Prime signs, wandering clocks, and the Jeffreys connection

Put two wandering insects on separate number lines. Both begin at zero. The first has walls at $-1$ and $+1$: when it touches a wall, ring a bell. Wherever the second insect happens to be when the bell rings, record its position.

After the right rescaling, those readings follow the same rapidity distribution as a Jeffreys belief. Their magnitude moments contain an alternating sum over odd integers. Factor those integers into primes, and the alternating signs turn into a precise rule about orientation. Draw integer points on a square lattice, and the same rule counts which circles contain them.

This is a field trip with several maps. Probability supplies the wandering clock, arithmetic supplies prime factorizations, and Epistemic Physics supplies the reason to care about Jeffreys rapidity. We will keep the maps aligned, so that each bridge has a visible equation beneath it.

Three betas, three jobs

The similar names can conceal different objects:

For positive parameters,

$$\mathrm{B}(\alpha ,\beta )={\int }_0^1{b}^{\alpha -1}(1-b{)}^{\beta -1}db.$$

Choose the Jeffreys state, $B\sim \mathrm{Beta}(1/2,1/2)$. Here the two half-weights are prior contributions; they do not describe an observed half-trial. Define a centered random value $U=2B-1$ and its random rapidity $\mathrm{\Phi }=\mathrm{artanh}U$. These are random variables drawn from the full belief distribution. The state’s centered mean is $\mathbb{E}U=0$.

The corresponding densities line up beautifully:

Here $\mathrm{sech}\varphi =1/\cosh \varphi$. The last line follows from $u=\tanh \varphi$ and $du/d\varphi ={\mathrm{sech}}^2\varphi$: stretch the horizontal ruler, and the density compensates. This is the full-distribution version of the rapidity transformation in the Bayesian relativity foundations.

Why arcsine? The centered cumulative probability is

$$Pr(U\le u)=\frac{1}{2}+\frac{\arcsin u}{\pi }.$$

Equivalently, choose an angle uniformly on $(-\pi /2,\pi /2)$ and project it by $U=\sin \theta$. Equal angular steps crowd together near the endpoints, producing the two high shoulders of the density.

Odd numbers emerge from the rapidity curve

Let $R=|\mathrm{\Phi }|$, the nonnegative rapidity magnitude. For , a geometric-series expansion gives

$$\mathrm{sech}r=2(e^{-r}-e^{-3r}+e^{-5r}-\cdots ).$$

Every odd decay rate appears, with alternating signs. Multiply by $r^{s-1}$ and integrate. The elementary integral ${\int }_0^{\mathrm{\infty }}{r}^{s-1}{e}^{-ar}dr=\mathrm{\Gamma }(s)a^{-s}$ uses the Gamma function, the continuous extension of factorials. We obtain

$$\mathbb{E}[R^{s-1}]=\frac{4}{\pi }\mathrm{\Gamma }(s){\beta }_D(s),{\beta }_D(s)=1-3^{-s}+5^{-s}-7^{-s}+\cdots .$$

Absolute interchange of the integrals and sum works for ; analytic continuation extends this identity to , where the original moment integral exists. In particular,

$${\beta }_D(1)=\frac{\pi }{4},\mathbb{E}R=\frac{4G}{\pi },\mathbb{E}{R}^2=\frac{{\pi }^2}{4},$$

where $G={\beta }_D(2)$ is Catalan’s constant. These constants describe ordinary magnitudes in a probability distribution. The arithmetic appears because the rapidity curve has those particular exponential modes.

Prime factors carry orientation bits

Give each positive integer a label:

The notation $n\equiv 3(\mathrm{mod}4)$ means that division by four leaves remainder three. Multiplication preserves the rule: ${\chi }_4(mn)={\chi }_4(m){\chi }_4(n)$. A negative sign acts like a flip; two flips restore the starting orientation. Thus

$$45=3^2\cdot 5⟶(-1{)}^2(+1)=+1,$$

while $15=3\cdot 5$ has negative orientation.

Unique prime factorization now reorganizes the sum into an Euler product, one factor per prime:

The prime two is absent because its character is zero. This is a specialization of the Dirichlet-character Euler product.

