# In the classic viscosity definition, why does doubling the plate gap cause the force to halve (intuitive)?

I am puzzled by an artifact of the definition of viscosity and need an intuitive picture to help explain it. I know $\tau_{yx}=-\mu{dv_x \over dy}$ but I am looking for an intuitive picture of the specific question below.

Take the prototypical definition/model of viscosity: Imagine two very large plates separated by a constant small distance with a fluid sandwiched inside. Apply a force $F$ parallel to the bottom plate so as to start moving the bottom plate at a speed of $V$. Once steady-state is reached, $F$ and $V$ are constant. The gap between plates is $Y$ and the flow is presumed/constrained to be laminar. One finds that the velocity gradient with respect to distance along $Y$ is linear with velocity at the top plate zero and the velocity at the bottom plate $V$ (no-slip condition). In equation form:

$${F \over A}=\mu{V \over Y}$$

where $\mu$ is the viscosity. So far, so good.

Now, consider the force $F$. It is proportional to $A$ which makes sense as a bigger plate will require more force to move due to the viscous contact with the liquid. It is proportional to $\mu$ which really is the definition of viscosity (higher viscosity fluid requires more force). It is proportional to $V$ which make sense since moving the plate at a higher velocity will require more force. Finally it is inversely proportional to $Y$, the plate gap. I can't quite understand this one intuitively. Why should doubling the plate gap cause the force to be halved everything else being equal? Conceptually, I get that the velocity gradient is half what it was and that this is what causes the force to be half but my intuitive picture is lacking. I just can't imagine myself exerting a smaller force on a doubled gap system. Taken to the extreme, at very large $Y$, $F$ is zero. This must be why my text Transport Phenomena by Bird, Stewart, and Lightfoot says the gap must be "small". But what really determines what "small" is?

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"I just can't imagine myself exerting a smaller force on a doubled gap system" Just take sandwich cookies en.wikipedia.org/wiki/Oreo and try to shear them. Then make double sandwich cookie (split two and make one with twice cream) and try again. It will be easier in the second case. Or triple, or quadruple... –  Yrogirg Sep 11 '12 at 8:06

The reason is that the flow of momentum is proportional to the momentum gradient. When you double the spacing, keeping the velocity fixed, the gradient is halved.

The momentum is doing diffusion, if you make a constant momentum in the gas, it doesn't flow anywhere, if you have lower momentum somewhere and higher momentum elsewhere (for some fixed component), the momentum diffuses to equalize the momentum. The law of diffusion applies because momentum is conserved and transmitted locally through many tiny collisions.

As for intuition, if you have a conserved linear momentum going both up and down, randomly, it is clear that the net drift of momentum up, the friction force you need to compensate for, is less the smaller the gradient.

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@JasonWaldrop Indeed it is like heat conduction --- doubling thermal insulation reduces heat flow twice. –  Yrogirg Sep 11 '12 at 7:58
but instead of energy (heat) you conduct momentum. –  Yrogirg Sep 11 '12 at 8:07

I think that deep inside it is simply turning the 'gradient halves' concept into a pretty picture, but the classical intuitive image for viscosity is a stack of paper...

The force you apply tangentially is applied by each sheet to the next one. For a given speed of the top sheet, the more sheets you have, the lower the speed difference between adjacent sheets will be. And force happens to be proportional to the speed difference between adjacent sheets, which you said makes sense to you.

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The implication of this analogy is that it takes less effort (force) to move a thicker stack of paper than a thinner stack at the same top sheet velocity?!? The bulk system momentum would seem to be the same in both cases since I have twice the mass each "new" piece with half the velocity of the old piece. But my force has decreased per the equation so it would seem I got something (same momentum) for nothing (half the force). Sorry, I can't tell whether this means your analogy fails this thought experiment or my thinking is just wrong (the probable case). –  Jason Waldrop Sep 10 '12 at 20:35