Read · Special relativity: from clock operations to electrodynamics

§4 · rods, clocks, and the limit of the model

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Letters

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

A derivation

Use the same measurement protocol

What do a moving clock and rod actually report?

The 6 printed paragraphs this passage explains

§4, paragraph 1: A rigid sphere of radius R at rest in the moving system is, seen from the stationary system at one moment, an ellipsoid whose axis along the motion is shortened.
§4, footnote 1: A footnote specifies that the sphere means a body that has the shape of a sphere when examined at rest.
§4, paragraph 2: Dimensions across the motion are unchanged, while the one along it is shortened, the more so the faster the body moves. At v = V moving bodies would shrink to flat figures, speeds above V make the considerations meaningless, and in the theory the speed of light plays the part of an infinitely great speed.
§4, paragraph 3: The same results hold the other way round, for bodies at rest in the stationary system viewed from a uniformly moving one.
§4, paragraph 4: Einstein asks how fast a clock at rest at the origin of the moving system runs, as judged from the stationary system.
§4, paragraph 5: Using the transformation at the clock's place, he finds that it falls behind by a fixed fraction of a second each second, to a first approximation half of (v/V) squared.

\Delta t = \gamma\,\Delta\tau

L = \frac{L_0}{\gamma}

Shown in modern letters; Einstein's are in §4 of the German source face.

Coordinate time between a moving clock’s ticks is gamma times its proper elapsed time; simultaneous moving-rod length is rest length over gamma.

At $0.6c$ the factor is 0.8 for both: the moving clock records 0.8 s while the resting frame’s clocks advance 1 s, and a rod 1 m long at rest measures 0.8 m. The same number comes out of two different measurements.

For the clock, the two ticks happen at one place in the clock’s own frame. For the rod, the two end marks happen at one time in the measuring frame. The moving-clock and simultaneity laboratories let you set up each measurement on its own.

1 - \frac{1}{\gamma} \approx \frac{1}{2}\,\frac{v^{2}}{c^{2}}

1 - \frac{1}{\beta} \approx \frac{1}{2}\,\frac{v^{2}}{V^{2}}

At small speed, the clock’s fractional loss per coordinate second approaches one half v squared over c squared.

Open the foundation: Events and distant clocks

Show every step here: Use the same measurement protocol

Letters: in §4 the paper's rigid sphere has radius R at rest in the moving system k; a clock at k's origin shows k's time τ; V is the speed of light. The formula below writes Δτ for the time that clock itself shows between two of its ticks and Δt for the time K's clocks give between the same two ticks; $L_0$ is a rod's length at rest and L its length measured in K.

The rod. A sphere of radius R at rest in k has the surface $ξ^2+η^2+ζ^2=R^2$ .
Express it in K at the time t = 0, marking every point of the surface at that one instant of K: then $ξ=\beta x$ , η = y and ζ = z, so $\beta^2x^2+y^2+z^2=R^2$ .
Read the shape: an ellipsoid with semi-axes $R/\beta$ along the motion and R, R across it. The paper prints the long axis as $R\sqrt{1-(v/V)^2}$ , the same thing. Only the dimension along the motion is shortened.
So a rod of rest length $L_0$ along the motion, with both ends marked at one time of K, measures $L=L_0/\gamma$ .
The clock. The clock at k's origin shows $τ=\beta(t-vx/V^2)$ and sits at x = vt.
Substitute x = vt: $τ=\beta t\left(1-\frac{v^2}{V^2}\right)=t\sqrt{1-(v/V)^2}$ , since β times $1-v^2/V^2$ is its square root.
So between two of its ticks, at one place in k, the moving clock shows Δτ while K's clocks advance $\Delta t=\gamma\Delta\tau$ . The two formulas below, one for the clock and one for the rod.

\Delta t = \gamma\,\Delta\tau

L = \frac{L_0}{\gamma}

Shown in modern letters; Einstein's are in §4 of the German source face.

Coordinate time between a moving clock’s ticks is gamma times its proper elapsed time; simultaneous moving-rod length is rest length over gamma.

