What Happens to the Lorentz Factor at Low Speeds?

A jet travels at \(300\ \mathrm{m\,s^{-1}}\), and you calculate its Lorentz factor. Your calculator displays \(1.000000\). Does that mean time dilation has disappeared?

No. At low speeds, the Lorentz factor approaches 1 from above. It equals exactly 1 only when the relative speed is zero. At ordinary speeds, the difference is so tiny that a calculator may round it away. But a sufficiently precise clock can still reveal the effect.

01Why the Lorentz factor approaches 1

The Lorentz factor tells us how strongly relative motion affects measurements of time and space.

It is given by

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

where:

• \(\gamma\) (gamma) is the Lorentz factor, which has no units.
• \(v\) is the relative speed between two inertial reference frames, in metres per second.
• \(c\) is the speed of light in a vacuum, approximately \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).

Consider what happens when \(v\) becomes very small compared with \(c\).

The ratio \(v/c\) approaches zero, so

\[ \gamma\longrightarrow \frac{1}{\sqrt{1-0}}=1 \]

For any non-zero speed below \(c\), however, \((v/c)^2\) is positive. The square root in the denominator is then slightly smaller than 1, making its reciprocal slightly larger than 1.

Therefore,

\[ \boxed{\gamma\geq1} \]

How close to 1 is it?

When \(v\ll c\), we can estimate the difference using

\[ \gamma-1\approx\frac12\left(\frac vc\right)^2 \]

This is a small-speed approximation, not an exact equation. It works because the higher-order corrections become negligible when \(v/c\) is very small.

The useful insight is that the correction depends approximately on the square of the speed. Double a sufficiently small speed, and \(\gamma-1\) becomes approximately four times as large.

These are theoretical values, not experimental measurements. Values in the last row are rounded from the exact Lorentz-factor equation.

Notice that the table gives \(\gamma-1\), not \(\gamma\) itself. At \(300\ \mathrm{m\,s^{-1}}\), the Lorentz factor is approximately \(1+5.00\times10^{-13}\), not \(5.00\times10^{-13}\).

02The experimental problem: are we measuring the clock or the signal?

Suppose we wanted to detect this extremely small effect.

Our first idea might be to put an atomic clock aboard an aircraft, make it transmit a radio pulse every second of its own time, and record those pulses using a receiver on the ground.

This is a hypothetical experiment, not a description of an actual historical trial.

The first design

Imagine the apparatus arranged along a straight flight path:

1. The aircraft carries a clock connected to a radio transmitter.
2. The transmitter emits a pulse whenever the clock reaches its next one-second mark.
3. A stationary receiver sits ahead of the approaching aircraft.
4. The receiver records when each radio pulse arrives.

What should happen?

You might predict that time dilation makes the aircraft clock tick more slowly, so the receiver should record pulses slightly more than one second apart.

But that prediction overlooks something important.

While the aircraft approaches, each successive pulse has a shorter distance to travel.

Radio waves travel at approximately \(c\) through vacuum. In our idealised experiment, the interval between received signals is

\[ \Delta t_{\mathrm{rec}} =\Delta t_{\mathrm{emit}} +\frac{d_2-d_1}{c} \]

Here:

• \(\Delta t_{\mathrm{rec}}\) is the interval between pulse arrivals at the receiver.
• \(\Delta t_{\mathrm{emit}}\) is the interval between emissions measured in the laboratory frame.
• \(d_1\) and \(d_2\) are the respective signal-travel distances measured in that frame.

If the aircraft approaches the receiver, \(d_2<d_1\), so the second pulse spends less time travelling.

For an aircraft approaching at \(300\ \mathrm{m\,s^{-1}}\), successive pulses emitted roughly one second apart have a path-length difference of approximately \(300\ \mathrm{m}\).

The resulting change in light-travel time is

\[ \begin{aligned} \Delta t_{\mathrm{path}} &\approx\frac{300}{3.00\times10^8}\\ &=1.00\times10^{-6}\ \mathrm{s} \end{aligned} \]

That’s approximately one microsecond.

