Do All Observers Measure the Same Time Interval?

Imagine an astronaut travelling past Earth at a very high speed.

Inside the spacecraft, a clock measures:

10 seconds

between two events.

Will an observer on Earth also measure exactly:

10 seconds?

At everyday speeds, the difference would be far too small to notice.

At relativistic speeds, however, the answer can be:

no.

Time intervals depend on the:

reference frame.

To understand how they transform, we need to distinguish carefully between:

proper time

and:

observed coordinate time.

---

Events and Time Intervals

A time interval is measured between:

two events.

Suppose Event 1 occurs at:

t₁

and Event 2 occurs at:

t₂.

The time interval is:

Δt = t₂ − t₁.

In another reference frame:

Δt′ = t′₂ − t′₁.

Special Relativity tells us that:

Δt and Δt′ are not generally equal.

---

The Lorentz Transformation for Time

For a single event:

t′ = γ(t − vx/c²)

where:

γ = 1/√(1 − v²/c²).

For two events, subtract the two time transformations.

This gives:

Δt′ = γ(Δt − vΔx/c²).

This is the general transformation equation for:

time intervals.

---

The Variables

The speed of light is:

c ≈ 3.00 × 10⁸ m/s.

---

Why Does Position Appear in a Time Equation?

Notice:

Δt′ = γ(Δt − vΔx/c²).

The transformed time interval depends not only on:

Δt

but also on:

Δx.

This is a major difference from classical physics.

In Special Relativity:

space and time are interconnected.

Two observers can disagree about the time between events partly because they also disagree about the:

spatial relationship between those events.

---

Proper Time

The proper time between two events is the time measured by a clock that is present at:

both events.

Equivalently, the two events occur at the same spatial location in the clock's:

rest frame.

Proper time is usually represented by:

Δτ.

If the two events occur at the same location in S′:

Δx′ = 0,

then S′ measures the:

proper time.

So:

Δt′ = Δτ.

---

A Simple Example of Proper Time

Imagine an astronaut starts a stopwatch while sitting in a spacecraft.

Ten seconds later, according to the same stopwatch, the astronaut stops it.

In the spacecraft frame:

• Event 1: stopwatch starts
• Event 2: stopwatch stops

Both events occur at the same place relative to the astronaut and stopwatch.

Therefore, the spacecraft clock measures:

proper time.

---

Coordinate Time

An observer in another frame sees the spacecraft:

moving.

The two stopwatch events therefore occur at:

different positions

in that observer's frame.

The time interval measured using synchronized clocks in that frame is a:

coordinate time interval.

For the standard time-dilation situation:

Δt = γΔτ.

The coordinate interval is larger than the:

proper time.

---

Time Dilation

The time-dilation equation is:

Δt = γΔτ

where:

• Δτ = proper time
• Δt = time interval measured in a frame where the clock is moving
• γ = Lorentz factor

Because:

γ ≥ 1

we have:

Δt ≥ Δτ.

The proper time is the shortest time interval between the same pair of timelike-separated events.

---

Why Does Time Dilation Follow from the Lorentz Transformation?

Suppose a clock is at rest in S′.

The two events occur at the same position in S′:

Δx′ = 0.

Use the inverse time transformation:

Δt = γ(Δt′ + vΔx′/c²).

Since:

Δx′ = 0,

we obtain:

Δt = γΔt′.

Since S′ measures proper time:

Δt′ = Δτ.

Therefore:

Δt = γΔτ.

So time dilation is not a separate rule.

It follows directly from the:

Lorentz transformations.

---

The Lorentz Factor

Recall:

γ = 1/√(1 − v²/c²).

Some useful values are:

At low velocities:

γ ≈ 1.

Near the speed of light:

γ increases rapidly.

---

Worked Example 1: Basic Time Dilation

An astronaut measures:

Δτ = 8.0 s

between two events on the spacecraft.

The spacecraft moves at:

0.60c

relative to Earth.

