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Mass and energy: two accounts, one subtraction

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

The paper's results

Page 639

The imported light-energy transformation

Light of energy $l$ travelling at angle $\varphi$ to the x-axis has the energy $l^*$ in a system moving along x at speed $v$ ; the paper takes this from §8 of the relativity paper and derives nothing new here.

As printed

Gestützt auf diese Grundlagen²⁾ leitete ich unter anderem das nachfolgende Resultat ab (l. c. § 8):
Ein System von ebenen Lichtwellen besitze, auf das Koordinatensystem $(x, y, z)$ bezogen, die Energie $l$ ; die Strahlrichtung (Wellennormale) bilde den Winkel $\varphi$ mit der $x$ -Achse des Systems. Führt man ein neues, gegen das System $(x, y, z)$ in gleichförmiger Paralleltranslation begriffenes Koordinatensystem $(\xi, \eta, \zeta)$ ein, dessen Ursprung sich mit der Geschwindigkeit $v$ längs der $x$ -Achse bewegt, so besitzt die genannte Lichtmenge — im System $(\xi, \eta, \zeta)$ gemessen — die Energie: $l^* = l \frac{1 - \frac{v}{V} \cos \varphi}{\sqrt{1 - \left(\frac{v}{V}\right)^2}},$ Energy of one light pulse, moving frameEnergy of one light pulse, rest frameSigned observer speedSpeed of lightPropagation angle (stationary system)Explain this equation wobei $V$ die Lichtgeschwindigkeit bedeutet. Von diesem Resultat machen wir im folgenden Gebrauch.

Text on page 639 of the German source, paragraph 4 · Text on page 639 of the German source, paragraph 5

In modern notation

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

Premise: Imported, not derived: the paper cites its own relativity paper (l. c. § 8) and uses the result as given. Everything after it depends on it.

Decoder

$l$: The energy of a system of plane light waves, measured in the system (x, y, z).
$l^*$: The energy of the same light, measured in the system (ξ, η, ζ), whose origin moves with speed $v$ along the x-axis.
$\varphi$: The angle between the direction of the rays (the wave normal) and the x-axis.
$V$: The speed of light, printed V; today it is written c.
$\sqrt{1-(v/V)^2}$: The square root the paper writes out in full. Later texts write its reciprocal as γ; the paper does not use that letter.

Try it

ME-01: How much energy does each pulse carry for an observer moving along x at 0.6 of the speed of light? Choose the preset “Collinear emission (v = 0.6c, phi = 0 deg)” there.

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

Argument: An assumption of this paper, admitted from the relativity paper
Model: Exact within the electrodynamics of the relativity paper
Evidence: A premise, not evidence
History: A 1905 result, introduced in the relativity paper

Page 640

A symmetric emission

The body at rest sends out light of energy $L/2$ in one direction and an equal amount in the opposite direction, and it stays at rest.

As printed

Dieser Körper sende in einer mit der $x$ -Achse den Winkel $\varphi$ bildenden Richtung ebene Lichtwellen von der Energie $L/2$ (relativ zu $(x, y, z)$ gemessen) und gleichzeitig eine gleich große Lichtmenge nach der entgegengesetzten Richtung. Hierbei bleibt der Körper in Ruhe in bezug auf das System $(x, y, z)$ .

Text on page 640 of the German source, paragraph 7

In modern notation

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

Premise: The body stays at rest because the emission is symmetric. With unequal pulses it would recoil, and the energy accounting would have to follow the change in its motion.

Decoder

$L$: The total energy of the emitted light, measured in the system (x, y, z) in which the body is at rest.
$L/2$: The energy of each of the two equal pulses in that system.
"the opposite direction": The second pulse travels at angle $\varphi$ plus a half turn to the x-axis, so its cosine has the opposite sign.
"remains at rest": The paper states that the body stays at rest in (x, y, z). With two equal pulses in opposite directions neither direction is favoured, so the emission gives the body no motion; in later language the two pulses carry equal and opposite momentum, which the paper does not mention.

Try it

ME-01: With the observer at rest, what does each pulse carry and what does the body lose? Choose the preset “Stationary observer (v = 0)” there.

Where this stops: Unequal pulses or external forces require recoil and momentum bookkeeping. Read the argument

Where this is used later

A later reading treats the two pulses as one system with an invariant mass, which the paper does not use.

Common wrong turns

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

Argument: A chosen arrangement of the thought experiment
Model: Exact within the model
Evidence: A construction, not evidence
History: Introduced in this paper

Page 640

Two energy balances

The energy principle holds in both systems: at rest the body's energy falls by $L$ , and in the moving system it falls by $L/\sqrt{1-(v/V)^2}$ , whatever the angle of emission.

