§4 · rods, clocks, and the limit of the model · Special relativity: from clock operations to electrodynamics

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

Use the same measurement protocol

What do a moving clock and rod actually report?

\Delta t=\gamma\Delta\tau,\qquad L=\frac{L_0}{\gamma}

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}

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

Return to what a moving clock and rod actually report. (included in this file)

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,\qquad L=\frac{L_0}{\gamma}

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}

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.

Return to what a moving clock and rod actually report. (included in this file)

A moving rod, measured at one time

Modern teaching equation, not the equation as printed.

L = \frac{L_0}{\gamma}

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.

At everyday speeds, the loss is tiny

Modern teaching equation, not the equation as printed.

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

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.

A moving clock's ticks, timed from outside

Modern teaching equation, not the equation as printed.

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

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.

Assumptions

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.

Limits

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.

A reunion compares whole paths

Can reciprocal rate descriptions settle a reunion?

\Delta\tau=\int_{t_0}^{t_1}\sqrt{1-\frac{u(t)^2}{c^2}}\,dt

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.

Return to whether each clock seeing the other run slow can settle a reunion. (included in this file)

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(t)^2}{c^2}}\,dt

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.

Return to whether each clock seeing the other run slow can settle a reunion. (included in this file)

Time on a clock carried along a path

Modern teaching equation, not the equation as printed.

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

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.

Assumptions

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.

Limits

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.

Foundations included in this file

Reading a graph

How does a curve show the relationship between two physical quantities?

Walk for a minute and note how far you have gone every ten seconds: 0 m at the start, 14 m after ten seconds, 28 m after twenty, and so on. The first drawing below puts time along the bottom and distance up the side, with a dot for each pair. That is a graph: each dot is one moment, read across for the time and up for the distance.

The dots of a steady walk fall on a straight line, and the steeper the line, the faster the walk. This one climbs 14 metres for every 10 seconds: 1.4 metres per second. A flat line would mean standing still, and a line that bends upward would mean speeding up.

The papers draw other pairs the same way. Along the bottom goes what you choose or wait for, such as a position or a time; up the side goes what you then find, such as how many particles sit there. A bell-shaped curve, high in the middle and falling away on both sides, says that most particles are near the centre and fewer are farther out.

Worked example: Four particles that average to zero

Four particles end at −3, −1, +1 and +3 micrometres from where they started.
The second drawing puts those positions along the bottom, with a bar one particle tall above each.
The bars stand two on each side of zero, at the same distances out. The picture is symmetric, so the average position is 0.
Yet no bar stands at zero: every particle moved. The graph shows at a glance what the average hides.

Where this lesson stops

Before reading a curve, read its two labels and their units, and ask whether its height is a count, a density or a running total.

This lesson builds on

A sign records direction

Return to the chapter

A sign records direction

How can movement add up to zero?

On the ruler below, the starting point is zero. A final mark three units to the right has displacement +3; one three units to the left has displacement −3. Adding the signed displacements gives zero. Adding the distances from the start gives six.

Displacement compares where something ends with where it started, whatever it did in between. A walker who goes three units right and comes back has displacement zero after walking six, which is the lower pair of arrows in the drawing.

Worked example: Two walkers, opposite directions, average zero

Two walkers both finish one unit from their start, one at −1 and the other at +1. The signed average is zero. The average distance is one.

Where this lesson stops

The sign names a direction relative to a chosen axis; it does not mean a negative distance.

This lesson builds on

Adding and averaging

Return to the chapter

Adding and averaging

What does an average keep, and what does it lose?

Four jars hold 3, 1, 1 and 3 marbles. Pour them together and there are 8. Share the 8 equally between the four jars and each holds 2. Two is the average, although no jar held 2 to begin with.

The average keeps the total and the count and nothing else, so 3, 1, 1, 3 and 2, 2, 2, 2 have the same average. Listing every value twice doubles both the total and the count, and the average stays at 2.

Worked example: Three mornings with the same average

One morning the four jars hold 3, 1, 1 and 3 marbles. Add them: 8. Divide by the four jars: an average of 2.
Another morning they hold 8, 0, 0 and 0. Add them: 8 again. Divide by four: 2 again, and now three jars are empty.
A third morning every jar holds 2, which is the average itself. Add them: 8. Divide by four: 2.
Write any of those three lists out twice over. The total doubles to 16 and the count doubles to 8 jars, and 16 divided by 8 is still 2.
Four descriptions and one average. It keeps the total and the count, and the differences between the mornings are what it leaves behind.

Where this lesson stops

An average is the total shared equally among the observations, and two very different lists can have the same one.

Return to the chapter

Density is not probability

What does the height of a probability curve mean?

Mark where 100 particles end up on a ruler divided into 2 μm bins. If 30 land in one bin, that bin holds a probability of about 0.3, and the density there is 0.3 divided by 2 μm, or 0.15 per micrometre. Measured in metres the same density is 150,000 per metre, while the bin still holds 0.3: the height depends on the unit of length, and the probability does not.

P(a\le X\le b)=\int_a^b p(x)\,dx

The probability of the interval from a to b is the integral of the density over that interval.

Under a smooth curve, any single exact position has probability zero, since an interval of zero width has no area. At the starting instant of the Brownian problem every particle sits at one point. No curve can draw that: the whole probability, 1, sits on the single point, and a curve appears only once the particles have begun to spread.

