Why Moving Clocks Run Slow: Proper Time and Light Delays

Imagine carrying an atomic clock on an aircraft. When it lands, should it always show less time than an identical clock that stayed on the ground?

The answer seems obvious if you’ve learnt that moving clocks run slow. But in the 1971 Hafele-Keating experiment, clocks flown eastward finished behind the ground clocks, while clocks flown westward finished ahead.

That result makes more sense once you understand exactly what time dilation compares.

In a standard HSC special relativity calculation, a moving clock records less time between two events than an observer in whose reference frame the clock is moving. This happens because the two events are separated by a longer time interval in that observer’s frame. A light-clock thought experiment makes the reason visible: light must travel a longer diagonal path while maintaining the same speed, \(c\).

The clock isn’t mechanically slowing down. Its own mechanism works normally in its rest frame. The difference is between the time intervals measured in different reference frames.

01What does it actually mean for a clock to run slow?

Start with two events.

Suppose a clock aboard a spacecraft produces a tick at one moment, then another tick a little later. Each tick is an event: something happening at a particular position and time.

There are two important ways to describe the interval between them.

Proper time, written as \(\Delta\tau\), is the time measured by a clock that is present at both events. In that clock’s rest frame, the two events happen at the same position.

Coordinate time, written as \(\Delta t\), is the interval assigned by another reference frame. If that frame sees the clock moving, the two events happen at different positions.

Consider a spacecraft travelling past Earth.

• In the spacecraft’s frame, the onboard clock stays in one position. It measures the proper time between its ticks.
• In Earth’s frame, the spacecraft moves between those same tick events. Earth uses synchronised clocks to determine the interval between them.
• Earth’s measured interval is longer than the onboard clock’s reading.

This is what physicists mean when they say the moving clock runs slow.

The important distinction is that we must compare the same two events in both frames.

The third row is particularly important. Receiving information about a clock’s ticks is not necessarily the same as measuring the time between the ticks themselves.

02Why does the moving clock run slow?

A useful way to understand the effect is to imagine a very simple clock made from two parallel mirrors.

A pulse of light travels between the mirrors, reflecting off each one. Define one complete tick as the time taken for the pulse to travel from the lower mirror to the upper mirror and back to the lower mirror.

This is called a light clock.

First consider the clock at rest

Imagine sitting beside the light clock.

The mirrors are stationary relative to you, so the light pulse travels vertically between them.

If the distance between the mirrors is \(L\), the time for one complete tick is:

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

Here:

• \(\Delta\tau\) is the proper time for one complete tick, in seconds.
• \(L\) is the separation between the mirrors, in metres.
• \(c\) is the speed of light in a vacuum, approximately \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).

The factor of two appears because light travels up and then back down.

Now watch the clock move

Imagine the same clock travelling horizontally past you at a constant speed \(v\).

What happens to the light pulse?

It cannot travel straight up and down from your perspective.

While the light travels towards the upper mirror, that mirror moves horizontally. The light must therefore follow a diagonal path to reach it.

The light travels further than it did in the clock’s rest frame.

But there is a catch: you still measure its speed as \(c\).

Light does not slow down to compensate for the longer path.

Longer distance at the same speed means a longer travel time.

The moving clock therefore takes longer to complete a tick according to your measurements.

This is the physical reasoning behind time dilation.

Turning the light path into an equation

We can use Pythagoras to calculate the difference.

Consider just half a tick, from the lower mirror to the upper mirror.

In the laboratory frame:

• The light travels a diagonal distance of \(c\Delta t/2\).
• The clock moves horizontally by \(v\Delta t/2\).
• The vertical mirror separation is \(L\).

The mirror separation is perpendicular to the direction of motion, so it is unchanged by length contraction.

These distances form a right-angled triangle.

