How to Identify Simultaneous Train Events in a Given Frame

A camera beside a railway track records a flash from one end of a moving train before a flash from the other. Which lamp actually flashed first?

You can’t tell from the camera footage alone. The nearer flash may have had a shorter journey to the camera, even if both lamps switched on together.

To find which train events are simultaneous in a stated reference frame, compare the times the events occurred according to clocks synchronised in that frame, not simply when their light was received. If two events have the same event time, they are simultaneous in that frame. If you change frames, you must compare the event times again.

01First identify what the clock actually measured

A flash of light creates two different kinds of events:

• Emission: the lamp produces the flash.
• Reception: the flash reaches a camera, detector, or observer.

These events happen at different locations and, generally, different times.

Imagine standing near one end of a school oval while two students at different distances clap. Hearing one clap first doesn’t prove it happened first. Sound needs time to reach you.

Light has the same basic delay problem, although the analogy has a limit: sound travels relative to its surrounding medium, while light in a vacuum travels at the same speed \(c\) in every inertial reference frame.

Einstein’s measurement problem

Einstein’s train-and-embankment thought experiment asked how observers moving relative to one another could establish whether distant events happened together.

The difficulty is that a single observer cannot directly measure when a distant event occurred. They can measure when information about it arrives.

A tempting experimental design would use one camera beside the track, record two flashes, and compare their arrival times.

What would this design show if both flashes occurred simultaneously but were different distances from the camera?

The farther flash would arrive later. The camera’s readings would correctly describe the receptions but would give a misleading impression of the emission times.

An improved measurement design uses synchronised clocks distributed along the platform. A detector close to each flash location records its local event time. The results can then be compared using the same platform time system.

This is a modernised hypothetical measurement setup, not a claim that Einstein performed a physical experiment with a relativistic train. His train argument was a thought experiment about how time is assigned to distant events.

The crucial improvement is separating the time an event happens from the time information about it arrives.

02Correct for light-travel time before deciding simultaneity

Suppose a detector is stationary in the platform frame.

If a flash occurs a distance \(d\) from the detector, its light takes \(d/c\) to arrive, assuming it travels directly through a vacuum.

Therefore,

\[ t_{\mathrm{event}} =t_{\mathrm{arrival}}-\frac{d}{c} \]

Here:

• \(t_{\mathrm{event}}\) is the time the flash occurs in the platform frame.
• \(t_{\mathrm{arrival}}\) is the time the stationary detector receives it, measured using the platform clock system.
• \(d\) is the distance from the emission position to the detector, measured in metres in that frame.
• \(c\) is the speed of light in a vacuum.

For our illustrative problems, use

\[ c=3.00\times10^8\ \mathrm{m\,s^{-1}} \]

or, more conveniently,

\[ c=300\ \mathrm{m}/\mu\mathrm{s} \]

This means light travels 300 metres in one microsecond, where \(1\,\mu\mathrm{s}=10^{-6}\,\mathrm{s}\).

For example, a flash travelling 600 metres takes \(600/300=2.00\,\mu\mathrm{s}\).

The decision process

Don’t subtract \(d/c\) from a timestamp that already records the event locally. That would count the light delay twice.

Also, if the receiver moves, its position changes while the light travels. You cannot automatically use its initial distance from the source as \(d\). You must determine where and when the light reaches it, or work consistently in the receiver’s frame.

03What changes when the question specifies the train frame?

The platform calculation establishes which events were simultaneous according to the platform.

It does not establish that passengers on the train will assign the events the same time.

This is where relativity differs from the older idea of universal time.

Einstein’s two postulates of special relativity state that:

1. The laws of physics are the same in all inertial reference frames.
2. Light in a vacuum has the same speed \(c\) in every inertial frame.

An inertial frame is one moving at constant velocity without acceleration.

Each frame can construct a network of synchronised clocks. However, clocks synchronised in the platform frame are not generally synchronised according to the moving train frame.

For the full physical argument, see Einstein’s Train Thought Experiment: Relativity of Simultaneity.

Here, the useful question is more specific: given a set of recorded event times and positions, how can we identify the simultaneous events in the moving frame?

A quantitative extension: converting between frames

Define:

• \(S\): the platform frame.
• \(S'\): the train frame.
• \(+x\): the direction the train moves.
• \(v\): the train’s velocity relative to the platform, positive when moving right.
• \(\Delta x=x_B-x_A\): the spatial separation of events A and B in the platform frame.
• \(\Delta t=t_B-t_A\): their platform-frame time difference.
• \(\Delta t'=t'_B-t'_A\): their train-frame time difference.

