Read · Mass and energy: two accounts, one subtraction

Mass and energy: two accounts, one subtraction

Follow the light-energy input, the two balances, the unchanged-offset premise, and the low-speed coefficient without assigning an absolute rest energy in advance.

Newly authored explanatory preview in modern notation; editorial and physics review are pending. This is not a German source transcription, this edition’s English translation, or a complete critical edition. The source faces and pinned facsimile remain in preparation. The paper has no numbered sections: s0 is an editorial route identifier, and the headings below organize this explanation, not a verified printed inventory.

Show me one example before the notation →

Detail

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First encounter · No algebra required

Can a body lose energy of motion without changing speed?

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

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

Choose the traveler

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

Beside the body

Before: unknown

After: that unknown amount minus 10 J

Light: 5 J in each of two opposite directions.

Total light: 10 J

The traveler’s account

Before: another unknown

After: that unknown amount minus 12.5 J

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

Total light: 12.5 J

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

Compare the accounts

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

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

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

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

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

60 percent of light speed

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

1 percent of light speed

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

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

Why the slower traveler matters

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

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

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

Why are the body energies unknown?

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

Why do the two observers disagree about the light?

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

What is an unchanged offset?

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

From two accounts to inertia

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

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

More guidance: energy of motion and inertia →

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

Continue with the slow traveler in the coefficient laboratory →

Follow the complete explanatory argument →

The unsectioned argument · explanatory reading

assumption · Within the stated model

One result is borrowed, not rediscovered

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

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(1-\beta\cos\varphi),\qquad\beta=v/c,\qquad\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, reference frames, and the coordinate grid

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 a reviewed transcription.

Explore the equation · Modern model notation

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.

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.

Input · Model result · Constant

Symbolic equation

No numerical binding is declared.

Lorentz factor: No accepted value is available here.
Signed observer speed: No accepted value is available here.
Speed of light in SI: No accepted value is available here.

Read the equation aloud in words

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 reviewed 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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

Connect the result terms to the coefficient laboratory →

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

Specify the relative speed v, with magnitude smaller than light speed c.
Write β for v divided by c, and γ for one divided by the square root of one minus β squared. These are modern teaching symbols.
Measure the pulse direction in the body’s rest frame. Its cosine determines whether the traveler’s account assigns more or less energy.
Multiply the pulse energy by γ(1 − β cos φ). Do this for each pulse of the same emission.
Keep the body’s energies unspecified. A rule for light is not a measurement of the body’s absolute internal energy.

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

This preview does not establish the imported transformation from first principles.
No absolute body energy or mass–energy formula is assigned.

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

derivation · Within the stated model

Choose the emission that removes recoil

Why send equal amounts of light in opposite directions?

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.

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

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 the total emitted light energy in the body’s rest frame L.
Give each pulse energy L/2 in that frame.
Apply the same imported transformation to each pulse, with opposite signs of cos φ.
Add the two energies. The negative and positive angle terms cancel.
The sum is γL for every emission angle, although each pulse can change with angle.
The cancellation concerns the light totals. It does not assert any absolute energy for the body.

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

derivation · Within the stated model

Two accounts of the same loss

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

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,\qquad 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.

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 a reviewed transcription.

Explore the equation · Modern model notation

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.

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.

Input · Model result · Constant

Symbolic equation

No numerical binding is declared.

Body energy before emission, moving description: No accepted value is available here.
Body energy after emission, moving description: No accepted value is available here.
Lorentz factor: No accepted value is available here.
Emitted energy in the body's rest frame: No accepted value is available here.

Read the equation aloud in words

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 reviewed 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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

Explore the equation · Modern model notation

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.

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.

Input · Model result · Constant

Symbolic equation

No numerical binding is declared.

Body energy before emission, rest description: No accepted value is available here.
Body energy after emission, rest description: No accepted value is available here.
Emitted energy in the body's rest frame: No accepted value is available here.

Read the equation aloud in words

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 reviewed 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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

Connect the result terms to the coefficient laboratory →

Show every step here: Two accounts of the same loss

Draw the boundary around the body alone.
Record its energy before and after in its rest frame; the loss is L.
Write a second before-and-after account for the same body in the traveler’s frame; the light loss is γL.
Do not solve for an unknown absolute energy. Only the changes have been specified.
Including the light in each account restores its original total. Changing the observer is not a source of extra 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

derivation · Within the stated model

Subtract what you cannot measure

Which difference survives when the accounts are subtracted?