Therefore:

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

From the clock’s rest frame, we already know:

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

Substitute this into the triangle equation:

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

Rearranging gives:

\[ c^2\Delta t^2 = c^2\Delta\tau^2+v^2\Delta t^2 \]

Move the velocity term to the left:

\[ \Delta t^2 \left(1-\frac{v^2}{c^2}\right) = \Delta\tau^2 \]

Taking the positive square root:

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

The expression multiplying the proper time is the Lorentz factor:

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

So the time-dilation equation is:

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

Because \(\gamma\) is always equal to or greater than one for speeds below the speed of light:

\[ \boxed{\Delta t\geq\Delta\tau} \]

The time interval in the frame where the clock moves is longer.

This is why the clock accumulates fewer seconds than the stationary-frame clocks during the same interval.

An important detail about the light clock

There is a subtle distinction between a half-tick and a complete tick.

In the diagram, event A occurs when the light leaves the lower mirror. Event B occurs when it reaches the upper mirror. Event C occurs when it returns to the lower mirror.

• A and C occur at the same position in the light clock’s rest frame.
• A and B occur at different positions in that frame.

Therefore, A to C is the proper-time interval for the clock’s complete tick.

A to B is not a proper-time interval. It is a light-travel interval between two spatially separated events.

You can calculate the time from A to B using a network of synchronised clocks in the clock’s rest frame, but that does not make it proper time.

This distinction becomes important when a question gives you the time for a light pulse to travel somewhere and asks you to identify the proper time.

03Why watching a moving clock can give you a different result

There is another complication.

Suppose you stand beside a road and watch a moving clock through a telescope. The clock sends you a light signal every time it ticks.

You might assume that the time between signals arriving at your telescope must equal the time between the clock’s ticks.

It doesn’t.

Light takes time to travel from the clock to your telescope. If the clock changes its distance from you between emissions, the second signal has a different travel time from the first.

This is a light-travel delay, not an additional time-dilation effect.

The arrival time of one signal is:

\[ t_{\mathrm{rec}}= t_{\mathrm{emit}}+\frac{d}{c} \]

where:

• \(t_{\mathrm{rec}}\) is the time the receiver detects the signal.
• \(t_{\mathrm{emit}}\) is the time the signal was emitted.
• \(d\) is the distance from the clock to the receiver at emission.
• \(c\) is the speed of light.

For two successive signals:

\[ \Delta t_{\mathrm{rec}} = \Delta t_{\mathrm{emit}} + \frac{\Delta d}{c} \]

Here, \(\Delta d\) is the change in the source-to-receiver distance between emissions.

If the clock is moving away, \(\Delta d\) is positive. The second signal travels further, so the received interval increases.

If the clock is approaching, \(\Delta d\) is negative. The second signal has less distance to travel, so the received interval decreases.

This is closely related to the Doppler effect.

The important point is that there are now two separate effects:

1. Time dilation changes the interval between the actual emission events.
2. Changing distance changes the interval between the signal reception events.

A telescope measures the second. To determine the clock’s actual ticking rate, you must account for both.

04A real experiment: Do physical clocks actually run at different rates?

The light clock is a thought experiment. It makes the geometry of time dilation easy to understand, but a real clock doesn’t need two mirrors and a light pulse.

An important historical test was carried out by physicists Joseph Hafele and Richard Keating in 1971.

Their question was straightforward: if relativistic effects really change the time accumulated by moving clocks, could they measure the difference using actual atomic clocks?

The apparatus and measurement

Hafele and Keating used four cesium-beam atomic clocks.

The experiment followed this sequence:

1. Reference atomic clocks remained at the United States Naval Observatory (USNO).
2. Two cesium clocks travelled on commercial aircraft around the world in an eastward direction.
3. Another two clocks travelled around the world in a westward direction.
4. The researchers recorded the actual flight paths, including the relevant speeds and altitudes.
5. After the flights, they compared the airborne clocks with the reference time scale at the observatory.

The researchers wanted to test whether the clocks accumulated different amounts of time during their journeys.

What should the experiment show?

A simple special-relativity prediction suggests that a clock moving relative to a chosen inertial reference frame accumulates less proper time than a clock at rest in that frame.

So you might predict that both aircraft clocks should finish behind the ground clocks.

But this prediction is incomplete.

There are two important complications.

First: Earth is rotating.

The ground clocks at the observatory are not actually stationary in an Earth-centred inertial frame. They move eastward because Earth rotates.

