The Kármán Spacing Ratio of the Vortex Street

Derives Lemma A61.1. Let \(F(\varphi):=\left(\pi/\sin\pi\zeta\right) \ee^{\ii\left(\pi-\varphi\right)\zeta}\) on \((0,2\pi)\) and compute its Fourier coefficients:

the elementary integral of \(\ee^{-\ii\left(\zeta+n\right)\varphi}\). Since \(n\) is an integer, \(\ee^{-2\pi\ii\left(\zeta+n\right)} =\ee^{-2\pi\ii\zeta}\), and \(1-\ee^{-2\pi\ii\zeta} =\ee^{-\ii\pi\zeta}\left(\ee^{\ii\pi\zeta}-\ee^{-\ii\pi\zeta}\right) =2\ii\,\ee^{-\ii\pi\zeta}\sin\pi\zeta\) by Equation (12.3). The two exponentials and the two sines cancel, leaving \(c_{n}=1/(n+\zeta)\), which is Equation (A61.3).

The series Equation (A61.4) converges absolutely and uniformly in \(\zeta\) on compact subsets of \(\C\setminus\Z\), so it may be obtained from Equation (A61.3) by differentiating term by term, using \(\dd\left(n+\zeta\right)^{-1}/\dd\zeta=-\left(n+\zeta\right)^{-2}\) and the product rule on the right-hand side. For the limit, set \(\varphi=0\) in Equation (A61.4): the bracket becomes \(\pi\left(\cos\pi\zeta-\ii\sin\pi\zeta\right)/\sin^{2}\pi\zeta =\pi\,\ee^{-\ii\pi\zeta}/\sin^{2}\pi\zeta\), which cancels the prefactor \(\ee^{\ii\pi\zeta}\) exactly.

Derives Corollary A61.2. The function \(\cot\left(\pi z/a\right)\) is meromorphic with simple poles exactly at \(z=na\), where \(\sin\left(\pi z/a\right)\) has its simple zeros, and near \(z=na\) its principal part is \(a/\pi\left(z-na\right)\). Hence \(\dd w/\dd z\) in Equation (A61.5) behaves near \(z=na\) as \(-\left(\ii\Gamma/2a\right)\cdot a/\pi\left(z-na\right) =-\ii\Gamma/2\pi\left(z-na\right)\), which by Equation (A61.2) is precisely a vortex of circulation \(\Gamma\) there, and it is holomorphic elsewhere. As \(\operatorname{Im}z\rightarrow\pm\infty\), \(\cot\left(\pi z/a\right)\rightarrow\mp\ii\), so \(\dd w/\dd z\rightarrow\mp\Gamma/2a\); since \(\dd w/\dd z=v_{x}-\ii v_{y}\), the row drives the fluid backwards above it and forwards below it, at the stated speed. Any other candidate differs from Equation (A61.5) by a function that is entire (the singularities coincide) and bounded (both velocities are), hence constant by Liouville's theorem (Theorem 12.18).

Derives Proposition A61.4. Take the vortex \(A_{0}=0\). Its own row induces nothing there: the contributions of \(A_{n}\) and \(A_{-n}\) in Equation (A61.2) are equal and opposite. Row \(B\), of circulation \(-\Gamma\) and origin \(a/2-\ii h\), contributes by Equation (A61.5)

using \(\cot\left(\theta-\pi/2\right)=-\tan\theta\) and \(\tan\left(\ii s\right)=\ii\tanh s\). The result is real and positive, so \(v_{x}=\Gamma t/2a\) and \(v_{y}=0\). Repeating at \(B_{0}=a/2-\ii h\) under row \(A\) gives \(-\left(\ii\Gamma/2a\right)\cot\left(\pi/2-\ii s\right) =-\left(\ii\Gamma/2a\right)\tan\left(\ii s\right) =\Gamma t/2a\), the same velocity; and again the vortex's own row contributes nothing. Every vortex therefore moves with Equation (A61.7).

