How to Compare Two Moving Atomic Clocks Using Time Dilation

Two atomic clocks leave the same laboratory. Clock A travels at \(0.60c\), while clock B travels at \(0.80c\), where \(c\) is the speed of light. After 10.0 seconds have passed in the laboratory, which clock has recorded less time?

Clock B has recorded less: 6.00 seconds compared with 8.00 seconds on clock A. But there’s a catch. If you watch their displays through a telescope, you won’t necessarily see those readings at the same instant. The clocks are different distances away, so their light takes different amounts of time to reach you.

To compare two moving atomic clocks correctly, you need to identify the proper time recorded by each clock, choose a reference frame for the comparison, and separate genuine time dilation from light-travel delays.

01Which clock measures the proper time?

Imagine an atomic clock fixed inside a spacecraft.

It emits a light pulse when its display reads 0.00 seconds. Later, it emits another pulse when its display reads 8.00 seconds.

Both emissions occur at the same position inside the spacecraft: the clock’s location. Although the spacecraft has moved relative to Earth, the clock hasn’t moved relative to itself.

The time between those two events, measured by the clock travelling with them, is called proper time.

Now consider an observer in the laboratory. They record the first emission at one position and the second emission at another. They use synchronised laboratory clocks to determine the time between those emission events.

That interval is the coordinate time in the laboratory frame. It isn’t the proper time for the moving spacecraft clock.

The equation that compares them

For a clock moving at constant speed \(v\) in an inertial reference frame,

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

where:

• \(\Delta t\) is the elapsed time between the events in the chosen reference frame, measured in seconds (s).
• \(\Delta\tau\) is the proper time recorded by the moving clock, in seconds (s).
• \(\gamma\) is the Lorentz factor, which has no units.
• \(v\) is the clock’s speed relative to the chosen frame, in metres per second (m/s).
• \(c=3.00\times10^8\text{ m/s}\) is the speed of light in a vacuum.

The Lorentz factor is

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

For comparing clocks, it’s often easier to rearrange:

\[
\Delta\tau
=\Delta t\sqrt{1-\frac{v^2}{c^2}}
\]

Notice what happens as \(v\) increases. The quantity inside the square root decreases, so the moving clock accumulates less proper time during the same laboratory interval.

This comparison assumes both clocks have constant speeds relative to the same inertial frame, and that gravitational differences can be neglected.

One important limit: if the clocks are at different positions at the end, we’re comparing their readings at events that are simultaneous in the chosen laboratory frame. Another inertial frame may select different events as simultaneous.

There is no universal frame that declares one clock absolutely stationary.

02Why observing a clock is not the same as comparing it

Suppose you stand at the starting laboratory and watch both spacecraft through a telescope.

At laboratory time 10.0 s, spacecraft A is 6.00 light-seconds away. Spacecraft B is 8.00 light-seconds away.

A light-second is a distance of \(3.00\times10^8\) metres: the distance light travels in one second.

Here’s the question: if the two spacecraft send their clock readings at laboratory time 10.0 s, do the messages arrive together?

No. Light from A takes 6.00 s to reach the laboratory, while light from B takes 8.00 s. The messages arrive at laboratory times 16.0 s and 18.0 s respectively.

This delay is not itself time dilation.

The first measurement design: watch the incoming signals

Imagine that an investigator wants to test time dilation by receiving one radio pulse every time a moving atomic clock advances by one second.

It sounds reasonable. If the travelling clock runs slower, the pulses should arrive farther apart.

But this design has a flaw.

When a spacecraft moves away, each new pulse must travel farther than the previous pulse. Even if the emission interval were unchanged, that increasing distance would spread out the arrival times.

When the spacecraft approaches, successive pulses have less distance to travel. The received pulses bunch together.

This is closely related to the relativistic Doppler effect.

Think of watching a delayed live stream of a sports scoreboard. The image you receive shows what the scoreboard displayed earlier, not necessarily what it’s displaying now. The difference is that light-travel delays are determined by a measurable distance and the speed of light, rather than by unpredictable network delays.

