Mass–energy · An explanatory investigation

Can you reach the conclusion without assuming it?

A body emits equal flashes in opposite directions and does not recoil. Two observers assign different energies to those flashes. Build the argument that connects their accounts to a change in the body’s inertia, keeping every extra premise visible.

A route you could take: one argument that is enough to reach the result, not an account of Einstein’s private thinking. Editorial and physics review remain pending. The symbols are modern: c, β = v/c and γ. The September paper writes V for the speed of light and the square root out in full.

What is being admitted?

Energy is conserved in each inertial frame. The symmetric emission keeps the body’s speed unchanged. A light-energy transformation is imported from §8 of the special-relativity paper; it is not smuggled into a supposed 1904 starting point. No value is assigned to the body’s absolute internal energy.

Use subscripts 0 and 1 for before and after emission. E is body energy in the rest frame, H is body energy in the moving frame, K is kinetic energy, and L is the total light energy emitted in the rest frame. C denotes the additive offset in the stated identification H − E = K + C. Whether that offset changes is a premise to inspect, not an algebraic detail.

InstrumentInspect the imported finite-light-complex transformation

InstrumentRead the two-ledger equations and their source context

Build an argument, then change a premise

Add cards in the order you would use them. Each step names what it needs; move an earlier premise into place or keep a weaker conclusion. Several orders are valid. This checks the dependencies of these authored steps, not arbitrary mathematics or the truth of a premise.

No score, timer or answer gate is used. The worked derivation and every explanation remain available whether or not you assemble a chain.

What follows from the current order?

The selected order does not yet reach a two-ledger conclusion. A blocked step cannot supply a premise to later steps. You may reorder cards or read the worked route; there is no score or locked content.

Your selected chain

No cards selected. Start with a setup or a premise, or load either route above.

Available argument cards

You can add a card before its prerequisites to see precisely what is missing. Selecting a premise admits it; it does not prove it.

