How to Convert Relativistic Journey Times Between Frames

A spacecraft reaches a distant checkpoint 10.0 seconds after leaving Earth. Its pilot’s clock has advanced only 8.00 seconds. Yet the mission controller back at the launch site doesn’t receive the arrival message until 16.0 seconds after departure.

Which time is the actual journey time: 8, 10, or 16 seconds?

All three measurements can be correct. The spacecraft experiences 8.00 s, Earth assigns 10.0 s between departure and arrival, and the remaining 6.00 s is the time taken for the arrival message to travel back to Earth.

To convert a relativistic journey time between reference frames, identify the departure and arrival events, determine which frame measures their proper time, and apply:

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

where \(\Delta\tau\) is the proper time, \(\Delta t\) is the longer time interval measured in a frame where the two events occur at different positions, and \(\gamma\) is the Lorentz factor.

The difficult part usually isn’t calculating \(\gamma\). It’s deciding which measurements belong in the equation.

01Identify proper time before using the equation

Imagine a spacecraft moving past two stationary checkpoints.

At the first checkpoint, its pilot starts a stopwatch. At the second, the pilot stops it.

Here’s the question: which reference frame measures the proper time between these events?

The spacecraft does, because both events happen at the same position in the spacecraft’s reference frame: where the pilot is sitting.

The Earth-based checkpoints see something different. The spacecraft passes the first checkpoint at one position and the second checkpoint somewhere else.

This gives us the most useful test for proper time:

Find the reference frame in which both events occur at the same position. A clock at rest at that position measures the proper time between them.

For a constant-speed spacecraft journey, that is the spacecraft’s frame.

The same journey, described in two frames

Return to our opening example.

Assume the checkpoint is stationary relative to Earth, at a distance of \(1.80\times10^9\ \mathrm{m}\). The spacecraft moves at a constant \(0.600c\), where \(c\) is the speed of light in a vacuum.

We will treat Earth as an inertial reference frame, neglect gravity, and assume the spacecraft passes the checkpoints without accelerating.

The measurements are:

In the spacecraft frame, its clock remains at the same position while the Earth checkpoints move past it.

Neither description is an illusion. The frames assign different elapsed times to the same two physical events.

Calculate the time-dilation factor

For constant relative speed, the Lorentz factor is:

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

Here:

• \(\gamma\) is the Lorentz factor, with no units.
• \(v\) is the relative speed between the two frames, in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\) is the speed of light in a vacuum.

At \(v=0.600c\):

\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-(0.600)^2}}\\ &=\frac{1}{0.800}\\ &=1.25 \end{aligned} \]

The Earth-frame interval is therefore 1.25 times the spacecraft’s proper time.

\[ \begin{aligned} \Delta t_{\mathrm E} &=\gamma\Delta\tau\\ &=1.25(8.00\ \mathrm{s})\\ &=10.0\ \mathrm{s} \end{aligned} \]

To convert in the opposite direction:

\[ \Delta\tau=\frac{\Delta t_{\mathrm E}}{\gamma} \]

The spacecraft’s elapsed time is shorter, but its clock isn’t malfunctioning. It measures the elapsed time along a different path through spacetime.

Which time belongs to which measurement?

The symbol \(\Delta\) means an interval or change.

For our spacecraft journey:

• \(\Delta t_{\mathrm E}\) is the Earth-frame time between departure and arrival.
• \(\Delta\tau\) is the proper time recorded by the spacecraft.
• \(t_{\mathrm{receive}}\) is the time when Earth receives a message about arrival.

That third quantity is not automatically a journey time.

Before calculating, ask: are these timestamps for the actual events, or for when someone learned about them?

02Converting journey times: the worked method

For a spacecraft moving at constant velocity between two stationary checkpoints, use this sequence:

1. Identify the departure and arrival events. The spacecraft is present at both, so its clock measures proper time.
2. If the Earth-frame distance \(L\) and speed \(v\) are given, calculate the Earth-frame journey time using \(\Delta t_{\mathrm E}=L/v\).
3. Calculate \(\gamma\) using the spacecraft’s speed relative to Earth.
4. Convert with \(\Delta\tau=\Delta t_{\mathrm E}/\gamma\), or multiply by \(\gamma\) when converting from proper time to Earth-frame time.
5. Check whether any supplied timestamps include light-signal travel time.

