How is negative pressure created in a fluid system?Isn't it counter intuitive that we are reducing the pressure of a system containing no molecules(zero pressure) to a lower value?

  • $\begingroup$ Negative pressures can be achieved, but such states are metastable. For example the vdW eqn of state has negative-pressure states, but these are actually all resolved through the Maxwell equal area rule to be positive (when the system equilibrates i.e. phase separates). This is to say that the negative pressure is caused by forcing the system to be uniform where in reality it "wants" to phase separate. Finite size effects typically manifest in analogous ways. $\endgroup$
    – alarge
    Commented Apr 3, 2015 at 13:58
  • $\begingroup$ I think you are refering to negative pressure created by scalar fields, right? If that is the case, the argument of Maxwell equal area is not valid because this wouldn't be a poper thermodynamical system $\endgroup$
    – Hydro Guy
    Commented Apr 3, 2015 at 14:03
  • $\begingroup$ @alarge I see some point in your answer I'cant say that I've grasped everything you've just said.Could you provide some relevant links supporting your answer? $\endgroup$ Commented Apr 3, 2015 at 14:07

2 Answers 2


Suppose you have a box filled with water in a uniform fashion. Now if you try to stretch the box in the $z$-direction, say, while keeping the other dimensions constant, what is the energy required? Well if the water distribution remains uniform, you can approximate this by the Hookean law $E_\text{elastic} = \frac{1}{2}k(z-z_0)^2$. Note that the constant $k$ is actually huge as water really, really doesn't want to stretch as it has a certain density it likes to have.

Other than decreasing in density, how can the system react to the stretch? Well, it can phase separate so that the liquid water keeps at the optimal density and the other part of the box becomes gas (let's imagine it is basically a vacuum). The energy penalty you pay is then approximately the surface tension $\gamma$ times the area of the liquid-gas interface $xy$.

Because $k$ is so large, the latter effect is almost always preferable. However, to go from a uniform system to one with two phases, something must break the symmetry (uniformity). This is often called a nucleation site, and it can be for example an air bubble. Certain geometries of the wall can also act as sites.

If the system is very pure (and/or small) and uniform when you start the stretch, nucleation will not happen, which means that phase separation will not happen. This means that water has to stretch. There's an energy penalty associated with this and thus you will be creating a pressure (due to the stretch alone) of $-kz$, which is to say that you can get to negative pressures if the strain is large enough. Remember though, that this is a metastable state, as all states with negative pressures are, and the system will phase separate if given the opportunity.

Finally, to relate this to my earlier comment about the van der Waals equation: In the physical derivation of vdW you assume that the system is uniform. You will get some very strangely behaving lines (e.g. pressure increases as the box size increases). These are only obeyed if the system remains uniform (which it does if nothing can break the symmetry). However, as I explained above, if you stretch the system, uniformity is not what gets you the minimal free energy: If you assume that the system is actually made up of two phases (with no interfacial tension between them), and find the composition that minimizes the free energy, you'll arrive at Maxwell's equal area rule.


'Negative ' pressure is strictly a relative state; relative to what one may wish to define as zero pressure, and here on earth we chose to define that as one standard atmosphere of pressure which is about 760 mm Hg absolute pressure. If you are capable of removing all gas particles from a space, then you will achieve -760 mm Hg gauge pressure, but you cannot reduce pressure beyond that point

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    $\begingroup$ I think you need to know how the water is transported in trees through xylem tube that are ~100 m long.For more information on the application of negative pressure, you may watch the Veritasium video:youtube.com/watch?v=BickMFHAZR0&index=2&list=WL $\endgroup$ Commented Apr 3, 2015 at 14:04
  • $\begingroup$ @akshay 760 mm Hg (1 atm) is roughly a water column height of 34 feet ( 10 meters). Xylem transport of fluid by negative pressure alone would be limited to this column (-10m) height, but differential pressure is unlimited. Positive pressure applied at the base of the tree can be increased to any pressure to push the fluid up as long as the xylem can sustain the pressure without bursting. And how or whether xylem transport systems can do that is a question or matter for the biology forum. Physics limits you to a hard vacuum. $\endgroup$
    – docscience
    Commented Apr 3, 2015 at 23:48

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