Questions scratchpad
Follow-ups to the math page. Same notation: hue $h$, difference $d = h_i - h_j$, $g(r) = 1/r^2$.
A symmetric pair is flying along with velocity $\mathbf v$. How does it get charged back up into a chase pair?
Short answer: the pair's bulk motion is what charges it, and it happens whenever the pair is moving against the (tiny) chase force. Braking the pair converts its kinetic energy into hue mismatch. But see Question 2: it needs a seed mismatch to start.
Setup
"Symmetric pair" means hues nearly equal, $d \approx 0$, so $\Lambda \approx 0$ and the battery is nearly empty. Both particles move with a common velocity $\mathbf v$.
Step 1: nothing is exactly zero
Near $d = 0$ everything is linear in $d$:
$$ \Lambda \approx b\,d, \qquad F^{\text{chase}} \approx \frac{\kappa b\,d}{r^2}, \qquad W_i \approx \frac{(a + 2\lambda b^2)\,d}{r} $$The $a$ comes from the bond term $-S'(d)/r$, the $2\lambda b^2$ from the battery term. Both the chase power $P_i = F^{\text{chase}}\,(\hat{\mathbf u}\cdot\mathbf v)$ and the gradient $W_i$ are proportional to $d$.
Step 2: the hue rate does not vanish
$$ \dot h_i = -\frac{P_i}{W_i} \approx -\frac{\kappa b}{a + 2\lambda b^2}\cdot\frac{\hat{\mathbf u}\cdot\mathbf v}{r} $$The $d$ cancels. Even a nearly matched pair changes hue at a finite rate, and the sign is set entirely by $\hat{\mathbf u}\cdot\mathbf v$: is the pair moving toward $i$'s side of the line or $j$'s?
Step 3: the pair moving against the chase
Say $d > 0$ slightly, so $\Lambda > 0$ and the chase pushes both particles along $+\hat{\mathbf u}$. If the pair happens to be drifting along $-\hat{\mathbf u}$:
- $P_i = P_j < 0$: the chase is braking the pair. Bulk KE drops.
- $\alpha_i, \alpha_j < 0$: both hues run uphill on $E$.
- Uphill means $|d|$ grows: $h_i$ rises, $h_j$ falls, $\Lambda^2$ grows, the battery $\lambda\Lambda^2/r$ fills by exactly the KE lost.
- As $\Lambda$ grows the chase force grows, brakes harder, charges faster. Self-reinforcing until the pair's velocity along $\hat{\mathbf u}$ is spent (then $P \to 0$, hold).
Step 4: what if it was moving with the chase instead?
Then $P > 0$, both burn, $d$ shrinks toward zero. But because $\dot h$ stays finite at $d = 0$, it does not stop there: $d$ overshoots through zero, $\Lambda$ flips sign, the chase force flips to $-\hat{\mathbf u}$, and now the pair is moving against it. It is in the recharge case of step 3. A pair sliding along its own axis oscillates: burn, cross zero, recharge on the other side.
Step 5: an orbiting pair
If the velocity is not along $\hat{\mathbf u}$ but the pair is bound and orbiting, $\hat{\mathbf u}$ rotates relative to $\mathbf v$ continuously and $\hat{\mathbf u}\cdot\mathbf v$ changes sign twice per orbit. Same thing, just periodic: that is the burn / hold / recharge cycle on the math page.
But if it is sliding along at a constant velocity it will not change colour, right?
Right. If it is genuinely at constant velocity, nothing happens. There are two cases and they need to be kept apart.
Exactly symmetric, $d = 0$
Then $\Lambda = 0$, so the chase force is zero, so the pair does coast at constant velocity (only the bond and core act, and they are internal to the pair). $P = 0$ and $W = 0$, so the code's guard fires and freezes hue. Correction (see Question 4): this is a numerical artifact, not a fixed point of the equations. $\dot h = -P/W$ is $0/0$ at $d = 0$ with a finite nonzero limit, so in the continuous model even an exactly matched moving pair starts changing hue immediately. Velocity alone charges nothing; it is the chase doing work on that velocity that charges, and the work-per-hue-change stays finite as $\Lambda o 0$.
