The unsectioned argument · explanatory reading

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

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.

Open the foundation: Events and distant clocks

Show every step here: One result is borrowed, not rediscovered

Two frames describe one emission. In the first, the body is at rest. The second moves at a constant speed v along a line called the motion axis, and v is smaller than the speed of light c. Describing the body from a second frame does nothing to it: these are two accounts of the same event.

Divide the relative speed by the speed of light: β = v/c. It is a pure number, 0 when the frames share a speed and closer to 1 the faster they move apart.
Form γ = 1/√(1 − β²). At β = 0 it is exactly 1, and it grows without bound as β approaches 1. For β = 0.6: 1 − β² = 1 − 0.36 = 0.64, and √0.64 = 0.8, so γ = 1/0.8 = 1.25.
Let e be the energy of one pulse of light, measured in the body’s rest frame, and φ the angle between the pulse’s direction and the motion axis, measured in the same frame.
The imported rule gives the energy e′ that the moving frame assigns to that pulse:

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.

Check it at β = 0.6, where γ = 1.25. A pulse along the motion axis has cos φ = 1, so e′ = e × 1.25 × (1 − 0.6) = 0.5e. A pulse against it has cos φ = −1, so e′ = e × 1.25 × 1.6 = 2e. A pulse across it has cos φ = 0, so e′ = 1.25e.
The rule is borrowed from the relativity paper, where it is derived for light. It says nothing yet about the body’s own energy before or after the emission, and nothing in it assumes a loss of mass.

A tool for this step

Events and distant clocks

What does it mean to say that two distant events happen at the same time?

An event is something that happens at one place at one instant: a spark, a collision, a flash leaving a lamp. To give it a time, you read a clock at that place. Reading a distant clock by looking at it would mix in the time the light takes to reach you.

A reference frame is a set of rigid measuring rods and clocks at rest, with a clock at every place, so that each event gets three coordinates of position and one of time from the rods and the clock beside it. Before that works, the clocks must agree, and for clocks far apart agreeing needs a rule.

§1 of the relativity paper gives the rule. Send a flash of light from clock A to clock B and back. Note A's reading when it leaves, B's reading when it arrives, and A's reading when it returns. The clocks are called synchronous if the trip out takes as long, by the clocks, as the trip back:

t_B - t_A = t'_A - t_B

t B minus t A equals t A prime minus t B: the trip out takes as long as the trip back.

The paper calls this a definition. It cannot be checked by measurement, because checking it would need clocks that are already synchronous. §1 adds two assumptions: that the rule is consistent, so that if B agrees with A then A agrees with B, and two clocks that agree with a third agree with each other; and that light's speed measured out and back is a universal constant, which the paper writes V.

Worked example: Setting a distant clock with one light signal

Clock A reads 0 s as a flash leaves it for clock B.
The flash is reflected at B and arrives back at A when A reads 10 s.
By the rule the two trips take equal times, so the reflection happened at 5 s on A's time.
If B's clock read 5 s at the reflection, it is synchronous with A. If it read 7 s, it is 2 s fast, and setting it back 2 s makes the two agree.

t_B = \frac{1}{2}\,\left(t_A + t'_A\right)

The time at B equals one half of the sum of A's reading when the flash left and A's reading when it came back.

The rule needs no distance. If B is known to be 1.5 × 10⁹ metres away, the same round trip also gives light's speed: 2 × 1.5 × 10⁹ m divided by 10 s is 3 × 10⁸ metres per second.

One flash, out and back

A flash leaves clock A, turns round at clock B and comes back. By the rule the two trips take equal times, so the turn-round happened half way between leaving and returning: at 5 on A's clock.

B reads 5 at the turn-round, which is what A's clock says the moment was. The two are synchronous.

Now the same flash, against a clock that is running ahead.

B reads 7 where the rule says 5, so it is 2 seconds fast. Set it back 2 and the two agree. Nothing about the flash changed between the drawings; only what B's hands were showing.

What it shows, in words

Two clocks stand apart, drawn as vertical lines with time running down the page. A flash leaves A when A reads 0, reaches B, is reflected, and returns to A when A reads 10. The rule says the trip out took as long as the trip back, so the turn-round was half way between: 5 by A's clock. In the first drawing B reads 5 at that moment and the clocks are synchronous. In the second B reads 7, which is 2 seconds more, so B is 2 seconds fast and setting it back 2 makes them agree.

The same round trip also measures light, if the distance is known separately: with B at 1.5 × 10⁹ metres, a there-and-back of 10 seconds gives 3 × 10⁸ metres per second.

What it does not show, and this is the lesson's own point rather than a limitation of the drawing: the rule is a definition, and no picture can show it being tested. Checking that the trip out really took as long as the trip back would need two clocks already synchronised, which is the thing the rule exists to supply. Nor is there a second observer anywhere here. Only one frame is drawn, so nothing in it bears on how someone moving past would time the same three events; that question needs the paper's §2, and the geometric aid usually reached for is Minkowski's of 1908, which is later than the argument and not its presentation.

Where this lesson stops

An event is something that happens at one place and one time. Whether two distant events are simultaneous depends on a rule for setting clocks, which §1 of the relativity paper supplies.

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A pulse's energy seen from the moving frame

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

is times times one minus times the cosine of .

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e prime equals e times gamma times one minus beta cosine phi.

The relativity paper's section 8 result, taken here as a borrowed premise: a pulse emitted forward loses energy in the moving frame's account, one emitted backward gains it.

Model assumptions and every term’s meaning

Paper 3, section 8, taken as a premise, not re-derived here.
Plane pulses of light.
These are modern teaching equations in SI notation, not a transcription of the printed paper.

Beta cos phi: The Doppler-like share that depends on direction. Read the prerequisite
The direction factor: Everything that depends on where the pulse points. Read the prerequisite
Cosine of the angle: Plus one for a pulse sent straight ahead, minus one straight back. Read the prerequisite
One minus beta cos phi: Smaller forward, larger backward. Read the prerequisite
e times gamma times the factor: The moving frame's energy. Read the prerequisite
What it asserts: The same pulse carries different energies in the two frames. Read the prerequisite
Emission angle: The angle between the pulse and the direction of motion. Read the prerequisite
Speed ratio: v over c. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. Read the prerequisite
Pulse energy, moving frame: The pulse's energy in the frame where the body moves. Read the prerequisite
Pulse energy, rest frame: The same pulse's energy in the body's own frame. Read the prerequisite

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Gamma written with beta

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

is one over the square root of one minus squared.

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Gamma equals one over the square root of one minus beta squared.

The same factor as one over the root of one minus v squared over c squared, written with beta.

Model assumptions and every term’s meaning

These are modern teaching equations in SI notation, not a transcription of the printed paper.

One minus beta squared: Positive for any speed below light. Read the prerequisite
One over the root: At least one. Read the prerequisite
What it asserts: A definition. Read the prerequisite
Its square root: The number that squares to it. Read the prerequisite
Beta squared: The quantity multiplied by itself. Read the prerequisite
Speed ratio: v over c. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. Read the prerequisite

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The factor borrowed from relativity

\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.

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

The Lorentz factor is defined here in modern notation. Using it for the light energy imports the relativity result; it does not independently derive that result.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.

Same observer speed: v is the signed relative speed between inertial descriptions. Its magnitude is less than c. It is held fixed while the body emits. Read the prerequisite
Modern SI speed of light: c denotes the speed of light in modern SI notation. A normalized model with c = 1 is a different unit convention and cannot supply values labeled metres per second. Read the prerequisite
Compare the speed with c: v/c is dimensionless. Its sign gives the relative direction; squaring removes that direction. Read the prerequisite
Square the speed ratio: Both positive and negative observer speeds give the same square. The low-speed expansion uses this small dimensionless number. Read the prerequisite
One minus the squared speed ratio: This quantity stays positive for an admitted observer and approaches zero at the excluded light-speed boundary. Read the prerequisite
Take the positive square root: For |v| < c the radicand is positive. The boundary |v| = c is not an allowed inertial observer. Read the prerequisite
Modern Lorentz factor: γ depends on the observer speed. In this modern teaching notation γ is the factor, not the speed ratio v/c. The transformation of the light energy is imported from relativity, not derived from these energy accounts. Read the prerequisite
Take the reciprocal: The positive reciprocal is at least one. This definition alone says nothing about a body’s rest energy. Read the prerequisite
What this relation asserts: The Lorentz factor is defined here in modern notation. Using it for the light energy imports the relativity result; it does not independently derive that result. Read the prerequisite

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Beta is speed as a fraction of light speed

\beta = \frac{v}{c}

is over .

