Is there a logical explanation why Stokes's drag
$F_d=6\pi R \eta v$
is proportional to the radius, $R$ of the sphere?
Naively I would have expected that it is proportional to the cross section, i.e. to $R^2$.
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$\begingroup$ If the force solely depends on viscosity, velocity, and sphere geometry, what does dimensional analysis tell you about the required exponent of R? $\endgroup$– Chet MillerCommented Jun 3, 2018 at 2:36
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$\begingroup$ @ChesterMiller: Well, yes. But how do I know that it depends on these quantities (only)? $\endgroup$– user1583209Commented Jun 3, 2018 at 7:44
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$\begingroup$ You asked how it depends on R and not R^2, so I answered in that context. What other parameters do you think might be involved? Do you know how to do dimensional analysis using the Buckingham Pi Theorem? $\endgroup$– Chet MillerCommented Jun 3, 2018 at 11:55
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2 Answers
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The answer to your question can be found in section 2.3 of this document by Lagree.
Essentially the drag on the sphere is given by $F_d = 6\pi R \eta v$ because the boundary conditions on the velocity at the surface of the sphere and at infinity ensure that pressure and shear stress scales as $1/r^2$. When the pressure and shear stress are integrated over the surface of the sphere, one finds that the drag force will scale with $R$, the radius of the sphere.
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$\begingroup$ This answer summarises the derivation of Stokes' Law but IMO it does not give any insight why the result is different for low and high Reynolds Number. $\endgroup$ Commented Jun 3, 2018 at 17:55
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$\begingroup$ @sammygerbil unless i missed an edit, i didn't think an explanation for high Re was included in the question. $\endgroup$– RagnarCommented Jun 4, 2018 at 3:26
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$\begingroup$ For high Re drag is proportional to area, as per user's intuition. User is asking for insight why that intuition does not hold the same for low Re. $\endgroup$ Commented Jun 4, 2018 at 4:16
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Consider this alternate form of Stokes drag: $$F_d=6\pi\mu Rv=6\pi\mu \frac{v}{R}R^2 \sim \tau_w A$$ where $\tau_w\sim\mu \frac{v}{R}\propto R^{-1}$ is roughly the magnitude of the shear stress at the surface of the sphere and $A\propto R^{2}$ is the surface area of the sphere.
Clearly, its an algebraic combination of the shear stress and surface area which leads to a linear dependence on $R$.
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$\begingroup$ Not clear. Where does $C_d \sim Re^{-1}$ come from? Isn't that just saying $F_d \propto R$ without explanation, which is what the question asks for? $\endgroup$ Commented Jun 3, 2018 at 16:03
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$\begingroup$ I think the user is asking for intuition, physical insight. In your edit I can understand $\tau=F/A$ but I don't get why $\tau \sim \mu\frac{v}{R}$. $\endgroup$ Commented Jun 3, 2018 at 18:00
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$\begingroup$ @sammygerbil - shear stress is defined as $\tau_w = -\mu\frac{\partial u}{\partial r}_w$, given that $v$ and $R$ are the characteristic velocity and length scales it follows that by approximation $\tau_w \sim \mu\frac{v}{R}$. $\endgroup$– nluigiCommented Jun 3, 2018 at 18:05
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$\begingroup$ That makes a lot of sense. I had forgotten about the velocity gradient. I do like your approximation $\frac{v}{R}$. I am not sure that the first part of your answer (for large Re) is actually needed. $\endgroup$ Commented Jun 3, 2018 at 19:43