Why Both Inertial Observers See the Other’s Clock Run Slow

Two spacecraft pass at the same point. Both pilots set their clocks to 0.00 s at that instant, then coast apart at a constant relative speed of \(0.60c\).

Pilot A says B’s clock is running slow. Pilot B says exactly the same about A’s clock.

How can both be right? Neither pilot has made a mistake. Each is comparing the other clock using a different definition of which distant events are happening at the same time. This is the relativity of simultaneity, and it is the missing piece that makes reciprocal time dilation logically consistent.

01Why both observers can say the other clock is slow

Start with a simpler question.

Imagine you are sitting inside spacecraft A. Your clock ticks normally, and one second on your clock is one second on your wristwatch. Nothing about your local experience suggests that time has slowed down.

Now look at spacecraft B moving past you.

According to your reference frame, B’s clock accumulates less time than yours over the same interval. Its clock is running slow relative to yours.

But B’s pilot can make exactly the same argument.

There is no universal stationary reference frame that gives A the correct answer and B the incorrect one. Both spacecraft are moving inertially, meaning neither is accelerating, and the laws of physics work identically in both frames.

The important distinction is that each observer uses their own reference frame to compare events at different locations.

First, what does a clock actually measure?

An event is simply something that happens at a particular position and time.

Examples include:

• A spacecraft clock recording a tick.
• A light pulse being emitted.
• A detector receiving a signal.
• A particle decaying.

Now consider two consecutive ticks on spacecraft B’s clock.

Both ticks occur at the same position relative to B because the clock is attached to the spacecraft. B can measure the time between them using its own clock.

This interval is called the proper time, represented by \(\Delta\tau\).

Proper time is the time measured by a single clock that is present at both events. The two events occur at the same position in that clock’s rest frame.

A different observer, such as A’s pilot, sees B moving. The two ticks therefore occur at different positions in A’s reference frame.

To determine the time between them, A uses a network of synchronised clocks positioned throughout A’s frame. These clocks record when B’s two ticks pass their respective positions.

The time interval assigned by A is called the coordinate time, represented by \(\Delta t\).

This gives us three quantities worth keeping separate:

The third quantity will become important when we discuss telescopes and light signals.

The time-dilation relationship

For a clock moving at constant speed relative to an inertial reference frame:

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

The Lorentz factor is:

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

Here:

• \(\Delta t\) is the coordinate time in the reference frame where the clock is moving, measured in seconds (s).
• \(\Delta\tau\) is the proper time recorded by the moving clock, measured in seconds (s).
• \(v\) is the relative speed between the clock and the reference frame, in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(c\) is the speed of light in a vacuum, approximately \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).
• \(\gamma\) is the dimensionless Lorentz factor.

Since \(\gamma\) is always greater than or equal to 1, the moving clock accumulates less proper time than the coordinate time assigned to the same pair of events.

Rearranging:

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

Notice that nothing in this equation identifies a permanently stationary spacecraft.

If A is the reference frame, B is moving.

If B is the reference frame, A is moving.

The relative speed is the same in either description, so the Lorentz factor is also the same.

That is why the time-dilation calculation is reciprocal.

02The missing ingredient: relativity of simultaneity

The reason the two observers disagree about distant clock readings is that simultaneity depends on the reference frame.

This idea was central to Albert Einstein’s development of special relativity in 1905.

Before considering the consequence for moving clocks, it helps to understand the problem Einstein was addressing.

Einstein’s train thought experiment

Imagine a long train travelling to the right past a railway platform.

Two lightning strikes occur simultaneously at opposite ends of the train, according to an observer standing at the midpoint of the platform.

The platform observer is exactly halfway between the strikes. If the flashes are emitted simultaneously, light from both ends travels equal distances and reaches the observer together.

Now consider an observer sitting at the midpoint of the moving train.

At the instant the strikes occur, the train observer is moving towards the light coming from the front of the train and away from the light coming from the rear.

What happens?

The light from the front reaches the train observer first.

The important point is not simply that the train observer receives one flash first.