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Beta equals v over c.

A pure number between zero and one for any body that can carry a clock.

Model assumptions and every term’s meaning

These are modern teaching equations in SI notation, not a transcription of the printed paper.

v over c: Metres per second over metres per second. Read the prerequisite
What it asserts: A definition. Read the prerequisite
Speed ratio: v over c. Read the prerequisite
Speed of light: The paper writes V. Read the prerequisite
Speed of the body: How fast the body moves in the second frame. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

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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

Call L the total energy of the light the body emits, measured in its rest frame. It sends L/2 one way and L/2 the opposite way.
The first pulse travels at angle φ to the motion axis. By the imported rule the moving frame assigns it (L/2)γ(1 − β cos φ).
The second pulse travels the opposite way. Turning a direction around reverses its cosine, so its cosine is −cos φ, and the moving frame assigns it (L/2)γ(1 + β cos φ).
Add the two energies. Both terms share the factor (L/2)γ, so take it out: (L/2)γ times the sum (1 − β cos φ) + (1 + β cos φ).
In that sum −β cos φ and +β cos φ cancel, leaving 1 + 1 = 2. Then (L/2)γ × 2 = γL:

\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.

Check it at β = 0.6, where γ = 1.25, with φ = 0: the moving frame assigns the pulses 0.25L and 1.0L, and 0.25L + 1.0L = 1.25L = γL. Another angle gives a different pair with the same sum.
Why two opposite pulses: each pushes the body back as it leaves, and two equal pushes in opposite directions cancel. The body stays at rest in its own frame, so the moving frame assigns it the same speed before and after the emission.
The cancellation concerns the light’s energy totals only. It assigns no absolute energy to the body.

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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.

Explore the equation · Modern model notation

Two opposite pulses add up to gamma L

\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}

times times one minus cos , plus the same with a plus, is times .

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L over two gamma times one minus beta cos phi, plus L over two gamma times one plus beta cos phi, equals gamma L.

The direction terms are equal and opposite, so they cancel: the pair carries gamma L in the moving frame whatever the angle.

Model assumptions and every term’s meaning

Two equal pulses in opposite directions, so the body does not recoil.
These are modern teaching equations in SI notation, not a transcription of the printed paper.

The second pulse: Its energy in the moving frame. Read the prerequisite
Beta cos phi: Equal and opposite to the first pulse's. Read the prerequisite
Beta cos phi: The part that depends on direction. Read the prerequisite
In brackets: The second pulse's direction factor. Read the prerequisite
In brackets: The first pulse's direction factor. Read the prerequisite
Cosine of the angle: The same angle; the opposite pulse flips the sign in front of it. Read the prerequisite
Cosine of the angle: Plus one straight ahead, minus one straight back. Read the prerequisite
One plus beta cos phi: The pulse sent the opposite way. Read the prerequisite
One minus beta cos phi: The pulse sent along phi. Read the prerequisite
The first pulse: Its energy in the moving frame. Read the prerequisite
Gamma L: The pair's energy in the moving frame. Read the prerequisite
L over 2: The other pulse's share, the same size. Read the prerequisite
L over 2: This pulse's share of the light in the rest frame. Read the prerequisite
What it asserts: The pair carries gamma L, independent of the angle. Read the prerequisite
Both pulses: The beta cos phi terms cancel. Read the prerequisite
Emission angle: The angle between the pulse and the direction of motion. Read the prerequisite
Speed ratio: v over c. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. Read the prerequisite
Emitted light energy: The total energy of the two pulses in the body's rest frame; the paper writes L. Read the prerequisite
Half the light: Each pulse carries L over 2 in the rest frame. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Connect the result terms to the coefficient laboratory →

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

A tool for this step

An equals sign states a balance

Is an equals sign an instruction to calculate, or a statement?

In school, 7 + 5 = asks for an answer. In physics the equals sign more often states a fact: two ways of counting the same thing agree.

Here is an account in made-up units. A body starts with 100 units of energy, sends 30 away as light and keeps 70. Energy before equals energy after plus energy sent out: 100 = 70 + 30. The line is a balance to check, not a sum to do, and it is just as true written 100 − 70 = 30.

The mass-energy paper writes the same kind of balance: E₀ − E₁ = L. The body's energy before, minus its energy after, equals the energy L that the light carries away. It is a statement about the two accounts. Nothing in it says to work out E₀ first.

Because an equation states a balance, it can be read from either side and rearranged without changing what it says. E₁ = E₀ − L tells the same story.

Worked example: Rearranging a balance without changing it

Start from 100 = 70 + 30, in made-up units.
Take 70 from both sides: 100 − 70 = 30. Still true, and still the same account.
Put the paper's letters in place of the numbers: E₀ = E₁ + L, or E₀ − E₁ = L.
Whichever form is written, it states one fact: what the body lost is what the light carried away.

The two sides of the account, drawn at one scale

The upper bar is the energy the body started with. The lower bar is the same length, built from two pieces: what it kept, and what the light carried away.

What it shows, in words

Both bars run from the same starting edge to the same finishing edge, so both stand for 100 units. The lower one is cut at 70, leaving a stretch of 30 beside it. Read on the lower bar that stretch is the energy sent out; read on the upper bar it is what is left of the 100 after 70 is set aside. One length, two readings, which is why 100 = 70 + 30 and 100 − 70 = 30 say the same thing.

Nothing in the drawing says which piece to work out first, and there is no step in it that has to be done in an order. That is the difference between a statement and an instruction.

Where this lesson stops

An equals sign states that two quantities are the same. Which quantities, measured how, is what the paper around it has to say.

Draw the boundary around the body alone. Name its energies without giving them values: E₀ before the emission and E₁ after, both in the body’s rest frame; H₀ before and H₁ after, both in the moving frame. These are unknowns, not hidden values of mass times the speed of light squared.
In the rest frame energy is conserved once the light is counted. Before, the body has E₀. After, the body has E₁ and the light carries L. So E₀ = E₁ + L.
In the moving frame the same emission carries γL of light energy (the symmetric emission). Counting the light again gives H₀ = H₁ + γL.

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.

Move the unknowns to one side to read the changes: E₀ − E₁ = L and H₀ − H₁ = γL. Only these changes are known. No absolute energy has been found, and none is needed.
With β = 0.6 the moving account says the body lost 1.25L, while the rest account says L. That does not break conservation: conservation compares before and after inside one frame, never the numbers two frames assign.

A tool for this step

Conservation and symmetry

When an experiment is turned, moved or watched from a moving frame, what should stay the same?

Turn an apparatus to face another way, move it to another bench, or run it tomorrow instead of today, and it gives the same result. Each of those changes is a symmetry of the experiment: a change of setting that leaves the measured outcome unchanged. The relativity paper's first postulate adds one more: describing the experiment from a frame in uniform motion leaves the laws in the same form.

What stays the same is often a relation, not a number. The second postulate is a relation kept in both frames: light moves at the same speed c. §3 of the paper checks it on a flash from the origin. A sphere of light, $x^2 + y^2 + z^2 = c^2t^2$ , is also a sphere in the moving frame, with the moving frame's own coordinates.

\begin{gathered} x^2 + y^2 + z^2 = c^2 t^2 \\ \text{and} \\ x'^2 + y'^2 + z'^2 = c^2 t'^2 \end{gathered}

x squared plus y squared plus z squared equals c squared t squared, and the same with every coordinate primed.

