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The Four Pi That Catches Molecules Is the Four Pi That Spreads a Field ── Steady Diffusion Is Laplace's Equation: the Flux Through Every Enclosing Sphere Is 12.566370614359, and in Two Dimensions No Steady Absorber Exists ── [Paper 249]

Aug 2026 · Zenodo (CERN European Organization for Nuclear Research)
Cancer and biochemical research

Abstract

When four pi turns up in a biological setting, is it a coincidence of numbers or the same root? This paper treats exactly one case: the four pi in the rate at which a cell captures molecules is the same solid angle as in Gauss's law. Not by analogy, but because it is the same equation. No new mathematical theorem and no new law is claimed. Scope of this paper (scope note): no new mathematical theorem and no new law is claimed. The Smoluchowski capture rate, the receptor count of Berg and Purcell, and the reduction of steady diffusion to Laplace's equation are all standard. No measured value is cited; every number is computed from a definition, and no measurement on a real cell is used. No physiology is discussed; nothing is claimed about the actual size, number or arrangement of receptors. The ratio of one to a thousand in Section 7 is a value put there for the sake of the calculation and belongs to no particular cell. No biological conclusion is drawn; nothing is said about why cells are the size they are or about how evolution acted. The reaction-limited case is not treated; only the diffusion-limited steady state appears, with no binding or unbinding rates. No general account of four pi in biology is attempted; one case is treated. Joining things because their numbers agree is what this paper most wants to avoid: the claim of a shared root rests on the equation being the same, not on the value being the same. The relation to earlier papers. Paper 2 treated the pure solid angle appearing in an inverse-square field; this paper shows the same solid angle in diffusion, from the identity of the equations. Paper 1 established that the exponent is the dimension minus one; the failure in two dimensions in Section 6 is where that exponent becomes zero. Paper 156 traced the flatness inside a spherical shell to harmonicity; this paper uses the same harmonicity. Papers 36 and 37 dissected the six pi of Stokes drag; Section 8 places that six pi beside this four pi as a different root. Paper 44 showed that the power of four pi is a mass dimension and that fixing it is a convention of electromagnetic units; Section 8 puts that conventional four pi third. The setting. Let a sphere of radius a absorb on contact the molecules drifting around it. With a distant concentration and a diffusion coefficient, the amount captured per unit time in the steady state is four pi times the diffusion coefficient times the radius times the concentration (Smoluchowski, 1917). The question is what that four pi is. First, steady diffusion is Laplace's equation. The diffusion equation relates the change of concentration in time to the Laplacian, and in the steady state the time derivative vanishes, leaving only the vanishing of the Laplacian. That is the same equation as electrostatics, so the solutions take the same form and one over r appears in spherical symmetry. The four pi enters when the flux through a sphere is counted, for exactly the reason it appears in Gauss's law: the full solid angle of a sphere is four pi. The harmonicity to which Paper 156 traced the flatness inside a shell is the same harmonicity used here. Second, the flux is the same through every enclosing sphere. Setting the diffusion coefficient, the concentration and the radius to one and measuring the flux from the steady solution, moving the radius from one and a half times to a thousand times gives 12.566370614359172 throughout, the six values spreading by 3.55 times ten to the minus fifteenth. And four pi is 12.566370614359172. That the flux does not depend on the radius is Gauss's law itself; this is not a resemblance but the same consequence of the same equation. Third, solving numerically without using the form of the solution gives the same value. Integrating the radial equation over two million points with the outer boundary at four hundred times the radius gives 12.597865194, which differs from the infinite formula 12.566370614 by 3.1 times ten to the minus second. That is not an error: the ratio is 1.002506259, agreeing with R over R minus a, namely 1.002506266, to 6.7 times ten to the minus ninth. What differs is not the method but the outer boundary being finite, and stretching the boundary tenfold shrinks the gap to exactly a tenth. When a number and a formula disagree, first ask what each of them assumed. Fourth, this is the core of the paper. Capture is proportional to the radius and not to the area. Moving the radius from a tenth to ten, the capture rate grows a hundredfold from 1.256637061 to 125.663706144 while the surface grows ten thousandfold from 0.125663706 to 1256.637061436. Capture per unit area falls from 10.000000000 