Receiving a signal first does not automatically mean the event happened first. Light takes time to travel.

The deeper reasoning is that both observers use the same speed of light, \(c\), to synchronise clocks and assign times to distant events.

The platform observer assigns the two lightning strikes the same time.

The train observer, using the train’s own synchronised clocks, assigns different times to the strikes.

The two observers therefore disagree about whether the distant events happened simultaneously.

This is a thought experiment, not a physical experiment Einstein performed with a relativistic train. Its purpose was to expose an assumption that classical physics had treated as obvious: that everyone shares the same definition of now.

You can explore the event-timing method in more detail in How to Identify Simultaneous Train Events in a Given Frame.

Why this resolves reciprocal time dilation

Return to spacecraft A and B.

A’s synchronised clocks define one set of simultaneous events throughout A’s reference frame.

B’s synchronised clocks define a different set throughout B’s reference frame.

When the spacecraft are far apart, their pilots cannot simply point to one distant event and agree that it is happening right now.

Each pilot uses a different simultaneity convention.

That allows both to describe the other’s clock as slow without contradicting the other’s description.

Neither clock has malfunctioned. Neither observer has discovered an absolute rest frame.

The disagreement concerns how each frame assigns times to events at different locations.

03How physicists separated time dilation from ordinary Doppler shifts

There is another important distinction.

A clock can be assigned a slower rate in another reference frame, but that does not mean it will necessarily look slower through a telescope by exactly the same factor.

Light takes time to travel from the clock to the observer. If the source is moving, the distance changes between successive emissions.

This creates a Doppler effect.

A historical experiment by Herbert Ives and G. R. Stilwell in 1938 provides a useful example of how physicists investigated the small frequency shift associated with the rate of a moving source.

The Ives-Stilwell experiment

The measurement problem was straightforward to state but difficult to solve.

If light from moving atoms has a different observed frequency, how can we determine whether the difference comes from the motion of the source or from the change in its clock rate?

Ives and Stilwell investigated this using hydrogen canal rays and a spectrograph.

The apparatus worked in the following sequence.

1. A hydrogen discharge produced positive hydrogen particles inside the tube.
2. An electric field between electrodes accelerated the particles into a fast-moving beam.
3. The moving particles emitted a characteristic blue-green hydrogen spectral line, with a wavelength close to \(486.1\text{ nm}\).
4. A spectrograph recorded the emitted light and compared it with an undisplaced reference line from stationary hydrogen.
5. A mirror provided a second optical path, allowing the experimenters to compare light associated with opposite directions of observation.

The spectrograph recorded the positions of the spectral lines on a photographic plate.

Why the original measurement idea was difficult

The simplest proposed arrangement was to observe the moving particles exactly perpendicular to their direction of travel.

At exactly \(90^\circ\), the ordinary first-order Doppler shift disappears.

That should leave the much smaller second-order frequency shift associated with time dilation.

However, two problems made this difficult.

First, conventional canal-ray tubes produced particles with a spread of velocities. This broadened the spectral lines, making a tiny shift difficult to identify.

Second, observing at exactly \(90^\circ\) required extremely accurate alignment. Even a small angular error could introduce a first-order Doppler shift large enough to obscure the effect being investigated.

How the revised arrangement helped

The experiment used a canal-ray design that produced a narrower range of particle velocities.

Instead of relying on a perfectly perpendicular viewing angle, Ives and Stilwell observed light in opposite directions using the direct and reflected optical paths.

The two directions produced opposite first-order Doppler shifts.

By examining the centre of the pair of displaced spectral lines, the experimenters could isolate the much smaller second-order shift.

What did the experiment find?

In one reported measurement series, the mean observed shift was approximately:

\[ 0.0468\text{ Å} \]

The calculated shift for the measured particle velocity was:

\[ 0.0472\text{ Å} \]

The agreement was within the experimental uncertainty.

The original authors also discussed limitations involving the small size of the shift and possible effects from overlapping spectral-line components and changing line intensities.

The experiment was an important early test of the moving-source frequency change associated with time dilation. The 1938 paper interpreted the result within the Larmor-Lorentz framework; the same second-order effect is predicted by special relativity.

