How to Find Speed from a Time-Dilation Factor | HSC Physics

A clock aboard a spacecraft advances \(8.00\ \mu\text{s}\) between two flashes. A camera back on Earth receives those flashes \(16.0\ \mu\text{s}\) apart. Has the camera measured a time-dilation factor of \(2\), meaning the spacecraft is travelling at \(0.866c\)?

Not necessarily. Predict what happens to the second flash if the spacecraft is moving away from the camera. It has farther to travel, so some of that extra waiting time comes from the light reaching Earth, not from time dilation.

To find speed from a genuine time-dilation factor, \(D\), use:

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

Here, \(v\) is the relative speed, \(c=3.00\times10^8\ \text{m s}^{-1}\) is the speed of light in a vacuum, and \(D\) is the Lorentz factor, usually written as \(\gamma\).

The important step comes before substituting: establish that \(D\) compares the Earth-frame time between two events with the proper time between those same events. A camera’s pulse-arrival interval is not automatically either of those quantities.

01How to turn a time-dilation factor into speed

Suppose two events occur at the same position inside a spacecraft moving at constant velocity. They might be two flashes from a lamp bolted to the wall.

A clock fixed beside that lamp measures the elapsed time between the flashes. In the spacecraft’s inertial reference frame, both flashes happen at the same position.

This is the proper time, written as \(\Delta\tau\).

An Earth observer assigns different positions to the two flashes because the spacecraft has moved between them. Using synchronised clocks in the Earth frame, the observer measures a longer interval, \(\Delta t\).

The time-dilation factor is:

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

where:

• \(\Delta t\) is the Earth-frame time interval between the events, in seconds.
• \(\Delta\tau\) is the proper time measured on the spacecraft, in seconds.
• \(D\) or \(\gamma\) is dimensionless.

If \(\gamma=1.50\), Earth assigns \(1.50\) times as much time between the flashes as the spacecraft clock records.

It doesn’t mean the spacecraft’s clock is faulty. The two reference frames assign different time intervals to the same events.

Rearranging the Lorentz factor

The Lorentz factor is defined by:

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

We want \(v\), rather than \(\gamma\).

Square both sides, take the reciprocal, and rearrange:

\[ \begin{aligned} \frac{1}{\gamma^2} &=1-\frac{v^2}{c^2}\\[4pt] \frac{v^2}{c^2} &=1-\frac{1}{\gamma^2}\\[4pt] \frac{v}{c} &=\sqrt{1-\frac{1}{\gamma^2}} \end{aligned} \]

Therefore:

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

This gives the magnitude of the relative velocity. The factor alone cannot tell you whether the spacecraft is moving left, right, towards Earth, or away from Earth.

Two useful checks:

• If \(\gamma=1\), the relative speed is zero.
• If \(\gamma>1\), the calculated speed is positive but always less than \(c\).

A calculated Lorentz factor below \(1\) means the intervals have been identified incorrectly, the measurements are inconsistent with the assumptions, or a quantity other than the Lorentz factor has been supplied.

Watch for a different meaning of “factor”

Some questions give the moving clock’s rate as a fraction of the Earth-frame rate rather than giving \(\gamma\).

For instance, if the moving clock accumulates \(0.80\ \text{s}\) while Earth assigns \(1.00\ \text{s}\) between the same events, then:

\[ \gamma=\frac{1.00}{0.80}=1.25 \]

The speed is \(0.600c\), not \(0.80c\).

The equations are straightforward once you know what the measurements represent.

02Designing a measurement that actually gives \(\gamma\)

Imagine an experimental arrangement with two stationary ground gates and a small spacecraft flying between them.

The spacecraft travels from gate A to gate B in a straight line at constant speed.

• Gate A is at \(x=0\).
• Gate B is \(1.80\ \text{km}\) to the right.
• The spacecraft carries one clock and one radio transmitter.
• The transmitter emits a flash of radio energy as the spacecraft passes each gate.
• A detector at each gate records the local crossing time using ground clocks synchronised in the Earth frame.

The same spacecraft clock records the instant of both emissions. Its readings are stored aboard the spacecraft and can be retrieved later.

Because the clock and transmitter remain at the same position relative to each other, the spacecraft clock measures the proper time between emissions.

