Read · Special relativity: from clock operations to electrodynamics

§10 · electron force, work, and closing scope

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Detail

Letters

Show modern qualifications

Einstein's letters change 12 of this paper's 40 formulas. 28 stay in today's letters, with a note saying so.

A derivation

A force-to-acceleration ratio needs two frame labels

Why are there two transverse coefficients?

The 9 printed paragraphs this passage explains

§10, paragraph 1: A small particle carrying an electric charge ε, called an electron, moves in an electromagnetic field, and Einstein makes only the following assumption about its law of motion.
§10, paragraph 2: When the electron is at rest at some moment, its motion in the next instant follows the ordinary law: its mass μ times its acceleration equals ε times the electric force, as long as it moves slowly.
§10, paragraph 3: Now the electron has a speed v at some moment, and Einstein seeks the law of its motion in the next instant.
§10, paragraph 4: Without loss of generality the electron is taken to be at the origin at that moment, moving along the x-axis, so that it is momentarily at rest in a system k moving with the same speed.
§10, paragraph 5: In k it therefore obeys the law for a resting electron, and the transformations of sections 3 and 6 connect the coordinates, time and forces of k with those of the stationary system.
§10, paragraph 6: Transforming the equations of motion back to the stationary system gives the equations Einstein labels (A).
§10, paragraph 7: Following the usual approach, Einstein asks for the electron's longitudinal and transverse mass. Taking the force to be the one measured in a system moving with the electron, for example with a spring balance, and measuring accelerations in the stationary system, he obtains one mass along the motion and another across it.
§10, paragraph 8: Other definitions of force and acceleration would give other numbers for these masses, so different theories of the electron's motion must be compared with great care.
§10, paragraph 9: These results about mass hold for any ponderable material point too, since adding an arbitrarily small charge makes it an electron in this sense.

\frac{F_x}{a_x} = m\,\gamma^{3}

\frac{F_y'}{a_y} = m\,\gamma^{2}

\frac{F_y}{a_y} = m\,\gamma

Shown in modern letters; Einstein's are in §10 of the German source face.

Longitudinal laboratory force over acceleration is m gamma cubed. Rest-frame transverse force over laboratory acceleration is m gamma squared. Laboratory transverse force over laboratory acceleration is m gamma.

For an electron moving along $x$ , a force across the motion is $\gamma$ times larger in the electron’s momentary rest frame than in the laboratory. Einstein divides the rest-frame force by the laboratory acceleration and gets $m\gamma^2$ . Dividing the laboratory force by the same acceleration gives $m\gamma$ .

At $0.6c$ , $\gamma = 1.25$ . The longitudinal mass is $m\gamma^3 = 1.953125\,m$ , Einstein’s transverse mass is $m\gamma^2 = 1.5625\,m$ , and the all-laboratory transverse ratio is $m\gamma = 1.25\,m$ . The electron-dynamics laboratory shows each number with the frames it uses, and leaves Einstein’s result as he printed it.

Open the foundation: Energy of motion and inertia

Show every step here: A force-to-acceleration ratio needs two frame labels

Letters: in §10 the paper's electron is a point particle of charge ε and mass μ, where μ is its mass while it moves slowly; X, Y, Z are the electric force and M, N magnetic components, in Gaussian units; β is the modern γ and V the speed of light. This explanation writes m for μ, and F with a frame label for each force.

The only assumption about the law of motion: when the electron is at rest at some moment, in the next instant it moves by $\mu\frac{d^2x}{dt^2}=ε X$ , and likewise for y and z, ordinary mechanics for slow motion.
Now let the electron have speed v along x at the moment t = 0, at the origin of K. It is then at rest in the system k moving at v, so in k, for small times, $\mu\frac{d^2ξ}{d\tau^2}=ε X'$ and likewise for η and ζ.
Transform back to K with §3 and §6, using $X'=X$ , $Y'=\beta(Y-\frac{v}{V}N)$ , $Z'=\beta(Z+\frac{v}{V}M)$ . The paper's equations (A): $\frac{d^2x}{dt^2}=\frac{ε}{\mu}\frac{1}{\beta^3}X$ , $\frac{d^2y}{dt^2}=\frac{ε}{\mu}\frac{1}{\beta}(Y-\frac{v}{V}N)$ , and the same for z.
Rewrite them as the paper does: $\mu\beta^3\frac{d^2x}{dt^2}=ε X=ε X'$ and $\mu\beta^2\frac{d^2y}{dt^2}=ε\beta(Y-\frac{v}{V}N)=ε Y'$ .
Read $εX'$ , $εY'$ , $εZ'$ as the force on the electron measured in a system moving with it at that moment; the paper says it could be measured with a spring balance at rest in that system.
Keep mass times acceleration equal to force, with that force and with accelerations measured in K. Then the ratio of force to acceleration is $\mu\beta^3$ along the motion and $\mu\beta^2$ across it: the paper's longitudinal mass $\mu/(\sqrt{1-(v/V)^2})^3$ and transverse mass $\mu/(1-(v/V)^2)$ .
Measure the transverse force in K as well: it is smaller than the co-moving force by the factor β, so that ratio is $m\gamma$ . The three ratios are below.