You can read Ives and Stilwell’s original 1938 paper.

A quick scale estimate

Why was the measurement so difficult?

For a particle travelling at approximately \(0.005c\), define:

\[ \beta=\frac{v}{c}=0.005 \]

The fractional change associated with the Lorentz factor is approximately:

\[ \begin{aligned} \gamma-1 &=\frac{1}{\sqrt{1-\beta^2}}-1\\ &\approx\frac{\beta^2}{2}\\ &=1.25\times10^{-5} \end{aligned} \]

For a reference wavelength of \(486.1\text{ nm}\), the magnitude of the corresponding second-order wavelength shift is approximately:

\[ \begin{aligned} \Delta\lambda &\approx\lambda_0(\gamma-1)\\ &=(486.1)(1.25\times10^{-5})\\ &=0.00608\text{ nm}\\ &=0.0608\text{ Å} \end{aligned} \]

This is a tiny displacement.

It explains why narrow spectral lines, accurate alignment, and careful measurements were essential.

There is also an important limitation to remember: the Ives-Stilwell experiment investigated the frequency shift from a moving source in a laboratory frame. It did not directly demonstrate that two inertial observers each describe the other’s clock as slow.

That reciprocal conclusion follows from the symmetry of inertial reference frames and the relativity of simultaneity.

04What you see through a telescope is not the same as time dilation

Let’s return to the spacecraft.

Suppose B is moving away from A and sends a light pulse every time its own clock advances by one second.

A receives those pulses at increasingly separated intervals.

Is this entirely due to time dilation?

No. There are two effects involved.

1. Time dilation increases the interval between successive emissions in A’s frame.
2. Because B is moving away, each new pulse has a greater distance to travel.

The second effect stretches the interval between arrivals even further.

This is the relativistic Doppler effect.

What changes if B approaches A?

If B approaches instead, the later pulse has a shorter distance to travel.

The time-dilation contribution remains 1.25 s, but the changing distance now reduces the arrival interval.

\[ \begin{aligned} \Delta t_{\mathrm{arrival}} &=1.25-0.75\\ &=0.50\text{ s} \end{aligned} \]

The comparison is:

Notice something interesting.

A moving clock can be assigned a slower rate because of time dilation, while its incoming signals can appear faster because it is approaching.

There is no contradiction.

The three effects must be kept separate:

• Time dilation changes the emission interval assigned by another inertial frame.
• Light-travel delay determines how long an individual signal takes to reach the receiver.
• Doppler effect changes the spacing between received signals when the source-observer distance changes.

This distinction is essential whenever an HSC question describes an observer watching a moving clock, rather than comparing its readings using synchronised clocks.

05What changes if the spacecraft turn around?

So far, A and B have both remained inertial.

What happens if they separate, then one spacecraft turns around and returns?

This is a useful contrasting case because the two clocks can now meet again and have their displays compared directly.

Imagine:

• A remains stationary in an inertial frame.
• B travels away at \(0.80c\).
• B reverses direction after 5.0 years of A-frame time.
• B returns at \(0.80c\), taking another 5.0 years in A’s frame.

Assume the reversal is instantaneous and ignore gravitational effects.

A measures a total journey time of:

\[ \Delta t_A=5.0+5.0=10.0\text{ years} \]

B’s clock measures proper time on each constant-speed section.

The Lorentz factor is:

\[ \gamma=\frac{1}{\sqrt{1-0.80^2}} =\frac{1}{0.60} =1.67 \]

The proper time for each section is:

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

Therefore, B accumulates:

\[ \Delta\tau_{B,\mathrm{total}} =3.0+3.0 =6.0\text{ years} \]

When they reunite:

A has accumulated four more years.

Why doesn’t this contradict reciprocal time dilation?

Because B did not remain in one inertial reference frame throughout the journey.

During the outbound section, B uses one inertial frame to assign times to distant events on A.

During the return section, B uses a different inertial frame.

The change of frame changes which distant events B considers simultaneous with its own clock readings. This is the crucial difference from the two spacecraft that simply coast apart.