For a stationary laboratory receiver, the arrival time of a light signal is

\[
t_{\mathrm{rec}}
=t_{\mathrm{emit}}+\frac{d}{c}
\]

where:

• \(t_{\mathrm{rec}}\) is the laboratory time when the signal is received.
• \(t_{\mathrm{emit}}\) is the laboratory time when the signal was emitted.
• \(d\) is the source-to-receiver distance at emission, measured in metres.
• \(c\) is the speed of light, in metres per second.

The travel distance must be measured from the clock’s position when it emitted the pulse, not from its position when the pulse arrives.

03Repairing the measurement: compare emission times, not just arrivals

The problem with the first design wasn’t the atomic clocks. It was the measurement method.

The receiver mixed two effects:

1. The rate at which the moving clock accumulated proper time.
2. The changing distance travelled by its light signals.

A better experiment has to account for both.

There are two useful approaches.

Method A: Correct time-stamped signals

Equip the laboratory with a time reference and position measurements. Equip each spacecraft with an atomic clock and a transmitter that includes the clock’s reading in every message.

The measurement proceeds like this:

1. Set the travelling clocks at a known initial event.
2. Record their positions and velocities in the chosen laboratory frame.
3. Receive messages containing their local atomic-clock readings.
4. Calculate each signal’s light-travel time from the source position at emission.
5. Subtract that delay from the laboratory reception time to reconstruct the emission time.
6. Compare clock readings associated with equal laboratory coordinate times.

If the receiver is stationary and the source-receiver distance at emission is known,

\[
t_{\mathrm{emit}}
=t_{\mathrm{rec}}-\frac{d}{c}
\]

The result is a comparison in the laboratory reference frame, not a universal definition of simultaneity.

This procedure also assumes the transmitted timestamps and position measurements are calibrated correctly. An unexplained transmission delay or an incorrect distance can produce a false clock difference.

Method B: Bring the clocks back together

An even more direct comparison is possible if the two clocks start at the same event and later reunite at the same event.

Set both clocks to zero while they’re side by side. Send them along different routes, then bring them back together.

When they reunite, read their displays directly.

There’s no light-travel correction between separated endpoints because the final readings are taken at the same place and time.

However, returning clocks may accelerate or change speed. The simple constant-speed equation must then be applied to each constant-velocity section of the journey, with acceleration intervals handled appropriately.

For short, idealised changes of speed, we can approximate the accumulated proper time by

\[
\Delta\tau
=\sum_i\Delta t_i
\sqrt{1-\frac{v_i^2}{c^2}}
\]

Here, \(i\) labels each section of the journey, \(\Delta t_i\) is that section’s laboratory duration, and \(v_i\) is the clock’s speed during that section.

You calculate how much time each clock accumulates along its full path.

That is more reliable than asking which clock was moving faster at one particular moment.

04What the real atomic-clock experiments established

Flying portable atomic clocks is one way to make the comparison physical rather than hypothetical.

In October 1971, Joseph Hafele and Richard Keating carried four caesium-beam atomic clocks aboard commercial aircraft around the world, once eastwards and once westwards.

Their aim was to test whether the clocks accumulated the elapsed times predicted by relativity when compared with the atomic time reference at the US Naval Observatory.

The experiment had a clear measurement sequence.

First, the portable clocks were compared with the reference time before the flights. The clocks were then carried on aircraft, while their changing flight paths, speeds, and altitudes were used in the theoretical predictions.

After the journeys, the portable clocks were compared with the reference again. The result wasn’t simply a delayed signal from a distant aircraft. It was a difference in accumulated clock readings.

The reported results were:

• Eastward journey: the flying clocks finished approximately 59 nanoseconds behind the ground reference.
• Westward journey: the flying clocks finished approximately 273 nanoseconds ahead.

A nanosecond is \(10^{-9}\) s. These observations were consistent with the corresponding relativistic predictions within the reported uncertainties (Hafele and Keating, Science, 1972).

