If we have two fluids $1$ on top of $2$, I know that the absolute pressure of a fluid $2$ is $p_2 = p_1 + \rho gh$ where $h$ is the height of the second fluid, and $p_1$ is the absolute pressure at the bottom of fluid $1$. In other words, we add the pressures.

Now, consider a thin closed off pipe filled with water as shown, such that the Rayleigh-Taylor instability does not apply:

enter image description here

However, looking at the drawing, why would the absolute pressure at $P_1$ be $P_1=p_0 + \rho gh$ and not $2p_0 + 2 \rho gH$, and similarly, why is $P_2 = p_0+2\rho gh$ and not $2p_0 + 2\rho gh$.

Why don't we add atmospheric pressure the same we do pressure from other liquids?


1 Answer 1


We do "add the pressures," but only when the effect is big enough to matter.

The density of air at sea level is about $1.2\,\text{kg}/\text{m}^3$.

So the pressure change in a column of air $1\,\text{m}$ high is about $1.2\times 9.8 \approx 12\,\text{Pa}$.

Compare that with change the atmospheric pressure at sea level of about $100,000\,\text{Pa}$. In most situations a change of $0.012\%$ over a height of $1\,\text{m}$ can be ignored.

However if the "column of air" is $1\,\text{km}$ or $10\,\text{km}$ high, the pressure change is not negligible, and this is the reason why atmospheric pressure changes with altitude!

  • $\begingroup$ +1 for understanding the question. $\endgroup$
    – my2cts
    Jun 21, 2021 at 21:43
  • $\begingroup$ I still don't understand why $P_2=p_0+2\rho gh$. To calculate $P_1$ we add $p_0$ to the pressure exerted by the liquid. Now to calculate $P_2$ we only consider $P_1$ that is added, and we completely forget about $p_0$. $\endgroup$
    – Neox
    Jun 20, 2023 at 13:22

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