Mass and energy · The small-speed coefficient

A smaller energy of motion
at the same speed.

Under the unchanged-offset premise, emitting energy L reduces energy of motion by L(γ − 1). What does that drop tell you about the body's inertia, and why does the conclusion come from low speeds rather than from a slogan assigned in advance?

Read the coefficient argument (explanatory preview) →

ME-02 · An executable model

Inertia from the small-speed coefficient

Ideal model, host calculation

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At low speed the drop in energy of motion looks exactly like a body that became lighter by the energy it sent out divided by the speed of light squared.

The exact difference is L(gamma - 1). The quadratic estimate is half L times (v/c) squared. The finite-speed proxy is 2 L (gamma - 1) / v^2, never labeled the exact mass loss. The limiting coefficient L/c^2 is an analytic limit at vanishing speed.

Type a speed, choose what to read, and compare 0.6c with 0.01c. The plot and table change together after you apply valid settings. No dragging is required.

Predict before the numbers

At 0.6c, is the exact energy difference larger or smaller than the quadratic estimate?

Three relations the model could have Larger: The exact difference lies above the quadratic estimate. Smaller: The exact difference lies below the quadratic estimate. Equal: The two curves coincide at 0.6c.

Show me the reasoning

Accepted snapshot

Outputs from the host calculation. Presentation does not recompute them. Normalized energy and mass units (c = 1).
Exact difference L(γ−1)	0.25
Quadratic estimate	0.18
Finite-speed proxy	1.3889
Limiting coefficient L/c²	1 (analytic limit)
Signed mass change	-1

Named-speed comparison (worked example)

At 0.6c versus 0.01c, with L = 1 in normalized units. These two columns were calculated at build time.

Quantity	0.6c	0.01c
Exact difference	0.25	0.000050004
Quadratic estimate	0.18	0.00005
Finite-speed proxy	1.3889	1.0001

Printed mass change 1 g (einstein-1905-mass-energy-printed) for L = 9e20 erg. Modern mass change 1.0014 g (modern-si-2019). the printed factor is 0.1385 percent larger than the modern c^2.

Explore the accepted result, term by term

Only declared result terms read this laboratory's accepted snapshot. Body energies, the unknown offset and unbound inputs stay symbolic. A draft edit does not change these values.

The current example uses normalized units (c = 1). These SI equations remain symbolic. Choose joule or erg and apply settings to attach physical-unit results.

The conditional exact kinetic-energy drop

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

Static worked example

Values describe this accepted snapshot, not unsaved input edits.

Exact drop in energy of motion: SI substitution is unavailable for normalized units. Apply joule or erg settings in this laboratory first.
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.

Why a finite-speed quotient is not the mass decrease

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

Static worked example

Values describe this accepted snapshot, not unsaved input edits.

Finite-speed mass proxy: SI substitution is unavailable for normalized units. Apply joule or erg settings in this laboratory first.
Exact drop in energy of motion: SI substitution is unavailable for normalized units. Apply joule or erg settings in this laboratory first.
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.

Identify the positive inertia decrease

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

Static worked example

Values describe this accepted snapshot, not unsaved input edits.

Positive inertial mass decrease: SI substitution is unavailable for normalized units. Apply joule or erg settings in this laboratory first.
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.

Keep only the second-order energy term

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

Static worked example

Values describe this accepted snapshot, not unsaved input edits.

Exact drop in energy of motion: SI substitution is unavailable for normalized units. Apply joule or erg settings in this laboratory first.
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.

A signed change is the negative of a decrease

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

Static worked example

Values describe this accepted snapshot, not unsaved input edits.

Signed change in body mass: SI substitution is unavailable for normalized units. Apply joule or erg settings in this laboratory first.
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.

Return to the unchanged-offset premise →

Not modeled: the premises themselves (explained, not simulated); accelerated motion; modern momentum formulations; any body not covered by the paper's argument; uncertainty in the printed factor's rounding (shown as a labeled comparison, not modeled).

Open the coefficient argument

The exact drop, then the Newtonian coefficient

The two-ledger subtraction (ME-01) gives the change in energy of motion at speed v. This laboratory never assigns the body a rest energy Mc² or γMc² to begin with. The printed glyph Einstein used for the Lorentz factor in paper 4 is UNKNOWN until the facsimile is pinned; the formulas below use the modern γ.

K_0-K_1=L(\gamma-1)=L\frac{\gamma^2\beta^2}{\gamma+1},\qquad \beta=\frac{v}{c}

At small speed the Newtonian energy of motion is ½mv². Matching the second-order term identifies a mass change L/c². The quadratic estimate is that second-order piece alone.

\tfrac12 L\beta^2\qquad\text{(quadratic)}\qquad\lim_{v\to 0}\frac{2L(\gamma-1)}{v^2}=\frac{L}{c^2}

The finite-speed proxy is not the limit

At a finite speed the ratio 2L(γ − 1)/v² is a proxy for the coefficient, never labeled the exact mass loss. At 0.6c with L = 1 (normalized, c = 1) the exact drop is 0.25 L, the quadratic estimate is 0.18 L, and the proxy is 1.3888889 L/c². The analytic limit at vanishing speed remains 1 × L/c². The claim that the proxy equals the limit at every speed fails at 0.6c for that reason. The same numbers hold at −0.6c: every output is even in v.

At v = 0 the proxy is not applicable, because it divides by v². The identified coefficient is then the analytic limit L/c², obtained without a 0/0 division.

Einstein's printed factor is a rounded V²

Paper 4 converts energy in erg with V² = 9 × 10²⁰. That printed factor is 0.1385 percent larger than the modern c². The two constant sets are never combined into one number. The laboratory shows both conversions of L = 9 × 10²⁰ erg as a labeled comparison, not as a percentage computed in the page.

Return to the small-speed argument →Compare two energy accounts without algebra