An eastbound aircraft increases its speed relative to that frame. A westbound aircraft can reduce its speed relative to that frame.

Consequently, the kinematic time-dilation contribution is different for the two journeys.

Second: gravity also affects clock rates.

The aircraft are at higher altitudes than the observatory.

Clocks at higher gravitational potential run faster than clocks at lower gravitational potential, according to general relativity.

The experiment therefore involved both:

• Special-relativistic time dilation due to motion.
• Gravitational time dilation due to differences in altitude.

The simplified prediction that both aircraft clocks must lose time because they are moving ignores these additional effects.

What did the researchers observe?

The predicted and measured differences were:

The sign convention is airborne clock reading minus the USNO reference reading.

A negative result means the airborne clock finished behind. A positive result means it finished ahead.

The measurements agreed with the relativistic predictions within the reported uncertainties.

The original prediction and observation papers were published in Science in 1972:

• Hafele and Keating: Predicted relativistic time gains
• Hafele and Keating: Observed relativistic time gains

A useful modern historical account is also available from NIST.

What was the important correction?

The clocks weren’t faulty, and the experiment didn’t disprove time dilation.

The problem was the oversimplified prediction that both aircraft clocks must lose time simply because they were moving relative to the ground.

The correct analysis had to account for the ground clocks’ own motion, the aircraft’s motion relative to an inertial frame, and the gravitational effects of altitude.

The researchers also compared the clocks after their journeys rather than relying simply on the times at which distant observers received signals. This avoided confusing the accumulated clock readings with changing light-travel delays.

The experiment demonstrated that real clocks can accumulate different amounts of time along different journeys.

For an ordinary HSC special-relativity calculation, however, you will usually be asked to isolate the kinematic effect and neglect gravity. Don’t apply the combined aircraft-experiment results to a question that specifies only constant relative velocity.

A useful scale estimate

Why did the experiment need atomic clocks?

Consider a simplified example.

A clock moves at \(250\ \mathrm{m\,s^{-1}}\) relative to an inertial reference frame for \(40.0\) hours. Assume constant speed and ignore gravity.

First convert the time:

\[ T=40.0(3600) =1.44\times10^5\ \mathrm{s} \]

Calculate the Lorentz factor:

\[ \gamma= \frac{1}{\sqrt{1- \left(\frac{250}{3.00\times10^8}\right)^2}} \]

The result is extremely close to one:

\[ \gamma-1\approx3.47\times10^{-13} \]

The moving clock records:

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

The difference between the reference-frame interval and the moving clock’s reading is approximately:

\[ \Delta t_{\mathrm{loss}} = T-\frac{T}{\gamma} \]

\[ \approx (1.44\times10^5) (3.47\times10^{-13}) \]

\[ \boxed{\Delta t_{\mathrm{loss}}\approx50.0\ \mathrm{ns}} \]

A difference of just 50 nanoseconds can build up over a long journey.

This is an illustrative estimate, not a reconstruction of the Hafele-Keating flight data. Their actual calculations required the changing aircraft velocities and gravitational effects.

The experiment also shows why precise atomic clocks are useful for testing relativistic predictions.

05How to choose the correct time in an HSC calculation

Before substituting numbers into the time-dilation equation, work through these decisions.

For the standard constant-speed time-dilation calculation:

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

Remember:

• \(\Delta\tau\) is the proper time, measured in seconds.
• \(\Delta t\) is the longer interval in the frame where the clock is moving, measured in seconds.
• \(\gamma\) is the dimensionless Lorentz factor.
• \(v\) is the relative speed between the frames, in \(\mathrm{m\,s^{-1}}\).
• \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\) is the speed of light in a vacuum.

If you know the proper time and want the longer frame interval, multiply by \(\gamma\).

If you know the frame interval and want the moving clock’s reading, divide by \(\gamma\).

For a more detailed treatment of identifying the correct interval, see How to Identify the Two Events Before Choosing Proper Time.

For further practice with the equation itself, see Time Dilation in HSC Physics: Using the Lorentz Factor.

06Questions and solutions

Question 1

A spacecraft travels at \(0.60c\) relative to an Earth laboratory.

