How to Identify the Two Events Before Choosing Proper Time
Learn how to identify the two physical events first, then determine which clock measures proper time. Explore lunar laser ranging, moving spacecraft, light-travel delays, and challenging worked HSC Physics questions.
A laser pulse leaves an observatory on Earth, travels to a reflector on the Moon, and returns to the same telescope about 2.5 seconds later. Which time is the proper time: the time between the pulse leaving and returning, or the time it spends travelling to the Moon?
The answer depends on which two events you’re timing.
This is the key to identifying proper time in special relativity: choose the two physical events first, then find the clock that can measure both. Don’t choose a time interval simply because it is the smallest number in the question, or because an observer says they saw something happen.
01Identify the two events before choosing proper time
An event is a physical happening at a particular position and time.
Examples include:
- A laser pulse being emitted.
- A light pulse being reflected.
- A detector receiving a signal.
- A spacecraft clock recording a tick.
- A particle being created or decaying.
These are separate events, even when they are all part of one experiment.
Proper time, written as \(\Delta\tau\), is the elapsed time measured by one clock that is present at both events.
In the clock’s rest frame, the two events occur at the same position. The clock doesn’t need to move between them.
For the usual HSC time-dilation problems, use this decision process.
Step 1: Name the first event.
Write down exactly what happens. For example, a transmitter emits a laser pulse.
Step 2: Name the second event.
Is it the pulse being reflected, the pulse arriving at a detector, or a later emission from the same transmitter?
Don’t treat these as interchangeable.
Step 3: Find where both events occur.
Consider the position of each event in the frame you’re investigating.
Step 4: Find a clock that is present at both events.
If one clock can record both events at the same position in its rest frame, its reading is the proper time for that pair.
Step 5: Only then choose your equation.
For a clock moving at constant speed relative to another inertial frame:
\[ \Delta t=\gamma\Delta\tau \]
where:
- \(\Delta\tau\) is the proper time, measured by the clock present at both events, in seconds (s).
- \(\Delta t\) is the time interval assigned by another inertial frame, in seconds (s).
- \(\gamma\) is the Lorentz factor.
- \(v\) is the relative speed between the frames, in metres per second (\(\mathrm{m\,s^{-1}}\)).
- \(c\) is the speed of light in a vacuum, approximately \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).
The Lorentz factor is:
\[ \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \]
Some HSC questions use \(\Delta t_0\) instead of \(\Delta\tau\) for proper time. Both notations describe the same quantity.
Remember that the time-dilation equation compares the same two events in different frames. It cannot be used to compare an emission interval with an unrelated reception interval.
02A real experiment: measuring the distance to the Moon
Lunar laser ranging provides a useful example of why identifying events matters.
In 1969, Apollo 11 astronauts deployed a laser retroreflector on the lunar surface. A retroreflector contains corner-cube mirrors that send incoming light back towards its source.
On 1 August 1969, the Lick Observatory in California successfully received a laser pulse returned from the Apollo 11 reflector.
The experiment answered a practical question: how accurately can we measure the distance between Earth and the Moon?
How the apparatus works
Follow the pulse in chronological order.
- An observatory on Earth contains a laser, a telescope, and a timing system.
- The laser emits a short pulse. Call this event A.
- The pulse travels towards the Moon.
- It reaches the lunar retroreflector and is sent back. Call this event C.
- The returning light reaches the original observatory. Call this event B.
- The observatory records the elapsed time between transmission and reception.
Now consider the original measurement.
If the reflector is stationary at a distance \(d\), the pulse travels a total distance of \(2d\). Since light travels at speed \(c\), the expected round-trip time is:
\[ \Delta t=\frac{2d}{c} \]
The timing system should therefore record a delay of approximately 2.5 seconds for the Earth-Moon distance.
The first successful measurement was indeed about 2.5 seconds. NIST records that the original experiment achieved a distance measurement accurate to approximately 25 cm. Modern lunar laser ranging systems have improved the precision to approximately 1 mm.
