P=J\boldsymbol\sigma\boldsymbol F^{-T}$ yields the power of internal forces $$ - \boldsymbol P:\dot{\boldsymbol F}\mathrm dV_0 = - \boldsymbol\sigma:(\dot{\boldsymbol F}{\boldsymbol F}^{-1})\mathrm dV $$ Conclusion: the Cauchy stress conjugates with the (symmetric part of the) velocity gradient $\boldsymbol l=\dot{\boldsymbol F}{\boldsymbol F}^{-1}$.

From there using $\boldsymbol\sigma=-p\boldsymbol I$ the hydrostatic part of the power of internal forces reads $$ - \boldsymbol\sigma:\boldsymbol l\mathrm dV = p (\mathrm{tr}\,\boldsymbol l)\mathrm dV + (\ldots)_\text{deviatoric}. $$ Conclusion: the pressure conjugates with the trace of the velocity gradient.

Note: we have $\mathrm{tr}\,\boldsymbol l = \mathrm{div}\, \boldsymbol v$ and $(\mathrm{div}\, \boldsymbol v\mathrm) dV= \dot J\mathrm dV_0$ therefore $p(\mathrm{tr}\,\boldsymbol l)\mathrm dV = p\dot J\mathrm dV_0$. The next section brings to the same result coming from a different angle.

What is obtained when the deformation is spheric?

This is to show where the missing $1/3$ would come from in a more general case.

The original post implies a deformation where only the volume changes. This is a particular case where $\boldsymbol F$ reduces to $\boldsymbol F^\text{vol}=J^{1/3}\boldsymbol I$. We apply chain rule to find the time derivative: $$ \dot{\boldsymbol F}^\text{vol} = \frac13 J^{-2/3}\dot J\boldsymbol I $$ and the corresponding velocity gradient $$ \boldsymbol l = \dot{\boldsymbol F}^\text{vol}\boldsymbol F^{\text{vol},-1} = \frac13 J^{-1}\dot J $$ The power of internal forces becomes $$ p (\mathrm{tr}\,\boldsymbol l)\mathrm dV = p \frac{\mathrm{tr}\,\boldsymbol I}3 \big(J^{-1}\dot J\big)\mathrm dV\\ = p \dot J\mathrm dV_0 $$ The time-and-space integration of $\dot J\mathrm dV_0$ is the total volume.


I don't completely understands your first expression. We normally use a density of thermodynamic potential. In fluid mechanics it is common to give them relative to unit mass and in solid mechanics it is more common to give them per unit volume. Thus, if you are given the expression relative to unit volume (for example) it reads

$$H = U - \mathbf{P}:\mathbf{F}\, ,$$

if, on the other hand, this expression is the total value on your body/control volume $\Omega$, it would be

$$H = U - \int\limits_\Omega \mathbf{P}:\mathbf{F} dV\, .$$

  • $\begingroup$ Yeah, you are correct. I am not very familiar with the conventions of fluid mechanics but I guess I can treat H and U as thermodynamic densities. The original equation has been edited. $\endgroup$
    – oukore
    Commented Oct 27, 2021 at 17:33
  • $\begingroup$ @nicoguaro Any thoughts on the edited question? I arranged a bounty because I'm curious about the resolution of this problem. $\endgroup$ Commented Nov 5, 2021 at 20:46
  • $\begingroup$ @Chemomechanics, there should be a mistake somewhere. I spent some time trying to find it, but I could not find it. I might have some time tomorrow to ponder it. $\endgroup$
    – nicoguaro
    Commented Nov 5, 2021 at 21:10

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