Notice that both distance and speed in Step 2 must belong to the same reference frame.

03The hidden problem: receiving a signal isn’t the same as timing an event

Return to the mission controller who receives an arrival message 16.0 seconds after launch.

Would it be reasonable to divide 16.0 seconds by the spacecraft’s proper time to calculate \(\gamma\)?

It might feel reasonable. Both timestamps appear on the same mission control clock.

But the 16.0-second interval ends when the message reaches Earth, not when the spacecraft reaches its destination.

The arrival event occurs at the distant checkpoint. The message reception occurs later, back at the launch station.

These are different events.

Correct the signal travel time first

For a light signal travelling directly through a vacuum to a stationary receiver:

\[ t_{\mathrm{receive}} =t_{\mathrm{emit}}+\frac{\ell}{c} \]

where:

• \(t_{\mathrm{receive}}\) is the time the receiver detects the signal, in seconds.
• \(t_{\mathrm{emit}}\) is the time the signal was emitted, in the same reference frame, in seconds.
• \(\ell\) is the signal’s travel distance in that frame, in metres.
• \(c\) is the speed of light in a vacuum.

For the opening journey, the message travels \(1.80\times10^9\ \mathrm{m}\) back to Earth.

\[ \begin{aligned} t_{\mathrm{light}} &=\frac{1.80\times10^9}{3.00\times10^8}\\ &=6.00\ \mathrm{s} \end{aligned} \]

So the message arrives at:

\[ t_{\mathrm{receive}}=10.0+6.00=16.0\ \mathrm{s} \]

The 16.0 seconds contains two separate intervals: 10.0 seconds for the spacecraft’s journey in Earth’s frame, followed by 6.00 seconds for the return signal.

A contrasting case: What if the two events happen on Earth?

Suppose a lamp fixed at the Earth launch station flashes twice, 8.00 seconds apart according to a clock beside it.

A spacecraft moves past at \(0.600c\).

Which frame measures proper time now?

Earth does. Both flashes occur at the same position in Earth’s frame.

The spacecraft frame assigns a longer coordinate-time interval:

\[ \begin{aligned} \Delta t_{\mathrm{ship}} &=\gamma\Delta\tau_{\mathrm E}\\ &=1.25(8.00)\\ &=10.0\ \mathrm{s} \end{aligned} \]

Notice the change. In the journey example, the spacecraft measured proper time. In the lamp example, Earth measures proper time.

The equation hasn’t changed. The pair of events has.

A single spacecraft clock isn’t present at both Earth-lamp flashes. The 10.0 seconds is the interval between those events in the spacecraft’s reference frame, determined using synchronised clocks at the relevant positions in that frame.

This is why you should never memorise a rule such as “Earth time is always dilated”. Proper time depends on the events.

04How an experiment tested whether the time conversion was real

The equations predict that a fast-moving clock accumulates less elapsed time between suitable events than clocks in Earth’s frame.

But how could scientists test that without taking an impossibly fast spacecraft into space?

One useful natural clock is the muon.

A muon is an unstable subatomic particle that decays after a random amount of time. At rest, its mean lifetime is approximately \(2.20\ \mu\mathrm{s}\), where one microsecond is \(10^{-6}\) seconds.

Cosmic rays produce muons high in Earth’s atmosphere. Many travel downwards at speeds close to \(c\).

That creates an experimental puzzle: how can so many muons survive the journey down through the atmosphere when their mean lifetime at rest is so short?

The prediction: too much travel time without dilation

In a 1963 experiment, David Frisch and James Smith investigated this effect using cosmic-ray muons. Their work was a refined test building on earlier muon experiments, including work by Bruno Rossi and David Hall.

