Where zeta waves fall silent

Imagine a room full of spinning clock hands. Every clock works. Every hand moves. Yet the average arrow can be zero, because the arrows point in directions that cancel.

What would a sensor report? If it measures the average arrow, silence. If it counts clocks, a crowded room.

This is a useful entrance to the zeros of the Riemann zeta function. We can build probability distributions whose phase responses vanish exactly where zeta vanishes. The population remains present; a particular collective measurement falls silent. Along the way, the mysterious fraction $1/2$ acquires a surprisingly concrete role as a statistical compromise between two rival descriptions of scale.

Our route begins with an ordinary Beta belief, visits a Brownian bridge, and ends with a sensor that loses a pattern and then recovers it by asking a different question.

Choose what to measure

Let $B\sim \mathrm{Beta}(\alpha ,\beta )$, with positive, prior-inclusive weights $\alpha ,\beta$. Its centered mean is $v=(\alpha -\beta )/(\alpha +\beta )$, and the rapidity of that mean is $\varphi =\frac{1}{2}\log (\alpha /\beta )$. Transforming an individual draw gives a different quantity:

$$U=2B-1,\mathrm{\Phi }=\mathrm{artanh}U=\frac{1}{2}\log \frac{B}{1-B}.$$

Capital $\mathrm{\Phi }$ is random. In general, $\mathbb{E}[\mathrm{\Phi }]\ne \varphi$. We will measure this whole distribution, rather than identify its mean with every possible draw.

Three instruments can look at the same belief and see different mathematics:

Here $i^2=-1$, $t$ is a dimensionless frequency, and $s=\sigma +it$ is a complex exponent. A phase is the angle of the rotating arrow $e^{it\mathrm{\Phi }}$. It becomes physical time only after a model supplies a time calibration.

A uniform belief hides an alternating ladder

Choose $B\sim \mathrm{Beta}(1,1)$, uniform between zero and one. Changing variables gives

where $\mathrm{sech}x=1/\cosh x$. These are normalized probability densities.

Now take a possibly complex moment:

$$\boxed{M(s)=\mathbb{E}[R^s]=2\mathrm{\Gamma }(s+1)\eta (s)=2\mathrm{\Gamma }(s+1)(1-2^{1-s})\zeta (s).}$$

The Gamma function extends factorials, with $\mathrm{\Gamma }(n+1)=n!$. The alternating zeta function is $\eta (s)=\sum _{n\ge 1}(-1{)}^{n-1}{n}^{-s}$ in its convergent half-plane. The identity follows from the classical Mellin integral; our expectation exists for , using removable limits in the continued formula.

Where did the integers come from? Expand the density into decaying exponentials:

$$\frac{e^r}{(1+e^r{)}^2}=\sum _{n\ge 1}(-1{)}^{n-1}n{e}^{-nr}.$$

Integrating $r^s{e}^{-nr}$ contributes $\mathrm{\Gamma }(s+1)/n^{s+1}$. The extra $n$ leaves an alternating ladder of $n^{-s}$. Termwise absolute integration works for ; the integral identity extends the result further.

For a quick reality check,

$$M(0)=1,M(1)=2\log 2,M(2)=\frac{{\pi }^2}{3}.$$

Inside the critical strip, , the factors multiplying $\zeta (s)$ are nonzero. Our moment therefore vanishes at precisely the nontrivial zeta zeros in that strip. The Riemann hypothesis says they all have $\sigma =1/2$.

To see what a complex moment measures, separate its two jobs:

$$R^{\sigma +it}=\underset{\text{positive weight}}{\underbrace{R^{\sigma }}}\underset{\text{rotating arrow}}{\underbrace{e^{it\log R}}}.$$

Weight each magnitude by $R^{\sigma }$, divide by $M(\sigma )$, and you have another probability law. Its average arrow is $M(\sigma +it)/M(\sigma )$. A zero means exact phase cancellation across logarithmic scales. The real normalizer remains positive.

Multiplying $R$ by a fixed multiplies $M(s)$ by $c^s$, which never vanishes. A change of units preserves these zeros. Changing the shape of the magnitude distribution can move them.

