Some time ago i saw this question Why does water falling slowly from a tap bend inwards?

Which explains why the water flowing from a tap bend inwards while falling, using an inviscid flow model. However all calculations there assume steady flow, so that the continuity equation and bernoulli can be used in the usual forms.

But i don't see how can a free fall flow ever can be steady. I know it is the case, but i struggle to understand how a free fall flow would transition from unsteady to steady.

I imagine, for instance, a large volume of water with the exit closed at the bottom. At a initial moment, the bottom exit opens, and the water starts moving from rest. Since the fluid is now unconstrained, the pressure reduces to zero (or simply atmospheric pressure) and the water falls due to gravity, but every single point of the water should fall with the same acceleration g, therefore there is no relative motion between the fluid parts, so it should fall as an unique body. The flow is unsteady of course and would simply be that $\frac{\partial v}{\partial t} = g$.

How would this flow ever become steady?

  • $\begingroup$ Do you think that all the water exits at the same time, or does water exiting sequentially replace water that exited immediately before it? $\endgroup$ Feb 3 at 12:25
  • $\begingroup$ Every element of water replaces the other below it, that's for sure, but that doesn't guarantee deformation. A solid is a continuous body yet all parts fall with the same velocity. I don't see why water falling from rest would bend, since acceleration at all parts would be equal, it should fall exactly like a rigid body. And those slow motion balloon popping videos, the water falls at the exact shape of the balloon. I my mind gravity by itself can't cause deformations unless the field is not uniform. So what exactly causes the deformation in steady flow? $\endgroup$
    – Klaus3
    Feb 3 at 16:14

2 Answers 2


If the acceleration of the fluid particles is expressed as $\frac{dv}{dt}=g$, we can use the chain rule to write $$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=g$$where x is the distance measured downward from the tank exit. Integrating this with respect to x yields: $$\frac{v^2}{2}=\frac{v_0^2}{2}+gx$$where $v_0$ is the exit velocity from the tank. This shows that, from the perspective of a stationary observer, the downward velocity is a function only of x and not time. This is the definition of a steady flow.

  • $\begingroup$ absolutely agree. But notice i said in my example that water moves from rest, so the speed is: $v=\sqrt{2gx}$. But notice that the model gives the equation for the radius: $r(h)=\frac{r_0\sqrt{v_0}}{(v_0^2-2gh)^{1/4}}$ . And the model collapses when $v_0 = 0$ outputting $r(h)=0$ , which obviously isn't correct, but if you take the limit $v_0->0 and h->0$, $r(h)=r_0$. So, the only reason the area changes is because of the initial velocity? $\endgroup$
    – Klaus3
    Feb 3 at 15:51
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    $\begingroup$ The jet is able to contract and remain intact because of surface tension surrounding the jet as well as by small viscous forces causing a small radial pressure gradient. $\endgroup$ Feb 3 at 16:28
  • $\begingroup$ Gotcha. I was inspecting the euler equations and, for the incompressibility condition to be satisfied, since there are axial gradients of velocity, there must be radial gradients of velocity, which implies radial gradients of pressure. Thanks. $\endgroup$
    – Klaus3
    Feb 5 at 16:17

The reason slowly falling water contracts as it falls (up to a point at least) is that the center of the stream is falling faster than the periphery of the stream. This tends to draw the sides in. When viscous dissipation erases this velocity difference, the stream stops contracting.

NOTE ADDED: This model presupposes a velocity profile that exists across the diameter of the jet at its point of origin, and viscous coupling via the shear gradient across that cross-section which would act to accelerate the circumference of the jet while decelerating its faster-moving core- all while the jet itself is being pulled out in length by gravity. I am going to pose a separate question on this topic to determine if this effect is real and if not, I will delete my answer to this question. -NN

  • $\begingroup$ Thanks for the answer, but look at the linked question. The accepted answer model is fully inviscid and it predicts both the initial contraction and the moment it stops, staying with constant area. And why does the center falls faster? In this situation, isn't the acceleration of all points equal to the gravity? $\endgroup$
    – Klaus3
    Feb 3 at 3:57
  • $\begingroup$ It doesn't stay constant area. As it falls, the velocity speeds up and the cross sectional area correspondingly decreases. $\endgroup$ Feb 3 at 12:28
  • $\begingroup$ Of course, but as the height of fall increases, the radius tends to a constant value. At least in the scale of the graph. $\endgroup$
    – Klaus3
    Feb 3 at 15:56
  • $\begingroup$ @Klaus3, at the origin of the jet the velocity of flow around the circumference of the slug of water exiting the hole is zero (the no-slip condition at the edge of the hole) but along the central axis the water has already begun to accelerate and has a (pre-existing) nonzero flow velocity there at the point where it passes through the hole. this means there is a velocity profile across the cross-section of the jet at its point of origin, in which the center of the jet is moving faster than the exterior surface of the jet. I will pose this effect as a separate question. $\endgroup$ Feb 4 at 6:32

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