\frac{F_x}{a_x} = m\,\gamma^{3}

\frac{F_y'}{a_y} = m\,\gamma^{2}

\frac{F_y}{a_y} = m\,\gamma

Shown in modern letters; Einstein's are in §10 of the German source face.

Worked at 0.6c, γ = 1.25: longitudinal 1.953125 m, Einstein's transverse 1.5625 m, and the all-laboratory transverse ratio 1.25 m.
The paper's own warning, kept with its result: with other definitions of force and of acceleration, other numbers for these masses would be obtained, so theories of the electron's motion must be compared with great care.
The transverse mass $m\gamma^2$ is Einstein's, under his convention: force measured in the co-moving system, acceleration in K. The later habit of defining force as the rate of change of momentum gives $m\gamma$ for the same transverse case; it is a different convention, and it does not correct his result.
The paper adds that these results hold for any ponderable material point too, since a vanishingly small charge turns it into an electron in its sense.
The electron's mass m is one number; each ratio depends on the direction of the force and on the frames used, and m does not.

A tool for this step

Energy of motion and inertia

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

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

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

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

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

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

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

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

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

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

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

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

Try it: a before-and-after ledger

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

Mass

Speed before the push

Force

Distance pushed

Count every energy from

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

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

What it shows, in words

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

Where this lesson stops

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

Explore the equations in this step

Read each operation, check its units and assumptions, or open the mathematical step behind it. These are modern teaching equations, not the equations as printed.

Explore the equation · Modern model notation

Along the motion: m gamma cubed

\frac{F_x}{a_x} = m\,\gamma^{3}

Shown in modern letters; Einstein's are in §10 of the German source face.

over is times cubed.

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Longitudinal laboratory force over acceleration gives m gamma cubed.

Pushing along the motion, the ratio of force to acceleration is m times gamma cubed. At 0.6c that is 1.953125 m.

Model assumptions and every term’s meaning

An electron moving along x at speed v in the laboratory, with |v| < c, pushed by a small force.
A force-over-acceleration ratio means nothing until both are labelled with their frames. The paper's transverse mass uses the force measured in the frame moving with the electron.
Modern teaching notation: gamma is the paper's beta, m the electron's mass at rest.

m times gamma cubed: The longitudinal mass. Read the prerequisite
Gamma cubed: Grows fast: 1.95 at 0.6c. Read the prerequisite
Force over acceleration: What the old electron theories called a mass. It depends on the direction of the push. Read the prerequisite
What it asserts: The ratio for a push along the motion. Pushes in other directions give other ratios. Read the prerequisite
Acceleration along the motion: Measured in the laboratory. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. At 0.6c it is 1.25. Read the prerequisite
Force along the motion: Along x, in the laboratory. For this direction the force is the same in the electron's own frame. Read the prerequisite
Electron mass: The electron's mass at rest. Read the prerequisite

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

Explore the equation · Modern model notation

Across the motion, both in the laboratory: m gamma

\frac{F_y}{a_y} = m\,\gamma

Shown in modern letters; Einstein's are in §10 of the German source face.

over is times .

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Each term and operation, in words

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Read the equation aloud in words

Laboratory transverse force over acceleration gives m gamma.

With force and acceleration both measured in the laboratory, the sideways ratio is m times gamma: 1.25 m at 0.6c. Changing the frame of the force changes the ratio by gamma.