An onboard clock measures \(4.0\ \mu\mathrm{s}\) between two flashes generated at the same position inside the spacecraft.

Calculate the time interval between the flash-generation events measured in Earth’s reference frame.

Assume constant velocity and negligible gravitational effects.

Solution 1

The Earth-frame interval is \(5.0\ \mu\mathrm{s}\). The onboard clock measures proper time because both flash-generation events occur at the same position in the spacecraft’s rest frame.

Step 1: Identify the proper time.

The onboard clock is present at both events.

Therefore:

\[ \Delta\tau=4.0\ \mu\mathrm{s} \]

Step 2: Calculate the Lorentz factor.

\[ \gamma= \frac{1}{\sqrt{1-(0.60)^2}} \]

\[ =\frac{1}{\sqrt{0.64}} \]

\[ \gamma=1.25 \]

Step 3: Calculate the Earth-frame interval.

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

\[ =1.25(4.0) \]

\[ \boxed{\Delta t=5.0\ \mu\mathrm{s}} \]

Earth measures \(5.0\ \mu\mathrm{s}\) between the same events, while the spacecraft clock records \(4.0\ \mu\mathrm{s}\).

The distinction is between the time recorded by the moving clock and the time assigned by the Earth frame.

Question 2

A transmitter carrying an atomic clock moves directly away from a stationary receiver at \(0.60c\).

The receiver detects two consecutive light pulses \(4.0\ \mu\mathrm{s}\) apart.

Assuming the transmitter emits one pulse per clock tick, determine the proper time between its emissions.

Assume constant velocity, and that the signals travel through a vacuum.

Solution 2

The transmitter’s proper tick interval is \(2.0\ \mu\mathrm{s}\). The receiver’s measurement includes both time dilation and the extra light-travel time caused by the transmitter moving away.

Step 1: Calculate the Lorentz factor.

\[ \gamma= \frac{1}{\sqrt{1-(0.60)^2}} =1.25 \]

Let the proper time between emissions be \(\Delta\tau\).

The laboratory emission interval is:

\[ \Delta t_{\mathrm{emit}} = 1.25\Delta\tau \]

Step 2: Calculate the change in distance.

The transmitter moves away during the time between emissions.

Its change in distance is:

\[ \Delta d=v\Delta t_{\mathrm{emit}} \]

Substitute:

\[ \Delta d= 0.60c(1.25\Delta\tau) \]

\[ \Delta d=0.75c\Delta\tau \]

The additional light-travel time is:

\[ \frac{\Delta d}{c} = 0.75\Delta\tau \]

Step 3: Relate the received interval to proper time.

The receiver measures:

\[ \Delta t_{\mathrm{rec}} = \Delta t_{\mathrm{emit}} + \frac{\Delta d}{c} \]

Therefore:

\[ 4.0=1.25\Delta\tau+0.75\Delta\tau \]

\[ 4.0=2.00\Delta\tau \]

Rearranging:

\[ \boxed{\Delta\tau=2.0\ \mu\mathrm{s}} \]

The clock’s own tick interval is \(2.0\ \mu\mathrm{s}\).

A tempting mistake would be to divide the received interval by the Lorentz factor:

\[ \frac{4.0}{1.25}=3.2\ \mu\mathrm{s} \]

But this is incorrect because the \(4.0\ \mu\mathrm{s}\) is a reception interval, not the interval between emission events in the laboratory frame.

The extra travel time must be removed before applying the time-dilation relationship.

Question 3

A stationary detector on a railway platform records two events at the same position, separated by \(12.0\ \mu\mathrm{s}\) according to the platform’s clock.

A train passes the detector at \(0.80c\).

Calculate the time interval between the same two events in the train’s reference frame.

Identify which interval is the proper time, and explain why it would be incorrect to describe this as the train’s clock running slow.

Solution 3

The train frame assigns an interval of \(20.0\ \mu\mathrm{s}\), and the platform’s \(12.0\ \mu\mathrm{s}\) is the proper time. The two events occur at the same position in the platform frame, not in the train frame.