Worked at v = 0.6c, γ = 1.25: the moving clock shows 0.8 second while K's clocks advance 1 second, and a rod 1 metre long at rest measures 0.8 metre. The same number comes out of two different measurements.
Keep the two protocols apart. For the clock, the two ticks happen at one place in the clock's frame. For the rod, the two end marks happen at one time in the measuring frame.
How much the clock loses per second of K: $1-\sqrt{1-(v/V)^2}$ . At 0.6c that is 1 − 0.8 = 0.2 second in each second.
The paper also gives the loss to second order, dropping terms of fourth and higher order in v/V: $\sqrt{1-ε}$ is about $1-ε/2$ for small ε, so the loss is about $\frac{1}{2}(v/V)^2$ . That is the formula below.

1 - \frac{1}{\gamma} \approx \frac{1}{2}\,\frac{v^{2}}{c^{2}}

1 - \frac{1}{\beta} \approx \frac{1}{2}\,\frac{v^{2}}{V^{2}}

At small speed, the clock’s fractional loss per coordinate second approaches one half v squared over c squared.

At 0.6c the second-order formula gives $\frac{1}{2}\times0.36=0.18$ , close to the exact 0.2 but not equal; at 3 metres per second the two would agree to many decimal places.
The effect runs both ways: the paper notes that bodies and clocks at rest in K, examined from k, show the same results.
The paper adds that at v = V every moving object would shrink to a flat figure, as seen from K, and that for speeds greater than light the considerations lose their meaning.
A measurement is not a photograph. The contracted length is found by marking both ends at one instant of K; a camera records light that arrives together, which left different parts of the body at different times.

A tool for this step

Events and distant clocks

What does it mean to say that two distant events happen at the same time?

An event is something that happens at one place at one instant: a spark, a collision, a flash leaving a lamp. To give it a time, you read a clock at that place. Reading a distant clock by looking at it would mix in the time the light takes to reach you.

A reference frame is a set of rigid measuring rods and clocks at rest, with a clock at every place, so that each event gets three coordinates of position and one of time from the rods and the clock beside it. Before that works, the clocks must agree, and for clocks far apart agreeing needs a rule.

§1 of the relativity paper gives the rule. Send a flash of light from clock A to clock B and back. Note A's reading when it leaves, B's reading when it arrives, and A's reading when it returns. The clocks are called synchronous if the trip out takes as long, by the clocks, as the trip back:

t_B - t_A = t'_A - t_B

t B minus t A equals t A prime minus t B: the trip out takes as long as the trip back.

The paper calls this a definition. It cannot be checked by measurement, because checking it would need clocks that are already synchronous. §1 adds two assumptions: that the rule is consistent, so that if B agrees with A then A agrees with B, and two clocks that agree with a third agree with each other; and that light's speed measured out and back is a universal constant, which the paper writes V.

Worked example: Setting a distant clock with one light signal

Clock A reads 0 s as a flash leaves it for clock B.
The flash is reflected at B and arrives back at A when A reads 10 s.
By the rule the two trips take equal times, so the reflection happened at 5 s on A's time.
If B's clock read 5 s at the reflection, it is synchronous with A. If it read 7 s, it is 2 s fast, and setting it back 2 s makes the two agree.

t_B = \frac{1}{2}\,\left(t_A + t'_A\right)

The time at B equals one half of the sum of A's reading when the flash left and A's reading when it came back.

The rule needs no distance. If B is known to be 1.5 × 10⁹ metres away, the same round trip also gives light's speed: 2 × 1.5 × 10⁹ m divided by 10 s is 3 × 10⁸ metres per second.

One flash, out and back

A flash leaves clock A, turns round at clock B and comes back. By the rule the two trips take equal times, so the turn-round happened half way between leaving and returning: at 5 on A's clock.

B reads 5 at the turn-round, which is what A's clock says the moment was. The two are synchronous.

Now the same flash, against a clock that is running ahead.

B reads 7 where the rule says 5, so it is 2 seconds fast. Set it back 2 and the two agree. Nothing about the flash changed between the drawings; only what B's hands were showing.

What it shows, in words

Two clocks stand apart, drawn as vertical lines with time running down the page. A flash leaves A when A reads 0, reaches B, is reflected, and returns to A when A reads 10. The rule says the trip out took as long as the trip back, so the turn-round was half way between: 5 by A's clock. In the first drawing B reads 5 at that moment and the clocks are synchronous. In the second B reads 7, which is 2 seconds more, so B is 2 seconds fast and setting it back 2 makes them agree.

The same round trip also measures light, if the distance is known separately: with B at 1.5 × 10⁹ metres, a there-and-back of 10 seconds gives 3 × 10⁸ metres per second.