to 0.100000000, by a factor of a hundred. One naturally expects a wider absorbing surface to catch more, but that is surface-limited thinking. In the diffusion-limited case what decides is not the surface but the rate at which molecules arrive from far away, which is proportional to the radius, since a gradient of one over r multiplied by an area of r squared leaves one power of r. That multiplication is exactly the exponent of Paper 1. Fifth, in two dimensions no steady absorber exists. The same calculation in d dimensions gives a solution containing r to the power two minus d, and at d equal to two that power is zero, the power-law solution disappears and a logarithm takes its place. With the outer boundary at ten times the radius the flux is 0.434294482, at a thousand times 0.144764827, at a million times 0.072382414, at ten to the twelfth 0.036191207, and even at ten to the hundredth 0.004342945 remains. It falls as one over the logarithm and reaches zero only in the limit, yet in that limit the steady solution itself does not exist. A creature in a flat world cannot gather food by diffusion alone; more precisely, it needs a wall at a finite distance. This is where the exponent of Paper 1 becomes one at two dimensions and a logarithm replaces one over r. Sixth, the surface may be left almost bare. The whole surface need not absorb: with small receptors scattered over the sphere, the capture rate is a fraction of that of a perfect absorber (Berg and Purcell, 1977). Putting the relative radius of a receptor at one thousandth, 3142 receptors reach half the rate of a perfect absorber while covering 0.078540 percent of the surface. Ninety percent needs 28275 receptors covering 0.706858 percent, and ninety-nine percent needs 311018 covering 7.775442 percent. Zero point zero eight percent of the surface gives half the rate, and more than ninety-nine percent may be left bare, because a molecule strikes the surface many times by diffusion: missing once, it wanders off and comes back. The ratio of one thousandth is a value put there for the calculation and is not a measurement on any cell. Seventh, three kinds of pi appear in this neighbourhood. The four pi of diffusion capture, of the volume of a sphere and of Gauss's law are all solid angles. In the six pi of Stokes drag the pi is a solid angle but the six has another source (Papers 36 and 37), and the six pi in the Einstein relation is inherited from it. The four pi in Poisson's equation in Gaussian units is a convention of units (Paper 44). The same characters sometimes mean a solid angle and sometimes a convention, and six pi is not one and a half times four pi. The only case in which a shared root is claimed here is diffusion capture, and the claim rests on the same Laplace equation, not on the values agreeing. Closing. The four pi in the rate at which a cell captures molecules is the same solid angle as in Gauss's law. Not because the values agree, but because steady diffusion is Laplace's equation. The same equation gives the same one over r, multiplied by the same spherical area, leaving the same four pi, and the check is that the flux does not depend on the radius. Two things followed: capture scales with the radius and not the area, and no steady absorber exists in two dimensions. Last, the surface may be left almost bare, zero point zero eight percent giving half the rate, because a molecule, by diffusion, comes back again and again. The separator is whether it comes from the same equation; that two values agree shows nothing about a shared root. 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. ----- 生物の話に 4π が出てきたとき、それは数字の一致なのか、同じ根なのか。本稿が扱うのは一つだけである——細胞が分子を捕らえる速さに現れる 4π は、ガウスの法則の 4π と同じ立体角である。類推ではなく、同じ方程式だからである。新しい数学定理も新しい法則も主張しない。 本稿の射程(射程注記):新しい数学定理も新しい法則も主張しない。スモルコフスキーの拡散捕捉率、ベルクとパーセルによる受容体の勘定、定常拡散がラプラス方程式に帰着することは、いずれも標準的である。測定値を引かない——本稿の数はすべて定義から計算したものであり、実在の細胞の測定値は一つも使っていない。生理学を論じない——受容体の実際の大きさ・数・分布については何も主張しない。第7節の 1000 分の 1 という比は仮に置いた値であって、特定の細胞のものではない。生物学的な結論を引き出さない——細胞がなぜその大きさなのか、進化がどう働いたかについては何も述べない。反応律速の場合を扱わない——本稿が扱うのは拡散律速の定常状態だけであり、結合速度や解離は入っていない。生物に現れる 4π を一般に論じない——扱うのは拡散捕捉の一件だけである。数字が一致していることを根拠に何かを結ぶことは、本稿がもっとも避けたいことである——本稿が同根だと言えるのは、同じ方程式から出ているからであって、値が同じだからではない。 既刊との関係。論文2 は逆二乗場に純粋な立体角が現れることを扱った——本稿は同じ立体角が拡散にも現れることを、方程式の同一性から示す。論文1 は n = d−1 を示した——第6節の二次元の破れはその指数がゼロになる場所である。論文156 は球殻の内部が平らな理由を調和性に帰した——本稿が使うのは同じ調和性である。論文36・37 はストークス抵抗の 6π を解剖した——第8節はその 6π と本稿の 4π が別根であることを並べる。論文44 は 4π の冪が質量次元であり、a = 4π を固定するのが電磁単位の規約だと示した——第8節はその規約としての 4π を三つ目に置く。 設定。半径 a の球が、周囲にただよう分子を触れた瞬間に吸収するとする。遠方の濃度をC0、拡散係数を D とし、定常状態で単位時間に捕らえる量を求めると、答は 4πDaC0 である(スモルコフスキー 1917)。問いは、この 4π が何かである。 第一に、定常拡散はラプラス方程式である。拡散方程式は濃度の時間変化を D 掛ける濃度のラプラシアンと結ぶが、定常状態では時間変化が消えるので、残るのはラプラシアンがゼロという式だけである。これは静電場の方程式と同じ式であり、したがって解も同じ形になり、球対称なら 1 / r が出る。4π が現れるのは球面を通る流束を数えるときで、ガウスの法則で4π が出るのとまったく同じ理由——球の全立体角が 4π だからである。論文156 が球殻の内部の平らさを帰した調和性は、ここで使っている調和性と同じものである。 第二に、どの半径で測っても流束は変わらない。D も C0 も a も 1 と置き、定常解 C(r) = C0(1 − a/r) から半径 r の球面を通る流束を測ると、r を a の 1.5 倍から 1000 倍まで動かして 12.56637061435917

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