This experiment also exposes a limitation of our simplified spacecraft model.

Earth rotates, so a clock fixed on the ground is not stationary in a non-rotating Earth-centred frame. Aircraft also fly at higher altitudes, where gravitational time dilation becomes relevant.

The original predictions therefore needed both motion and gravity, not just one Lorentz factor.

The experiment supported relativity because it predicted different clock shifts for the two journeys and the observations agreed with those predictions. It did not show that all westward-moving clocks always run faster.

For the fuller treatment of the aircraft experiment, including Earth’s rotation, altitude corrections, and uncertainty, see Atomic-Clock Tests of Relativity: HSC Physics Explained.

For comparing two clocks in a special-relativity question, however, you usually begin with a simpler controlled situation: one inertial laboratory frame, specified speeds, a common laboratory interval, and negligible gravitational differences.

05The comparison method to use in an HSC question

Before calculating anything, decide what the question has actually measured.

One subtle point: faster relative to what?

Suppose two clocks move in opposite directions at \(0.60c\) relative to the same inertial laboratory.

One travels left and the other travels right.

Which records more proper time during 10.0 s of laboratory time?

Neither. Both record 8.00 s.

The Lorentz factor depends on \(v^2\), so the sign of velocity doesn’t change the time-dilation calculation.

But this doesn’t mean two observers moving relative to one another must agree about which separated events are simultaneous. Each inertial observer can describe the other’s moving clock as time-dilated.

For a comparison between separated clocks, the reference frame and the chosen event pairs matter. For clocks that reunite, their accumulated proper times can be compared directly at their common event.

These ideas become especially important in the more demanding problems below.

06Questions and solutions

All numerical values in the following questions are illustrative. Unless stated otherwise, use \(c=3.00\times10^8\text{ m/s}\), ignore gravity, and treat the laboratory as an inertial reference frame. Instantaneous changes of velocity are idealisations with negligible duration.

Question 1

A spacecraft carrying an atomic clock travels at a constant speed of \(0.80c\) relative to a laboratory.

Two flashes are emitted from a lamp fixed beside the clock. Synchronised laboratory clocks determine that the emissions occur 5.00 s apart.

Calculate the interval recorded by the spacecraft’s atomic clock. Identify which interval is the proper time and explain why.

Solution 1

The spacecraft records 3.00 s of proper time. The 5.00 s laboratory interval is the dilated coordinate-time interval.

Both emissions occur at the same position in the spacecraft’s own reference frame, because the lamp is fixed beside its atomic clock.

Using time dilation,

\[
\Delta\tau
=\Delta t\sqrt{1-\frac{v^2}{c^2}}
\]

Substitute \(\Delta t=5.00\text{ s}\) and \(v=0.80c\):

\[
\begin{aligned}
\Delta\tau
&=5.00\sqrt{1-0.80^2}\\
&=5.00(0.60)\\
&=3.00\text{ s}
\end{aligned}
\]

The spacecraft clock records 3.00 s between the emissions, while the laboratory measures 5.00 s between the same events.

The tempting mistake is to call the larger interval the proper time because it was measured using stationary laboratory clocks. Proper time belongs to the clock for which the two events occur at the same position.

Question 2

Two atomic clocks, A and B, begin together at laboratory position \(x=0\) and time \(t=0\).

Clock A travels at a constant \(0.60c\) for 10.0 s of laboratory time.

Clock B travels at \(0.80c\) for the first 5.00 s, then slows to \(0.40c\) for the next 5.00 s. Both velocities are in the positive \(x\)-direction.

The clocks meet again at laboratory time 10.0 s.

A student argues that their readings must agree because both have an average laboratory velocity of \(0.60c\).

Determine whether the student is correct. Calculate the proper time accumulated by each clock and explain the physical reason for any difference.

Solution 2

The student is incorrect. Clock A accumulates 8.00 s, while clock B accumulates approximately 7.58 s. Clock A is ahead by about 0.417 s despite both clocks having the same average laboratory velocity.

