Free falling fluid pressure distribution

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: https://physics.stackexchange.com/questions/4619/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?

• Have you included the surface tension forces? If not the blob will come apart... Sep 27 '18 at 5:26
• @Solar_Mike, it's a tank of fluid, so it's being held together by the tank. Sep 29 '18 at 7:26
• So the tank is completely full with no free surface? Sep 29 '18 at 8:08
• Yup, if there was a free surface then this would complicated things, the fluid would disperse into the free volume. I should have clarified but I'm just keeping this example as a tank with no free surface, so that the focus can be on the effect of downward gravitational acceleration on pressure distribution. Sep 29 '18 at 14:33
• While in your simplified example it's really simple (c'mon, since there's no designated 'z' direction anymore, your equations need to be unchanging due to rotation of the system of coordinates). This thing is much more complex a problem in reality though. Keyword: ullage - the set of problems surrounding mechanics of liquid in freefall, as experienced by rocket fuel in tanks of a rocket. It's a huge set of problems with a whole lot of creative and often rather risky solutions.
– SF.
Oct 29 '18 at 0:28

In free fall, any segment of a control volume for any material experiences absolutely no gravity. As noted, Euler's equation yields a statement that pressure is identically isotropic at that point. In essence, there is no "down" to the control volume. Apply this throughout the control volume, and you find that pressure is identically isotropic throughout the entire body of the control volume.

Before we continue, the above statements are the founding principles behind Drop Tube Experiments to measure phase transformations without the influence of gravity.

Now consider a control volume with a surface. That surface has a surface tension. The interesting note here happens when the external surroundings have a static fluid. We find that each point on the surface of the drop should theoretically experience a different pressure because the external fluid is static. In essence, the external surface knows which way is "down".

The other issue we face is to account for the drag affects of the external fluid on the falling body. A pressure difference will arise top to bottom of the body from this factor.

The escape out of having to consider the surface tension and drag affects is to invoke a perfect vacuum around the falling body.

lets try not to complex out the answer, so consider a volume of liquid freely falling.

Now lets talk in a frame of reference attached to that liquid. Effective weight of every infintesimally small element would be zero during a free fall.

Also since we are in frame of the liquid a=0 , this a hydrostatic condition. And how does pressure difference in z direction arise in hydrostatic condition?

due to the fluids own weight , which is effectively zero in this case.

So no pressure difference should arise in this case.

And yes Euler's equation yields the same(in ground frame), and you have done it correctly.

Tanklike objects freefall towards a system barycenter. We don't know the spacial distribution of masses that comprise the system, so we can't say much about the gravitational field in the area of the tank. But if the barycenter lies outside the tank, then there is some gradient to the gravitational field, and there will be a corresponding pressure gradient in the fluid in the tank. The rigid tank will accelerate at some average value of the field bounded by the tank, with the fluid at the "front" end at higher pressure, the fluid in the middle at lower pressure, and the fluid at the "back" at higher pressure. These pressures cause tides on a free surface.

Inside the tank, a fluid also self-gravitates. This turns out to be a surprisingly thorny problem to solve for a general self-gravitating fluid mass. The unbounded, self-gravitating, incompressible, rotating fluid solution was finally put to bed by Chandrasekhar in 1967 after the world's best mathematicians having worked on it for about 400 years (Newton started the whole business trying to explain the ocean's tides). The bounded case still appears to be an open problem.

So no more hand waving.

Here's a fun paper - http://faculty.georgetown.edu/tl48/lsfinal.pdf

But you might want to warm up with Chandra - https://onlinelibrary.wiley.com/doi/10.1002/cpa.3160200203

... and a model for compressible fluids based on Chandra's work - https://www.researchgate.net/publication/23881101_Ellipsoidal_figures_of_equilibrium_-_Compressible_models