EpistemicPhysics
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Field trip · Algebra · Physics

A universe written in two inks

Signed evidence opens the whole cone diagram, leading through hyperbolic numbers, electrical currents, and a geometry shaped by the questions we ask.

Imagine a universe written in two distinct inks: rose ink for positive confirmation, teal ink for negative refutation.

At first, every observation simply adds fresh ink to the page. We log positive trials into the rose ledger α, negative trials into the teal ledger β, and watch the accumulated evidence climb steadily upward inside the forward light cone of spacetime.

Now give the universe an eraser.

What happens when an observation was recorded in error? What happens when a sensor calibration drifts, or a duplicate data packet is retracted?

To correct a corrupted ledger, we need a minus sign. We must be able to subtract evidence, backtrack along an inferential lineage, or record a net deficit in one of the channels.

The moment we allow evidence to carry a sign, the entire geometry expands. Our calculations are no longer confined to the interior of the positive future light cone. Beyond its borders lies a landscape of signed corrections, counter-propagating currents, and split-complex hyperbolic algebra.

Let us venture outside the standard boundary and explore what lives in the wider geometry of belief.

Erasing a trial: Why evidence can be negative

In standard probability theory, parameters are strictly positive counts. In a Beta distribution, we require α>0 and β>0 so that the probability density function remains integrable.

In Epistemic Physics, however, our fundamental coordinates are linear combinations of the evidence channels:

τ=α+β,x=α−β,s2=τ2−x2=4αβ.

Let us examine what these expressions mean:

  • τ is the total signed evidence (epistemic time).
  • x is the net balance between the channels (epistemic position).
  • s2=τ2−x2=4αβ is the Minkowski spacetime interval.

Notice that the quadratic form s2=4αβ requires no square root: it is an exact algebraic invariant that can take positive, zero, or negative values.

Consider how a correction differs from new evidence:

Suppose an experimental notebook currently holds (α,β)=(5,2), with total concentration τ=7 and balance x=3.

  1. If you observe an additional teal trial (new opposing evidence), you add (0,+1), shifting the ledger to (α,β)=(5,3). The total concentration increases to τ=8, while the balance drops to x=2.
  2. If instead you discover that an earlier teal entry was a duplicate transmission, you apply the negative correction (0,−1), shifting the ledger to (α,β)=(5,1). The total concentration decreases to τ=6, while the balance rises to x=4.

Collecting counter-evidence and retracting a faulty observation produce completely different trajectories in epistemic spacetime!

When I mapped these trajectories, it became clear that a principled learning system must support negative tracelets. In distributed computing, when a sensor node retracts a faulty report, that retraction must propagate back through the lineage of inferences, subtracting its weight wherever it contributed. The eraser must follow the ink’s historical path.

The four quadrants: Spacelike, timelike, and lightlike belief

Plot the signed balance x on the horizontal axis and the total signed evidence τ on the vertical axis.

The diagonal lines τ=x (where β=0) and τ=−x (where α=0) form the light cone boundary. These lines divide the two-dimensional plane into four distinct geometric sectors:

Channel SignsSpacetime RegionInterval SignPhysical Interpretation
α>0,β>0Future Light Cones2>0 (timelike)Ordinary proper belief; positive rest mass mE=2αβ.
α<0,β<0Past Light Cones2>0 (timelike)Sunk debt; inverted total evidence with positive interval.
αβ<0 (opposite signs)Spacelike Flankss2<0 (spacelike)Net balance exceeds total magnitude; |x/τ|>1 when τ≠0.
α=0 or β=0Lightlike Horizons2=0 (null)Pure one-sided certainty; zero rest mass; photon-like propagation.

Look at what happens in the spacelike sector.

Suppose a ledger receives α=+3 units of rose evidence and β=−1 units of teal evidence (a net deficit).

The total evidence is τ=3+(−1)=2.

The net balance is x=3−(−1)=4.