Conservation works the same way. An energy is not the same number in every frame, but in each frame the energy before equals the energy after. The mass-energy paper uses exactly this. A body at rest sends out two equal pulses of light, L in all, in opposite directions. Seen from a frame moving along the pulses, the pulses carry different energies, and their total is γL rather than L, where γ = 1/√(1 − v²/c²) is the factor the relativity paper prints as β. Each frame keeps its own balance, and comparing the two balances is the whole argument.

Modern texts add one more quantity that every frame agrees on: $c^2\Delta t^2 - \Delta x^2$ for any two events, the spacetime interval. It is a useful check on a calculation, and its home is Minkowski's geometry of 1908. It is not a premise of the 1905 route, which starts from the two postulates.

Worked example: Two pulses from 0.6c, and one interval checked

In the body's frame, two pulses of energy $L/2$ leave in opposite directions along x: $L/2 + L/2 = L$ leaves the body.
From a frame moving at 0.6c along x, where γ = 1.25, a pulse sent along the motion carries $\frac{L}{2}\gamma\left(1 - \frac{v}{c}\right) = 0.5L \times 1.25 \times 0.4 = 0.25L$ , and the pulse sent back carries $0.5L \times 1.25 \times 1.6 = L$ .
Their total is $1.25L = \gamma L$ , not L. Each frame's account still balances, because each is compared only with itself, before and after.
The interval as a check: two events at the same time 10 light-seconds apart have $c^2\Delta t^2 - \Delta x^2 = 0 - 100 = -100$ square light-seconds. From 0.6c they are 7.5 seconds and 12.5 light-seconds apart, and $7.5^2 - 12.5^2 = 56.25 - 156.25 = -100$ again.

JavaScript is off, so the controls below cannot change anything. The construction shows its first setting, and its closing paragraph, “What it shows, in words”, describes the rest.

Try it: a change of description, or a change of the world?

A body at rest sends out two equal pulses of light, L in all, in opposite directions. Pick something to do to this experiment, and read whether it changes how the experiment is described or the experiment itself.

Pick a changeDescribe it from a frame moving at 0.6c along the pulsesCall the old y axis x, and the old x axis yTurn the whole apparatus so the pulses go along y instead of xHeat the body halfway through, before it emitsRecord the pulses with a detector that averages over a longer timeEstimate the pulse energy by fitting several readings instead of taking oneDraw the pulses in another colour

This is an observer change: a change of description. Nothing in the experiment changes, only how it is described or recorded.

What changes: The numbers: the pulses carry 0.25L and L instead of 0.5L each, 1.25L in all.

What stays: The events, the body, and the fact that each frame's energy balance holds before and after the emission.

What it shows, in words

Of the seven changes, two change the world: turning the whole apparatus, which is a new run and a symmetry because its outcome is the same, and heating the body, which changes the outcome. The other five change only the description: a moving observer sees the pulses carry 0.25L and L instead of 0.5L each, with each frame's energy balance still holding; a relabelling, a slower detector, a different estimate and a new colour change the numbers, the readings or the picture, never the experiment. These are the same classes the site's laboratories use to keep an observer change from ever starting a new experiment.

Where this lesson stops

A symmetry is a change that leaves the measured outcome unchanged. This lesson stops at naming what stays the same; showing why it must is the papers' work.

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The same emission in the moving description

H_0 - H_1 = \gamma\,L

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

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

The two opposite pulse energies acquire opposite angular terms, which cancel in their sum. Their total is γL in the moving description; the light transformation is an imported premise.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.
The moving-frame energy law for light is imported from special relativity.

Moving-frame account before: H₀ is the same body’s pre-emission energy described by the moving observer, not a second physical emission. Read the prerequisite
Moving-frame account after: H₁ is the body’s post-emission energy in that same moving frame. The observer speed is unchanged between the two accounts. Read the prerequisite
The moving-frame decrease: This is the energy lost by the same body according to the same moving observer before and after. Read the prerequisite
Modern Lorentz factor: γ depends on the observer speed. In this modern teaching notation γ is the factor, not the speed ratio v/c. The transformation of the light energy is imported from relativity, not derived from these energy accounts. Read the prerequisite
Emitted energy L: Total energy of both opposite pulses, in the body’s rest frame. It is not the body’s unknown total energy. Read the prerequisite
Multiply the factors: Apply the imported light-energy factor to the total rest-frame emitted energy after the opposite angular terms cancel. Read the prerequisite
What this relation asserts: The two opposite pulse energies acquire opposite angular terms, which cancel in their sum. Their total is γL in the moving description; the light transformation is an imported premise. Read the prerequisite

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The energy removed in the rest description

E_0 - E_1 = L

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

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

Conservation in the rest description gives a difference of body energies. It does not assign the body an absolute rest energy.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.

Rest-frame account before: E₀ is the body’s pre-emission energy in its rest description. Its absolute value remains unknown: no formula involving an assigned rest mass is inserted. Read the prerequisite
Rest-frame account after: E₁ is the body’s remaining energy in the same rest description after both pulses leave. Before and after refer to the same body boundary. Read the prerequisite
Before minus after: The difference is the positive energy leaving the selected body boundary. Read the prerequisite
Emitted energy L: Total energy of both opposite pulses, in the body’s rest frame. It is not the body’s unknown total energy. Read the prerequisite
What this relation asserts: Conservation in the rest description gives a difference of body energies. It does not assign the body an absolute rest energy. Read the prerequisite

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Connect the result terms to the coefficient laboratory →

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

Start from the two accounts written as changes: H₀ − H₁ = γL in the moving frame, and E₀ − E₁ = L in the rest frame.
Subtract the second equation from the first, side by side: (H₀ − H₁) − (E₀ − E₁) = γL − L.
Remove the brackets on the left. A minus sign in front of a bracket changes the sign of every term inside it, which gives H₀ − H₁ − E₀ + E₁.
Put the two terms from before the emission together, and the two from after: (H₀ − E₀) − (H₁ − E₁). Expanding this again gives H₀ − E₀ − H₁ + E₁, the same four terms, so nothing was lost.
On the right, L is a common factor: γL − L = L(γ − 1). The whole chain reads:

\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.

Read each bracket. H₀ − E₀ is how much the moving account exceeds the rest account before the emission, and H₁ − E₁ is the same comparison after it. Their difference is L(γ − 1), which is 0.25L at β = 0.6.
No absolute energy was needed. The algebra does not say what the surviving quantity is physically: calling it a change in energy of motion needs a premise, which the next passage states.

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Subtract the two ledgers

\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}

minus , minus minus , is minus , which is times minus one.

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H zero minus E zero, minus H one minus E one, equals gamma L minus L, which equals L times gamma minus one.

Each bracket is the kinetic energy plus the same unknown constant, so the unknowns cancel and what is left depends only on L and the speed.

Model assumptions and every term’s meaning

Each frame's energy differs from the other's by the kinetic energy plus a constant that does not change on emission.
These are modern teaching equations in SI notation, not a transcription of the printed paper.

H one minus E one: Kinetic energy after, plus the same constant. Read the prerequisite
The after bracket: Kinetic energy after emission, plus the same constant; subtracted as a whole. Read the prerequisite
H zero minus E zero: Kinetic energy before, plus the constant. Read the prerequisite
The before bracket: Kinetic energy before emission, plus the unknown constant. Read the prerequisite
L times gamma minus one: Factored. Read the prerequisite
Using the two ledgers: Each frame's energy balance replaces the brackets by the light it gave off. Read the prerequisite
In brackets: The light's energy in the moving frame, kept together. Read the prerequisite
Gamma L: The light energy in the moving frame. Read the prerequisite
In brackets: Gamma minus one, multiplied as a whole by L. Read the prerequisite
Gamma minus one: Zero at rest, growing with speed. Read the prerequisite
Gamma L minus L: The two ledgers' light, subtracted. Read the prerequisite
Before minus after: The constant cancels. Read the prerequisite
What it asserts: The drop in kinetic energy is L times gamma minus one, with no unknown left. Read the prerequisite
Body energy before, rest frame: The body's energy before emission, in its own frame. Read the prerequisite
Body energy after, rest frame: The body's energy after emission, in its own frame. Read the prerequisite
Emitted light energy: The total energy of the two pulses in the body's rest frame; the paper writes L. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. Read the prerequisite
Body energy before, moving frame: The body's energy before emission, in the frame where it moves. Read the prerequisite
Body energy after, moving frame: The body's energy after emission, in the frame where it moves. Read the prerequisite

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Subtract the two energy accounts

\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.

