Energy, momentum, and mass

From a stored state to an exchange

Energy is an amount a system can carry and exchange. Momentum also keeps track of direction. A right-moving pulse and a left-moving pulse both carry positive energy, while their momenta have opposite signs.

We begin in one space dimension and choose units where the speed of light is $c=1$. Energy, momentum, and mass can then be measured in compatible units.

Our notebook holds an accumulated state: ${\tau }_{\mathrm{E}}=\alpha +\beta$ and $x_{\mathrm{E}}=\alpha -\beta$. An exchange makes a change in that state. The symbol $\mathrm{\Delta }$ means the amount added or removed: $\mathrm{\Delta }{\tau }_{\mathrm{E}}=\mathrm{\Delta }\alpha +\mathrm{\Delta }\beta$ and $\mathrm{\Delta }{x}_{\mathrm{E}}=\mathrm{\Delta }\alpha -\mathrm{\Delta }\beta$.

The energy-momentum cone describes this exchange. Choose a positive conversion factor $\epsilon$ (epsilon), the energy represented by one unit of trial weight. A received packet carries energy $E_{\mathrm{packet}}$ and momentum $p_{\mathrm{packet}}$ proportional to those increments. Absorbing it changes the receiver's energy and momentum by exactly those amounts.

This gives us two drawings to keep side by side: a point for the stored belief and an arrow for what arrives. The arrow starts at the old point and ends at the new one.

Mass is the part that survives a change of frame

A reference frame is an observer's system of clocks and rulers. A body can have different energies and momenta for observers moving relative to one another. Its rest mass stays the same.

For a received packet, the square root of the energy-momentum cone norm gives its invariant mass. Positive contributions in both directions put the total inside the cone. A single isolated light pulse lies on its boundary and has zero mass.

Try a packet with $\mathrm{\Delta }\alpha =4$ and $\mathrm{\Delta }\beta =1$, choosing $\epsilon =1$. Its energy is 5, momentum is 3, and invariant mass is 4. Its center-of-energy velocity is $p_{\mathrm{packet}}/E_{\mathrm{packet}}=3/5=0.6$. If all that energy is retained in one composite body, this is the body's velocity.

The accumulated belief has its own cone norm, $m_{\mathrm{E}}^2={\tau }_{\mathrm{E}}^2-x_{\mathrm{E}}^2$. The packet uses the increments instead. An old notebook and its next incoming packet can therefore have very different masses and directions, while both follow the same difference-of-squares geometry.

Rest energy and the energy of motion

The Lorentz factor $\gamma$ measures how much total energy a massive body carries relative to its rest energy. In our units, $E=\gamma m$ and $p=\gamma mv$. Kinetic energy, written $K$, is the extra energy associated with motion: $K=E-m$.

For the next few examples, write $E$, $p$, and $m$ for the physical energy, momentum, and mass of the body or packet. Our 3-4-5 packet has $\gamma =5/4$ and $K=1$. Change the reference frame to move with the body: its momentum becomes zero and its energy becomes 4. That remaining amount is its rest energy.

There are two different experiments here. Holding mass fixed and changing frame increases energy as the speed grows. Holding total energy fixed and redistributing it between the two channels changes the system's mass. The collision game lets you compare energy, momentum, and the mass of the whole system.

In physical units these familiar relations read $E=\gamma m{c}^2$, $p=\gamma mv$, and $K=E-m{c}^2$. The evidence dictionary acquires physical units only when we choose what carries a trial and how much energy each unit represents.

A collision keeps two totals

For a closed collision, add the incoming energies and add the signed momenta. The outgoing arrangement must retain both totals. Closed means we include every body or pulse carrying energy away.

Let $E_{\mathrm{tot}}$ and $p_{\mathrm{tot}}$ denote these sums. The system mass $M$ is the invariant mass of the whole collection. It includes the energy of its constituents' relative motion, so it generally exceeds the sum of their individual rest masses.

Two equal opposing light pulses make a striking example. Give each energy 1 and momenta $+1$ and $-1$. Each pulse has zero rest mass; together they have $E_{\mathrm{tot}}=2$, $p_{\mathrm{tot}}=0$, and $M=2$. The collection already has this invariant mass before anything merges. If an interaction binds all that energy into one body, the body's rest mass is 2.

Merge or bounce?

The game supplies two interaction rules. When the pucks stick together, the incoming bodies form one composite and all their energy remains inside the system. Motion relative to the center becomes internal energy: perhaps vibration, heat, or the motion of trapped constituents. The composite can therefore have more rest mass than the incoming rest masses added together.

Set both incoming pucks to mass 1 and speed 0.6. Their energies are 1.25 apiece. The stationary merged body has rest mass 2.5, including 0.5 of retained internal energy. The visible motion has become part of what the body carries at rest.

In an elastic bounce, each body's rest mass stays fixed. We move to the center-of-momentum frame, where the total momentum is zero, reverse the two momenta, then return to the original frame. This exchanges motion while preserving both totals.

Predict what happens to the joined puck when the two incoming momenta cancel. Then switch to a bounce and launch again. The same energy and momentum totals can produce a different outgoing arrangement. Conservation constrains the possible outcomes; the interaction rule chooses which one happens.

The arriving packet and the belief it changes

The packet's direction is its own ratio, $\mathrm{\Delta }{x}_{\mathrm{E}}/\mathrm{\Delta }{\tau }_{\mathrm{E}}$. The belief's centered mean depends on everything already in the notebook as well as the new arrival. Add the increments to the existing coordinates before taking the new ratio.

For example, begin with the symmetric Jeffreys prior: $\alpha =\beta =1/2$, so ${\tau }_{\mathrm{E}}=1$ and $x_{\mathrm{E}}=0$. Now receive our four-support, one-opposition packet. Its direction is $3/5$. The updated notebook has ${\tau }_{\mathrm{E}}^{\prime }=6$ and $x_{\mathrm{E}}^{\prime }=3$, giving centered mean $v_{\mathrm{E}}^{\prime }=1/2$ and ordinary mean probability ${\mu }^{\prime }=3/4$.

The game uses physical packet totals for its scorecard. The prior belongs to the belief's starting state and contributes no energy to a physical collision. If we later describe the same belief ledger from another frame, its original prior travels through that geometric transformation too.

Combining two records about one latent question retains their shared prior once. Keeping the accumulated state, the incoming change, and the prior visible lets us use the connection without losing what each quantity means.