Model assumptions and every term’s meaning

An electron moving along x at speed v in the laboratory, with |v| < c, pushed by a small force.
A force-over-acceleration ratio means nothing until both are labelled with their frames. The paper's transverse mass uses the force measured in the frame moving with the electron.
Modern teaching notation: gamma is the paper's beta, m the electron's mass at rest.

m times gamma: 1.25 m at 0.6c. Read the prerequisite
Force over acceleration, one frame: Both measured in the laboratory. Read the prerequisite
What it asserts: A different convention from the paper's, giving a different number. Neither is a correction of the other. Read the prerequisite
Acceleration across the motion: Measured in the laboratory. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. At 0.6c it is 1.25. Read the prerequisite
Sideways force, laboratory: Measured in the laboratory: the electron's-frame force divided by gamma. Read the prerequisite
Electron mass: The electron's mass at rest. Read the prerequisite

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

Explore the equation · Modern model notation

Across the motion, the paper's convention: m gamma squared

\frac{F_y'}{a_y} = m\,\gamma^{2}

Shown in modern letters; Einstein's are in §10 of the German source face.

over is times 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.

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Each term and operation, in words

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Read the equation aloud in words

Comoving transverse force over laboratory acceleration gives m gamma squared.

With the sideways force measured in the electron's own frame and the acceleration in the laboratory, the ratio is m times gamma squared: 1.5625 m at 0.6c. This is the paper's transverse mass.

Model assumptions and every term’s meaning

An electron moving along x at speed v in the laboratory, with |v| < c, pushed by a small force.
A force-over-acceleration ratio means nothing until both are labelled with their frames. The paper's transverse mass uses the force measured in the frame moving with the electron.
Modern teaching notation: gamma is the paper's beta, m the electron's mass at rest.

m times gamma squared: The transverse mass as the paper defines it. Read the prerequisite
Gamma squared: 1.5625 at 0.6c. Read the prerequisite
Force over acceleration, two frames: The force is from one frame and the acceleration from another; that choice is what gives gamma squared. Read the prerequisite
What it asserts: The printed transverse mass follows from this convention. The paper's result is not wrong; it answers a question stated with these frames. Read the prerequisite
Acceleration across the motion: Measured in the laboratory. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. At 0.6c it is 1.25. Read the prerequisite
Sideways force, electron's frame: Measured in the frame moving with the electron at this moment. It is gamma times the laboratory's sideways force. Read the prerequisite
Electron mass: The electron's mass at rest. 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.

Modern qualifications

Einstein himself warns that other definitions of force and acceleration would give other values for these masses. The modern habit of defining force as the rate of change of momentum gives $m\gamma$ for the transverse ratio. It is a later convention, and this edition does not substitute it for his printed result.

Assumptions and limits: A force-to-acceleration ratio needs two frame labels

Assumed here

In its momentary rest frame the electron obeys ordinary mechanics for slow motion.
Every force and every acceleration is labelled with the frame it is measured in.

What this does not establish

The two masses are ratios of force to acceleration under stated conventions; the electron’s own mass does not change with the observer’s speed.
The electron is treated as a point driven slowly, without the effects of its own radiation or field.

Earlier step: Electric and magnetic components mix together

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

A derivation

The work integral has an observable endpoint

What energy is required to accelerate an electron?

The 9 printed paragraphs this passage explains

§10, paragraph 10: Einstein finds the electron's kinetic energy. An electron starting from rest and accelerated slowly by an electrostatic force, so that it loses no energy as radiation, gains energy equal to the work done on it, which he calculates.
§10, paragraph 11: That energy grows without limit as the speed approaches V, so, as in the earlier results, speeds greater than light have no possibility of existing.
§10, paragraph 12: By the argument given earlier, the same expression for kinetic energy holds for ponderable masses.
§10, paragraph 13: Einstein turns to the properties of the electron's motion that follow from equations (A) and can be tested by experiment, and lists three.
§10, paragraph 14: First, an electric force and a magnetic force deflect a moving electron equally when the electric force is v/V times the magnetic one, so the ratio of its magnetic to its electric deflectability is v/V and gives its speed, at any speed.
§10, paragraph 15: That relation can be tested, because the electron's speed can also be measured directly, for example with rapidly oscillating electric and magnetic fields.
§10, paragraph 16: Second, the kinetic energy fixes how the potential difference an electron passes through is related to the speed it gains.
§10, paragraph 17: Third, when a magnetic force perpendicular to its velocity is the only force deflecting it, the second of equations (A) gives the radius of curvature of the electron's path.
§10, paragraph 18: These three relations, he says, completely express the laws by which the electron must move according to the theory.