Two other passengers: learning and thermal physics

For a balanced $\mathrm{Beta}(a,a)$ belief, the signed-rapidity response has a different structure:

$${\chi }_a(t)=\frac{\mathrm{\Gamma }(a+it/2)\mathrm{\Gamma }(a-it/2)}{\mathrm{\Gamma }(a{)}^2}=\prod _{n=0}^{\mathrm{\infty }}{(1+\frac{t^2}{4(n+a{)}^2})}^{-1}.$$

The Gamma product reveals an additive construction: ${\mathrm{\Phi }}_a$ has the same distribution as a sum of independent, centered Laplace variables with scales $1/[2(n+a)]$. A Laplace density of scale $b$ is $e^{-|x|/b}/(2b)$. Their variances sum to a finite value, so the series converges in mean square.

A balanced pair of new trials sends $a$ to $a+1$. Under this coupling,

$${\mathrm{\Phi }}_a\stackrel{d}{=}{\mathrm{\Phi }}_{a+1}+L_0,L_0\sim \mathrm{Laplace}(0,1/(2a)),$$

with independent terms on the right. Learning removes the largest remaining fluctuation component in this representation. It is an appealing cooling picture: each balanced update takes one broad mode out of the cloud.

The same ladder gives $\mathrm{Var}({\mathrm{\Phi }}_a)=\frac{1}{2}\sum _{n\ge 0}(n+a{)}^{-2}$, a Hurwitz-zeta value. But ${\chi }_a(t)$ stays positive for real $t$. The logarithmic-magnitude instrument is what brought us to cancellation zeros. Adding independent rapidity variables is also a different operation from pooling evidence about one rate, where the prior is included once and repeated lineage does not create fresh independent trials.

There is a second surprise hiding in the uniform case. Write the Fermi occupation at dimensionless energy $ϵ=(E-\mu )/(k_{\mathrm{B}}T)$ as $n_F(ϵ)=1/(1+e^{ϵ})$. Here $\mu$ is chemical potential, $T$ temperature, and $k_{\mathrm{B}}$ Boltzmann’s constant. Its derivative supplies a normalized thermal window:

$$-\frac{d{n}_F}{dϵ}=\frac{1}{4}{\mathrm{sech}}^2(ϵ/2)=f_{2\mathrm{\Phi }}(ϵ).$$

Thus absolute energy in this ideal window and absolute log-odds of a uniform belief share the same moment formula. The occupancy derivative, rather than occupancy itself, plays the probability-density role. Tong’s quantum-gas lectures develop the thermal setting.

Could a rapidity cloud describe the susceptibility of a two-state medium? That possibility suggests measuring response windows directly. A pure temperature rescaling preserves the ideal moment zeros; an energy-dependent density of states changes the window and can shift them.

A Brownian bridge makes the half special

The uniform-Beta construction gives zeta zeros, but does not make its log-magnitude density symmetric. For symmetry, take another path.

A standard Brownian bridge $b(u)$ is Brownian motion on $0\le u\le 1$ conditioned to return to zero. Its full vertical range gives the positive variable

$$Q=\sqrt{\frac{2}{\pi }}(\underset{u}{max}b(u)-\underset{u}{min}b(u)).$$

Biane, Pitman, and Yor describe the remarkable identity

$$\mathbb{E}[Q^s]=2\xi (s),$$

valid for every complex $s$. Here $\xi$ is completed zeta:

$$\xi (s)=\frac{1}{2}s(s-1){\pi }^{-s/2}\mathrm{\Gamma }(s/2)\zeta (s),\xi (s)=\xi (1-s).$$

This is a whole-path range, distinct from the Beta magnitude $R$. The completion’s extra factors arise with a new probability construction. Brownian source, Sections 1 and 2; functional equation.

Let $L=\log Q$, with density $p(\ell )$. The reflection equation becomes

$$p(-\ell )=e^{\ell }p(\ell ).$$

Now imagine two rival hypotheses about an observed log scale: one uses $p(\ell )$; the other uses its mirror $p(-\ell )$. Their likelihood ratio is $e^{\ell }$. With equal prior odds, the posterior probability of the mirrored hypothesis is $e^{\ell }/(1+e^{\ell })$, and its centered value is $\tanh (\ell /2)$. We have landed back in rapidity: this experiment’s posterior rapidity is $\ell /2$.