Step 1: Identify the proper-time frame.

Both events occur at the stationary detector.

The platform clock is present at both events, so:

\[ \Delta\tau=12.0\ \mu\mathrm{s} \]

The train moves relative to the platform.

Step 2: Calculate the Lorentz factor.

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

\[ =\frac{1}{\sqrt{0.36}} \]

\[ \gamma=1.667 \]

Step 3: Calculate the interval in the train frame.

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

\[ =1.667(12.0) \]

\[ \boxed{\Delta t=20.0\ \mu\mathrm{s}} \]

The train frame assigns a longer interval because the two events occur at different positions in that frame.

The trap is assuming that the moving train must always be the clock experiencing the shorter interval.

Proper time belongs to the particular pair of events being considered. Here, the platform detector measures the proper time.

This is a useful reminder that you must identify the event pair before deciding which clock is moving and which time is proper.

Question 4

The Hafele-Keating experiment compared atomic clocks flown around the world with reference clocks at the US Naval Observatory.

The following results use the convention:

\[ \Delta R=R_{\mathrm{airborne}}-R_{\mathrm{ground}} \]

A student argues:

“The westward clock gained time, so time dilation cannot be correct. A moving clock should always run slower than a stationary clock.”

Evaluate this argument.

In your answer:

(a) Explain the two relativistic effects involved in the experiment.

(b) Explain why the westward clock could finish ahead of the ground reference clocks.

(c) State what the experiment demonstrated.

Solution 4

The student’s argument is incorrect. The westward result does not contradict time dilation because the experiment involved both motion and gravitational effects, and the ground reference clocks were also moving in an inertial frame.

Step 1: Identify the kinematic effect.

Special relativity predicts that a clock moving relative to an inertial reference frame accumulates less proper time than a clock at rest in that frame.

However, the ground clocks at the observatory were not stationary in an Earth-centred inertial frame. They were moving eastward because Earth rotates.

An eastbound aircraft moves faster eastward relative to this frame, while a westbound aircraft can move more slowly than the ground clocks in that frame.

Consequently, the kinematic contribution differs between the two flights.

Step 2: Identify the gravitational effect.

The aircraft travelled at higher altitudes than the observatory.

Clocks at higher gravitational potential run faster than clocks at lower gravitational potential.

This produces a positive contribution to the airborne clocks’ accumulated time relative to the ground clocks.

Step 3: Explain the westward result.

For the westward flight, the aircraft’s lower speed relative to an Earth-centred inertial frame favoured a gain in accumulated time compared with the ground clocks.

The gravitational effect also favoured a gain.

Together, these effects explain why the westward clock finished ahead.

The phrase “moving clocks run slow” is incomplete unless you specify the reference frame and the effects being considered.

Step 4: Evaluate the experimental evidence.

The eastward clock was predicted to lose \(40\ \mathrm{ns}\), and actually lost \(59\ \mathrm{ns}\).

The westward clock was predicted to gain \(275\ \mathrm{ns}\), and actually gained \(273\ \mathrm{ns}\).

Both measurements agreed with the predictions within their reported uncertainties.

The experiment therefore provided evidence supporting the combined predictions of special and general relativity.

It was not a simple isolated test of special-relativistic time dilation, because gravitational effects were also significant.

Question 5

A light clock consists of two parallel mirrors separated by \(0.60\ \mathrm{m}\) in its rest frame.

The clock moves at \(0.80c\) towards a stationary detector in a laboratory. Its lower mirror contains a transmitter that emits a short radio pulse towards the detector every time the light pulse completes one full journey and returns to the lower mirror.

Consider three events:

• Event A: the light pulse leaves the lower mirror and the first radio signal is emitted.
• Event B: the light pulse reflects from the upper mirror.
• Event C: the light pulse returns to the lower mirror and the second radio signal is emitted.

The laboratory detector receives both radio signals.

Assume:

• The clock moves at constant velocity.
• The detector is ahead of the clock, in its direction of motion.
• Radio signals travel through a vacuum at speed \(c\).
• The initial separation is large enough for both signals to reach the detector.
• Gravitational effects are negligible.