Calculate the time interval measured by Earth.

First:

γ = 1.25.

Use:

Δt = γΔτ.

Therefore:

Δt = 1.25(8.0)

Δt = 10.0 s.

The astronaut measures:

8.0 s.

Earth measures:

10.0 s.

---

Interpreting the Result

The result does not mean the astronaut's clock is:

malfunctioning.

Nor does it mean Earth's clocks are:

incorrect.

Each observer measures time normally within their own inertial frame.

The difference arises from the:

geometry of spacetime.

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Worked Example 2: Faster Spacecraft

A process takes:

20.0 s

according to a clock travelling with a spacecraft moving at:

0.80c.

Find the time interval measured by Earth.

At:

0.80c

the Lorentz factor is:

γ ≈ 1.667.

Therefore:

Δt = γΔτ

Δt = 1.667(20.0)

Δt ≈ 33.3 s.

Earth measures approximately:

33.3 seconds.

---

Worked Example 3: Finding Proper Time

Earth measures a moving process lasting:

50 s.

The object carrying the clock moves at:

0.60c.

Find the proper time.

Use:

Δt = γΔτ.

Rearrange:

Δτ = Δt/γ.

Therefore:

Δτ = 50/1.25

Δτ = 40 s.

The clock travelling with the process measures:

40 seconds.

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Which Time Is Proper Time?

This is one of the most important questions to ask.

Do not decide that the smaller number is proper time simply because it is smaller.

Instead ask:

Which frame has both events occurring at the same place?

That frame measures:

proper time.

For a single clock that records both events, the clock's own rest frame measures:

Δτ.

---

Example: A Spacecraft Clock

Suppose:

Event A: spacecraft clock reads 0 s.

Event B: spacecraft clock reads 12 s.

Both events occur at:

the spacecraft clock.

Therefore:

12 s

is the proper time interval.

An Earth observer sees the spacecraft move between Event A and Event B.

Earth therefore measures a:

dilated coordinate time interval.

---

Example: Events at an Earth Laboratory

Now suppose:

Event A: a machine in an Earth laboratory switches on.

Event B: the same machine switches off.

Both events occur at the same location in the:

Earth frame.

Therefore, Earth measures:

proper time.

A passing spacecraft does not.

Proper time is not automatically associated with:

spacecraft.

It belongs to whichever inertial frame places both events at:

the same spatial coordinate.

---

General Time Transformation

Time dilation applies to a particular arrangement of events.

For the general case, use:

Δt′ = γ(Δt − vΔx/c²).

If:

Δx ≠ 0

then you cannot usually use the simple time-dilation equation:

Δt′ = γΔt.

The spatial separation must also be considered.

---

Worked Example 4: General Time Transformation

Suppose two events in S have:

Δt = 5.0 s

and:

Δx = 6.0 × 10⁸ m.

Frame S′ moves at:

0.60c.

Find:

Δt′.

We know:

γ = 1.25.

Use:

Δt′ = γ(Δt − vΔx/c²).

Calculate:

vΔx/c²

= (0.60c)(6.0 × 10⁸)/c²

Since:

c = 3.0 × 10⁸ m/s,

this gives:

vΔx/c² = 1.2 s.

Therefore:

Δt′ = 1.25(5.0 − 1.2)

Δt′ = 1.25(3.8)

Δt′ = 4.75 s.

So S′ measures:

Δt′ = 4.75 s.

---

Why Isn't the Answer 6.25 s?

Someone might incorrectly calculate:

1.25 × 5.0 = 6.25 s.

But that assumes the simple time-dilation situation.

Here:

Δx ≠ 0.

Therefore, we must use:

Δt′ = γ(Δt − vΔx/c²).

This distinction is extremely important.

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Time Dilation vs General Time Transformation

Time dilation

Use:

Δt = γΔτ

when one of the frames measures the two events at the:

same location.

General time transformation

Use:

Δt′ = γ(Δt − vΔx/c²)

when transforming arbitrary events between:

reference frames.

