Why Relativity of Simultaneity Matters When Comparing Clocks

Two identical clocks can tick at exactly the same rate and still disagree about which of their distant readings count as happening at the same time. This is why relativity of simultaneity matters when comparing distant clocks: a clock comparison requires a rule for synchronising clocks at different positions, and that rule depends on the reference frame.

Imagine two clocks fixed 600 m apart on a platform. Both are synchronised and display zero at the same instant according to someone standing on the platform. A spacecraft passes them, travelling to the right at \(0.60c\).

Predict what someone travelling in the spacecraft will conclude. Will they agree that both clocks displayed zero at the same time?

No. In the spacecraft’s reference frame, the event where the right-hand clock displays zero occurs earlier than the event where the left-hand clock displays zero.

Neither clock is faulty. Neither has suddenly changed its ticking rate. The disagreement comes from the fact that observers moving relative to one another do not agree on which events at different locations are simultaneous.

This has a practical consequence. If you compare distant clocks without considering relativity of simultaneity, you can mistake a difference in synchronisation for a difference in how quickly the clocks tick.

To understand why, we first need to look at how distant clocks are synchronised in the first place.

01The problem: how do you synchronise two distant clocks?

A single clock can measure the time between two events that occur at its location. For example, a stopwatch can measure the time between a runner starting and finishing a race if the stopwatch remains at the finish line only when the finish event occurs? Not quite: it must be present at both events to measure their proper-time interval.

With two distant clocks, the problem is different.

Suppose clock A is at one end of a laboratory and clock B is 900 m away. You want both clocks to display the same time whenever an observer in the laboratory says the events are simultaneous.

How do you arrange that?

You cannot simply stand halfway between them and set both clocks by hand. You need a method that works using signals travelling at a finite speed.

Einstein’s method of synchronising distant clocks

In 1905, Einstein’s paper On the Electrodynamics of Moving Bodies established an operational approach to defining simultaneity using light signals.

Consider two clocks:

• Clock A is at position \(x=0\).
• Clock B is at position \(x=L\).
• Both clocks are stationary relative to the laboratory.
• A light pulse travels from A to B, reflects immediately, and returns to A.

The sequence is:

1. Clock A records the time \(t_1\) when it sends the pulse.
2. The pulse reaches B and is reflected at time \(t_2\).
3. The pulse returns to A at time \(t_3\).

Clock A measures both the departure and return times. The synchronisation procedure assigns the reflection event at B the time halfway between those two readings.

Therefore,

\[ t_2=\frac{t_1+t_3}{2} \]

Clock B is set to display \(t_2\) at the reflection event.

Why use the midpoint?

In an inertial reference frame, special relativity takes the speed of light in a vacuum to be \(c\) in both directions. The outgoing and returning journeys therefore take equal amounts of time in this frame.

The reflection event must occur halfway through the total round-trip time.

A tempting shortcut would be to set B’s clock to A’s reading when the outgoing pulse arrives. That would ignore the time the light takes to travel between the clocks. B would not be synchronised with A.

Einstein’s procedure avoids needing B’s clock to be synchronised before the measurement begins. It uses the round-trip time measured by A and an equal-travel-time rule to assign B’s time.

There is an important limitation: the round-trip measurement alone does not independently measure the one-way speed of light. Assigning equal travel times to the outgoing and returning signals is part of the standard Einstein synchronisation convention for an inertial frame.

The result is a consistent network of synchronised clocks in the laboratory’s reference frame.

But does another observer, moving relative to the laboratory, agree that these clocks are synchronised?

Not necessarily.

02Why a moving observer disagrees

Let’s return to clocks A and B.

They are stationary in the laboratory reference frame, which we will call \(S\). They are synchronised according to Einstein’s procedure.

A spacecraft travels to the right at constant speed \(v\) relative to the laboratory. Its reference frame is \(S'\).

We will use the following conventions:

• Positive \(x\) points to the right.
• The spacecraft moves in the positive \(x\)-direction.
• Clock A is at \(x=0\).
• Clock B is at \(x=L\).
• \(t\) is the time coordinate in the laboratory frame.
• \(t'\) is the time coordinate in the spacecraft frame.

The spacecraft’s frame is moving to the right relative to the laboratory. From the spacecraft’s perspective, the laboratory clocks are moving to the left.

Both frames are inertial, meaning they move at constant velocity relative to one another.