You can read more about the experiment in NIST’s historical account of lunar laser ranging.
Which two events does the observatory actually time?
Here is the important distinction.
| Event | What happens? | Position in the Earth-station frame |
|---|---|---|
| A | Laser pulse is emitted | Observatory |
| C | Pulse is reflected | Moon |
| B | Returning pulse is received | Observatory |
The observatory’s timing system measures A to B.
Both events occur at the same observatory. One local clock can record the emission and reception.
Therefore, the elapsed time between A and B is the proper time measured by the observatory clock.
But A to C is different.
The emission occurs on Earth, while the reflection occurs on the Moon. The observatory clock is not present at both events. The interval A to C is a light-travel interval, not the proper time measured by a clock travelling with the photon.
The same applies to C to B.
This gives us three distinct intervals:
- A to B: proper time on the observatory clock.
- A to C: outward light-travel time in the chosen Earth-frame model.
- C to B: return light-travel time in the chosen Earth-frame model.
The fact that all three intervals belong to the same experiment doesn’t make them the same measurement.
Worked example: Calculate the Earth-Moon distance from a laser return
An observatory records a round-trip laser travel time of \(2.56\ \mathrm{s}\). Assume the reflector is stationary during the measurement and the laser pulse travels at \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).
Calculate the distance between the observatory and the reflector.
Step 1
The events are:
- A: the laser pulse leaves the observatory.
- B: the returning pulse reaches the same observatory.
The observatory clock is present at both events, so the measured interval of \(2.56\ \mathrm{s}\) is its proper time.
Step 2
Using distance = speed × time:
\[ \begin{aligned} s&=c\Delta t\\ &=(3.00\times10^8)(2.56)\\ &=7.68\times10^8\ \mathrm{m} \end{aligned} \]
This is the total out-and-back distance.
Step 3
The reflector is assumed stationary, so the outward and return distances are equal.
\[ \begin{aligned} d&=\frac{7.68\times10^8}{2}\\ &=3.84\times10^8\ \mathrm{m} \end{aligned} \]
The estimated distance is:
\[ \boxed{d=3.84\times10^8\ \mathrm{m}} \]
This is an illustrative calculation, not a reconstruction of an actual lunar ranging observation.
Notice what we have calculated. The observatory measures a proper-time interval of 2.56 seconds, but the one-way light-travel time is 1.28 seconds in this simplified Earth-frame model.
The smaller number is not automatically the proper time. The event pair determines that.
What the simple experiment model leaves out
The original calculation assumes a fixed distance and an ideal timing system.
Real measurements are more complicated. The Earth-Moon separation changes, and precise ranging requires increasingly accurate timing and modelling.
The experimental method was refined rather than abandoned. The central measurement remained the time between transmission and reception at the observatory.
That is the useful lesson: improving a measurement doesn’t change which two events the local clock actually records.
03A moving spacecraft creates a different event pair
Now consider a more complicated situation.
An Earth station sends a laser pulse towards a spacecraft. The spacecraft receives the pulse, waits for a short time according to its onboard clock, and then sends a reply.
The reply eventually reaches Earth.
There are four events:
- A: Earth emits the original pulse.
- C: The spacecraft receives it.
- D: The spacecraft sends its reply.
- B: Earth receives the reply.
Which events belong to the same clock?
| Event pair | Clock present at both events | Proper time? |
|---|---|---|
| A to B | Earth station clock | Yes |
| C to D | Spacecraft clock | Yes |
| A to C | No physical clock at both endpoints | No clock’s proper-time reading |
| D to B | No physical clock at both endpoints | No clock’s proper-time reading |
Notice that both A to B and C to D have proper times.
They belong to different pairs of events and different clocks.
The signal travelling between the spacecraft and Earth introduces additional delays. Those delays must be calculated separately.
