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Mass and energy: two accounts, one subtraction

Three pages that end with a body’s mass as a measure of its energy. Follow the two energy accounts, and the one subtraction that leads there.

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Detail

Show modern qualifications

First encounter · No algebra required

Can a body lose energy of motion without changing speed?

A body sends equal flashes of light in opposite directions. One observer stays beside it; a traveler keeps moving past. Both write a before-and-after energy account for the same event. The body does not recoil. Compare their accounts without guessing the energy inside the body.

Authored teaching example, calculated when the site was built. Each energy unit here is one joule (J). Numbers use the modern SI constant set, not an observation or a reviewed source transcription.

Choose the traveler

The traveler keeps the same speed before and after: 60 percent of light speed. Equal opposite emission prevents recoil in this model.

Beside the body

Before: unknown

After: that unknown amount minus 10 J

Light: 5 J in each of two opposite directions.

Total light: 10 J

The traveler’s account

Before: another unknown

After: that unknown amount minus 12.5 J

Light: 2.5 J one way; 10 J the other way.

Total light: 12.5 J

Use the button, or drag the “Beside the body” card here. Both actions perform the same alignment.

Compare the accounts

Each account conserves energy. The body loses exactly what its departing light carries.
Align the two before entries, then the two after entries. The unknown body energies remain unknown.
Subtract the losses: 12.5 J minus 10 J leaves 2.5 J. This is a difference between accounts, not an absolute body energy.
With the unchanged-offset premise, the body’s energy of motion decreases by 2.5 J at the same speed.

Test the interpretation, not the arithmeticAssume the additive offset is unchanged during emission

Admit the light-energy transformation and, separately, the unchanged-offset premise. These are model inputs, not conclusions of the table.

Read the full worked example without controls (also works without JavaScript)

The unchanged-offset premise is assumed in this static worked route.

60 percent of light speed

Each account conserves energy. The body loses exactly what its departing light carries.
Align the two before entries, then the two after entries. The unknown body energies remain unknown.
Subtract the losses: 12.5 J minus 10 J leaves 2.5 J. This is a difference between accounts, not an absolute body energy.
With the unchanged-offset premise, the body’s energy of motion decreases by 2.5 J at the same speed.

1 percent of light speed

Each account conserves energy. The body loses exactly what its departing light carries.
Align the two before entries, then the two after entries. The unknown body energies remain unknown.
Subtract the losses: 10.00050004 J minus 10 J leaves 0.0005000375031 J. This is a difference between accounts, not an absolute body energy.
With the unchanged-offset premise, the body’s energy of motion decreases by 0.0005000375031 J at the same speed.

Relaxing the unchanged-offset premise leaves the difference between accounts known but the energy-of-motion change underdetermined.

Why the slower traveler matters

Compare a traveler at sixty percent of light speed with one at one percent. The slow-speed approximation must be tested at slow speed, not certified by the easier large-number example.

Same emission, two speeds. The exact-within-model drop and its low-speed approximation are different quantities. Kinetic interpretation assumes the unchanged offset.
Traveler speed	Rest-frame light	Moving-frame light	Exact drop	Low-speed approximation
60 percent of light speed	10 J	12.5 J	2.5 J	1.8 J
1 percent of light speed	10 J	10.00050004 J	0.0005000375031 J	0.0005 J

Same emission, two speeds. The exact-within-model drop and its low-speed approximation are different quantities. Kinetic interpretation assumes the unchanged offset.

60 percent of light speed

Rest-frame light: 10 J
Moving-frame light: 12.5 J
Exact drop: 2.5 J
Low-speed approximation: 1.8 J

1 percent of light speed

Rest-frame light: 10 J
Moving-frame light: 10.00050004 J
Exact drop: 0.0005000375031 J
Low-speed approximation: 0.0005 J

The large-number example makes the subtraction easy to see. To identify inertia, compare at low speed: the energy of motion then follows the ordinary speed-squared rule. The high-speed result cannot make that approximation exact.

Why are the body energies unknown?

Only the amount transferred to the light is specified. The boxes are not hiding a mass-times-light-speed-squared formula; no absolute body energy has been supplied.

Why do the two observers disagree about the light?

The light-energy transformation is an input borrowed from relativity. It gives different energies in different frames. Each observer conserves energy within their own account.

What is an unchanged offset?

The difference between the two accounts is interpreted as energy of motion plus an offset. This route assumes that the offset is the same before and after emission. Without that assumption, subtraction alone cannot isolate the change in energy of motion.

From two accounts to inertia

The new skill: Subtract two accounts of the same event to eliminate what neither account determines.

Why it helps here: This isolates a change without guessing how much internal energy the body had. The slow-speed comparison then identifies the change in inertia.

More guidance: energy of motion and inertia →

Less guidance: open these exact two-ledger settings in ME-01 →

Continue with the slow traveler in the coefficient laboratory →

Follow the complete explanatory argument →

The unsectioned argument · explanatory reading

An assumption

One result is borrowed, not rediscovered

Where does the traveler’s account of the light come from?

The 7 printed paragraphs this passage explains

Paragraph 1: Einstein announces that his recent paper on the electrodynamics of moving bodies leads to an interesting consequence, which this short paper derives.
Footnote 1: The footnote identifies that earlier paper: Annalen der Physik, volume 17, page 891, 1905.
Paragraph 2: He recalls what the earlier paper assumed: the Maxwell-Hertz equations for empty space, Maxwell's expression for electromagnetic energy, and a principle stated next.
Paragraph 3: The principle of relativity: the laws by which physical states change are the same whichever of two uniformly moving coordinate systems they are referred to.
Paragraph 4: On those foundations he had derived, among other things, a result in section 8 of that paper, which he now quotes.
Footnote 2: The footnote adds that the constancy of the speed of light, also used there, is contained in Maxwell's equations.
Paragraph 5: Plane light waves with energy l in one coordinate system have a different energy in a second system moving along the x-axis, depending on the speed and on the light's angle to the motion; this transformation is used in what follows.

A frame is a way of assigning measurements with rods and clocks moving together. Here the body is at rest in one frame; the other frame moves uniformly relative to it. These are two accounts of one emission, not two experiments.

The argument imports a transformation of light energy. For a pulse of rest-frame energy e traveling at angle φ to the relative-motion axis, the other frame assigns the energy below. This is the admitted electromagnetic result, not a consequence of a mass loss already assumed.

e' = e\,\gamma\,\left(1 - \beta\,\cos\left(\varphi\right)\right)

\beta = \frac{v}{c}

\gamma = \frac{1}{\sqrt{1 - \beta^{2}}}

The transformed pulse energy is its original energy times gamma times one minus beta cosine phi; beta is speed divided by light speed.

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Modern qualifications

The 1905 argument starts from electromagnetic theory and relativity, not the light-quantum hypothesis. Modern γ and β here are explanatory notation, not a transcription of paper 4’s printed glyphs. A later four-momentum derivation would be a different route.