What it does not show, and this is the lesson's own point rather than a limitation of the drawing: the rule is a definition, and no picture can show it being tested. Checking that the trip out really took as long as the trip back would need two clocks already synchronised, which is the thing the rule exists to supply. Nor is there a second observer anywhere here. Only one frame is drawn, so nothing in it bears on how someone moving past would time the same three events; that question needs the paper's §2, and the geometric aid usually reached for is Minkowski's of 1908, which is later than the argument and not its presentation.

Where this lesson stops

An event is something that happens at one place and one time. Whether two distant events are simultaneous depends on a rule for setting clocks, which §1 of the relativity paper supplies.

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A moving rod, measured at one time

L = \frac{L_0}{\gamma}

Shown in modern letters; Einstein's are in §4 of the German source face.

is divided by .

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Simultaneous moving-rod length is rest length over gamma.

Locating both ends of a moving rod at the same time in the stationary system gives its rest length divided by gamma. At 0.6c a 1 m rod measures 0.8 m.

Model assumptions and every term’s meaning

A clock and a rod at rest in the moving system, which moves at speed v relative to the stationary system, with |v| < c.
The clock's two ticks happen at the same place in its own rest frame; the rod's two ends are located at the same time in the stationary system.
Modern teaching notation: Delta tau is the modern proper time, the time one clock records. It is not the paper's tau, which is the moving system's time coordinate.

Rest length over gamma: Gamma is at least 1, so the measured length is never longer than the rest length. Read the prerequisite
What it asserts: The result of one stated measurement procedure. Locate the two ends at different times and the number is different; that is a different measurement, not a different rod. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared: 1 at rest, larger at any speed, without limit as v approaches c. The paper writes it as beta. Read the prerequisite
Length of the moving rod: The distance between where the two ends are at one time of the stationary system. Read the prerequisite
Rest length: The length a ruler at rest with the rod gives. 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

At everyday speeds, the loss is tiny

1 - \frac{1}{\gamma} \approx \frac{1}{2}\,\frac{v^{2}}{c^{2}}

1 - \frac{1}{\beta} \approx \frac{1}{2}\,\frac{v^{2}}{V^{2}}

The fraction of each second a moving clock falls behind, one minus one over , is about half of squared over squared.

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At small speed, the clock's fractional loss per coordinate second approaches one half v squared over c squared.

The moving clock falls behind by 1 − 1/γ of each stationary-system second. For speeds much smaller than c that fraction is about half of v²/c²: at 300 m/s, about five parts in ten million million.

Model assumptions and every term’s meaning

A clock and a rod at rest in the moving system, which moves at speed v relative to the stationary system, with |v| < c.
Speeds much smaller than the speed of light; the approximation drops terms in the fourth power of v/c.

Half of it: The first term of the expansion of 1 − 1/γ for small v/c. Read the prerequisite
c squared: The square of the speed of light. Read the prerequisite
What it falls behind: One minus its rate: the part of each second the moving clock does not record. Read the prerequisite
The clock's rate: One over gamma: the fraction of a stationary-system second that passes on the moving clock. Read the prerequisite
v squared over c squared: For everyday speeds this is a very small number. Read the prerequisite
Approximately equal: Exact only in the limit of small speed. At 0.6c the left side is 0.2 and the right side 0.18. Read the prerequisite
v squared: The loss does not depend on the direction of motion. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared: 1 at rest, larger at any speed, without limit as v approaches c. The paper writes it as beta. Read the prerequisite
Speed of light: The paper's V. It enters because both systems must find the same speed for light. Read the prerequisite
Speed of the moving system: The speed of k relative to K. Its sign gives the direction. 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.

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A moving clock's ticks, timed from outside

\Delta t = \gamma\,\Delta\tau

Shown in modern letters; Einstein's are in §4 of the German source face.

is times .

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Coordinate time between a moving clock's ticks is gamma times its proper elapsed time.

Between two ticks of a moving clock, the stationary system's clocks record gamma times what the moving clock itself records. At 0.6c a clock shows 0.8 s while 1 s passes in the stationary system.

Model assumptions and every term’s meaning

A clock and a rod at rest in the moving system, which moves at speed v relative to the stationary system, with |v| < c.
The clock's two ticks happen at the same place in its own rest frame; the rod's two ends are located at the same time in the stationary system.
Modern teaching notation: Delta tau is the modern proper time, the time one clock records. It is not the paper's tau, which is the moving system's time coordinate.