The two ground detectors, meanwhile, use their synchronised clocks to establish the Earth-frame interval.

What should the clocks record?

If classical physics were correct about time being identical in every reference frame, the onboard and ground intervals would match.

Special relativity predicts that the moving onboard clock accumulates less time between these two events.

Here is an illustrative set of ideal measurements.

These are invented data for a controlled example, not measurements from a historical experiment. Assume ideal clocks, synchronisation in the ground frame, constant spacecraft velocity, and negligible gravitational effects.

03Why receiving a signal is different from timing its emission

Now remove the ground detector at gate B.

Instead, place a single camera or radio receiver at gate A. It records when both spacecraft flashes arrive.

Would that be enough to measure the Earth-frame interval directly?

Not without accounting for signal travel.

The first flash is emitted right beside the receiver at A, so it arrives immediately. The second flash is emitted at B, \(1.80\ \text{km}\) away, and must travel back to A.

Radio waves in a vacuum travel at \(c\). The second flash therefore needs an extra:

\[ \begin{aligned} t_{\text{light}} &=\frac{1.80\times10^3}{3.00\times10^8}\\ &=6.00\times10^{-6}\ \text{s}\\ &=6.00\ \mu\text{s} \end{aligned} \]

The camera’s arrival interval is:

\[ \Delta t_{\text{arrival}} =10.0+6.00 =16.0\ \mu\text{s} \]

But the Earth-frame emission interval remains \(10.0\ \mu\text{s}\).

Dividing the camera’s \(16.0\ \mu\text{s}\) by the onboard \(8.00\ \mu\text{s}\) gives \(2\), but that number is not the Lorentz factor.

The surprising result: Put the camera at B instead

Suppose a second camera is fixed at gate B.

The first flash, emitted at A at Earth time \(0\), travels \(1.80\ \text{km}\) towards B and arrives at Earth time \(6.00\ \mu\text{s}\).

The second flash is emitted at B at Earth time \(10.0\ \mu\text{s}\) and arrives immediately.

This second camera records a gap of only:

\[ 10.0-6.00=4.00\ \mu\text{s} \]

So two cameras measuring the same pair of flashes report very different intervals.

The camera at A receives the second pulse later because the spacecraft is moving away. The camera at B receives the pulses closer together because the spacecraft is approaching it.

Neither camera has disproved time dilation. They have measured different light-propagation delays.

This is also why a moving source can appear to send pulses faster or slower depending on whether it approaches or recedes. That observed pulse-rate change is related to the relativistic Doppler effect. It includes both time dilation and changing light-travel distance.

04Three checks before trusting a calculated speed

A time-dilation factor contains enough information to calculate relative speed only when its meaning and the motion are clear.

1. Are the events co-located in the proper-time frame?

For a spacecraft moving uniformly, two flashes from one fixed onboard lamp occur at the same position in the spacecraft’s rest frame. The onboard clock gives the proper time.

If one flash comes from the front of a spacecraft and another from its rear, two onboard timestamps cannot automatically be treated as a proper-time interval. The emission locations differ in that frame.

The rule is about the positions of the events, not simply which clock readings are smaller.

2. Are the times emission times or arrival times?

A radio receiver records when information reaches it.

A synchronised ground clock network can record when events happen at different ground positions.

Those are different measurements. A single receiver’s data may still allow you to recover speed, but only with a model of how the signals travel.

3. Has the speed remained constant?

The formula:

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

with one fixed \(\gamma\) describes two events on a clock moving uniformly relative to an inertial frame.

If the spacecraft changes speed, different parts of its journey have different Lorentz factors.

A ratio of total Earth time to total onboard time does not necessarily represent the Lorentz factor associated with any actual speed during that journey.

Likewise, a spacecraft that turns around and returns to its starting point requires more care. Both the travelling clock and a stationary Earth clock may be present at departure and reunion, but they have followed different paths through spacetime. The traveller cannot use one inertial rest frame for the entire round trip.

These limits become especially important when a question gives impressive-looking clock data but does not establish what was measured.

05Questions and solutions

Assume special relativity, ideal clocks, negligible gravity, and \(c=3.00\times10^8\ \text{m s}^{-1}\) unless stated otherwise. Times and distances in each question are illustrative.