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.

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

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.

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 a reviewed transcription.

Explore the equation · Modern model notation

Subtract the two energy accounts

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

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

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.

Input · Model result · Constant

Symbolic equation

No numerical binding is declared.

Body energy before emission, moving description: No accepted value is available here.
Body energy before emission, rest description: No accepted value is available here.
Body energy after emission, moving description: No accepted value is available here.
Body energy after emission, rest description: No accepted value is available here.
Emitted energy in the body's rest frame: No accepted value is available here.
Lorentz factor: No accepted value is available here.

Read the equation aloud in words

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 reviewed 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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

Explore the equation · Modern model notation

Subtract the rest account from the moving account

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

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

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.

Input · Model result · Constant

Symbolic equation

No numerical binding is declared.

Body energy before emission, moving description: No accepted value is available here.
Body energy after emission, moving description: No accepted value is available here.
Body energy before emission, rest description: No accepted value is available here.
Body energy after emission, rest description: No accepted value is available here.
Lorentz factor: No accepted value is available here.
Emitted energy in the body's rest frame: No accepted value is available here.
Emitted energy in the body's rest frame: No accepted value is available here.

Read the equation aloud in words

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 reviewed 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
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
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
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
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
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
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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

Connect the result terms to the coefficient laboratory →

Show every step here: Subtract what you cannot measure

Start with H₀ − H₁ = γL and E₀ − E₁ = L.
Subtract the second equation from the first: H₀ − H₁ − E₀ + E₁ = γL − L.
Regroup the left as (H₀ − E₀) − (H₁ − E₁).
Factor L on the right to obtain L(γ − 1).
Keep the result as a difference between accounts until the relation to energy of motion has been stated.

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

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

assumption · Within the stated model

The premise that the subtraction needs

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

H_0-E_0=K_0+C,\qquad 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(\gamma-1)

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

Open the foundation: Energy of motion and inertia

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 a reviewed transcription.

Explore the equation · Modern model notation

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.

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.

Input · Model result · Constant

Symbolic equation

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

Exact drop in energy of motion: No accepted value is available here.
Emitted energy in the body's rest frame: No accepted value is available here.
Lorentz factor: No accepted value is available here.

Read the equation aloud in words

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 reviewed 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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

Explore the equation · Modern model notation

Name the before-minus-after difference

\Delta K := K_0 - K_1

Delta K is defined as K before minus K after.

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.

Input · Model result · Constant

Symbolic equation

No numerical binding is declared.

Exact drop in energy of motion: No accepted value is available here.
Energy of motion before emission: No accepted value is available here.
Energy of motion after emission: No accepted value is available here.

Read the equation aloud in words

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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

Explore the equation · Modern model notation

Cancel the shared offset

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

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

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.

Input · Model result · Constant

Symbolic equation

No numerical binding is declared.

Energy of motion before emission: No accepted value is available here.
Energy of motion after emission: No accepted value is available here.
Emitted energy in the body's rest frame: No accepted value is available here.
Lorentz factor: No accepted value is available here.

Read the equation aloud in words

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 reviewed 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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

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.

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.

Input · Model result · Constant

Symbolic equation

No numerical binding is declared.

Body energy after emission, moving description: No accepted value is available here.
Body energy after emission, rest description: No accepted value is available here.
Energy of motion after emission: No accepted value is available here.
Unchanged additive energy offset: No accepted value is available here.

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 reviewed 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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

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.

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.

Input · Model result · Constant

Symbolic equation

No numerical binding is declared.

Body energy before emission, moving description: No accepted value is available here.
Body energy before emission, rest description: No accepted value is available here.
Energy of motion before emission: No accepted value is available here.
Unchanged additive energy offset: No accepted value is available here.

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 reviewed 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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

Explore the equation · Modern model notation

Substitute both offset premises before cancelling anything

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

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

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.

Input · Model result · Constant

Symbolic equation

No numerical binding is declared.

Energy of motion before emission: No accepted value is available here.
Unchanged additive energy offset: No accepted value is available here.
Energy of motion after emission: No accepted value is available here.
Unchanged additive energy offset: No accepted value is available here.
Emitted energy in the body's rest frame: No accepted value is available here.
Lorentz factor: No accepted value is available here.

