How to Solve Relativity Questions Using Time, Distance and Speed

A probe leaves Earth for a research station 10.0 light-years away, measured in Earth’s frame. It travels at \(0.80c\). Earth observers calculate a journey time of 12.5 years, but the probe’s clock records only 7.5 years. From the probe’s perspective, the station is also just 6.0 light-years away.

Now try dividing the original distance by the onboard time:

\[ v=\frac{10.0}{7.5}=1.33c \]

That looks like the probe has travelled faster than light. It hasn’t. The calculation has mixed a distance measured in Earth’s frame with a time measured in the probe’s frame.

The key to solving relativity questions involving time, distance, and speed is to keep every measurement attached to its reference frame. Identify the proper time and proper length, apply the Lorentz factor, and only then use the familiar speed equation with quantities from the same frame.

01Start by identifying the reference frames

A reference frame is the coordinate system an observer uses to measure positions, distances, and times.

In most HSC special relativity questions, you’ll compare two frames:

• The stationary frame, often called the Earth or laboratory frame.
• The moving frame, attached to a spacecraft, particle, or other moving object.

Consider our probe travelling from Earth to the station.

We’ll use the following conventions:

• Earth and the station are stationary relative to each other.
• The probe travels in the positive \(x\)-direction.
• Its speed relative to Earth is \(v=0.80c\).
• The probe travels at constant velocity, so we can treat both reference frames as inertial.
• Acceleration during departure and arrival, and gravitational effects, are ignored.

The two observers describe the same journey differently.

Notice that the probe measures a shorter distance and a shorter journey time.

This is not a contradiction. The observers are measuring the same physical journey using different rulers and clocks.

Identify the proper quantities

Two quantities are particularly important.

Proper length, \(L_0\)

The proper length is the length measured in the reference frame where the two endpoints are stationary.

In our example, Earth and the station are stationary relative to each other. Therefore, their separation of 10.0 light-years is the proper length.

Proper time, \(\Delta\tau\)

Proper time is the elapsed time measured by a single clock that is present at both events.

The probe’s onboard clock records both its departure from Earth and its arrival at the station. Therefore, the onboard time of 7.5 years is the proper time.

Earth observers, on the other hand, use synchronised clocks at different locations to determine the interval between departure and arrival.

This distinction is important. Proper time is not simply whichever time appears in the question, or automatically the smallest number. It depends on the events and the clock measuring them.

For a more detailed method of identifying the correct clock, see How to Identify Proper Time in HSC Special Relativity.

02Apply the Lorentz factor

Once you’ve identified the reference frames and the proper quantities, you can use the Lorentz factor.

The Lorentz factor, represented by \(\gamma\), describes how strongly relativistic effects influence measurements at a particular relative speed.

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

Where:

• \(\gamma\) is the Lorentz factor (dimensionless).
• \(v\) is the relative speed between the reference frames, in \(\text{m s}^{-1}\).
• \(c\) is the speed of light in a vacuum, approximately \(3.00\times10^8\ \text{m s}^{-1}\).

For our probe travelling at \(0.80c\):

\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.80^2}}\\ &=\frac{1}{\sqrt{0.36}}\\ &=1.67 \end{aligned} \]

This gives us the factor needed for both time dilation and length contraction.

Time dilation

The time dilation equation is:

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

Here:

• \(\Delta t\) is the time interval measured in the frame where the clock is moving.
• \(\Delta\tau\) is the proper time measured by the clock present at both events.

The moving clock records a shorter interval between the same two events than an observer who sees that clock moving.

For our journey:

\[ \Delta t=1.67(7.5) \]

\[ \boxed{\Delta t=12.5\text{ years}} \]

Earth observers measure 12.5 years, while the probe’s clock records 7.5 years.

Length contraction

The corresponding length contraction equation is:

\[ \boxed{L=\frac{L_0}{\gamma}} \]

Where:

• \(L\) is the length measured in the frame where the object or endpoints are moving.
• \(L_0\) is the proper length, measured in their rest frame.
• \(\gamma\) is the Lorentz factor.