Worked example: A flat density across four micrometres

Spread the probability evenly over 4 μm. The density is 1 divided by 4 μm, or 0.25 per micrometre.
The area over the whole 4 μm is 0.25 per μm × 4 μm = 1.
A 1 μm piece has probability 0.25 × 1 = 0.25, and a 2 μm piece has 0.5.
Squeeze the same probability into 2 μm and the height doubles to 0.5 per micrometre, while the total stays 1. The height was never a probability.

Where this lesson stops

The height of the curve is a density, a probability per unit length. Only the area over an interval is a probability.

This lesson builds on

Adding and averaging

Return to the chapter

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.

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.

This lesson builds on

Functions and graphs

Return to the chapter

Functions and graphs

How does a graph record the way one quantity depends on another?

A bath fills from a tap. After one minute it holds 10 litres, after two 20, after three 30. For each time there is exactly one amount of water, and that pairing, one output for each input, is a function. Plot the pairs with time along the bottom and litres up the side and they fall on a straight line that climbs 10 litres for every minute. The slope, in litres per minute, is how fast the bath fills.

In the papers the pairs are physical: the mean square displacement of a Brownian particle for each elapsed time, or the concentration for each position along a tube. A function says how one thing responds when another changes.

A graph puts the input along the horizontal axis and the output up the vertical one, so each point is one pair. Both axes carry units, and so does everything read off the graph: the slope of mean square displacement against time is in square micrometres per second, and the area under a probability density is a plain probability.

A formula such as ⟨x²⟩ = 2Dt describes an average over many particles. One particle's squared displacement scatters widely around the line; the average over many lies close to it. The Brownian paper predicts the average, and ends by hoping that a researcher will soon decide the question by observation.

Worked example: Mean square displacement plotted against time

\langle x^2\rangle = 2Dt

The mean square displacement equals two times the diffusion coefficient times the time.

Take D = 0.5 square micrometres per second, so 2D = 1 square micrometre per second.
At t = 1 s the mean square displacement is 1 µm²; at t = 4 s it is 4 µm²; at t = 9 s, 9 µm².
Plotted against time, these points lie on a straight line through the origin whose slope is 2D, 1 µm² per second.
Their square roots, the root-mean-square displacements, are 1, 2 and 3 µm: four times the time gives twice the distance.

Where this lesson stops

A graph is a record of paired quantities, each with its units. A slope read off it has units too, and a slope stated without them explains nothing.

This lesson builds on

Reading a graph

Return to the chapter

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.

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.

This lesson builds on

Density is not probability

Return to the chapter

Static laboratory examples

sr-03: a built-in worked result

Static host-calculated example, not a running experiment or an observation. Canonical units and owner-supplied statuses are retained. Open sr-03 online

Build-time scalar readouts
Quantity	Value	Unit	Status
spatialSeparationK	10	ls	value
temporalSeparationK	0	s	value
spatialSeparationKPrime	12.5	ls	value
temporalSeparationKPrime	-7.5	s	value
simultaneityK	0	1	value
simultaneityKPrime	-1	1	value
measuredLength	10	ls	value
rodLengthK	8	ls	value
rodLengthKPrime	10	ls	value
readingsOnRodEnds	0	1	value
spacetimeIntervalSquared	100	ls^2	value
causalOrder	1	1	value
gammaFactor	1.25	1	value
ellipsoidAxisLongitudinal	0.8	ls	value
ellipsoidAxisTransverseY	1	ls	value
ellipsoidAxisTransverseZ	1	ls	value

0 array outputs are not reproduced. This table does not claim to reproduce the interactive plot.

Settings (canonical units)

L0: 10
R: 1
endpointPairId: platform-simultaneous
measuringFrame: K
rodRestFrame: k
v: 0.6

Evaluator: source:sha256:5aca8eaa733b21262f272915553a148b6395ceb2c24be06a26d30c2487a9acd4

sr-05: a built-in worked result

Static host-calculated example, not a running experiment or an observation. Canonical units and owner-supplied statuses are retained. Open sr-05 online

Build-time scalar readouts
Quantity	Value	Unit	Status
speed	0.6	1	value
coordinateDuration	10	s	value
lightClockArm	1	ls	value
frameOfDescription	0	1	value
properTime	8	s	value
coordinateTime	10	s	value
dilationLossExact	0.19999999999999998	1	value
dilationLossPrintedSecondOrder	0.18	1	value
dilationLossDifference	0.01999999999999999	1	value
reunionExactLag	2	s	value
reunionPrintedApproxLag	1.7999999999999998	s	value
reciprocalDilationFactor	1.25	1	value
lightClockProperTick	2	s	value
lightClockCoordinateTick	2.5	s	value
equatorFractionalRate	1.2034312297602602e-12	1	value
equatorApproxNanosecondsPerDay	103.97645825128649	ns	value
dailyLossSpeedBeta	0.0048112383217792775	1	value

0 array outputs are not reproduced. This table does not claim to reproduce the interactive plot.

Settings (canonical units)

coordinateDuration: 10
equatorMode: false
frameOfDescription: 0
lightClockArm: 1
showPrintedSecondOrder: true
speed: 0.6
worldlinePreset: out-and-back

Evaluator: source:sha256:a99370713c537ee2faa0cb196020e2ecce73ba4a0a70832e19ee889d055fe87f

Source references

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