Assumptions and limits: One result is borrowed, not rediscovered

Assumed here

Energy conservation in each inertial frame.
The light-energy transformation imported from the relativity paper, §8.

What this does not establish

The light-energy transformation is taken from the relativity paper, not derived here.
No absolute body energy or mass–energy formula is assigned.

Source context: German source · English · Interlinear gloss · Facsimile

A derivation

Choose the emission that removes recoil

Why send equal amounts of light in opposite directions?

The 2 printed paragraphs this passage explains

Paragraph 6: A body at rest in the first system has energy E₀ there, and energy H₀ in the moving system.
Paragraph 7: The body sends equal amounts of light in opposite directions and stays at rest. Balancing energy in both systems and subtracting shows that its kinetic energy in the moving system falls by a fixed amount, since the constant C in H − E = K + C does not change during the emission.

A body sends half the total light energy each way. The choice is deliberate: the opposed emissions remove the recoil complication in the idealized experiment. The body stays at rest in its own frame, so the traveler assigns it the same speed before and after.

The two pulse energies need not be equal in the traveler’s frame. Reversing a direction reverses its cosine. The direction-dependent contributions therefore cancel when the two energies are added.

\begin{aligned}& \frac{L}{2}\,\gamma\,\left(1 - \beta\,\cos\left(\varphi\right)\right) \\ &\qquad {} \mathbin{+} \frac{L}{2}\,\gamma\,\left(1 + \beta\,\cos\left(\varphi\right)\right) \\ &\qquad \mathrel{=} \gamma\,L\end{aligned}

The opposite-direction terms cancel, leaving total moving-frame light energy gamma times L.

Show every step here: Choose the emission that removes recoil

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Modern qualifications

Equal opposite radiation has zero total momentum in the body’s rest frame in the admitted model. A modern treatment with unequal pulses must follow recoil explicitly; it cannot reuse this same-speed comparison unchanged.

Assumptions and limits: Choose the emission that removes recoil

Assumed here

Two equal opposite pulses in the body’s rest frame.
The body remains at rest in that frame; both observers are inertial.

What this does not establish

Unequal pulses or external forces require recoil and momentum bookkeeping.

Earlier step: One result is borrowed, not rediscovered

Source context: German source · English · Interlinear gloss · Facsimile

A derivation

Two accounts of the same loss

How can both accounts conserve energy while disagreeing about its amount?

The 2 printed paragraphs this passage explains

Paragraph 6: A body at rest in the first system has energy E₀ there, and energy H₀ in the moving system.
Paragraph 7: The body sends equal amounts of light in opposite directions and stays at rest. Balancing energy in both systems and subtracting shows that its kinetic energy in the moving system falls by a fixed amount, since the constant C in H − E = K + C does not change during the emission.

Let E₀ and E₁ stand for the body’s rest-frame energies before and after emission. H₀ and H₁ are the corresponding moving-frame energies. The symbols are unknown quantities, not covered-up values of mass times light speed squared.

E_0 - E_1 = L

H_0 - H_1 = \gamma\,L

The initial body energy equals its final energy plus the emitted light, separately in each frame.

The light is outside the body after emission. Both accounts conserve energy when body and emitted light are included. Different frame totals are compatible with conservation; conservation compares before and after within one frame, not the numerical energies of different frames.

Open the foundation: Conservation and symmetry

Show every step here: Two accounts of the same loss

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Modern qualifications

Einstein leaves the body’s energies unknown: $E_0$ and $E_1$ before and after in the rest frame, $H_0$ and $H_1$ in the moving one. Only their differences enter the argument. An offset added to all four cancels, which is why the paper can say nothing about an absolute rest energy.

Assumptions and limits: Two accounts of the same loss

Assumed here

No energy enters from outside during the emission.
The same body and the same emission are compared in both frames.

What this does not establish

The accounts do not determine E₀, E₁, H₀, or H₁ separately.

Earlier step: Choose the emission that removes recoil

Source context: German source · English · Interlinear gloss · Facsimile

A derivation

Subtract what you cannot measure

Which difference survives when the accounts are subtracted?

The printed paragraph this passage explains

Paragraph 7: The body sends equal amounts of light in opposite directions and stays at rest. Balancing energy in both systems and subtracting shows that its kinetic energy in the moving system falls by a fixed amount, since the constant C in H − E = K + C does not change during the emission.

Subtract the rest-frame balance from the moving-frame balance and regroup the terms. The first bracket compares the two accounts before emission; the second compares them afterward.

\begin{aligned}\left(H_0 - E_0\right) - \left(H_1 - E_1\right) &\mathrel{=} \left(\gamma\,L\right) - L \\ &\mathrel{=} L\,\left(\gamma - 1\right)\end{aligned}

The before frame-energy difference minus the after frame-energy difference equals L times gamma minus one.

This step needs no value for the absolute internal energy. It is an algebraic consequence of the two balances. It has not yet established that the surviving quantity is the change in energy of motion: that identification needs a physical premise.

Show every step here: Subtract what you cannot measure

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Modern qualifications

A cancellation does not prove that the terms have the physical interpretation we would like. Keeping the algebraic subtraction separate from its kinetic interpretation makes a changed premise testable in the two-ledgers laboratory.

Assumptions and limits: Subtract what you cannot measure

Assumed here

Both energy balances describe the same emission.

What this does not establish

Calling the remaining difference kinetic energy requires the next premise.

Earlier step: Two accounts of the same loss

What is getting in the way?

An unfamiliar word or symbol
E is the body's energy counted in its own rest frame, and H is the same body's energy as a traveler moving past it counts it. The subscript 0 means before the light leaves, 1 after. L is the light's energy in the rest frame, and γ, 1.25 at sixty percent of light speed, is how much larger the traveler finds it.
Open the explanation that addresses this
An algebraic move
Two balances, H₀ − H₁ = γL and E₀ − E₁ = L, taken one from the other: H₀ − H₁ − E₀ + E₁ = γL − L. Grouping the left as (H₀ − E₀) − (H₁ − E₁) and taking L out on the right gives L(γ − 1). The unknown energies are never given values, only compared.
Open the explanation that addresses this
The physical reason for a step
Each observer's account balances on its own: the body's energy drops by exactly the energy the light carries away, in that observer's reckoning. The two observers only assign the light different amounts, so the body's drop differs between the accounts. Neither account is wrong.
Open the explanation that addresses this
The connection to the picture
Picture two ledgers for one emission. In the rest ledger the body's energy falls by 10 units; in the traveler's by 12.5. Taking one ledger's change from the other's needs only those two losses, not the balances themselves, which nobody can measure.
Open the explanation that addresses this
The purpose of the calculation
It isolates the one quantity both accounts agree how to compute, L(γ − 1), without knowing the body's total energy in either frame. The passage stops there on purpose: that this difference is a change in the body's energy of motion is a separate premise, which the next passage states.
Open the explanation that addresses this
Too much at once
One example: the light carries 10 units in the rest frame, and the traveler moves at sixty percent of light speed, so γ = 1.25. The rest account loses 10, the traveler's loses 12.5, and the difference is 2.5, which is 10 × (1.25 − 1). Nothing more of the passage is needed yet.
Open the explanation that addresses this

Source context: German source · English · Interlinear gloss · Facsimile

An assumption

The premise that the subtraction needs

When may we call the difference a loss of energy of motion?