Choose an emission without recoil
Role: setup.
$L/2+L/2=L$
Two equal pulses leave in opposite directions in the body's rest frame. Their momenta cancel. Compare the same body before and after; its speed in either inertial frame is unchanged. Asymmetric emission needs a recoil model and is outside this reconstruction.
Requires: No earlier card; this is an explicitly admitted starting point.
Keep an energy account in each frame
Role: premise.
$\begin{aligned}&\text{energy before}=\text{energy after}\\&\qquad+\text{emitted light energy}\end{aligned}$
Assume energy conservation separately in each inertial frame. Do not equate the numerical energies assigned by different observers. No value for the body's absolute internal energy is assumed.
Requires: No earlier card; this is an explicitly admitted starting point.
Import the light-energy transformation
Role: import.
$\begin{gathered}l^*=l\gamma(1-\beta\cos\varphi),\\ \gamma=(1-\beta^2)^{-1/2},\quad\beta=v/c\end{gathered}$
This is an admitted result of the special-relativity paper, §8, not knowledge supplied by a 1904 shelf and not a consequence of the mass–energy relation being sought. Here c and gamma are modern notation; the September paper uses V and an explicit radical. The observer must satisfy |v| < c.
Requires: No earlier card; this is an explicitly admitted starting point.
Write the rest-frame balance
Role: algebra.
$E_0-E_1=L$
The body loses energy L in its rest frame. E₀ and E₁ remain unknown symbols. This balance by itself makes no statement about the body's mass.
Requires: Choose an emission without recoil; Keep an energy account in each frame
Add the two transformed pulses
Role: algebra.
$\begin{aligned}&\frac{L\gamma}{2}(1-\beta\cos\varphi)+\frac{L\gamma}{2}(1+\beta\cos\varphi)\\&\quad=\gamma L\end{aligned}$
Opposite directions give opposite cosine terms. The individual pulse energies depend on angle, but their sum does not. This cancellation does not say that the two pulses have equal energies for a moving observer.
Requires: Choose an emission without recoil; Import the light-energy transformation
Write the moving-frame balance
Role: algebra.
$H_0-H_1=\gamma L$
The moving observer's account loses gamma L in light. H₀ and H₁, like the rest-frame body energies, stay symbolic. A body-energy formula of the form gamma M c² has not entered the calculation.
Requires: Add the two transformed pulses; Keep an energy account in each frame
Subtract the two balances
Role: algebra.
$(H_0-E_0)-(H_1-E_1)=L(\gamma-1)$
Subtract the rest-frame balance from the moving-frame balance. This determines a difference of energy differences without assigning either absolute internal energy. It has not yet identified that difference as a kinetic-energy loss.
Requires: Write the rest-frame balance; Write the moving-frame balance
State how frame-energy differences relate to motion
Role: premise.
$H_i-E_i=K_i+C_i\quad(i=0,1)$
Admit the paper's premise relating frame-energy differences to kinetic energy, with a possible additive offset before and after emission. This is a physical identification, not a consequence of subtracting two equations.
Requires: No earlier card; this is an explicitly admitted starting point.
Keep the possible offset change visible
Role: algebra.
$K_0-K_1+(C_0-C_1)=L(\gamma-1)$
A valid, weaker result. If C₀ − C₁ is unspecified, the ledgers determine this combination, not K₀ − K₁ on its own. More algebra cannot determine an unconstrained offset.
Requires: Subtract the two balances; State how frame-energy differences relate to motion
Admit the unchanged-offset premise
Role: premise.
$C_0=C_1$
Einstein states that the additive constant is unchanged by the emission. Make that premise explicit. Selecting it does not prove it, and a simulator programmed with it is not independent evidence for it.
Requires: State how frame-energy differences relate to motion
Identify the kinetic-energy decrease
Role: algebra.
$K_0-K_1=L(\gamma-1)$
With the offset change removed, the body has less kinetic energy at the same speed. This relation is exact within the admitted model, for |v| < c. At v = 0 both sides are zero; that single evaluation does not identify a mass.
Requires: Keep the possible offset change visible; Admit the unchanged-offset premise
Define inertia by the small-speed coefficient
Role: premise.
$\lim_{v\to0}\frac{K_0-K_1}{v^2/2}=M_0-M_1$
Use the Newtonian leading coefficient of kinetic energy to identify inertial mass. This is a limit across small nonzero speeds, not division by v² at exactly zero, and it does not assume a rest-energy formula.
Requires: No earlier card; this is an explicitly admitted starting point.
Take the limit, not a finite-speed shortcut
Role: approximation.
$\gamma-1=\tfrac12(v/c)^2+O((v/c)^4)$
Expand the imported Lorentz factor at small |v|/c. The leading term is an approximation at nonzero speed; the coefficient obtained as v tends to zero is a limit. At 0.6c the leading term must not be presented as the exact kinetic-energy difference.
Requires: Identify the kinetic-energy decrease; Import the light-energy transformation
Read off the inertia lost
Role: conclusion.
$M_0-M_1=\frac{L}{c^2},\qquad\Delta M=-\frac{L}{c^2}$
Comparing the limiting coefficients gives the positive mass lost and the signed mass change. The conclusion follows conditionally on the listed premises for symmetric emission. It does not assign a numerical value to either absolute body energy.
Requires: Take the limit, not a finite-speed shortcut; Define inertia by the small-speed coefficient
Alternative: start by assuming rest energy equals M c²
Role: Assumption of the target relation.
$E_0=M_0c^2,\qquad E_1=M_1c^2$
A useful modern starting point for checking consistency, but it already assumes the mass–energy relation this exercise asks you to establish. A conclusion depending on this card is labeled a consistency check, never an independent derivation.
Requires: No earlier card; this is an explicitly admitted starting point.
Check the rest balance using that assumption
Role: conclusion.
$(M_0-M_1)c^2=L$
This is valid substitution under the declared rest-energy assumption. It agrees with the sought relation, but agreement after assuming that relation cannot establish it independently. This branch is not described as a physical contradiction.
Requires: Write the rest-frame balance; Alternative: start by assuming rest energy equals M c²

Predict, change something, explain

What survives when the offset premise is removed? Why is assuming E = M c² a different kind of argument?