What it asserts: A comparison of two measurements of the same pair of ticks. It does not say the clock's mechanism has slowed in its own frame. Read the prerequisite
Gamma times the clock's own time: Gamma is at least 1, so the stationary system's time between the ticks is never shorter than the clock's. Read the prerequisite
Time between the ticks, stationary system: Found from two stationary clocks at the two places the ticks happen, synchronized by the section 1 rule. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared: 1 at rest, larger at any speed, without limit as v approaches c. The paper writes it as beta. Read the prerequisite
Time the moving clock records: What the moving clock itself shows between the two ticks, which happen at one place in its own rest frame. 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 light clock, a flash bouncing between two mirrors, is a later teaching device; the 1905 paper derives these results from the transformation. Einstein prints the moving clock’s lag only to second order, $\frac{1}{2}t\,v^2/V^2$ ; at $0.6c$ that gives 0.18 of each second where the exact value is 0.2.

Assumptions and limits: Use the same measurement protocol

Assumed here

The clock is ideal and the rod rigid, both moving at constant velocity, and each measurement uses the right pair of events.
γ is computed from the relative speed of the two frames.

What this does not establish

Each of two clocks in relative motion runs slow as judged from the other, because the two judgments use different distant clocks.
Transforming one pair of events does not by itself give a length or a clock reading; the pair has to be the right one.

Earlier step: Light constraints leave a scale to determine

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

A qualification

A reunion compares whole paths

Can reciprocal rate descriptions settle a reunion?

The 3 printed paragraphs this passage explains

§4, paragraph 6: If, of two synchronized clocks at A and B, the one at A is carried to B at speed v, on arrival it lags behind the clock at B by about half of t times (v/V) squared seconds, where t is the time the journey took.
§4, paragraph 7: The result still holds when the clock travels from A to B along any path made of straight pieces, even when A and B are the same point.
§4, paragraph 8: Assuming the result also holds for a smooth closed curve, a clock sent round such a curve and back to A lags behind the one that stayed. Einstein concludes that a balance-wheel clock at the equator must run very slightly slower than an identical clock at a pole.

\Delta\tau = \int_{t_{0}}^{t_{1}} \sqrt{1 - \frac{u\left(t\right)^{2}}{c^{2}}}\,\mathrm{d}t

Shown in modern letters; Einstein's are in §4 of the German source face.

An ideal clock’s elapsed proper time is the path integral of the square root of one minus its speed squared over c squared.

Send one clock out at $0.6c$ and back at the same speed, 10 seconds in all by the clocks at rest, and ignore the turnaround. The travelling clock records 8 seconds and the clock that stayed records 10. To get those numbers you have to name each leg and the two meetings.

Each of two clocks in steady relative motion finds the other running slow, and there is no contradiction until they meet again; the reunion is decided by the two paths. Einstein’s §4 remark that a clock at the equator should run slower than one at a pole is a different case: on the real rotating Earth gravity enters, and this flat-spacetime model does not decide it. The moving-clock laboratory keeps the two kinds of comparison apart.

Open the foundation: Adding continuously

Show every step here: A reunion compares whole paths

Letters: the paper's clocks sit at points A and B of the stationary system K, in step with each other; one is moved from A to B at speed v, taking a time t; V is the speed of light. The formula below writes u(t) for a clock's speed at each moment and Δτ for the time that clock itself shows along its path.

The paper's first case: two clocks at A and B, in step in K. Move the clock at A to B at constant speed v. It takes a time t by K's clocks.
While moving, the clock loses $1-\sqrt{1-(v/V)^2}$ of each second, as the previous passage found. So on arrival it is behind the clock that was at B all along by $t\left(1-\sqrt{1-(v/V)^2}\right)$ , which the paper gives to second order as $\frac{1}{2}tv^2/V^2$ seconds.
The paper extends the result to a clock moved along any polygon from A to B, and to the case where A and B coincide, so that the clock comes back to where it started.
Assuming the same holds for a continuously curved path, the paper states the theorem: of two clocks in step at A, one is moved around a closed curve at constant speed and returns after t seconds; it is then behind the one that stayed by $\frac{1}{2}t(v/V)^2$ seconds.
In modern notation the clock's own elapsed time along any path is the sum of its small steps, each slowed by the factor for the speed it has at that moment. That is the formula below.