Read the equation aloud in words

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 reviewed 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
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
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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

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 · editorial and physics review pending. 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.
$H_0 - H_1 - \left(E_0 - E_1\right) = \gamma\,L - L$
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.
$H_0 - E_0 - \left(H_1 - E_1\right) = L\,\left(\gamma - 1\right)$
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.
$K_0 + C - \left(K_1 + C\right) = L\,\left(\gamma + -1\right)$
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 move. 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 →

Show every step here: The premise that the subtraction needs

Before emission write the moving account as the rest account plus energy of motion plus C₀.
After emission write the corresponding relation with C₁.
The subtraction gives (K₀ − K₁) + (C₀ − C₁) = L(γ − 1).
Admit C₀ = C₁ to cancel the offsets and identify the kinetic-energy drop.
Relax that premise in ME-01: the light totals and algebraic subtraction remain known, but the kinetic interpretation becomes underdetermined.

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

derivation · Model approximation

Read the slow-speed coefficient

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

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\to0}\frac{2L(\gamma-1)}{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: A local expansion

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 a reviewed transcription.

Explore the equation · Modern model notation

Why a finite-speed quotient is not the mass decrease

m_{\mathrm{proxy}} := \frac{2\,\Delta K}{\left(v\right)^{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.

Input · Model result · Constant

Symbolic equation

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

Finite-speed mass proxy: No accepted value is available here.
Exact drop in energy of motion: No accepted value is available here.
Signed observer speed: No accepted value is available here.

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 reviewed 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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

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.

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.

Input · Model result · Constant

Symbolic equation

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

Exact drop in energy of motion: No accepted value is available here.
Emitted energy in the body's rest frame: No accepted value is available here.
Signed observer speed: No accepted value is available here.
Speed of light in SI: No accepted value is available here.

Read the equation aloud in words

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 reviewed 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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

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 · editorial and physics review pending. 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.
$m_{\mathrm{proxy}} := \frac{2\,\Delta K}{\left(v\right)^{2}}$
The finite-speed proxy is defined as twice the exact kinetic-energy drop divided by v squared.
$\frac{2\Delta K}{v^2}=\frac{L}{c^2}\,\frac{2(\gamma-1)}{\beta^2},\qquad \beta=\frac{v}{c},\quad v\ne 0$
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.
Now identify inertia: this is the additional physical step
Supported under the stated conditions.
$m_{\mathrm{loss}} = \frac{L}{\left(c\right)^{2}}$
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, source alignment and editorial review 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.

Show every step here: Read the slow-speed coefficient

Calculate the exact difference L(γ − 1) from the admitted two-ledger model.
Expand γ − 1 at zero speed: β²/2 + 3β⁴/8 plus higher even powers.
Keep only the quadratic term for the low-speed comparison.
Replace β² by v²/c² and compare with the Newtonian coefficient (m₀ − m₁)/2.
At a nonzero finite speed the ratio 2L(γ − 1)/v² still contains higher-order terms. It is not the limiting coefficient.
At exactly zero speed do not calculate 0/0. Evaluate the analytic limit separately.
ME-02 displays the exact drop, quadratic estimate, finite-speed proxy, and limiting coefficient as distinct quantities.

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

derivation · Within the stated model

A decrease, a signed change, and a conversion

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

m_0-m_1=\frac{L}{c^2},\qquad\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. ME-02 keeps that historical constant set separate from the modern SI value. A unit conversion changes the description of the result, not the physical event.

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 a reviewed transcription.

Explore the equation · Modern model notation

Identify the positive inertia decrease

m_{\mathrm{loss}} = \frac{L}{\left(c\right)^{2}}

The positive inertial mass decrease equals L divided by c 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.

Input · Model result · Constant

Symbolic equation

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

Positive inertial mass decrease: No accepted value is available here.
Emitted energy in the body's rest frame: No accepted value is available here.
Speed of light in SI: No accepted value is available here.

Read the equation aloud in words

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 reviewed 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: m_loss is 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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

Explore the equation · Modern model notation

A signed change is the negative of a decrease

\Delta m = -\left(\frac{L}{\left(c\right)^{2}}\right)

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

Input · Model result · Constant

Symbolic equation

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

Signed change in body mass: No accepted value is available here.
Emitted energy in the body's rest frame: No accepted value is available here.
Speed of light in SI: No accepted value is available here.