Time dilation is therefore a:

special case

of the Lorentz time transformation.

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Worked Example 5: Events at the Same Position in S

Suppose two events occur at the same position in S.

Therefore:

Δx = 0.

Let:

Δt = 12 s

and:

v = 0.80c.

Then:

γ = 1.667.

Use:

Δt′ = γ(Δt − vΔx/c²).

Because:

Δx = 0,

we get:

Δt′ = γΔt

Δt′ = 1.667(12)

Δt′ ≈ 20.0 s.

Here, S measures the:

proper time.

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Proper Time Is Frame-Specific

Notice what happened.

Earlier, S′ measured proper time.

In this example:

S measures proper time.

There is no universal frame that always measures proper time.

The defining condition is:

the two events occur at the same location in that frame.

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A Spacetime View of Proper Time

A clock traces a path through spacetime called its:

worldline.

Two readings of the same clock correspond to two events on that:

worldline.

The time recorded directly by that clock between the events is:

proper time.

---

The Spacetime Interval

For two events separated along one spatial dimension:

c²Δτ² = c²Δt² − Δx²

when the separation is timelike.

Divide by c²:

Δτ² = Δt² − Δx²/c².

Therefore:

Δτ = √(Δt² − Δx²/c²).

This provides another way to calculate the:

proper time.

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Worked Example 6: Finding Proper Time from Two Events

Suppose two events are separated by:

Δt = 5.0 s

and:

Δx = 9.0 × 10⁸ m.

Since:

c = 3.0 × 10⁸ m/s,

we have:

Δx/c = 3.0 s.

Therefore:

Δτ = √(5.0² − 3.0²)

Δτ = √(25 − 9)

Δτ = √16

Δτ = 4.0 s.

So the proper time between the events is:

4.0 seconds.

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Why Proper Time Is Invariant

Different inertial observers may disagree about:

Δt

and:

Δx.

However, they agree on the combination:

c²Δt² − Δx².

Therefore they agree on:

Δτ.

Proper time is related to the invariant:

spacetime interval.

This is why proper time has such an important role in relativity.

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Worked Example 7: Particle Lifetime

A particle has a proper lifetime of:

2.2 μs.

It travels at:

0.90c.

How long does its lifetime appear in the laboratory frame?

At:

0.90c

γ ≈ 2.294.

Use:

Δt = γΔτ.

Therefore:

Δt = 2.294(2.2 μs)

Δt ≈ 5.05 μs.

The laboratory measures approximately:

5.05 μs.

---

Particle Lifetimes and Evidence for Time Dilation

High-speed unstable particles provide important experimental tests of:

time dilation.

For example, muons produced in Earth's atmosphere can reach detectors at Earth's surface in greater numbers than a simple nonrelativistic lifetime argument would suggest.

In the Earth frame, their moving decay clocks are:

time-dilated.

In the muon's frame, the atmospheric travel distance is:

length-contracted.

Both descriptions are consistent.

---

Worked Example 8: A Longer Journey

A spacecraft travels at:

0.80c.

According to astronauts aboard the spacecraft, a certain phase of the journey lasts:

3.0 years.

How much time passes in Earth's frame?

Use:

γ = 1.667.

Then:

Δt = γΔτ

Δt = 1.667(3.0)

Δt ≈ 5.0 years.

The astronauts measure:

3.0 years.

Earth measures:

5.0 years.

---

What Does "Moving Clocks Run Slow" Mean?

You may often hear:

moving clocks run slow.

This is a useful shorthand, but it must be interpreted carefully.

It means that when comparing appropriate clock readings between inertial frames, a clock moving relative to an inertial observer accumulates less proper time between the relevant events.

It does not mean that someone looking at their own clock notices it:

physically ticking strangely.

Every observer sees their own local clock behaving normally.

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The Symmetry Question

If Earth says the spacecraft clock is moving, couldn't the spacecraft say:

Earth is moving?