Worked example: A spacecraft receives a pulse and sends a delayed reply
An Earth station is located at \(x=0\). A spacecraft is initially \(1.20\ \mathrm{km}\) away and moves directly away from Earth at \(0.60c\).
At Earth time \(t=0\), the station sends a laser pulse towards the spacecraft.
The spacecraft receives the pulse and waits \(2.00\ \mu\mathrm{s}\), measured by its onboard clock, before transmitting a reply.
Calculate:
(a) the Earth-frame time between receiving the original pulse and sending the reply,
(b) the position of the spacecraft when it sends the reply, and
(c) the total time measured by the Earth station between sending the original pulse and receiving the reply.
Assume constant spacecraft speed, instantaneous signal processing, and light travelling through a vacuum. Take the positive \(x\)-direction to be away from Earth.
Step 1
The spacecraft receives the original pulse at C and sends the reply at D.
Both events occur at the same onboard transponder. Therefore:
\[ \Delta\tau_{CD}=2.00\ \mu\mathrm{s} \]
This is the proper time.
Step 2
The spacecraft moves at \(0.60c\).
\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.60^2}}\\ &=\frac{1}{\sqrt{0.64}}\\ &=1.25 \end{aligned} \]
Step 3
The initial separation is:
\[ d_0=1.20\times10^3\ \mathrm{m} \]
The laser moves at \(c\), while the spacecraft moves away at \(0.60c\).
Their closing speed is:
\[ c-v=0.40c \]
Therefore:
\[ \begin{aligned} \Delta t_{AC} &=\frac{d_0}{c-v}\\ &=\frac{1200}{0.40(3.00\times10^8)}\\ &=1.00\times10^{-5}\ \mathrm{s} \end{aligned} \]
The pulse reaches the spacecraft after \(10.0\ \mu\mathrm{s}\).
Its position at C is:
\[ \begin{aligned} x_C&=ct_C\\ &=(3.00\times10^8)(10.0\times10^{-6})\\ &=3.00\times10^3\ \mathrm{m} \end{aligned} \]
Step 4
The spacecraft clock measures the proper time, so:
\[ \begin{aligned} \Delta t_{CD}&=\gamma\Delta\tau_{CD}\\ &=1.25(2.00\ \mu\mathrm{s})\\ &=2.50\ \mu\mathrm{s} \end{aligned} \]
Earth assigns \(2.50\ \mu\mathrm{s}\) to the same two events.
Step 5
During the waiting period, the spacecraft continues moving.
\[ \begin{aligned} \Delta x&=v\Delta t_{CD}\\ &=0.60(3.00\times10^8)(2.50\times10^{-6})\\ &=450\ \mathrm{m} \end{aligned} \]
Therefore:
\[ x_D=3000+450 \]
\[ x_D=3450\ \mathrm{m} \]
Step 6
The reply travels from \(x=3450\ \mathrm{m}\) back to the station at \(x=0\).
\[ \begin{aligned} \Delta t_{DB} &=\frac{3450}{3.00\times10^8}\\ &=11.5\ \mu\mathrm{s} \end{aligned} \]
The complete Earth-frame time from transmission to reception is:
\[ \begin{aligned} \Delta t_{AB} &=\Delta t_{AC}+\Delta t_{CD}+\Delta t_{DB}\\ &=10.0+2.50+11.5\\ &=24.0\ \mu\mathrm{s} \end{aligned} \]
The Earth station measures \(24.0\ \mu\mathrm{s}\) on its own clock.
Because A and B occur at the same station, this is also the proper time for the A-to-B event pair.
We have now identified two different proper-time intervals:
- C to D: \(2.00\ \mu\mathrm{s}\), measured by the spacecraft clock.
- A to B: \(24.0\ \mu\mathrm{s}\), measured by the Earth station clock.
The other intervals describe the light travelling between the two locations.
The crucial point is that the station’s \(24.0\ \mu\mathrm{s}\) measurement is not the spacecraft’s elapsed time. It includes the outward and return journeys of the signals.