Nearly symmetric, $d$ small but nonzero
Then it is not at constant velocity. There is a small chase force $\kappa b d/r^2$ accelerating or braking it. Small force, so the power $P \propto d$ is small: the pair loses KE slowly. But the hue rate is finite, because near $d = 0$ the battery is flat-bottomed ($\lambda\Lambda^2 \approx \lambda b^2 d^2$), so shifting hue costs almost nothing per radian:
$$ \dot h = -\frac{P}{W} = -\frac{\text{(small power)}}{\text{(small cost per radian)}} = \text{finite} $$Tiny braking buys a substantial hue shift while the battery is nearly empty. As $d$ grows the battery walls steepen, the same power moves hue less, and the chase force grows: the pair slows down noticeably and the hue rate settles.
So the honest statement is: a symmetric pair cannot start on its own if it is at rest. If it is moving, see Question 4: under the current battery it does not stay symmetric. It needs either a seed mismatch, or a third particle with a different hue (which gives both $\Lambda \ne 0$ and $W \ne 0$ immediately). Once there is any $\Lambda$ at all and any velocity component along the pair axis, the chase brakes it and the KE goes into the battery.
So there is a 3rd particle. The pair approaches it and maybe repels it. How do we know where the energy goes: into the KE of the 3rd particle, or into the colour energy of the pair? What decides the rate of transfer?
Two separate channels, and the bookkeeping never mixes them. That is the thing to hold on to.
Channel 1: the reciprocal forces (core, bond, battery gradient)
Ordinary potential forces. As the pair approaches $k$, the repulsion between $i$–$k$ and $j$–$k$ does work $\mathbf F^{\text{rec}}\cdot\mathbf v$ on each particle; energy comes out of KE and goes into $U$, then back out of $U$ into $k$'s KE as it is pushed away. Rate: force times velocity, same as a billiard-ball collision. Hue is not involved except that $U$'s depth depends on it. This channel conserves energy on its own. With the chase off ($\kappa = 0$) it is the whole story, and the 3rd particle's KE comes entirely from the pair's KE via the potential.
Channel 2: the chase
Now there are three chase pairs: $(i,j)$, $(i,k)$, $(j,k)$. Each puts the same force on both of its members. Per particle, add them up:
$$ \mathbf F^{\text{chase}}_i = \kappa\frac{\Lambda(h_i - h_j)}{r_{ij}^2}\hat{\mathbf u}_{ij} + \kappa\frac{\Lambda(h_i - h_k)}{r_{ik}^2}\hat{\mathbf u}_{ik}, \qquad P_i = \mathbf F^{\text{chase}}_i\cdot\mathbf v_i $$and likewise for $j$ and $k$. Three separate powers, three separate bills. Each particle pays its own $P$ from its own hue, nobody else's.
Does the energy go into $k$'s KE or the pair's colour energy?
The chase force on $k$ does work $P_k$ on $k$'s KE. That is paid by $k$ moving its hue, $\dot h_k = -P_k/W_k$. The chase force on $i$ does work $P_i$ on $i$'s KE, paid by $i$'s hue. The chase never moves energy from one particle's hue to another particle's motion in a single step. It only converts a particle's own hue-gradient energy into that same particle's own motion (or back). Energy then travels between particles through channel 1.
What decides the rate?
| Channel | Rate | Set by |
|---|---|---|
| 1: reciprocal | $\mathbf F^{\text{rec}}\cdot\mathbf v$ | ordinary mechanics: how hard they push, how fast they move |
| 2: chase | $P_i = \mathbf F^{\text{chase}}_i\cdot\mathbf v_i$ | chase strength $\kappa b\sin d$, distance $1/r^2$, alignment of $\mathbf v_i$ with the chase direction |
$\alpha_i$ does not set a rate of energy transfer. The chase already fixed that as $P_i$. $\alpha_i$ is the exchange rate: how many radians of hue it takes to pay $P_i$, given how steep the energy is in hue right now ($W_i$).