For the probe’s journey:

\[ L=\frac{10.0}{1.67} \]

\[ \boxed{L=6.0\text{ light-years}} \]

The probe measures the distance to the station as 6.0 light-years rather than 10.0 light-years.

Check that your speed is consistent

Now we can calculate the speed in either reference frame.

In Earth’s frame:

\[ v=\frac{10.0\text{ ly}}{12.5\text{ y}} =0.80c \]

In the probe’s frame:

\[ v=\frac{6.0\text{ ly}}{7.5\text{ y}} =0.80c \]

Both observers obtain the same relative speed magnitude.

This is exactly what we want.

The important point is that each calculation uses a distance and time measured in the same frame.

A light-year is the distance light travels in one year, so \(1\text{ ly/year}=c\). This makes light-years and years particularly convenient in space-travel questions.

03Check the direction of length contraction

There is another detail that can catch students out: length contraction only occurs in the direction of relative motion.

Imagine a rectangular solar panel attached to a spacecraft travelling horizontally.

Its proper dimensions are:

• 5.0 m parallel to the direction of motion.
• 2.0 m perpendicular to the direction of motion.

If the spacecraft travels at \(0.80c\), its Lorentz factor is 1.67.

The station observer measures:

\[ L_{\parallel}=\frac{5.0}{1.67}=3.0\text{ m} \]

However, the perpendicular dimension remains:

\[ L_{\perp}=2.0\text{ m} \]

The station measures a panel that is 3.0 m long and 2.0 m wide.

A common mistake is to divide both dimensions by the Lorentz factor.

That is incorrect because the relative motion is horizontal. Only the horizontal component contracts.

For an object at an angle, resolve its length into parallel and perpendicular components first. Contract only the parallel component.

Also remember that a length measurement requires the two endpoints to be recorded simultaneously in the measuring observer’s frame. Otherwise, the object moves between the two measurements.

04Worked examples: putting time, distance, and speed together

05Where the experimental evidence comes from

The equations aren’t arbitrary corrections to ordinary motion. They follow from the behaviour of light and have been tested experimentally.

Two important developments help explain why physicists needed a different model of space and time.

Michelson and Morley: testing the old model

In the nineteenth century, many physicists believed light travelled through an invisible medium called the luminiferous ether.

If Earth moved through this stationary medium, physicists expected the measured travel times of light travelling in different directions to differ.

Michelson and Morley designed an interferometer to test this prediction.

Follow the light through the apparatus:

1. A light source sends a beam towards a half-silvered mirror.
2. The mirror splits the beam into two perpendicular paths.
3. Each beam travels along an arm to a mirror at the far end.
4. Both beams reflect back towards the centre.
5. They recombine and produce an interference pattern that can be observed through a telescope.

The original prediction was that the relative travel times would change as the apparatus rotated. This should produce a shift in the interference fringes.

An earlier version of the apparatus was highly sensitive to vibration. Michelson and Morley’s improved 1887 design used a larger interferometer mounted on a massive stone block floating in mercury, allowing the apparatus to be rotated more smoothly.

The expected systematic fringe shift was still not observed.

This was an important result, but there is a distinction worth remembering: the experiment did not directly measure time dilation or prove every part of Einstein’s theory. It challenged the simple stationary-ether model and contributed to the development of new ideas about light and motion.

You can explore the apparatus in Harvard’s explanation of the Michelson interferometer and the historical experiment in the American Institute of Physics historical account.

Einstein’s light-clock thought experiment

Einstein’s thought experiments provide a way to understand why time measurements change.

Imagine a clock made from two parallel mirrors, one above the other. A pulse of light bounces between them.

Each complete journey of the light pulse marks a clock tick.

First, consider the clock in its own rest frame.