The printed paragraph this passage explains

Paragraph 7: The body sends equal amounts of light in opposite directions and stays at rest. Balancing energy in both systems and subtracting shows that its kinetic energy in the moving system falls by a fixed amount, since the constant C in H − E = K + C does not change during the emission.

H_0 - E_0 = K_0 + C

H_1 - E_1 = K_1 + C

Each frame-energy difference equals the corresponding kinetic energy plus the same additive constant C.

Because the same C occurs twice, it cancels. The kinetic-energy drop is then L(γ − 1), even though the body’s speed is unchanged. The unchanged offset is an asserted premise of this route, not something the subtraction independently measured.

K_0 - K_1 = L\,\left(\gamma - 1\right)

The kinetic energy before minus the kinetic energy after equals L times gamma minus one.

Open the foundation: Energy of motion and inertia

Show every step here: The premise that the subtraction needs

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Follow the checked ledger subtraction step by step

Why does the shared unknown disappear?

Follow one checked elimination through five small steps. The equations below are the same semantic records used by the term inspector. The checker verifies the algebra exactly; it does not prove conservation, the light-energy transformation, or the unchanged-offset premise.

Modern teaching derivation. No rest energy is assigned to the body, and no mass-energy conclusion is used as an input.

Which premises should this route be allowed to use? Conservation in the body's rest description Conservation with the imported light-energy transformation The same offset C before and after emission Delta K means before minus afterPresentation

5 of 5 steps supported by this selection. The unchanged-offset premise is in use.

What is assumed, and what is only a definition?

Conservation in the body's rest description · assumption

The energy removed from this body is the total energy L in the two equal, opposite light pulses.

E_0 - E_1 = L

The body’s energy before minus its energy after equals the total emitted light energy L.

Conservation with the imported light-energy transformation · assumption

The same emission removes gamma L in the moving description. The radiation transformation comes from relativity; this subtraction does not derive it.

H_0 - H_1 = \gamma\,L

The body’s moving-frame energy before minus after equals gamma times L.

The same offset C before and after emission · assumption

Both frame differences equal energy of motion plus one unchanged additive offset. This is an additional physical premise, not something proved by conservation.

H_0 - E_0 = K_0 + C

The moving-frame body energy minus the rest-frame energy equals the energy of motion plus the unchanged additive offset C.

H_1 - E_1 = K_1 + C

The moving-frame body energy minus the rest-frame energy equals the energy of motion plus the unchanged additive offset C.

Delta K means before minus after · definition

This is only a definition of notation. It does not assume a value for the kinetic-energy difference.

\Delta K := K_0 - K_1

Delta K is defined as K before minus K after.

1. Subtract the complete accounts
Algebra checked; supported under the selected premises.
$\begin{aligned}& H_0 - H_1 - \left(E_0 - E_1\right) \\ &\qquad \mathrel{=} \gamma\,L - L\end{aligned}$
H before minus H after, less E before minus E after, equals gamma L minus L.
Subtract the rest-frame equality from the moving-frame equality, including both right-hand sides.
Why this step is allowed
The outer minus sign applies to E before minus E after as a whole. Its expansion is minus E before plus E after. No absolute body energy is assigned.
Open the mathematical tool behind this step →
Inspect the exact algebra certificate
For each equation, form left minus right. The result at this step is exactly the following linear combination of earlier residuals, with no floating-point tolerance:
- 1 × The same emission in the moving description
- -1 × The energy removed in the rest description
Matching these residuals proves a conditional implication, not the truth of the starting equations.
2. Put the two descriptions beside each other
Algebra checked; supported under the selected premises.
$\begin{aligned}& H_0 - E_0 - \left(H_1 - E_1\right) \\ &\qquad \mathrel{=} L\,\left(\gamma - 1\right)\end{aligned}$
H before minus E before, less H after minus E after, equals L times gamma minus one.
Reorder the left-hand terms into before and after frame differences. Factor gamma L minus L on the right.
Why this step is allowed
This is still only a comparison between frame accounts. Nothing in these two algebraic steps identifies either difference as energy of motion.
Open the mathematical tool behind this step →
Inspect the exact algebra certificate
For each equation, form left minus right. The result at this step is exactly the following linear combination of earlier residuals, with no floating-point tolerance:
- 1 × Subtract the rest account from the moving account
Matching these residuals proves a conditional implication, not the truth of the starting equations.
3. Use the additional offset premise · the additional premise
Algebra checked; supported under the selected premises.
$\begin{aligned}& K_0 + C - \left(K_1 + C\right) \\ &\qquad \mathrel{=} L\,\left(\gamma - 1\right)\end{aligned}$
K before plus C, less K after plus C, equals L times gamma minus one.
Replace each H minus E with its corresponding K plus C. Keep the same C in both places.
Why this step is allowed
This is the consequential premise. If the offset can change with emission, the two frame differences cannot be replaced by kinetic energies plus a shared constant. The earlier conservation comparison remains valid without it.
Open the mathematical tool behind this step →
Inspect the exact algebra certificate
For each equation, form left minus right. The result at this step is exactly the following linear combination of earlier residuals, with no floating-point tolerance:
- 1 × Subtract the two energy accounts
- -1 × State the offset premise before emission
- 1 × State the offset premise after emission
Matching these residuals proves a conditional implication, not the truth of the starting equations.
4. Now cancel the shared unknown
Algebra checked; supported under the selected premises.
$K_0 - K_1 = L\,\left(\gamma - 1\right)$
K before minus K after equals L times gamma minus one.
Expand the brackets. The first C is added and the same C is subtracted, so they cancel exactly.
Why this step is allowed
The checker matches canonical quantity identities, not similar-looking letters. Different before and after offsets would remain in the equation rather than disappear.
Open the mathematical tool behind this step →
Inspect the exact algebra certificate
For each equation, form left minus right. The result at this step is exactly the following linear combination of earlier residuals, with no floating-point tolerance:
- 1 × Substitute both offset premises before cancelling anything
Matching these residuals proves a conditional implication, not the truth of the starting equations.
5. Give the result a short name
Algebra checked; supported under the selected premises.
$\Delta K = L\,\left(\gamma - 1\right)$
The kinetic-energy drop, K before minus K after, equals L times gamma minus one.
Use the declared definition Delta K = K before minus K after.
Why this step is allowed
The exact kinetic-energy drop is established under the stated premises. Identifying the mass decrease still requires the separate low-speed coefficient argument; a finite-speed ratio is not that limit.
Open the mathematical tool behind this step →
Inspect the exact algebra certificate
For each equation, form left minus right. The result at this step is exactly the following linear combination of earlier residuals, with no floating-point tolerance:
- 1 × Cancel the shared offset
- 1 × Name the before-minus-after difference
Matching these residuals proves a conditional implication, not the truth of the starting equations.

