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Consider a free falling tank of fluid. The goal is to find the pressure distribution.

My intuition says that there should be no pressure distribution; the pressure should be uniform since the container is also accelerating at the same rate and direction as gravity, so there is no upward reacting force on the fluid to create a hydrostatic pressure distribution. This is also previously supported by others: Water pressure in free fall

Previous questions and answers, however, are entirely handwavy and also based on intuition. How can we prove this mathematically?

I start with Euler's equation in the downward direction of acceleration, let's say the $z$ direction:

$\rho a_z = -\frac{\partial P}{\partial z} - \rho g$

and the fluid is accelerating downward with acceleration $a_z = -g$.

Substituting into Euler's equation yields

$ \frac{\partial P}{\partial z} = 0$

thus proving that the pressure in a free-falling fluid is uniform.

Is this correct? Can I apply $a_z = -g$ in the fluid equation of motion even though gravity is already accounted for?

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    $\begingroup$ It is correct. $a_z$ is the acceleration of fluid as seen in an inertial reference frame (earth frame). Since the fluid is in free fall $a_z=-g$. $\endgroup$ – Deep Sep 27 '18 at 5:19
  • $\begingroup$ This post really should be an answer in the question you linked, rather than a question on its own. $\endgroup$ – DrSheldon Sep 27 '18 at 6:29
  • $\begingroup$ @Deep what's confusing to me is that Euler's equation, essentially a force balance, already includes a gravitational pressure force $-\rho g$ on the right hand side. Isn't using $a_z=g$ double counting gravity? What if we choose a noninertial reference frame following the fluid? $\endgroup$ – Drew Sep 27 '18 at 11:10
  • $\begingroup$ @Dr_Sheldon, I see what you mean but I wanted to give some attempted thought myself. $\endgroup$ – Drew Sep 27 '18 at 11:11
  • $\begingroup$ $a_z$ is the acceleration of the fluid which is proportional to the net force acting on it, which is vector sum of pressure gradient force and gravity force. You are talking about the case where acceleration of the fluid has a particular value viz. $-g$, but physically it is different from gravity force term. So no, you are not double-counting gravity. In a non-inertial frame you must introduce appropriate pseudo-forces. $\endgroup$ – Deep Sep 27 '18 at 11:40

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