On the critical line, write
$$ \zeta(1/2+it)=e^{-i\theta(t)}Z(t), $$
with \(Z(t)\) real. Then
$$ \boxed{ \frac{\zeta'}{\zeta}(1/2+it) = -\theta'(t)-i\frac{Z'(t)}{Z(t)}. } $$
The normal logarithmic flux is the gamma-controlled term \(-\theta'(t)\). The tangential coordinate \(Z'/Z\) carries the arithmetic motion.
It is tempting to combine the normal flux with the exact zero-counting function and hope that the horizontal positions of the zeros are forced. They are not. There is a tiny quartic counterfeit that removes them while preserving every one of those data.
Move a double zero sideways#
Put
$$ z=s-\frac12, \qquad 0<a<\frac12, \qquad \alpha=a+ib. $$
Compare
$$ E_{\mathrm{line}}(z)=(z^2+b^2)^2 $$
with
$$ E_{\mathrm{off}}(z) =(z^2-\alpha^2)(z^2-\overline\alpha^{\,2}). $$
The first has a double upper zero at \(z=ib\). The second has two upper zeros at
$$ z=a+ib,\qquad z=-a+ib. $$
Both therefore have exactly two upper zeros at ordinate \(b\), counting multiplicity. Their zero-counting functions by height are identical.
On the symmetry axis,
$$ E_{\mathrm{line}}(it)=(b^2-t^2)^2\ge0, $$
while
$$ E_{\mathrm{off}}(it) =(t^2+a^2-b^2)^2+4a^2b^2>0. $$
Both have zero phase contribution there, and
$$ \boxed{ \operatorname{Re}\frac{E_j'(it)}{E_j(it)}=0. } $$
The zeros moved horizontally. The normal flux did not notice.
Give both functions the zeta shell#
Let
$$ \mathcal C(s) = \frac12s(s-1)\pi^{-s/2}\Gamma(s/2) $$
and define
$$ \boxed{ F_j(s) = \frac{ \cosh(\pi(s-1/2))E_j(s-1/2) }{ \mathcal C(s) }. } $$
Both \(F_j\) are order-one, real-type meromorphic functions with the same zeta functional equation. The common cosh factor supplies only critical-line zeros. Consequently the two functions have the same exact global nontrivial zero count at every height.
One member has only critical-line nontrivial zeros. The other has an off-line symmetric pair.
Nevertheless,
$$ \boxed{ \operatorname{Re} \frac{F_j'}{F_j}(1/2+it) = -\theta'(t) } $$
for both.
What the flux forgot#
Differentiating the Hardy representation gives
$$ \zeta'(1/2+it) = -e^{-i\theta(t)} \left(\theta'(t)Z(t)+iZ'(t)\right). $$
The full Speiser boundary curve is therefore
$$ \boxed{ \mathcal W(t)=\theta'(t)Z(t)+iZ'(t). } $$
Its winding sees the relocation. Normal flux and exact counting do not, because they discarded the Hardy-amplitude derivative.
Put the curve on the Riemann-Siegel clock#
There is a beautifully exact way to see what the full curve is measuring. On an interval where \(\theta'(t)\neq0\), set
$$ u=\theta(t), \qquad Y(u)=Z(t(u)). $$
Then
$$ \boxed{ \mathcal W(t) = \theta'(t)(Y+iY_u). } $$
So, where \(\theta'>0\), the phase velocity is
$$ \boxed{ \frac{d}{du}\arg\mathcal W = \frac{YY_{uu}-Y_u^2}{Y^2+Y_u^2}. } $$
The first Laguerre expression
$$ Y_u^2-YY_{uu} $$
is not merely analogous to the Speiser winding. It is its exact local angular velocity in zeta's own gamma clock.
At a nonzero stationary point \(Z'(t)=0\),
$$ \frac{d}{dt}\arg\mathcal W = \frac{Z''(t)}{\theta'(t)Z(t)}. $$
Thus the expected-extremum rule
$$ Z(t)Z''(t)<0 $$
is precisely the stationary trace of one-way Speiser-Dini phase.
That is much stronger than the normal-flux coordinate. It is still not enough.
A carrier can hide the off-line pair from every stationary sign#
Keep the same line and off-line quartics, but multiply both by a sufficiently rapid common carrier \(\cosh(\omega z)\). After division by the zeta completion factor, the corresponding real Hardy amplitudes are
$$ H_{\mathrm{line}}(t) = \frac{\cos(\omega t)(t^2-b^2)^2} {|\mathcal C(1/2+it)|}, $$
and
$$ H_{\mathrm{off}}(t) = \frac{\cos(\omega t) ((t-b)^2+a^2)((t+b)^2+a^2)} {|\mathcal C(1/2+it)|}. $$
Let
$$ G(t)=\log|\mathcal C(1/2+it)|. $$
The archimedean curvature \(G''\) is bounded below, while
$$ \left( \log((t-b)^2+a^2)((t+b)^2+a^2) \right)'' \leq\frac4{a^2}. $$
Therefore a qualitative choice
$$ \omega^2> -\inf_{\mathbb R}G''+\frac4{a^2} $$
makes
$$ \boxed{ H_{\mathrm{off}}'(t)^2 - H_{\mathrm{off}}(t)H_{\mathrm{off}}''(t) >0 } $$
on the entire real axis. The line member has the same global first-Laguerre positivity. At every nonzero stationary point of either amplitude,
$$ H'(t)=0 \quad\Longrightarrow\quad H(t)H''(t)<0. $$
The two functions still have identical exact zero counts by height. One still has an off-line pair.
This counterfeit is rude in exactly the useful way: even the global real-axis curvature sign, not just a sample of stationary extrema, can be overwhelmed by a common critical-line carrier. The missing information is interior and arithmetic. A Speiser proof needs the literal Euler-specific winding, complete Hermite--Biehler or Pick positivity, or some other rigidity that forbids the hidden positive multiplier. Boundary curvature cannot do that job by itself.
This is a useful kind of obstruction: not "the estimate is too weak," but "the proposed data are mathematically incapable of remembering the horizontal divisor." A Speiser route survives only if it controls the literal arithmetic winding of \(\mathcal W\), equivalently the missing \(Z'/Z\) coordinate.
Notebook references: K-0169, O-0314, K-0173, O-0318