What would a changed offset leave behind?

If the two offsets differ, substitution leaves their before-minus-after difference in the account. It cannot be cancelled. The kinetic-energy drop alone is then not fixed by the two conservation sheets.

For an authored arithmetic example, let the frame differences be 6 before and 5.5 after, so their decrease is 0.5. Offsets of 2 before and 3 after correspond to kinetic energies of 4 and 2.5, whose decrease is 1.5. The conservation difference is still 0.5, but it is not the kinetic difference. These invented accounting numbers are not a claim about a realizable body or experimental evidence.

The mass conclusion is a further step. The exact energy drop is not yet a mass decrease. The low-speed comparison supplies the next premise and limit.

Share this premise selection

Continue through the checked low-speed limit → Continue to the low-speed coefficient laboratory → Read the coefficient argument →

Modern qualifications

This is a productive conditional argument, not a premise-free theorem about arbitrary interacting extended systems. The paper’s comparison with the electron kinetic energy from relativity §10 is a comparison, not an additional premise used here.

Assumptions and limits: The premise that the subtraction needs

Assumed here

For each body state, H − E equals K plus an additive offset.
The same offset C applies before and after emission.

What this does not establish

Without the unchanged-offset premise, the kinetic-energy drop is underdetermined.

Earlier step: Subtract what you cannot measure

Source context: German source · English · Interlinear gloss · Facsimile

A derivation, to an approximation

Read the slow-speed coefficient

Why is a fast-traveler example not enough to identify the mass decrease?

The printed paragraph this passage explains

Paragraph 8: Neglecting terms of fourth and higher order, the drop in kinetic energy becomes L divided by V squared, times v squared over 2.

For small β = v/c, γ − 1 begins with β²/2. The next term is 3β⁴/8. Keeping the second-order term gives a kinetic-energy drop of one half times L/c² times v².

K_0 - K_1 \approx \frac{1}{2}\,\frac{L}{c^{2}}\,v^{2}

At low speed, the kinetic-energy drop is approximately one half times L over c squared times v squared.

At the same low speed Newtonian mechanics writes the kinetic-energy drop as (m₀ − m₁)v²/2. Comparing the coefficients, rather than assuming a rest-energy formula, identifies m₀ − m₁ = L/c².

\lim_{v \to 0} \frac{2\,L\,\left(\gamma - 1\right)}{v^{2}} = \frac{L}{c^{2}}

The limit as speed approaches zero of twice the kinetic-energy drop divided by speed squared is L over c squared.

Open the foundation: Approximating a curve near a point

Show every step here: Read the slow-speed coefficient

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Follow the low-speed limit all the way to the mass conclusion

Why does the coefficient become a mass decrease?

The ledger subtraction gives an energy difference. To identify inertia, we must take a low-speed limit and add the Newtonian meaning of mass. A single finite-speed calculation cannot replace either step.

Modern teaching derivation. Exact algebra and local Taylor coefficients are checked, not the truth of the physical premises.

Fix L > 0 and c > 0, hold both fixed, and take v/c toward zero through |v/c| < 1. Division by v² uses v ≠ 0.

The complete second-order, all-premise explanation is readable without JavaScript. Controls and shared premise selections require JavaScript; no prediction or answer is required to read the conclusion.

Examine the premises and approximation order

Conservation in the body's rest description

Conservation with the imported light-energy transformation

The same offset C before and after emission

Delta K means before minus after

Newtonian meaning of inertial mass at low speed

Powers retained in the modern Taylor comparison

The mass-decrease conclusion follows under all selected premises. The mathematical limit remains distinct from its physical interpretation. Changing the retained order does not change that limit.

Read the physical premises in full

Conservation in the body's rest description. The energy removed from this body is the total energy L in the two equal, opposite light pulses.

Conservation with the imported light-energy transformation. The same emission removes gamma L in the moving description. The radiation transformation comes from relativity; this subtraction does not derive it.

The same offset C before and after emission. Both frame differences equal energy of motion plus one unchanged additive offset. This is an additional physical premise, not something proved by conservation.

Delta K means before minus after. This is only a definition of notation. It does not assume a value for the kinetic-energy difference.

Newtonian meaning of inertial mass at low speed. For the before and after body at the same small speed, the kinetic-energy difference has coefficient one half times the mass decrease; the remaining terms vanish faster than v squared. This is an additional physical premise, not proved by the Taylor calculation.

Expand the factor, not an assumed body energy

Supported under the stated conditions.

\gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^{2}}}

The Lorentz factor equals one divided by the square root of one minus the squared speed ratio.

The expansion variable is the dimensionless ratio β = v/c. The positive radical is analytic near zero. Exact rational arithmetic gives the following coefficients of γ − 1; these are not fitted to plotted samples.

Exact coefficients of each power of β, through order eight
Power	Coefficient	Current truncation
β⁰	0	Retained
β¹	0	Retained
β²	1/2	Retained
β³	0	Beyond selected order
β⁴	3/8	First nonzero term omitted
β⁵	0	Beyond selected order
β⁶	5/16	Beyond selected order
β⁷	0	Beyond selected order
β⁸	35/128	Beyond selected order

\gamma-1\approx \frac{1}{2}\beta^{2}

Keep powers through 2 in the dimensionless speed ratio beta, which means v divided by c. This is an approximation, not a finite-speed equality.

The first nonzero omitted term has coefficient 3/8 and power 4. A local order statement is not a numerical error bound at an arbitrary finite speed.