The light travels vertically, covering a distance \(d\) in half a tick. If the complete tick takes proper time \(\Delta\tau\), then:

\[ d=\frac{c\Delta\tau}{2} \]

Now imagine an observer watching the clock move horizontally.

The light no longer appears to travel straight up and down. It follows a diagonal path because the clock moves sideways while the light travels.

The observer measures a longer light path.

But Einstein’s second postulate says that the speed of light in a vacuum is the same for all inertial observers.

So if the light travels a longer distance at the same speed, it must take more time.

Using Pythagoras on one half of the light’s journey:

\[ \left(\frac{c\Delta t}{2}\right)^2 = d^2+ \left(\frac{v\Delta t}{2}\right)^2 \]

Substituting \(d=c\Delta\tau/2\):

\[ \frac{c^2\Delta t^2}{4} = \frac{c^2\Delta\tau^2}{4} + \frac{v^2\Delta t^2}{4} \]

Rearranging gives:

\[ \Delta t^2(c^2-v^2)=c^2\Delta\tau^2 \]

Therefore:

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

This is the time dilation equation.

The light clock is a thought experiment, not a physical measurement of a moving clock. It demonstrates the geometrical reason for time dilation.

Length contraction follows from the same relativistic framework. The distance measured parallel to relative motion must also change to maintain consistent measurements of light and motion.

A real experimental check: cosmic-ray muons

A particularly useful experimental example involves muons produced high in Earth’s atmosphere.

Muons are unstable subatomic particles. Their mean lifetime when at rest is approximately:

\[ \tau_0=2.2\times10^{-6}\text{ s} \]

If a muon travelled at almost the speed of light but only survived for its ordinary rest lifetime, it would cover approximately:

\[ d=c\tau_0 \]

\[ d=(3.00\times10^8)(2.2\times10^{-6}) \]

\[ d=660\text{ m} \]

Yet cosmic-ray muons are detected at Earth’s surface, even though they can be produced several kilometres above it.

How is that possible?

A muon experiment can measure the relevant quantities using two separate measurements.

First, scintillation counters record the passage of muons at known positions. Their time-of-flight measurements allow researchers to determine particle speed.

Next, muons are stopped in a large plastic scintillator. Electronic detectors record the delays between the stopping signal and subsequent decay signals. Repeating this measurement for many particles allows researchers to determine the mean lifetime at rest.

The measurements of speed and lifetime can then be compared with the distance travelled through the atmosphere.

This is the approach described in MIT’s cosmic-ray muon laboratory.

Consider an illustrative muon travelling at \(0.998c\).

Step 1: Calculate the Lorentz factor.

\[ \gamma= \frac{1}{\sqrt{1-0.998^2}} \]

\[ \gamma\approx15.8 \]

Step 2: Calculate its mean lifetime in Earth’s frame.

\[ \begin{aligned} \Delta t&=\gamma\tau_0\\ &=15.8(2.2\times10^{-6})\\ &=3.48\times10^{-5}\text{ s} \end{aligned} \]

Earth observers measure a mean lifetime of approximately 35 microseconds.

Step 3: Calculate the mean distance travelled.

\[ \begin{aligned} d&=v\Delta t\\ &=(0.998)(3.00\times10^8) (3.48\times10^{-5})\\ &=1.04\times10^4\text{ m} \end{aligned} \]

\[ \boxed{d\approx10.4\text{ km}} \]

The extended lifetime gives the muon enough time to travel the required distance.

There is another way to describe the same result.

In the muon’s rest frame, the atmosphere moves towards it. A 10.4 km separation is contracted to:

\[ L=\frac{10.4}{15.8} \]

\[ L\approx0.66\text{ km} \]

The muon covers approximately 0.66 km in its mean proper lifetime:

\[ d=v\tau_0 \]

\[ d=(0.998)(3.00\times10^8) (2.2\times10^{-6}) \]

\[ d\approx0.66\text{ km} \]

Both descriptions agree.