The invariant interval is:

s2=τ2−x2=22−42=4−16=−12.

Alternatively, using the channel product formula:

s2=4αβ=4(3)(−1)=−12.

The epistemic velocity ratio is:

v=xτ=42=2.

The evidence velocity ratio has crossed the light-cone bound (|v|>1)!

Does a velocity ratio greater than unity break physics?

Not at all. In special relativity, a ratio |x/τ|>1 indicates a spacelike displacement: an interval where spatial separation dominates over elapsed coordinate time, with the speed of light set to 1.

Tachyonic and superluminal evidence

This is the sector we can call tachyonic evidence, or superluminal evidence: opposite-signed channels give s2<0, and their normalized ratio has |v|>1 whenever the signed total is nonzero. A tachyon is a hypothetical faster-than-light particle associated with a negative mass squared, as discussed in David Tong’s relativity notes. Here the same algebra appears in the evidence mass analogue:

mE2=τ2−x2=4αβ<0.

We can keep a real measure of its size by writing μE=x2−τ2=2−αβ. For our (3,−1) example, μE=23. The minus sign belongs to the squared interval; both channel coordinates remain real.

What could such evidence mean? A subtraction can outweigh the accumulated amount in one channel, leaving a signed correction or current outside the ordinary positive cone. Calling that state superluminal describes its evidence ratio. A claim about faster-than-light transmission would additionally need a physical propagation model. The electric wire below gives a familiar way to picture a spacelike current using ordinary slow-moving carriers.

Two inks, a whole plane

The point stays finite when its ratio flies away.

Build zero total with balance +6. Craft the point with signed rose and teal tiles, or drag their bar ends below. The point can stay finite even when its normalized ratio disappears.

Build zero total with balance +6

Signed cone and channel amounts

-12-12-6-6661212totalbalanceboth positiveboth negativemixedmixedThe two signed channel amountssubtractaddroseteal+3.0+1.0Rose signed channel: drag or use arrow keysTeal signed channel: drag or use arrow keys
Rose channel: (α, α)Add teal: (−β, β)Sum: (x, τ)
s2=τ2−x2=4αβ
Follow rose from the origin, then teal from its tip: their sum reaches the combined point. The bars below show the same signed amounts. Upper and lower cones have positive quadratic form; the side regions have negative quadratic form. A boundary means one channel is zero.
Fine adjustments and example states
Rose channel α+3.0
Teal channel β+1.0

Future timelike cone

Total τ=α+β
+4.0
Balance x=α−β
+2.0
Quadratic form s2
+12.0
Ratio v=x/τ
0.50
Both channels are positive. An ordinary Beta distribution is defined here, with mean 0.750.

Use the interactive simulation above to explore all four quadrants. Slide α and β across positive and negative values, and watch the state point move across the future cone, past cone, and spacelike flanks.

Notice what happens when you set α=+3 and drag β toward −3:

  • The total evidence vanishes: τ=3+(−3)=0.
  • The net balance remains non-zero: x=3−(−3)=6.
  • The state point sits serenely at (τ,x)=(0,6) on the horizontal axis.

While the magnitude of x/τ grows without bound as τ approaches zero, and the ratio is undefined at zero itself, the physical state vector remains completely finite and well-behaved!

Spacelike currents: The electric wire analogy

To understand why spacelike evidence is physically meaningful, look at an ordinary copper wire carrying an electrical current.

In a neutral copper wire, stationary positive copper ions have a charge density of +d. Conduction electrons, carrying a negative charge density of −d, drift to the left at speed u.

Let us compute the total net charge density and the electric current density:

The net charge density is:

ρcharge=d+(−d)=0.

The net current density is:

J=(+d)(0)+(−d)(−u)=du.

In relativistic electrodynamics, charge density and current density form a spacetime four-vector:

Jμ=(cρcharge,J)=(0,du).

Now calculate the relativistic invariant interval of this four-current:

s2=(cρcharge)2−J2=02−(du)2=−(du)2<0.