To make a probability model, normalize positive weights first and measure the sign afterward. For a real exponent , choose an odd integer $N$ with

$$Pr(N=n)=\frac{n^{-\sigma }}{Z_{\mathrm{o}\mathrm{d}\mathrm{d}}(\sigma )},Z_{\mathrm{o}\mathrm{d}\mathrm{d}}(\sigma )=(1-2^{-\sigma })\zeta (\sigma ).$$

Here $\zeta (\sigma )=\sum _{n\ge 1}{n}^{-\sigma }$. Large $\sigma$ favors small integers. The orientation $S={\chi }_4(N)$ has mean

$$r(\sigma )=\mathbb{E}S=\frac{{\beta }_D(\sigma )}{(1-2^{-\sigma })\zeta (\sigma )}.$$

Factor $N=\prod p^{K_p}$. The multiplicities $K_p$ are independent geometric random variables: $Pr(K_p=k)=(1-p^{-\sigma })p^{-k\sigma }$. For a flipping prime,

$$\mathbb{E}[(-1{)}^{K_p}]=\frac{1-p^{-\sigma }}{1+p^{-\sigma }}=\tanh \left(\frac{\sigma \log p}{2}\right).$$

Therefore

$$r(\sigma )=\prod _{p\equiv 3(4)}\tanh \left(\frac{\sigma \log p}{2}\right).$$

There is our velocity-rapidity function again! Each factor is the mean of a binary orientation. Independent orientations compose by multiplying their means; likelihood ratios compose by adding their log odds. Those are different measurements of a composite record.

At $\sigma =2$, $r=8G/{\pi }^2$. As $\sigma$ approaches one from above, $r$ approaches zero while the normalizer diverges. The finite signed sum ${\beta }_D(1)=\pi /4$ survives, but the infinite odd-integer probability ensemble has no normalization there.

Try the prime buttons before collecting any journeys. Can you predict which repeated factor restores positive orientation? Then lower $\sigma$ and enlarge the prime cutoff: more flipping modes should reduce the expected sign.

Arithmetic meets probability

A prime switchboard and a wandering clock

Reset lab

Build an odd integer with prime tiles. Then reveal a wandering clock one piece at a time: its first wall crossing decides when the independent reader stops.

Walk the clockPrime switchboard

Let one wandering coordinate stop another

The rose clock X starts at 0 and stops on reaching either wall ±1. The independent sky reader W stops at the same time T. Start a journey, predict the reader's sign, then uncover the path until the clock stops it. The reading is rapidity Φ = πWₜ/2; U = tanh Φ.

Start a journey Start and step Collect 128 journeysClear journeys

Latest stopped journey

● X: clock and walls● W: independent reader

Readings and the exact distribution

Rapidity ΦCentered value U

■ Simulated readings━ Exact bin probability

Completed journeys

0

Mean exit time, target 1

Waiting

Revealed rapidity Φ

Waiting

Simulation limits and exact comparison

Simulation: independent Gaussian increments with Δt = 1/512 (0.001953), first detected grid exit, and time cap 12. Grid monitoring misses some between-step crossings and biases exit times upward. 0 capped journeys excluded; 0 completed readings lie outside this chart. Bars divide by all completed journeys, including values outside the chart. The latest 2,048 journeys are retained. The gold marks integrate the exact continuous law. Seeded runs are reproducible; collecting more journeys reduces sampling noise but leaves the time-step approximation.

A prime cutoff $P$ keeps all odd primes up to $P$, with unlimited multiplicities. It still permits arbitrarily large integers. An integer cutoff $N\le M$ defines a different ensemble. This distinction matters especially near $\sigma =1$, where convergence slows.

A bell rung by Brownian motion

Now give the insects an exact mathematical motion. Let $X_t$ and $W_t$ be independent standard Brownian motions, continuous random paths whose increments over a duration $h$ are Gaussian with mean zero and variance $h$. Both start at zero. Define the clock’s first exit time

$$T=inf\{t\ge 0:|X_t|=1\}.$$

The clock has a simple Laplace transform:

$$\mathbb{E}[e^{-\lambda T}]=\mathrm{sech}\sqrt{2\lambda },\lambda \ge 0.$$

You can derive it by solving the backward equation $u^{″}/2=\lambda u$ with $u(-1)=u(1)=1$ and evaluating at zero. Given $T$, the independent reader has law $W_T\sim N(0,T)$. Thus, using a Fourier frequency $k$,

$$\mathbb{E}[e^{ik(\pi /2)W_T}]=\mathbb{E}[e^{-{\pi }^2{k}^{2}T/8}]=\mathrm{sech}\frac{\pi k}{2}.$$

This is the characteristic function of our Jeffreys rapidity density. Consequently,

$$\mathrm{\Phi }\stackrel{d}{=}\frac{\pi }{2}{W}_T\stackrel{d}{=}\frac{\pi }{2}\sqrt{T}Z,Z\sim N(0,1),Z\text{ independent of }T.$$

The symbol $\stackrel{d}{=}$ means equal distributions. Transform the reading by $U=\tanh \mathrm{\Phi }$, and it acquires the centered arcsine law. The clock’s own endpoint is always a wall, $X_T=\pm 1$; using the independent reader is essential. These hyperbolic transforms and Brownian exit laws appear together in Biane, Pitman, and Yor’s probability constructions.

