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

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:

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?

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!

• +1 for understanding the question. Commented Jun 21, 2021 at 21:43
• 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 $p_0$.
– Neox
Commented Jun 20, 2023 at 13:22