The experiment compared muons travelling from approximately 1,907 m above sea level to sea level, selecting particles with speeds near \(0.995c\).

Start with a simplified constant-speed estimate.

Using \(v=0.9952c\):

\[ \begin{aligned} \Delta t_{\mathrm E} &=\frac{1907}{0.9952(3.00\times10^8)}\\ &=6.39\times10^{-6}\ \mathrm{s}\\ &=6.39\ \mu\mathrm{s} \end{aligned} \]

Earth assigns about 6.39 microseconds to the descent.

Now convert that interval into the muon’s proper time.

\[ \begin{aligned} \gamma &=\frac{1}{\sqrt{1-(0.9952)^2}}\\ &\approx10.2 \end{aligned} \]

Therefore:

\[ \begin{aligned} \Delta\tau &=\frac{6.39}{10.2}\ \mu\mathrm{s}\\ &\approx0.625\ \mu\mathrm{s} \end{aligned} \]

The same descent takes approximately 6.39 microseconds in Earth’s frame but just 0.625 microseconds of proper time.

The muon’s mean rest lifetime is 2.20 microseconds. A proper flight time of about 0.625 microseconds makes substantial survival plausible.

This does not mean every muon survives. Radioactive decay is random, and 2.20 microseconds is a mean lifetime, not a fixed expiry time.

The calculation also assumes constant speed and a vertical path. Real muons lose some energy as they travel through the atmosphere, so an accurate experimental prediction needs corrections.

The apparatus: selecting and counting muons

Frisch and Smith needed to compare the number of suitably selected muons at two altitudes.

Their detector used a physical arrangement that could select particles according to their ability to penetrate matter.

From top to bottom:

1. A thick layer of iron slowed incoming particles and stopped those that could not penetrate far enough.
2. Beneath the iron was a plastic scintillator, a material that produces flashes of light when charged particles pass through it.
3. A photomultiplier detected those flashes and converted them into electrical pulses.
4. Electronic circuits and an oscilloscope helped identify events in which muons stopped in the scintillator and subsequently decayed.

The time between the initial pulse and a later decay pulse gave information about the lifetime of muons after they had stopped.

The iron and scintillator also helped select muons in a narrow range of incoming speeds.

Why the detector design needed an adjustment

A simple plan would be to count muons high on a mountain, move the same detector to sea level, and compare the readings.

But there is a problem.

Muons lose energy as they pass through air. A muon reaching sea level is generally slower than it was when passing the mountain’s altitude.

If the same thickness of iron were used at both locations, the detector would select different initial ranges of muon energies.

That makes the count comparison unreliable.

Frisch and Smith used less iron at sea level to compensate for the additional slowing in the atmosphere. This helped them compare muons selected from a similar range of energies at the mountain’s altitude.

The experiment also had to account for particle trajectories, detector selection, and other physical effects.

The observation and its meaning

In their reported measurements, Frisch and Smith obtained average hourly decay-event counting rates of approximately 563 at the mountain and 408 at sea level, using their selected muon population.

That was far more survival than would be expected if the particles’ short rest-frame mean lifetime were simply used as their Earth-frame mean lifetime.

After allowing for the experimental conditions, they reported an observed time-dilation factor of approximately \(8.8\pm0.8\), consistent within uncertainty with their relativistic prediction.

That experimental factor is not identical to the simple \(\gamma\approx10.2\) estimate above. The simple calculation assumes exactly constant speed and an ideal vertical path, while the experiment involved energy loss, a range of particle speeds, and detector geometry.

The important result is that the observed survival was consistent with muons accumulating less proper time than the Earth-frame travel time.

The detectors did not directly follow one muon and read its personal clock at both altitudes. They used many muons and their statistical decay behaviour.

What the experiment established: the decay behaviour of moving muons was consistent with the elapsed-time relationship predicted by special relativity.

What it did not establish: that each muon lives for a fixed time, that all muons survive a given journey, or that signal delays explain the result.