A possible Epistemic Physics interpretation is deliciously concrete: one internal process accumulates toward a decision threshold while another explores possible rapidities. A random commitment time mixes narrow and broad Gaussian explorations. Independence, Brownian increments, and symmetric thresholds are the chosen ingredients. An actual lineage of accumulating trials would need an update rule producing them. Repeated testimony does not become an independent tracelet merely because a diffusion model is convenient.

Turn the clock upside down

The same clock reaches farther into number theory. Its survival probability is a sum of odd modes:

Integrating the survival probability against $q{t}^{q-1}$ yields

For example, $\mathbb{E}T=1$ and $\mathbb{E}{T}^2=5/3$. Both tails of the exit law decay fast enough that all complex-order moments exist; the displayed formula extends with apparent singularities interpreted by limits.

Define an inverse exit scale and a completed arithmetic function:

$$Q=\sqrt{\frac{2}{\pi T}},{\mathrm{\Lambda }}_{\beta }(s)={\left(\frac{4}{\pi }\right)}^{(s+1)/2}\mathrm{\Gamma }\left(\frac{s+1}{2}\right){\beta }_D(s).$$

The surprising correspondence is

$$\mathbb{E}[Q^s]={\mathrm{\Lambda }}_{\beta }(s)={\mathrm{\Lambda }}_{\beta }(1-s).$$

The Gamma and scale factors complete the function so that it reflects about $s=1/2$. The reciprocal law and completion are given in Biane, Pitman, and Yor, equations 39 through 41. As quick checks, $\mathbb{E}Q=1$ and $\mathbb{E}{Q}^2=4G/\pi$.

Let $L=\log Q$ have density $p$. Reflection takes the concrete form $p(-\ell )=e^{\ell }p(\ell )$. Weight by $e^{\ell /2}$ and normalize:

$$g(\ell )=\frac{e^{\ell /2}p(\ell )}{{\mathrm{\Lambda }}_{\beta }(1/2)}=\frac{\sqrt{p(\ell )p(-\ell )}}{\int \sqrt{p(u)p(-u)}du}.$$

This density is symmetric. Its Fourier response is

$${\mathbb{E}}_g[e^{ikL}]=\frac{{\mathrm{\Lambda }}_{\beta }(1/2+ik)}{{\mathrm{\Lambda }}_{\beta }(1/2)}.$$

Picture one tiny compass arrow per log-scale reading, rotated through angle $kL$. At certain frequencies their average disappears. A zero on the critical line is precisely such a cancellation in a positive probability ensemble. Requiring every nontrivial Dirichlet-beta zero to lie on that line is this function’s instance of the generalized Riemann hypothesis. Symmetry supplies the mirror, while a theorem about all zeros requires more.

Which quantity gets the square root?

A half-moment needs a named random variable. Here are three different weights when the arithmetic argument is $s=1/2+ik$:

The index shifts matter. A square root of velocity is another observable altogether. Substituting $z=u^2$ in the arcsine density gives

This time Euler’s Beta function appears. Taking $\sqrt{|U|}$ discards direction and compresses magnitudes. Taking the principal complex square root preserves the sign as a phase:

For a symmetric law, $\mathbb{E}\sqrt{U}=(1+i)\mathbb{E}\sqrt{|U|}/2$. Both components are nonnegative, so these two directions cannot cancel like the rotating log-scale arrows. If we call $A=\sqrt{U}$ an amplitude, then $A^2=U$ and $|A{|}^2=|U|$ answer different questions. In a plane, square-rooting $r{e}^{i\theta }$ halves its angle: a continuous full turn changes the root’s sign. That invites a spinor analogy; a physical spin model needs a specified representation and dynamics.

Likewise, ordinary cumulants have integer orders. A useful alternative to an ambiguous “half-cumulant” is $F(\sigma )=\log \mathbb{E}[Q^{\sigma }]$. Its first two derivatives are the mean and variance of $\log Q$ under the normalized $Q^{\sigma }$ weighting. At $\sigma =1/2$, the mean is zero by reflection symmetry. The half appears in the selected ensemble, while the derivatives retain their ordinary meaning.

A square lattice hiding in the prime signs

Write a lattice point $(a,b)$ as a Gaussian integer $a+ib$. Its squared distance from the origin is $a^2+b^2$. A prime congruent to one modulo four splits into conjugate Gaussian factors, such as $5=(2+i)(2-i)$. A prime congruent to three modulo four stays prime in this larger number system. Conrad’s Gaussian-integer notes develop this geometry through factorization.