The Lorentz transformation explains the disagreement

The time-coordinate transformation between the two frames is

\[ t'=\gamma\left(t-\frac{vx}{c^2}\right) \]

where:

• \(t'\) is the time coordinate in the spacecraft frame, in seconds.
• \(t\) is the time coordinate in the laboratory frame, in seconds.
• \(v\) is the velocity of the spacecraft relative to the laboratory, in \(\mathrm{m\,s^{-1}}\).
• \(x\) is the position of the event in the laboratory frame, in metres.
• \(c\) is the speed of light, approximately \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).
• \(\gamma\) is the Lorentz factor.

The Lorentz factor is

\[ \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \]

For two events, we can subtract their time coordinates to obtain

\[ \Delta t'= \gamma\left( \Delta t-\frac{v\Delta x}{c^2} \right) \]

Here, we define:

• \(\Delta t=t_B-t_A\).
• \(\Delta x=x_B-x_A\).
• \(\Delta t'=t'_B-t'_A\).

The signs matter. If B is to the right of A, then \(\Delta x\) is positive.

Now consider the two events where clocks A and B each display zero.

In the laboratory frame, these events are simultaneous because the clocks are synchronised.

Therefore,

\[ \Delta t=0 \]

However, the clocks are separated by distance \(L\), so

\[ \Delta x=L \]

Substituting into the Lorentz transformation gives

\[ \begin{aligned} \Delta t' &=\gamma\left(0-\frac{vL}{c^2}\right)\\ &=-\frac{\gamma vL}{c^2} \end{aligned} \]

This is the mathematical expression of relativity of simultaneity for these two events.

Because \(v\), \(L\), and \(\gamma\) are positive, \(\Delta t'\) is negative.

That means the event at B occurs earlier than the event at A according to the spacecraft frame.

The two events are simultaneous in \(S\), but not in \(S'\).

What does the spacetime diagram tell us?

The horizontal line represents one instant in the laboratory frame. Both clock events lie on that line, so they are simultaneous in \(S\).

The tilted line represents one instant in the spacecraft frame. To be simultaneous in \(S'\), the event at B must occur at a later laboratory time than the event at A.

This is the key idea: each inertial frame has its own definition of simultaneity for events at different positions.

The disagreement is not caused by light taking different amounts of time to reach the observers. It remains even after the light-travel times have been accounted for.

It is also not a physical malfunction. Both frames are describing the same two events using different time coordinates.

04Proper time is not the same as a comparison of distant clocks

There is another concept you need to keep separate: proper time.

Proper time, written as \(\Delta\tau\), is the elapsed time measured by a single clock that is present at both events.

In the clock’s rest frame, those two events occur at the same position.

For example, imagine clock A records two ticks:

• Event 1: clock A displays 5.0 s.
• Event 2: clock A displays 8.0 s.

The proper time between those events is

\[ \Delta\tau=8.0-5.0=3.0\ \mathrm{s} \]

The same clock records both events at the same location.

Now consider a different pair of events:

• Event A: clock A displays 5.0 s.
• Event B: clock B, 600 m away, displays 5.0 s.

Even if these events are simultaneous in the laboratory frame, they do not occur at the same location.

No single clock is present at both events.

Therefore, the interval between these two events is not a proper-time measurement.

This gives us an important distinction:

Each quantity answers a different question.

Where time dilation fits in

The time-dilation equation is

\[ \Delta t=\gamma\Delta\tau \]

Here:

• \(\Delta\tau\) is the proper time measured by the moving clock.
• \(\Delta t\) is the time-coordinate interval between the same two events in another inertial frame.
• \(\gamma\) is the Lorentz factor.

The phrase the same two events is essential.

Suppose a spacecraft clock records two ticks. You can compare the time between those ticks with the time assigned to those same events by the laboratory.

You cannot take one event on clock A and another event on clock B, call the interval proper time, and substitute it into the time-dilation equation.

The events must first be identified.

For a more systematic method of selecting the correct event pair, see How to Identify the Two Events Before Choosing Proper Time.

05Do not confuse relativity of simultaneity with light-travel delays

So far, we have compared the times assigned to events by different reference frames.

There is another reason an observer might receive information from two distant clocks at different times: light needs time to travel.

These are two separate effects.

