Annus Mirabilis · Interactive critical edition in preparation
Inertia from the small-speed coefficient
Inspect what the low-speed coefficient tells you about inertia.
An executable model
Inertia from the small-speed coefficient
Static worked example
CurrentThese numbers match the current settings.
Model note
- Primary output kineticEnergyDifference: Host calculation (massEnergy.exactDifference). Owner massEnergy.exactDifference.
- Primary output quadraticKineticDifference: Host calculation (massEnergy.quadraticApproximation). Owner massEnergy.quadraticApproximation.
- Primary output quadraticRelativeDiscrepancy: Host calculation (massEnergy.quadraticDiscrepancy). Owner massEnergy.quadraticDiscrepancy.
- Primary output finiteSpeedMassProxy: Host calculation (massEnergy.finiteSpeedProxy). Owner massEnergy.finiteSpeedProxy.
- Primary output inertialMassDecrease: Host calculation (massEnergy.limitingCoefficient). Owner massEnergy.limitingCoefficient.
- Primary output proxyExcessOverLimit: Host calculation (massEnergy.proxyExcess). Owner massEnergy.proxyExcess.
- Primary output massChangeSigned: Host calculation (massEnergy.massChangeSigned). Owner massEnergy.massChangeSigned.
- Primary output naiveGammaMinusOne: Host calculation (massEnergy.naiveGammaMinusOne). Owner massEnergy.naiveGammaMinusOne.
- Accepted input revision 1.
- Snapshot version 1.
- 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).
A body that sends out energy as light loses a little of its energy of motion, as seen by anyone it moves past, and at low speed the loss is exactly what a lighter body would lose. The mass it has lost is the energy it sent out divided by the speed of light squared.
The last page of the paper compares the body's energy before and after it emits two equal pulses of light, in its own frame and in a frame where it moves at v. The two descriptions differ by the body's energy of motion, and the emission lowers it by K0 − K1 = L{1/√(1 − (v/V)2) − 1}, an amount that does not depend on what the body is made of. Neglecting quantities of fourth and higher order, this is (L/V2)(v2/2), the drop in ½mv2 for a mass that has fallen by L/V2. Einstein concludes that a body giving off energy L as radiation loses mass L/V2, that it does not matter that the energy leaves as radiation, and that the mass of a body is a measure of its energy content. The lab shows the exact drop and the quadratic estimate parting company: at 0.6c they are 0.25 L and 0.18 L, and dividing the exact drop by ½v2 gives 1.39 L/V2. At 0.1c the ratio is 1.0076, and it reaches L/V2 only as the speed goes to zero, so the lab reports the mass decrease as an analytic limit, not as a value at any finite speed.
Call E0 and E1 the body's energy before and after the emission in its own frame, and H0 and H1 the same energies in a frame where it moves at v. In its own frame the light carries away L. In the moving frame, by §8 of the relativity paper, a plane wave's energy changes by the factor (1 − (v/V) cos φ)/√(1 − (v/V)2); for two equal pulses sent in opposite directions the cosine terms cancel and the pair carries L/√(1 − (v/V)2). Energy is conserved in each frame, so E0 = E1 + L and H0 = H1 + L/√(1 − (v/V)2). Subtract the first from the second: (H0 − E0) − (H1 − E1) = L(1/√(1 − (v/V)2) − 1). Each difference H − E is the body's energy of motion K plus a constant C that the emission does not change, so the left side is K0 − K1. Now expand for small speeds, with β = v/V: 1/√(1 − β²) = 1 + ½β² + ⅜β⁴ + …, so subtracting 1 leaves ½β² + ⅜β⁴ + …, and dropping the fourth-order term gives K0 − K1 = ½(L/V2)v2. A body of mass m moving at v has energy of motion ½mv2, so this drop at the same speed is the drop a mass decrease of L/V2 would cause. At 0.6c the neglected terms are not small: γ = 1.25, the exact drop is 0.25 L, the quadratic 0.18 L, and 2(K0 − K1)/v2 = 2 × 0.25/0.36 = 1.389 L/V2. The series for that ratio is 1 + ¾β² + …, so at 0.1c it is 1.0076 and at 0.01c 1.000075, approaching 1 only in the limit.
Paper 4 writes V for the speed of light and prints the factor as the explicit radical 1/√(1 − (v/V)2), never as β. It neglects 'Größen vierter und höherer Ordnung' and converts the result as L/9·1020, energy in erg and mass in grams, with V rounded to 3·1010 cm/s; with today's c the factor is 8.988·1020. The identification of the mass rests on the Newtonian ½mv2 at low speed. The paper adds that bodies whose energy content varies greatly, radium salts for instance, might test the theory, and that if the theory corresponds to the facts, radiation carries inertia between the bodies that emit and absorb it. It is dated Bern, September 1905, and was received on 27 September. Ives (1952) argued that the derivation assumes what it proves; Stachel and Torretti (1982) answered that it does not, once the body's energy in the moving frame is taken as the text defines it.
Predict before the numbers
At 0.6c, is the exact energy difference larger or smaller than the quadratic estimate?
Predict before the numbers
As the speed gets smaller and smaller, what happens to the drop in energy of motion?
The result appears when you choose, say you have one in mind, or skip.
Values at these settings
| 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 (the paper's printed constants) for L = 9 × 10²⁰ erg; modern mass change 1.0014 g (2019 SI): the printed factor is 0.1385 percent larger than the modern c².
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
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.
Each term and operation, in words
Static worked example
These values are for the settings last applied, not for edits you have not applied yet.
- 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
Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper, and its review is pending.
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
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.
Each term and operation, in words
Static worked example
These values are for the settings last applied, not for edits you have not applied yet.
- 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
Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper, and its review is pending.
Identify the positive inertia decrease
Explore the equation · Modern model notation
Identify the positive inertia decrease
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.
Each term and operation, in words
Static worked example
These values are for the settings last applied, not for edits you have not applied yet.
- 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
- 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
Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper, and its review is pending.
Keep only the second-order energy term
Explore the equation · Modern model notation
Keep only the second-order energy term
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.
Each term and operation, in words
Static worked example
These values are for the settings last applied, not for edits you have not applied yet.
- 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
Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper, and its review is pending.
A signed change is the negative of a decrease
Explore the equation · Modern model notation
A signed change is the negative of a decrease
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.
Each term and operation, in words
Static worked example
These values are for the settings last applied, not for edits you have not applied yet.
- 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
Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper, and its review is pending.
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).
The explanation
Full explanation
Compare the exact change in energy of motion with its small-speed approximation. A limiting coefficient and a finite-speed numerical estimate answer different questions.
Show every step of the investigation
Inspect the unchanged-offset premise, the exact kinetic drop, and the coefficient of the squared speed. Vary the speed to see the approximation's range; do not divide a zero-speed measurement by zero to manufacture an inertia estimate.
An explanatory model, not an observation of nature. This embed starts from the laboratory’s worked defaults, not a saved run. Presentation options change the surrounding guide, never the numerical inputs.