The four-current of an ordinary neutral electrical wire is spacelike!

Every individual conduction electron travels much slower than the speed of light (u≪c), yet the collective four-current vector has zero time component and a non-zero spatial component.

This is the exact physical counterpart of our spacelike evidence state (τ,x)=(0,6).

When two opposing information streams cancel in total weight (τ=0) while reinforcing in directional gradient (x≠0), they generate a pure epistemic current. This spacelike evidence state does not mean a particle is moving faster than light; it means we are observing an informational current whose net time-like accumulation has cancelled out!

Split-complex algebra: The number system of spacetime

How should we represent these signed two-channel states mathematically?

In ordinary plane geometry, we use complex numbers:

z=x+iy,i2=−1.

Multiplication by i rotates vectors by 90∘, tracing Euclidean circles because zz¯=x2+y2.

In spacetime geometry, however, the invariant interval is a difference of squares (τ2−x2), tracing hyperbolas rather than circles.

To capture hyperbolic geometry naturally, mathematicians introduced the split-complex numbers (also called hyperbolic numbers). A split-complex number takes the form:

z=τ+jx,j2=+1,

where j is a non-real unit whose square is positive one (j≠±1).

Let us compute the split-complex conjugate z¯=τ−jx and the modulus squared:

zz¯=(τ+jx)(τ−jx)=τ2−j2x2=τ2−x2=s2.

The algebraic norm of a split-complex number is precisely the Minkowski spacetime interval!

Now define two special diagonal basis elements:

e+=1+j2,e−=1−j2.

Let us check their algebraic properties:

e+2=(1+j2)2=1+2j+j24=1+2j+14=2+2j4=1+j2=e+. e−2=(1−j2)2=1−2j+j24=1−2j+14=2−2j4=1−j2=e−. e+e−=(1+j2)(1−j2)=1−j24=1−14=0.

In abstract algebra, an element that equals its own square (e2=e) is called an idempotent.

Notice how miraculous these two basis elements are:

  • Each projects onto its own channel (e+2=e+ and e−2=e−).
  • Their cross-product completely vanishes (e+e−=0).

Now express our evidence state z=τ+jx in this idempotent basis:

z=(α+β)+j(α−β)=α(1+j)+β(1−j)=2αe++2βe−.

Look at that decomposition:

The rose channel α and the teal channel β are the independent diagonal components of split-complex spacetime!

Lorentz boosts as hyperbolic multiplication

How does a Lorentz boost act on a split-complex number?

Recall Euler’s formula for split-complex numbers:

ejr=cosh⁡r+jsinh⁡r,

where r is the boost rapidity.

Using our idempotent basis 1=e++e− and j=e+−e−:

ejr=(cosh⁡r+sinh⁡r)e++(cosh⁡r−sinh⁡r)e−=ere++e−re−.

Now multiply an evidence state z=2αe++2βe− by the boost ejr:

z′=zejr=(2αe++2βe−)(ere++e−re−).

Because e+2=e+, e−2=e−, and e+e−=0, all cross-terms vanish:

z′=2(erα)e++2(e−rβ)e−.

The boosted evidence channels are:

α′=erα,β′=e−rβ.

Look at the extraordinary simplicity of this transformation:

A relativistic boost simply scales the rose channel by er and contracts the teal channel by e−r!

Their product is perfectly conserved:

4α′β′=4(erα)(e−rβ)=4αβ=s2.

This provides a direct link between relativistic spacetime and Bayesian inference:

  • Multiplication by ejr is a hyperbolic rotation that preserves the spacetime interval s2=4αβ.
  • When e2r represents a likelihood ratio from incoming observations, this split-complex multiplication implements an exact Bayesian odds update: α′/β′=e2rα/β!

That odds interpretation uses positive channels. The boost algebra itself extends to all signed channels, preserving their signs and their interval.