Yes.

For observers in uniform relative motion, each can describe the other's moving clocks as:

time-dilated.

This is not a contradiction because comparisons of distant clocks depend on:

simultaneity.

The simple symmetry changes if one observer turns around or accelerates, as in the:

twin paradox.

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Time Dilation and the Twin Paradox

Suppose one twin remains on Earth while another travels away at high speed and later returns.

The travelling twin changes inertial frames during the journey.

Their paths through spacetime are:

different.

When reunited, the twins can directly compare clocks at the same location.

The elapsed proper times along their worldlines can therefore:

differ.

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Transforming Simultaneous Events

Suppose two events are simultaneous in S.

Then:

Δt = 0.

But if they occur at different positions:

Δx ≠ 0.

The transformation becomes:

Δt′ = γ(0 − vΔx/c²)

or:

Δt′ = −γvΔx/c².

Therefore:

Δt′ ≠ 0.

The events are not simultaneous in S′.

This is the:

relativity of simultaneity.

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Worked Example 9: Simultaneous in One Frame

Two flashes occur simultaneously in S:

Δt = 0.

They are separated by:

Δx = 3.0 × 10⁸ m.

Frame S′ moves at:

0.60c.

Since:

γ = 1.25,

use:

Δt′ = −γvΔx/c².

Now:

vΔx/c² = 0.60 s.

Therefore:

Δt′ = −1.25(0.60)

Δt′ = −0.75 s.

The two events are separated by:

0.75 s

in S′.

The negative sign tells us that their:

time order in the chosen coordinate labeling

is opposite to the order implied by positive Δt′.

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Can Event Order Change?

For some pairs of events:

yes.

If two events are separated enough in space that light cannot travel between them during their time separation, they are:

spacelike separated.

Different observers may disagree about which occurred:

first.

For events that can be causally connected:

timelike or lightlike separated events,

all inertial observers preserve the causal ordering.

Cause cannot become:

effect after its own consequence.

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Transforming Back

The inverse time-interval transformation is:

Δt = γ(Δt′ + vΔx′/c²).

This can be used to:

verify a calculation.

If you transform from S to S′ and then back again, you should recover the original:

time interval.

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Worked Example 10: Verification

Suppose:

Δt = 5.0 s

Δx = 6.0 × 10⁸ m

v = 0.60c

Earlier we found:

Δt′ = 4.75 s.

The corresponding position transformation is:

Δx′ = γ(Δx − vΔt).

Calculate:

vΔt = (1.8 × 10⁸)(5.0)

vΔt = 9.0 × 10⁸ m.

Therefore:

Δx′ = 1.25(6.0 × 10⁸ − 9.0 × 10⁸)

Δx′ = −3.75 × 10⁸ m.

Now transform the time back:

Δt = γ(Δt′ + vΔx′/c²).

Here:

vΔx′/c² = −0.75 s.

Therefore:

Δt = 1.25(4.75 − 0.75)

Δt = 1.25(4.00)

Δt = 5.00 s.

The original value is recovered.

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Verification Using the Spacetime Interval

Another check is:

c²Δt² − Δx² = c²Δt′² − Δx′².

If both sides agree, the transformed measurements are:

consistent.

This is often a powerful check in more advanced relativity problems.

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Dimensional Check

Consider:

vΔx/c².

Its units are:

(m/s)(m)/(m²/s²).

This simplifies to:

seconds.

Therefore:

Δt − vΔx/c²

is dimensionally valid because both terms are:

time intervals.

Always checking units can catch:

calculation errors.

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Choosing the Correct Equation

A common challenge is deciding which equation to use.

If you know proper time:

Use:

Δt = γΔτ.

If you know the dilated interval:

Use:

Δτ = Δt/γ.

If the two events occur at different positions in the known frame:

Use:

Δt′ = γ(Δt − vΔx/c²).

If transforming back:

Use:

Δt = γ(Δt′ + vΔx′/c²).