The Lorentz time-coordinate transformation gives

\[ \Delta t' =\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right) \]

where the Lorentz factor is

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

The Lorentz factor \(\gamma\) has no units.

This frame-conversion equation is a quantitative extension to the usual HSC treatment of Einstein’s thought experiments. The key HSC idea is that simultaneity depends on the chosen inertial frame; the equation lets us test more complicated timestamp data.

The train frame considers A and B simultaneous when

\[ \Delta t'=0 \]

Because \(\gamma\) cannot be zero, this requires

\[ \Delta t=\frac{v\Delta x}{c^2} \]

This is a particularly useful test.

It says that two events can occur at different platform times and still be simultaneous in the train frame. The platform time difference must compensate for the effect of motion and spatial separation.

With the train moving right, an event farther to the right must occur later in the platform frame for the two events to have equal train-frame times.

04Proper time: check whether the events happen in one place

A common HSC mistake is treating any time difference involving a train as a proper-time interval.

Proper time has a specific meaning.

For two events on a clock’s worldline, the proper-time interval is the elapsed time measured by that clock between the events. In an inertial frame where the clock is stationary, both events occur at the same position.

Think of a clock bolted to one carriage.

If the clock ticks twice, it is present at both tick events. In the train’s rest frame, both ticks occur at the same location.

Now compare two lamps at opposite ends of the train flashing together.

Those flashes occur at different locations in the train frame. No single stationary train clock is present at both emission events.

Therefore, their zero train-frame time difference is not a proper-time interval between the emissions.

The proper-time test

For a clock moving at speed \(v\), time dilation gives

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

Here \(\Delta\tau\) is the proper-time interval between two ticks of that clock, and \(\Delta t\) is their time separation in the frame where the clock moves.

This relationship applies because the two events occur at the same location in the clock’s rest frame.

It is not a shortcut for changing the frame of two spatially separated flash events. Those require simultaneity reasoning, rather than simply multiplying or dividing a time difference by \(\gamma\).

One more subtlety: if a stationary detector receives two separate light pulses, its reception events occur in the same place. The time between those receptions can be proper time for the detector, even when the original emissions happened in different places.

You must keep the emission pair and the reception pair separate.

05Questions and solutions

Use \(c=300\,\mathrm{m}/\mu\mathrm{s}\) throughout. All speeds are constant, light travels directly through a vacuum, clocks are ideal, and detector response delays are negligible. The train speeds and distances are illustrative, chosen to make relativistic effects measurable.

Unless otherwise stated, the train moves in the positive \(x\)-direction relative to the platform.

Question 1

A stationary platform detector is located at \(x=0\).

Two lamps on a passing train produce flashes:

• Lamp A flashes at \(x=-150\,\mathrm{m}\). Its light arrives at \(7.00\,\mu\mathrm{s}\).
• Lamp B flashes at \(x=+450\,\mathrm{m}\). Its light arrives at \(8.00\,\mu\mathrm{s}\).

Determine whether the lamps flashed simultaneously in the platform frame.

Is the time interval between their receptions a proper-time interval for the stationary detector?

Solution 1

The lamps flashed simultaneously in the platform frame, and the \(1.00\,\mu\mathrm{s}\) interval between receptions is proper time for the stationary detector.

First, reconstruct the emission times using \(t_{\mathrm{event}}=t_{\mathrm{arrival}}-d/c\).

For lamp A:

\[ \begin{aligned} t_A&=7.00-\frac{150}{300}\\ &=6.50\,\mu\mathrm{s} \end{aligned} \]

For lamp B:

\[ \begin{aligned} t_B&=8.00-\frac{450}{300}\\ &=6.50\,\mu\mathrm{s} \end{aligned} \]

Both values are in microseconds, with the distance-to-speed ratios giving the light-travel times.

Since \(t_A=t_B\), the emission events are simultaneous in the platform frame.

The stationary detector receives the pulses one microsecond apart:

\[ \Delta\tau_D=8.00-7.00 =1.00\,\mu\mathrm{s} \]

Both reception events occur at \(x=0\), so the detector’s clock measures the proper time between those receptions.

