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A similar question was asked here but mine is a bit different.

In thermodynamics, a mechanical equilibrium is defined as a uniform pressure (for a fluid).

In classical mechanics, equilibrium is defined by: sum of external forces equals to zero.

The link between the two is that no external forces work at an equilibrium. Why do we use pressure for the definition of thermodynamics, while it does not cover solid mechanics (because stress tensor is not a scalar tensor)? Would it be possible to give a global definition, compatible in both frameworks? I'm thinking of "an equilibrium is a state where all the macroscopic velocities are 0 with respect to an inertial frame". Unfortunately, this does not constitute a definition similar to common definitions of equilibrium in thermodynamics, which involve conjugate intensive and extensive quantities.

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  • $\begingroup$ Do a search for "gibbs free energy solid mechanics" and you should pick up some pointers on how it can be done. $\endgroup$ – CuriousOne May 4 '15 at 2:47
  • $\begingroup$ @CuriousOne Seems promising, thank you. I'll have a deeper look tomorrow but from what I understand it is possible to add a term $\sigma:\varepsilon$, on top of $p\mathrm{d}V$, in the energy, and $\sigma$ and $\varepsilon$ would be the extensive variable, $\sigma$ of course intensive. I have to think about it, thank you. $\endgroup$ – anderstood May 4 '15 at 2:55
  • $\begingroup$ @CuriousOne This seems to confirm that as such, both definitions of mechanical equilibrium are different, they just have the same name. In a comment I was indicated that a mechanical equilibrium in the sense of thermodynamics could have non-zero velocities. $\endgroup$ – anderstood May 4 '15 at 2:58
  • $\begingroup$ Of course the definitions are different, they both treat different types of systems. One can reduce the more general case of continuums mechanics to the simple high school level force case. I have no idea, though, where you see velocities entering. If something is in dynamic motion, it's not in thermal equilibrium and thermodynamics is not the right way to think about it. $\endgroup$ – CuriousOne May 4 '15 at 4:34
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The definitions are equal: Sum of external forces zero, sum of external torques is zero. This comes from classical mechanics.

For a perfect ideal fluid, the external force density is the pressure gradient: $\mathbf f = -\nabla p$, and therefore, uniform pressure in a fluid means no external force on it, and then it is in mechanical equilibrium. So, its more convinient for thermodynamics, to define mechanical equilibrium in terms of pressure. But this definition is completely equivalent to the definition from classical mechanics. The definition of mechanical equilibrium is also valid for continuum mechanics.

There is a catch therefore. We can do thermodynamics of "everything". So, in general, the state of a system includes: generalized displacement, generalized force, temperature. May include others. For an hydrostatic system this becomes volume, pressure and temperature. Therefore, the way you define mechanical equilibrium for general thermodynamical systems, might result in a different definition from classical mechanics. And more, since this is too general, there may not be a definition of mechanical equilibrium which remains valid for every possible general thermodynamical system.

Also, what you said about the velocities is not correct. For example: A gas on thermodynamical equilibrium has a velocity distribution of particles/molecules/whatever but it is on mechanical equilibrium too: Sum of external torques and forces in all molecules are zero. So, I cannot take a reference frame such that the whole system has zero velocity, but system is in mechanical equilibrium.

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  • $\begingroup$ Regarding your last paragraph, I meant macroscopic velocity. Even though every particle has a velocity, its macroscopic velocity is 0 at equilibrium, right? Also, what about a hydrostatic fluid, e.g. a bottle of water on Earth: the pressure is not uniform (hydrostatic pressure) while in mechanics the state is at equilibrium. Am I wrong? $\endgroup$ – anderstood May 4 '15 at 14:27
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    $\begingroup$ You are right. You are not wrong. How interesting... My answer is wrong then. I'll give some research and thinking before updating my answer. Very interesting indeed. By the way, A wrong part I just saw: I said definition is no external forces and torques. That's wrong. That's just a consequence in some cases of the real definition: Momentum component of the system is conserved. Which means, the net force is null, not the external one. Now agrees with your bottle of water: $\mathbf f_{net} = \rho\mathbf a = \mathbf f -\nabla p$. $\endgroup$ – Physicist137 May 4 '15 at 23:34

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