Annus Mirabilis · Interactive critical edition in preparation
Opposite pulses and two energy ledgers
Describe the same balanced emission in two frames.
An executable model
Two ledgers and opposite pulses
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CurrentThese numbers match the current settings.
Model note
- Primary output lightComplexEnergyMoving: Host calculation (massEnergy.movingBalanceLight). Owner massEnergy.movingBalanceLight.
- Primary output emittedEnergyRestFrame: Host calculation (massEnergy.restBalanceLight). Owner massEnergy.restBalanceLight.
- Primary output bodyEnergyRestBefore: Host calculation (massEnergy.restBodyBefore). Owner massEnergy.restBodyBefore.
- Primary output bodyEnergyRestAfter: Host calculation (massEnergy.restBodyAfter). Owner massEnergy.restBodyAfter.
- Primary output bodyEnergyMovingBefore: Host calculation (massEnergy.movingBodyBefore). Owner massEnergy.movingBodyBefore.
- Primary output bodyEnergyMovingAfter: Host calculation (massEnergy.movingBodyAfter). Owner massEnergy.movingBodyAfter.
- Primary output kineticEnergyDifference: Host calculation (massEnergy.kineticEnergyDifference). Owner massEnergy.kineticEnergyDifference.
- Primary output additiveEnergyConstant: Host calculation (massEnergy.additiveEnergyConstant). Owner massEnergy.additiveEnergyConstant.
- Accepted input revision 1.
- Snapshot version 1.
- Not modeled: recoil from asymmetric emission; finite pulse duration and shape; the emission mechanism; radiation pressure on the body during emission; gravity; the quantum nature of light; the body's absolute rest energy (kept symbolic).
Predict before the numbers
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 result appears when you choose, say you have one in mind, or skip.
Experiment settings emitted energy, the premise, how internal energy is written, notation, a link to these settings
Worked example: a body emits 1 L of light in its rest frame; seen from a frame moving at 0.6c the two pulses carry 1.25 L together, and the body's kinetic energy falls by 0.25 L.
Seen from a frame moving at v = 0.6c
The body sends out two equal pulses in opposite directions, at φ = 0° to the direction of motion.
The two energy accounts
- In the body's rest frame (K₀)
- Light sent out: L/2 + L/2 = 1.0000 L
- E₀ − E₁ = L
- In the moving frame (k, speed v)
- Light sent out: 1.25 L (= 1/√(1 − v²/V²)·L)
- H₀ − H₁ = 1/√(1 − v²/V²)·L
Subtracting one account from the other
(H₀ − E₀) − (H₁ − E₁) = L(1/√(1 − v²/V²) − 1)
Drop in energy of motion: 0.25 L
The body's internal energies at rest cancel.
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;Two equal flashes of light leave a body at rest in opposite directions. Seen by someone moving past, the flashes carry more energy than they do for the body, and the extra can only come from the body's energy of motion, which falls as if the body had lost mass.
The mass–energy paper imports one result from §8 of the relativity paper: light of energy l, seen from a frame moving at v along x, has energy l* = l(1 − (v/V) cos φ)/√(1 − (v/V)²), where φ is the angle between the light's direction and the x-axis. A body at rest in (x, y, z) sends out light of energy L/2 at the angle φ and an equal amount the opposite way, and stays at rest. The energy principle must hold in both frames: in the rest frame E₀ − E₁ = L, and in the moving one H₀ − H₁ = γL, writing γ for 1/√(1 − (v/V)²), because the two direction factors, 1 − (v/V) cos φ and 1 + (v/V) cos φ, add to 2. The body's energies E and H stay symbolic, and the lab never gives them values. Subtracting the two balances removes them: (H₀ − E₀) − (H₁ − E₁) = L{1/√(1 − (v/V)²) − 1}. Each H − E is the body's kinetic energy K in the moving frame plus a constant C, and C does not change during the emission, so K₀ − K₁ = L{1/√(1 − (v/V)²) − 1}. At the lab's default, L = 1 J and v = 0.6c, the moving observer counts 1.25 J of light, and the body's kinetic energy falls by 0.25 J whatever the angle. Drop the premise that C is unchanged and the difference is underdetermined, and the lab says so. To lowest order the drop is (L/V²)(v²/2), the kinetic energy of a mass L/V², and the paper concludes that a body giving off energy L as radiation loses mass L/V².
Start with the factor 1/√(1 − (v/V)²). At v = 0.6c it is 1/√0.64 = 1.25. Take L = 1 J, so each flash carries 0.5 J in the body's frame. Along the x-axis, φ = 0: the forward flash, going the same way as the moving observer, has 0.5 × 1.25 × (1 − 0.6) = 0.25 J in the moving frame, and the backward flash has 0.5 × 1.25 × (1 + 0.6) = 1.0 J, together 1.25 J. At φ = 60° the factors are 1 − 0.3 and 1 + 0.3, giving 0.4375 J and 0.8125 J, again 1.25 J: the angle cancels because the two factors always sum to 2, the flashes going in opposite directions. Now write both energy balances. In the body's frame the body loses exactly what the light carries, E₀ − E₁ = 1 J. In the moving frame it loses what the light carries there, H₀ − H₁ = 1.25 J. Subtract the first from the second: (H₀ − E₀) − (H₁ − E₁) = 1.25 − 1 = 0.25 J. Nothing about the body's internal energy was needed, because E₀, E₁, H₀ and H₁ appear only in differences. What is H − E? It is the same body's energy seen from two frames, one in which it rests and one in which it moves, so it is its kinetic energy K in the moving frame plus a constant C that depends only on where each frame puts its zero of energy. With C unchanged, the constants cancel, and K₀ − K₁ = L(γ − 1) = 0.25 J. If C could change during the emission, the two constants would not cancel, and the subtraction would leave K₀ − K₁ tied to an unknown change in C; that is why the lab reports the result as underdetermined when the premise is dropped. The body moves at the same speed before and after, since it stays at rest in its own frame, so a smaller kinetic energy at the same speed means a smaller mass. For slow motion, 1/√(1 − (v/V)²) − 1 ≈ (v/V)²/2, so K₀ − K₁ ≈ (L/V²)(v²/2), which is ½mv² with m = L/V². At 0.6c the approximation gives 0.18 J against the exact 0.25 J. For L = 1 J the mass lost is 1/(2.998 × 108)² = 1.11 × 10−17 kg.
The paper, received on 27 September 1905, writes the factor in full as 1/√(1 − v²/V²), V for the speed of light, L for the emitted energy and l* for the transformed light energy. It cites §8 of the June paper for that transformation and notes in a footnote that the constancy of the speed of light used there is contained in Maxwell's equations. It states the result as L/V², and in cgs units as a mass change of L/9 · 1020 grams for L ergs, and suggests that radium salts might test it. The mass statement rests on the lowest-order term in v/V and on the source premise that C is unchanged. Planck gave a more general treatment in 1907, and Ives argued in 1952 that the 1905 argument was circular, a reading others have disputed. The paper itself says only that the mass of a body is a measure of its energy content.
The explanation
Full explanation
Change the emission angle and compare the individual pulse energies with their sum. Then inspect the subtraction of the two energy balances and the role of the offset premise.
Show every step of the investigation
Follow the moving pulses, their angle-free sum, both balances, the subtraction, and the kinetic-energy interpretation. The light-energy transformation is an imported premise; a mass-energy formula is not installed as the starting answer.
An explanatory model, not an observation of nature. This embed starts from the laboratory’s worked defaults, not a saved run. Presentation options change the surrounding guide, never the numerical inputs.