The tempting mistake is to say that lamp A flashed one microsecond earlier because its signal arrived earlier. That ignores lamp B’s extra light-travel distance.

There is no proper-time interval between the simultaneous, spatially separated emission events.

Question 2

A single lamp fixed to a train travels at \(0.60c\).

It flashes twice:

• Event A: \(x=0\,\mathrm{m}\), \(t=0\,\mu\mathrm{s}\).
• Event B: \(x=900\,\mathrm{m}\), \(t=5.00\,\mu\mathrm{s}\).

These are platform-frame coordinates.

A stationary platform detector is positioned at \(x=450\,\mathrm{m}\).

Determine:

a. The reception time of each flash at the platform detector.

b. The proper-time interval between the emissions, as measured by a clock travelling with the lamp.

c. The proper-time interval between the receptions, as measured by the platform detector.

d. Whether any inertial frame can consider the emission events simultaneous.

Solution 2

The flashes reach the platform detector at \(1.50\,\mu\mathrm{s}\) and \(6.50\,\mu\mathrm{s}\). The train lamp’s proper-time interval is \(4.00\,\mu\mathrm{s}\), while the platform detector’s proper-time interval between receptions is \(5.00\,\mu\mathrm{s}\). No inertial frame considers the emissions simultaneous.

Part a: Reception times

Both emission positions are 450 metres from the detector, so both light signals take

\[ t_{\mathrm{travel}}=\frac{450}{300} =1.50\,\mu\mathrm{s} \]

Event A’s signal arrives at

\[ t_{\mathrm{arrival},A} =0+1.50=1.50\,\mu\mathrm{s} \]

Event B’s signal arrives at

\[ t_{\mathrm{arrival},B} =5.00+1.50=6.50\,\mu\mathrm{s} \]

Part b: Proper time for the lamp

Both emissions occur at the same position in the train frame because the same lamp produces them.

The platform records a \(5.00\,\mu\mathrm{s}\) interval between emissions.

With \(v=0.60c\),

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

Using time dilation:

\[ \begin{aligned} \Delta\tau_{\mathrm{lamp}} &=\frac{\Delta t}{\gamma}\\ &=\frac{5.00}{1.25}\\ &=4.00\,\mu\mathrm{s} \end{aligned} \]

This is what a clock travelling with the lamp measures between its flashes.

Part c: Proper time for the detector

Both receptions occur at \(x=450\,\mathrm{m}\), where the detector is stationary.

Therefore,

\[ \begin{aligned} \Delta\tau_D&=6.50-1.50\\ &=5.00\,\mu\mathrm{s} \end{aligned} \]

These proper-time intervals describe different pairs of events. One belongs to the two lamp emissions; the other belongs to the two detector receptions.

Part d: Simultaneity

No inertial frame can assign equal times to these two emission events.

They occur at the same location in the lamp’s rest frame, separated by \(4.00\,\mu\mathrm{s}\). A change of inertial frame cannot make two distinct events on the same subluminal clock’s worldline simultaneous.

The trap is to assume that a time interval measured by a platform detector automatically equals the proper-time interval between events on the train.

Here the equal light-path lengths happen to make the platform emission and reception intervals both \(5.00\,\mu\mathrm{s}\). That does not make either interval the lamp’s proper time.

Question 3

A train moves right at \(0.60c\).

Two lamps produce the following emission events, measured in the platform frame:

A stationary receiver at \(x=900\,\mathrm{m}\) detects both light signals.

Determine:

a. Which lamp flashed first in the platform frame.

b. Whether the emissions are simultaneous in the train frame.

c. Which signal the platform receiver detects first, and by how much time.

d. Whether the reception order contradicts the event-time conclusions.

Solution 3

A flashes first in the platform frame, A and B are simultaneous in the train frame, and the receiver detects B first by \(0.80\,\mu\mathrm{s}\). All three statements are consistent.

Part a: Platform emission order

From the supplied event times,

\[ \Delta t=2.20-1.00=1.20\,\mu\mathrm{s} \]

So A occurs \(1.20\,\mu\mathrm{s}\) before B in the platform frame.

Part b: Train-frame simultaneity

The platform separation is

\[ \Delta x=600\,\mathrm{m} \]

The simultaneity correction is

\[ \frac{v\Delta x}{c^2} =\frac{0.60(600)}{300} =1.20\,\mu\mathrm{s} \]

With \(\gamma=1.25\),

\[ \begin{aligned} \Delta t' &=1.25(1.20-1.20)\\ &=0\,\mu\mathrm{s} \end{aligned} \]

The train frame considers the emissions simultaneous.