First, confirm that the clocks reach the same final position.

Clock A travels

\[
x_A=(0.60c)(10.0\text{ s})
\]

or 6.00 light-seconds.

Clock B travels 4.00 light-seconds during its first 5.00 s, then 2.00 light-seconds during its second 5.00 s.

Its total displacement is also 6.00 light-seconds.

For A, the proper time is

\[
\begin{aligned}
\Delta\tau_A
&=10.0\sqrt{1-0.60^2}\\
&=8.00\text{ s}
\end{aligned}
\]

For B, calculate the two sections separately.

\[
\begin{aligned}
\Delta\tau_B
&=5.00\sqrt{1-0.80^2}\\
&\quad+5.00\sqrt{1-0.40^2}\\
&=3.00+4.5826\\
&=7.5826\text{ s}
\end{aligned}
\]

The difference is

\[
\begin{aligned}
\Delta\tau_A-\Delta\tau_B
&=8.0000-7.5826\\
&=0.4174\text{ s}
\end{aligned}
\]

A accumulates approximately 0.417 s more proper time.

The trap is substituting average velocity into the time-dilation equation.

Clock B’s faster first stage doesn’t cancel its slower second stage. Proper time depends on the time spent at each speed, and the relationship between speed and time dilation is nonlinear.

The clocks share both their starting event and final event, so this is a genuine reunion comparison rather than a dispute about which distant events are simultaneous.

Question 3

A detector is stationary at laboratory position \(x=0\).

Two spacecraft, R and S, carry identical atomic clocks and transmit light pulses exactly 1.00 s apart according to their own clock readings.

Spacecraft R starts at \(x=0\) and moves away from the detector at \(0.60c\).

Spacecraft S starts 20.0 light-seconds to the right of the detector and approaches it at \(0.80c\).

Both travel at constant velocities. Consider the first two pulses from each spacecraft, beginning with a pulse emitted at laboratory time \(t=0\).

(a) Calculate the laboratory time between the two emissions from each spacecraft.

(b) Calculate the interval between the pulses received from R and the interval between those received from S.

(c) Calculate the proper time accumulated by each clock during the first 10.0 s of laboratory time.

(d) A student concludes that S’s atomic clock is running faster because its pulses arrive more frequently. Explain why this conclusion is incorrect.

Solution 3

R’s pulses arrive 2.00 s apart, while S’s arrive only \(0.333\) s apart. Nevertheless, over 10.0 s of laboratory time, R accumulates 8.00 s of proper time and S accumulates just 6.00 s.

The more frequent received pulses from S result from its decreasing distance to the detector.

(a) Emission intervals

For R,

\[
\begin{aligned}
\Delta t_R
&=\frac{1.00}{\sqrt{1-0.60^2}}\\
&=\frac{1.00}{0.80}\\
&=1.25\text{ s}
\end{aligned}
\]

For S,

\[
\begin{aligned}
\Delta t_S
&=\frac{1.00}{\sqrt{1-0.80^2}}\\
&=\frac{1.00}{0.60}\\
&=1.667\text{ s}
\end{aligned}
\]

S’s emissions are actually farther apart in laboratory time.

(b) Reception intervals

During R’s 1.25 s emission interval, it travels

\[
\Delta x_R=(0.60c)(1.25\text{ s})
\]

or 0.750 light-seconds farther from the receiver.

The second pulse therefore takes an additional 0.750 s to arrive.

\[
\begin{aligned}
\Delta t_{\mathrm{rec},R}
&=1.25+0.750\\
&=2.00\text{ s}
\end{aligned}
\]

During S’s emission interval, it travels towards the receiver by

\[
\begin{aligned}
d_S&=(0.80c)(1.667\text{ s})\\
&=1.333\text{ light-seconds}
\end{aligned}
\]

Its second pulse has 1.333 light-seconds less distance to travel.