Question 1

A spacecraft moving at constant velocity emits two flashes from the same onboard transmitter.

Its onboard clock records \(3.00\ \mu\text{s}\) between emissions. An Earth-frame network of synchronised clocks records \(3.75\ \mu\text{s}\) between those same events.

Calculate:

a) The Lorentz factor.

b) The spacecraft’s speed relative to Earth.

c) Whether the clock measurements alone establish the spacecraft’s direction of motion.

Solution 1

The Lorentz factor is \(1.25\), and the spacecraft’s speed is \(0.600c\), or \(1.80\times10^8\ \text{m s}^{-1}\). Its direction cannot be determined from time dilation alone.

a) Find the factor.

The onboard clock is at both emission events in the spacecraft’s inertial rest frame, so \(3.00\ \mu\text{s}\) is the proper time.

\[ \gamma=\frac{3.75}{3.00}=1.25 \]

b) Find the speed.

\[ \begin{aligned} \frac{v}{c} &=\sqrt{1-\frac{1}{1.25^2}}\\ &=0.600 \end{aligned} \]

Therefore:

\[ \begin{aligned} v&=0.600(3.00\times10^8)\\ &=1.80\times10^8\ \text{m s}^{-1} \end{aligned} \]

c) Interpret the result.

The Lorentz factor depends on \(v^2\), so reversing the velocity does not change it.

The spacecraft could be moving in either direction at this speed. Positions or other observations are needed to determine the direction.

Question 2

A spacecraft travels directly away from a stationary radio receiver at constant speed.

Two radio pulses are emitted \(4.00\ \mu\text{s}\) apart according to the spacecraft’s single onboard clock. The Earth receiver detects the pulses \(12.0\ \mu\text{s}\) apart.

The spacecraft remains on the same side of the receiver throughout.

a) Determine the spacecraft’s speed relative to Earth.

b) Calculate the actual Earth-frame interval between emissions.

c) Explain why the receiver’s \(12.0\ \mu\text{s}\) interval must not be divided directly by \(4.00\ \mu\text{s}\) to obtain \(\gamma\).

Solution 2

The spacecraft travels at \(0.800c\), and Earth assigns \(6.67\ \mu\text{s}\) between emissions. The receiver’s longer \(12.0\ \mu\text{s}\) interval includes the extra time taken by the second radio pulse to reach it.

a) Connect the received interval to the speed.

Let \(\beta=v/c\), where \(v\) is the positive speed away from the receiver.

In the Earth frame, the time between emissions is:

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

During this time, the spacecraft moves a distance \(v\Delta t\) farther from the receiver.

The second pulse therefore needs an additional light-travel time:

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

So:

\[ \begin{aligned} \Delta t_{\text{arrival}} &=\Delta t(1+\beta)\\ &=\gamma\Delta\tau(1+\beta) \end{aligned} \]

Substituting the measured intervals:

\[ 12.0=4.00\gamma(1+\beta) \]

The time units cancel, giving:

\[ 3.00=\gamma(1+\beta) \]

Using the Lorentz factor equation:

\[ 3.00=\frac{1+\beta}{\sqrt{1-\beta^2}} \]

Square and simplify:

\[ \begin{aligned} 9.00&=\frac{1+\beta}{1-\beta}\\ 9.00(1-\beta)&=1+\beta\\ 8.00&=10.0\beta\\ \beta&=0.800 \end{aligned} \]

Thus:

\[ v=0.800c=2.40\times10^8\ \text{m s}^{-1} \]

b) Find the Earth-frame emission interval.

\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.800^2}}\\ &=\frac{5}{3} \end{aligned} \]

Hence:

\[ \begin{aligned} \Delta t &=\gamma\Delta\tau\\ &=\frac{5}{3}(4.00)\\ &=6.67\ \mu\text{s} \end{aligned} \]

c) Account for the difference.

The additional light-travel delay is:

\[ \begin{aligned} \Delta t_{\text{extra}} &=\beta\Delta t\\ &=0.800(6.67)\\ &=5.33\ \mu\text{s} \end{aligned} \]

The receiver therefore records approximately:

\[ 6.67+5.33=12.0\ \mu\text{s} \]

The tempting calculation is \(\gamma=12.0/4.00=3\). It fails because \(12.0\ \mu\text{s}\) is a reception interval, not the Earth-frame interval between emissions.

