# 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?

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!