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H before minus E before, less H after minus E after, equals L times gamma minus one.

Subtracting the rest-frame conservation equation from the moving-frame equation eliminates the unknown absolute energies. This arithmetic alone does not yet identify a kinetic-energy change.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.

Moving-frame account before: H₀ is the same body’s pre-emission energy described by the moving observer, not a second physical emission. Read the prerequisite
Rest-frame account before: E₀ is the body’s pre-emission energy in its rest description. Its absolute value remains unknown: no formula involving an assigned rest mass is inserted. Read the prerequisite
Compare the pre-emission descriptions: H₀ − E₀ compares two frame descriptions of the pre-emission body, not two different bodies. Read the prerequisite
Moving-frame account after: H₁ is the body’s post-emission energy in that same moving frame. The observer speed is unchanged between the two accounts. Read the prerequisite
Rest-frame account after: E₁ is the body’s remaining energy in the same rest description after both pulses leave. Before and after refer to the same body boundary. Read the prerequisite
Compare the post-emission descriptions: H₁ − E₁ compares those same frames after the same emission. Read the prerequisite
Subtract the frame differences: Taking the difference of these differences removes the unknown absolute energy levels. A kinetic interpretation still needs an additional premise. Read the prerequisite
Emitted energy L: Total energy of both opposite pulses, in the body’s rest frame. It is not the body’s unknown total energy. Read the prerequisite
Modern Lorentz factor: γ depends on the observer speed. In this modern teaching notation γ is the factor, not the speed ratio v/c. The transformation of the light energy is imported from relativity, not derived from these energy accounts. Read the prerequisite
Subtract the rest-frame factor: At zero relative speed γ is one. Subtracting one isolates the extra light energy in the moving account; it is not an independent mass-energy premise. Read the prerequisite
Multiply the factors: The two radiation totals differ by γL − L = L(γ − 1). Read the prerequisite
What this relation asserts: Subtracting the rest-frame conservation equation from the moving-frame equation eliminates the unknown absolute energies. This arithmetic alone does not yet identify a kinetic-energy change. Read the prerequisite

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Subtract the rest account from the moving account

\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.

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H before minus H after, less E before minus E after, equals gamma L minus L.

Subtract the whole rest-frame conservation relation from the moving-frame relation. Keep parentheses so the second minus sign also reverses the after-emission term.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.

Moving-frame account before: H₀ is the same body’s pre-emission energy described by the moving observer, not a second physical emission. Read the prerequisite
Moving-frame account after: H₁ is the body’s post-emission energy in that same moving frame. The observer speed is unchanged between the two accounts. Read the prerequisite
Subtract the whole expression: The negative sign applies to every term in the second expression. Both accounts use the same body boundary. Read the prerequisite
Rest-frame account before: E₀ is the body’s pre-emission energy in its rest description. Its absolute value remains unknown: no formula involving an assigned rest mass is inserted. Read the prerequisite
Rest-frame account after: E₁ is the body’s remaining energy in the same rest description after both pulses leave. Before and after refer to the same body boundary. Read the prerequisite
Modern Lorentz factor: γ depends on the observer speed. In this modern teaching notation γ is the factor, not the speed ratio v/c. The transformation of the light energy is imported from relativity, not derived from these energy accounts. Read the prerequisite
Emitted energy L: Total energy of both opposite pulses, in the body’s rest frame. It is not the body’s unknown total energy. Read the prerequisite
Moving-frame emitted energy: Use the imported light-energy transformation. Read the prerequisite
Subtract the rest account from the moving account: Subtract the whole rest-frame conservation relation from the moving-frame relation. Keep parentheses so the second minus sign also reverses the after-emission term. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Connect the result terms to the coefficient laboratory →

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.
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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.
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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.
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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

Let K₀ and K₁ be the body’s energy of motion, its kinetic energy, in the moving frame before and after the emission. In its rest frame the body does not move, so there it has none.
The premise: the difference between the two accounts is the kinetic energy plus a constant C that does not change when the body emits light. The same C before and after:

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.

Subtract the second relation from the first: (H₀ − E₀) − (H₁ − E₁) = (K₀ + C) − (K₁ + C).
On the right, +C and −C cancel: K₀ + C − K₁ − C = K₀ − K₁.
The left side is the subtraction already worked out, L(γ − 1). So the kinetic-energy drop is:

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

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

If the offsets were allowed to differ, C₀ before and C₁ after, the same steps would give (K₀ − K₁) + (C₀ − C₁) = L(γ − 1), and the kinetic-energy drop could not be separated from the change in offset. The unchanged C is what makes the identification possible, and it is asserted by this route, not measured.
The body moves at the same speed before and after, yet its kinetic energy drops. The small-speed passage compares this drop with Newton’s kinetic energy to find what changed.

A tool for this step

Energy of motion and inertia

If a body moves at the same speed but carries less energy of motion, what has changed?

A force that pushes a body through a distance does work on it, and the work becomes the body's energy of motion, its kinetic energy. That is separate from energy stored inside the body, as heat or in its chemistry.

For one body at everyday speeds, doubling the speed multiplies the energy of motion by four. At a fixed speed, doubling the mass doubles it. The mass-energy paper's title asks about a body's Trägheit, its inertia: its resistance to a change in its motion, which is measured by pushing it, not by weighing it.

K = \frac{1}{2}\,m\,v^{2}

In Newtonian mechanics, kinetic energy is one half times inertial mass times speed squared.

This formula holds at speeds small compared with the speed of light, and it is not exact close to that speed. So the mass-energy argument works with the exact expression and reads the mass from its low-speed limit, where it takes this form.

Worked example: Two bodies at the same speed, different masses

Two bodies move at 2 metres per second.
The 3 kg body has ½ × 3 × 2² = 6 joules of energy of motion; the 2 kg body has ½ × 2 × 2² = 4 joules.
The speed is the same, and the energies differ by 2 joules because the masses differ by 1 kg.

K_0 - K_1 = \frac{1}{2}\,\left(m_0 - m_1\right)\,v^{2}

At the same low speed, the kinetic-energy drop equals one half times the mass decrease times the squared speed.

Here the two masses are given, so the example proves nothing about mass and energy. The 1905 argument runs the other way: it computes the energy difference from the light the body gives off, finds that it has this form, and reads the mass difference from the coefficient of ½v².

JavaScript is off, so the controls below cannot change anything. The construction shows its first setting, and its closing paragraph, “What it shows, in words”, describes the rest.

Try it: a before-and-after ledger

A steady push along the motion does work equal to the force times the distance, and the work becomes energy of motion. Choose the body, how fast it is already going, and the push. The last choice changes where every energy is counted from.

Mass

Speed before the push

Force

Distance pushed

Count every energy from

The push does 3 N × 2 m = 6 J of work, and the 3 kg body goes from 0 to 2 m/s. Its energy of motion rises by exactly the work.

The ledger, in joules
Entry	Energy of motion	Counted from zero
Before the push, ½mv²	0	0
Work done, F × d	+ 6	+ 6
After the push	6	6

What it shows, in words

A push of 3 N through 2 m does 6 J of work. A 3 kg body starting at rest ends with 6 J of energy of motion, which at ½mv² is a speed of 2 m/s, the lesson's worked example read the other way. From rest, doubling the distance doubles the work but raises the speed only by a factor of √2, about 1.41, because energy of motion grows with the square of the speed. Counting every energy from 100 J or from a million joules changes both totals by the same amount and changes no difference, so nothing that can be measured changes. That is why a comparison of energies gives a difference in mass, and never the total energy stored inside a body.

Where this lesson stops

At the same low speed, less energy of motion means less mass. The comparison gives only a difference in mass, never the total energy stored inside the body.