Calculate:

(a) The proper time between A and C.

(b) The laboratory-frame interval between the two radio-signal emissions.

(c) The interval between the signals arriving at the detector.

(d) Explain why the time \(L/c\) is not the proper time between A and B, and why the detector’s measured interval cannot be used directly as the clock’s proper tick.

Solution 5

The proper time between A and C is \(4.00\ \mathrm{ns}\), the laboratory emission interval is \(6.67\ \mathrm{ns}\), and the detector receives the signals \(1.33\ \mathrm{ns}\) apart.

The key is to keep three different measurements separate: the complete light-clock tick, the laboratory interval between emissions, and the interval between signal receptions.

Step 1: Identify the proper-time events.

Events A and C both occur at the lower mirror.

In the light clock’s rest frame, these events happen at the same position.

Therefore, the proper time is the time for one complete round trip:

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

Substitute:

\[ \Delta\tau= \frac{2(0.60)}{3.00\times10^8} \]

\[ \boxed{\Delta\tau=4.00\ \mathrm{ns}} \]

Step 2: Explain why A to B is not proper time.

Event A occurs at the lower mirror, while event B occurs at the upper mirror.

They are separated by \(0.60\ \mathrm{m}\) in the clock’s rest frame.

Therefore, they are not co-located in that frame.

The time between A and B is:

\[ \frac{L}{c} = \frac{0.60}{3.00\times10^8} \]

\[ =2.00\ \mathrm{ns} \]

This is half the duration of a complete tick, but it is not the proper-time interval between A and B.

The distinction matters because the radio transmitter only emits signals at A and C. It does not emit one at B.

Using \(2.00\ \mathrm{ns}\) as the proper tick would describe the wrong pair of events.

Step 3: Calculate the laboratory emission interval.

The clock travels at \(0.80c\).

Its Lorentz factor is:

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

\[ =\frac{1}{0.60} \]

\[ \gamma=1.667 \]

The laboratory interval between the emission events is:

\[ \Delta t_{\mathrm{emit}} = \gamma\Delta\tau \]

\[ =1.667(4.00) \]

\[ \boxed{\Delta t_{\mathrm{emit}}=6.67\ \mathrm{ns}} \]

The laboratory assigns \(6.67\ \mathrm{ns}\) between the two emissions.

Step 4: Calculate how far the clock moves towards the detector.

During this interval:

\[ \Delta d=v\Delta t_{\mathrm{emit}} \]

\[ =0.80c(6.67\times10^{-9}) \]

\[ \boxed{\Delta d=1.60\ \mathrm{m}} \]

The second radio signal is emitted \(1.60\ \mathrm{m}\) closer to the detector.

Step 5: Calculate the change in signal travel time.

The second signal has \(1.60\ \mathrm{m}\) less distance to travel.

The reduction in travel time is:

\[ \Delta t_{\mathrm{delay}} = \frac{1.60}{3.00\times10^8} \]

\[ =5.33\ \mathrm{ns} \]

Step 6: Calculate the reception interval.

Because the second signal travels a shorter distance:

\[ \Delta t_{\mathrm{rec}} = \Delta t_{\mathrm{emit}} - \Delta t_{\mathrm{delay}} \]

\[ =6.67-5.33 \]

\[ \boxed{\Delta t_{\mathrm{rec}}=1.33\ \mathrm{ns}} \]

The detector receives the signals much closer together than the laboratory emission events.

This does not mean the clock’s proper tick is \(1.33\ \mathrm{ns}\).

The three intervals describe different event pairs:

The tempting mistakes are to use the one-way mirror transit as the proper tick, or to treat the received signal interval as the actual emission interval.

Both mistakes mix different pairs of events.

The correct approach is to identify the events first, determine where they occur in each reference frame, and then account for light-travel time separately.

The central idea is simple: a moving clock runs slow in a chosen inertial frame because its own proper-time interval between ticks is shorter than the corresponding coordinate-time interval in that frame.

Once you can distinguish proper time, coordinate time, and signal reception time, the next challenge is usually simultaneity: deciding which events at different locations an observer considers to happen at the same time.