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I'm so confused. I see both of these equations used for equilibrium states, and my textbook actually references the second equation for finding the first, but I don't understand how that works except if the pressure at the liquid's surface was 0 due to being a vacuum. The closest I am seeing to something consistent (I think?) is that it looks like P = ρgh is getting used for situations where you have to integrate due to looking at pressure over a range. So, where does P0 go in that case? Why not integrate P = P0 + ρgh for the range of area?

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If you have a range, the $p_0$ cancels out when you find a pressure difference: $\Delta p=\rho g h_1 +p_0- \rho g h_2-p_0$, so you can get away with not writing it.

$p_0$ might also be omitted if you are using gauge pressure. That is, in the case where you are ignoring atmospheric air pressure near some denser fluid. For example, filling a tire to 35 PSI is commonly understood to 35 PSI above atmospheric pressure, not above vacuum. So if someone quotes pressure at a bottom of a pool in casual conversation, they're probably counting air pressure as 0 and quoting a difference because the difference is what a swimmer would feel. You could also leave off $p_0$ if it were actually 0 as would be the case if there were a true vacuum at $h=0$ or if you were only interested in taking derivatives of pressure.

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  • $\begingroup$ That's really clear, thanks! Is there any non-range scenario where it would be appropriate to leave it out? $\endgroup$
    – Zefyr
    Commented Nov 7, 2016 at 7:12
  • $\begingroup$ For example here, P0 is left out but the examples are definitely not using a range: softschools.com/formulas/physics/pressure_formula/94 $\endgroup$
    – Zefyr
    Commented Nov 7, 2016 at 7:32
  • $\begingroup$ If there's actually a vacuum at $h=0$, if there's air pressure at $h=0$ but it's understood that you're measuring pressure above air pressure (common for filling tires,etc. where 30 PSI means 30 PSI above atmospheric pressure), or if you only care about the derivative. $\endgroup$
    – EL_DON
    Commented Nov 7, 2016 at 17:39
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    $\begingroup$ You should add the term "gauge pressure" somewhere in here, since that is the name of what you're describing. $\endgroup$
    – tpg2114
    Commented Nov 7, 2016 at 22:16

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