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The conditional exact kinetic-energy drop

\Delta K = L\,\left(\gamma - 1\right)

The kinetic-energy drop, K before minus K after, equals L times gamma minus one.

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The kinetic-energy drop, K before minus K after, equals L times gamma minus one.

Under the unchanged-offset premise the C terms cancel, so the frame-account difference becomes the exact kinetic-energy drop. This still does not identify the mass decrease at a finite speed.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.
The relation H − E = K + C uses the same unknown additive C before and after emission. This is a premise; conservation and subtraction do not prove it.

Positive kinetic-energy drop: ΔK means K₀ − K₁ in this card: before minus after. It is identified from the ledger subtraction only under the unchanged-offset premise. Read the prerequisite
Emitted energy L: Total energy of both opposite pulses, in the body’s rest frame. It is not the body’s unknown total energy. Read the prerequisite
Modern Lorentz factor: γ depends on the observer speed. In this modern teaching notation γ is the factor, not the speed ratio v/c. The transformation of the light energy is imported from relativity, not derived from these energy accounts. Read the prerequisite
Subtract the rest-frame factor: At zero relative speed γ is one. Subtracting one isolates the extra light energy in the moving account; it is not an independent mass-energy premise. Read the prerequisite
Multiply the factors: Multiply the known emitted energy by the excess of the moving-frame light factor over one. Read the prerequisite
What this relation asserts: Under the unchanged-offset premise the C terms cancel, so the frame-account difference becomes the exact kinetic-energy drop. This still does not identify the mass decrease at a finite speed. Read the prerequisite

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Name the before-minus-after difference

\Delta K := K_0 - K_1

Delta K is defined as K before minus K after.

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Delta K is defined as K before minus K after.

Delta K is shorthand for a difference, not an additional energy law. Its before-minus-after sign is the reverse of a signed after-minus-before change.

Model assumptions and every term’s meaning

Delta K is defined here as kinetic energy before minus after. The definition alone supplies no physical value.
This is modern explanatory notation, not a printed source transcription.

Positive kinetic-energy drop: ΔK means K₀ − K₁ in this card: before minus after. It is identified from the ledger subtraction only under the unchanged-offset premise. Read the prerequisite
Energy of motion before: K₀ is energy of motion before emission, relative to the moving observer. The subtraction does not determine its absolute value. Read the prerequisite
Energy of motion after: K₁ is energy of motion after emission at the same speed. Comparing the two kinetic energies identifies a drop without inventing their absolute values. Read the prerequisite
Subtract the whole expression: The negative sign applies to every term in the second expression. Both accounts use the same body boundary. Read the prerequisite
Name the before-minus-after difference: Delta K is shorthand for a difference, not an additional energy law. Its before-minus-after sign is the reverse of a signed after-minus-before change. Read the prerequisite

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Cancel the shared offset

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

K before minus K after equals L times gamma minus one.

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K before minus K after equals L times gamma minus one.

Expanding the parentheses gives K before plus C minus K after minus C. The C terms cancel because they represent the same unchanged offset, leaving a difference of kinetic energies.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.
The relation H − E = K + C uses the same unknown additive C before and after emission. This is a premise; conservation and subtraction do not prove it.

Energy of motion before: K₀ is energy of motion before emission, relative to the moving observer. The subtraction does not determine its absolute value. Read the prerequisite
Energy of motion after: K₁ is energy of motion after emission at the same speed. Comparing the two kinetic energies identifies a drop without inventing their absolute values. Read the prerequisite
Subtract the whole expression: The negative sign applies to every term in the second expression. Both accounts use the same body boundary. Read the prerequisite
Emitted energy L: Total energy of both opposite pulses, in the body’s rest frame. It is not the body’s unknown total energy. Read the prerequisite
Modern Lorentz factor: γ depends on the observer speed. In this modern teaching notation γ is the factor, not the speed ratio v/c. The transformation of the light energy is imported from relativity, not derived from these energy accounts. Read the prerequisite
One less than the Lorentz factor: This factor comes from subtracting the rest account from the moving account, not from a rest-energy formula. Read the prerequisite
The difference in emitted energy: Factor the difference gamma L minus L as L times gamma minus one. Read the prerequisite
Cancel the shared offset: Expanding the parentheses gives K before plus C minus K after minus C. The C terms cancel because they represent the same unchanged offset, leaving a difference of kinetic energies. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

State the offset premise after emission

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.

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Each term and operation, in words

No live values

This equation states a relation. No instrument on this page computes its terms.

Read the equation aloud in words

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

This relation is assumed, not proved by conservation. Both before and after use the same C; if the offset changes, the two kinetic energies cannot be identified by cancelling it.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.
The relation H − E = K + C uses the same unknown additive C before and after emission. This is a premise; conservation and subtraction do not prove it.

Moving-frame account after: H₁ is the body’s post-emission energy in that same moving frame. The observer speed is unchanged between the two accounts. Read the prerequisite
Rest-frame account after: E₁ is the body’s remaining energy in the same rest description after both pulses leave. Before and after refer to the same body boundary. Read the prerequisite
Compare the two descriptions: The energy difference between frames is related to kinetic energy only with the stated additive-offset premise. Read the prerequisite
Energy of motion after: K₁ is energy of motion after emission at the same speed. Comparing the two kinetic energies identifies a drop without inventing their absolute values. Read the prerequisite
The unchanged offset is a premise: C is an unknown additive energy offset. Using the same C in both equations assumes that emission does not change it. It is not set to zero or measured by the subtraction. Read the prerequisite
Keep the unknown offset: The additive C is left symbolic. It is neither silently set to zero nor inferred from the conservation equations. Read the prerequisite
What this relation asserts: This relation is assumed, not proved by conservation. Both before and after use the same C; if the offset changes, the two kinetic energies cannot be identified by cancelling it. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

State the offset premise before emission

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.

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Explain this equation

Each term and operation, in words

No live values

This equation states a relation. No instrument on this page computes its terms.

Read the equation aloud in words

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

This relation is assumed, not proved by conservation. Both before and after use the same C; if the offset changes, the two kinetic energies cannot be identified by cancelling it.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.
The relation H − E = K + C uses the same unknown additive C before and after emission. This is a premise; conservation and subtraction do not prove it.

Moving-frame account before: H₀ is the same body’s pre-emission energy described by the moving observer, not a second physical emission. Read the prerequisite
Rest-frame account before: E₀ is the body’s pre-emission energy in its rest description. Its absolute value remains unknown: no formula involving an assigned rest mass is inserted. Read the prerequisite
Compare the two descriptions: The energy difference between frames is related to kinetic energy only with the stated additive-offset premise. Read the prerequisite
Energy of motion before: K₀ is energy of motion before emission, relative to the moving observer. The subtraction does not determine its absolute value. Read the prerequisite
The unchanged offset is a premise: C is an unknown additive energy offset. Using the same C in both equations assumes that emission does not change it. It is not set to zero or measured by the subtraction. Read the prerequisite
Keep the unknown offset: The additive C is left symbolic. It is neither silently set to zero nor inferred from the conservation equations. Read the prerequisite
What this relation asserts: This relation is assumed, not proved by conservation. Both before and after use the same C; if the offset changes, the two kinetic energies cannot be identified by cancelling it. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

Substitute both offset premises before cancelling anything

\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.

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Each term and operation, in words

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K before plus C, less K after plus C, equals L times gamma minus one.

Replace H before minus E before by K before plus C, and the after-emission frame difference by K after plus the same C. Retain both copies before simplifying.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.
The relation H − E = K + C uses the same unknown additive C before and after emission. This is a premise; conservation and subtraction do not prove it.