Where in hue-space does the payment come from?
The one genuinely non-obvious part. "The pair's colour energy" is not something $i$ can draw on separately. When $i$ moves its hue by $\delta h$, every pair potential involving $i$ changes:
$$ \delta E = \underbrace{\frac{\partial U_{ij}}{\partial h_i}}_{\text{bond + battery with } j}\delta h \;+\; \underbrace{\frac{\partial U_{ik}}{\partial h_i}}_{\text{bond + battery with } k}\delta h \;=\; W_i\,\delta h $$There is no choice about the split. $i$ moves along the single steepest-descent direction, and how much comes out of the $(i,j)$ battery versus the $(i,k)$ battery versus the two bonds is fixed by their partial derivatives. If $i$ and $j$ are nearly matched, $\partial U_{ij}/\partial h_i \approx 0$ and almost all of $W_i$ comes from $k$: the pair's own battery barely participates. $i$ is really trading against its mismatch with $k$.
A concrete chain: battery energy reaching $k$'s motion
- $k$'s hue differs from $i$'s, so the $(i,k)$ chase is nonzero. It pushes $i$ along $\hat{\mathbf u}_{ik}$. If $i$ is moving that way, $P_i > 0$: $i$'s KE rises.
- $i$ pays: $h_i$ moves downhill on $E$. That lowers the $(i,k)$ battery (and, if $j$ is not matched, the $(i,j)$ one). Battery → $i$'s KE.
- $i$, now faster, closes on $k$. The reciprocal repulsion (channel 1) converts $i$'s KE into potential and then into $k$'s KE as $k$ is shoved away.
Net: battery energy ended up in $k$'s motion, but through two accounting steps: chase (hue → own KE), then ordinary force (KE → KE via $U$). Meanwhile the same $(i,k)$ chase force is also acting on $k$; if $k$ is moving against it, $P_k < 0$ and $k$'s hue is charging from $k$'s own motion at the same time. Each ledger closes on its own:
$$ \dot E = \sum_{n\in\{i,j,k\}} \big(P_n - \alpha_n W_n^2\big) = 0 + 0 + 0 $$A constantly moving inert pair should be stable. A chase pair should settle into a symmetric pair moving at constant velocity. The question is how to recharge it: where does that energy come from?
Agreed, and the current model does not do it. It sloshes. But it is one change to the battery shape away from doing exactly what you describe, and that change is what makes the recharge question well posed.
What the current model does
Start a chase pair from rest at $d_0 = 0.8$ with the code's real envelopes and validated parameters. The chase accelerates it, the battery drains, $d$ passes through zero, and then the pair is moving against the (now reversed) chase, so it recharges from its own translation and turns round. Forever.
Why: near $d = 0$ the chase power $P \propto \Lambda \propto d$ and the battery gradient $W \propto \Lambda\Lambda' \propto d$ vanish at the same rate, so
$$ \dot d \;\propto\; -\frac{P}{W} \;\to\; \text{finite, nonzero, at } d = 0. $$The burn does not slow down as it reaches $d = 0$; it overshoots. Linearised: $\dot d = -C u$, $\dot u = K d$, a harmonic oscillator. The "dead pair" at $d = 0$ described in Question 2 was really the guard freezing $\alpha$ at exactly $W = 0$, a numerical artifact rather than a fixed point of the equations.