Therefore,

\[
\begin{aligned}
\Delta t_{\mathrm{rec},S}
&=1.667-1.333\\
&=0.334\text{ s}
\end{aligned}
\]

Using unrounded values gives exactly \(1/3\) s, or approximately 0.333 s.

S’s pulses reach the detector six times as frequently as R’s.

(c) Proper times after 10.0 laboratory seconds

For R,

\[
\begin{aligned}
\Delta\tau_R
&=10.0\sqrt{1-0.60^2}\\
&=8.00\text{ s}
\end{aligned}
\]

For S,

\[
\begin{aligned}
\Delta\tau_S
&=10.0\sqrt{1-0.80^2}\\
&=6.00\text{ s}
\end{aligned}
\]

S accumulates 2.00 s less proper time.

(d) Why the received pulse rate is misleading

The student has confused reception intervals with emission intervals.

S’s faster approach shortens the light-travel distance for each successive pulse. That effect more than compensates for the greater separation between its emissions in laboratory time.

R’s motion produces the opposite effect. Every new pulse travels farther, spreading out the reception times.

Neither spacecraft experiences its own atomic clock as ticking abnormally. The comparison is between elapsed proper times measured along different paths and signal intervals measured at the detector.

Question 4

Two identical atomic clocks, P and Q, begin together at \(x=0\) with readings of zero.

Clock P travels in the positive \(x\)-direction at \(0.60c\), while clock Q travels in the same direction at \(0.80c\).

After 10.0 s of laboratory time, both spacecraft instantaneously reverse their velocities and return towards \(x=0\) at their original speed magnitudes.

At the turnaround event, each spacecraft transmits a light pulse containing its current atomic-clock reading.

The laboratory receiver is stationary at \(x=0\).

(a) Determine each clock’s reading at its turnaround.

(b) Determine when each turnaround pulse reaches the laboratory receiver.

(c) Determine when the spacecraft reunite and the difference between their atomic-clock readings at that event.

(d) A student argues that, because each travelling spacecraft can regard the other as moving, their elapsed proper times at reunion must be equal. Evaluate this argument.

Solution 4

At turnaround, P reads 8.00 s and Q reads 6.00 s. Their turnaround signals arrive at laboratory times 16.0 s and 18.0 s respectively. They physically reunite at laboratory time 20.0 s, when P is ahead by 4.00 s.

(a) Turnaround readings

Each spacecraft has travelled for 10.0 s in the laboratory frame.

For P,

\[
\begin{aligned}
\tau_{P,\mathrm{out}}
&=10.0\sqrt{1-0.60^2}\\
&=8.00\text{ s}
\end{aligned}
\]

For Q,

\[
\begin{aligned}
\tau_{Q,\mathrm{out}}
&=10.0\sqrt{1-0.80^2}\\
&=6.00\text{ s}
\end{aligned}
\]

(b) Arrival times of the turnaround signals

P travels 6.00 light-seconds from the laboratory before reversing.

Its signal takes 6.00 s to travel back.

\[
t_{\mathrm{rec},P}=10.0+6.00
\]

so

\[
t_{\mathrm{rec},P}=16.0\text{ s}
\]

Q travels 8.00 light-seconds before reversing.

\[
t_{\mathrm{rec},Q}=10.0+8.00
\]

so

\[
t_{\mathrm{rec},Q}=18.0\text{ s}
\]

The messages arrive at different times despite being emitted simultaneously in the laboratory frame.

(c) Reunion readings

Both spacecraft return along their original paths at the same speed magnitudes.

Because each return leg takes 10.0 s of laboratory time, both arrive back at \(x=0\) at \(t=20.0\text{ s}\).

Each return leg contributes the same proper time as its outward leg.

For P,

\[
\begin{aligned}
\tau_P&=8.00+8.00\\
&=16.0\text{ s}
\end{aligned}
\]

For Q,

\[
\begin{aligned}
\tau_Q&=6.00+6.00\\
&=12.0\text{ s}
\end{aligned}
\]

Their difference is

\[
16.0-12.0=4.00\text{ s}
\]

P is older by 4.00 s when they reunite.