K = \int_{0}^{v} m\,\gamma\left(u\right)^{3}\,u\,\mathrm{d}u

K = m\,c^{2}\,\left(\gamma - 1\right)

Shown in modern letters; Einstein's are in §10 of the German source face.

Integrating the longitudinal work from rest gives m c squared times gamma minus one.

The integral follows from one derivative, $d\gamma/du = \gamma^3 u/c^2$ : the integrand $m\gamma^3 u$ is $mc^2$ times it, so the integral is $mc^2(\gamma - 1)$ . At low speed this approaches the familiar $mv^2/2$ . At $0.6c$ the exact value is $0.25\,mc^2$ and the low-speed formula gives $0.18\,mc^2$ , a difference large enough to see.

|q|\,U = K

r_B = \frac{\gamma\,m\,v}{|q|\,B}

Shown in modern letters; Einstein's are in §10 of the German source face.

For acceleration from rest, charge magnitude times voltage magnitude supplies kinetic energy; in a transverse magnetic field the circular radius is gamma m v over charge magnitude B.

The third relation compares bending by the two kinds of field. In an electric field across the motion the path curves with radius $\gamma mv^2/(|q|E_y)$ at that instant; the field also speeds the electron up as the path turns, so the radius keeps changing. In a magnetic field the speed stays constant and the path is a circle of radius $\gamma mv/(|q|B)$ . The electron-dynamics laboratory keeps the two cases apart.

Open the foundation: Adding continuously

Show every step here: The work integral has an observable endpoint

Letters: the paper's electron has charge ε and mass μ (m here); it starts at rest at the origin of K and moves along the X axis under an electrostatic force X; W is its energy of motion, P the potential difference it has passed through, its magnetic and electric deflectabilities $A_m$ and $A_e$ , R the radius of curvature of its path, N a magnetic force across its motion, and V the speed of light.

The energy taken from the electrostatic field is $\intε X\,dx$ . The electron is accelerated slowly and so gives off no energy as radiation, so that energy all becomes its energy of motion W.
The first of equations (A) holds throughout: $ε X=\mu\beta^3\frac{d^2x}{dt^2}$ . Along the path, acceleration times a small distance is $v\,dv$ , so $ε X\,dx=\mu\beta^3v\,dv$ .
Integrate from rest to speed v, using $\frac{d}{dv}\frac{1}{\sqrt{1-v^2/V^2}}=\frac{v}{V^2}\beta^3$ : the paper's result $W=\mu V^2\left\{\frac{1}{\sqrt{1-(v/V)^2}}-1\right\}$ . In the explanatory letters, the formula below.

K = \int_{0}^{v} m\,\gamma\left(u\right)^{3}\,u\,\mathrm{d}u

K = m\,c^{2}\,\left(\gamma - 1\right)

Shown in modern letters; Einstein's are in §10 of the German source face.

Integrating the longitudinal work from rest gives m c squared times gamma minus one.

At low speed, $1/\sqrt{1-ε}$ is about $1+ε/2$ , so W approaches $\frac{1}{2}mv^2$ .
Worked at 0.6c: exact $(1.25-1)mc^2=0.25\,mc^2$ ; low-speed $0.5\times0.36\,mc^2=0.18\,mc^2$ . For an electron, with modern SI constants, $mc^2$ is about 511 kiloelectronvolts, so W is about 127.7 kiloelectronvolts.
W becomes infinite at v = V. The paper concludes, as from its earlier results, that speeds greater than light have no possibility of existence.
The paper lists three properties of the motion that experiment can reach. First: an electric force Y and a magnetic force N deflect an electron moving at v equally when $Y=N\,v/V$ , so the ratio of magnetic to electric deflectability gives the speed, $A_m/A_e=v/V$ , and the speed can be measured independently.
Second: the potential difference passed through and the speed gained are related by $P=\int X\,dx=\frac{\mu}{ε}V^2\left\{\frac{1}{\sqrt{1-(v/V)^2}}-1\right\}$ . In SI, charge times voltage equals W. Worked: to reach 0.6c from rest, an electron needs about 128 kilovolts, with no energy lost to radiation.
Third: the radius of curvature R when a magnetic force N across the motion is the only deflecting force. From the second of (A), $v^2/R=\frac{ε}{\mu}\frac{v}{V}N\sqrt{1-(v/V)^2}$ , so $R=V^2\frac{\mu}{ε}\frac{v/V}{\sqrt{1-(v/V)^2}}\frac{1}{N}$ . In SI that is $r_B=\gamma mv/(|q|B)$ , the second formula below.