Read the equation aloud in words

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 reviewed 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

Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.

Connect the result terms to the coefficient laboratory →

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

Use the kinetic-energy coefficient to identify the positive mass decrease.
For final minus initial mass, change the sign.
Keep L positive for energy emitted out of the body.
For the printed conversion, divide energy in erg by the printed factor 9 × 10²⁰ to obtain grams.
For modern SI, divide joules by c squared in metres-squared per second-squared to obtain kilograms.
Never combine an erg input with an SI denominator without converting the units.
Do not label a finite-speed proxy as the exact mass decrease.

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.
Printed and modern constants must not be mixed.

Earlier step: Read the slow-speed coefficient

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

qualification · Within the stated model

The light has left the body, not vanished

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

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 reviewed transcription of those sentences.

ME-03 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 interpreting any reported change.
Body only: energy L has crossed outward in the rest frame.
Radiation only: the emitted energy is present in the departing pulses.
Body and all light: no energy has crossed the larger isolated boundary.
Separate the conditional prediction from an independently calibrated observation.
Return to the subtraction and name its two essential ingredients: the imported transformation and the unchanged-offset premise, followed by the low-speed identification.

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

Assumed here

For the combined account, include the body and all emitted light.
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

References and source status

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

10 foundation readings sit behind this argument.

Functions and graphs

How does a mathematical function record relationships between physical quantities?

A physical function is an unambiguous rule that assigns to each value of an independent input (such as elapsed time t or coordinate position x) a specific value of a dependent quantity (such as particle displacement, local concentration, or field energy). It describes how one aspect of a physical system responds when another changes.

Plotting a function creates a curve on a coordinate plane. The horizontal axis represents the input, and the vertical axis represents the output. Every point on the curve represents a simultaneously paired measurement. Because both axes correspond to physical measurements, any slope or area derived from the graph carries the combined dimensions of those axes.

Distinguishing functional laws from empirical scatter is essential. A theoretical function predicts the ideal expectation of an ensemble, such as the mean square displacement growing linearly with time. Individual experimental trials fluctuate around this expectation, but the functional form governs the ensemble average.

One worked example

\langle x^2\rangle = 2Dt

Mean square coordinate displacement equals two times the diffusion coefficient times time.

For a diffusion coefficient D = 0.5 square micrometres per second, plotting mean square displacement against time t produces a straight line through the origin with slope 2D = 1.0 square micrometres per second. At t = 1 second the value is 1.0 square micrometre; at t = 4 seconds it is 4.0 square micrometres.

A stopping point: A graph is a record of paired quantities with physical units; a slope without units is not an explanation.

Interactive construction: from measurement table to curve

In section 5 of the Brownian motion paper, Einstein shows that while the mean displacement is zero, the mean squared displacement grows linearly with time: ⟨x²⟩ = 2Dt. Consequently, the observable root-mean-square displacement grows as the square root of time: √⟨x²⟩ ∝ √t.

Showing 5 of 5 points plotted.

Dashed line: continuous √t trajectory. Dots: discrete observations.

Table of paired measurements (Brownian §5)
Time t (s)	⟨x²⟩ (µm²)	√⟨x²⟩ (µm)	Status
0	0	0.0	Plotted
1	4	2.0	Plotted
4	16	4.0	Plotted
9	36	6.0	Plotted
16	64	8.0	Plotted

Textual summary of the construction

Each observation pairs an elapsed time in seconds with an accumulated squared displacement in square micrometers. Plotting these pairs demonstrates that displacement does not scale proportionally with time (which would indicate constant drift velocity), but rather with the square root of time (the hallmark of diffusive random walks).

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A local expansion

Why can a smooth function be replaced by a few terms?

The first derivative is local change per unit of the input. The second derivative describes how that first rate changes. A Taylor expansion writes the nearby value using those rates. The dots stand for a remainder; they are not automatically negligible.

f(x+\Delta)=f(x)+\Delta f'(x)+\frac{\Delta^2}{2}f''(x)+\ldots

Nearby value equals the value plus displacement times slope plus half the squared displacement times curvature, followed by a remainder.

For a diffusion limit, the mean square step shrinks in proportion to the time interval, while the retained variance per unit time stays finite. Higher moments and the variation scale of the density must justify neglecting higher terms. A real tracer does not have independent steps at arbitrarily short times.

