§1 · operational simultaneity · Special relativity: from clock operations to electrodynamics

§1 · operational simultaneity

Time at a distant clock is an operation

How can clocks at different places share a time coordinate?

t_B-t_A=t_{A,\mathrm{return}}-t_B

The outward assigned travel time equals the return assigned travel time.

t_B=\frac{t_A+t_{A,\mathrm{return}}}{2}

The assigned reflection time is the midpoint of departure and return times.

If clock A sends a flash at time 0 and the reflection comes back at time 10, the rule says the flash reached B at time 5. Seeing B later, when its light arrives, is a separate event and does not change that assignment.

The rule assumes light takes as long to go as to come back, and Einstein states that as a definition rather than a measurement. It sets clocks within one frame; §2 shows that clocks in another frame, set by the same rule, disagree about which distant events happen at the same time. The clock-synchronization laboratory keeps departure, reflection, return and each clock reading as separate records.

Return to how clocks in different places can share a time. (included in this file)

Letters: a light ray leaves A at the A-time $t_A$ , is reflected at B at the B-time $t_B$ , and arrives back at A at the A-time $t'_A$ . This explanation writes that return reading $t_{A,\mathrm{return}}$ . V is the speed of light, c here.

Start from one coordinate system in which Newton's equations of mechanics hold; the paper calls it the stationary system. A point at rest in it has a position measured with rigid rods by the methods of Euclidean geometry.
Describing a motion means giving the coordinates as functions of time, and that has a physical meaning only once it is said what time means. For an event at the place of a clock, the time is the position of the clock's hand at that event.
A clock at A times the events at A, and a clock at B times the events at B. So far there is an A-time and a B-time, but no time common to A and B. Comparing an event at A with an event at B needs a further rule.
The rule, set by definition: the time light takes to go from A to B equals the time it takes to come back from B to A. With the three readings above, that is the first formula below.

t_B-t_A=t_{A,\mathrm{return}}-t_B

The outward assigned travel time equals the return assigned travel time.

Solve for $t_B$ . Add $t_B$ to both sides and $t_A$ to both sides: $2t_B=t_A+t_{A,\mathrm{return}}$ . Divide by 2.

t_B=\frac{t_A+t_{A,\mathrm{return}}}{2}

The assigned reflection time is the midpoint of departure and return times.

Worked case: A sends the flash at time 0 and the reflection comes back at time 10, so the reflection is given the time 5. If B's clock read 5.3 at the reflection, it is 0.3 ahead and is set back.
The paper assumes that this definition can be applied without contradiction to any number of points: if the clock at B runs in step with the clock at A, then A's runs in step with B's; and if A's runs in step with both B's and C's, then B's and C's run in step with each other.
The speed of light comes out of the same readings: twice the distance AB between the stations, divided by $t'_A-t_A$ , is V, a universal constant, in agreement with experience. With A and B $3\times10^{8}$ metres apart and a round trip of 2 seconds, V is $3\times10^{8}$ metres per second.
The time of an event is then the reading, at the event, of a clock at rest at its place and in step with the chosen clock. Since it is set with clocks at rest in one system, the paper calls it the time of the stationary system.
Keep two things apart: the time this rule gives to a distant event, and the moment its light reaches you. Seeing the reflection from A at time 10 does not move it from time 5.
The status: equal travel times out and back are a definition, not a measurement, and a round trip read on one clock cannot measure the two one-way times separately. §2 shows that clocks set this way in a moving system disagree with the stationary system about which distant events are simultaneous.

Return to how clocks in different places can share a time. (included in this file)

The rule: out and back take equal times

Modern teaching equation, not the equation as printed.

t_B - t_A = t_{A,\mathrm{return}} - t_B

The outward assigned travel time equals the return assigned travel time.

This is a definition of what B's clock should read, not a law found by measurement: it sets the time light takes from A to B equal to the time it takes from B back to A.

So B reads the midpoint

Modern teaching equation, not the equation as printed.

t_B = \frac{t_A + t_{A,\mathrm{return}}}{2}

The assigned reflection time is the midpoint of departure and return times.

Solving the equal-times rule for B's reading: B's clock is set to read halfway between the moments the signal leaves A and returns to A.

Assumptions

The clocks and the endpoints are at rest in one inertial frame.
The rule assumes light takes as long to go as to come back.

Limits

A round trip read on one clock cannot measure the two one-way times separately.
Setting clocks does not move them or change their rates.

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

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

Static laboratory examples

sr-01

This laboratory is not included in this file. The chapter's authored foundation examples remain available above. Open sr-01 online

Source references

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

Special relativity: from clock operations to electrodynamics: §1 · operational simultaneity