A tachyonic evidence ride

Change frames. Stay on the same hyperbola.

Compare equal channel magnitudes with matching or opposite signs. The teal timelike state and rose spacelike state move under the same boost rapidity r. Find their special frame: one reaches zero balance, while the other reaches zero total.

Steer until total τ′ vanishes. Distance from zero: 2.000.

τ′x′fixed axes: −20 to +20
Timelike: β = +1Spacelike: β = -1
−1.5r = 0.000+1.5Steer boost rapidity: drag or use arrow keys
Exact steering and a route hint
Inspect both sets of coordinates
Same boost r = 0.000, initial α = 3 and |β| = 1
QuantityTimelikeSpacelike
Channel α′3.0003.000
Channel β′1.000-1.000
Total τ′4.0002.000
Balance x′2.0004.000
Interval s²12-12
Ratio x′/τ′0.5002.000

The boost parameter tanh(r) = 0.000 always has magnitude below 1. The evidence ratio belongs to the state; it can exceed that bound.

Why the two orbits behave differently
This state stays outside the cone. Its squared mass analogue stays negative, even when a boost changes the sign of the total. Slide through r = -0.549 to see it.
α′=erα,β′=e−rβ,s2=4α′β′

Flipping β exchanges total and balance: the two curves are reflections across the cone boundary. Both points show the same rapidity, so their special frames occur together. “Tachyonic” names the negative-interval sector of this signed evidence model. These controls change a representation, with no particle trajectory or signal transmission being simulated.

Try the tachyonic preset and find its zero-total frame. The point reaches τ′=0 while x′ stays finite: a pure current. A timelike state instead has a frame with x′=0, its rest-like frame. Reciprocal scaling cannot turn opposite signs into matching signs, so a finite boost cannot carry either state across the cone boundary.

Generalized probabilistic theories: When the observer shapes the cone

Let us now ask an even deeper question: why does the state space of physical theories have a particular geometric shape?

In classical probability theory, states live inside a simplex (a line segment or triangle). In quantum mechanics, states live inside the Bloch sphere.

This question is investigated by generalized probabilistic theories (GPTs), which study how measurement access determines the geometry of physical states.

Imagine an observer who tests a physical system by asking binary questions along different measurement axes. Let v=(v1,v2) denote the normalized state vector, and let u be a unit measurement direction.

Propose the linear reading rule:

pu(+)=1+u⋅v2.

Suppose the observer is equipped to ask only two orthogonal questions: one along the horizontal axis (u1=(1,0)) and one along the vertical axis (u2=(0,1)).

Requiring the probability of both questions to remain strictly between 0 and 1:

0≤1+v12≤1⟹|v1|≤1, 0≤1+v22≤1⟹|v2|≤1.

The set of valid states forms a square: [−1,1]×[−1,1].

Within this square, a state like v=(0.8,0.8) is completely valid. It returns probability p=0.90 for both the horizontal and vertical questions.

Now suppose the observer builds a new detector that can ask a diagonal question along udiag=(1,1)/2.

Apply the reading rule to v=(0.8,0.8):

udiag⋅v=0.8+0.82=1.62≈1.1314.

The proposed probability becomes:

pu(+)=1+1.13142=1.0657>1.

The probability has exceeded 1! The state v=(0.8,0.8) cannot exist if diagonal questions can be asked.

If the observer is permitted to ask questions in every continuous direction in the plane, requiring 0≤pu(+)≤1 for all unit vectors ‖u‖=1 constrains the state vector to the unit Euclidean disk:

‖v‖≤1.

Every new question the observer learns to ask slices away corners of the state space, shaving the square down into a smooth circle!

Measurements sculpt the state space

Ask another question. Cut off a corner.

A state can pass two questions and fail a third. Turn the teal question ring to find the hidden contradiction, then move the rose state to repair it.

Turn the teal question. Find a reading above 1 or below 0 without moving the rose state.