One worked example

(x+\Delta)^2=x^2+2x\Delta+\Delta^2

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

At x = 2 and Δ = 0.1, the three terms are 4, 0.4 and 0.01, adding to 4.41. Keeping only the value and slope gives 4.4, with an error of 0.01.

A stopping point: Name what is held fixed, which terms are retained, and why the remainder is small.

Extension · Planned callers: Special Relativity §4, Mass–Energy

Binomial Series Expansion for the Relativistic Lorentz Factor γ

In relativistic mechanics, the Lorentz factor γ = (1 − v²/c²)^−1/2 governs time dilation, length contraction, and the relativistic kinetic energy E_k = (γ − 1)mc². Setting x = (v/c)², the function is expanded using Newton’s generalized binomial theorem:

(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. The progressive partial sums illustrate how higher-order terms accumulate:

Partial sums of γ − 1 at x = 0.36 against exact 0.25
Order	Term added	Term value	Partial sum	Difference to exact (0.25)
1st (linear in x)	½ x	0.180000	0.180000	0.070000
2nd (quadratic)	⅜ x²	0.048600	0.228600	0.021400
3rd (cubic)	⁵/₁₆ x³	0.014580	0.243180	0.006820
4th (quartic)	³⁵/₁₂₈ x⁴	0.0045927	0.247773	0.002227

Notice that the leading quadratic term 0.18 corresponds to the 0.18L mass–energy fixture beside the exact 0.25L, making both the retained classical kinetic energy (½ mv²) and the neglected relativistic remainder visible.

2. Low speeds: catastrophic cancellation and the cancellation-free rule

At everyday velocities such as v/c = 10⁻⁴ (where x = 10⁻⁸), evaluating γ − 1 by naive floating-point subtraction suffers catastrophic loss of precision. Because γ ≈ 1.000000005, subtracting 1 throws away the leading digits and amplifies round-off error.

Comparison at v/c = 10⁻⁴ (x = 10⁻⁸)

First series term (½ x):: 5.0000000000e-9
Cancellation-free route (trustworthy):: 5.0000000375e-9 (computed: 5.0000000375e-9)
Relative diff to first term:: 7.500000e-9 (exact theoretical 0.75x = 7.500000e-9)
Planted naive route 1/√(1−x) − 1:: 5.0000001917e-9 (computed: 5.0000001917e-9)
Naive relative difference:: 3.833145e-8 (approx 3.833e-8; over five times too large)

Cancellation-free evaluation rule: The cancellation-free route usingexpm1(-0.5 * log1p(-x)) or x / (√(1−x) · (1 + √(1−x)))is required for trustworthy numerical evaluation. Naive subtraction in IEEE 754 double precision yields 5.0000001917e-9 instead of the true 5.0000000375e-9, corrupting relativistic corrections with false precision artifacts.

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Original journal scan

Beside the body

The traveler’s account

Compare the accounts

60 percent of light speed

1 percent of light speed

Why the slower traveler matters

From two accounts to inertia

The unsectioned argument · explanatory reading

One result is borrowed, not rediscovered

What this relies on

Assumed here

What this does not establish

Choose the emission that removes recoil

What this relies on

Assumed here

What this does not establish

Two accounts of the same loss

What this relies on

Assumed here

What this does not establish

Subtract what you cannot measure

What this relies on

Assumed here

What this does not establish

The premise that the subtraction needs

Why does the shared unknown disappear?

Conservation in the body's rest description · assumption

Conservation with the imported light-energy transformation · assumption

The same offset C before and after emission · assumption

Delta K means before minus after · definition

1. Subtract the complete accounts

2. Put the two descriptions beside each other

3. Use the additional offset premise · the additional premise

4. Now cancel the shared unknown

5. Give the result a short name

What this relies on

Assumed here

What this does not establish

Read the slow-speed coefficient

Why does the coefficient become a mass decrease?

Expand the factor, not an assumed body energy

Read the energy-of-motion coefficient

Divide only away from zero speed

Take the limit, not the value at zero

Now identify inertia: this is the additional physical step

What this relies on

Assumed here

What this does not establish

A decrease, a signed change, and a conversion

What this relies on

Assumed here

What this does not establish

The light has left the body, not vanished

What this relies on

Assumed here

What this does not establish

References and source status

Keep the explanation for offline reading