Part c: Reception order

The receiver is 900 metres from A’s emission position and 300 metres from B’s.

Their light-travel times are

\[ \begin{aligned} t_{\mathrm{travel},A}&=\frac{900}{300} =3.00\,\mu\mathrm{s}\\ t_{\mathrm{travel},B}&=\frac{300}{300} =1.00\,\mu\mathrm{s} \end{aligned} \]

Therefore,

\[ \begin{aligned} t_{\mathrm{arrival},A} &=1.00+3.00\\ &=4.00\,\mu\mathrm{s}\\[4pt] t_{\mathrm{arrival},B} &=2.20+1.00\\ &=3.20\,\mu\mathrm{s} \end{aligned} \]

The receiver detects B first, by \(0.80\,\mu\mathrm{s}\).

Part d: Why this isn’t a contradiction

The three comparisons concern different quantities:

• Platform event times: A happens before B.
• Train event times: A and B happen together.
• Platform reception times: B’s signal arrives before A’s.

The receiver gets B’s light first because B’s signal travels a much shorter distance, even though B was emitted later in the platform frame.

The receiver’s \(0.80\,\mu\mathrm{s}\) detection interval is proper time for its own two reception events. It is not a time interval between the two train emission events.

This question combines two distinct effects that students sometimes merge into one: light-travel delays within a frame, and differences in event-time assignments between frames.

Question 4

A train moves right at \(0.60c\).

A rear train lamp emits a light pulse towards the right at event A:

\[ x_A=0\,\mathrm{m},\qquad t_A=0\,\mu\mathrm{s} \]

A second lamp fixed farther forward on the train flashes at event B:

\[ x_B=600\,\mathrm{m},\qquad t_B=1.20\,\mu\mathrm{s} \]

A stationary platform detector at \(x=600\,\mathrm{m}\) directly records the B flash as it passes that position.

The same platform detector later receives the rightward light pulse emitted at A. Call this reception event C.

Determine:

a. The platform time of C.

b. Which of A, B, and C are simultaneous in the train frame.

c. Which pair among these events has a proper-time interval measured directly by the stationary platform detector, and what that interval is.

d. Why calculating when A’s light catches the moving lamp B would give a different result from event C.

Solution 4

A and B are simultaneous in the train frame, while C occurs \(1.00\,\mu\mathrm{s}\) later in that frame. The platform detector measures \(0.80\,\mu\mathrm{s}\) of proper time between B and C.

Part a: Time of C

The pulse from A travels 600 metres to the stationary platform detector.

\[ \begin{aligned} t_C&=t_A+\frac{600}{300}\\ &=0+2.00\\ &=2.00\,\mu\mathrm{s} \end{aligned} \]

This is the reception time in the platform frame.

Part b: Train-frame times

With \(v=0.60c\), the Lorentz factor is \(1.25\).

Take A as the reference event. For B,

\[ \begin{aligned} \Delta t'_{BA} &=1.25(1.20-1.20)\\ &=0\,\mu\mathrm{s} \end{aligned} \]

So A and B occur simultaneously in the train frame.

For C, the platform separation from A is also 600 metres, but the time difference is \(2.00\,\mu\mathrm{s}\):

\[ \begin{aligned} \Delta t'_{CA} &=1.25(2.00-1.20)\\ &=1.00\,\mu\mathrm{s} \end{aligned} \]

C occurs one microsecond after A and B according to train-frame clocks.

The light pulse cannot arrive before it is emitted in either inertial frame. C is causally connected to A by a light signal, so their order cannot reverse.

Part c: Proper time

B and C both occur at the stationary platform detector’s position, \(x=600\,\mathrm{m}\).

Their platform time difference is

\[ \begin{aligned} \Delta\tau_{BC} &=2.00-1.20\\ &=0.80\,\mu\mathrm{s} \end{aligned} \]

That is proper time measured by the platform detector’s clock.

In the train frame, B and C occur at different positions. Their train-frame time difference is \(1.00\,\mu\mathrm{s}\), but this is not their proper-time interval.

Neither A-C nor B-C can be simultaneous in another inertial frame: A-C are connected by light, and B-C happen successively at the same fixed platform location.