(d) Why the apparent symmetry fails

The student’s argument applies a statement about relative motion in inertial frames to a journey that includes direction changes.

During a constant-velocity stage, each spacecraft can use its own inertial frame to describe the other’s clock as time-dilated. But those comparisons refer to events selected as simultaneous in different reference frames.

The turnaround events occur at different positions. They’re simultaneous in the laboratory frame, but not necessarily in either travelling spacecraft’s frame.

Neither spacecraft remains in one inertial frame for the complete journey because both reverse their velocities.

The proper times found by adding the segments remain valid. They are readings that can be checked directly at their shared reunion event, independent of which coordinate system is used.

The tempting mistake is to treat the two viewpoints during the outward journey as if they specified the same pair of distant events and could be extended unchanged through the turnarounds.

Question 5

A laboratory has two stationary light detectors: a western detector at \(x=-20.0\) light-seconds and an eastern detector at \(x=+20.0\) light-seconds.

Two spacecraft, A and B, travel eastwards at different, unknown constant speeds. Both start at \(x=0\), and their positions remain between the detectors throughout the measurements.

Each spacecraft has an atomic clock and transmits consecutive numbered light pulses. The pulses are programmed to be emitted every 1.00 s according to that spacecraft’s own atomic clock.

Every emitted pulse travels to both detectors.

The laboratory obtains the following idealised results:

The detector clocks measure intervals reliably, but their absolute time readings have an unknown constant offset relative to one another.

(a) Using time dilation and light-travel distance, derive expressions for the received pulse intervals at the western and eastern detectors for a spacecraft with eastward speed \(v\).

(b) Show how the spacecraft’s speed can be found using only those two received intervals, without first knowing its emission interval.

(c) Determine the speeds of A and B, and verify that both transmit every 1.00 s of proper time.

(d) After 10.0 s of laboratory time, which spacecraft has accumulated less proper time, and by how much?

(e) A student claims that the experiment cannot establish either spacecraft’s speed because the detector clocks are not synchronised. Another student claims the eastern detector proves that B’s clock is ticking faster than A’s. Explain why both claims are incorrect, and identify an assumption under which the two-detector method would fail.

Solution 5

A travels at \(0.60c\), and B travels at \(0.80c\). Both transmit once per 1.00 s of proper time, but B accumulates 2.00 s less proper time than A over 10.0 s of laboratory time.

The crucial insight is that motion changes the signal spacing in opposite ways at the two detectors. Combining both measurements allows the light-travel effects to be separated from the clock’s own emission period.

(a) Derive the two reception intervals

Let \(P_0\) be the proper-time interval between successive emissions, measured in seconds.

Let

\[
\beta=\frac{v}{c}
\]

where \(v\) is the positive eastward speed of the spacecraft and \(\beta\) has no units.

The corresponding laboratory emission interval is

\[
\Delta t_{\mathrm{emit}}=\gamma P_0
\]

During that interval, the spacecraft moves east by

\[
\Delta x=v\gamma P_0
\]

The western detector is behind the moving spacecraft.

For the western detector, the second pulse travels an additional distance \(\Delta x\). Its reception interval is therefore

\[
P_W=\gamma P_0(1+\beta)
\]

The eastern detector is ahead of the moving spacecraft.

For the eastern detector, the second pulse travels a shorter distance by the same amount.

Its reception interval is

\[
P_E=\gamma P_0(1-\beta)
\]

Here, \(P_W\) and \(P_E\) are the received intervals in seconds at the western and eastern detectors.

Both expressions contain time dilation through \(\gamma\), but their light-travel corrections have opposite signs.

(b) Recover the velocity without knowing the proper emission period

Divide the two expressions:

\[
\frac{P_W}{P_E}
=\frac{1+\beta}{1-\beta}
\]

The factors \(\gamma\) and \(P_0\) cancel.

Rearranging gives

\[
\boxed{\beta=\frac{P_W-P_E}{P_W+P_E}}
\]

The result is dimensionless. It determines the eastward speed as a fraction of \(c\).