|q|\,U = K

r_B = \frac{\gamma\,m\,v}{|q|\,B}

Shown in modern letters; Einstein's are in §10 of the German source face.

For acceleration from rest, charge magnitude times voltage magnitude supplies kinetic energy; in a transverse magnetic field the circular radius is gamma m v over charge magnitude B.

Worked: an electron at 0.6c in a field of 0.1 tesla circles with radius $1.25\times m\times0.6c/(|q|\times0.1)$ , about 1.28 centimetres, with modern SI constants.
Keep the magnetic case, a circle at constant speed, apart from bending in an electric field, which also changes the speed and so gives a radius that holds only at one instant.
The paper calls these three relations a complete expression of the laws by which, according to the theory, the electron must move. They hold within its assumptions: the fields as specified, and no energy lost to radiation.
The kinetic energy formula alone does not give the September paper's result that a body's mass changes with its energy content; that needs the argument from the light complex of §8.

A tool for this step

Adding continuously

How do you add up something that varies continuously?

Chance spread along a line is measured per unit length: a density. Where the density is the same all along a stretch, the chance of the stretch is the density times its length. A particle equally likely to be anywhere in a 4 µm stretch has density 0.25 per micrometre, and the chance of finding it in the first 1.5 µm is 0.25 × 1.5 = 0.375.

Where the density changes from place to place, cut the line into narrow strips. Each strip contributes its height times its width, as if the density were flat across it. Add the strips, then cut them narrower and add again. For a smooth curve the totals settle on one value, and that value is the integral:

\int_a^b p(x)\,dx = \lim_{\Delta x \to 0} \sum_i p(x_i)\,\Delta x

The integral of p of x from a to b is the value that the sum of p of x i times delta x settles on as delta x shrinks.

The units multiply. A probability density is a probability per metre, so a strip of it, times a width in metres, is a plain probability. For a density the whole area must be 1, because the particle has to be somewhere. The Brownian paper's §4 writes exactly that for the chances of a jump Δ, ∫φ(Δ) dΔ = 1, and the light-quanta paper's §3 adds entropy over every frequency, S = v∫φ dν.

One more tool recurs in the papers: integration by parts. It is the product rule for derivatives added up over an interval, and it moves a derivative from one factor to the other at the price of an endpoint term:

\int_a^b u\frac{dv}{dx}\,dx = [uv]_a^b - \int_a^b v\frac{du}{dx}\,dx

The integral of u times the derivative of v equals u v at b minus u v at a, minus the integral of v times the derivative of u.

The endpoint term, uv at b minus uv at a, is part of the answer. It disappears only when uv is zero at both ends, as it is for a density that falls to zero far out on both sides.

Worked example: A rising density, added up in narrower strips

A particle's density rises steadily across a 4 µm stretch, from 0 at the left end to 0.5 per micrometre at the right. The whole area is a triangle, ½ × 4 × 0.5 = 1.
Four strips 1 µm wide, each taking the height at its left edge: 0 + 0.125 + 0.25 + 0.375 = 0.75.
Eight strips 0.5 µm wide: the heights add to 1.75, and times 0.5 that is 0.875.
Sixteen strips 0.25 µm wide: 0.9375. The shortfall halves each time the strips do: 0.25, then 0.125, then 0.0625.
The totals settle on 1, the area of the triangle. That settled value is the integral.

A flat density gives the same total for every cut, which is why the uniform stretch above needed no strips. A curved one, such as the Brownian bell curve, does, and the integral is the value the totals settle on.

The same triangle, cut two ways

A density rising from 0 to 0.5 per micrometre across 4 μm. Each strip is as tall as the curve at its own left edge, which is why every strip falls short and why the shortfall shrinks when the strips do.

