From a coin to a compass · Epistemic Physics

Up to this point, our beliefs were binary: a coin lands heads or tails, a patient recovers or does not. This is one-dimensional kinematics, with only two spatial directions: left ( $- 1$ ) and right ( $+ 1$ ).

What happens when a question has multiple answers, or when evidence points along continuous headings on a 2D plane?

Our 1D light cone opens up into a full relativistic cone, and belief becomes a compass.

Averaging directional unit vectors

Suppose evidence arrives as directional observations. Let $n_{k}$ be a unit vector pointing along heading $k$ , with evidence weight $w_{k} \geq 0$ .

Total evidence is the scalar sum of all weights: $τ_{E} = \sum_{k} w_{k}$ . Assume this total is positive.

The epistemic velocity vector $v_{E}$ is the weighted vector sum divided by total evidence:

Because all $‖ n_{k} ‖ = 1$ , the resulting velocity vector lies in the closed unit disk: $‖ v_{E} ‖ \leq 1$ . Equality occurs when all positively weighted directions align. A spread of directions, including our neutral directional prior, pulls it strictly inside!

\begin{aligned} τ_{E} & = \sum_{k} w_{k}, \\ v_{E} & = \frac{\sum_{k} w_{k} n_{k}}{τ_{E}} . \end{aligned}

From discrete polygons to the continuous disk

If you only allow four discrete compass headings (North, South, East, West), any weighted average is trapped inside a square diamond.

The diamond only covers $2 / π \approx 63.66 %$ of the unit disk, leaving over a third of valid relativistic velocities completely unreachable!

As you increase the number of allowable headings $K$ , the reachable polygon expands to fill more of the disk, eventually covering all possible directional beliefs.

Your turn

Start with four headings in the lab, then increase the number of bins. Watch the reachable polygon fill the disk while the total prior weight stays fixed at 1.

Give your evidence more directions

From four headings to a whole horizon.

Mix the available headings and you can reach the polygon. Add directions to fill more of the disk.

K = 4, α_{0, k} = \frac{1}{4}

total prior = K \times \frac{1}{K} = 1

Uncovered disk area: 36.34%. More resolution, the same amount of prior evidence.

Two analytic densities · one prior

See the exact distribution of a compass mean

Every point inside the disk is a possible average of a directional allocation. Start with the same uniform prior of total one. Spread the added weight evenly around the horizon, or point all fresh trials in one direction. These heatmaps evaluate the exact continuous-circle densities.

Added weight: 2 · total τ_E = 3

The left model adds a continuous, uniform evidence ledger. The right adds 2 fresh unit trials at a shared heading. Rotate that heading by dragging the orange handle, using its arrow keys, or choosing a compass direction.

Spread the added weight uniformly

Mean vector: (0.00, 0.00)
Sum of coordinate variances: 0.2500

Point the added trials one way

Mean vector: (0.67, 0.00)
Sum of coordinate variances: 0.1389

04.3327.46

Mean vectorFresh trial heading

Shared color scale: density per unit disk area, using log(1 + density) to show both peaks and tails. Colors are evaluated from formulas at pixel centers; no sampled or smoothed density is used. Pointed trials create an integrable peak at the rim, beyond the finite color scale shown here.

\begin{matrix} \begin{aligned} f_{τ_{E}} (v) & = \frac{τ_{E}}{π} \\ \times (1 - | v |^{2})^{τ_{E} - 1} \end{aligned} \\ tr Cov (V) = \frac{1 - | v_{E} |^{2}}{τ_{E} + 1} \end{matrix}

$V$ is the random directional average and $v_{E} = E [V]$ is its mean. The density formula on the left applies to uniformly distributed weight; the variance formula applies to both panels. $v$ labels a possible location in the disk.

At zero added weight, the mean is uniformly distributed over the disk. Equal total concentration does not imply equal shape: where the added evidence points changes both the mean and its spread.

The Dirichlet process: resolution without manufactured evidence

If you divide the circle into $K$ directional bins, how do you set a neutral prior without artificially inflating the total trial weight?

Give each bin a weight of $1 / K$ . The sum of all prior weights is always exactly 1, regardless of whether $K = 4$ or $K = 4,000$ .

Taking $K \to \infty$ leads to a directional Dirichlet process over all possible angles, preserving unit prior concentration. Its neutral base measure is continuous and uniform around the circle, while each random probability measure drawn from the process is almost surely discrete: a countable collection of weighted directions. Infinite resolution does not require manufacturing extra evidence!

\begin{aligned} α_{0, k} & = 1 / K, \\ \sum_{k = 1}^{K} α_{0, k} & = 1. \end{aligned}