Energy of motion before: K₀ is energy of motion before emission, relative to the moving observer. The subtraction does not determine its absolute value. Read the prerequisite
The unchanged offset is a premise: C is an unknown additive energy offset. Using the same C in both equations assumes that emission does not change it. It is not set to zero or measured by the subtraction. Read the prerequisite
Before: energy of motion plus offset: This is the pre-emission offset premise, not an algebraic identity. Read the prerequisite
Energy of motion after: K₁ is energy of motion after emission at the same speed. Comparing the two kinetic energies identifies a drop without inventing their absolute values. Read the prerequisite
After: the same offset: The second C is the same quantity as the first. An independently changed offset would not cancel. Read the prerequisite
Subtract the whole expression: The negative sign applies to every term in the second expression. Both accounts use the same body boundary. Read the prerequisite
Emitted energy L: Total energy of both opposite pulses, in the body’s rest frame. It is not the body’s unknown total energy. Read the prerequisite
Modern Lorentz factor: γ depends on the observer speed. In this modern teaching notation γ is the factor, not the speed ratio v/c. The transformation of the light energy is imported from relativity, not derived from these energy accounts. Read the prerequisite
One less than the Lorentz factor: This factor comes from subtracting the rest account from the moving account, not from a rest-energy formula. Read the prerequisite
The difference in emitted energy: Factor the difference gamma L minus L as L times gamma minus one. Read the prerequisite
Substitute both offset premises before cancelling anything: Replace H before minus E before by K before plus C, and the after-emission frame difference by K after plus the same C. Retain both copies before simplifying. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Connect the result terms to the coefficient laboratory →

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

Start from the exact result: K₀ − K₁ = L(γ − 1), with γ = 1/√(1 − β²) and β = v/c.
Write γ as a power: γ is (1 − β²) raised to the power −1/2. For small x, (1 − x) to the power −1/2 equals 1 + x/2 + 3x²/8 + and so on: the binomial series, a Taylor expansion at x = 0. Put x = β².
So γ − 1 = β²/2 + 3β⁴/8 + and so on, every later term carrying a higher even power of β.
Check the series at β = 0.1: γ = 1/√0.99 = 1.0050378, so γ − 1 = 0.0050378, and the first two terms give 0.005 + 0.0000375 = 0.0050375.
At low speed keep only the first term: K₀ − K₁ is close to Lβ²/2. Replace β² by v²/c²:

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.

Newton’s kinetic energy for a body of mass m at speed v is mv²/2. The speed is the same before and after, so Newton writes the drop as m₀v²/2 − m₁v²/2 = (m₀ − m₁)v²/2.
Both expressions are a coefficient times v²/2: (m₀ − m₁) in Newton’s, and L/c² in the one above. Setting the coefficients equal gives m₀ − m₁ = L/c². No rest-energy formula was assumed to get here.
The same identification, stated exactly: divide the exact drop by v²/2 and let v go to zero. From the series, 2(γ − 1)/v² = 1/c² + 3v²/(4c⁴) + and so on, so every later term vanishes in the limit:

\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.

At exactly v = 0 the quotient is 0/0 and cannot be computed; the limit is taken instead. At any nonzero speed the quotient still contains the later terms, so it is not the limiting coefficient itself.

A tool for this step

Approximating a curve near a point

Why can a curve be replaced, close to one point, by a few simple terms?

What is 2.1 squared, without a calculator? Start from what you know: 2 squared is 4. Picture a square 2 wide growing to 2.1 wide. It gains two strips, each 2 long and 0.1 thick, which add 0.4, and a small corner, 0.1 by 0.1, which adds 0.01. So 2.1 squared is 4.41. The first correction, 0.4, did almost all the work; the second, 0.01, is forty times smaller.

The same recipe works for any smooth curve, close to a point where you know it. Start with its value there. Add its slope times the step Δ: the slope, or first derivative, is the change in output per unit of input. Then add half its curvature times Δ²: the curvature, or second derivative, says how fast the slope itself changes. Further terms carry higher powers of Δ, so they are small only when Δ is:

\begin{aligned}f(x+\Delta)&=f(x)+\Delta f'(x)\\&\quad+\frac{\Delta^2}{2}f''(x)+\ldots\end{aligned}

The value a step delta away equals the value, plus delta times the slope, plus half of delta squared times the curvature, plus further terms.

The Brownian paper does exactly this in §4. It expands the number of particles at x + Δ in powers of the jump Δ and keeps the terms up to Δ², because only very small jumps contribute and each further surviving term is much smaller than the one before. The Δ term averages away, since jumps to the left and to the right are equally likely, and the Δ² term becomes the diffusion coefficient. A real particle does not make independent jumps over arbitrarily short times, so the expansion holds for times long enough for successive jumps to be independent and short enough for each jump to be small.

Worked example: What the dropped term costs

\left(x + \Delta\right)^{2} = x^{2} + 2\,x\,\Delta + \Delta^{2}

For the square function, the local expansion is exact through the quadratic term.

Keep only the value and the slope term, 4 + 2 × 2 × Δ, and see what dropping Δ² costs as the step grows:

Δ = 0.1: 4.4 instead of 4.41. The error, 0.01, is about 0.2 per cent.
Δ = 0.5: 6 instead of 6.25. The error, 0.25, is 4 per cent.
Δ = 2: 12 instead of 16. The error, 4, is a quarter of the answer.
Halving the step quarters the error, because the dropped term is Δ².

The square, its two strips and its corner

A square 2 wide grown to 2.1, drawn to scale. The strips are a twentieth of the side and the corner is what the straight-line estimate throws away.

Keeping the square and the strips gives 4.4, which is the straight-line estimate. The whole grown square is 4.41. The difference is the corner, 0.01, and it did about a fiftieth of the work the strips did.

Now the same square at each step the lesson lists, all three at one scale. The corner grows as the square of the step, so it catches up quickly.

What it shows, in words

The first drawing is a square of side 2 and area 4, grown to 2.1. It gains a strip along its right edge and another along its top, each 2 long and 0.1 thick, which together add 0.4, and a small square corner of 0.1 by 0.1, which adds 0.01. Together that is 4.41. Everything is at its true size, so the corner really is the speck it looks like: a four-hundredth of the area.

The second drawing is the same square grown by 0.1, 0.5, 2 in turn, all at one scale. The dropped corners are 0.01, 0.25, 4. At the smallest step the corner is a speck and the straight-line estimate is almost the whole answer; at a step of 2 the corner is a quarter of the square and the estimate has stopped being useful. Halving the step quarters the corner, because the corner is the step squared.

What it does not show: squaring is the case where the series stops of its own accord. A square has one corner and nothing after it, so keeping the value, the strips and the corner is not an approximation at all, it is the whole answer. For a curve that is not a square there are further terms behind the corner, each smaller than the last near the point and each growing faster away from it, and nothing here draws them.

Extension · The step the mass–energy paper takes

Try it: the Lorentz factor at low speed, term by term

The factor 1/√(1 − v²/c²) runs through relativity. It is now called γ; the relativity paper writes it β, with V for the speed of light. The mass–energy paper finds that a body's kinetic energy falls by L(1/√(1 − v²/V²) − 1) when it emits light of energy L, and then expands that bracket. Writing x = v²/c², Newton's binomial series gives:

(1 − x)−1/2 = 1 + ½ x + ⅜ x² + ⁵/₁₆ x³ + ³⁵/₁₂₈ x⁴ + …   (|x| < 1)

1. Moderate speeds: partial sums at v/c = 0.6 (x = 0.36)

At sixty percent the speed of light, x = 0.36. The exact value of γ − 1 is 1/√(1 − 0.36) − 1 = 1/0.8 − 1 = 0.25. Add the terms of the series one at a time and the total closes in on it:

Partial sums of γ − 1 at x = 0.36 against exact 0.25
Term added	Its value	Running total	Short of 0.25
½ x	0.180000	0.180000	0.070000
⅜ x²	0.048600	0.228600	0.021400
⁵/₁₆ x³	0.014580	0.243180	0.006820
³⁵/₁₂₈ x⁴	0.004593	0.247773	0.002227

Keeping only the first term is the paper's own step: "neglecting quantities of fourth and higher order", it writes the loss of kinetic energy as (L/V²)(v²/2). At v = 0.6c that gives 0.18L where the exact bracket gives 0.25L. The paper needs only low speeds, where the neglected terms are tiny.

