Mass–Energy · The Two-Ledger Derivation

Opposite pulses
and two energy ledgers.

If a body at rest emits two equal pulses of light in opposite directions, what do the energy accounting books of two different inertial observers force you to conclude about the body's energy of motion?

Read the two-ledger argument (explanatory preview) →

Ideal model, host calculation

ME-01: Two ledgers and opposite pulses

Two equal flashes sent opposite ways: someone moving past measures a larger total, and comparing the two accounts shows the body's energy of motion must drop.

Notation:

Presets:

Predict before changing the angle:

If the body emits the two opposite pulses at an angle φ = 60° rather than along the direction of motion (φ = 0°), what happens to the total energy of the two light pulses measured by the moving observer?

The total light energy in the moving frame is unchanged (still γL), because the angle terms cancel in the sum.The total light energy decreases because the forward pulse is less blue-shifted.The total light energy increases due to the transverse Doppler effect.

Show the reference code & kernel bindings

Reference evaluator: src/physics/reference/massEnergy.ts

// evaluatePulseEnergies
const g = 1 / Math.sqrt(1 - frameSpeed * frameSpeed);
const p1 = (emittedEnergyRestFrame / 2) * g * (1 - frameSpeed * Math.cos(phi));
const p2 = (emittedEnergyRestFrame / 2) * g * (1 + frameSpeed * Math.cos(phi));
const pulseSumMoving = g * emittedEnergyRestFrame; // invariant under phi!

// evaluateSubtraction
const subtractionDifference = emittedEnergyRestFrame * (g - 1);
const kineticEnergyDifference = premise === "unchanged" ? subtractionDifference : null;

Open the derivation

The imported light-energy transformation

Einstein imports a result proven in §8 of his third 1905 paper (Special Relativity). When light of energy l is emitted in the stationary system at an angle φ to the direction of relative motion, an observer moving past at speed v measures its energy l* as:

l^* = l \frac{1 - \frac{v}{V}\cos\varphi}{\sqrt{1 - \left(\frac{v}{V}\right)^2}}

In modern notation with β = v/c and the Lorentz factor γ = 1/√(1 - β²):

l^* = l\,\gamma\,(1 - \beta\cos\varphi)

The two accounting sheets: Rest frame and moving frame

Let the body at rest have initial internal energy E₀. It emits two equal light pulses of energy L/2 in opposite directions (φ and φ + 180°). Conservation of energy in the stationary frame requires:

E_0 = E_1 + \frac{1}{2}L + \frac{1}{2}L = E_1 + L \implies E_0 - E_1 = L

Now consider the same physical event as measured by an observer moving at speed v. The initial energy of the body in this frame is H₀. The two pulses have energies:

\text{Pulse 1} = \frac{1}{2}L\,\gamma\,(1 - \beta\cos\varphi), \qquad \text{Pulse 2} = \frac{1}{2}L\,\gamma\,(1 + \beta\cos\varphi)

When the two pulse energies are added together, the angle terms −β cos φ and +β cos φ cancel identically:

\text{Total moving light} = \frac{1}{2}L\,\gamma\,(1 - \beta\cos\varphi) + \frac{1}{2}L\,\gamma\,(1 + \beta\cos\varphi) = \gamma L

Energy conservation in the moving frame therefore gives:

H_0 = H_1 + \gamma L \implies H_0 - H_1 = \gamma L

The Subtraction Move: Eliminating the unknown internal energies

Neither E₀ nor H₀ is known. But subtracting the stationary-system balance from the moving-system balance completely eliminates the body's unknown internal rest energy:

(H_0 - E_0) - (H_1 - E_1) = \gamma L - L = L\,(\gamma - 1)

Identifying the kinetic energy drop

The difference between a body's energy in a moving system and its energy in the rest system differs from its kinetic energy K only by an additive constant C:

H - E = K + C

Under Einstein's source premise that the constant C does not alter upon the emission of light (C = C'), substituting this relation yields:

(K_0 + C) - (K_1 + C) = K_0 - K_1 = L\,(\gamma - 1)

The body's energy of motion drops by L(γ − 1) while its speed remains unchanged.

Explore the small-speed coefficient (ME-02) →Start with two concrete energy accounts