Part d: Why the moving-lamp calculation is different

Event C is defined as light reaching the stationary platform detector, not catching the moving train lamp.

Those are different targets.

At platform time \(t=0\), lamp B is not yet at \(x=600\,\mathrm{m}\). Since it reaches that location \(1.20\,\mu\mathrm{s}\) later, its earlier position was

\[ \begin{aligned} x_{B,0} &=600-(180)(1.20)\\ &=384\,\mathrm{m} \end{aligned} \]

Here the speed is \(180\,\mathrm{m}/\mu\mathrm{s}\), so the product gives metres.

For A’s light to catch the moving lamp, their positions must satisfy

\[ 300t=384+180t \]

Solving,

\[ \begin{aligned} 120t&=384\\ t&=3.20\,\mu\mathrm{s} \end{aligned} \]

The meeting position would be \(960\,\mathrm{m}\), not \(600\,\mathrm{m}\).

That catch-up is a different event from C, which occurred at \(2.00\,\mu\mathrm{s}\) at the fixed detector.

The essential constraint is the identity and motion of the receiver. You cannot switch between a stationary detector and a moving lamp halfway through a light-travel calculation.

Question 5

An idealised train travels to the right at \(0.80c\). Two lamps are fixed at different positions along the train.

The platform assigns the following coordinates to two emission events:

At A, a light pulse travels rightward. At B, another pulse travels leftward. The pulses continue travelling through a vacuum and pass through one another.

An engineer wants to place a stationary detector on the platform where both pulses will arrive at exactly the same event.

A second detector is attached to the train, exactly halfway between the two lamps in the train’s rest frame.

Determine:

a. Where and when the two light pulses meet in the platform frame.

b. Whether the emission events A and B are simultaneous in the train frame.

c. The separation between the lamps measured at one common platform time, and their separation in the train’s rest frame.

d. Whether the moving midpoint detector can receive both light pulses at the same event as the stationary platform detector.

e. Whether the following claim is correct:

“Because the platform detector and train detector can both register the two pulses together, both frames must agree that the lamps emitted them simultaneously.”

Explain carefully, distinguishing emission events, reception events, and proper time.

Solution 5

The light pulses meet at \(x=675\,\mathrm{m}\), \(t=2.25\,\mu\mathrm{s}\) in the platform frame. The emissions are simultaneous in the train frame, and both a stationary platform detector and the moving train-midpoint detector can receive the two pulses at that same meeting event. Yet the platform frame does not consider the emissions simultaneous.

Part a: Find the meeting event

Let \(t\) be platform time in microseconds and \(x\) position in metres.

The pulse emitted at A moves rightward from \(x=0\), so its position is

\[ x_A(t)=300t \]

The pulse emitted at B starts at \(x=750\,\mathrm{m}\) when \(t=2.00\,\mu\mathrm{s}\) and moves leftward:

\[ x_B(t)=750-300(t-2.00) \]

The pulses meet when their positions are equal:

\[ \begin{aligned} 300t&=750-300(t-2.00)\\ 600t&=1350\\ t&=2.25\,\mu\mathrm{s} \end{aligned} \]

Substitute into the first position equation:

\[ x=300(2.25)=675\,\mathrm{m} \]

Thus the two light signals intersect at one spacetime event:

\[ x=675\,\mathrm{m},\qquad t=2.25\,\mu\mathrm{s} \]

A stationary detector at this position receives both pulses together.

Notice that \(675\,\mathrm{m}\) is not halfway between the two emission positions. The later-emitted pulse has less distance to travel before the signals meet.

Part b: Are the emissions simultaneous in the train frame?

The platform assigns a time separation of \(2.00\,\mu\mathrm{s}\) and a spatial separation of 750 metres.

For \(v=0.80c\),

\[ \gamma=\frac{1}{\sqrt{1-0.80^2}} =\frac53 \]

The simultaneity correction is

\[ \frac{v\Delta x}{c^2} =\frac{0.80(750)}{300} =2.00\,\mu\mathrm{s} \]

Therefore,

\[ \begin{aligned} \Delta t' &=\frac53(2.00-2.00)\\ &=0\,\mu\mathrm{s} \end{aligned} \]

The emission events are simultaneous in the train frame.

The platform records B two microseconds later than A, but the two emissions share the same train-frame time.

Part c: Find the actual train length between the lamps

The most tempting wrong answer is that the lamps are 750 metres apart on the platform.

