# How does Newtonian viscosity not depend on depth?

Good morning, StackExchange.

I am currently trying to self-study basic Classical Mechanics and right now I am trying to understand fluid dynamics. I have read about Newtonian viscosity: as my book says, the force exerted on a viscous fluid depends on the coefficient of viscosity, the cross-section and the velocity gradient, but, to my amazement, not on the fluid's height, or in any other way on its total amount.

Usually I'd be more than willing to trust such a result even if not proven, and the way it is used in deriving something like Poiseuille's law, for example, almost makes sense to me. But I still, to my best efforts, fail to understand or even get an intuition on what should be the most basic case of application, i.e. a fluid that can be approximated as many rectangular plates flowing on top of each other. If I try to apply the same method I would use for a cylindrical tube, I get that considering a section of any height $$h$$, if the velocity gradient in the rectangle is constant at every depth, the force I need to apply to that section is the same no matter the value of $$h$$. In particular, dividing the rectangle in an arbitrarily large number of sections of arbitrarily small value of $$h$$, keeping the gradient the same I get that the total force I need to apply to maintain it is arbitrarily large.

Also, it sounds unintuitive that keeping the same gradient and halving the amount of fluid I would need the same force (and double the pressure). In addition to this, even with the tube, if instead of applying the viscosity law to a full cylinder I try applying it to a cylindrical ring, I get the same problems.

Can someone explain? Am I doing this wrong (probably)? Does the viscosity law somehow only work for the "top" section of the liquid? If so, what does top mean? How can the fluid "know" what is the top and what isn't? Are different pressures really needed depending on the amount of fluid to maintain the same gradient? What's an intuitive explanantion of this? Is there something else I'm missing that would make the explanation work?

• " In particular, dividing the rectangle in an arbitrarily large number of sections of arbitrarily small value of h, keeping the gradient the same I get that the total force I need to apply to maintain it is arbitrarily large. " Why would it be arbitrarily large ? Even if you take small value of h, the velocity gradient stays the same. So, force should stay the same Apr 27, 2021 at 13:10

The dynamic viscosity $$\eta$$ of a Newtonian fluid is a constant coefficient in the relation between the stress tensor and the velocity gradient. It is a property of the material and the relation between stress tensor $$\sigma$$ and velocity gradient $$\nabla u$$ is valid at every point in the fluid: $$\sigma = \eta\, \text{sym}\nabla u$$ (sym denotes the symmetrisation and the above relation is only valid in incompressible flow, but don't worry about those details right now.) You might want to read up on what a stress is - essentially a force per unit area.
Now consider the most simple example: A fluid sheared between two horizontal plates separated by a vertical distance h, the lower plate being at rest, the upper one moving with velocity v. The force you need to exert on the upper plate to keep the shear flow going is then determined by multiplying the stress on the upper plate by its area. With the velocity gradient in the fluid being uniform and equal to $$\frac{v}{h}$$ you find from the above relation that the force is proportional to $$\eta \frac{v}{h}$$. So if for example you keep the same velocity gradient while halving the height, as you suggest above, you indeed need to exert the same force on the upper plate to keep up the shear flow, but the upper plate will be moving at half the velocity.