Open the Taylor-expansion tool →

Read the energy-of-motion coefficient
Supported under the stated conditions.
$\Delta K = L\,\left(\gamma - 1\right)$
The kinetic-energy drop, K before minus K after, equals L times gamma minus one.
Using the already checked ledger argument, divide by the fixed positive emitted energy L. The first nonzero coefficient is exactly 1/2. Multiplying back by L makes the leading term one half (L/c²)v². This remains conditional on the unchanged-offset premise.
$\Delta K \approx 0.5\,L\,\left(\frac{v}{c}\right)^{2}$
The exact kinetic-energy drop is approximately one half L times the squared ratio v over c, at small speed.
Inspect how the unknown body energies and shared offset were removed →
Divide only away from zero speed
Supported under the stated conditions.
$m_{\mathrm{proxy}} := \frac{2\,\Delta K}{v^{2}}$
The finite-speed proxy is defined as twice the exact kinetic-energy drop divided by v squared.
$\begin{aligned}&\frac{2\Delta K}{v^2}=\frac{L}{c^2}\,\frac{2(\gamma-1)}{\beta^2},\\ &\beta=\frac{v}{c},\quad v\ne 0\end{aligned}$
For nonzero v, twice the kinetic-energy drop over v squared equals L over c squared times the dimensionless quotient twice the difference gamma minus one, divided by beta squared. The unchanged-offset premise is required for the kinetic interpretation.
This exact rescaling uses β = v/c and v ≠ 0. It does not cancel a zero denominator. The finite-speed proxy retains its speed dependence; it is not yet the mass decrease.
Take the limit, not the value at zero
Supported under the stated conditions.
$\lim_{\beta\to0}\frac{2(\gamma-1)}{\beta^2}=1$
The two-sided limit of the dimensionless quotient is exactly 1. The original quotient has no value at beta equal to zero.
The constant and first-order coefficients of γ − 1 vanish exactly. Dividing by β² shifts the series by two powers; multiplying by two leaves constant term 1. All remaining terms tend to zero locally. This establishes the two-sided limit.
At β = 0 the original quotient is still undefined. The analytic continuation has a value there, but it is not a measurement of the finite-speed proxy at zero.
Inspect the exact regularized coefficients
Coefficients from power zero through power 6: 1, 0, 3/4, 0, 5/8, 0, 35/64. The nonzero β² coefficient is why the finite-speed ratio is not identically its limit.
The move: read the coefficient as a lost mass
Supported under the stated conditions.
$m_{\mathrm{loss}} = \frac{L}{c^{2}}$
The positive inertial mass decrease equals L divided by c squared.
The Newtonian low-speed meaning of inertial mass says that the coefficient of v² in the kinetic-energy decrease is one half the mass decrease. Comparing it with one half L/c² identifies the positive mass loss. It does not assign either absolute body energy or assume a rest-energy formula.
Without that Newtonian premise, the mathematical limit survives but the identification as a mass decrease does not. Removing the unchanged-offset premise instead breaks the connection between the ledger difference and kinetic energy.

Compare the exact drop, finite-speed proxy and limit at 0.6c in the laboratory →

Share this premise selection and retained order

What the check establishes, and what it does not

The algebra and local limit are checked. Physical premises and source alignment are not certified; no finite-speed error bound is asserted.

The binomial-series identity used by the checker is documented in NIST DLMF 4.6.7. This is a modern mathematical reference, not an item on the 1904 historical shelf.

Modern qualifications

Einstein keeps only the leading term: at small speed the energy of motion is $\frac{1}{2}mv^2$ , so he reads the loss of mass, $L/V^2$ , from the coefficient of $v^2$ . The higher terms matter at large speed, as the drop at 60 percent of light speed shows.

Assumptions and limits: Read the slow-speed coefficient

Assumed here

The unchanged-offset premise.
Newtonian kinetic energy supplies the second-order coefficient at low speed.

What this does not establish

The quadratic approximation is not exact at finite speed.
At zero speed the ratio of the energy drop to speed squared is undefined; its limit is meaningful.

Earlier step: The premise that the subtraction needs

Source context: German source · English · Interlinear gloss · Facsimile

A derivation

A decrease, a signed change, and a conversion

What has the argument identified, and which way does the change point?

The 2 printed paragraphs this passage explains

Paragraph 9: So a body that gives off energy as radiation loses mass, by an amount set by the energy it gives off, and Einstein argues that it does not matter that the energy leaves as radiation.
Paragraph 10: The general conclusion: the mass of a body is a measure of its energy content. When the energy changes, the mass changes in the same direction, and the paper gives the size of that change in grams for an energy in ergs.

m_0 - m_1 = \frac{L}{c^{2}}

\Delta m = m_1 - m_0 = -\frac{L}{c^{2}}

The mass before minus the mass after is positive L over c squared; the signed final-minus-initial change is negative L over c squared.

The conclusion here is a difference. It does not retroactively fill the unknown energy boxes with mc². The paper goes on to infer a broader relation between energy content and inertia; that generalization should remain distinguishable from the calculation that led to it.

Its numerical conversion uses a rounded light-speed-squared factor of 9 × 10²⁰ in centimetre-second units, so energy in erg gives mass in grams. The kinetic-energy laboratory keeps that historical constant set separate from the modern SI value. A unit conversion changes the description of the result, not the physical event.

Show every step here: A decrease, a signed change, and a conversion

Read every step on this section’s own page, where they are part of the page.

Explore the equations in this step

Read each operation, check its units and assumptions, or open the mathematical step behind it. These are modern teaching equations, not the equations as printed.

Open them on this section’s own page, where they are part of the page.

Connect the result terms to the coefficient laboratory →

Modern qualifications

In modern theory invariant mass depends on the chosen system boundary; total energy also depends on the frame. The modern rest-energy relation is not used as a premise of this explanatory route.

Assumptions and limits: A decrease, a signed change, and a conversion

Assumed here

Positive emitted energy L.
The low-speed coefficient identification from the preceding step.

What this does not establish

This calculation determines a change, not an absolute internal energy.
Einstein’s printed constants and modern values are kept in separate calculations.

Earlier step: Read the slow-speed coefficient

Source context: German source · English · Interlinear gloss · Facsimile

A qualification

The light has left the body, not vanished

Does the combined isolated system lose energy when the body emits?

The 2 printed paragraphs this passage explains

Paragraph 11: Bodies whose energy content varies strongly, such as radium salts, might allow the theory to be tested.
Paragraph 12: If the theory agrees with the facts, radiation carries inertia from the body that emits it to the body that absorbs it.

An account of the body alone loses the emitted energy. An account including the body and both pulses retains its total energy in each frame. Losing energy from one subsystem does not mean energy has disappeared from the isolated whole.

The paper closes by proposing systems with strongly varying energy content, such as radium salts, as possible tests, and leaves the physical claim conditional on agreement with facts. This preview supplies neither a new experiment nor a transcription of those sentences.

The system-boundaries laboratory lets you distinguish the body, radiation, and combined-system boundaries. Its modern invariant-mass interpretation and dated 1906 box extension are additional explanations, not steps secretly inserted into the 1905 inference.

Show every step here: The light has left the body, not vanished

Read every step on this section’s own page, where they are part of the page.

Modern qualifications

Two oppositely directed light pulses can have nonzero invariant mass as a combined system even though individual photons have zero rest mass. That statement uses modern energy–momentum accounting; it is not a premise of the 1905 two-ledger route.

Assumptions and limits: The light has left the body, not vanished

Assumed here

The larger account includes the body and all the light it emitted.
No energy crosses that larger boundary.

What this does not establish

A successful model calculation is not an observed experimental confirmation.
The 1906 box argument and modern four-momentum are separate extensions.

Earlier step: A decrease, a signed change, and a conversion

Source context: German source · English · Interlinear gloss · Facsimile

Common wrong turns

Things often said about this paper, what makes each one tempting, and what the paper and later evidence support.

A common wrong turn: Einstein was the first to connect mass and energy.

Einstein was the first to connect mass and energy.

The paper's conclusion is the version everyone remembers, and it carries his name.

The 1905 paper states the relation as a general conclusion about the energy content of a body. Whether that makes it first is a judgment the record here does not make.