Imagine two stations, A and B, separated by 900 m. Both emit a light pulse at exactly the same time according to their stationary reference frame.

A receiver, R, is positioned 300 m from A and 600 m from B.

Which pulse will the receiver detect first?

The pulse from A.

But does that mean A emitted first?

No. The receiver is closer to A, so the light from A has a shorter distance to travel.

For a stationary receiver in a chosen reference frame, the relationship is

\[ t_{\text{arrival}}= t_{\text{emission}}+\frac{d}{c} \]

where:

• \(t_{\text{arrival}}\) is the time the signal reaches the receiver.
• \(t_{\text{emission}}\) is the time the signal was emitted.
• \(d\) is the signal’s travel distance in that frame.
• \(c\) is the speed of light.

To find when the event occurred, rearrange:

\[ t_{\text{emission}}= t_{\text{arrival}}-\frac{d}{c} \]

06Why the distinction matters in physics

It is tempting to think that if two clocks are identical and tick at the same rate, they must remain synchronised for everyone.

That assumption is incorrect.

Two clocks can:

• Have identical mechanisms.
• Tick at the same rate in their shared rest frame.
• Be synchronised using Einstein’s procedure in that frame.
• Still be judged unsynchronised by an observer moving relative to them.

The important distinction is between a clock’s ticking rate and the time assigned to events at different locations.

Time dilation concerns the elapsed time between two events on a clock’s worldline.

Relativity of simultaneity concerns whether two spatially separated events are assigned the same time in different reference frames.

Light-travel delay concerns when an observer receives information about an event.

These ideas are related, but they are not interchangeable.

A reliable method for HSC questions

When a question involves distant clocks, use this sequence.

1. Identify the events.

Write down exactly what happens at each location. For example, clock A displays zero and clock B displays zero.

2. Identify their positions.

Choose a positive direction and calculate \(\Delta x\).

3. Identify the reference frame.

Which frame has the clocks synchronised? Which frame is moving?

4. Decide whether you are comparing event times or clock readings.

A time-coordinate difference between two events is not automatically the difference between clock displays at one instant.

5. Check for light-travel delays.

If the question gives reception times, account for the distance travelled by the signals before deciding when the events occurred.

6. Check whether proper time is involved.

Only use the proper-time version of the time-dilation equation when you have identified two events that can be recorded by the same clock.

Getting these decisions right is often more important than the arithmetic.

07Questions and solutions

Question 1

Two stationary clocks, A and B, are separated by 500 m in reference frame \(S\). Clock A is at \(x=0\), and clock B is at \(x=+500\ \mathrm{m}\).

Both clocks are synchronised and display zero at the same instant in \(S\).

A spacecraft moves in the negative \(x\)-direction at \(0.60c\) relative to \(S\).

Calculate the time-coordinate difference between the two zero-reading events in the spacecraft frame \(S'\). State which event occurs first in \(S'\).

Solution 1

The event at A occurs 1.25 microseconds earlier than the event at B in the spacecraft frame.

The key is to use the correct sign for the spacecraft’s velocity.

Step 1: Identify the event separation.

Because both clocks display zero at the same instant in \(S\),

\[ \Delta t=0 \]

The position difference is

\[ \Delta x=+500\ \mathrm{m} \]

The spacecraft moves in the negative direction, so

\[ v=-0.60c \]

Step 2: Calculate the Lorentz factor.

\[ \gamma=\frac{1}{\sqrt{1-0.60^2}} =1.25 \]

Step 3: Apply the Lorentz transformation.

\[ \begin{aligned} \Delta t' &=\gamma\left( \Delta t-\frac{v\Delta x}{c^2} \right)\\ &=1.25\left( 0-\frac{(-0.60c)(500)}{c^2} \right)\\ &=1.25\left( \frac{300}{3.00\times10^8} \right)\\ &=1.25\times10^{-6}\ \mathrm{s} \end{aligned} \]

Therefore,

\[ \boxed{\Delta t'=+1.25\ \mu\mathrm{s}} \]

Since \(\Delta t'=t'_B-t'_A\) is positive, the event at A occurs first.

Notice what changed compared with a spacecraft travelling to the right. Reversing the direction of relative motion reverses which distant clock event is assigned the earlier time.

The sign is not something to memorise separately. It follows directly from the Lorentz transformation.

Question 2

Two independent detectors are stationary in reference frame \(S\).