Notice that speed could still be recovered from one receiver. We had enough extra information: the onboard proper interval, the spacecraft’s constant velocity, its receding direction, and the signal speed. We needed a propagation model rather than a simple division.

Question 3

An experimental spacecraft flies between two ground stations \(3.36\ \text{km}\) apart.

Synchronised Earth clocks measure \(16.00\ \mu\text{s}\) between departure and arrival. The onboard clock measures \(11.20\ \mu\text{s}\).

An engineer calculates:

\[ \gamma_{\text{eff}}=\frac{16.00}{11.20} \]

and concludes that the spacecraft’s speed was approximately \(0.714c\) throughout the journey.

However, the distance and Earth-frame travel time appear to imply a different speed.

a) Calculate both speeds and explain the contradiction.

b) Test this proposed motion: the spacecraft travels at \(0.600c\) for the first \(8.00\ \mu\text{s}\) of Earth time, then at \(0.800c\) for the next \(8.00\ \mu\text{s}\). Treat the change of speed as instantaneous.

c) Explain what the total time ratio can and cannot establish.

Solution 3

The observations are inconsistent with a single constant spacecraft speed, but they agree with the proposed two-stage journey. The factor \(16.00/11.20\) is an effective total-time ratio, not the Lorentz factor of one constant velocity.

a) Compare the two inferred speeds.

Using the clock ratio:

\[ \gamma_{\text{eff}} =\frac{16.00}{11.20} =\frac{10}{7} \]

If we incorrectly assume one constant speed:

\[ \begin{aligned} \frac{v_{\text{ratio}}}{c} &=\sqrt{1-\left(\frac{7}{10}\right)^2}\\ &=\sqrt{0.510}\\ &\approx0.714 \end{aligned} \]

But the distance and Earth-frame time give:

\[ \begin{aligned} v_{\text{avg}} &=\frac{3.36\times10^3}{16.00\times10^{-6}}\\ &=2.10\times10^8\ \text{m s}^{-1}\\ &=0.700c \end{aligned} \]

For uniform motion, average speed and constant speed would be the same.

A spacecraft travelling at a constant \(0.700c\) would accumulate:

\[ \begin{aligned} \Delta\tau &=16.00\sqrt{1-0.700^2}\\ &\approx11.43\ \mu\text{s} \end{aligned} \]

That does not match the measured \(11.20\ \mu\text{s}\). Assuming ideal measurements, constant-speed motion cannot explain all the data.

b) Test the two-stage motion.

At \(0.600c\):

\[ \gamma_1=\frac{1}{\sqrt{1-0.600^2}}=1.25 \]

The proper time accumulated during the first \(8.00\ \mu\text{s}\) is:

\[ \Delta\tau_1=\frac{8.00}{1.25} =6.40\ \mu\text{s} \]

At \(0.800c\):

\[ \gamma_2=\frac{1}{\sqrt{1-0.800^2}} =\frac{5}{3} \]

The second stage contributes:

\[ \Delta\tau_2=\frac{8.00}{5/3} =4.80\ \mu\text{s} \]

Total onboard time:

\[ \Delta\tau_{\text{total}} =6.40+4.80 =11.20\ \mu\text{s} \]

This matches the measurement.

Now calculate the distances.

First stage:

\[ \begin{aligned} d_1&=0.600c(8.00\times10^{-6})\\ &=1.44\times10^3\ \text{m} \end{aligned} \]

Second stage:

\[ \begin{aligned} d_2&=0.800c(8.00\times10^{-6})\\ &=1.92\times10^3\ \text{m} \end{aligned} \]

Therefore:

\[ d_{\text{total}}=1.44+1.92=3.36\ \text{km} \]

The proposed journey satisfies the Earth-frame distance, Earth-frame time, and onboard proper time simultaneously.

c) Interpret the factor.

The overall ratio is still a valid comparison of elapsed times:

\[ \frac{16.00}{11.20}\approx1.429 \]

But it combines two different speeds and therefore two different Lorentz factors.

Inverting this total ratio as though it represented uniform motion produces \(0.714c\), a number that was not the spacecraft’s speed during either stage.