For more on how proper lifetime is used to predict travel distance, see How to Calculate the Distance a Muon Travels Before Decaying.

05A decision table for HSC journey-time questions

Once you recognise the measurement, choosing the equation becomes much easier.

Three checks can catch most mistakes:

• The Lorentz factor must satisfy \(\gamma\geq1\).
• For two events co-located in one inertial frame, that frame measures the shortest time interval between those events.
• A signal reception time is not automatically the time of the event that produced the signal.

The first two points apply directly to inertial frames and constant-speed motion. If a spacecraft accelerates or turns around, its accumulated proper time must be calculated along its journey, rather than using one constant-velocity reference frame for the whole trip.

All the questions below use special relativity without gravitational corrections. Unless stated otherwise, assume ideal clocks, vacuum light propagation, and an inertial Earth frame.

06Questions and solutions

Question 1

A spacecraft travels in a straight line between two stationary space stations separated by \(2.40\times10^9\ \mathrm{m}\), measured in Earth’s reference frame.

Its constant speed is \(0.800c\).

(a) Calculate the journey time in Earth’s reference frame.

(b) Calculate the elapsed time recorded by the spacecraft.

(c) Identify the proper time and explain why.

Solution 1

The Earth-frame journey takes \(10.0\ \mathrm{s}\), while the spacecraft records \(6.00\ \mathrm{s}\) of proper time.

(a) Earth-frame journey time

The stations are stationary in Earth’s frame, so use their separation with the spacecraft’s Earth-frame speed.

\[ \begin{aligned} \Delta t_{\mathrm E} &=\frac{L}{v}\\ &=\frac{2.40\times10^9}{0.800(3.00\times10^8)}\\ &=10.0\ \mathrm{s} \end{aligned} \]

(b) Spacecraft elapsed time

At \(0.800c\):

\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.800^2}}\\ &=1.667 \end{aligned} \]

Then:

\[ \begin{aligned} \Delta\tau&=\frac{\Delta t_{\mathrm E}}{\gamma}\\ &=\frac{10.0}{1.667}\\ &=6.00\ \mathrm{s} \end{aligned} \]

(c) Proper time

The \(6.00\ \mathrm{s}\) interval is proper time because the spacecraft clock is present at both the departure and arrival events.

The \(10.0\ \mathrm{s}\) interval is the Earth-frame coordinate-time interval between events at different Earth-frame positions.

The common mistake is to call the Earth interval proper time just because Earth’s clocks are stationary. The location of the two events determines which inertial frame measures proper time.

Question 2

A spacecraft leaves a stationary Earth receiver at \(t=0\), travelling in the positive \(x\)-direction at \(0.800c\).

Ahead of it is an automated station, initially \(1.80\times10^9\ \mathrm{m}\) away. The station is also moving in the positive \(x\)-direction, at \(0.200c\) relative to Earth.

When the spacecraft catches the station, it immediately transmits a light signal directly back to the Earth receiver.

Calculate:

(a) the Earth-frame time until the spacecraft catches the station,

(b) the proper time experienced by the spacecraft up to the meeting, and

(c) the time after departure when Earth receives the signal.

Assume constant velocities and vacuum light propagation.

Solution 2

The spacecraft catches the moving station after \(10.0\ \mathrm{s}\) in Earth’s frame, experiences \(6.00\ \mathrm{s}\) of proper time, and Earth receives the signal at \(18.0\ \mathrm{s}\).

(a) Find the meeting time

The station is moving away, so the spacecraft must travel farther than the station’s initial distance.

Let \(t\) be the meeting time measured in Earth’s frame.

The spacecraft’s position is:

\[ x_{\mathrm{ship}}=(0.800c)t \]

The station’s position is:

\[ x_{\mathrm{station}} =1.80\times10^9+(0.200c)t \]

At the meeting, their positions are equal.

\[ \begin{aligned} (0.800c)t &=1.80\times10^9+(0.200c)t\\ (0.600c)t&=1.80\times10^9\\ t&=\frac{1.80\times10^9}{1.80\times10^8}\\ &=10.0\ \mathrm{s} \end{aligned} \]

The spacecraft takes 10.0 Earth-frame seconds to catch the station.

