Forces from a network of messages

Imagine a town where every house can speak only to its immediate neighbors. One house keeps a lamp lit. Each evening, everyone looks at the four houses around them and adjusts their own lamp to the average brightness they see.

The news travels. A house two streets away responds, then one three streets away. Eventually, a whole landscape of brightness surrounds the original lamp. No house needs a map of the town.

Now make every relay lose four percent of the signal it receives. The same local conversations build a much smaller island of light.

That change has a close mathematical cousin in physics: a long-range potential acquiring a finite range. We can build the connection one neighbor at a time, then ask what else would be needed to turn it into a theory of forces.

Give every house one rule

Let ${\varphi }_i$ be the signed signal at house $i$. Positive values favor one answer to a question; negative values favor the other. The quantity $q_i$ is a fixed local source, usually zero. A positive source keeps supplying a positive message; a negative one supplies its opposite.

At each round, every house replaces its signal with

$$\boxed{{\varphi }_i^{(t+1)}=q_i+\frac{r}{4}\sum _{j\sim i}{\varphi }_j^{(t)}.}$$

Here $t$ counts relay rounds, $j\sim i$ means that $j$ is one of $i$’s four neighbors, and the retention $r$ is the fraction of their average signal that survives. With $r=1$, the relays retain all of it. With $r=0.96$, each relay retains 96 percent.

This is a chosen linear rule for influence. In an EP reading, $\varphi$ can play the role of rapidity, the half-log-odds coordinate introduced in One rapidity, many worlds. Averaging this coordinate is a modeling choice; ordinary pooling of trial counts follows a different rule. We will work with the signed signal directly, so the relay map does not claim to give calibrated posterior probabilities.

Predict before trying: will losing just four percent per relay mainly change the brightness near the source, or the distance its influence can reach?

Two-dimensional relay experiment

Build a field, one neighbor at a time

Reset experiment

Every square reads its four neighbors. Compare the same sources with all their signal retained and with some lost on each relay. Tap either map to move the selected source; the arrow buttons work too.

One source Two agreeing sources Two opposed sources

Signal retained per relay 90% retained96% retained99% retained

Move source

A (+)

← ↑ ↓ →

2D · no forgetting

r = 1.00 · same sources, same color scale

121 cells have signal magnitude at least 0.1

2D · forgetting

r = 0.96 · same sources, same color scale

49 cells have signal magnitude at least 0.1

Positive signalNegative signalZero-valued outer ring

Source A adds +1. Source B adds +1 or −1. Color grows darker up to a signal magnitude of 3; white is zero. These maps show a finite 2D lattice, not a slice of a 3D field. The fixed outer ring lets the lossless field settle.

Start from silence Send one round Send 12 rounds Settle the field

48 rounds. Selected source A is at column 10, row 10. Largest pending change: 1.3e-2.

$${\varphi }_i^{(t+1)}=q_i+\frac{r}{4}\sum _{j\sim i}{\varphi }_j^{(t)}$$

φ is signal, q is the fixed source, t counts rounds, and j ∼ i means a neighboring cell. Moving a source redraws both maps from silence for the displayed number of rounds.

A different geometry: the 3D continuum prediction

In three dimensions, the lossless potential falls as 1/R. Forgetting adds an exponential factor. These curves are the analytic prediction for one point source, scaled to agree at R = 1; they are separate from the 2D simulation above.

3D · no forgetting3D · forgetting

2D screening length, continuum approximation

2.45 spacings

3D screening length, continuum approximation

2.00 spacings

The same retained fraction has different ranges because four neighbors share the signal in 2D and six in 3D. Continuum formulas work best when the field varies over many lattice spacings.

Open the message relay in the playground.

Try three small investigations:

1. Choose Start from silence, then Send one round. Only the source responds. On the next round, its immediate neighbors respond. Each round opens one more possible hop.
2. Choose Settle the field, then compare 99 percent and 90 percent retention. Small local losses accumulate over the many routes needed to reach distant houses.
3. Choose Two opposed sources and move them together. Their positive and negative contributions cancel. Separate them again and a ridge and a valley appear.

The two maps share a color scale. Their outer ring stays at zero, like the grounded edge of a conducting sheet. This boundary gives even the lossless model somewhere to discharge its signal.

The landscape is solving an equation

Once the field settles, another round changes nothing. Write $S\varphi$ for the vector of neighbor averages. The steady state obeys

$$\varphi =q+rS\varphi ,(I-rS)\varphi =q.$$

$I$ is the identity operation: it leaves every value alone. When the inverse exists, the solution is

$$\varphi =(I-rS{)}^{-1}q.$$

That compact inverse contains a travel diary. In our finite grid with its zero boundary, it can be expanded as

$$\varphi =q+rSq+r^2{S}^{2}q+r^3{S}^{3}q+\cdots .$$

The first term is the source itself. The next contains one-hop journeys, the next two-hop journeys, and so on. Each extra hop contributes another factor of $r$. The averaging operation spreads a signal among possible paths; the boundary removes paths that leave the town.