04A quick event-pair audit for HSC questions
Before using time dilation, write down the event pair and the clock that can record it.
| Situation | Events to identify | Correct proper-time clock |
|---|---|---|
| Two ticks of a spacecraft clock | First tick and second tick | Spacecraft clock |
| Two particle detections at a fixed laboratory detector | First detection and second detection | Laboratory detector clock |
| Lunar laser ranging | Transmission and return reception | Observatory clock |
| A pulse travels from Earth to the Moon | Earth emission and lunar reflection | No physical clock measures both events |
| A moving transponder receives and later sends a reply | Reception and reply transmission | Transponder clock |
Three common mistakes are worth avoiding.
Mistake 1: Treating emission and reception as the same event.
They are different physical events at different locations. A signal takes time to travel between them.
Mistake 2: Assuming the observer who receives a signal measures the time of emission.
The receiver measures when the signal arrives. To reconstruct when it was emitted, you must account for the signal’s travel time.
Mistake 3: Choosing the smallest time interval as proper time.
Proper time is identified by the event pair and the clock present at both events. It isn’t identified by comparing the numbers first.
A useful final check is:
\[ \Delta t\geq\Delta\tau \]
for the same two timelike-separated events, when comparing the proper time with the interval in another inertial frame.
But don’t use this inequality as your first decision. Identify the events before checking the numbers.
For a broader frame-by-frame approach, see How to Identify Proper Time in HSC Special Relativity.
05Questions and solutions
Question 1
A detector is fixed to a laboratory bench. It records two particle arrivals \(5.0\ \mu\mathrm{s}\) apart according to its own clock.
A spacecraft passes the laboratory at \(0.70c\).
Identify the proper time between the two arrival events and explain which clock measures it.
Solution 1
The proper time is \(5.0\ \mu\mathrm{s}\), measured by the laboratory detector’s clock.
Both events occur at the same physical location in the laboratory frame: the detector.
The detector’s clock is present at both events, so it directly records the proper-time interval:
\[ \Delta\tau=5.0\ \mu\mathrm{s} \]
The spacecraft’s speed does not change which clock measures the proper time for this pair.
An observer travelling with the spacecraft would describe the detector as moving. The two detection events would then occur at different positions in the spacecraft frame.
The important distinction is that the proper time belongs to the two detection events, not to whichever observer is moving through the laboratory.
Question 2
A detector is fixed inside a spacecraft travelling at \(0.80c\) relative to Earth.
The detector records two events, A and B. Earth-frame synchronised clocks at the positions of those events measure a time interval of \(6.0\ \mu\mathrm{s}\).
Calculate the proper time between A and B.
Explain why the spacecraft’s onboard detector can measure the proper time even though Earth observes the spacecraft moving between the events.
Solution 2
The proper time is \(3.6\ \mu\mathrm{s}\), measured by the spacecraft’s onboard detector.
The detector remains fixed relative to the spacecraft. Therefore, it is present at both events in the spacecraft’s rest frame.
The given \(6.0\ \mu\mathrm{s}\) is the Earth-frame coordinate time, not the proper time.
Step 1: Calculate the Lorentz factor.
\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.80^2}}\\ &=\frac{1}{\sqrt{0.36}}\\ &=1.67 \end{aligned} \]
Step 2: Rearrange the time-dilation equation.
\[ \Delta t=\gamma\Delta\tau \]
Therefore:
\[ \Delta\tau=\frac{\Delta t}{\gamma} \]
Step 3: Substitute the values.
\[ \begin{aligned} \Delta\tau&=\frac{6.0}{1.67}\\ &=3.59\ \mu\mathrm{s} \end{aligned} \]
To two significant figures:
\[ \Delta\tau=3.6\ \mu\mathrm{s} \]
Earth assigns \(6.0\ \mu\mathrm{s}\) to the two events, while the spacecraft clock records \(3.6\ \mu\mathrm{s}\).