Using the initial separation divided by \(0.800c\) would be wrong because the destination moves during the journey.

(b) Convert to proper time

The spacecraft’s speed relative to Earth is \(0.800c\), not the closing speed of \(0.600c\).

Therefore:

\[ \begin{aligned} \Delta\tau &=10.0\sqrt{1-(0.800)^2}\\ &=10.0(0.600)\\ &=6.00\ \mathrm{s} \end{aligned} \]

The closing speed determines when the two objects meet. The spacecraft’s speed relative to Earth determines the time-dilation factor.

(c) Calculate reception time

The spacecraft’s position when it meets the station is:

\[ \begin{aligned} x&=(0.800c)(10.0)\\ &=2.40\times10^9\ \mathrm{m} \end{aligned} \]

The signal travels this distance back to the Earth receiver.

\[ \begin{aligned} t_{\mathrm{light}} &=\frac{2.40\times10^9}{3.00\times10^8}\\ &=8.00\ \mathrm{s} \end{aligned} \]

Therefore:

\[ t_{\mathrm{receive}}=10.0+8.00=18.0\ \mathrm{s} \]

The 18.0-second reception timestamp doesn’t measure the spacecraft’s journey. It includes an additional eight seconds of signal propagation.

This problem requires three different speeds to be handled correctly: the spacecraft’s speed, the closing speed, and the return signal’s speed.

Question 3

An experimental team wants to estimate the proper time experienced by cosmic-ray muons travelling between two laboratories separated vertically by \(1.80\ \mathrm{km}\).

The selected muons have a speed of approximately \(0.9950c\) at the upper laboratory. Their mean lifetime at rest is \(2.20\ \mu\mathrm{s}\).

For parts (a) and (b), assume the muons travel straight down at constant speed and suffer no absorption.

(a) Calculate their Earth-frame descent time and elapsed proper time.

(b) Two students make the following predictions:

• Student A: “The Earth-frame flight time is longer than 2.20 microseconds, so no muons should survive.”
• Student B: “The proper flight time is shorter than 2.20 microseconds, so every muon should survive.”

Explain why neither prediction is correct.

(c) The team uses a detector with an iron absorber above a plastic scintillator at each altitude. Muons lose energy while travelling through the air. Explain why keeping the iron absorber thickness identical at both sites could make the count comparison misleading, and describe a suitable adjustment.

Solution 3

The Earth-frame descent takes approximately \(6.03\ \mu\mathrm{s}\), while the muons experience approximately \(0.602\ \mu\mathrm{s}\) of proper time. Neither student can predict certain survival or certain decay, and the lower detector must account for energy lost in the atmosphere.

(a) Calculate both time intervals

The distance is:

\[ L=1.80\times10^3\ \mathrm{m} \]

First calculate the Earth-frame flight time:

\[ \begin{aligned} \Delta t_{\mathrm E} &=\frac{L}{v}\\ &=\frac{1.80\times10^3}{0.9950(3.00\times10^8)}\\ &=6.03\times10^{-6}\ \mathrm{s}\\ &=6.03\ \mu\mathrm{s} \end{aligned} \]

The Lorentz factor is:

\[ \begin{aligned} \gamma &=\frac{1}{\sqrt{1-(0.9950)^2}}\\ &\approx10.0 \end{aligned} \]

Therefore:

\[ \begin{aligned} \Delta\tau &=\frac{\Delta t_{\mathrm E}}{\gamma}\\ &\approx\frac{6.03}{10.0}\ \mu\mathrm{s}\\ &=0.602\ \mu\mathrm{s} \end{aligned} \]

The proper flight time is about one-tenth of the Earth-frame flight time.

(b) Evaluate the predictions

Both students are incorrect.

Student A treats the mean rest lifetime as though every muon must decay after exactly 2.20 microseconds.

