Atmospheric pressure in non-nertial frame?

Question

Any object kept in an accelerating container of water feels different pressure than unaccelerated. Because if we go into the frame of water the g effective changes. Since air is also a fluid, a container of liquid accelerating upwards should experience more atmospheric pressure than it feels at rest, but intuitively it does not feel so. Am I correct in assuming that liquid feels more atmospheric pressure? (Quantitatively, $P×(g+a)/g$)

Answer 1

Existing answer is correct but I will present it in less technical terms.

If the air is not moving relative to the ground, then the atmospheric pressure at the ground is whatever is enough to prevent the air from the next layer up from falling down towards the ground.

If the whole system (ground and air) is accelerating upwards then in order to prevent the layer above from moving towards the ground, the pressure at the ground must be higher. Therefore the pressure is higher under that condition. It goes up in proportion to the "effective gravity" ($g+a$) as you suspected.

Answer 2

This question may be based on an expression for pressure in a column in hydrostatic equilibrium, like $P = \rho g h$ (or its integral generalisation for variable density $P = \int_{z_0}^{\infty} \rho g dz$). It's important to bear in mind the assumptions underlying these expressions. Your equation for pressure would hold subject to the following:

• Your liquid is subject to acceleration $a$ upward.
• A column of atmosphere above that liquid accelerates upward at the same rate.
• There is a barrier which prevents pressure inside the column from equilibrating with the atmosphere which is at rest outside the column.

This seems like a rather contrived case, which may be why the kind of dependence of $P$ on $a$ you're talking about seems counterintuitive.

To look at the more realistic case where the atmosphere is not in uniform motion we need to solve the Euler equations. The one for pressure can be stated: $$ \frac{D \mathbf{u}}{D t} - \mathbf{g} = \frac{\nabla p}{\rho}$$ This shows how pressure gradient $\nabla p$ is related to gravitational field g and (Lagrangian) acceleration $\frac{D \textbf{u}}{D t}$ of a fluid parcel.