2. Low speeds: a computer is also better off with the series

At v/c = 10⁻⁴, about 30 kilometres per second, roughly the Earth's speed round the Sun, x = 10⁻⁸ and γ = 1.000000005. A computer keeps about 16 significant digits, so subtracting 1 from γ throws away the first eight and leaves an answer made mostly of rounding error. The first term of the series, or the bracket rearranged so that nothing close to 1 is subtracted, keeps every digit:

Comparison at v/c = 10⁻⁴ (x = 10⁻⁸)

First series term (½ x):: 5.0000000000 × 10⁻⁹
Rearranged, nothing near 1 subtracted:: 5.0000000375 × 10⁻⁹
Its difference from the first term, relative:: 7.500000 × 10⁻⁹, which the next term predicts: 0.75x
Direct subtraction, 1/√(1 − x) − 1:: 5.0000001917 × 10⁻⁹
Its difference from the first term, relative:: 3.833145 × 10⁻⁸, about five times the true difference

The rearrangement: γ − 1 = x / (√(1 − x) · (1 + √(1 − x))). It is the same number as 1/√(1 − x) − 1, written so that no two nearly equal quantities are subtracted. At low speed, direct subtraction gets the eighth significant figure wrong; the rearrangement and the series do not.

Where this lesson stops

An expansion is a claim about size. Say which terms you keep, and why the ones you drop are smaller: here, because they carry higher powers of a small step.

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.

Explore the equation · Modern model notation

Why a finite-speed quotient is not the mass decrease

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.

Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.

Explain this equation

Each term and operation, in words

No live values

Symbolic here. Open the linked laboratory for a worked example and live values.

Read the equation aloud in words

The finite-speed proxy is defined as twice the exact kinetic-energy drop divided by v squared.

This definition is useful for approaching the low-speed coefficient. At finite speed the proxy is generally larger than L/c². At v = 0 the quotient is not applicable; the analytic limit must be evaluated separately.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.
The relation H − E = K + C uses the same unknown additive C before and after emission. This is a premise; conservation and subtraction do not prove it.
v is nonzero for this quotient; no zero-over-zero value is supplied.

A finite-speed proxy, not the mass loss: This is 2ΔK/v² evaluated at a nonzero speed. Its speed dependence is exactly why the low-speed limit is needed. At v = 0 this quotient is not applicable, not zero. Read the prerequisite
Positive kinetic-energy drop: ΔK means K₀ − K₁ in this card: before minus after. It is identified from the ledger subtraction only under the unchanged-offset premise. Read the prerequisite
Multiply the factors: Twice the exact kinetic-energy drop is the numerator, not twice the quadratic approximation. Read the prerequisite
Same observer speed: v is the signed relative speed between inertial descriptions. Its magnitude is less than c. It is held fixed while the body emits. Read the prerequisite
Square the dimensional speed: v² carries square metres per square second. Energy divided by v² therefore has units of mass. Read the prerequisite
Compare with the Newtonian coefficient: Divide by one half v² to inspect the coefficient a Newtonian kinetic-energy form would have. Taking this ratio at finite speed is not taking its limit. Read the prerequisite
What this relation asserts: This definition is useful for approaching the low-speed coefficient. At finite speed the proxy is generally larger than L/c². At v = 0 the quotient is not applicable; the analytic limit must be evaluated separately. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

At low speed, the drop looks like a kinetic energy

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

minus is approximately one half times over squared times squared.

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K zero minus K one is approximately one half times L over c squared times v squared.

The drop has the shape of one half m v squared, with L over c squared standing where a mass stands. That shape, not a coincidence of numbers, is the argument.

Model assumptions and every term’s meaning

Speeds small compared with light; higher powers of v over c dropped.
These are modern teaching equations in SI notation, not a transcription of the printed paper.

The drop: Before minus after. Read the prerequisite
One half: The half in one half m v squared. Read the prerequisite
c squared: The quantity multiplied by itself. Read the prerequisite
One half times L over c squared times v squared: The form of a kinetic energy. Read the prerequisite
L over c squared: In the place of a mass. Read the prerequisite
What it asserts: Approximately, to leading order in v over c. Read the prerequisite
v squared: The quantity multiplied by itself. Read the prerequisite
Kinetic energy after: Measured in the same moving frame, after emission. Read the prerequisite
Kinetic energy before: Measured in the moving frame. Read the prerequisite
Emitted light energy: The total energy of the two pulses in the body's rest frame; the paper writes L. Read the prerequisite
Speed of light: The paper writes V. Read the prerequisite
Speed of the body: How fast the body moves in the second frame. Read the prerequisite

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Explore the equation · Modern model notation

The low-speed limit: the mass the body loses

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

As the shrinks, twice the kinetic-energy drop over its square settles on divided by the square of .

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The limit, as v approaches zero, of 2 L times gamma minus one, over v squared, equals L over c squared.

At any finite speed the quotient is a little larger than L/c². As the speed shrinks it settles on L/c², the coefficient Newtonian mechanics reads as the mass the body lost. The limit is taken, not set at v = 0, where the quotient is not defined.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The same unchanged additive energy offset before and after emission, so the kinetic-energy drop is exactly L(γ − 1).
These are authored modern teaching equations, not a transcription or printed-notation concordance.

What it asserts: The quotient settles on L/c² as the speed shrinks, whatever the size of L. Read the prerequisite
The limit as v shrinks: The value the quotient settles on as v gets closer and closer to zero. It is not the quotient at v = 0, which divides by zero. Read the prerequisite
Twice the drop over v²: The finite-speed quotient. At every finite speed it is a little larger than L/c². Read the prerequisite
Twice the kinetic-energy drop: 2L(γ − 1): twice the exact drop, so that it compares with the Newtonian ½ m v². Read the prerequisite
The factor's excess over one: γ − 1, zero at rest. For small speeds it begins with v²/(2c²). Read the prerequisite
Speed squared: Divide by v² to leave a mass. Read the prerequisite
The coefficient: L/c²: the energy given off, divided by the square of the speed of light. Read the prerequisite
c squared: The speed of light, squared. Read the prerequisite
Observer speed: The speed of the second observer relative to the body, which the limit shrinks toward zero. Read the prerequisite
Observer speed: The same v. Read the prerequisite
Emitted energy L: Total energy of both opposite pulses, in the body’s rest frame. Read the prerequisite
Emitted energy L: The same L. Read the prerequisite
Lorentz factor: γ for the observer speed v: one at rest, larger at any other speed. Read the prerequisite
Speed of light: c, the speed of light in empty space. Read the prerequisite

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Explore the equation · Modern model notation

Keep only the second-order energy term

\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.

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The exact kinetic-energy drop is approximately one half L times the squared ratio v over c, at small speed.

The approximation sign is essential. Retaining only second order in v/c drops fourth and higher orders; it is not an exact equality at a finite speed.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.
The relation H − E = K + C uses the same unknown additive C before and after emission. This is a premise; conservation and subtraction do not prove it.
The observer speed is small compared with c; fourth and higher orders in v/c are omitted.

Positive kinetic-energy drop: ΔK means K₀ − K₁ in this card: before minus after. It is identified from the ledger subtraction only under the unchanged-offset premise. Read the prerequisite
Emitted energy L: Total energy of both opposite pulses, in the body’s rest frame. It is not the body’s unknown total energy. Read the prerequisite
Multiply the factors: The coefficient one half comes from the quadratic term in the Lorentz-factor expansion. Read the prerequisite
Same observer speed: v is the signed relative speed between inertial descriptions. Its magnitude is less than c. It is held fixed while the body emits. Read the prerequisite
Modern SI speed of light: c denotes the speed of light in modern SI notation. A normalized model with c = 1 is a different unit convention and cannot supply values labeled metres per second. Read the prerequisite
Compare the speed with c: v/c is dimensionless. Its sign gives the relative direction; squaring removes that direction. Read the prerequisite
Square the speed ratio: Both positive and negative observer speeds give the same square. The low-speed expansion uses this small dimensionless number. Read the prerequisite
Multiply the factors: This second-order expression is a low-speed approximation to the exact drop, not an identity. Read the prerequisite
What this relation asserts: The approximation sign is essential. Retaining only second order in v/c drops fourth and higher orders; it is not an exact equality at a finite speed. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Connect the result terms to the coefficient laboratory →

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

From the small-speed comparison, m₀ − m₁ = L/c². L is the light energy emitted, measured in the rest frame, and it is positive, so the difference is positive: the body’s mass after the emission is the smaller one.
The signed change is the final mass minus the initial mass, Δm = m₁ − m₀. It is the negative of the difference above:

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.