That is not what a mean lifetime means. Individual muons decay at random times, so some survive much longer than the mean, while others decay sooner.

Student B makes the opposite mistake. A proper flight time shorter than the mean lifetime does not guarantee survival. Some muons decay before 0.602 microseconds of proper time has elapsed.

Time dilation changes the decay-time distribution as measured in Earth’s frame. It does not remove the randomness of decay.

(c) Evaluate and improve the apparatus

Keeping identical iron thicknesses is not necessarily a fair comparison.

A muon reaching the lower detector has passed through more atmosphere and lost additional energy. The same thickness of iron may therefore select a different population of muons than it selected at the upper laboratory.

A suitable adjustment is to reduce the lower detector’s iron absorber thickness to compensate for the extra slowing in the air, while controlling the detector’s acceptance and background.

The exact adjustment requires knowledge of energy loss in the materials.

The important experimental principle is to compare equivalent populations. Otherwise, a change in counting rate might partly reflect detector selection rather than radioactive decay.

The calculation in part (a) is an idealised scale estimate. A full experimental analysis must also account for the muons’ changing speeds and paths.

Question 4

A spacecraft travels at an unknown constant speed in the positive \(x\)-direction.

Two stationary receivers, L and R, are positioned at \(x=-4.00\) light-seconds and \(x=+4.00\) light-seconds in Earth’s frame.

One light-second is the distance light travels through a vacuum in one second:

\[ 1\ \text{light-second}=3.00\times10^8\ \mathrm{m} \]

The spacecraft emits its first flash at \(x=0\) and \(t=0\). It emits a second flash later, somewhere between the receivers.

The receivers record the following Earth-frame timestamps:

Determine:

(a) the Earth-frame position and emission time of the second flash,

(b) the spacecraft’s speed relative to Earth,

(c) the proper time between the flashes, and

(d) why the receivers record different time intervals for the same pair of flashes.

Assume light travels directly to each receiver at speed \(c\).

Solution 4

The second flash is emitted at \(t=5.00\ \mathrm{s}\), at \(x=+3.00\) light-seconds. The spacecraft travels at \(0.600c\), and its proper time between flashes is \(4.00\ \mathrm{s}\).

The receivers disagree about the arrival interval because the second flash travels different distances to them.

(a) Reconstruct the actual second emission

Let \(t_2\) be the second flash’s Earth-frame emission time.

Define:

\[ \xi=\frac{x_2}{c} \]

where \(x_2\) is the second emission’s Earth-frame position in metres, and \(\xi\) is measured in seconds.

Since the second emission occurs between the receivers, its light-travel times are \(4.00\ \mathrm{s}+\xi\) to L and \(4.00\ \mathrm{s}-\xi\) to R.

Receiver L records:

\[ 12.0\ \mathrm{s} =t_2+4.00\ \mathrm{s}+\xi \]

Therefore:

\[ t_2+\xi=8.00\ \mathrm{s} \]

Receiver R records:

\[ 6.00\ \mathrm{s} =t_2+4.00\ \mathrm{s}-\xi \]

Therefore:

\[ t_2-\xi=2.00\ \mathrm{s} \]

Add the equations:

\[ \begin{aligned} 2t_2&=10.0\ \mathrm{s}\\ t_2&=5.00\ \mathrm{s} \end{aligned} \]

Then:

\[ \begin{aligned} \xi&=3.00\ \mathrm{s}\\ x_2&=c\xi\\ &=9.00\times10^8\ \mathrm{m} \end{aligned} \]

The second flash occurs 3.00 light-seconds to the right of the origin.

(b) Determine the spacecraft speed

The spacecraft moves 3.00 light-seconds in 5.00 Earth-frame seconds.

\[ \begin{aligned} v&=\frac{9.00\times10^8}{5.00}\\ &=1.80\times10^8\ \mathrm{m\,s^{-1}}\\ &=0.600c \end{aligned} \]

(c) Convert to proper time

Both flashes come from the same source fixed aboard the spacecraft. They are co-located in its reference frame.