Importantly, B’s own clock does not jump when it turns around. Its accumulated proper time remains continuous. It is the assigned time of distant events on A that changes when B switches frames.

Acceleration is not a magical process that directly makes a clock lose four years. The difference comes from the proper time accumulated along the two different journeys.

This is the central idea behind the twin paradox.

06A reliable way to approach reciprocal time-dilation questions

When a question asks why both observers describe the other clock as slow, work through these distinctions.

A useful rule is to never use the phrase at the same time in a relativity problem without specifying the reference frame.

That small habit prevents a surprising number of mistakes.

For a more detailed refresher on identifying the correct event pair before choosing proper time, read How to Identify the Two Events Before Choosing Proper Time.

07Questions and solutions

Question 1

A spacecraft carries an atomic clock and moves at a constant speed of \(0.60c\) relative to an inertial laboratory.

The spacecraft clock records 8.0 s between two ticks.

Calculate the time interval assigned to those same two tick events by an observer in the laboratory.

Solution 1

The laboratory assigns a time interval of 10.0 s between the two ticks.

The spacecraft clock measures the proper time because both events occur at the same position in the spacecraft’s rest frame.

Therefore:

\[ \Delta\tau=8.0\text{ s} \]

Calculate the Lorentz factor:

\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.60^2}}\\ &=\frac{1}{0.80}\\ &=1.25 \end{aligned} \]

Apply the time-dilation equation:

\[ \begin{aligned} \Delta t&=\gamma\Delta\tau\\ &=1.25(8.0)\\ &=10.0\text{ s} \end{aligned} \]

The laboratory assigns a longer interval than the spacecraft clock records.

The trap is reversing the relationship. The moving clock measures the proper time, which is the shorter interval.

Question 2

Two spacecraft, A and B, pass each other at a constant relative speed of \(0.80c\). Both clocks are reset to zero at the instant they pass.

An observer in A’s reference frame determines the reading on B’s clock at the event simultaneous with A’s clock reading 12.0 s.

A second observer in B’s reference frame determines the reading on A’s clock at the event simultaneous with B’s clock reading 12.0 s.

Calculate both readings.

A student argues that the two results are contradictory because both observers cannot be correct.

Explain why the student’s argument is incorrect.

Solution 2

Both observers assign a reading of 7.20 s to the other spacecraft’s clock, and there is no contradiction.

The key is that they are selecting different events as simultaneous.

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: Consider A’s reference frame.

A is stationary in its own frame.

The coordinate time is:

\[ \Delta t_A=12.0\text{ s} \]

B’s clock measures proper time:

\[ \begin{aligned} \Delta\tau_B&=\frac{\Delta t_A}{\gamma}\\ &=\frac{12.0}{1.67}\\ &=7.20\text{ s} \end{aligned} \]

A therefore assigns B a reading of 7.20 s at the event simultaneous with A = 12.0 s.

Step 3: Consider B’s reference frame.

Now B is stationary, and A is moving.

The coordinate time is:

\[ \Delta t_B=12.0\text{ s} \]

A’s clock records:

\[ \begin{aligned} \Delta\tau_A&=\frac{\Delta t_B}{\gamma}\\ &=\frac{12.0}{1.67}\\ &=7.20\text{ s} \end{aligned} \]

B also assigns the other clock a reading of 7.20 s.

Step 4: Identify the incorrect assumption.

The student assumes both observers are referring to the same distant events when they use the phrase at the same time.

They are not.

A’s synchronised clocks and B’s synchronised clocks define different sets of simultaneous events.

Both observers correctly describe the other clock as slow within their own inertial reference frame.

Question 3

Spacecraft B passes stationary receiver A at event O. Immediately afterwards, B moves directly away from A at a constant speed of \(0.60c\).

B emits a light pulse whenever its own clock advances by 2.00 s.

Determine:

• The interval between successive emissions according to A’s reference frame.
• The distance B travels between emissions.
• The interval between successive pulse arrivals at A.

Explain why the arrival interval is not the time-dilation interval.