The precise interaction is between speed changes and accumulated proper time. Proper time is calculated separately for each constant-speed segment and then added. It is not generally found by applying one Lorentz factor to the total journey.

Question 4

Two ideal clocks meet at a stationary ground marker and are set to zero.

One remains at the marker. The other travels away, eventually returns, and reunites with the first clock.

At reunion:

• Ground clock: \(20.0\ \mu\text{s}\).
• Travelling clock: \(16.0\ \mu\text{s}\).

A technician argues that the ratio \(20.0/16.0=1.25\) proves the travelling clock moved at \(0.600c\).

Consider two possible journeys, with changes of velocity occurring over negligible time:

Journey A: Travel away for \(10.0\ \mu\text{s}\) of Earth time at \(0.600c\), then return for another \(10.0\ \mu\text{s}\) at the same speed.

Journey B: Travel away for \(5.00\ \mu\text{s}\) at \(0.800c\), stop for \(10.0\ \mu\text{s}\) at the distant turning point, then return for \(5.00\ \mu\text{s}\) at \(0.800c\).

a) Show that both journeys reproduce the clock readings.

b) Explain why the measured total ratio cannot establish a unique travelling speed.

c) Both clocks were present at the departure and reunion events. Does this mean both readings are the same proper-time interval? Explain.

d) Identify one additional measurable quantity that would distinguish the journeys.

Solution 4

Both journeys produce \(20.0\ \mu\text{s}\) on the ground clock and \(16.0\ \mu\text{s}\) on the travelling clock. The ratio does not uniquely determine the spacecraft’s speed because the travelling clock follows different velocity histories.

a) Calculate each journey’s proper time.

For Journey A, the speed magnitude is \(0.600c\), so \(\gamma=1.25\).

Each leg takes \(10.0\ \mu\text{s}\) in the Earth frame.

\[ \begin{aligned} \Delta\tau_A &=\frac{10.0}{1.25}+\frac{10.0}{1.25}\\ &=8.00+8.00\\ &=16.0\ \mu\text{s} \end{aligned} \]

For Journey B, each moving leg has speed \(0.800c\), so \(\gamma=5/3\).

Each \(5.00\ \mu\text{s}\) moving leg contributes:

\[ \frac{5.00}{5/3}=3.00\ \mu\text{s} \]

While stopped relative to Earth, the travelling clock accumulates the same \(10.0\ \mu\text{s}\) as Earth clocks in that frame.

Therefore:

\[ \begin{aligned} \Delta\tau_B &=3.00+10.0+3.00\\ &=16.0\ \mu\text{s} \end{aligned} \]

The two journeys have identical total clock readings despite different speeds.

b) Explain why the speed is not unique.

The spacecraft cannot leave a ground marker and return to it while maintaining one constant nonzero velocity throughout the entire journey.

Both journeys involve changes in velocity. Journey B also includes a stationary period.

Consequently, \(20.0/16.0\) summarises the accumulated effect of the whole journey. Treating it as the Lorentz factor for one uniform velocity assumes something the journey does not satisfy.

c) Resolve the proper-time issue.

Both clocks measure their own elapsed proper time along their respective paths through spacetime.

The ground clock follows a stationary, inertial path between the two meetings. Its \(20.0\ \mu\text{s}\) is the proper-time interval associated with the inertial frame in which the two meeting events occur at the same position.

The travelling clock follows a different path involving changes of velocity. It accumulates \(16.0\ \mu\text{s}\) along that path.

Being present at both endpoints does not make two different clock paths accumulate identical proper times.

The simple co-located-events rule works directly for finding the rest frame of an inertially moving clock. A journey containing turnarounds must be divided into suitable sections rather than assigned one inertial rest frame.

d) Distinguish the journeys.

Measure the maximum distance from the ground marker.

Journey A reaches:

\[ \begin{aligned} d_A&=0.600c(10.0\times10^{-6})\\ &=1.80\ \text{km} \end{aligned} \]

Journey B reaches:

\[ \begin{aligned} d_B&=0.800c(5.00\times10^{-6})\\ &=1.20\ \text{km} \end{aligned} \]

The different maximum distances distinguish the two motion histories.