Detector A is at \(x=0\) and records an event at \(t_A=0\).

Detector B is at \(x=1200\ \mathrm{m}\) and records an unrelated event at \(t_B=+2.0\ \mu\mathrm{s}\).

A spacecraft frame \(S'\) moves in the positive \(x\)-direction at \(0.80c\) relative to \(S\).

(a) Calculate the time-coordinate difference between the two events in \(S'\).

(b) Which event occurs first in \(S'\)?

(c) Could either event have caused the other? Justify your answer using the speed of light.

Solution 2

The event at B occurs 2.0 microseconds earlier in \(S'\), but neither event can have caused the other.

This is an important result: relativity of simultaneity can reverse the assigned order of sufficiently separated events without violating causality.

Step 1: Identify the event differences in frame \(S\).

The time difference is

\[ \Delta t=2.0\times10^{-6}\ \mathrm{s} \]

The position difference is

\[ \Delta x=+1200\ \mathrm{m} \]

The spacecraft velocity is

\[ v=+0.80c \]

Step 2: Calculate the Lorentz factor.

\[ \begin{aligned} \gamma &=\frac{1}{\sqrt{1-0.80^2}}\\ &=\frac{1}{\sqrt{0.36}}\\ &=\frac{5}{3} \end{aligned} \]

Step 3: Calculate the time-coordinate difference in \(S'\).

\[ \Delta t'= \gamma\left( \Delta t-\frac{v\Delta x}{c^2} \right) \]

First calculate the second term:

\[ \begin{aligned} \frac{v\Delta x}{c^2} &=\frac{0.80c(1200)}{c^2}\\ &=3.20\times10^{-6}\ \mathrm{s} \end{aligned} \]

Now substitute:

\[ \begin{aligned} \Delta t' &=\frac{5}{3}(2.0-3.2)\ \mu\mathrm{s}\\ &=\frac{5}{3}(-1.2)\ \mu\mathrm{s}\\ &=-2.0\ \mu\mathrm{s} \end{aligned} \]

Therefore,

\[ \boxed{\Delta t'=-2.0\ \mu\mathrm{s}} \]

The negative result means B’s event occurs first in \(S'\), even though A’s event occurs first in \(S\).

Step 4: Determine whether either event could cause the other.

For an event to cause another event, a signal or physical influence must be able to travel between their locations.

The distance between the detectors is 1200 m.

The minimum time needed for light to travel that distance is

\[ \begin{aligned} t_{\text{light}} &=\frac{1200}{3.00\times10^8}\\ &=4.0\times10^{-6}\ \mathrm{s}\\ &=4.0\ \mu\mathrm{s} \end{aligned} \]

However, the events are separated by only 2.0 microseconds in \(S\).

Even light could not travel from A to B within that time.

Therefore, neither event can causally influence the other.

Step 5: Explain why the reversal is allowed.

These events are called spacelike-separated because their spatial separation is too large for a light-speed signal to travel between them during their time separation.

For spacelike-separated events, different inertial frames can assign different time orders.

This does not create a causality problem because neither event can be the cause of the other.

A frame-dependent order is not the same as reversing cause and effect.

Question 3

Two identical clocks, A and B, are stationary in reference frame \(S\), separated by 1.2 km.

They are synchronised and both display zero at \(t=0\).

Clock B records two successive ticks that are separated by 6.0 microseconds.

A spacecraft frame \(S'\) moves in the positive \(x\)-direction at \(0.75c\) relative to \(S\).

(a) Determine the proper time between the two ticks recorded by clock B.

(b) Calculate the time-coordinate interval between those same two tick events in \(S'\).

(c) Calculate the time-coordinate difference in \(S'\) between the two events where clocks A and B each display zero.

(d) Explain why the answers to parts (b) and (c) describe different quantities.

Solution 3

The proper time between B’s ticks is 6.0 microseconds. Their time-coordinate separation in \(S'\) is approximately 9.07 microseconds, while the two distant zero-reading events have a time-coordinate separation of approximately \(-4.54\) microseconds.

These results are different because they refer to different pairs of events.

Step 1: Identify the proper time.

Both tick events occur at clock B.

The same clock records both events, so the interval measured by B is the proper time.

Therefore,

\[ \boxed{\Delta\tau=6.0\ \mu\mathrm{s}} \]

Step 2: Calculate the Lorentz factor.