The fix: a battery with a finite gradient at $d = 0$
Replace $\lambda\Lambda^2 w$ with $\lambda|\Lambda|\,w$. Now $W$ is finite at $d = 0$ while $P$ still vanishes, so
$$ \dot d \;\propto\; -u\,d \qquad\Longrightarrow\qquad d(t) \sim d_0\,e^{-k u t}. $$The burn finishes exponentially and the pair coasts. Same start, same parameters:
The coasting state is an attractor, not merely a fixed point. Kick it against its own faint chase ($d_0 = 0.05$, $u_0 = -0.25$): the mismatch grows, the chase reverses the pair, and it burns back down and coasts the other way. The chase always accelerates the pair toward the side on which it burns, so $d\cdot u > 0$ is self-correcting.
| Battery shape | $W$ at $d\to0$ | $\dot d$ at $d\to0$ | Chase pair ends as |
|---|---|---|---|
| $\lambda\Lambda^2 w$ (current) | $\to 0$ | finite | sloshing, never coasts |
| $\lambda|\Lambda|\,w$ | finite | $\to 0$ | symmetric pair at constant $v$ |
| $\lambda\sqrt{\Lambda^2+\varepsilon^2}\,w$ | $\to 0$ over width $\varepsilon$ | finite inside $|d|\lesssim\varepsilon$ | coasts with residual slosh $\sim\varepsilon$ (smooth version) |
So where does the recharge come from?
A coasting symmetric pair has $\Lambda = 0$ with its partner, so its own KE is locked away: the chase cannot touch it. That is what makes the state stable, and it also means recharge must come from contact with something hue-different. There are exactly two sources.
Source 1: collisions (closed system)
The pair meets a particle $k$ with $h_k \neq h_i$. Now $\Lambda_{ik} \neq 0$, so there is a chase between $i$ and $k$. Wherever that chase brakes $i$ ($P_i < 0$), $i$'s hue climbs and the lost KE goes into $i$'s mismatch: mostly with $k$, but partly with $j$ too, because $W_i$ sums over both. After the encounter the pair is no longer symmetric and it burns again. The energy came from the KE of the encounter, whichever body was braked. The pair eats relative motion.
Source 2: an external feed (open system)
The code already has EnergySource: a region that feeds power $\phi$ straight into the colour balance. The net the battery must supply becomes $P_i - \phi$; where the feed wins this is negative, $\alpha$ flips, and hue climbs with no partner needed. That is "light". Pair it with damping $-\eta\mathbf v$ as the sink and you have the dissipative-structure loop:
Does there need to be a rate constant between KE transfer and hue transfer? If the pair meets a third particle, what decides between energy going into that particle's KE versus into hue change? Is it like regenerative braking, some linear constant based on $v$?
No rate constant, and yes, it is regenerative braking. The analogy is almost exact, and it shows why there is no free constant to choose.
The regen-braking dictionary
| Electric car | This model | Formula |
|---|---|---|
| motor / generator | the chase | $\mathbf F^{\text{chase}}_i = \kappa\Lambda(d)/r^2\;\hat{\mathbf u}$ |
| motor force on the wheels | chase force on particle $i$ | set by hue difference and distance, not a constant |
| electrical power in or out | chase power | $P_i = \mathbf F^{\text{chase}}_i\cdot\mathbf v_i$, linear in $v$ |
| battery | hue mismatch with neighbours | $\lambda|\Lambda|/r$ (or $\lambda\Lambda^2/r$) |
| state of charge | how far the hues disagree | $|d|$ |
| battery voltage (energy per unit charge) | energy per radian of hue | $W_i = \partial E/\partial h_i$ |
| current (charge per second) | hue rate | $\dot h_i = -P_i / W_i$ |
| driving vs braking | burn vs recharge | sign of $\mathbf F\cdot\mathbf v$ |
| round-trip efficiency | the conserve dial | $s$; at $s = 1$ it is lossless |
| the road, other cars, collisions | reciprocal forces | ordinary $-\partial U/\partial r$, never touch the battery directly |
In a car, motor power is force times velocity; there is no separate "rate constant" for how fast the battery charges during braking. The motor force and the speed fix the power, and the battery absorbs exactly that (times efficiency). Here it is the same: $P_i$ is fixed by the chase force and the velocity, the battery absorbs exactly $P_i$, and $\alpha_i$ is just power divided by voltage, i.e. current. Nothing is left to choose.
So what decides the KE-vs-hue split in an encounter?