As printed

Für diesen Vorgang muß das Energieprinzip gelten und zwar (nach dem Prinzip der Relativität) in bezug auf beide Koordinatensysteme.
$E_0 = E_1 + \left[ \frac{L}{2} + \frac{L}{2} \right],$ Body energy before emission, rest descriptionBody energy after emission, rest descriptionEmitted energy in the body's rest frameExplain this equation
$\begin{aligned} H_0 &= H_1 + \left[ \frac{L}{2} \frac{1 - \frac{v}{V} \cos \varphi}{\sqrt{1 - \left(\frac{v}{V}\right)^2}} + \frac{L}{2} \frac{1 + \frac{v}{V} \cos \varphi}{\sqrt{1 - \left(\frac{v}{V}\right)^2}} \right] \\ &= H_1 + \frac{L}{\sqrt{1 - \left(\frac{v}{V}\right)^2}}. \end{aligned}$ Body energy before emission, moving descriptionBody energy after emission, moving descriptionEmitted energy in the body's rest frameSigned observer speedSpeed of lightEmission angleExplain this equation

Text on page 640 of the German source, paragraph 7 · Display on page 640 of the German source, § 0, display 2 · Display on page 640 of the German source, § 0, display 3

In modern notation

E_0 - E_1 = L

H_0 - H_1 = \gamma\,L

Premise: The energy principle is required to hold in both systems, by the principle of relativity; that is what the paper states before writing the two balances.

Decoder

$E_0$ , $E_1$: The body's energy before and after the emission, measured in the system (x, y, z) in which it is at rest.
$H_0$ , $H_1$: The same body's energy before and after, measured in the moving system (ξ, η, ζ).
the bracket in $H_0$: The two pulses' energies in the moving system, each from the imported formula, with $+\cos\varphi$ for one and $-\cos\varphi$ for the other.
why $\cos\varphi$ cancels: The two numerators are $1-\frac{v}{V}\cos\varphi$ and $1+\frac{v}{V}\cos\varphi$ ; they add to 2, so the sum is $L$ divided by the square root for every angle.

Try it

ME-01: Turn the emission axis: does either system's loss change with the angle? Choose the preset “Tilted axis (v = 0.6c, phi = 60 deg)” there.

Where this stops: The accounts do not determine E₀, E₁, H₀, or H₁ separately. Read the argument

Argument: Two applications of the energy principle
Model: Exact within the model
Evidence: A derivation from the imported formula
History: Introduced in this paper

Pages 640, 641

The drop in kinetic energy, and its premise

Subtracting the two balances leaves $L\{1/\sqrt{1-(v/V)^2}-1\}$ , and if the difference $H-E$ is the kinetic energy plus a constant that the emission does not change, the body's kinetic energy in the moving system falls by exactly that much.

As printed

$(H_0 - E_0) - (H_1 - E_1) = L \left\{ \frac{1}{\sqrt{1 - \left(\frac{v}{V}\right)^2}} - 1 \right\}.$ Body energy before emission, moving descriptionBody energy before emission, rest descriptionBody energy after emission, moving descriptionBody energy after emission, rest descriptionEmitted energy in the body's rest frameSigned observer speedSpeed of lightExplain this equation
Es ist also klar, daß die Differenz $H - E$ sich von der kinetischen Energie $K$ des Körpers in bezug auf das andere System (System $(\xi, \eta, \zeta)$ ) nur durch eine additive Konstante $C$ unterscheiden kann, welche von der Wahl der willkürlichen additiven Konstanten der Energien $H$ und $E$ abhängt. Wir können also setzen: $\begin{aligned} H_0 - E_0 &= K_0 + C, \\ H_1 - E_1 &= K_1 + C, \end{aligned}$ Body energy before emission, moving descriptionBody energy before emission, rest descriptionEnergy of motion before emissionUnchanged additive energy offsetBody energy after emission, moving descriptionBody energy after emission, rest descriptionEnergy of motion after emissionExplain this equation da $C$ sich während der Lichtaussendung nicht ändert. Wir erhalten also: $K_0 - K_1 = L \left\{ \frac{1}{\sqrt{1 - \left(\frac{v}{V}\right)^2}} - 1 \right\}.$ Energy of motion before emissionEnergy of motion after emissionEmitted energy in the body's rest frameSigned observer speedSpeed of lightExplain this equation Die kinetische Energie des Körpers in bezug auf $(\xi, \eta, \zeta)$ nimmt infolge der Lichtaussendung ab, und zwar um einen von den Qualitäten des Körpers unabhängigen Betrag. Die Differenz $K_0 - K_1$ hängt ferner von der Geschwindigkeit ebenso ab wie die kinetische Energie des Elektrons (l. c. § 10).