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A Reliable Problem-Solving Method

For each problem:

Step 1: Identify the two events.

Step 2: Identify the reference frames.

Step 3: Determine whether either frame sees both events at the same position.

Step 4: If yes, identify the proper time.

Step 5: Calculate γ.

Step 6: Choose the appropriate equation.

Step 7: Substitute with consistent units.

Step 8: Interpret the result physically.

Step 9: Check whether the answer is reasonable.

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Common Error: Assuming Δt Is Always Proper Time

The symbol:

Δt

does not automatically mean proper time.

Proper time is normally written:

Δτ.

But what really determines proper time is the physical condition:

both events occur at the same position in that frame.

Always examine the:

events and frame.

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Common Error: Multiplying by γ in Every Problem

You cannot automatically use:

Δt′ = γΔt.

The general transformation is:

Δt′ = γ(Δt − vΔx/c²).

The simple time-dilation equation applies only when the events satisfy the appropriate:

same-location condition.

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Common Error: Thinking Proper Time Belongs to Earth

Proper time is not automatically:

Earth time.

If both events occur at the same position on a spacecraft, then the spacecraft measures proper time.

If both occur at the same Earth laboratory, Earth measures proper time.

Proper time depends on:

the pair of events.

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Common Error: Thinking Proper Time Is "True Time"

Proper time is not a universal time that is more correct than all others.

It is the time measured along a particular:

worldline.

Other inertial frames can measure different coordinate time intervals without being:

incorrect.

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Common Error: Ignoring Relativity of Simultaneity

Comparing distant clocks requires deciding which distant events are:

simultaneous.

Different inertial observers generally disagree about this.

This is essential for understanding why time dilation does not create a contradiction between:

moving observers.

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Common Error: Confusing Time Dilation with Signal Delay

Suppose you look through a telescope at a distant spacecraft.

What you literally see is affected by the:

travel time of light.

Time dilation is a different physical effect.

Relativistic measurements are defined after accounting for:

signal propagation.

So time dilation is not simply caused by light taking longer to:

reach an observer.

---

Everyday Speeds

Suppose an aircraft travels at:

250 m/s.

Compared with:

c = 3.00 × 10⁸ m/s,

this is extremely slow.

Therefore:

γ ≈ 1.

The time-dilation effect is extremely small.

For ordinary activities, we can safely treat:

Δt ≈ Δτ.

---

Relativistic Speeds

At:

0.99c

the Lorentz factor is approximately:

7.09.

Suppose a process lasts:

1.0 hour

in its own rest frame.

Another inertial frame in which the process moves at 0.99c measures:

Δt = γΔτ

Δt ≈ 7.09 hours.

At very high speeds, the difference becomes:

dramatic.

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Time Intervals in Particle Physics

Relativistic time transformations are essential when studying:

• muons
• unstable particles
• particle accelerators
• cosmic rays
• high-energy collisions

Particles may have extremely short:

proper lifetimes.

But because they travel near c, laboratory observers can measure significantly longer:

coordinate lifetimes.

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Time Intervals in Space Travel

Relativistic time dilation would also matter for hypothetical spacecraft travelling at a substantial fraction of:

c.

Travellers could experience less elapsed proper time than observers remaining in another frame between appropriately defined departure and reunion events.

This is not because their biological processes somehow escape physics.

Their clocks, chemical reactions and biological processes all evolve according to their own:

proper time.

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Connecting Time Transformations to Previous Topics

The time transformation:

Δt′ = γ(Δt − vΔx/c²)

connects several major ideas in Special Relativity.

If:

Δx′ = 0

we obtain:

time dilation.

If:

Δt = 0

we obtain:

relativity of simultaneity.

Combined with the spatial transformation:

Δx′ = γ(Δx − vΔt),

we obtain the complete transformation of:

spacetime intervals.

These effects are therefore not separate phenomena.

They are different consequences of the same:

Lorentz transformations.