At \(0.600c\), \(\gamma=1.25\).

\[ \begin{aligned} \Delta\tau&=\frac{5.00}{1.25}\\ &=4.00\ \mathrm{s} \end{aligned} \]

(d) Explain the different reception intervals

Receiver L records:

\[ 12.0-4.00=8.00\ \mathrm{s} \]

Receiver R records:

\[ 6.00-4.00=2.00\ \mathrm{s} \]

The first flash travels 4.00 light-seconds to each receiver.

The second flash travels:

• 7.00 light-seconds to L, because the spacecraft has moved away from L.
• 1.00 light-second to R, because the spacecraft has moved towards R.

The Earth-frame interval between the two emissions is still 5.00 seconds.

The different signal paths change the recorded arrival intervals, not the elapsed time between the actual emission events.

The tempting route is to choose one receiver’s measured interval and use it directly in the time-dilation equation.

That would make the inferred proper time depend on which receiver you chose, which is physically wrong.

The correct method reconstructs the emission events first, then converts between frames.

Question 5

A spacecraft leaves Earth at \(t=0\), travels to a stationary outpost 3.00 light-seconds away, reverses direction, and returns to the same Earth station.

Its outward speed is \(0.600c\). Its return speed is \(0.800c\), both measured in Earth’s frame.

Assume the reversal takes negligible time, all speeds are constant between events, and gravity can be ignored.

The spacecraft transmits a light pulse towards Earth at the instant it reaches the outpost.

Three events are relevant:

• Event A: spacecraft departs Earth.
• Event B: spacecraft reaches the outpost and sends its pulse.
• Event C: spacecraft returns to Earth and reunites with a stationary Earth clock.

Two students disagree.

Student 1 argues:

“The spacecraft begins and ends at the same Earth position, so its average velocity over the entire journey is zero. Its Lorentz factor must be 1, and its clock must agree with Earth’s clock at reunion.”

Student 2 argues:

“The spacecraft is present at both A and C, so those events are co-located on the spacecraft. Therefore, the time measured by the spacecraft must be the single proper-time interval between A and C, and Earth’s clock cannot also measure proper time.”

(a) Calculate the Earth-frame times of A, B, and C.

(b) Calculate the total elapsed time recorded by the spacecraft and by the stationary Earth clock at reunion.

(c) Determine when Earth receives the pulse emitted at B, and how long before reunion the pulse arrives.

(d) Evaluate both students’ arguments. Explain how the two clocks can start together, reunite at the same event, and accumulate different elapsed times without contradicting the definition of proper time.

Solution 5

The spacecraft records \(6.25\ \mathrm{s}\), while Earth’s stationary clock records \(8.75\ \mathrm{s}\). Both students are incorrect because they have applied the constant-speed proper-time rule to a journey in which the spacecraft changes reference frames.

The light pulse from the turnaround reaches Earth at \(t=8.00\ \mathrm{s}\), just 0.750 seconds before the spacecraft returns.

(a) Determine the Earth-frame event times

Event A occurs at:

\[ t_A=0 \]

The outpost is 3.00 light-seconds away. Travelling at \(0.600c\), the spacecraft takes:

\[ \begin{aligned} \Delta t_{AB} &=\frac{3.00\ \text{light-seconds}}{0.600c}\\ &=5.00\ \mathrm{s} \end{aligned} \]

Here, one light-second divided by \(c\) equals one second.

Therefore:

\[ t_B=5.00\ \mathrm{s} \]

For the return at \(0.800c\):

\[ \begin{aligned} \Delta t_{BC} &=\frac{3.00\ \text{light-seconds}}{0.800c}\\ &=3.75\ \mathrm{s} \end{aligned} \]

So reunion occurs at:

\[ t_C=5.00+3.75=8.75\ \mathrm{s} \]

The Earth-frame event times are A at 0, B at 5.00 s, and C at 8.75 s.