Display on page 640 of the German source, § 0, display 4 · Text on pages 640–641 of the German source, paragraph 7

In modern notation

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

H_0 - E_0 = K_0 + C

H_1 - E_1 = K_1 + C

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

Premise: The identification of $H-E$ as kinetic energy plus $C$ , with $C$ unchanged by the emission, is a premise the paper states, not a result it derives.
A comparison, not a premise: The paper notes that the decrease does not depend on the properties of the body, and that it depends on the speed as the electron's kinetic energy does in §10 of the relativity paper. That is a comparison, not a premise of the argument.

Decoder

$H-E$: The same body's energy in the two systems, taken as a difference; it contains the body's energy of motion in the system in which the body moves.
$K_0$ , $K_1$: The body's kinetic energy in the moving system (ξ, η, ζ), before and after the emission.
$C$: An additive constant: energies are fixed only up to a constant, and $C$ depends on how the constants of $H$ and $E$ were chosen.
why the subtraction removes $E_0$ and $H_0$: Neither the body's energy at rest nor in motion is known. Only the changes enter the two balances, and subtracting one balance from the other leaves the change of $H-E$ , which needs neither absolute value.
why the unchanged constant is a premise: Nothing in the two balances says that $C$ is the same before and after; the paper states it, at the end of the sentence that sets $H-E = K + C$ , and without it the difference is not a kinetic energy.

Try it

ME-01: Starting from the collinear preset, relax the premise that the constant is unchanged: what can the subtraction still say about the kinetic energy? Choose the preset “Collinear emission (v = 0.6c, phi = 0 deg)” there.

Where this stops: Calling the remaining difference kinetic energy requires the next premise. Read the argument

Where this stops: Without the unchanged-offset premise, the kinetic-energy drop is underdetermined. Read the argument

Argument: A derivation resting on a stated premise
Model: Exact within the model, given the premise
Evidence: A derivation, not evidence
History: Introduced in this paper

Page 641

The low-speed approximation

Neglecting quantities of fourth and higher order in the speed, the drop in kinetic energy is $\frac{L}{V^2}\frac{v^2}{2}$ , the form of a kinetic energy for a mass $L/V^2$ .

As printed

Unter Vernachlässigung von Größen vierter und höherer Ordnung können wir setzen: $K_0 - K_1 = \frac{L}{V^2} \frac{v^2}{2}.$ Energy of motion before emissionEnergy of motion after emissionEmitted energy in the body's rest frameSpeed of lightSigned observer speedExplain this equation

Text on page 641 of the German source, paragraph 11

In modern notation

\Delta K \approx 0.5\,L\,\left(\frac{v}{c}\right)^{2}

Approximation: The formula holds to second order in $v/V$ , as the paper says. At a finite speed the exact drop is larger; the coefficient that is a mass is the limit as the speed goes to zero.

Decoder

$\frac{v^2}{2}$: Half the square of the speed: a kinetic energy is a mass times this.
$\frac{L}{V^2}$: The coefficient of $v^2/2$ , which therefore plays the part of a mass: the mass the body's kinetic energy has lost.
fourth and higher order: The exact drop, expanded in powers of $v/V$ , has terms in $(v/V)^2$ , $(v/V)^4$ and so on; the paper keeps the first and states in words that it drops the rest.

Try it

ME-02: Compare the exact drop, its second-order form, the finite-speed ratio and the limit: which one is the mass? Choose the preset “Observer at 0.6c” there.

Where this stops: 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. Read the argument

Common wrong turns

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

Argument: An approximation of the exact result
Model: An approximation, exact in the limit of small speed
Evidence: A derivation, not evidence
History: Introduced in this paper

Page 641

The mass decrease

A body that gives off the energy $L$ as radiation loses the mass $L/V^2$ ; the paper then generalizes that the mass of a body is a measure of its energy content, and gives the change as $L/9\cdot10^{20}$ grams for $L$ in erg.

As printed

Gibt ein Körper die Energie $L$ in Form von Strahlung ab, so verkleinert sich seine Masse um $L/V^2$ . Hierbei ist es offenbar unwesentlich, daß die dem Körper entzogene Energie gerade in Energie der Strahlung übergeht, so daß wir zu der allgemeineren Folgerung geführt werden:
Die Masse eines Körpers ist ein Maß für dessen Energieinhalt; ändert sich die Energie um $L$ , so ändert sich die Masse in demselben Sinne um $L/9 \cdot 10^{20}$ , wenn die Energie in Erg und die Masse in Grammen gemessen wird.