One final detail: 2.2 microseconds is a mean lifetime, not a guaranteed countdown for every muon. Particle decay is statistical, so some muons decay earlier and others survive longer.

06Don’t confuse invariant mass with energy

Some relativity questions include an object’s mass as well as its speed. Before using the number, work out whether the question is asking about mass, rest energy, total energy, or kinetic energy.

These are related, but they are not interchangeable.

Invariant mass, \(m\)

The invariant mass (also called rest mass) is a property of the object that does not change simply because another observer sees it moving.

Rest energy, \(E_0\)

The energy associated with an object’s invariant mass is:

\[ E_0=mc^2 \]

Total relativistic energy, \(E\)

For an object moving at speed \(v\), its total energy in that reference frame is:

\[ E=\gamma mc^2 \]

Relativistic kinetic energy, \(K\)

The kinetic energy is the difference between total energy and rest energy:

\[ K=E-E_0 \]

Therefore:

\[ \boxed{K=(\gamma-1)mc^2} \]

A numerical comparison

Consider an illustrative object with invariant mass \(2.0\text{ kg}\), travelling at \(0.80c\).

We already know:

\[ \gamma=1.67 \]

Its rest energy is:

\[ \begin{aligned} E_0&=mc^2\\ &=2.0(3.00\times10^8)^2\\ &=1.80\times10^{17}\text{ J} \end{aligned} \]

Its total energy in the moving frame is:

\[ \begin{aligned} E&=\gamma mc^2\\ &=1.67(1.80\times10^{17})\\ &=3.00\times10^{17}\text{ J} \end{aligned} \]

Its kinetic energy is:

\[ \begin{aligned} K&=E-E_0\\ &=3.00\times10^{17}-1.80\times10^{17}\\ &=1.20\times10^{17}\text{ J} \end{aligned} \]

The object’s invariant mass remains 2.0 kg.

Its total energy is greater when measured by an observer who sees it moving.

Some older treatments use the term relativistic mass for \(\gamma m\). The modern invariant-mass convention keeps mass unchanged and accounts for the speed dependence through energy and momentum.

For HSC problem-solving, the practical distinction is:

• Use the invariant mass when a question gives rest mass.
• Use \(E_0=mc^2\) for rest energy.
• Use \(E=\gamma mc^2\) for total energy in a frame where the object is moving.
• Use \(K=(\gamma-1)mc^2\) for relativistic kinetic energy.

Mass does not need to be included when calculating time dilation or length contraction. The Lorentz factor depends on relative speed, not mass.

07Questions and solutions

Question 1

A spacecraft travels between two beacons that are stationary relative to Earth. Their separation is 4.0 light-years, measured in Earth’s frame.

The spacecraft travels towards the second beacon at \(0.80c\).

Calculate:

a) The time taken according to Earth observers.

b) The elapsed time recorded by the spacecraft’s onboard clock.

c) The separation between the beacons in the spacecraft’s reference frame.

Solution 1

The Earth-frame journey takes 5.0 years, the spacecraft clock records 3.0 years, and the contracted distance is 2.4 light-years.

Part a: Earth-frame time

The given distance is measured in the reference frame where the beacons are stationary.

Use:

\[ v=\frac{d}{t} \]

Rearrange:

\[ t=\frac{d}{v} \]

Substitute:

\[ \begin{aligned} t&=\frac{4.0\text{ ly}}{0.80c}\\ &=5.0\text{ years} \end{aligned} \]

Part b: Spacecraft clock time

Calculate the Lorentz factor:

\[ \gamma=\frac{1}{\sqrt{1-0.80^2}} =1.67 \]

The spacecraft clock measures proper time:

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

\[ \begin{aligned} \Delta\tau&=\frac{5.0}{1.67}\\ &=3.0\text{ years} \end{aligned} \]

The onboard clock records 3.0 years.