4 strips add to 0.75; 8 strips add to 0.875. The triangle itself is 1. Cut narrower and the total climbs toward it without ever passing it, and the value it climbs toward is the integral.

What it shows, in words

Both drawings are the same density, rising in a straight line from 0 at the left end to 0.5 per micrometre at the right, across 4 micrometres. The first is cut into 4 strips of 1 micrometre each and the second into 8 strips of 0.5.

Each strip is as tall as the line at its own left edge, so each leaves a wedge uncovered above it. The 4 strips add to 0.75 and the 8 strips to 0.875, against the triangle's area of 1. Narrower strips leave smaller wedges, so the totals climb toward 1 and never past it. That limit is what the integral names.

What it does not show: these strips take the height at their left edge, which is one choice among several. Taking the right edge would overshoot 1 and climb down to it instead, and taking the middle of each strip would land much closer from the start. All three settle on the same 1, and that they agree is the fact the integral rests on. Nothing in either drawing demonstrates it.

Where this lesson stops

An integral is a total of height times width over a range, in the units of both. The widths and any endpoint term are part of the calculation, not decoration on the integral sign.

Explore the equations in this step

Read each operation, check its units and assumptions, or open the mathematical step behind it. These are modern teaching equations, not the equations as printed.

Explore the equation · Modern model notation

The voltage supplies the energy

|q|\,U = K

Shown in modern letters; Einstein's are in §10 of the German source face.

times equals .

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Charge magnitude times voltage magnitude supplies kinetic energy.

Accelerated from rest through a voltage U, the electron gains energy |q| times U, and all of it is energy of motion. With K = mc²(γ − 1) this fixes the electron's speed.

Model assumptions and every term’s meaning

An electron accelerated from rest, then moving at speed v with |v| < c. In section 10 the electron is momentarily at rest in the moving system, so its speed is that system's speed v.
Modern teaching notation: gamma is the paper's beta, |q| the size of the electron's charge, U the size of the voltage.

What it asserts: Energy conservation for acceleration from rest: the work done becomes energy of motion. Read the prerequisite
Charge times voltage: The work the field does on the electron: coulombs times volts is joules. Read the prerequisite
Size of the electron's charge: The elementary charge. The sign of the charge decides the direction of the force, not these magnitudes. Read the prerequisite
Kinetic energy: The electron's energy of motion afterwards: m c squared times gamma minus one, not one half m v squared, which is only its low-speed form. Read the prerequisite
Accelerating voltage: The voltage the electron falls through, starting from rest. Read the prerequisite

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

Explore the equation · Modern model notation

Add up the work from rest

K = \int_{0}^{v} m\,\gamma\left(u\right)^{3}\,u\,\mathrm{d}u

Shown in modern letters; Einstein's are in §10 of the German source face.

is the integral from 0 to of times of u cubed times , over .

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Integrating the longitudinal work from rest gives the kinetic energy.

The work done on the electron as it speeds up from rest to v: the longitudinal mass m gamma cubed times the speed, added up over every speed on the way.

Model assumptions and every term’s meaning

An electron accelerated from rest to speed v along x, with |v| < c, by a longitudinal force.
Modern teaching notation: gamma is the paper's beta; m is the electron's mass at rest.

Gamma cubed: The longitudinal mass of section 10 is m gamma cubed. Read the prerequisite
From rest to v: Adding the work over every speed from 0 to v. Read the prerequisite
Longitudinal mass times speed: Force times distance, rewritten as mass times speed times the change in speed. Read the prerequisite
What it asserts: The energy of motion is the work done to reach speed v from rest. Read the prerequisite
Kinetic energy: The electron's energy of motion at speed v. Read the prerequisite
Gamma at each speed: The Lorentz factor evaluated at the speed u reached so far, not at the final speed. Read the prerequisite
Final speed: The speed the electron ends at. Read the prerequisite
Electron mass: The electron's mass at rest. Read the prerequisite
Speed on the way: The electron's speed at an intermediate stage, from 0 up to v. Read the prerequisite

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

Explore the equation · Modern model notationEinstein's letters

The result: m c squared times gamma minus one

K = m\,c^{2}\,\left(\gamma - 1\right)

W = \mu\,V^{2}\,\left(\beta - 1\right)

is times squared times minus one.