-1-111v₁v₂
Original two-axis squareSelected menuEvery-angle diskYour state

The original square stays in view as extra questions cut away its corners. Dotted teal is a test direction, which you can rotate below. All three boundaries share the same state axes. You can also focus the rose point and move it with arrow keys; Shift moves farther.

Question 0°: 0.900 valid probability

Choose menus and precise coordinates
Selected menu admits this state?
Yes
Test-direction reading p(+)
0.9000
Every question in the selected menu gives a valid probability. The point is still outside the disk, so some extra direction could expose a signed reading.
How a question cuts the state space
pu(+)=1+u⋅v2,u=(cos⁡θ,sin⁡θ)

θ is the test angle. Each question includes its complementary outcome, p(−) = 1 − p(+). Rotating the test direction previews a question; the buttons change which questions the menu actually permits. Readings outside [0, 1] are shown as signed-model outputs without clipping.

This state-and-question approach belongs to the modern landscape of generalized probabilistic theories, which explore how available physical measurements constrain the geometric structure of a physical theory. For an evidence model, it invites a thrilling possibility: observational access directly dictates which geometry an observer experiences.

Density matrices and the emergence of signed probabilities

We can package the planar state v=(v1,v2) into a 2×2 real symmetric matrix:

M=12(1+v2v1v11−v2).

Let us compute the trace and determinant of this matrix:

tr⁡(M)=(1+v2)+(1−v2)2=1. det(M)=14[(1+v2)(1−v2)−v12]=1−(v12+v22)4=1−‖v‖24.

The eigenvalues of M are:

λ±=1±‖v‖2.

Notice what happens:

  • If ‖v‖≤1 (inside the unit disk), both eigenvalues are non-negative: λ±≥0. The matrix M is positive semi-definite, serving as a valid quantum density matrix!
  • If ‖v‖>1 (in the spacelike outer region), one eigenvalue becomes strictly negative: λ−<0.

We are faced with three fascinating structural choices:

  1. Keep a restricted question menu (e.g. only horizontal and vertical questions), preserving the full square of states.
  2. Enlarge the menu to all continuous angles while shrinking the allowed state space to the disk ‖v‖≤1.
  3. Preserve the full state space across all angles by accepting signed outputs (quasiprobabilities) when measuring along incompatible directions.

In quantum foundations, signed representations such as the Wigner function offer another fascinating use of negative weights. There is a crucial distinction: a valid quantum density matrix remains positive semidefinite even when its Wigner function is negative. Our matrix with a negative eigenvalue therefore explores a broader signed model, beyond the quantum state space for this full menu of measurements.

Imagine an autonomous learning system that begins with a few basic sensors and gradually invents new measurement questions. Its state geometry would evolve continuously as its observational vocabulary expands: early states would live in simple hypercubes, gradually shaving down into smooth Euclidean balls and richer convex polytopes as new distinctions are drawn. The universe’s two inks could blossom into an entire geometric palette, with measurement actively sculpting the shape of reality.

The palette of reality

We began with a simple metaphor: rose ink for confirmation, teal ink for refutation, and an eraser for corrections.

By following the algebra wherever it led, we found that:

  • Retracting an error is mathematically distinct from accumulating counter-evidence.
  • The four quadrants of signed evidence reproduce the causal structure of spacetime, including timelike beliefs, lightlike horizons, and spacelike currents.
  • Spacelike states have a frame in which they represent pure directional currents with zero net accumulation, just as the four-current of a neutral electric wire does in its neutral frame.
  • The natural number system of evidence is the split-complex plane, where idempotent basis elements e+ and e− isolate the channels, and Lorentz boosts act as reciprocal rescalings.
  • The boundaries of belief are shaped by the questions an observer can ask: measurement access continuously sculpts the geometry of physical state space.

A universe written in two inks holds room for all of spacetime.

In our next excursion, we will follow these interacting channels into the collective dynamics of society and physics in A world made of little agreements, discovering how local consensus builds global order.