The spacecraft moves at \(0.75c\), so

\[ \begin{aligned} \gamma &=\frac{1}{\sqrt{1-0.75^2}}\\ &=\frac{1}{\sqrt{0.4375}}\\ &\approx1.512 \end{aligned} \]

Step 3: Calculate the coordinate time between B’s ticks in \(S'\).

The clock moves relative to \(S'\).

Using time dilation,

\[ \Delta t'=\gamma\Delta\tau \]

Substitute:

\[ \begin{aligned} \Delta t' &=1.512(6.0)\\ &=9.072\ \mu\mathrm{s} \end{aligned} \]

Therefore,

\[ \boxed{\Delta t'\approx9.07\ \mu\mathrm{s}} \]

The moving clock records less elapsed time than the coordinate time assigned to the same two tick events in \(S'\).

Step 4: Calculate the time difference between the two zero-reading events.

Clock A is at \(x=0\), while clock B is at

\[ x=1200\ \mathrm{m} \]

The two zero-reading events are simultaneous in \(S\), so

\[ \Delta t=0 \]

Using the Lorentz transformation:

\[ \begin{aligned} \Delta t' &=\gamma\left( 0-\frac{v\Delta x}{c^2} \right)\\ &=-1.512\left( \frac{0.75c(1200)}{c^2} \right)\\ &=-1.512(3.00\ \mu\mathrm{s})\\ &=-4.536\ \mu\mathrm{s} \end{aligned} \]

Therefore,

\[ \boxed{\Delta t'\approx-4.54\ \mu\mathrm{s}} \]

The zero-reading event at B occurs earlier than the zero-reading event at A in \(S'\).

Step 5: Explain the difference between the two results.

Part (b) compares two events on the same clock, B. They occur at the same spatial position in B’s rest frame, so B can measure their proper-time interval.

Part (c) compares two events at different positions: one at A and one at B.

No single clock records both of those events.

The \(-4.54\ \mu\mathrm{s}\) result is therefore a difference in assigned event times, not a proper-time measurement.

The trap is to assume that every time difference involving two clocks must be a time-dilation interval. The event locations determine whether that is true.

Question 4

Two clocks, A and B, are stationary in reference frame \(S\), separated by 1200 m.

A light pulse leaves A when its clock reads \(2.0\ \mu\mathrm{s}\). The pulse travels to B, reflects immediately, and returns to A when its clock reads \(10.0\ \mu\mathrm{s}\).

Assume the pulse travels through a vacuum at \(c\).

(a) Using Einstein’s synchronisation procedure, determine the time that should be assigned to the reflection event at B.

(b) A second observer uses reference frame \(S'\), which moves in the positive \(x\)-direction at \(0.60c\) relative to \(S\). Calculate the time-coordinate interval between the emission event at A and the reflection event at B in \(S'\).

(c) Explain why the interval in part (b) is not a proper-time measurement.

Solution 4

The reflection event at B is assigned a time of \(6.0\ \mu\mathrm{s}\) in \(S\). The time-coordinate interval between emission and reflection is \(2.0\ \mu\mathrm{s}\) in \(S'\), and it is not a proper-time measurement.

Step 1: Apply Einstein’s synchronisation procedure.

The pulse leaves A at

\[ t_1=2.0\ \mu\mathrm{s} \]

It returns to A at

\[ t_3=10.0\ \mu\mathrm{s} \]

The reflection event at B is assigned the midpoint time:

\[ t_2=\frac{t_1+t_3}{2} \]

Substitute:

\[ \begin{aligned} t_2 &=\frac{2.0+10.0}{2}\\ &=6.0\ \mu\mathrm{s} \end{aligned} \]

Therefore,

\[ \boxed{t_2=6.0\ \mu\mathrm{s}} \]

Clock B should display \(6.0\ \mu\mathrm{s}\) at the reflection event.

Step 2: Identify the event separation in \(S\).

The emission event occurs at A, while the reflection event occurs at B.

Their time-coordinate difference is

\[ \Delta t=6.0-2.0 =4.0\ \mu\mathrm{s} \]

Their position difference is

\[ \Delta x=+1200\ \mathrm{m} \]

Step 3: Calculate the Lorentz factor.

