It is said that for local thermodynamic equilibrium the local entropy production needs to be 0.

Now, I am reading the following from the book by de Groot and Mazur "Non-Equilibrium Thermodynamics". It says that the local entropy production = $\sigma = J_{\nu}.grad(\frac{1}{T})$, where $J_{\nu}$ is the heat flow and $J_{\nu} \propto grad(\frac{1}{T})$.

Finally, the book goes on to say that to minimize entropy production, a) $div. grad\frac{1}{T} = 0$, or equivalently, b) $div. J_{\nu}=0$.

My questions are then as follows:

1) In an adiabatic system, there is no net inflow or outflow of heat. This should imply that $div. J_{\nu}=0$. Then, does this mean that all adiabatic systems would be in local thermodynamic equilibrium since entropy production is zero?

2) Let me also take a tub filled with huge quantity of mercury at temperature $T_1$. I gently add a drop of mercury at temperature $T_2$, with $T_2 > T_1$. Slowly the additional heat from the drop of mercury is going to dissipate into the mercury heat tub. There is no other external heat flow into or from the mercury heat bath. Hence, again $div. J_{\nu}=0$. And does this necessarily have to be a system in local thermodynamic equilibrium, during the process when the heat is being dissipated from the mercury droplet through the mercury heat bath? Does this have to be in local thermodynamic equilibrium, even if $T_2$ is >>> $T_1$ (as this may create some huge gradients in the temperature)?


1 Answer 1


The local entropy production is a volume quantity : $\frac{d{{S}_{c}}}{dt}=\iiint\limits_{V}{\sigma d\tau }$

For an adiabatic evolution, you just require that $\overrightarrow{{{J}_{\nu }}}=\overrightarrow{0}$ at the frontier of the system. But the entropy production can be non zero inside the system. So an adiabatic system can be the siege of entropy creation.

I don't understand why you should have $\overrightarrow{\nabla }\overrightarrow{{{J}_{\nu }}}=0$ in the second case (and the system is an open system, which could complicate the entropy balance)

edit :

The equation $\overrightarrow{\nabla }\cdot \overrightarrow{\nabla }(1/T)=0$ is the equation for the minimum entropy production. Not for zero entropy production.

In my memory, this minimum entropy production correspond to a stationary state of the system : temperature independent of time but non zero thermal flux. The created entropy is compensated by the flux of entropy at the boundaries.

For example, a bar withe fixed temperature at the two ends. It evolves a state for which temperature vary linearly along the bar.

There is a well know problem with this formulation : the law $\overrightarrow{{{J}_{\nu }}}=K\overrightarrow{\nabla }(1/T)$ is not the Fourier law $\overrightarrow{{{J}_{\nu }}}=-\lambda \overrightarrow{\nabla }(T)$ and so you find an equation that is different from the Laplace equation ${{\overrightarrow{\nabla }}^{2}}(T)=0$ which is usually considered in these problems.

Sorry for my english !

  • $\begingroup$ I agree with your assessment for the first case. In the 2nd case, the new drop of mercury plus the mercury in the tub can be regarded as a closed system. So, as you point out, certainly in this case there are temperature gradients in the system giving rise to entropy generation. $\endgroup$ Commented Feb 11, 2019 at 17:41
  • $\begingroup$ @Vincent: Consider the mercury and the droplet to be inside a temperature chamber, so that it becomes a closed/isolated system. $\endgroup$
    – Angela
    Commented Feb 11, 2019 at 17:45
  • $\begingroup$ I agree with the comment by @ChesterMiller : the temperature gradient is not zero and so there is a non zero entropy production ? $\endgroup$ Commented Feb 11, 2019 at 17:48
  • $\begingroup$ @Chester: From the book, what I see is that div.grad{1/T} = 0 for no entropy production. So, grad{1/T} can still be non-zero. Just the fact that grad{1/T} is non-zero does not seem to imply that local entropy production is 0. Is there any other factor that needs to be considered to call the mercury and the droplet system to be in local thermodynamic equilibrium? $\endgroup$
    – Angela
    Commented Feb 11, 2019 at 17:48
  • $\begingroup$ I have completed the answer with an edit. It was a little too long for a comment. Sorry for my poor english ! $\endgroup$ Commented Feb 11, 2019 at 18:05

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