Two things happen at once when the pair meets $k$, and each is completely determined on its own:
- The reciprocal forces (core, bond, battery gradient) do ordinary elastic scattering. KE moves between $i$, $j$, $k$ and the potential $U$ at rate $\mathbf F^{\text{rec}}\cdot\mathbf v$. This is the collision. It is exactly as in any conservative $N$-body problem and hue is not involved.
- The chase acts as a motor/generator on each particle separately. On $i$ it delivers $P_i = \mathbf F^{\text{chase}}_i\cdot\mathbf v_i$ and $i$'s hue pays it; on $k$ it delivers $P_k$ and $k$'s hue pays it. Linear in each particle's own velocity.
There is no dial that steers energy from channel 1 into channel 2 or back, because energy conservation already fixes both: channel 1 conserves on its own, and channel 2 is forced to conserve by the $\alpha$ solve. If you added a rate constant to channel 2 ("only pay 70% of $P$ from hue"), the other 30% would be created from nothing. That constant exists in the code and is called conserve; setting it below 1 is a deliberate leak, booked as unpaid.
What does change the proportions
| Knob | Effect on an encounter |
|---|---|
| $\kappa$ (and $b$ relative to $S_0$, $a$) | How big the motor is compared to the springs. Larger chase means more of the encounter's energy goes through hue, less is plain elastic scattering. This is the closest thing to "the constant" you are asking about, and it is a property of the force law, not a transfer coefficient. |
| Geometry | Whether each particle happens to move with or against its chase force during the encounter. Decides the sign of each $P$: who charges, who burns. |
| $\lambda$ | Does not change how much energy goes into hue. It changes how many radians that energy buys (battery voltage). Larger $\lambda$ means the same encounter shifts hue less. |
| $s$ | Efficiency. Below 1, energy leaks. |
With self-repelling hues ($a_1 < 0$), a same-hue pair at small radius is high in $U$. How does that work with the potential energy?
It works exactly like two like charges pushed together. High $U$ is stored energy, not a contradiction. What is worth seeing is that the pair has two ways down the hill, and they are paid for differently.
The landscape
The pair potential is a function of two coordinates, distance and hue difference: $U(r,d) = k(r_0-r)^2 + S(d)V(r) + \lambda\Lambda(d)^2 w(r)$. With $S_0 = 0.5$, $a_1 = -0.8$, $S(0) = -0.3$ and $S(\pi) = 1.3$:
Way 1: move apart (ordinary)
$\partial U/\partial r > 0$ on the ridge, so the reciprocal force pushes the pair apart. $U$ falls, KE rises by the same amount, channel 1, no hue involved. Two same-hue particles simply fly apart until they leave the cutoff. This is the whole story if $d = 0$ exactly, because then $\Lambda = 0$, the chase is off, and hue has no reason to move.
Way 2: change hue (paid through the chase)
At $d = 0$ the bond energy $S(d)V(r)$ is at its maximum in $d$ (because $S$ is at its minimum and $V < 0$). So $d = 0$ is a hilltop in hue-space as well as in $r$: any small mismatch wants to grow, sliding the pair toward complementary hues where the bond flips from repulsive to attractive and $U$ drops by $(S(\pi) - S(0))\,|V| = 1.6\,|V|$.
But hue cannot slide for free. It only moves at $\dot h = -P/W$, i.e. exactly as fast as the chase is doing work on the particle. So the bond energy released by the hue change is not "lost into the landscape": it is delivered, joule for joule, as chase work on the particle's motion. The pair rolls off the hue hilltop by accelerating. That is the twist: in ordinary mechanics a potential drop becomes KE via the force that is the gradient of that potential; here the drop in the hue direction becomes KE via a different force, the chase, and $\alpha$ is what makes the amounts match.