The deeper point is that clock readings at reunion do not, by themselves, reveal everything that happened between departure and reunion.

Question 5

A spacecraft travels at constant velocity to the right along a straight track. It emits two brief radio pulses from the same onboard transmitter.

Unfortunately, its onboard clock data are lost.

Three stationary ground receivers independently record the interval between the arrivals of the same two pulses:

• Receiver P is at \(x=0\).
• Receiver M is at \(x=1.50\ \text{km}\).
• Receiver Q is at \(x=3.00\ \text{km}\).

Both emission events occurred somewhere strictly between P and Q.

The measured pulse-arrival intervals are:

Each receiver uses one local clock. The clocks at different receivers have not been synchronised, but their interval measurements are ideal.

Assume the radio signals travel at \(c\), and the spacecraft remains inertial between emissions.

a) Find the Earth-frame time between emissions and the spacecraft’s displacement.

b) Determine the spacecraft’s speed, Lorentz factor, and proper time between emissions, despite the missing onboard data.

c) Find the Earth-frame positions at which the first and second pulses were emitted.

d) An engineer argues that because M is halfway between P and Q, its received interval must be the average of \(12.0\ \mu\text{s}\) and \(3.00\ \mu\text{s}\), or \(7.50\ \mu\text{s}\). Explain precisely why that prediction fails.

Solution 5

The spacecraft moves at \(0.600c\), its Lorentz factor is \(1.25\), and the lost onboard clock would have recorded \(6.00\ \mu\text{s}\).

The emissions occurred at \(x_1=0.600\ \text{km}\) and \(x_2=1.95\ \text{km}\). The engineer’s midpoint prediction fails because M lies between the two emission positions, so the radio signals approach it from opposite sides.

The crucial insight is that two receivers on opposite sides of both emissions experience opposite changes in signal-travel distance.

a) Recover the Earth-frame interval and displacement.

Let:

• \(\Delta t\) be the Earth-frame time between emissions.
• \(x_1\) and \(x_2\) be the first and second emission positions.
• \(\Delta x=x_2-x_1\) be the positive spacecraft displacement.
• \(R_P\) and \(R_Q\) be the measured arrival intervals at P and Q.

Because both emissions are to the right of P, the second signal must travel an extra distance \(\Delta x\) to reach P.

Therefore:

\[ R_P=\Delta t+\frac{\Delta x}{c} \]

For Q, both emissions are to its left. The spacecraft approaches Q, so the second signal has a shorter journey by \(\Delta x\).

Therefore:

\[ R_Q=\Delta t-\frac{\Delta x}{c} \]

The measurements give:

\[ \begin{aligned} R_P&=12.0\ \mu\text{s}\\ R_Q&=3.00\ \mu\text{s} \end{aligned} \]

Adding the equations cancels the propagation difference:

\[ R_P+R_Q=2\Delta t \]

Thus:

\[ \begin{aligned} \Delta t&=\frac{12.0+3.00}{2}\\ &=7.50\ \mu\text{s} \end{aligned} \]

Subtracting them gives:

\[ R_P-R_Q=\frac{2\Delta x}{c} \]

Therefore:

\[ \begin{aligned} \frac{\Delta x}{c} &=\frac{12.0-3.00}{2}\\ &=4.50\ \mu\text{s} \end{aligned} \]

The displacement is:

\[ \begin{aligned} \Delta x &=c(4.50\times10^{-6})\\ &=1.35\times10^3\ \text{m}\\ &=1.35\ \text{km} \end{aligned} \]

Notice that the receivers did not need synchronised clocks. We used the interval between two arrivals at each receiver, not the difference between timestamps recorded at different locations.

b) Recover the speed and proper time.

The spacecraft covers \(1.35\ \text{km}\) in \(7.50\ \mu\text{s}\) in the Earth frame.

\[ \begin{aligned} v&=\frac{1.35\times10^3}{7.50\times10^{-6}}\\ &=1.80\times10^8\ \text{m s}^{-1}\\ &=0.600c \end{aligned} \]

The Lorentz factor is:

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

Since both emissions occur at the same position on the spacecraft in its inertial rest frame:

\[ \begin{aligned} \Delta\tau &=\frac{\Delta t}{\gamma}\\ &=\frac{7.50}{1.25}\\ &=6.00\ \mu\text{s} \end{aligned} \]

The missing onboard clock data can therefore be reconstructed from the receiver measurements.