The trap is thinking that events occurring at different Earth-frame positions cannot have a proper time. They can, provided there is another frame in which a single clock is present at both events.
Question 3
A light clock is fixed inside a spacecraft. Its two mirrors are separated by \(300\ \mathrm{m}\) in the spacecraft’s rest frame.
A light pulse leaves the lower mirror at event A, reflects from the upper mirror at event C, and returns to the lower mirror at event B.
The spacecraft travels at \(0.80c\) relative to Earth.
The onboard clock measures \(2.00\ \mu\mathrm{s}\) between A and B.
(a) Identify the proper-time event pair.
(b) Calculate the time interval between A and B in Earth’s frame.
(c) Explain why the interval from A to C is not the proper time measured by the spacecraft clock.
Solution 3
The proper-time pair is A to B, and the proper time is \(2.00\ \mu\mathrm{s}\). Earth assigns \(3.33\ \mu\mathrm{s}\) to the same pair.
Step 1: Identify the events.
Event A is the pulse leaving the lower mirror.
Event B is the pulse returning to the lower mirror.
Both events occur at the same position in the spacecraft’s rest frame. The onboard clock can measure both, so:
\[ \Delta\tau_{AB}=2.00\ \mu\mathrm{s} \]
Step 2: Calculate the Lorentz factor.
\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.80^2}}\\ &=\frac{1}{0.60}\\ &=1.67 \end{aligned} \]
Step 3: Calculate the Earth-frame interval.
\[ \begin{aligned} \Delta t_{AB}&=\gamma\Delta\tau_{AB}\\ &=1.67(2.00)\\ &=3.34\ \mu\mathrm{s} \end{aligned} \]
Using the unrounded Lorentz factor:
\[ \Delta t_{AB}=3.33\ \mu\mathrm{s} \]
Step 4: Consider the reflection event.
Event C occurs at the upper mirror, while A and B occur at the lower mirror.
Therefore, A and C are not co-located in the spacecraft frame.
The interval A to C is the time taken by the light pulse to travel between the mirrors. It is not the proper time measured by the lower-mirror clock.
This is a useful distinction: a light pulse can travel away from a clock and return to it. The clock’s proper time is measured between the departure and return events, not between departure and reflection.
Question 4
A spacecraft carries one atomic clock throughout a journey.
In Earth’s reference frame, the spacecraft travels:
- At \(0.60c\) for \(5.0\ \mu\mathrm{s}\).
- At \(0.80c\) for a further \(3.0\ \mu\mathrm{s}\).
Assume the change in speed is instantaneous, so the acceleration interval can be neglected.
Calculate the total proper time recorded by the spacecraft clock.
A student suggests averaging the two speeds and using one Lorentz factor for the entire \(8.0\ \mu\mathrm{s}\) journey. Explain why this approach is incorrect.
Solution 4
The spacecraft clock records a total proper time of \(5.8\ \mu\mathrm{s}\).
The two sections must be calculated separately because the spacecraft has a different speed during each section.
Step 1: Calculate the proper time for the first section.
The speed is \(0.60c\).
\[ \gamma_1=\frac{1}{\sqrt{1-0.60^2}}=1.25 \]
Using:
\[ \Delta\tau_1=\frac{\Delta t_1}{\gamma_1} \]
we obtain:
\[ \begin{aligned} \Delta\tau_1&=\frac{5.0}{1.25}\\ &=4.0\ \mu\mathrm{s} \end{aligned} \]
Step 2: Calculate the proper time for the second section.
The speed is \(0.80c\).
\[ \gamma_2=\frac{1}{\sqrt{1-0.80^2}}=1.67 \]
Therefore:
\[ \begin{aligned} \Delta\tau_2&=\frac{3.0}{1.67}\\ &=1.8\ \mu\mathrm{s} \end{aligned} \]
Step 3: Add the elapsed times recorded by the spacecraft clock.
\[ \begin{aligned} \Delta\tau_{\mathrm{total}} &=\Delta\tau_1+\Delta\tau_2\\ &=4.0+1.8\\ &=5.8\ \mu\mathrm{s} \end{aligned} \]
The spacecraft records \(5.8\ \mu\mathrm{s}\) between the starting and finishing events.