Three records sit beside the paper. In 1900 Poincaré published the paper on Lorentz's theory and the principle of reaction which Einstein himself, in 1906, said already contained in the main the formal considerations his centre-of-mass proof needed. In 1904 Hasenöhrl gave radiation enclosed in a moving cavity an apparent mass of $\frac{8}{3}hε_0/c^2$ , and in January 1905 corrected it to $\frac{4}{3}hε_0/c^2$ . In September 1905 this paper concluded that the mass of a body is a measure of its energy content. The historian's margin gives each from what was written, and the site does not rank them.

A matter of the historical record: see the historian's margin, the notes on Einstein's 1906 paper and on Hasenöhrl.

Sources: A. Einstein, Ann. Phys. (4) 18, 639–641 (1905), plate p. 641; A. Einstein, Ann. Phys. (4) 20, 627–633 (1906), p. 627; F. Hasenöhrl, Ann. Phys. (4) 15, 344–370 (1904), p. 363; (4) 16, 589–592 (1905), pp. 589–590; historian's margin: note-me-c-1906-poincare, note-me-d-hasenoehrl

A common wrong turn: The paper derives

E=mc^2

The paper derives $E=mc^2$ .

It is the one formula every account of this paper quotes, so a reader expects to find it on the page.

The paper's conclusion is the same relation written for a change: a body that gives off the energy $L$ loses the mass $L/V^2$ , where $V$ is the speed of light.

The paper states its result on p. 641 as a change: if a body gives off the energy $L$ as radiation, its mass decreases by $L/V^2$ , with $V$ the speed of light. It then prints the conversion for everyday units: the mass changes by $L/9\cdot10^{20}$ when the energy is measured in erg and the mass in grams. It never writes the familiar equation, and it never gives a whole body an energy equal to its mass times the square of the speed of light.

See it in the instrument

Sources: A. Einstein, Ann. Phys. (4) 18, 639–641 (1905), plate p. 641; scenario mass-energy-printed-factor: L = 9·10²⁰ erg gives 1 g

A common wrong turn: The low-speed formula gives the mass loss exactly, at any speed.

The low-speed formula gives the mass loss exactly, at any speed.

The paper's conclusion follows straight from an approximate equation, and $\frac{L}{V^2}\frac{v^2}{2}$ looks like an exact kinetic energy.

At speeds small compared with light, where the two agree up to the quantities of fourth and higher order that the paper neglects.

The paper's exact result is $K_0-K_1=L\left\{\frac{1}{\sqrt{1-(v/V)^2}}-1\right\}$ . Neglecting quantities of fourth and higher order, it writes $K_0-K_1=\frac{L}{V^2}\frac{v^2}{2}$ and reads the mass change off the coefficient of $v^2/2$ . That coefficient is a limit. Divide the exact difference by $v^2/2$ at a finite speed and the result is larger: at $0.6$ of the speed of light it is about $1.389\,L/V^2$ , as ME-02's preset for that speed shows.

See it in the instrument

Sources: A. Einstein, Ann. Phys. (4) 18, 639–641 (1905), plate p. 641; ME-02 preset me-02-0.6c: finiteSpeedMassProxy / limitingCoefficient = 1.38889 at β = 0.6

A common wrong turn: Mass is converted into energy.

Mass is converted into energy.

Popular accounts of nuclear energy say mass is converted, and the paper does tie a loss of mass to a loss of energy.

As bookkeeping for the emitting body alone: after giving off the energy $L$ it has less mass, by $L/V^2$ .

The paper's last sentence says that, if the theory agrees with the facts, radiation carries inertia between the bodies that emit it and the bodies that absorb it. So the body's mass falls because the energy $L$ left it, and the radiation takes that inertia along. Draw the boundary around the body and its light together and nothing has gone: ME-03's sealed lamp and mirror, where the light is emitted and absorbed inside the enclosure, shows a total mass change of zero.

See it in the instrument

Sources: A. Einstein, Ann. Phys. (4) 18, 639–641 (1905), plate p. 641; ME-03 preset me-03-sealed-lamp-and-mirror

A common wrong turn: Light has no mass, so it cannot carry mass away.

Light has no mass, so it cannot carry mass away.

In modern physics light is massless, so a pulse of light seems to have no mass to carry.

Of each pulse on its own: a single pulse of light has zero invariant mass.

In the modern formulation, the invariant mass of a system is not the sum of the masses of its parts. The paper's body sends out two equal pulses in opposite directions, with total energy $L$ . Their momenta cancel, so the pair has an invariant mass of $L/c^2$ , although each pulse alone has none. ME-03's four-momentum mode shows this. It is a later explanation that resolves the puzzle; the 1905 argument speaks of energy and inertia and uses neither idea.

See it in the instrument

Sources: A. Einstein, Ann. Phys. (4) 18, 639–641 (1905), plate p. 641; ME-03 preset me-03-four-momentum-opposite-pulses

A common wrong turn: Einstein measured the mass change.

Einstein measured the mass change.

The paper ends with a numerical conversion and names a material, radium salts.

There is no reading under which this is a correct thing to say.

The paper's second-to-last sentence is conditional. It is not ruled out ('Es ist nicht ausgeschlossen'), it says, that bodies whose energy content varies to a high degree, such as radium salts, will allow a test of the theory. No experiment is reported, and none is claimed. ME-03's radium card shows how small the effect is: one alpha decay of radium-226 releases 4.871 MeV, which carries off 5.229 mg of mass for every mole that decays.

See it in the instrument

Sources: A. Einstein, Ann. Phys. (4) 18, 639–641 (1905), plate p. 641; ME-03 preset me-03-card-radium

A common wrong turn:

E=mc^2

is about nuclear bombs.

$E=mc^2$ is about nuclear bombs.

The formula became famous with nuclear physics, where a small change of mass releases a very large energy.

Nuclear reactions are where the energy released per unit mass is large enough for the mass change to be easy to measure.

Nothing in the paper's argument depends on where the energy comes from: a body that gives off the energy $L$ loses the mass $L/V^2$ . ME-03's candle card makes it concrete. A flame giving off about 80 W for one hour releases 288 kJ, which carries off about 3.20 ng of mass. That is separate from the grams of wax that leave as combustion products, which is matter crossing the boundary. Nuclear reactions only make the energy released per unit mass large.

See it in the instrument

Sources: A. Einstein, Ann. Phys. (4) 18, 639–641 (1905), plate p. 641; ME-03 preset me-03-card-candle (Sundström 1995, candle heat release near 80 W)

A common wrong turn: Only light carries this mass; heating a body or storing chemical energy does not change its inertia.

Only light carries this mass; heating a body or storing chemical energy does not change its inertia.

Every step of the paper's argument is about light sent out by a body.

Of the derivation: it treats only energy given off as radiation.