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Each term and operation, in words

No live values

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

Read the equation aloud in words

Integrating the longitudinal work from rest gives m c squared times gamma minus one.

Because the rate of change of gamma with speed is gamma cubed u over c squared, the integral is m c squared times the change in gamma, from 1 at rest to gamma at v. At 0.6c that is 0.25 m c squared; the familiar one half m v squared gives 0.18 m c squared, so at this speed the difference is plain.

Model assumptions and every term’s meaning

An electron accelerated from rest to speed v along x, with |v| < c, by a longitudinal force.
Modern teaching notation: gamma is the paper's beta; m is the electron's mass at rest.

Gamma minus one: Zero at rest, 0.25 at 0.6c. For small speeds it is about half of v squared over c squared. Read the prerequisite
c squared: The square of the speed of light. Read the prerequisite
m c squared times that: Near zero speed this becomes one half m v squared. Read the prerequisite
What it asserts: The exact energy of motion. It grows without limit as v approaches c. Read the prerequisite
Kinetic energy: The electron's energy of motion at speed v. Read the prerequisite
Gamma at the final speed: The Lorentz factor at v. Read the prerequisite
Speed of light: The paper's V. Read the prerequisite
Electron mass: The electron's mass at rest. Read the prerequisite

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

Explore the equation · Modern model notation

The circle in a magnetic field

r_B = \frac{\gamma\,m\,v}{|q|\,B}

Shown in modern letters; Einstein's are in §10 of the German source face.

is times times , over times .

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In a transverse magnetic field the circular radius is gamma m v over charge magnitude B.

In a uniform magnetic field across its motion the electron goes round a circle. Its radius is gamma m v over |q| B: larger than m v over |q| B, the value without gamma, by gamma. At 0.6c that is 25 per cent.

Model assumptions and every term’s meaning

An electron accelerated from rest, then moving at speed v with |v| < c. In section 10 the electron is momentarily at rest in the moving system, so its speed is that system's speed v.
Modern teaching notation: gamma is the paper's beta, |q| the size of the electron's charge, U the size of the voltage.
A uniform magnetic field of strength B at right angles to the electron's velocity, in the laboratory.

Gamma m v: The quantity the field has to turn round. Without gamma this is the low-speed m v. Read the prerequisite
Charge times field: How strongly the field turns the electron. Read the prerequisite
The ratio: Metres: kilograms times metres per second, over coulombs times teslas. Read the prerequisite
What it asserts: A prediction that deflection experiments on fast electrons can test: the radius grows with gamma, not with v alone. Read the prerequisite
Size of the electron's charge: The elementary charge. The sign of the charge decides the direction of the force, not these magnitudes. Read the prerequisite
Lorentz factor: Of the electron's speed: one over the square root of one minus v squared over c squared. Read the prerequisite
Magnetic field: Across the motion, in the laboratory. Read the prerequisite
Electron mass: The electron's mass at rest. Read the prerequisite
Radius of the circle: The radius of the electron's circular path in the laboratory. Read the prerequisite
Electron speed: The speed of the system in which the electron is momentarily at rest, which is the electron's own 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.

Modern qualifications

The three relations hold within the model’s assumptions: fields as specified, and no energy lost to radiation. After §10 the paper closes with thanks to M. Besso and the date line, Bern, June 1905. Those belong to the German source text, and this explanation does not repeat them.

Assumptions and limits: The work integral has an observable endpoint

Assumed here

The electron has a fixed mass m and moves slower than light, and force is defined as stated.
For the voltage relation, the electron starts at rest and all the work of the field becomes its energy of motion.

What this does not establish

Losses by radiation and interactions with matter are left out.
The kinetic energy formula alone does not give the September paper’s result that a body’s mass changes with its energy.

Earlier step: A force-to-acceleration ratio needs two frame labels

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

References

A. Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt. Annalen der Physik (4), 17, 132–148 (1905).

A. Einstein, On the motion of particles suspended in liquids at rest required by the molecular-kinetic theory of heat. Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

A. Einstein, Zur Elektrodynamik bewegter Körper. Annalen der Physik (4), 17, 891–921 (1905).

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 translation or its pinned facsimile.

28 foundation readings sit behind this argument.

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