

The Impact of the Accelerator on the Environment
Question: Silvia, Mexico
Answer:
The short answer is that a system with negative pressure must be unstable,
and thus, there are no thermodynamic states of a system with negative
pressure. The reason for this is simple: The second law of thermodynamics
tells us that all thermodynamic states of a system (aka your fluid
with negative pressure) are such as to maximize the entropy of the system
subject to any external constraints (fixed temperature, volume, etc.).
However, a system with negative pressure can always increase its entropy
by decreasing its volume. Mathematically this is: So, decreasing the volume ( dV < 0 ) increases the entropy ( dS > 0 ) if the pressure is negative ( P = P ). Such a system will spontaneously collapse until the pressure again becomes positive. It's useful to invert the reasoning here and see why *positive* pressures are not a problem. For positive pressures ( P = P ), the system will try to expand ( dV > 0 ) and thereby (again) increase its entropy ( dS > 0). But in this case, the container holding the fluid constrains it to have a fixed, maximum volumethe fluid is not free to expand anymore than to fill the container, which is what it does. For negative pressure, on the other hand, no container can keep the fluid from collapsing in on itself. Physically, what happens when you reach a region of negative pressure in, for example, the PV equation of state is that the system undergoes a phase transition. The initially homogeneous fluid filling a volume V spontaneously splits into two (or more) phases such that each phase occupies a fraction of the total volume and is in a state with positive pressure. These two phases will then each be described by their own, separate (and stable) portion of the equation of state. None of this is to say that negative pressures don't exist. They do. But the evolution of a system with a negative pressure is unstable and thus can't be fully described just by thermodynamics. It has to be described by following more degrees of freedom of the fluid than just the usual 3 thermodynamic variables. I hope I haven't gotten too technical with this answer; from the phrasing of your question it looked like you would be able to follow this approach. If you need further clarification, don't hesitate to ask. Best regards,
Patrick Greene 
last modified 12/16/2003 physicsquestions@fnal.gov 