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Can the derivative defining pressure $dU \over dV$ or ${∂S \over ∂V}|_{E,N} $ be negative in processes occuring in system not cosmological but statistical (gases or solids or liquids - I mean the statistical study of systems). Altough I have read about negative temperatures and negative pressures, could we have for a system at positive temperature, a negative price of pressure (absolute)?

Note: I have read Are negative temperatures typically associated with negative absolute pressures?

Volker Siegel
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Constantine Black
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1 Answers1

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Depending on how "real" you want the system to be. In a Casimir plate setup in a way the vacuum gets a negative pressure. But for any system of real particles a negative pressure means, that the system will be unstable and collapse, so you cannot have negative pressure in equilibrium. For example, negative pressure can formally occur in the van der Waals model of non-ideal gases, but there it is only an artifact as the uniform phase is unstable (even when the pressure does not drop below zero), and instead there is an equilibrium between gas and liquid at a constant temperature and pressure.

Sebastian Riese
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    It is worth mentioning that the metastable state (before the collapse), while not an equilibrium state, can sometimes be maintained for very long times if the substance is extremely pure (similar to supercooled purified water), or if the system is very small so that nucleation centers don't take. In fact here it was argued that trees would not be able to grow as tall as they do were they not able to upkeep negative pressures. – alarge Jun 12 '15 at 01:48
  • Hi, and thanks for the answer. Could you elaborate or give me a clue on how to understand why from negative pressure follows an unstable system that will collapse?Is there a way to see this statistically nad thermodynamically?Thank you. – Constantine Black Jun 13 '15 at 08:17
  • @alarge Hi. I have read your comment and the answer you gave in the question you mention, and I see that as the answer above, the negative pressure is interpreteted as a situation in a metastable state. Thus, couldn't in any way one have a negative pressure in a stable state? Thanks. – Constantine Black Jun 13 '15 at 08:26
  • @ConstantineBlack No, a negative pressure state cannot be an equilibrium state (because the entropy could be made higher by contracting the body, so this would happen spontaneously. The reason there is a metastability rather than a strict requirement for pressures to be positive is that contracting involves creating new surface area which means there would first be an increase in free energy before the eventual decrease). That said, if the state is stable enough, you can probably treat it as if it were a stable state and use the equations of thermodynamics to describe some secondary phenomena. – alarge Jun 13 '15 at 11:48
  • This answer might be right for solids and liquids, but OP explicitly mentions solids, which can be under tension at equilibrium without collapsing. – Emilio Pisanty Jul 06 '19 at 06:45
  • This is also true for solids in the thermodyamic limit. Imagine a large solid fixed at the walls to be stressed to a negative pressure. If the material tears (creating a vacuum in the cracks) and releasing the stress, the free energy of the system can be reduced. While there is no way to reach that lower energy state (because the local stress is to low and this happing due to the fluctuations will take $10^{100}$ life times of the universe), this still means that the state is only meta-stable, not stable. – Sebastian Riese Jul 26 '19 at 17:29