This is why two detectors provide more information than a single detector. The complementary distance changes reveal the motion independently of the clock’s emission interval.

(c) Determine the speeds and check the emission periods

For A,

\[
\begin{aligned}
\beta_A
&=\frac{2.00-0.500}{2.00+0.500}\\
&=\frac{1.50}{2.50}\\
&=0.600
\end{aligned}
\]

Therefore,

\[
v_A=0.600c
\]

or

\[
v_A=1.80\times10^8\text{ m/s}
\]

For B,

\[
\begin{aligned}
\beta_B
&=\frac{3.00-1/3}{3.00+1/3}\\
&=\frac{8/3}{10/3}\\
&=0.800
\end{aligned}
\]

Therefore,

\[
v_B=0.800c
\]

or

\[
v_B=2.40\times10^8\text{ m/s}
\]

Now multiply the two reception-interval equations:

\[
P_WP_E
=\gamma^2P_0^2(1-\beta^2)
\]

But

\[
\gamma^2=\frac{1}{1-\beta^2}
\]

so the velocity-dependent factors cancel exactly:

\[
P_WP_E=P_0^2
\]

Hence,

\[
P_0=\sqrt{P_WP_E}
\]

For A,

\[
\begin{aligned}
P_{0,A}
&=\sqrt{(2.00)(0.500)}\\
&=1.00\text{ s}
\end{aligned}
\]

For B,

\[
\begin{aligned}
P_{0,B}
&=\sqrt{(3.00)(1/3)}\\
&=1.00\text{ s}
\end{aligned}
\]

Both spacecraft really do emit pulses every 1.00 s of their own proper time.

The surprising result is that the geometric mean of the two received periods gives the proper emission period, even though neither detector individually measures that period.

(d) Compare the accumulated proper times

Both spacecraft are compared over the same 10.0 s of laboratory time.

For A,

\[
\begin{aligned}
\Delta\tau_A
&=10.0\sqrt{1-0.60^2}\\
&=8.00\text{ s}
\end{aligned}
\]

For B,

\[
\begin{aligned}
\Delta\tau_B
&=10.0\sqrt{1-0.80^2}\\
&=6.00\text{ s}
\end{aligned}
\]

Therefore,

\[
\Delta\tau_A-\Delta\tau_B=2.00\text{ s}
\]

B accumulates 2.00 s less proper time than A.

Even though B’s pulses reach the eastern detector more frequently, it is the clock that accumulates less proper time in the laboratory comparison.

(e) Evaluate the two claims and the limits of the method

The first student is incorrect because the calculation only uses intervals between successive receptions at each individual detector.

An unknown constant offset between the detectors’ clock readings doesn’t change those intervals. The method doesn’t require the two detector clocks to agree about the absolute time.

It does require the detectors to identify the same consecutive emitted pulses and measure their local reception intervals correctly.

The second student is also incorrect.

B’s shorter interval at the eastern detector occurs because B approaches that detector more quickly than A. The reduced light-travel distance makes successive signals arrive closer together.

B’s own proper emission period is still 1.00 s, as established by combining both measurements.

The method would fail in its present form if the spacecraft changed speed substantially between emissions, or if significant unaccounted transmission delays affected the pulses differently. Our equations assume constant velocity, light propagation at \(c\), stationary detectors, and a clear identification of the same emission pair.

A particularly important detail is that both spacecraft must remain between the detectors during the relevant emissions. If one passes a detector, the direction of the distance change reverses, and the original equations no longer describe that detector’s measurements.

The deeper lesson is that a received signal contains information about both the clock and the signal’s journey. With the right measurements, those effects can be separated mathematically.

For any new atomic-clock comparison, start by marking the emission, reception, and reunion events separately. Identify which event pairs belong to the moving clock’s own proper time, then account for the geometry of the measurement. That same event-based reasoning leads into relativity of simultaneity and the twin paradox, where the clocks’ complete paths through spacetime determine what they record.