Assume that the first pulse is emitted after B’s clock has advanced by 2.00 s. Ignore gravity and assume light travels directly through a vacuum.

Solution 3

The pulses arrive 4.00 s apart, even though their emission events are only 2.50 s apart in A’s reference frame.

The additional 1.50 s comes from the increasing distance that the light must travel.

Step 1: Calculate the Lorentz factor.

\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.60^2}}\\ &=1.25 \end{aligned} \]

Step 2: Calculate the emission interval in A’s frame.

B’s clock measures:

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

Therefore:

\[ \begin{aligned} \Delta t_{\mathrm{emit}} &=\gamma\Delta\tau_B\\ &=1.25(2.00)\\ &=2.50\text{ s} \end{aligned} \]

The emissions are 2.50 s apart according to A.

Step 3: Calculate B’s displacement between emissions.

\[ \begin{aligned} \Delta d&=v\Delta t_{\mathrm{emit}}\\ &=(0.60c)(2.50\text{ s})\\ &=1.50\text{ light-seconds} \end{aligned} \]

B moves 1.50 light-seconds farther away between emissions.

Step 4: Calculate the extra light-travel time.

The second pulse must travel an additional 1.50 light-seconds.

Since light travels one light-second per second:

\[ \Delta t_{\mathrm{extra}}=1.50\text{ s} \]

Step 5: Calculate the arrival interval.

\[ \begin{aligned} \Delta t_{\mathrm{arrival}} &=\Delta t_{\mathrm{emit}}+\Delta t_{\mathrm{extra}}\\ &=2.50+1.50\\ &=4.00\text{ s} \end{aligned} \]

The received pulses are 4.00 s apart.

The time-dilation effect accounts for the 2.50 s emission interval. The additional 1.50 s comes from the changing distance between B and A.

A common mistake is to treat the time between received pulses as the time interval assigned to the emission events. These are different pairs of events.

Question 4

Spacecraft A remains stationary in an inertial reference frame.

Spacecraft B starts beside A, travels away at \(0.80c\) for 5.0 years according to A’s clock, instantly reverses direction, and returns at \(0.80c\) for another 5.0 years according to A’s clock.

Both clocks initially read zero.

Assume gravitational effects are negligible and the turnaround takes no time.

Calculate the final reading on each clock when the spacecraft reunite.

Explain why the reciprocal time-dilation argument for two continuously inertial observers cannot be applied to B’s entire journey.

Solution 4

At reunion, A’s clock reads 10.0 years and B’s clock reads 6.0 years.

The two spacecraft can now compare their displays directly because they occupy the same position at the same event.

Step 1: Calculate the Lorentz factor.

\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.80^2}}\\ &=\frac{1}{0.60}\\ &=1.67 \end{aligned} \]

Step 2: Calculate the proper time accumulated by B on the outbound journey.

A’s clock measures 5.0 years.

B’s clock measures:

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

Step 3: Calculate the proper time accumulated by B on the return journey.

The speed and duration in A’s frame are unchanged.

Therefore:

\[ \Delta\tau_{\mathrm{return}}=3.0\text{ years} \]

Step 4: Calculate the final readings.

For B:

\[ \begin{aligned} \Delta\tau_{B,\mathrm{total}} &=3.0+3.0\\ &=6.0\text{ years} \end{aligned} \]

For A:

\[ \Delta\tau_A=5.0+5.0=10.0\text{ years} \]

The final comparison is:

Why is there no contradiction?

The crucial difference is that B accelerates and changes direction.

It cannot use one inertial reference frame for its entire journey.

During the outbound section, B describes A’s clock as slow using its outbound inertial frame. During the return section, it uses a different inertial frame.

The change in simultaneity assignments between these frames resolves the apparent contradiction.

The acceleration does not directly cause the four-year difference. The difference comes from the different proper times accumulated along the two complete paths.

Question 5

Two spacecraft, A and B, pass each other at event O and set both clocks to zero.

They then continue at a constant relative speed of \(0.60c\).