In the historical units energy is in erg, mass in grams, and c² is the rounded printed factor 9 × 10²⁰, in centimetres squared per second squared. An emission of L = 9 × 10²⁰ erg would lower the mass by 9 × 10²⁰ erg divided by 9 × 10²⁰ cm²/s², which is 1 gram.
In modern SI units energy is in joules and mass in kilograms, and c = 299 792 458 metres per second exactly, so c² is about 8.99 × 10¹⁶ m²/s². An emission of 1 joule lowers the mass by 1/c², about 1.11 × 10⁻¹⁷ kilogram.
Never divide an energy in erg by an SI value of c², or joules by the printed factor: convert the units first. A unit conversion changes how the result is described, not the physical event.
The result is a difference between two masses. It does not fill the unknown energies E₀ and E₁ with mc², and the paper’s broader relation between energy content and inertia is a further inference, kept separate from this calculation.
A finite-speed estimate such as 2L(γ − 1)/v² at some nonzero speed is not this mass decrease; only its zero-speed limit is.

Explore the equations in this step

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Explore the equation · Modern model notation

The same change, signed

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

is minus : minus over squared.

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Delta m equals m one minus m zero, which equals minus L over c squared.

After minus before: negative, because the body lost mass. The same fact as the positive loss, with the opposite sign convention.

Model assumptions and every term’s meaning

Low speed.
These are modern teaching equations in SI notation, not a transcription of the printed paper.

Its definition: Delta m means after minus before. Read the prerequisite
After minus before: Negative for a loss. Read the prerequisite
L over c squared: The loss, as a mass. Read the prerequisite
What it asserts: The mass change is minus L over c squared. Read the prerequisite
c squared: The quantity multiplied by itself. Read the prerequisite
Mass after: The body's inertial mass afterwards. Read the prerequisite
Mass before: The body's inertial mass before it emits the light. Read the prerequisite
Signed mass change: After minus before. Read the prerequisite
Emitted light energy: The total energy of the two pulses in the body's rest frame; the paper writes L. Read the prerequisite
Speed of light: The paper writes V. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

Identify the positive inertia decrease

m_{\mathrm{loss}} = \frac{L}{c^{2}}

The positive inertial mass decrease equals L divided by c squared.

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The positive inertial mass decrease equals L divided by c squared.

Matching the low-speed drop to the Newtonian form one half times the mass decrease times v squared identifies L/c². This coefficient identification needs the low-speed limit and the unchanged-offset premise; it is not obtained by assuming a body energy mc².

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.
The relation H − E = K + C uses the same unknown additive C before and after emission. This is a premise; conservation and subtraction do not prove it.
The Newtonian low-speed kinetic-energy coefficient is one half m times v squared.
The mass decrease is identified from the analytic v → 0 limit, not by equating the finite-speed proxy with that limit.

Positive decrease in inertia: The positive amount by which the body’s inertial mass decreases. It is identified from the limiting low-speed coefficient, using the Newtonian kinetic-energy premise. Read the prerequisite
Emitted energy L: Total energy of both opposite pulses, in the body’s rest frame. It is not the body’s unknown total energy. Read the prerequisite
Modern SI speed of light: c denotes the speed of light in modern SI notation. A normalized model with c = 1 is a different unit convention and cannot supply values labeled metres per second. Read the prerequisite
Convert energy units to mass units: Dividing joules by square metres per square second gives kilograms. This SI factor is separate from a rounded historical cgs conversion. Read the prerequisite
Read the limiting coefficient: After matching the common one half v² factor at low speed, L/c² is the positive mass-decrease coefficient. Neither absolute body mass is supplied. Read the prerequisite
What this relation asserts: Matching the low-speed drop to the Newtonian form one half times the mass decrease times v squared identifies L/c². This coefficient identification needs the low-speed limit and the unchanged-offset premise; it is not obtained by assuming a body energy mc². Read the prerequisite

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Explore the equation · Modern model notation

The mass lost is the energy over c squared

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

minus is over squared.

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m zero minus m one equals L over c squared.

Before minus after: a positive loss, the emitted energy divided by the speed of light squared.

Model assumptions and every term’s meaning

Low speed, where kinetic energy is one half times mass times speed squared.
These are modern teaching equations in SI notation, not a transcription of the printed paper.

Before minus after: Positive when mass is lost. Read the prerequisite
L over c squared: A mass. Read the prerequisite
What it asserts: Giving off energy L lowers the mass by L over c squared. Read the prerequisite
c squared: The quantity multiplied by itself. Read the prerequisite
Mass after: The body's inertial mass afterwards. Read the prerequisite
Mass before: The body's inertial mass before it emits the light. Read the prerequisite
Emitted light energy: The total energy of the two pulses in the body's rest frame; the paper writes L. Read the prerequisite
Speed of light: The paper writes V. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

A signed change is the negative of a decrease

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

Mass after minus mass before equals negative L divided by c squared.

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Mass after minus mass before equals negative L divided by c squared.

For emission, L is positive but the signed body-mass change is negative. This concerns the body that lost the light; the energy has not vanished from a larger system containing body and radiation.

Model assumptions and every term’s meaning

Two equal, opposite light pulses remove total positive energy L; their symmetry removes recoil in the body’s rest frame.
The two inertial descriptions refer to the same emission and the same body boundary; |v| < c.
These are authored modern teaching equations, not a transcription or printed-notation concordance.
The relation H − E = K + C uses the same unknown additive C before and after emission. This is a premise; conservation and subtraction do not prove it.
Δm is defined as after minus before, whereas the decrease magnitude is before minus after.

Emitted energy L: Total energy of both opposite pulses, in the body’s rest frame. It is not the body’s unknown total energy. Read the prerequisite
Modern SI speed of light: c denotes the speed of light in modern SI notation. A normalized model with c = 1 is a different unit convention and cannot supply values labeled metres per second. Read the prerequisite
Square the speed of light: The squared SI speed supplies the energy-to-mass conversion, not an arbitrary numerical scaling. Read the prerequisite
Convert the emitted energy: L/c² is the positive magnitude of the mass decrease for the same body boundary. Read the prerequisite
After minus before: Δm is the signed change: mass after minus mass before. Emission of positive L therefore gives a negative value. It is not the positive decrease magnitude. Read the prerequisite
Reverse the subtraction order: Changing from before-minus-after to after-minus-before reverses the sign. Positive emission therefore makes this signed change negative. Read the prerequisite
What this relation asserts: For emission, L is positive but the signed body-mass change is negative. This concerns the body that lost the light; the energy has not vanished from a larger system containing body and radiation. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

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

Choose a boundary before reading any change of energy. The same emission reads differently depending on what the boundary encloses.
Body only: energy L has crossed the boundary outward, in the rest frame, and the body’s account records a loss.
Radiation only: the two departing pulses carry the emitted energy, L in total in the rest frame, and this account records a gain.
Body and all of the light together: nothing has crossed the larger boundary, so its total energy is the same before and after, in each frame separately.
A loss from one subsystem is therefore not a disappearance from the isolated whole. The light has left the body; it has not vanished.
The paper proposes bodies whose energy content varies strongly, such as radium salts, as possible tests, and leaves the physical claim conditional on agreement with facts. A prediction is not an observation: keep the conditional prediction apart from any independently calibrated measurement.
Return to the chain of the argument and name its three ingredients: the imported transformation of light energy, the premise that the offset between the two accounts does not change, and the low-speed identification that turns the kinetic-energy drop into a mass difference.
The system-boundaries laboratory lets you move the boundary yourself. Its modern invariant-mass reading and the 1906 box extension are later explanations, not steps hidden inside the 1905 inference.

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

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