Part c: Contracted distance

The beacons are moving relative to the spacecraft, so their separation is contracted:

\[ L=\frac{L_0}{\gamma} \]

\[ \begin{aligned} L&=\frac{4.0}{1.67}\\ &=2.4\text{ ly} \end{aligned} \]

Check:

\[ v=\frac{2.4}{3.0}=0.80c \]

The spacecraft-frame distance and time produce the correct relative speed.

Question 2

A rectangular solar sail is stationary relative to its spacecraft.

Its proper dimensions are:

• 5.0 m parallel to the direction of motion.
• 2.0 m perpendicular to the direction of motion.

The spacecraft travels at \(0.80c\) relative to a space station.

Calculate the dimensions and area of the solar sail measured by an observer at the station.

A student claims that both dimensions must be divided by the Lorentz factor. Explain why this is incorrect.

Solution 2

The station measures the sail as 3.0 m long and 2.0 m wide, giving an area of \(6.0\text{ m}^2\).

The student’s mistake is contracting the perpendicular dimension as well as the parallel dimension.

Step 1: Calculate the Lorentz factor.

\[ \gamma=\frac{1}{\sqrt{1-0.80^2}} =1.67 \]

Step 2: Contract the parallel dimension.

\[ L_{\parallel}=\frac{L_{0,\parallel}}{\gamma} \]

\[ \begin{aligned} L_{\parallel}&=\frac{5.0}{1.67}\\ &=3.0\text{ m} \end{aligned} \]

Step 3: Determine the perpendicular dimension.

Length contraction only affects the direction parallel to relative motion.

Therefore:

\[ L_{\perp}=2.0\text{ m} \]

Step 4: Calculate the measured area.

\[ A=L_{\parallel}L_{\perp} \]

\[ A=(3.0)(2.0) \]

\[ \boxed{A=6.0\text{ m}^2} \]

If the student contracts both dimensions, they obtain:

\[ A_{\text{incorrect}}=(3.0)(1.2) =3.6\text{ m}^2 \]

That result is incorrect because the vertical dimension is perpendicular to the spacecraft’s relative motion.

Question 3

An object has an invariant mass of \(1.5\text{ kg}\) and travels at \(0.80c\) relative to a laboratory.

Calculate:

a) Its rest energy.

b) Its total energy measured in the laboratory frame.

c) Its relativistic kinetic energy.

A student argues that the object’s mass has increased to \(2.5\text{ kg}\). Evaluate this statement.

Solution 3

The rest energy is \(1.35\times10^{17}\text{ J}\), the laboratory-frame total energy is \(2.25\times10^{17}\text{ J}\), and the kinetic energy is \(9.0\times10^{16}\text{ J}\). The invariant mass remains 1.5 kg.

Step 1: Calculate the Lorentz factor.

\[ \gamma=\frac{1}{\sqrt{1-0.80^2}} =1.67 \]

Step 2: Calculate the rest energy.

Use:

\[ E_0=mc^2 \]

Substitute:

\[ \begin{aligned} E_0&=1.5(3.00\times10^8)^2\\ &=1.5(9.00\times10^{16})\\ &=1.35\times10^{17}\text{ J} \end{aligned} \]

Step 3: Calculate the total energy.

The object is moving relative to the laboratory, so:

\[ E=\gamma mc^2 \]

\[ \begin{aligned} E&=1.67(1.35\times10^{17})\\ &=2.25\times10^{17}\text{ J} \end{aligned} \]

Step 4: Calculate kinetic energy.

\[ K=E-E_0 \]

\[ \begin{aligned} K&=2.25\times10^{17} -1.35\times10^{17}\\ &=9.0\times10^{16}\text{ J} \end{aligned} \]

Step 5: Evaluate the student’s statement.

The student has confused energy with invariant mass.

The object has the same invariant mass of 1.5 kg in every inertial reference frame.

Its energy increases because it has kinetic energy in the laboratory frame.