What distribution sits halfway between the two hypotheses? Multiply their densities and take a square root:

$$\boxed{q(\ell )=\frac{\sqrt{p(\ell )p(-\ell )}}{\int \sqrt{p(u)p(-u)}du}=\frac{e^{\ell /2}p(\ell )}{2\xi (1/2)}.}$$

This midpoint density is even. Its phase response is

$$\boxed{A(t)={\int }_{\mathbb{R}}{e}^{it\ell }q(\ell )d\ell =\frac{\xi (1/2+it)}{\xi (1/2)}.}$$

Every real-frequency value is real because opposite log scales cancel their imaginary parts. RH now asks whether every complex zero of the entire response $A(z)$ lies at a real frequency $z$.

Try finding that midpoint yourself. Move the weighting exponent until the law and its mirror share one silhouette.

Find the symmetry

Balance two views of scale

Start again

Teal sees log scale L; rose sees its mirror −L. Move the weight marker until their whole shapes match.

Weight exponent σ 0.00

Less weightMore weightA clue

Mirror mismatch: 8.64%. The weighted law favors negative log scales.

What is being balanced?

The Brownian-range law p(L) is weighted by exp(σL) and normalized. At σ = ½ it equals the normalized geometric mean of p(L) and p(−L). Mismatch is total variation distance to the mirror. Computation uses 800 Simpson panels over −2 ≤ L ≤ 2; the plot shows −1 ≤ L ≤ 1. This symmetry explains the special half tilt; it does not locate every zeta zero.

The half also solves a statistical optimization. Consider all intermediate overlaps,

$$C(\sigma )=\int p(\ell {)}^{1-\sigma }p(-\ell {)}^{\sigma }d\ell =2\xi (\sigma ),0\le \sigma \le 1.$$

Differentiate the log normalizer twice. The result is the variance of $L$ under that tilt, so $\log C$ is strictly convex. Reflection gives $C(\sigma )=C(1-\sigma )$. Its minimum is therefore at $1/2$.

The minimum is called Hellinger affinity, a measure of overlap; its negative logarithm is Chernoff information for this hypothesis pair. The midpoint is the most difficult compromise between the two descriptions. This establishes an inferential reason for the half. Explaining why every zero stays on its corresponding axis remains the further adventure.

Make the zero into a blind sensor

Let’s build an instrument with a precise failure mode. A hidden phase $\mathrm{\Theta }$ lies on a circle, with density

The parameter $\epsilon$ says how strong its cosine pattern is. The sensor adds independent noise $tL$, with $L$ drawn from our even midpoint density, then wraps the result around the circle:

Averaging over the noise gives

$$g_{\epsilon }(y)=\frac{1+\epsilon A(t)\cos y}{2\pi }.$$

At a zero of $A$, every possible $\epsilon$ produces exactly the same flat output. Even an unlimited sample from this instrument cannot identify the original cosine amplitude.

Predict before trying: if you keep a label saying whether the noise came from $L\ge 0$ or $L\le 0$, will the invisible pattern remain invisible?

An interference experiment

Find the sensor’s blind frequency

Reset experiment

Find the first blind frequency, then recover the hidden pattern. Drag the tuning rail; watch the teal sensor wave flatten and change sign.

1. Hunt for silence

Start near frequency 12, then tune slowly upward. Cross a sign change and narrow it down.

Slide to tune, or use the arrow keys

Dimensionless frequency t

− +

Pool everyone Read L ≥ 0 separately Read L ≤ 0 separately

The population stays active

Gold: density q(L)Rose: q(L) cos(tL)

Positive and negative rose areas cancel in the pooled response. Shading marks a separately observed group.

What the sensor can see

Rose dashed: inputTeal: observed density

The gray line is a uniform output. The latent modulation stays at 0.6 in every experiment.

