Bernoulli's equation and negative pressure Here is Bernoulli's equation:
$$P_{{1}}+1/2\,{\rho v_{{1}}}^{2}+\rho {\it gy}_{{1}}=P_{2}+1/2\,{\rho v
_{{2}}}^{2}+\rho {\it gy}_{{2}}
$$
A friend of mine was looking at some lecture notes today and there was an interesting question included, without an answer.
Let's say the equation is rearranged as follows, to solve the pressure $P_{2}$:
$$P_{2}=P_{{1}}+1/2\,{\rho 
v_{{1}}}^{2}+\rho {\it gy}_{{1}}
-1/2\,{\rho v_{{2}}}^{2}-\rho {\it gy}_{{2}}
$$
The question is: Is it possible for the pressure $P_{2}$ to be less than or equal to zero? And if so, what would that mean? Apparently there was no picture included so I cannot say much more about the scenario in the question.
So what would, say, a negative pressure indicate in this equation? I spent some time with this question but I could not think of a sure answer. Does negative pressure here indicate a negative force on the fluid element? Is the fluid flowing backwards?
 A: 
Is it possible for the pressure P2 to be less than or equal to zero?

In this equation, yes.  The Bernoulli equation is essentially an energy balance for a fluid at two points.

And if so, what would that mean?

For starters, it means you aren't measuring absolute pressure.  You could be measuring gauge pressure, or you could have some arbitrary pressure defined as 0.  It isn't really a big deal for the Bernoulli equation; because it is the pressure difference that determines the potential of the system due to pressure.

So what would, say, a negative pressure indicate in this equation?

Nothing special.  It might indicate that they should have picked a different scale for pressure to avoid this.  It doesn't say much besides that the potential due to pressure is lower at point 2 than point 1.  
A: No, Pressure is defined positive.
The only term of pressure that has a reality in this equation here is $\Delta P = -\frac{\rho \Delta(v^2)}{2} - \rho g \Delta z$.
As an example, with most systems, Pressure is really close to atmospheric pressure $10^5 Pa$ which is really high to make the other term $P_2$ become zero or negative.
A: We know that, with certain constraints (steady state, no friction, laminar flow), this equation should be valid - it is a law.
Therefore, if all given values are real, the calculated pressure (assuming we are working with absolute pressure) should be positive. Conversely, if the calculated pressure is negative, at least some of the given values must be bogus.
A: It is perfectly possible for the pressure to be negative, giving a liquid under tension - and this goes on regularly in trees if they are tall enough Check this out, this will surely blow your mind..
I m writing this as an answer and not as a comment, so that it maybe noticable.