The value \(\gamma m=2.5\text{ kg}\) is sometimes called relativistic mass in older treatments. It is not the invariant mass used in the modern convention.

Question 4

A cosmic-ray muon is produced 10.4 km above a detector, measured in Earth’s frame.

The muon travels vertically towards the detector at \(0.998c\). Its mean proper lifetime is \(2.2\ \mu\text{s}\).

Calculate:

a) The Lorentz factor.

b) The mean lifetime measured by Earth observers.

c) The mean distance travelled in Earth’s frame.

d) The distance between the production point and detector in the muon’s frame.

Explain how these results support the observation of muons at Earth’s surface.

Solution 4

The Lorentz factor is approximately 15.8, the Earth-frame mean lifetime is \(34.8\ \mu\text{s}\), and the mean travel distance is approximately 10.4 km. The atmosphere is contracted to approximately 0.66 km in the muon’s frame.

Part a: Lorentz factor

\[ \gamma=\frac{1}{\sqrt{1-0.998^2}} \]

\[ \boxed{\gamma\approx15.8} \]

Part b: Earth-frame lifetime

The given lifetime is measured in the muon’s rest frame. It is therefore the proper time.

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

Substitute:

\[ \begin{aligned} \Delta t&=15.8(2.2\times10^{-6})\\ &=3.48\times10^{-5}\text{ s} \end{aligned} \]

Therefore:

\[ \boxed{\Delta t=34.8\ \mu\text{s}} \]

Part c: Mean travel distance

Use:

\[ d=v\Delta t \]

\[ \begin{aligned} d&=(0.998)(3.00\times10^8) (3.48\times10^{-5})\\ &=1.04\times10^4\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{d=10.4\text{ km}} \]

The mean travel distance is approximately equal to the altitude of the detector.

Part d: Contracted distance

In the muon’s frame, the atmosphere and detector move towards it.

The original Earth-frame distance is the proper length because the production point and detector are stationary relative to Earth.

Therefore:

\[ L=\frac{L_0}{\gamma} \]

\[ \begin{aligned} L&=\frac{10.4}{15.8}\\ &=0.658\text{ km} \end{aligned} \]

Check this against the distance travelled during the muon’s proper lifetime:

\[ \begin{aligned} d&=v\Delta\tau\\ &=(0.998)(3.00\times10^8) (2.2\times10^{-6})\\ &=0.659\text{ km} \end{aligned} \]

The two results agree within rounding.

The muon can reach the detector because the Earth-frame lifetime is extended, or equivalently, because the atmosphere is contracted in the muon’s frame.

Remember that these are mean values. Not every muon survives for exactly the mean lifetime.

Question 5

An idealised spacecraft leaves Earth and travels to a research station 8.0 light-years away, measured in Earth’s frame.

It travels outbound at \(0.80c\), instantly turns around at the station, and returns to Earth at \(0.60c\).

The spacecraft carries an onboard clock.

Assume:

• Earth and the station remain stationary relative to each other.
• Both journeys are straight, constant-velocity segments.
• The turnaround takes negligible time.
• Gravitational effects are ignored.

Calculate:

a) The total elapsed time measured by Earth.

b) The time recorded by the spacecraft clock on each leg.

c) The total elapsed time recorded by the spacecraft clock.

d) The difference between the elapsed times measured by Earth and the spacecraft.

A student argues that the spacecraft has zero average velocity over the complete trip, so its Lorentz factor must be 1 and no time difference should occur.

Explain why this reasoning fails.

Solution 5

Earth measures a total journey time of approximately 23.3 years. The spacecraft clock records approximately 16.7 years. Earth therefore measures an additional 6.7 years.

The student’s argument fails because the spacecraft changes its velocity and reference frame at the turnaround. A single Lorentz factor cannot describe the entire journey.