A pair flying straight apart does not change hue at all
Take the same-hue pair separating symmetrically: $\mathbf v_i = +u\hat{\mathbf u}$, $\mathbf v_j = -u\hat{\mathbf u}$, with a tiny $d$. The chase force is the same vector on both, so
$$ P_i = +\kappa\Lambda g\,u,\qquad P_j = -\kappa\Lambda g\,u,\qquad W_j = -W_i \quad\Longrightarrow\quad \dot h_i = -\frac{P_i}{W_i} = -\frac{P_j}{W_j} = \dot h_j . $$Both hues move the same way, $d$ is unchanged, and $P_i + P_j = 0$ so no net energy moves. One particle burns while the other recharges by the identical amount. A repelling pair just repels; it does not sort itself into complementary hues by flying apart. Sorting needs a net centre-of-mass velocity along the axis or a third body, same as in Question 1.
What changes globally
| $a_1 > 0$ (default) | $a_1 < 0$ (self-repelling) | |
|---|---|---|
| bound pairs | matching hues | complementary hues |
| "discharged" state ($\Lambda = 0$, lowest bond energy) | $d = 0$ | $d = \pi$ |
| $d = 0$ is | the well | a hilltop in both $r$ and $d$ |
| same-hue cluster | collapses to the core | spreads out, gas-like, as a negative plife diagonal does |
| energy conservation | identical; nothing in the derivation cares about the sign of $S$ | |
Is it the same idea where, if two particles with high repulsive energy are near each other, they could change colour to transfer that energy into colour instead of KE?
Yes, mostly, but never on its own. The released repulsive energy is split between the battery and KE in a fixed ratio, and the chase gates the rate. Q6 undersold this; here is the accounting.
Hue moves downhill on the sum, and the two terms move opposite ways
$W_i = \partial E/\partial h_i$ is the gradient of bond plus battery. Near $d = 0$ with $a_1 < 0$ (writing $s = 1 - r/R$):
$$ \frac{\partial U_{\text{bond}}}{\partial h_i} = -|a|\,s^2\,d, \qquad \frac{\partial U_{\text{batt}}}{\partial h_i} = +2\lambda b^2\,s^2\,d, \qquad W_i = \big(2\lambda b^2 - |a|\big)\,s^2\,d . $$The bond is a hilltop in $d$ (it wants $d$ to grow), the battery is a well (it wants $d$ to shrink). Which wins is the sign of $2\lambda b^2 - |a|$.
With the self-repelling defaults, the bond wins
$|a| = 0.8$, $2\lambda b^2 = 2 \times 0.55 \times 0.49 = 0.54$, so $W_i < 0$ for $d > 0$ and downhill means $d$ grows. When the chase is driving ($P_i > 0$), hue moves that way, and per unit of hue motion $\delta h$:
| Term | Change | Share of released bond energy |
|---|---|---|
| bond | $-0.80\,s^2 d\,\delta h$ | released |
| battery | $+0.54\,s^2 d\,\delta h$ | 67% into colour |
| KE (via chase, $= P\,dt$) | $+0.26\,s^2 d\,\delta h$ | 33% into motion |
So a close, mutually repulsive pair does spend its bond energy charging the battery, and it is the majority share. It walks off the repulsive hilltop toward complementary hues, storing most of the drop as mismatch and delivering the rest as chase thrust.
Why it cannot happen without the chase
The rate is $\dot h = -P/W$. With $d = 0$ exactly, or with the pair's velocity perpendicular to its axis, $P = 0$ and hue does not move: the spring stays compressed and the only exit is flying apart (way 1 of Q6). There is no direct bond → battery channel that bypasses the chase, because with one hue coordinate there is only one direction in hue-space, and moving along it always changes the sum bond + battery, and the sum can only change by exactly what the chase does.
(With two hue coordinates there is an energy-neutral direction, orthogonal to $\mathbf W$, along which bond and battery trade one-for-one at zero net cost. The main codebase's colourTarget: 'force' flow uses a direction of that kind. But it is free and rate-less, so it is not a transfer of KE either.)