The student’s average-speed method fails because the Lorentz factor depends non-linearly on speed.
More importantly, the clock accumulates time throughout its actual journey. Each constant-speed section contributes its own elapsed proper time.
You cannot replace two different motion conditions with one average speed and expect the same clock reading.
Question 5
An Earth-based radar station is located at \(x=0\). It sends two short laser pulses towards a spacecraft travelling directly away from it at \(0.60c\).
At the moment the first pulse is sent, the spacecraft is \(1.20\ \mathrm{km}\) from the station.
The station sends the pulses \(4.00\ \mu\mathrm{s}\) apart, measured by its own clock.
The spacecraft reflects each pulse immediately when it arrives. It then sends each reflected pulse back towards Earth.
Define:
- A1 and A2 as the two emission events at Earth.
- C1 and C2 as the two reflection events at the spacecraft.
- B1 and B2 as the two reception events at Earth.
Calculate:
(a) The proper time between A1 and A2.
(b) The Earth-frame time interval between C1 and C2.
(c) The proper time between C1 and C2, measured by the spacecraft.
(d) The time interval between B1 and B2, measured by the Earth station.
(e) Explain why the interval between the two reflections is not simply the \(4.00\ \mu\mathrm{s}\) emission interval multiplied or divided by the Lorentz factor.
Assume constant spacecraft speed, instantaneous reflection, and light travelling at \(3.00\times10^8\ \mathrm{m\,s^{-1}}\) in a vacuum. Ignore gravitational effects and electronic delays. Take the positive \(x\)-direction to be away from Earth.
Solution 5
The four requested intervals are:
- A1 to A2: \(4.00\ \mu\mathrm{s}\), proper time on the Earth station.
- C1 to C2: \(10.0\ \mu\mathrm{s}\), Earth-frame coordinate time.
- C1 to C2: \(8.00\ \mu\mathrm{s}\), proper time on the spacecraft.
- B1 to B2: \(16.0\ \mu\mathrm{s}\), proper time on the Earth station.
The surprising result is that the spacecraft’s proper time between reflections is longer than the station’s proper time between emissions.
There is no contradiction. These are different pairs of events.
Step 1: Identify the proper time between the emissions.
Both A1 and A2 occur at the same Earth station.
The station clock is present at both events, so:
\[ \Delta\tau_{A1A2}=4.00\ \mu\mathrm{s} \]
Step 2: Find the first reflection event, C1.
The first pulse starts \(1.20\times10^3\ \mathrm{m}\) from the spacecraft.
The spacecraft moves away at \(0.60c\), so the closing speed between the pulse and spacecraft is:
\[ c-v=0.40c \]
The time taken for the first pulse to reach the spacecraft is:
\[ \begin{aligned} t_{C1}&=\frac{1200}{0.40(3.00\times10^8)}\\ &=10.0\ \mu\mathrm{s} \end{aligned} \]
The reflection position is:
\[ \begin{aligned} x_{C1}&=ct_{C1}\\ &=(3.00\times10^8)(10.0\times10^{-6})\\ &=3000\ \mathrm{m} \end{aligned} \]
Step 3: Find the second reflection event, C2.
The second pulse leaves Earth \(4.00\ \mu\mathrm{s}\) after the first.
During this time, the spacecraft moves further away.
Its position when the second pulse is emitted is:
\[ \begin{aligned} x&=1200+(0.60)(3.00\times10^8)(4.00\times10^{-6})\\ &=1200+720\\ &=1920\ \mathrm{m} \end{aligned} \]
The second pulse must now catch a spacecraft that is \(1920\ \mathrm{m}\) away.