Define two events:

• Event \(A_8\): A’s clock reads 8.00 s.
• Event \(B_8\): B’s clock reads 8.00 s.

Consider these two events, not two separate comparisons of clock rates.

Determine:

(a) The time and position of both events in A’s reference frame, taking A’s position as \(x=0\).

(b) Which event occurs first according to A’s reference frame.

(c) Which event occurs first according to B’s reference frame.

(d) Whether either event could physically cause the other. Justify your answer using the separation between the events and the speed of light.

Solution 5

A’s frame orders \(A_8\) before \(B_8\), while B’s frame orders \(B_8\) before \(A_8\). Neither event can cause the other because they are separated by a spacelike interval.

This is a particularly important result. Two observers can disagree about the order of events without disagreeing about cause and effect.

Step 1: Calculate the Lorentz factor.

The relative speed is:

\[ v=0.60c \]

Therefore:

\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.60^2}}\\ &=1.25 \end{aligned} \]

Step 2: Find the coordinates of event \(A_8\) in A’s frame.

A is stationary in its own frame.

Therefore, when A’s clock reads 8.00 s:

\[ t_{A_8}=8.00\text{ s} \]

and

\[ x_{A_8}=0 \]

Step 3: Find the coordinates of event \(B_8\) in A’s frame.

B’s clock records 8.00 s of proper time.

The coordinate time assigned by A is:

\[ \begin{aligned} t_{B_8}&=\gamma\Delta\tau_B\\ &=1.25(8.00)\\ &=10.0\text{ s} \end{aligned} \]

B travels at \(0.60c\), so its position at this event is:

\[ \begin{aligned} x_{B_8}&=vt\\ &=(0.60c)(10.0\text{ s})\\ &=6.00\text{ light-seconds} \end{aligned} \]

The coordinates are:

Step 4: Determine the order in A’s frame.

A assigns:

\[ t_{A_8}=8.00\text{ s} \]

and

\[ t_{B_8}=10.0\text{ s} \]

Therefore, A says \(A_8\) occurs 2.00 s before \(B_8\).

Step 5: Determine the order in B’s frame.

Now reverse the reference frame.

B is stationary, and A moves at \(0.60c\).

B’s clock reads 8.00 s at event \(B_8\).

For the event \(A_8\), A’s clock has recorded 8.00 s of proper time. B assigns the coordinate interval:

\[ \begin{aligned} t'_{A_8}&=\gamma\Delta\tau_A\\ &=1.25(8.00)\\ &=10.0\text{ s} \end{aligned} \]

Therefore, in B’s frame:

• \(B_8\) occurs at 8.00 s.
• \(A_8\) occurs at 10.0 s.

B says \(B_8\) occurs 2.00 s before \(A_8\).

The event order has reversed.

Step 6: Check whether this violates causality.

In A’s frame, the spatial separation is:

\[ \Delta x=6.00\text{ light-seconds} \]

The time separation is:

\[ \Delta t=10.0-8.00=2.00\text{ s} \]

In two seconds, light can travel:

\[ c\Delta t=2.00\text{ light-seconds} \]

But the events are separated by 6.00 light-seconds.

Since:

\[ \Delta x>c\Delta t \]

there is not enough time for a light signal, or any slower physical signal, to travel between the events.

The events are spacelike separated.

This is why the reversal is physically acceptable.

If one event could cause the other, all inertial observers would agree on their time order. Cause must precede effect in every inertial reference frame.

But these two events cannot influence each other, so their order is allowed to depend on the reference frame.

The deeper lesson is that reciprocal time dilation and relativity of simultaneity are connected. Neither observer is wrong about the other’s clock, because they are not using the same definition of distant simultaneity.

08Where to take this next

Once you understand why two inertial observers can each describe the other’s clock as slow, the next useful step is to compare clocks that follow different paths and eventually reunite.

That is where proper time becomes especially important: instead of comparing distant events using synchronised clock networks, you can bring the clocks together and compare their accumulated readings directly.

For the experimental evidence behind these effects, continue with Atomic-Clock Tests of Relativity: HSC Physics Explained.