Part a: Earth-frame time

First calculate the outbound time:

\[ \Delta t_1=\frac{8.0}{0.80} \]

\[ \Delta t_1=10.0\text{ years} \]

Now calculate the return time:

\[ \Delta t_2=\frac{8.0}{0.60} \]

\[ \Delta t_2=13.33\text{ years} \]

Add the two intervals:

\[ \begin{aligned} \Delta t_{\text{total}} &=10.0+13.33\\ &=23.33\text{ years} \end{aligned} \]

Therefore:

\[ \boxed{\Delta t_{\text{Earth}}=23.3\text{ years}} \]

Part b: Spacecraft time on each leg

The outbound and return speeds are different, so they have different Lorentz factors.

For the outbound journey:

\[ \gamma_1=\frac{1}{\sqrt{1-0.80^2}} =1.67 \]

The spacecraft clock records:

\[ \begin{aligned} \Delta\tau_1&=\frac{\Delta t_1}{\gamma_1}\\ &=\frac{10.0}{1.67}\\ &=6.0\text{ years} \end{aligned} \]

For the return journey:

\[ \gamma_2=\frac{1}{\sqrt{1-0.60^2}} =1.25 \]

The spacecraft clock records:

\[ \begin{aligned} \Delta\tau_2&=\frac{\Delta t_2}{\gamma_2}\\ &=\frac{13.33}{1.25}\\ &=10.67\text{ years} \end{aligned} \]

Part c: Total spacecraft time

Add the proper times from both legs:

\[ \begin{aligned} \Delta\tau_{\text{total}} &=\Delta\tau_1+\Delta\tau_2\\ &=6.0+10.67\\ &=16.67\text{ years} \end{aligned} \]

Therefore:

\[ \boxed{\Delta\tau_{\text{spacecraft}}=16.7\text{ years}} \]

Part d: Difference in elapsed time

\[ \begin{aligned} \text{Difference} &=23.33-16.67\\ &=6.66\text{ years} \end{aligned} \]

\[ \boxed{\text{Difference}=6.7\text{ years}} \]

Why the student’s argument fails

The student uses the average velocity over the entire trip.

Because the spacecraft returns to its starting position, its net displacement is zero. Therefore, its average velocity is zero.

However, zero average velocity does not mean the spacecraft was stationary throughout the journey.

It travelled at \(0.80c\) outbound and \(0.60c\) on the return leg.

The Lorentz factor depends on the actual relative speed during each segment, not the average velocity over the complete journey.

Even using the average speed would not solve the problem, because the Lorentz factor is a nonlinear function of speed.

There is another important detail.

The spacecraft changes its velocity at the turnaround. This means its reference frame changes, so there is no single inertial frame attached to the spacecraft for the entire journey.

Earth’s clock is present at both the departure and return events. The spacecraft follows a different path through spacetime and accumulates a different amount of proper time.

The instantaneous turnaround is an idealisation. A real spacecraft would accelerate over a finite interval, and its proper time during that acceleration would also need to be included.

This is the central idea behind the twin paradox: observers can compare their clocks when they meet again, but the symmetry between their journeys is broken when one observer changes inertial frames.

08What to remember when solving a mixed relativity question

Before substituting numbers, run through these checks:

1. Identify the reference frames. Which observer is stationary relative to the object or endpoints?
2. Set the direction of motion. Identify which distances are parallel and perpendicular to relative motion.
3. Identify proper quantities. Find the rest-frame length and the time recorded by a single clock present at both events.
4. Calculate the Lorentz factor. Use the relative speed between the two frames.
5. Choose the correct transformation. Use time dilation for time intervals and length contraction for distances.
6. Check your units and frame consistency. Never divide a distance measured in one frame by a time measured in another.
7. Check the physical result. The relative speed must remain below \(c\), and the transformed quantities should agree with the relevant physical constraints.

Once you can keep reference frames, time intervals, and distances organised, the next useful step is relativistic momentum. It uses the same Lorentz factor but introduces another important relationship between invariant mass, speed, and momentum.

Continue with How to Calculate Relativistic Momentum from Mass and Speed.