Text on page 641 of the German source, paragraph 12 · Text on page 641 of the German source, paragraph 13

In modern notation

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

An inference beyond the argument: That the mass of a body is a measure of its energy content goes beyond the radiation argument: the paper says it is evidently unimportant that the energy leaves as radiation, and generalizes. The argument itself covers energy given off as light.
Premise: The paper gives a change of mass with a change of energy. It assigns no absolute energy to a body.

Decoder

$L/V^2$: The decrease of the body's mass, from the coefficient of $v^2/2$ in the previous result.
$9\cdot10^{20}$: The square of the speed of light in centimetres per second, $(3\cdot10^{10})^2$ : energy in erg divided by it gives mass in grams. The paper prints only the square.
"in demselben Sinne": In the same direction: when the energy decreases, so does the mass, and when it increases, so does the mass.

Printed checks

Printed value beside the reproduced value, separately labeled
As printed	Reproduced	Label
1 g	1 g	as printed: L divided by the paper's 9·10²⁰ (constant set einstein-1905-mass-energy-printed)
(not printed)	1.0013851 g	modern comparison: L divided by the square of today's defined speed of light (constant set modern-si-2019)

Compared: the printed factor is 0.1385 percent larger than the modern c².

Try it

ME-02: Read the printed factor beside the modern one, each labelled by its constant set. Choose the preset “Observer at 0.6c” there.

ME-03: Draw the system boundary around the body alone, then around the body and the light: which system loses mass? Choose the preset “Sealed lamp and mirror (Internal transfer)” there.

Where this stops: This calculation determines a change, not an absolute internal energy. Einstein’s printed constants and modern values are kept in separate calculations. Read the argument

Where this is used later

Common wrong turns

Argument: The paper's conclusion, and an inference beyond it
Model: Exact in the limit of small speed; the conversion uses the printed constant
Evidence: A conclusion, not a measurement
History: Introduced in this paper

Page 641

Two conditional closing remarks

The paper ends with two conditional remarks: a test may succeed with bodies whose energy content varies greatly, such as radium salts, and if the theory corresponds to the facts, radiation carries inertia from the body that emits it to the body that absorbs it.

As printed

Es ist nicht ausgeschlossen, daß bei Körpern, deren Energieinhalt in hohem Maße veränderlich ist (z. B. bei den Radiumsalzen), eine Prüfung der Theorie gelingen wird.
Wenn die Theorie den Tatsachen entspricht, so überträgt die Strahlung Trägheit zwischen den emittierenden und absorbierenden Körpern.

Text on page 641 of the German source, paragraph 14 · Text on page 641 of the German source, paragraph 15

Conditional: Both remarks are conditional. The paper reports no test, and it does not say that radium salts would confirm the theory.

Decoder

"Es ist nicht ausgeschlossen": It is not excluded: a possibility, not a prediction that the test will succeed.
Radiumsalzen: Radium salts, which give off energy continually; the paper names them as an example of a body whose energy content varies greatly.
"Wenn die Theorie den Tatsachen entspricht": If the theory corresponds to the facts: the second remark is conditional on the theory being right.

Try it

ME-03: Read the radium card's boundary note: what would a measurement have to compare? Choose the preset “Radium-226 Alpha Decay” there.

Where this stops: A successful model calculation is not an observed experimental confirmation. The 1906 box argument and modern four-momentum are separate extensions. Read the argument

Where this is used later

The measured checks the radium remark proposes came decades later, from nuclear physics.

Common wrong turns

Argument: Two conditional remarks
Model: Outside the calculation
Evidence: A proposed test, not evidence
History: Introduced in this paper

The unsectioned argument · explanatory reading

One result is borrowed, not rediscovered

The light-energy transformation is an input from relativity; the body’s mass change is not an input.

Read the argument

Choose the emission that removes recoil

Symmetric emission keeps the body’s speed unchanged, isolating an energy change.

Read the argument

Two accounts of the same loss

The rest account loses L and the moving account loses γL; each balances internally.

Read the argument

Subtract what you cannot measure

Subtracting the energy losses gives L(γ − 1), without choosing the body’s internal energy.

Read the argument

The premise that the subtraction needs

The kinetic interpretation assumes that the additive energy offset is unchanged during emission.

Read the argument

Read the slow-speed coefficient

The low-speed coefficient identifies inertia; a finite-speed ratio is only a proxy.

Read the argument

A decrease, a signed change, and a conversion

The positive mass lost is L/c²; the signed body-mass change is −L/c².

Read the argument

The light has left the body, not vanished

The body’s loss is the radiation’s gain; distinguish a calculation, a broader inference, and an experimental test.

Read the argument