This is also where lineage matters. A message that travels around a block and comes back still descends from the same source. The sum above describes repeated influence through a network. It does not turn those routes into independent trials. If we wanted to estimate how many independent observations support a claim, we would have to retain their origins, as in One trial, many echoes.

Neighbor disagreement becomes a Laplacian

There is another way to read the same equation. At every house, compare its value with the values around it:

$$(L\varphi {)}_i=4{\varphi }_i-\sum _{j\sim i}{\varphi }_j.$$

$L$ is the graph Laplacian. In this sign convention, it is positive at a local hill and negative at a local valley. If a house agrees with its neighbors’ average, its Laplacian is zero.

Since $S=I-L/4$, the steady relay equation becomes

$$\frac{r}{4}L\varphi +(1-r)\varphi =q,$$

or, for ,

$$\boxed{(L+{\kappa }^{2}I)\varphi =\frac{4}{r}q,{\kappa }^2=\frac{4(1-r)}{r}.}$$

The Greek letter $\kappa$ (kappa) measures inverse range. The $L$ term spreads the signal by reducing differences between neighbors. The ${\kappa }^2\varphi$ term continually removes signal. A stronger loss asks the source to work harder to maintain a distant response.

The same steps work on a cubic lattice in $d$ dimensions. Each site has $2d$ neighbors:

Here $\ell =1/\kappa$ is the screening length. The word screening means that the medium suppresses the distant influence of a source. With 99 percent retention, the continuum estimate is about 4.97 spacings in two dimensions and 4.06 in three.

These length formulas describe the continuum approximation. If the range shrinks to one or two lattice spacings, individual steps and grid directions become conspicuous, and the approximation becomes less accurate.

A familiar field appears

Now imagine a field that changes smoothly over many grid spacings. With spacing set to one, $L\varphi$ approximates $-{\mathrm{\nabla }}^2\varphi$, the negative Laplacian in space. The symbol ${\mathrm{\nabla }}^2$ adds the second derivatives along the spatial directions. The relay equation approaches

$$\boxed{(-{\mathrm{\nabla }}^2+{\kappa }^2)\varphi =s.}$$

The source density $s$ includes the factor $2d/r$ from the lattice equation. This is the screened Poisson equation. At zero loss, $r=1$ and $\kappa =0$, it becomes the ordinary Poisson equation.

Place a point source of total strength $Q$ in unbounded three-dimensional space and ask for a potential that vanishes far away. At distance , the two cases line up:

The field in the last row is $\mathbf{E}=-\mathrm{\nabla }\varphi$: the direction and rate at which the potential decreases. Differentiating $1/R$ gives the inverse-square law. Differentiating $e^{-\kappa R}/R$ adds both exponential attenuation and the extra $\kappa /R$ term.

This is the mathematical form associated with Yukawa’s short-range interaction. Here, a loss per relay supplies the inverse range. Identifying that parameter with the mass of a physical particle would require a further physical model, including units and dynamics.

The lower graph in the experiment draws these three-dimensional continuum potentials. It scales both curves to agree at $R=1$, so their different shapes are easy to compare.

Why the map above is two-dimensional

A sheet and a volume give a signal different numbers of places to go. In three dimensions, spheres grow in area as $R^2$. In two dimensions, circles grow in circumference as $R$. The same local equation therefore has different radial solutions.

For a lossless point source, the two-dimensional potential varies logarithmically with distance, while the three-dimensional potential varies as $1/R$. These are the familiar dimension-dependent solutions of the Laplace equation. A finite two-dimensional system also needs its boundary value specified, as our zero-valued outer ring does.

So the square maps demonstrate how relay loss shortens influence and how sources combine. They do not measure the three-dimensional inverse-square law. To study that law numerically, we would build a three-dimensional lattice, place its boundary far from the source, and compare distances much larger than one spacing but much smaller than the box.

The boundary is part of the experiment. On a finite grid whose edges wrap around, a nonzero net source and perfectly lossless relays would keep increasing the average signal. A steady field needs somewhere for that net signal to go, or a source arrangement whose net injection is zero together with a choice of reference level.

From a field to a force

Place a small test object in the landscape. If its energy is $U=q_{\mathrm{test}}\varphi$, where $q_{\mathrm{test}}$ measures how it couples to the field, then a force law that moves it downhill gives

$$\mathbf{F}=-\mathrm{\nabla }U=q_{\mathrm{test}}\mathbf{E}.$$

A positive test object moves away from a positive source’s hill. A negative one moves toward it. The signs and the spatial decay now look like electrostatics.

We supplied the energy and motion rule in this step. The relay equation builds a field; it does not, by itself, make objects accelerate. Likewise, the demo’s rounds are numerical updates. Connecting them to physical time would require a dynamical model of propagation.

That leaves an inviting question. Could a physical field be understood as a medium continually reconciling local messages? The relay construction offers a concrete place to explore the idea: specify what travels, how lineage is handled, what is conserved, and how matter couples to the resulting field. Each choice gives us an equation we can inspect.

For the experiment we already have, try moving the two opposed sources one cell apart. A strong local disagreement can leave very little signal far away. Separate the sources and the distant field returns. Geometry has changed what the rest of the town can hear.