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Some cubic thermodynamical equations of state predict negative pressures, have negative pressures any physical meaning? Could they be related to negative mass?

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  • $\begingroup$ More on negative pressure: physics.stackexchange.com/q/38907/2451 $\endgroup$
    – Qmechanic
    Mar 21, 2013 at 10:03
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    $\begingroup$ Negative pressure is just tension. $\endgroup$
    – Michael
    Mar 21, 2013 at 10:16
  • $\begingroup$ Since this was updated today: you can have a look at this paper in Nature Physics, about water at (very) negative pressures: nature.com/nphys/journal/v9/n1/full/nphys2475.html. I recall that they were looking for a student to join their experiment ;) $\endgroup$
    – Vibert
    May 6, 2013 at 23:11
  • $\begingroup$ @Rudie taht was a mistake sorry $\endgroup$
    – Deiknymi
    Dec 17, 2013 at 9:13

1 Answer 1

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Thermodynamically pressure is related to the work done in changing the volume. Assuming an adiabatic change the change in the internal energy of a system is given by the work done on it:

$$ dU = \delta W = PdV $$

If you take a volume of an ideal gas and let it expand by $dV$ it does work on it's surrounds and it's internal energy decreases. This is the usual (positive) pressure.

But consider the dark energy that we believe pervades the universe. If we take some volume of vacuum and let it expand by $dV$ then the internal energy goes up. This may seem odd, but it goes up because dark energy density is a constant everywhere so a bigger volume has more dark energy. Anyhow, if the internal energy of our system goes up when the volume increases we must have done work on it i.e. the sign of $PdV$ is negative. Since the sign of $dV$ is positive, because the volume increases, we conclude that the sign of $P$ must be negative i.e. dark energy has a negative pressure.

Dark energy is not equivalent to negative mass in the $E = mc^2$ sense. The energy density is positive. The reason it shows a negative pressure is due to its equation of state and not to it being in some sense negative energy.

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  • $\begingroup$ I had to think for a while to verify the final paragraph, so maybe it's worth explaining a little more. Dark energy obeys the weak energy condition, which means that in any frame, the mass-energy density is positive. However, dark energy violates the strong energy condition, meaning that it produces a gravitationally repulsive effect. $\endgroup$
    – user4552
    May 6, 2013 at 23:52
  • $\begingroup$ Hello. Could I ask if the derivative ${dU \over dV} $ or ${\partial S \over \partial V}|_{E,N} $ can 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)? Thank you. $\endgroup$ Jun 10, 2015 at 18:20
  • $\begingroup$ @ConstantineBlack: you might want to post that as a new question. $\endgroup$ Jun 10, 2015 at 18:59
  • $\begingroup$ Energy is not conserved in the standard cosmological model, so applying the law of conservation of energy to infer that pressure is negative for the speculative concept of dark energy seems a bit unfounded to me. $\endgroup$
    – juanrga
    Jan 23, 2023 at 18:33

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