The receiver at Q measured only \(3.00\ \mu\text{s}\), less than the proper time of \(6.00\ \mu\text{s}\). There is no contradiction: Q recorded pulse arrivals, not the time interval between emission events in its reference frame.

c) Find the emission positions using the middle receiver.

We already know:

\[ x_2-x_1=1.35\ \text{km} \]

Now use M’s measured interval of \(6.00\ \mu\text{s}\).

First, determine where M lies relative to the two emissions.

If M were to the left of both emission points, its arrival interval would equal P’s \(12.0\ \mu\text{s}\), because both signals would travel leftwards towards it.

If M were to the right of both emissions, its arrival interval would equal Q’s \(3.00\ \mu\text{s}\).

Neither matches the observed \(6.00\ \mu\text{s}\).

Therefore M must be between the two emission positions:

\[ x_1<x_M<x_2 \]

where \(x_M=1.50\ \text{km}\).

The first radio pulse travels rightwards from \(x_1\) to M, covering \(x_M-x_1\).

The second travels leftwards from \(x_2\) to M, covering \(x_2-x_M\).

So the interval at M is:

\[ R_M=\Delta t+ \frac{x_1+x_2-2x_M}{c} \]

Rearrange to find the sum of the emission positions:

\[ x_1+x_2 =2x_M+c(R_M-\Delta t) \]

Using \(c=0.300\ \text{km}\) per microsecond:

\[ \begin{aligned} x_1+x_2 &=3.00+0.300(6.00-7.50)\\ &=2.55\ \text{km} \end{aligned} \]

We now have two equations:

\[ \begin{aligned} x_1+x_2&=2.55\ \text{km}\\ x_2-x_1&=1.35\ \text{km} \end{aligned} \]

Solving:

\[ \begin{aligned} x_1&=\frac{2.55-1.35}{2}\\ &=0.600\ \text{km} \end{aligned} \]

and:

\[ \begin{aligned} x_2&=\frac{2.55+1.35}{2}\\ &=1.95\ \text{km} \end{aligned} \]

These positions are physically consistent: both lie between P and Q, and M lies between the emissions.

d) Explain why the midpoint prediction fails.

The engineer assumes that signal-arrival intervals vary uniformly with receiver position.

But the light-path geometry changes when a receiver lies between the two emission locations.

At M, the first pulse travels:

\[ 1.50-0.600=0.900\ \text{km} \]

The second pulse travels:

\[ 1.95-1.50=0.450\ \text{km} \]

The second signal therefore has a \(0.450\ \text{km}\) shorter journey.

That reduces the reception interval by:

\[ \begin{aligned} \Delta t_{\text{light}} &=\frac{0.450}{0.300}\\ &=1.50\ \mu\text{s} \end{aligned} \]

Thus M receives the signals:

\[ 7.50-1.50=6.00\ \mu\text{s} \]

apart, exactly as measured.

The midpoint in receiver position is not necessarily the midpoint in light-travel delay. That relationship depends on which side of each emission the receiver occupies.

This problem shows something stronger than the ordinary speed-from-\(\gamma\) calculation: with enough independent arrival-time measurements and known geometry, the receiver network can reconstruct the emission interval, relative speed, proper time, and emission positions even without the moving clock’s original data.

06What this method leads to next

Once you can identify the correct time-dilation factor and recover speed, the next useful question is how physicists obtain trustworthy time measurements when the moving object cannot carry a readable clock.

For the underlying relationship between moving clocks and the Lorentz factor, see Time Dilation in HSC Physics: Using the Lorentz Factor.

A particularly important application is the survival of fast-moving muons produced in Earth’s atmosphere. Unlike our hypothetical spacecraft, a muon cannot carry a stopwatch for us to read. Its lifetime must be studied statistically through particle detections. The reasoning is developed in Muon Evidence for Time Dilation: HSC Physics Explained.

For any measured time-dilation problem, make one decision before reaching for the calculator: are you comparing when two events occurred in different reference frames, or simply when signals from those events reached a detector?

Only after that distinction is settled does the measured factor tell you the speed.