Derives Lemma A61.6. Multiplying the matrices of Equation (A61.19),

whose trace is \(-K^{2}\left(-2Q^{2}+Y+X\right)\), i.e. the coefficient displayed. Its determinant is the product of the two determinants. Now \(\overline{S_{1}}=\ee^{\ii\varphi}S_{1}\) — replace \(n\) by \(-n-1\) in the series defining \(\overline{S_{1}}\) and compare — and likewise \(\overline{\tilde{S}_{1}}=\ee^{-\ii\varphi}\tilde{S}_{1}\), so the determinant of \(\mathsf{B}\) is \(-K^{2}\left(-Q^{2} +\overline{\tilde{S}_{1}}\,\overline{S_{1}}\right) =K^{2}\left(Q^{2}-\Pi\right)\), and that of \(\mathsf{C}\) is the same. The characteristic polynomial of a \(2\times2\) matrix is \(\lambda^{2}-\left(\text{trace}\right)\lambda +\left(\text{determinant}\right)\), which is Equation (A61.21).

Derives Proposition A61.7. Since \(\sigma^{2}=\lambda\), the four exponents are purely imaginary exactly when both roots of Equation (A61.21) are real and non-positive; if either root is complex, or real and positive, some \(\sigma\) has positive real part. Write \(\mathcal{T}=2Q^{2}-X-Y\) and \(\mathcal{D}=\left(Q^{2}-\Pi\right)^{2}\ge0\) for the two coefficients divided by \(K^{2}\) and \(K^{4}\). The discriminant factorizes,

because \(\mathcal{T}\pm2\left(Q^{2}-\Pi\right)\) are the two brackets. Put \(\pi_{+}:=\sqrt{X}\), \(\pi_{-}:=\sqrt{Y}\), so \(\Pi=\pm\pi_{+}\pi_{-}\) by \(XY=\Pi^{2}\), the sign being that of \(-MM'\). If \(MM'>0\) then \(\Pi=-\pi_{+}\pi_{-}\) and the factorization reads \(-\left(\pi_{+}+\pi_{-}\right)^{2} \left[4Q^{2}-\left(\pi_{+}-\pi_{-}\right)^{2}\right]\), non-negative exactly when \(2\abs{Q}\le\abs{\pi_{+}-\pi_{-}}\); if \(MM'<0\) then \(\Pi=+\pi_{+}\pi_{-}\) and it reads \(-\left(\pi_{+}-\pi_{-}\right)^{2} \left[4Q^{2}-\left(\pi_{+}+\pi_{-}\right)^{2}\right]\), non-negative exactly when \(2\abs{Q}\le\pi_{+}+\pi_{-}\). Both cases are covered by \(2\abs{Q}\le\abs{EM-M'/E}\pi/a^{2}\), since \(EM\) and \(M'/E\) have the same sign as \(M\) and \(M'\) respectively; and the closed form on the right of Equation (A61.22) follows from Equation (A61.15),

Finally the second requirement, \(\mathcal{T}\le0\), is implied by the first: in either case \(4Q^{2}\le\left(\pi_{+}\pm\pi_{-}\right)^{2} \le2\left(X+Y\right)\), so \(2Q^{2}\le X+Y\).

Derives Theorem A61.8. Necessity. Take \(\varphi=\pi\), that is \(\psi=0\): neighbouring vortices of a row are then displaced in opposite senses, the disturbance of longest wavelength that is not a rigid translation. Then \(u=0\), \(E=1\) and \(M=M'=\pi t/c\), so the right-hand side of Equation (A61.22) vanishes identically and the criterion demands \(Q=0\). By Equation (A61.17) with \(\psi=0\) this is \(\pi^{2}/2=\pi^{2}/c^{2}\), i.e.\ Equation (A61.23). Solving, \(\sinh s=\sqrt{c^{2}-1}=1\) and \(s=\log\left(c+\sinh s\right)=\log\left(1+\sqrt{2}\right)\), whence Equation (A61.1).