The derivation covers only energy given off as radiation. The paper then judges it 'offenbar unwesentlich', evidently inessential, that the energy taken from the body turns into radiation, and draws the more general conclusion that the mass of a body is a measure of its energy content. That step goes beyond what the calculation shows, and the paper presents it as a conclusion it is led to rather than as a second derivation. In the modern formulation every form of internal energy contributes: ME-03's heated sealed box gains mass equal to the energy fed in through its leads divided by $c^2$ , with no light leaving.

See it in the instrument

Sources: A. Einstein, Ann. Phys. (4) 18, 639–641 (1905), plate p. 641; ME-03 preset me-03-heated-sealed-box

Historian’s margin

What was written, by whom and when, from the sources cited. A note that is this site’s own reconstruction rather than the record says so under its heading.

Historian’s margin

The equation $E=mc^2$ does not appear in the paper. Its three pages state a change: a body that gives off the energy $L$ as radiation loses the mass $L/V^2$ , with $V$ the speed of light, and the paper prints the conversion $L/9\cdot10^{20}$ for energy in erg and mass in grams (p. 641). It gives no body a total energy equal to its mass times $V^2$ : its energies are fixed only up to arbitrary additive constants (pp. 640–641). Einstein's own later papers take the next steps. In 1906 he wrote that the contradiction he was examining disappears if every energy $E$ is given the inertia $E/V^2$ (p. 629). In 1907 he wrote the energy of a mass point of mass $\mu$ moving with speed $v$ as $\mu V^2/\sqrt{1-(v/V)^2}$ , and concluded that a system of mass points in uniform motion can be replaced, as regards its energy, by a single mass point of mass $\mu=E_0/V^2$ , where $E_0$ is the system's energy at rest (pp. 383–384). The familiar string $E=mc^2$ uses letters these papers do not.

Sources: A. Einstein, Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig? Annalen der Physik (4) 18, 639–641 (1905), pp. 639–641. A. Einstein, Das Prinzip von der Erhaltung der Schwerpunktsbewegung und die Trägheit der Energie. Annalen der Physik (4) 20, 627–633 (1906), p. 629. A. Einstein, Über die vom Relativitätsprinzip geforderte Trägheit der Energie. Annalen der Physik (4) 23, 371–384 (1907), pp. 383–384.

A disputed point, as the record stands

The step from energy to mass rests on two premises the paper states. The difference $H-E$ of the body's energies in the two frames differs from its kinetic energy only by an additive constant $C$ , which does not change during the emission ('da $C$ sich während der Lichtaussendung nicht ändert', p. 641); and at low speed that kinetic energy is compared with $\frac{1}{2}mv^2$ . In 1952 H. E. Ives argued that the reasoning of the 1905 derivation 'is defective' and that Einstein 'did not derive the mass-energy relation', which Ives traced instead to Poincaré's momentum of radiation (1900) and principle of relativity (1904). In 1982 J. Stachel and R. Torretti argued, 'contrary to what several authors have claimed', that the derivation 'was logically sound', its alleged fallacy lying only in the fact that its conclusion is entailed by its premises. The site records both positions and gives no verdict.

Sources: A. Einstein, Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig? Annalen der Physik (4) 18, 639–641 (1905), pp. 640–641. H. E. Ives, Derivation of the Mass-Energy Relation. Journal of the Optical Society of America 42, 540 (1952). J. Stachel and R. Torretti, Einstein’s first derivation of mass–energy equivalence. American Journal of Physics 50, 760–763 (1982).

Editorial note

Einstein returned to this question several times after 1905, and each return is later work: it neither replaces the three pages of 1905 nor widens what they establish. In 1906 he argued from the uniform motion of the centre of mass of a closed system that every energy $E$ carries the inertia $E/V^2$ , and wrote that on this supposition the contradiction with the elements of mechanics disappears (p. 629); that argument imports the momentum of radiation and the centre-of-mass theorem, which the 1905 argument does not use. In 1907 he wrote the energy of a mass point of mass $\mu$ moving with speed $v$ as $\mu V^2/\sqrt{1-(v/V)^2}$ and concluded that a system of mass points in uniform motion can be replaced, as regards its energy, by a single mass point of mass $\mu=E_0/V^2$ , with $E_0$ the system's energy at rest (pp. 383–384); that step reaches a rest energy, which 1905 does not, because 1905 fixes its energies only up to additive constants. In 1935 he published an elementary derivation in the Bulletin of the American Mathematical Society (41, 223–230). This site has read the 1906 and 1907 papers at the pages cited and says nothing about the 1935 argument beyond its title and place, because the text has not been read here.

Sources: A. Einstein, Das Prinzip von der Erhaltung der Schwerpunktsbewegung und die Trägheit der Energie. Annalen der Physik (4) 20, 627–633 (1906), p. 629. A. Einstein, Über die vom Relativitätsprinzip geforderte Trägheit der Energie. Annalen der Physik (4) 23, 371–384 (1907), pp. 383–384. A. Einstein, Elementary derivation of the equivalence of mass and energy. Bulletin of the American Mathematical Society 41, no. 4, 223–230 (1935), pp. 223–230.

Historian’s margin

In 1906 Einstein reached the same rule by another route. He argued that the rule, a change of energy $\Delta E$ going with a change of mass $\Delta E/V^2$ , is the necessary and sufficient condition for the law of the conservation of the motion of the centre of gravity to hold, at least to a first approximation, also for systems in which electromagnetic processes occur. He wrote that the simple formal considerations needed for the proof are 'in der Hauptsache bereits in einer Arbeit von H. Poincaré enthalten', citing the Lorentz Festschrift of 1900, and that for the sake of clarity he would not rely on that paper (p. 627).

Sources: A. Einstein, Das Prinzip von der Erhaltung der Schwerpunktsbewegung und die Trägheit der Energie. Annalen der Physik (4) 20, 627–633 (1906), p. 627. H. Poincaré, La théorie de Lorentz et le principe de réaction. Archives néerlandaises des sciences exactes et naturelles (2) 5, 252–278 (1900), p. 252.

Historian’s margin

In 1904, before the paper, F. Hasenöhrl argued that radiation enclosed in a moving cavity gives the cavity an apparent mass, as if the mass of the system had been increased ('als ob die Masse unseres Systems ... vermehrt worden wäre'). He gave its value as $\frac{8}{3}hε_0/c^2$ , where $hε_0$ is the radiation energy in the cavity at rest (p. 363, equation 32). In January 1905 he corrected it: M. Abraham had sent him by letter a method giving half that value, $\frac{4}{3}hε_0/c^2$ , and Hasenöhrl found that the difference came from a calculation error of his own (pp. 589–590). Both values concern radiation enclosed in a cavity, and both were obtained neglecting higher powers of the speed.

Sources: F. Hasenöhrl, Zur Theorie der Strahlung in bewegten Körpern. Annalen der Physik (4) 15, 344–370 (1904), p. 363. F. Hasenöhrl, Zur Theorie der Strahlung in bewegten Körpern. Berichtigung. Annalen der Physik (4) 16, 589–592 (1905), pp. 589–590.