\Delta\tau = \int_{t_{0}}^{t_{1}} \sqrt{1 - \frac{u\left(t\right)^{2}}{c^{2}}}\,\mathrm{d}t

Shown in modern letters; Einstein's are in §4 of the German source face.

An ideal clock’s elapsed proper time is the path integral of the square root of one minus its speed squared over c squared.

For a leg at constant speed u, the square root is constant and the integral is just the leg's duration in K times $\sqrt{1-u^2/c^2}$ , that is, divided by γ.
Worked: send a clock out at 0.6c and back at 0.6c, 10 seconds in all by K's clocks, and ignore the turnaround. Each leg is slowed by 0.8, so the travelling clock shows 10 × 0.8 = 8 seconds; the clock that stayed shows 10. The paper's second-order formula gives a lag of $\frac{1}{2}\times10\times0.36=1.8$ seconds, against the exact 2.
To get those numbers you have to name the two meetings and each clock's whole path between them. Each of two clocks in steady relative motion finds the other slow; there is no contradiction until they meet again, and then the paths decide.
Acceleration at the turnaround changes the path. In this formula it adds no slowing of its own: only the speed at each moment enters.
The paper concludes that a balance-wheel clock at the earth's equator must run slower, by a very small amount, than an identical clock at a pole under otherwise identical conditions. That remark stays as printed. On the real rotating earth gravity enters, and this model, which has no gravity, does not decide it.
Do not use the delay of light reaching you in place of a clock's elapsed time.

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.

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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

Time on a clock carried along a path

\Delta\tau = \int_{t_{0}}^{t_{1}} \sqrt{1 - \frac{u\left(t\right)^{2}}{c^{2}}}\,\mathrm{d}t

Shown in modern letters; Einstein's are in §4 of the German source face.

is the integral, from to of the trip, of the square root of one minus squared over squared, over .

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An ideal clock's elapsed proper time is the path integral of the square root of one minus its speed squared over c squared.

Add up, moment by moment, the rate at which the carried clock runs: one when it is at rest, less when it moves. Two legs at 0.6c lasting 10 stationary-system seconds give the travelling clock 8 seconds; a clock left at rest records 10.

Model assumptions and every term’s meaning

An ideal clock carried along a path through the stationary system, with speed u(t) below c at every moment.
Flat spacetime, no gravity; the turnaround of a round trip is treated as instantaneous.
Modern teaching notation: Delta tau is the modern proper time, the time the carried clock records, not the paper's tau.

Add it up over the trip: The sum of rate times time over every moment from start to end. Read the prerequisite
c squared: The square of the speed of light. Read the prerequisite
One minus that: Positive at every speed below c. Read the prerequisite
The clock's rate: One over gamma at that moment: 0.8 at 0.6c. Read the prerequisite
u squared over c squared: 0 at rest, 0.36 at 0.6c. Read the prerequisite
What it asserts: The time a carried clock records between two events depends on the path it took between them. This is not a statement about two separated inertial clocks each judging the other slow. Read the prerequisite
Speed squared: The rate does not depend on the direction of travel. Read the prerequisite
End of the trip: The stationary system's time when it ends. Read the prerequisite
Speed of light: The paper's V. Read the prerequisite
Time the carried clock records: The clock's own elapsed time between the start and the end of the trip. Read the prerequisite
The clock's speed: Its speed at each moment, in the stationary system. Read the prerequisite
Start of the trip: The stationary system's time when the trip begins. Read the prerequisite
Time in the stationary system: The stationary system's time, the variable the integral runs over. Read the prerequisite
At time t: u(t) is the clock's speed at stationary-system time t. 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 integral is modern notation; Einstein reasons with a clock carried around a closed curve and states the result to second order. His equator remark stays as he printed it rather than being quietly corrected, and the laboratory declines to make an Earth prediction from special relativity alone.

Assumptions and limits: A reunion compares whole paths

Assumed here

The clocks are ideal, move slower than light along given paths, and gravity is absent.
Both clocks start together and meet again at named events.

What this does not establish

Comparisons on the real Earth also need gravity.
Acceleration changes the path a clock takes; it adds no extra slowing of its own to the formula.

Earlier step: Use the same measurement protocol

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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