Your note stays in this tab and disappears on reload unless you download it. It is not automatically graded, uploaded, stored locally or included in a share link.

A worked route, available without assembling cards

The first useful subtraction determines a difference of frame-energy differences. The unchanged-offset premise then makes it a kinetic-energy decrease. The small-speed coefficient, not a finite-speed division, identifies the inertia lost.

Show every step and its reason

Choose an emission without recoil
$L/2+L/2=L$
Two equal pulses leave in opposite directions in the body's rest frame. Their momenta cancel. Compare the same body before and after; its speed in either inertial frame is unchanged. Asymmetric emission needs a recoil model and is outside this reconstruction.
Keep an energy account in each frame
$\begin{aligned}&\text{energy before}=\text{energy after}\\&\qquad+\text{emitted light energy}\end{aligned}$
Assume energy conservation separately in each inertial frame. Do not equate the numerical energies assigned by different observers. No value for the body's absolute internal energy is assumed.
Import the light-energy transformation
$\begin{gathered}l^*=l\gamma(1-\beta\cos\varphi),\\ \gamma=(1-\beta^2)^{-1/2},\quad\beta=v/c\end{gathered}$
This is an admitted result of the special-relativity paper, §8, not knowledge supplied by a 1904 shelf and not a consequence of the mass–energy relation being sought. Here c and gamma are modern notation; the September paper uses V and an explicit radical. The observer must satisfy |v| < c.
Write the rest-frame balance
$E_0-E_1=L$
The body loses energy L in its rest frame. E₀ and E₁ remain unknown symbols. This balance by itself makes no statement about the body's mass.
Add the two transformed pulses
$\begin{aligned}&\frac{L\gamma}{2}(1-\beta\cos\varphi)+\frac{L\gamma}{2}(1+\beta\cos\varphi)\\&\quad=\gamma L\end{aligned}$
Opposite directions give opposite cosine terms. The individual pulse energies depend on angle, but their sum does not. This cancellation does not say that the two pulses have equal energies for a moving observer.
Write the moving-frame balance
$H_0-H_1=\gamma L$
The moving observer's account loses gamma L in light. H₀ and H₁, like the rest-frame body energies, stay symbolic. A body-energy formula of the form gamma M c² has not entered the calculation.
Subtract the two balances
$(H_0-E_0)-(H_1-E_1)=L(\gamma-1)$
Subtract the rest-frame balance from the moving-frame balance. This determines a difference of energy differences without assigning either absolute internal energy. It has not yet identified that difference as a kinetic-energy loss.
State how frame-energy differences relate to motion
$H_i-E_i=K_i+C_i\quad(i=0,1)$
Admit the paper's premise relating frame-energy differences to kinetic energy, with a possible additive offset before and after emission. This is a physical identification, not a consequence of subtracting two equations.
Keep the possible offset change visible
$K_0-K_1+(C_0-C_1)=L(\gamma-1)$
A valid, weaker result. If C₀ − C₁ is unspecified, the ledgers determine this combination, not K₀ − K₁ on its own. More algebra cannot determine an unconstrained offset.
Admit the unchanged-offset premise
$C_0=C_1$
Einstein states that the additive constant is unchanged by the emission. Make that premise explicit. Selecting it does not prove it, and a simulator programmed with it is not independent evidence for it.
Identify the kinetic-energy decrease
$K_0-K_1=L(\gamma-1)$
With the offset change removed, the body has less kinetic energy at the same speed. This relation is exact within the admitted model, for |v| < c. At v = 0 both sides are zero; that single evaluation does not identify a mass.
Define inertia by the small-speed coefficient
$\lim_{v\to0}\frac{K_0-K_1}{v^2/2}=M_0-M_1$
Use the Newtonian leading coefficient of kinetic energy to identify inertial mass. This is a limit across small nonzero speeds, not division by v² at exactly zero, and it does not assume a rest-energy formula.
Take the limit, not a finite-speed shortcut
$\gamma-1=\tfrac12(v/c)^2+O((v/c)^4)$
Expand the imported Lorentz factor at small |v|/c. The leading term is an approximation at nonzero speed; the coefficient obtained as v tends to zero is a limit. At 0.6c the leading term must not be presented as the exact kinetic-energy difference.
Read off the inertia lost
$M_0-M_1=\frac{L}{c^2},\qquad\Delta M=-\frac{L}{c^2}$
Comparing the limiting coefficients gives the positive mass lost and the signed mass change. The conclusion follows conditionally on the listed premises for symmetric emission. It does not assign a numerical value to either absolute body energy.