Historian’s margin

The radium remark is a proposal, stated conditionally: it is not ruled out ('Es ist nicht ausgeschlossen'), the paper says, that bodies whose energy content varies to a high degree, such as radium salts, will allow the theory to be tested (p. 641). The paper reports no test. Measured checks came decades later, from nuclear physics; the reports cited here are J. D. Cockcroft and E. T. S. Walton's on the disintegration of elements by fast protons (1932) and K. T. Bainbridge's letter on the equivalence of mass and energy (1933).

Sources: A. Einstein, Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig? Annalen der Physik (4) 18, 639–641 (1905), p. 641. J. D. Cockcroft and E. T. S. Walton, Experiments with high velocity positive ions. II. The disintegration of elements by high velocity protons. Proceedings of the Royal Society of London A 137, 229–242 (1932). K. T. Bainbridge, The Equivalence of Mass and Energy. Physical Review 44, 123 (1933).

Editorial note

A route you could take

In the modern formulation, the paper's two pulses, sent out with equal energies $L/2$ in opposite directions (p. 640), have zero total momentum, so as a system they have the invariant mass $L/c^2$ , although each pulse alone has none. This is a later formal explanation that resolves a common confusion. The 1905 argument does not use invariant mass, and nothing in it depends on this reading.

Sources: A. Einstein, Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig? Annalen der Physik (4) 18, 639–641 (1905), p. 640.

Editorial note

A route you could take

Three arguments reach the conclusion of this paper from different premises, and a reader who meets only one of them cannot tell which of its features belong to the result and which belong to that route. The table below asks each of them the same five questions. It is a record of premises and not a ranking: no row of it says that one argument supersedes another, and the site does not call any of them refuted. The row that repays reading across is the last one. The premise this paper states about the additive constant is a restriction on which systems the reasoning addresses, and the later arguments reach the same conclusion for wider classes of system by importing more than 1905 does.

Three routes to the same conclusion, each asked the same five questions. The cells are this site’s own reading of the three arguments, not a verdict on which is better, and no physicist has reviewed them.
The question	The 1905 ledger	The 1906 box	A modern four-momentum derivation
What result is imported	The 1905 ledger: The transformation of the energy of a light complex from section 8 of the electrodynamics paper, shown with its provenance, and nothing else dated after 1904.	The 1906 box: That radiation carries momentum equal to its energy divided by the speed of light, and the centre-of-mass theorem for a closed system, neither of which the 1905 argument uses.	A modern four-momentum derivation: The four-momentum formalism itself, which postdates 1905 and is the largest import of the three.
What definition of mass	The 1905 ledger: The coefficient of the square of the speed in the body's energy of motion at low speed, an inertial coefficient read off in a limit rather than a rest energy.	The 1906 box: The inertia that must travel with the light if the centre of mass of a closed system is to stay where it was.	A modern four-momentum derivation: The invariant mass of a total four-momentum, a property of the system rather than of any part of it.
Which conservation law	The 1905 ledger: Energy conservation, applied separately in each of two inertial frames.	The 1906 box: Uniform motion of the centre of mass of a closed system with no external forces.	A modern four-momentum derivation: Conservation of the total four-momentum.
What approximation	The 1905 ledger: An expansion to second order in the speed ratio, the paper's neglect of magnitudes of fourth and higher order; the conclusion is the limiting coefficient, and the finite-speed proxy ME-02 reports at a chosen speed is not that limit.	The 1906 box: Leading order in the ratio of the emitted energy to the box's rest energy, within the bound the instrument declares for box systems, with a rigid box, non-relativistic box motion, and a transit time set by the box's length and the speed of light.	A modern four-momentum derivation: None inside the formalism, which is exact where it applies; what that costs is that the formalism is later.
Which class of systems Where the three routes differ in scope.	The 1905 ledger: A body emitting radiation symmetrically, under the paper's stated premise that the additive constant in each frame's energy difference is unchanged by the emission.	The 1906 box: A closed rigid container with emission and absorption inside it and no external forces.	A modern four-momentum derivation: Any system with a well-defined total four-momentum, including stressed and open systems the 1905 route does not reach.

The 1905 ledger: Open ME-01, and The 1905 ledger’s sources. The 1906 box: Open ME-03 and choose the 1906 box, and The 1906 box’s sources. A modern four-momentum derivation: Open ME-03 and choose the four-momentum formalism, and A modern four-momentum derivation’s sources.

Sources: A. Einstein, Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig? Annalen der Physik (4) 18, 639–641 (1905), pp. 639–641.

References

A. Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt. Annalen der Physik (4), 17, 132–148 (1905).

A. Einstein, On the motion of particles suspended in liquids at rest required by the molecular-kinetic theory of heat. Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

A. Einstein, Zur Elektrodynamik bewegter Körper. Annalen der Physik (4), 17, 891–921 (1905).

A. Einstein, Does the inertia of a body depend upon its energy content? Annalen der Physik (4), 18, 639–641 (1905). External 1923 Perrett–Jeffery translation, electronically transcribed by John Walker; its notation was modernized. A reference for this explanatory preview, not this edition’s translation or its pinned facsimile.

18 foundation readings sit behind this argument.

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Original journal scan

Beside the body

The traveler’s account

Compare the accounts

60 percent of light speed

1 percent of light speed

Why the slower traveler matters

From two accounts to inertia

The unsectioned argument · explanatory reading

One result is borrowed, not rediscovered

Modern qualifications

What this relies on

Assumed here

What this does not establish

Choose the emission that removes recoil

Modern qualifications

What this relies on

Assumed here

What this does not establish

Two accounts of the same loss

Modern qualifications

What this relies on

Assumed here

What this does not establish

Subtract what you cannot measure

Modern qualifications

What this relies on

Assumed here

What this does not establish

The premise that the subtraction needs

Why does the shared unknown disappear?

Conservation in the body's rest description · assumption

Conservation with the imported light-energy transformation · assumption

The same offset C before and after emission · assumption

Delta K means before minus after · definition

1. Subtract the complete accounts

2. Put the two descriptions beside each other

3. Use the additional offset premise · the additional premise

4. Now cancel the shared unknown

5. Give the result a short name

Modern qualifications

What this relies on

Assumed here

What this does not establish

Read the slow-speed coefficient

Why does the coefficient become a mass decrease?

Expand the factor, not an assumed body energy

Read the energy-of-motion coefficient

Divide only away from zero speed

Take the limit, not the value at zero

The move: read the coefficient as a lost mass

Modern qualifications

What this relies on

Assumed here

What this does not establish

A decrease, a signed change, and a conversion

Modern qualifications

What this relies on

Assumed here

What this does not establish

The light has left the body, not vanished

Modern qualifications

What this relies on

Assumed here

What this does not establish

Common wrong turns

Historian’s margin

Historian’s margin

A disputed point, as the record stands

Editorial note

Historian’s margin

Historian’s margin

Historian’s margin

Editorial note

Editorial note

References

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