Is 6 pi really 4 pi plus 2 pi? The same number, and the supports do not meet.
The viscous drag on a sphere is 6 pi mu U a with a no-slip boundary and 4 pi mu U a with a perfectly slipping one. It is tempting to read this as 6 pi equals 4 pi plus 2 pi, so the 4 pi is the essential part and the 2 pi is what sticking adds. This paper refutes that reading by dissecting the drag. NO NEW FLUID MECHANICS IS CLAIMED. Stokes flow, slip and no-slip boundary conditions and the Einstein-Stokes relation are all standard.
First, the no-slip six splits as two plus zero plus four, in units of pi mu U a. Computing the traction on the sphere from the stream function and the pressure and integrating, the pressure contributes two, the normal viscous stress contributes exactly zero, and the tangential viscous stress contributes four (computed here, with the boundary conditions and the residual of the equation all verified to vanish). Note that the normal viscous contribution is exactly zero – this matters later.
Second, the slip four splits as four thirds plus eight thirds plus zero. Perfect slip means zero normal velocity and zero tangential traction. Integrating in the same way, the pressure contributes four thirds, the normal viscous stress eight thirds, and the tangential viscous stress exactly zero (both boundary conditions and the Stokes residual verified to vanish).
Third, and this is the core. The two fours have opposite contents. The four inside the no-slip split is made of tangential stress alone, the normal viscous part being exactly zero. The four that is the whole slip drag is made of two thirds normal viscous stress and one third pressure, the tangential part being exactly zero. The same number, and the supports of its components do not meet. So the reading six pi equals four pi plus two pi, the slip answer plus a correction for sticking, is wrong. No component of the split is the slip flow: slip is a different solution, not a part of the no-slip one. The equation holds as arithmetic and fails as meaning. Numbers agreeing does not make roots the same (Papers 43 and 45).
Fourth, lifted to d dimensions the two coefficients share a factor. In units of the area of the unit sphere times mu U a to the d minus two, no-slip is d times d minus two over d minus one, and slip is two times the same quotient. At d from three to six the no-slip values are three halves, eight thirds, fifteen quarters and twenty-four fifths, and the slip values are one, four thirds, three halves and eight fifths; all agree with those expressions (computed here). The quotient d minus two over d minus one is common to both, so the difference between clinging to a surface and sliding along it – physically two quite different conditions – collapses to the single factor d against two. Their ratio is d over two.
Fifth, the two values of three halves that appear at d of three are also distinct roots. One is the ratio d over two, the other the no-slip coefficient d times d minus two over d minus one. They coincide at d of three only because twice d minus two over d minus one equals one there, and at d of four they part as two against eight thirds. Both carry d minus two, so both die at d of two: Stokes’ paradox does not care which boundary condition was chosen.
Sixth, the six in the Einstein-Stokes diffusion constant, k_B T over six pi mu a, is the no-slip coefficient. With perfect slip the same relation reads k_B T over four pi mu a, so the diffusion constant is larger by a factor of three halves. A factor of one and a half in a measured diffusion constant therefore rides on a modelling choice about the surface, not on a property of the fluid. The fluctuation side has not moved; only the mobility side has. At molecular scales the slip length is known not to vanish, and the effective coefficient sits between four pi and six pi. The six is not a fact but the name of a boundary condition that was chosen.
IMPORTANT (scope): a census of established standard material (Layer 2). What is not claimed: any new fluid mechanics (there is none). What is added: splitting the drag into three parts and showing the two fours have disjoint support, fencing the reading six pi equals four pi plus two pi by counterexample, showing that in d dimensions the two coefficients share a factor and differ only as d against two, and writing addresses for the two values of three halves at d of three. What remains: on a calculator, six pi equals four pi plus two pi is correct. But one of those four pis is the whole of a flow that slides past the sphere, and the other is a part of a flow that clings to it. Seeing the same number, the question to ask is not whether they are equal but what each is made of.
On the making of this work: The ideas and content of this work stem from the author’s own considerations. Assistance from an AI (a large language model) was used for structuring, English translation, and checking the algebra. Any remaining errors or misinterpretations are solely the author’s. Feedback and corrections are sincerely appreciated.
Reference
Yamagishi, Y. (2026). Is $6\pi$ Really $4\pi+2\pi$? [Paper 37]. Zenodo.
DOI: 10.5281/zenodo.22091500
ORCID: 0009-0006-8587-3450