But 750 metres is the separation between the emission locations at different platform times. Length measurement requires the positions of both objects at the same time in the measurement frame.

The train travels at

\[ v=0.80c=240\,\mathrm{m}/\mu\mathrm{s} \]

Between the emissions it moves

\[ \begin{aligned} v\Delta t&=240(2.00)\\ &=480\,\mathrm{m} \end{aligned} \]

At \(t=0\), lamp A was at \(x=0\). Lamp B was 480 metres behind its later emission position of \(750\,\mathrm{m}\).

Therefore, B’s position at \(t=0\) was

\[ 750-480=270\,\mathrm{m} \]

Both lamp positions are now compared at the same platform time.

Their separation in the platform frame is

\[ L=270\,\mathrm{m} \]

Using length contraction,

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

where \(L_0\) is their separation in the train’s rest frame:

\[ \begin{aligned} L_0&=\gamma L\\ &=\frac53(270)\\ &=450\,\mathrm{m} \end{aligned} \]

So the lamps are 270 metres apart when measured simultaneously on the platform, and 450 metres apart in the train’s rest frame.

This is the hidden constraint: a separation between two events is not automatically the length of the moving object on which those events occur.

Part d: Can the train midpoint detector receive both pulses at the same event?

Yes.

At platform time \(t=0\), the lamps are at \(x=0\) and \(x=270\,\mathrm{m}\).

The midpoint of the train section between them is therefore at

\[ x_{\mathrm{mid},0}=135\,\mathrm{m} \]

The midpoint detector moves with the train at \(240\,\mathrm{m}/\mu\mathrm{s}\).

At \(t=2.25\,\mu\mathrm{s}\), its platform position is

\[ \begin{aligned} x_{\mathrm{mid}} &=135+240(2.25)\\ &=675\,\mathrm{m} \end{aligned} \]

That is exactly where the two light pulses meet.

So the moving midpoint detector passes the stationary platform detector at precisely the event where both pulses arrive.

We can check this independently in the train frame.

The two emission events are simultaneous in that frame, and the lamps are 450 metres apart. The midpoint detector is 225 metres from each lamp.

Each light pulse therefore takes

\[ t'_{\mathrm{travel}} =\frac{225}{300} =0.75\,\mu\mathrm{s} \]

to reach the midpoint in the train frame.

Both pulses arrive together, as expected.

Part e: Do simultaneous receptions prove simultaneous emissions in both frames?

No. The statement confuses an agreement about one reception event with an agreement about two spatially separated emission events.

Both detectors can register the pulses together because they occupy the same position at the same instant when the light paths intersect.

That is one spacetime event. Every observer agrees that the pulses meet there, whatever coordinates they assign to it.

But the original emissions A and B are two different events.

The platform assigns:

\[ t_A=0,\qquad t_B=2.00\,\mu\mathrm{s} \]

The train assigns:

\[ \Delta t'=0 \]

There is no contradiction.

The platform detector receives the light from A after a \(2.25\,\mu\mathrm{s}\) journey over 675 metres. Light from B travels only 75 metres, taking \(0.25\,\mu\mathrm{s}\), after B occurs at \(2.00\,\mu\mathrm{s}\).

The train detector instead uses train-frame measurements: the emissions have equal times and are equally distant from the detector.

Both descriptions predict the very same reception coincidence.

There is also no proper-time interval between the two original emission events. They are spatially separated but simultaneous in the train frame, so no clock can be present at both events while remaining at rest in an inertial frame. Their separation is spacelike.

And because the two light pulses meet at one event, their coincidence is not itself a nonzero interval of proper time.

The deeper result is that observers can agree completely about where light signals meet while disagreeing about when the distant sources emitted them. They are assigning coordinates to the same physical events using different reference frames, not reporting different physical histories.

06The final check before using a time-dilation equation

When a train simultaneity question gives you two timestamps, identify what those timestamps describe before calculating anything.

If they are reception times, first account for the light paths in the stated frame. If they are already event times, do not invent an extra propagation delay. If the question changes from the platform frame to the train frame, do not carry over the platform’s definition of simultaneous.

Finally, check whether the two events occur at the same position in a clock’s rest frame before calling their interval proper time.

This last decision is what connects simultaneity questions to time dilation. For example, when studying how far a muon travels before decaying, the creation and decay events happen at the same place in the muon’s rest frame. That identifies the proper-time interval before any time-dilation calculation begins.