Let $r_2(n)$ count ordered integer pairs with $a^2+b^2=n$, including signs. The orientation character counts them:

$$r_2(n)=4\sum _{d\mid n}{\chi }_4(d).$$

For $n=5$ there are eight points, $(\pm 1,\pm 2)$ and $(\pm 2,\pm 1)$; for $n=3$ there are none. Summing by squared radius gives

Build a squared radius from prime factors, predict whether its circle contains integer points, and try repairing an empty shell.

Prime factors leave footprints

Repair a lattice shell

Reset puzzle

Restore an empty shell. Make at least one integer point land on the circle. Build a squared radius with prime tiles, predict, then reveal.

× 3× 5× 7× 13Undo to start

I predict pointsI predict an empty shell

Squared radius 15. The circle is visible; its integer points are waiting to be revealed.

Puzzle 1Puzzle 2Puzzle 3

The prime rule and the exact count

A prime congruent to 3 modulo 4 must occur an even number of times for an integer to be a sum of two squares. Pairing a lone 3 can restore an empty shell: 15 × 3 = 45. The number of ordered, signed pairs is r₂(n) = 4 Σ χ₄(d), summing over divisors d of n. Enumeration gives 0; the character formula gives 0. This is an arithmetic construction on a square lattice. Squared radius is capped at 325 to keep the playfield readable.

The factor four accounts for quarter-turn rotations. Combining this with the rapidity moment aligns the two worlds directly:

$$\zeta (s)\mathbb{E}[|\mathrm{\Phi }{|}^{s-1}]=\frac{\mathrm{\Gamma }(s)}{\pi }\sum _{(a,b)\ne (0,0)}(a^2+b^2{)}^{-s}.$$

One exploratory construction is to begin with the Jeffreys circle and impose an integer lattice on its plane. Then the lattice supplies a discrete inventory of permitted radii. That discreteness is an added ingredient; its Euclidean squared norm also differs from the Lorentz difference of squares in the evidence cone.

Zeros as rhythms in a prime census

The bridge back to primes comes from differentiating the logarithm of the Euler product:

The von Mangoldt weight ${\mathrm{\Lambda }}_{\mathrm{v}\mathrm{M}}(n)$ is $\log p$ when $n=p^k$ and zero otherwise. Thus this is a signed census of prime powers, including positive signs for even powers of flipping primes.

Give the census a finite size and a smooth taper:

$$J(X)=\sum _{n\le X}{\chi }_4(n){\mathrm{\Lambda }}_{\mathrm{v}\mathrm{M}}(n)(1-\frac{n}{X}).$$

Mellin inversion represents this using $-{\beta }_D^{\prime }/{\beta }_D$ times $X^s/[s(s+1)]$. When its integration contour moves past a zero $\rho$ of multiplicity $m_{\rho }$, that zero contributes the residue

$$-m_{\rho }\frac{X^{\rho }}{\rho (\rho +1)}.$$

Other residues and contour terms complete the census formula. This one term already reveals the geometry: $X^{\rho }=X^{\mathrm{\Re }\rho }{e}^{i\mathrm{\Im }\rho \log X}$. The real part controls growth; the imaginary part sets a rhythm along logarithmic size. The same zero that cancels a continuous probability response contributes an oscillation to a discrete arithmetic count.

Change the experiment and see what survives

Jeffreys has a special role here. A balanced $\mathrm{Beta}(a,a)$ gives a rapidity density proportional to ${\mathrm{sech}}^{2a}\varphi$. At $a=1/2$, the exponential expansion has unit coefficients and odd-integer rates. Other choices change those coefficients and rates, generally losing the multiplicative character that organizes the prime product.

Moving the clock walls from $\pm 1$ to $\pm a$ is gentler: Brownian scaling multiplies exit time by $a^2$, so its moment transform acquires a nonzero scale factor. Adding a drift $\mu$ changes the shape:

$$\mathbb{E}[e^{-\lambda T_{\mu }}]=\frac{\cosh \mu }{\cosh \sqrt{{\mu }^2+2\lambda }}.$$

Correlating the clock and reader is another experiment. At perfect correlation the reader stops at a wall too, and the continuous rapidity law collapses to two readings. These variations give a research program something to measure: which update rules preserve the Jeffreys law, which preserve reciprocal symmetry, and which retain arithmetic multiplication? The interesting part is how much structure the same small wandering-clock experiment manages to carry.

Further reading

• Dirichlet L-functions, NIST DLMF: character sums, Euler products, functional equations, and zeros.
• Biane, Pitman, and Yor, Probability laws related to the Jacobi theta and Riemann zeta functions, and Brownian excursions: the exit-law and reciprocal-moment connections.
• Keith Conrad, The Gaussian Integers: prime splitting and sums of two squares.