Using the same closing speed:
\[ \begin{aligned} \Delta t_{\mathrm{second\ chase}} &=\frac{1920}{0.40(3.00\times10^8)}\\ &=16.0\ \mu\mathrm{s} \end{aligned} \]
This is measured from the second emission event.
Therefore, the second reflection occurs at Earth-frame time:
\[ t_{C2}=4.00+16.0 \]
\[ t_{C2}=20.0\ \mu\mathrm{s} \]
The Earth-frame interval between reflections is:
\[ \begin{aligned} \Delta t_{C1C2}&=20.0-10.0\\ &=10.0\ \mu\mathrm{s} \end{aligned} \]
Step 4: Calculate the spacecraft’s proper time between reflections.
Both reflections occur at the same transponder fixed to the spacecraft.
Its onboard clock is present at C1 and C2.
Therefore, the proper time is the Earth-frame interval divided by the Lorentz factor.
First:
\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.60^2}}\\ &=1.25 \end{aligned} \]
Then:
\[ \begin{aligned} \Delta\tau_{C1C2} &=\frac{\Delta t_{C1C2}}{\gamma}\\ &=\frac{10.0}{1.25}\\ &=8.00\ \mu\mathrm{s} \end{aligned} \]
The spacecraft records \(8.00\ \mu\mathrm{s}\) between reflections.
Step 5: Calculate the two return journeys.
The first reflected pulse returns from \(x=3000\ \mathrm{m}\).
Its return travel time is:
\[ \begin{aligned} t_{\mathrm{return,1}} &=\frac{3000}{3.00\times10^8}\\ &=10.0\ \mu\mathrm{s} \end{aligned} \]
It reaches Earth at:
\[ t_{B1}=10.0+10.0=20.0\ \mu\mathrm{s} \]
The second reflected pulse returns from:
\[ \begin{aligned} x_{C2}&=c(t_{C2}-t_{A2})\\ &=(3.00\times10^8)(16.0\times10^{-6})\\ &=4800\ \mathrm{m} \end{aligned} \]
Its return travel time is:
\[ \begin{aligned} t_{\mathrm{return,2}} &=\frac{4800}{3.00\times10^8}\\ &=16.0\ \mu\mathrm{s} \end{aligned} \]
It reaches Earth at:
\[ t_{B2}=20.0+16.0 \]
\[ t_{B2}=36.0\ \mu\mathrm{s} \]
Step 6: Calculate the interval between receptions.
Both B1 and B2 occur at the same Earth station.
Therefore, the station clock measures the proper time between them.
\[ \begin{aligned} \Delta\tau_{B1B2} &=t_{B2}-t_{B1}\\ &=36.0-20.0\\ &=16.0\ \mu\mathrm{s} \end{aligned} \]
The station measures \(16.0\ \mu\mathrm{s}\) between the returning pulses.
Step 7: Explain why the intervals differ.
The tempting approach is to take the \(4.00\ \mu\mathrm{s}\) interval between emissions and immediately apply time dilation to find the interval between reflections.
That fails because the emission events and reflection events are not the same event pair.
The second pulse has to catch a spacecraft that has moved further away since the first pulse was emitted.
Consequently, the reflection events are \(10.0\ \mu\mathrm{s}\) apart in Earth’s frame, not \(4.00\ \mu\mathrm{s}\) apart.
Time dilation then converts that \(10.0\ \mu\mathrm{s}\) interval into \(8.00\ \mu\mathrm{s}\) on the spacecraft clock.
The returning pulses introduce another difference. The second pulse has a longer return journey, so the reception events are \(16.0\ \mu\mathrm{s}\) apart.
All four results are consistent because they describe different event pairs.
This is the important decision to make before choosing proper time: identify exactly what happened at each endpoint, and establish which single clock could have been present at both.
Once you’ve done that, the time-dilation calculation becomes much more straightforward.