(b) Calculate each clock’s elapsed time

The stationary Earth clock measures the entire interval from A to C:

\[ \Delta t_{\mathrm E}=8.75\ \mathrm{s} \]

For the spacecraft, calculate the outward and return proper times separately.

Outward journey at \(0.600c\):

\[ \begin{aligned} \Delta\tau_{AB} &=5.00\sqrt{1-(0.600)^2}\\ &=5.00(0.800)\\ &=4.00\ \mathrm{s} \end{aligned} \]

Return journey at \(0.800c\):

\[ \begin{aligned} \Delta\tau_{BC} &=3.75\sqrt{1-(0.800)^2}\\ &=3.75(0.600)\\ &=2.25\ \mathrm{s} \end{aligned} \]

Add the spacecraft’s elapsed proper times:

\[ \begin{aligned} \tau_{\mathrm{total}} &=4.00+2.25\\ &=6.25\ \mathrm{s} \end{aligned} \]

At reunion, the two clocks can be placed side by side and read directly.

The Earth clock reads 8.75 s, and the spacecraft clock reads 6.25 s.

(c) Determine when the turnaround pulse reaches Earth

The pulse is emitted at B, when Earth’s clock reads 5.00 s.

It must travel 3.00 light-seconds to Earth, taking 3.00 s.

Therefore:

\[ t_{\mathrm{receive}}=5.00+3.00=8.00\ \mathrm{s} \]

Reunion occurs at 8.75 s.

The interval between reception and reunion is:

\[ 8.75-8.00=0.750\ \mathrm{s} \]

The spacecraft does not complete its return journey in 0.750 s. It takes 3.75 s in Earth’s frame.

The 0.750-second interval begins when Earth receives the turnaround message, three seconds after the spacecraft actually turned around.

(d) Evaluate both students’ arguments

Student 1 is incorrect because average velocity cannot be substituted into a Lorentz factor for a changing-speed journey.

The spacecraft’s net displacement is zero, so its average Earth-frame velocity is zero. But it spends its journey moving at substantial speeds.

The Lorentz factor depends on the spacecraft’s speed during each section, not on the displacement divided by the total time.

Student 2 makes a subtler error.

It is true that the spacecraft is physically present at both A and C. Its clock measures proper time along its own journey.

However, the spacecraft reverses direction at B. There is no single inertial reference frame in which the spacecraft remains at rest throughout the entire A-to-C journey.

The Earth clock, by contrast, remains at the same Earth-frame position from A to C. Its \(8.75\ \mathrm{s}\) is the proper-time interval along the stationary, inertial Earth-clock path.

The spacecraft accumulates \(6.25\ \mathrm{s}\) of proper time along a different path through spacetime.

These are not two incompatible answers for one clock. They are elapsed times accumulated by two different clocks following different journeys between the same events.

For two events connected by an inertial clock’s worldline, that clock accumulates more proper time than a clock that leaves it, changes velocity, and reunites with it. Here, the two clocks differ by:

\[ \begin{aligned} 8.75-6.25 &=2.50\ \mathrm{s} \end{aligned} \]

The light pulse doesn’t change either result. It creates an additional reception event, with its own timestamp.

This is the deeper reason the two-event method needs care.

For a single constant-speed journey, identifying co-located events in the traveller’s inertial frame lets you apply one Lorentz factor.

For a changing-speed journey, the traveller’s elapsed time comes from adding the proper times along the separate sections of its path.

And for an arrival message, the first job is still to identify when the actual event happened.

07Where this method leads next

When an HSC question asks you to convert a journey time, identify the event pair before touching the calculator.

If the two events happen at the same position in a single inertial frame, that frame measures proper time. If the journey has multiple speeds, calculate the proper time accumulated along each part. If the times come from received signals, remove the signal travel delays before comparing frames.

The next useful problem is comparing clocks that follow different journeys but are read at a common event. That requires exactly the same care with events, speeds, and reference frames, as explored in How to Compare Two Moving Atomic Clocks Using Time Dilation.

The equation is the easy part. The physics is deciding what the clocks have actually measured.