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Check Your Understanding

1. What is a time interval?

2. Write the general Lorentz transformation for a time interval.

3. Define proper time.

4. What condition identifies the frame that measures proper time?

5. What symbol is commonly used for proper time?

6. State the time-dilation equation.

7. A spacecraft moves at 0.60c. Calculate γ.

8. A spacecraft clock measures 12 s while travelling at 0.60c. What interval does Earth measure in the standard time-dilation setup?

9. Earth measures 40 s for a moving process at 0.60c. Determine the proper time.

10. Explain why the proper time is not automatically the time measured on Earth.

11. Two events occur at different positions in S. Why can't you automatically use Δt′ = γΔt?

12. Two events have Δt = 4.0 s, Δx = 3.0 × 10⁸ m, and v = 0.60c. Calculate Δt′.

13. Explain physically what a transformed time interval represents.

14. How does the Lorentz transformation produce the time-dilation equation?

15. What happens to time dilation as v approaches c?

16. Why is time dilation not simply caused by the travel time of light?

17. If two events are simultaneous in S but occur at different positions, must they be simultaneous in S′? Explain.

18. Write the equation relating proper time to the spacetime interval.

19. Describe one method for verifying a transformed time interval.

20. Explain the relationship between time transformation, time dilation and relativity of simultaneity.

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Key Terms

• Event: Physical occurrence at a particular place and time.
• Time interval: Difference between the time coordinates of two events.
• Proper time (Δτ): Time measured by a clock present at both events; equivalently, the interval measured in the frame where the events occur at the same spatial position.
• Coordinate time: Time interval between events measured using the clocks of a particular reference frame.
• Time dilation: Relationship in which a moving clock accumulates less proper time between appropriate events than the coordinate time measured in another inertial frame.
• Lorentz transformation: Equations relating space and time coordinates between inertial frames.
• Lorentz factor (γ): Factor 1/√(1 − v²/c²).
• Reference frame: Coordinate system used to measure positions, times and motion.
• Inertial frame: Non-accelerating reference frame.
• Worldline: Path of an object through spacetime.
• Spacetime interval: Invariant combination of spatial and temporal separation between events.
• Invariant: Quantity that has the same value for all inertial observers.
• Relativity of simultaneity: Principle that events simultaneous in one inertial frame need not be simultaneous in another.
• Timelike separation: Separation between events that permits a slower-than-light causal connection.
• Spacelike separation: Separation between events for which no causal signal travelling at or below c can connect them.

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Key Takeaways

• Time intervals are measured between two events.
• Different inertial frames can measure different time intervals between the same events.
• The general transformation is Δt′ = γ(Δt − vΔx/c²).
• Time transformation therefore depends on both temporal and spatial separation.
• Proper time is written Δτ.
• Proper time is measured by a clock that is present at both events.
• Equivalently, proper time is measured in the frame where the two events occur at the same position.
• Proper time does not automatically belong to Earth, a spacecraft, or any preferred frame.
• For the standard time-dilation situation, Δt = γΔτ.
• Because γ ≥ 1, the coordinate time interval in that setup is at least as large as the proper time.
• Time dilation is a special case of the Lorentz time transformation.
• You should not automatically multiply every time interval by γ.
• When Δx ≠ 0, the full Lorentz time transformation may be required.
• If Δt = 0 but Δx ≠ 0, another inertial observer generally measures Δt′ ≠ 0.
• This produces the relativity of simultaneity.
• Proper time is related to the invariant spacetime interval by Δτ² = Δt² − Δx²/c² for timelike-separated events.
• Moving-particle lifetimes provide important experimental evidence for relativistic time dilation.
• Time dilation is not simply a visual effect caused by light-travel delay.
• At everyday speeds, γ ≈ 1, so relativistic time differences are usually extremely small.
• At speeds close to c, transformed time intervals can differ substantially.
• A strong solution should identify the events, identify the frames, determine which frame—if any—measures proper time, calculate the transformation, and interpret the result physically.