The ratio is a switch
| Regime | $W$ near $d = 0$ | What a close same-hue pair does |
|---|---|---|
| $|a| > 2\lambda b^2$ | points toward growing $d$ | sorts toward complementary hues, charging the battery with most of its repulsive energy, thrusting with the rest |
| $|a| < 2\lambda b^2$ | points toward $d = 0$ | stays same-hue; $d = 0$ is a hue minimum of the sum; repulsive energy can only leave as KE by scattering |
A particle flies at another, they bond, and the energy goes into colour instead of KE. Is that possible?
Capture is possible, and it is the one thing an ordinary potential cannot do. But the energy flows the other way: colour pays for the bond. "Energy into colour" and "binding" are opposite directions, and there is a two-line proof.
Why two bodies cannot capture each other in ordinary mechanics
Split the motion into centre of mass and relative. With a conservative pair force the relative energy $E_{\text{rel}} = \tfrac12\mu v_{\text{rel}}^2 + U(r)$ is constant. A particle that came in from infinity with $E_{\text{rel}} > 0$ still has $E_{\text{rel}} > 0$ after the encounter, so it climbs back out. You need a third body or friction.
What the chase can and cannot touch
The chase force is the same vector on both particles. So it accelerates the centre of mass and does nothing to the relative motion. The only way $E_{\text{rel}}$ can change is through the hue-dependence of $U(r, d)$: if the hues change while the pair is inside the well, the well changes depth under them. And the energy balance says exactly how much:
$$ \frac{\mathrm dE_{\text{rel}}}{\mathrm dt} = -\frac{\mathrm d\,\text{KE}_{\text{cm}}}{\mathrm dt} = -\,\mathbf F^{\text{chase}}\cdot\mathbf v_{\text{cm}} = -\tfrac12\,(P_i + P_j) $$The relative energy drops precisely when the chase does positive work on the centre of mass, and that is a burn: the colour store gives up energy, the bond deepens under the pair, and the released energy shows up as centre-of-mass motion. Recharge does the opposite: braking the centre of mass raises $E_{\text{rel}}$ and loosens the bond.
| Burn ($P > 0$) | Recharge ($P < 0$) | |
|---|---|---|
| colour store | drains | fills |
| centre of mass | speeds up | slows down |
| $E_{\text{rel}}$ | drops: binding | rises: unbinding |
Measured, in the lab
Tight well ($\rho = 0.15$, $\sigma = 0.06$), $|\Lambda|$ battery, chase and battery envelopes following the well. A particle approaches a target at rest at $u_0 = 0.2$.
| $d_0$ | Outcome | $d_{\text{end}}$ | KE$_{\text{cm}}$ before → after | What happened |
|---|---|---|---|---|
| $-0.8$ (burn side) | captured, $r \approx 0.03$–$0.15$ | $0.00$ | $0.010 \to 0.100$ | mismatch burned inside the well, bond deepened from $S = 1.65$ to $1.8$, pair leaves bonded and moving |
| $+0.8$ (recharge side) | escaped | $0.96$ | $0.010 \to 0.000$ | projectile braked to a stop, its KE went into mismatch, but the shallower potential let the pair separate again |
The second row is exactly "energy into colour", and it is exactly why the pair did not bond. Both are scenarios in the lab ("Capture" and "Braked approach").
Three things that had to be true for capture to work at all
- The chase must not outrange the well. With the linear envelope $g = 1 - r/R$ the chase burned the mismatch at $r \approx 0.7$, before the bond could deepen under anything, and the projectile just picked up speed and sailed through. With $g$ following the well, the burn happens where it counts.
- The battery envelope must not outrange the well either. $\lambda|\Lambda|w$ is repulsive, and with $w = (1 - r/R)^2$ a slow projectile bounced off the mismatch before reaching the well. Storing mismatch energy only where the bond is makes it genuinely relational.
- The $|\Lambda|$ battery. With $\Lambda^2$ the captured pair overshoots $d = 0$, recharges on the other side, $E_{\text{rel}}$ goes back up, and it escapes: the capture is undone by the slosh. With $|\Lambda|$ the burn stops at $d = 0$ and the capture sticks.