Signal remaining: 55.43% · original phase

Readout numbers and frequency shortcuts

No phase noiseStart at 12Reveal the first zero

Pooled real response A(t)

0.55429

Selected real response

0.55429

Selected imaginary response

≈ 0

Fisher information at ε = 0

0.15362

The sensor multiplies the input cosine amplitude by A(t). A negative response flips the pattern; a zero erases this mode. Try the first zeta zero to see the erasure.

Model and numerical limits

Deterministic numerical model, with Fourier convention E[exp(itL)]. Simpson quadrature uses 1,600 panels on −2 ≤ L ≤ 2 and 16 theta terms, then normalizes the grid. Plots show the central −1 ≤ L ≤ 1 region. The preset is a known zero near 14.134725; tiny residuals reflect finite arithmetic. Information is per observation in the chosen group, evaluated at zero input modulation. These are density curves, not sampled observations or a search over all complex zeros.

The separate groups have responses $A_{+}(t)$ and $A_{-}(t)=\overline{A_{+}(t)}$. Their imaginary parts cancel when pooled. At the first zero, each group can still carry a sine pattern even though their average is flat. Recovering it requires an additional observable, the group label.

This has an exact information measure. Fisher information measures how sensitively an output distribution changes with a parameter. At $\epsilon =0$, the information per pooled observation is

$${\mathcal{I}}_{\epsilon }(t)=\frac{|A(t){|}^2}{2}.$$

It vanishes at the blind frequency. The same formula with $A_{+}$ describes a separately observed positive group. A collapsed direction in the instrument’s information geometry need not mean the source population lost its structure.

Think of cells with phases depending on logarithmic size, or experts estimating quantities over many orders of magnitude. A pooled assay may erase a harmonic through cancellation. Measuring individual activity, retaining subpopulation labels, or changing the readout distinguishes that mechanism from inactive individuals. The promising prediction is recoverable organization under a changed measurement.

One response, three physical readings

The same $A(t)$ supports several deliberately chosen experiments. Matching their equations makes their differences visible:

For the quantum experiment, $K$ multiplies a wavefunction by $\ell$. At a real zero, the phase-shifted state is orthogonal to the starting state, although both have identical position densities. The change lives entirely in relative phase. This $K$ is a valid phase generator; treating it as the energy of a stable material system requires more structure because its spectrum extends in both directions.

An optical implementation could encode log scale as transverse position on a spatial light modulator. Finite aperture and pixel size would replace ideal nulls with finite extinction. Encoding $\log r$ in a radial aperture also requires the correct coordinate weighting.

There is experimental precedent for observing complex partition-function zeros through quantum coherence: Peng and colleagues accessed Lee-Yang zeros in a spin system. That makes this kind of measurement architecture tangible. An EP model might similarly attach internal phase to scale, with ordinary probabilities averaging it away. Deriving the particular state $q$ from the model would give that proposal its explanatory content.

Reversibility, selection, and prime energies

Reflecting log scale gives a probability ratio that looks like a fluctuation relation. Set $S=-L$. Directly from the density reflection,

$$\frac{p_S(s)}{p_S(-s)}=e^s,\mathbb{E}[e^{-S}]=1,\mathbb{E}[S]=D_{\mathrm{KL}}(p\Vert p^{\mathrm{rev}})\ge 0,$$

where $p^{\mathrm{rev}}(\ell )=p(-\ell )$ and $D_{\mathrm{KL}}$ is the average log likelihood ratio. In Crooks’ fluctuation theorem, related ratios compare forward and reverse physical trajectories. Our reciprocal-scale reflection is a different operation: reversing the time order of a Brownian bridge leaves its range unchanged.

This suggests constructing forward and reverse evidence-update processes and asking whether their path likelihood ratio produces this scale law. The half-weighted ensemble would then remove the directional bias of that comparison.

Selection offers another reading. If $Q$ is a relative growth factor, weighting lineages by $Q^{\sigma }$ gives an ensemble with

$$\frac{d}{d\sigma }{\mathbb{E}}_{\sigma }[L]={\mathrm{Var}}_{\sigma }(L).$$

Variation in log growth determines how rapidly selection shifts its mean. The Brownian choice makes half-selection symmetric between reciprocal scales, a possible model of competing expansion and contraction biases.