Three changes that test your explanation

Remove “Admit the unchanged-offset premise”. What survives?

The two balances and their subtraction survive. The combination K₀ − K₁ + (C₀ − C₁) is constrained, but K₀ − K₁ is not separately identified. An unspecified offset is not zero. With the premise relaxed, the laboratory below gives no number for the kinetic-energy loss, because the argument no longer fixes one.

Keep a finite observer speed. Is twice the kinetic-energy drop divided by v² the exact mass loss?

No. It is a finite-speed coefficient proxy. Identifying the mass loss uses its limit as v approaches zero. A zero-speed run alone gives zero kinetic-energy difference; dividing that result by zero is not the limit calculation.

Compare the proxy, the approximation and the limit

Include the emitted light inside your system boundary. What changed?

The emitting body and the body-plus-radiation system are different systems. Energy that has left the body may remain inside a larger closed boundary. Do not transfer a statement about energy lost by the body to that larger system without a new account. Unequal pulses would also require a recoil calculation absent from this reconstruction.

Move the system boundary

Test consequences with the existing two-ledger instrument

Predict whether tilting the emission axis changes each pulse energy, their sum, or both. Then vary the observer and relax the offset premise. These numbers are computed by the two-ledgers instrument, not by the argument checker above, and a computed consequence cannot check its own premises.

An executable model

Two ledgers and opposite pulses

Static worked example

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).

JavaScript is off. This is a complete worked example calculated when the site was built, and its values and explanations remain available. The controls need JavaScript to respond.

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?

Three relations the model could have The total light energy decreases because the forward pulse is less blue-shifted. The total light energy increases due to the transverse Doppler effect. The total light energy is unchanged, because what one pulse gains the other loses.

The result appears when you choose, say you have one in mind, or skip.

Observer speed v/c (signed fraction of c)

Emission angle φ (degrees from the direction of motion)

Try

Experiment settings emitted energy, the premise, how internal energy is written, notation, a link to these settings

Emitted energy L (in the body's rest frame)

The additive constant C

Internal energies written as

Notation

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.

What is being admitted?

Build an argument, then change a premise

What follows from the current order?

Your selected chain

Available argument cards

Predict, change something, explain

A worked route, available without assembling cards

Choose an emission without recoil

Keep an energy account in each frame

Import the light-energy transformation

Write the rest-frame balance

Add the two transformed pulses

Write the moving-frame balance

Subtract the two balances

State how frame-energy differences relate to motion

Keep the possible offset change visible

Admit the unchanged-offset premise

Identify the kinetic-energy decrease

Define inertia by the small-speed coefficient

Take the limit, not a finite-speed shortcut

Read off the inertia lost

Three changes that test your explanation

Test consequences with the existing two-ledger instrument

Two ledgers and opposite pulses

Predict before the numbers