The relative speed is \(0.60c\).

\[ \gamma=\frac{1}{\sqrt{1-0.60^2}} =1.25 \]

Step 4: Transform the time-coordinate difference.

\[ \Delta t'= \gamma\left( \Delta t-\frac{v\Delta x}{c^2} \right) \]

Substitute:

\[ \begin{aligned} \Delta t' &=1.25\left( 4.0-\frac{0.60c(1200)}{c^2} \right)\ \mu\mathrm{s}\\ &=1.25(4.0-2.4)\ \mu\mathrm{s}\\ &=2.0\ \mu\mathrm{s} \end{aligned} \]

Therefore,

\[ \boxed{\Delta t'=2.0\ \mu\mathrm{s}} \]

The reflection event occurs 2.0 microseconds after the emission event in \(S'\).

Step 5: Explain why this is not proper time.

The emission occurs at A, while the reflection occurs at B.

They are separated by 1200 m in the laboratory frame. No single clock is present at both events.

The \(2.0\ \mu\mathrm{s}\) interval is a coordinate-time difference in \(S'\), not a proper-time measurement.

This is also a useful check on the speed of light. In \(S'\), the separation between the emission and reflection events is 600 m, and the time-coordinate difference is 2.0 microseconds.

Therefore,

\[ \frac{600}{2.0\times10^{-6}} =3.00\times10^8\ \mathrm{m\,s^{-1}} \]

The light still travels at \(c\).

The two observers disagree about the time and distance between the events, but both calculate the same speed of light.

Question 5

Two stationary clocks, A and B, are separated by 900 m in reference frame \(S\).

Clock A is at \(x=0\), and clock B is at \(x=+900\ \mathrm{m}\).

The clocks have been synchronised using Einstein’s procedure. Both display zero at \(t=0\).

A probe travels in the positive \(x\)-direction at a constant speed of \(0.80c\) relative to \(S\).

The probe passes clock A at event P, exactly when A displays zero. Its onboard clock is also set to zero at P.

The probe continues towards B and meets clock B at event Q.

Assume constant relative velocities, ideal clocks, and negligible gravitational effects.

(a) Calculate the time between P and Q in the laboratory frame \(S\).

(b) Calculate the proper time recorded by the probe between P and Q. What does clock B display at Q?

(c) In the probe frame \(S'\), determine whether the event where B displays zero occurs before or after P. Calculate the time difference.

(d) Calculate the time-coordinate interval in \(S'\) between the event where B displays zero and Q.

(e) At Q, the probe clock reads less elapsed time than clock B’s display. A student concludes that B’s clock must have ticked faster than the probe clock. Another student argues that time dilation means B should display exactly the same reading as the probe. Explain why both arguments are incorrect.

Solution 5

The probe records 2.25 microseconds between P and Q, while clock B displays 3.75 microseconds at Q.

However, neither clock has run faster. The crucial detail is that the two clocks’ zero-reading events are not simultaneous in the probe frame.

This question combines proper time, time dilation, and relativity of simultaneity. The order in which you analyse the events matters.

Step 1: Calculate the laboratory time between P and Q.

The probe travels 900 m at a constant speed of \(0.80c\).

Using

\[ \Delta t=\frac{\Delta x}{v} \]

we obtain

\[ \begin{aligned} \Delta t &=\frac{900}{0.80(3.00\times10^8)}\\ &=3.75\times10^{-6}\ \mathrm{s} \end{aligned} \]

Therefore,

\[ \boxed{\Delta t=3.75\ \mu\mathrm{s}} \]

The laboratory frame assigns 3.75 microseconds between P and Q.

Step 2: Calculate the probe’s proper time.

Both P and Q occur at the probe’s location in its own reference frame.

The probe’s onboard clock is present at both events, so the interval it records is the proper time.

First calculate the Lorentz factor:

\[ \begin{aligned} \gamma &=\frac{1}{\sqrt{1-0.80^2}}\\ &=\frac{1}{\sqrt{0.36}}\\ &=\frac{5}{3} \end{aligned} \]

Using

\[ \Delta t=\gamma\Delta\tau \]

rearrange:

\[ \Delta\tau=\frac{\Delta t}{\gamma} \]

Substitute:

\[ \begin{aligned} \Delta\tau &=\frac{3.75}{5/3}\\ &=2.25\ \mu\mathrm{s} \end{aligned} \]

Therefore,

\[ \boxed{\Delta\tau_{\text{probe}}=2.25\ \mu\mathrm{s}} \]

Step 3: Determine what clock B displays at Q.