There is also a literal prime gas. For , choose $P(N=n)=n^{-\sigma }/\zeta (\sigma )$. Unique factorization and the Euler product make prime multiplicities independent geometric variables, with additive energies $\log p$. This ensemble stops being normalizable at $\sigma =1$. Reaching the Brownian representation at the critical half therefore calls for a new ensemble or a derivation of collective correlations. Analytic continuation alone does not sample the original gas there.

How to ask a sharper question about the zeros

For the even midpoint law, the field-dependent partition function is

$$\mathcal{Z}(h)={\mathbb{E}}_q[e^{hL}]=\frac{\xi (1/2+h)}{\xi (1/2)}.$$

A real field $h$ biases scale; an imaginary field probes phase. RH becomes a statement that every zero lies on the imaginary field axis. This resembles the geometry of Lee-Yang theorems. A productive program would construct finite physical systems independently, establish their zero restriction, and show their normalized responses converge to this one.

Symmetry alone leaves room to roam. Consider an even law with mass $3/4$ at zero and $1/8$ at each of $-4$ and $4$. Tilt it by $e^{-\ell /2}$ and normalize. Its moment transform is

$$D(s)=\frac{3+\cosh (4(s-1/2))}{3+\cosh 2}.$$

It has $D(0)=D(1)=1$, the same reflection $D(s)=D(1-s)$, and zeros at

$$s=\frac{1}{2}\pm \frac{\mathrm{arcosh}3}{4}+\frac{(2k+1)\pi }{4}i,k\in \mathbb{Z}.$$

Both real parts lie inside the critical strip, away from its middle. Smoothing the even law with a centered Gaussian preserves these zeros, since the Gaussian adds a nonvanishing transform factor. The Brownian law needs some more specific structure to confine its zeros. Finding it is a focused mathematical challenge.

We can also perturb the original belief. For $\mathrm{Beta}(a,a)$, let $M_a(s)=\mathbb{E}[(2|{\mathrm{\Phi }}_a|{)}^s]$. Near a simple root $s_0$, changing $a$ moves it according to

$$\frac{d{s}_0}{da}=-\frac{{\partial }_a{M}_a(s_0)}{{\partial }_s{M}_a(s_0)}.$$

A small loop around a simple zero winds the response phase through $2\pi$, much like circling an optical vortex. That winding preserves the local count of zeros under suitable small deformations; it does not fix their position. Tracking root motion distinguishes a coordinate rescaling from a change in the underlying belief.

Finally, perhaps an independently defined scale operator has these zeros as its eigenfrequencies. A self-adjoint operator has real spectral values, making this an enticing route. Choosing $\psi =\sqrt{q}$ and manufacturing a ground-state operator from it does not establish the needed link between excited eigenvalues and Fourier zeros. The sensor already supplies concrete null modes; understanding their dynamics could be a more directed starting point.

Another ladder to explore

Change the prior to Jeffreys’ $\mathrm{Beta}(1/2,1/2)$. Its rapidity density is $\mathrm{sech}\phi /\pi$, and

The odd-integer ladder brings in Dirichlet beta, a different function from both the Beta distribution and Euler’s Beta function. Its zeros belong to the Dirichlet $L$-function story. The next expedition follows those odd numbers toward primes and Brownian clocks.

For now, the lab leaves a concrete question to carry into other subjects: when a collective measurement is silent, which individual phases did it average together? Sometimes a second instrument, or one retained label, can make the hidden pattern speak again.

Reading further

• DLMF’s Mellin integral, completion, and zero geometry connect the transforms to standard zeta notation.
• Biane, Pitman, and Yor develop Brownian range, theta densities, and reciprocal size bias.
• Crooks supplies the physical forward/reverse setting for fluctuation relations.
• Peng and colleagues show how quantum coherence can reveal Lee-Yang zeros.
• Tong’s statistical-physics lectures introduce Fermi occupation and thermal response.