Sufficiency. Let \(c=\sqrt{2}\), so \(t=1/\sqrt{2}\), \(\sinh s=1\) and \(s=\log\left(1+\sqrt{2}\right)\). Then \(Q=-\psi^{2}/2a^{2}\) by Equation (A61.17), and Equation (A61.22) becomes, after multiplying by \(a^{2}\) and using \(c=\sqrt{2}\),

Both sides are even in \(\psi\), so take \(0\le\psi\le\pi\), where the bracket is negative and Equation (A61.24) is the assertion \(G(u)\ge0\) for \(0\le u\le s\), with \(\psi=\pi u/s\) and

obtained by dividing Equation (A61.24) by \(\pi^{2}\) and substituting \(\sqrt{2}=\cosh s\), \(1=\sinh s\). Now \(G(0)=0\) and \(G(s)=\cosh^{2}s-\sinh s-1=2-1-1=0\); moreover

Since \(2\cosh s/s>1\), every \(u\)-dependent term of \(G''\) is strictly increasing on \([0,s]\), so \(G''\) is; and \(G''(0)=-2/s^{2}<0\) while \(G''(s)=2\sqrt{2}/s+1-2/s^{2}>0\) numerically (\(s=0.88137\) gives \(3.209+1-2.575\)). Hence \(G''\) changes sign exactly once, so \(G'\) decreases and then increases. As \(G'(0)=\cosh s/s-1>0\) and \(G'(s)=\cosh^{2}s/s+\cosh s\sinh s-\cosh s-2/s=0\) — the two terms \(\cosh^{2}s/s=2/s\) and \(-2/s\) cancel, as do \(\cosh s\sinh s=\sqrt{2}\) and \(-\cosh s=-\sqrt{2}\) — the function \(G'\) is positive on \((0,u^{*})\) and negative on \((u^{*},s)\) for a single \(u^{*}\). Therefore \(G\) rises from \(G(0)=0\) and falls back to \(G(s)=0\) without crossing zero in between, so \(G\ge0\) on \([0,s]\), which is Equation (A61.24). Equality holds only at \(\psi=0\) and \(\psi=\pi\), the mode of Equation (A61.23) and the rigid translation.

Instability elsewhere. If \(c\neq\sqrt{2}\) then \(Q\neq0\) at \(\psi=0\), while \(X=Y=-\Pi=\pi^{4}t^{2}/a^{4}c^{2}\) there, so the discriminant computed in Proposition A61.7 equals \(\left[4\Pi\right]\left[4Q^{2}\right]=16\Pi Q^{2}<0\): the roots \(\lambda\) are a complex conjugate pair, their square roots are not purely imaginary, and the mode \(\varphi=\pi\) grows.

Derives Proposition A61.9. The only change is in the offset, so Lemma A61.1 is applied at \(\zeta=-\ii h/a\) instead of \(\tfrac{1}{2}-\ii h/a\), giving \(\sin\pi\zeta=-\ii\sinh s\) and \(\cos\pi\zeta=\cosh s\); hence \(S_{0}=-\pi^{2}/a^{2}\sinh^{2}s\), and \(S_{1}\), \(\tilde{S}_{1}\) come out real. At \(\psi=0\) they are equal, with \(\sqrt{X}=\pi^{2}\cosh s/a^{2}\sinh^{2}s\) and \(\Pi=+X\), so the second case of Proposition A61.7 applies and neutrality would require \(\abs{Q}\le\sqrt{X}\). Here \(Q=\left(\pi^{2}/a^{2}\right)\left(\tfrac{1}{2} +1/\sinh^{2}s\right)\), so the requirement reads \(\tfrac{1}{2}\sinh^{2}s+1\le\cosh s\), i.e.\ \(\left(\cosh s-1\right)^{2}\le0\) after using \(\sinh^{2}s=\cosh^{2}s-1\). That holds only at \(s=0\), where the two rows coincide. For every real separation the unstaggered arrangement therefore grows, and no spacing rescues it — which is why only the staggered pattern is ever photographed.