Clock B is stationary in \(S\).

It displays zero at \(t=0\), which is simultaneous with P in the laboratory frame.

The probe reaches B after 3.75 microseconds of laboratory time.

Therefore, B displays

\[ \boxed{3.75\ \mu\mathrm{s}} \]

at Q.

At Q, the probe clock reads \(2.25\ \mu\mathrm{s}\), while B displays \(3.75\ \mu\mathrm{s}\).

We now need to explain why these readings differ.

Step 4: Find when B’s zero-reading event occurs in the probe frame.

Call the event where B displays zero \(E_{B0}\).

In the laboratory frame, \(E_{B0}\) and P are simultaneous:

\[ \Delta t=0 \]

However, they occur at different positions.

Taking the difference from P to \(E_{B0}\):

\[ \Delta x=+900\ \mathrm{m} \]

The probe moves at \(+0.80c\), and

\[ \gamma=\frac{5}{3} \]

Apply the Lorentz transformation:

\[ \begin{aligned} \Delta t' &=\gamma\left( 0-\frac{v\Delta x}{c^2} \right)\\ &=-\frac{5}{3} \left( \frac{0.80c(900)}{c^2} \right)\\ &=-\frac{5}{3}(2.40\ \mu\mathrm{s})\\ &=-4.00\ \mu\mathrm{s} \end{aligned} \]

Therefore,

\[ \boxed{\Delta t'=-4.00\ \mu\mathrm{s}} \]

The event where B displays zero occurs four microseconds before P according to the probe frame.

This is the essential result.

The laboratory says B’s zero-reading event and P are simultaneous. The probe does not agree.

Step 5: Calculate the time-coordinate interval from B’s zero-reading event to Q.

These two events occur at clock B.

In the laboratory frame, B is stationary, so

\[ \Delta x=0 \]

The time between the events is 3.75 microseconds.

Therefore,

\[ \begin{aligned} \Delta t' &=\gamma\left( 3.75-\frac{v(0)}{c^2} \right)\\ &=\frac{5}{3}(3.75)\\ &=6.25\ \mu\mathrm{s} \end{aligned} \]

Thus,

\[ \boxed{\Delta t'=6.25\ \mu\mathrm{s}} \]

The probe frame assigns 6.25 microseconds between B’s zero-reading event and Q.

B’s own clock records only 3.75 microseconds over this interval, which is consistent with time dilation:

\[ 6.25=\frac{5}{3}(3.75) \]

The clock moving relative to the probe accumulates less proper time than the coordinate time assigned by the probe frame.

Step 6: Resolve the two students’ arguments.

The first student argues that B must have ticked faster because it displays 3.75 microseconds while the probe displays 2.25 microseconds at Q.

This is incorrect because the clocks did not start from zero at simultaneous events in the probe frame.

B’s zero-reading event occurred at \(t'=-4.00\ \mu\mathrm{s}\), while the probe’s zero-reading event occurred at \(t'=0\).

B therefore started accumulating its displayed time earlier according to the probe frame.

The second student argues that time dilation requires B to display 2.25 microseconds at Q.

This is also incorrect.

Time dilation compares elapsed times between the same two events. It does not say that two clocks must display identical readings whenever they meet, regardless of when their zero settings occurred.

In fact, the probe frame assigns 6.25 microseconds between B’s zero-reading event and Q. B accumulates 3.75 microseconds during that interval because it is moving relative to the probe.

Meanwhile, the probe accumulates 2.25 microseconds between P and Q.

The two readings at Q are therefore consistent with both time dilation and relativity of simultaneity.

The tempting mistake is to assume that the clocks started their measurements at the same instant in every reference frame. They did not.

08What to understand next

Relativity of simultaneity matters whenever you compare clocks at different positions. Before deciding that one clock has accumulated more time, you must establish which events are being compared and whether their starting points are simultaneous in the chosen frame.

Once two clocks are brought back together, their readings can be compared at the same physical location. You can then examine the proper time each clock accumulated between the shared departure and reunion events.

That is the natural next step: understanding how time dilation and proper-time differences appear in real atomic-clock experiments.

The key distinction to carry forward is simple: time dilation compares elapsed time along clock worldlines, while relativity of simultaneity determines how different frames assign times to events at separate locations.