Meanwhile, special-relativistic time dilation over one second at this speed is only about \(5.00\times10^{-13}\ \mathrm{s}\), or half a picosecond.

The change in signal-travel time is roughly two million times larger.

The receiver could therefore record pulses arriving closer together even though the aircraft clock is running slower relative to the laboratory frame.

This changing arrival rate is part of the Doppler effect. It must not be mistaken for the time-dilation factor alone.

03Improving the experiment: compare clocks at local events

We need a design that separates the clock’s elapsed time from the delay caused by travelling signals.

Imagine two laboratory stations, A and B, positioned along a straight flight path.

Each station contains a clock synchronised with the other station’s clock in the laboratory frame. Signal-propagation delays have already been accounted for during synchronisation.

An aircraft carries a third clock.

When the aircraft passes station A, a local sensor records the aircraft clock’s reading alongside the reading of station A’s clock. The same comparison happens at station B.

The measurements are stored and compared afterwards.

This revised design avoids treating the arrival time of a distant signal as the time of the original event.

Which clock measures proper time?

We now have two events:

• Event A: the aircraft passes the first station.
• Event B: the aircraft passes the second station.

The two events occur at different positions in the laboratory frame.

However, for an aircraft moving at constant velocity, both events occur at the same position in the aircraft’s reference frame: the position of its onboard clock.

That means the aircraft clock measures the proper time, \(\Delta\tau\).

The laboratory’s synchronised clocks measure the longer coordinate interval, \(\Delta t\).

The relationship is

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

This is the crucial distinction: proper time is identified by the locations of the two events in the relevant reference frame, not simply by choosing whichever clock looks stationary.

The complete derivation is covered in Time Dilation in HSC Physics: Using the Lorentz Factor. Here, we’re concentrating on what the equation predicts when \(\gamma\) is almost 1.

A practical decision rule

The last row is easy to miss. Two simultaneous events at different positions are separated by a space-like interval. No physical clock travelling below \(c\) can be present at both events.

There is another limitation. If the moving clock changes speed, it still accumulates proper time along its path, but no single inertial rest frame covers the whole journey. We must calculate the time accumulated during each constant-speed section.

04What actual aircraft-clock experiments found

In October 1971, Joseph Hafele and Richard Keating carried four caesium atomic clocks on commercial flights around Earth, first eastward, then westward.

The experimental question was whether real, macroscopic clocks would accumulate the different elapsed times predicted by relativity.

The apparatus was more complicated than our straight-flight model:

• Atomic clocks travelled aboard the aircraft.
• Reference atomic clocks remained at the US Naval Observatory.
• Flight paths, altitudes, and directions mattered.
• The travelling clocks were compared with the ground reference after returning.

The clocks were not judged simply by when their radio signals reached a distant receiver.

The researchers reported that the flying clocks finished approximately \(59\ \mathrm{ns}\) behind the reference after the eastward journey and \(273\ \mathrm{ns}\) ahead after the westward journey.

The observations were published in Science in 1972 (Hafele and Keating, “Around-the-World Atomic Clocks: Observed Relativistic Time Gains”, volume 177, pages 168-170).

Does the westward gain contradict the rule \(\gamma\geq1\)?

No. The Lorentz factor describes special-relativistic comparisons between inertial frames at constant relative velocity. A clock flying around the rotating Earth is not in that simple situation.

Earth’s rotation affects the aircraft’s speed relative to a suitable non-rotating reference frame. The aircraft’s altitude also changes its clock rate through gravitational time dilation.

Both effects were needed to predict the actual observations. The reported results agreed with the combined relativistic predictions within experimental uncertainty.

The experiment showed that tiny relativistic clock differences could be detected at aircraft speeds. It did not show that the Lorentz factor can become smaller than 1.

05Questions and solutions

The following situations are idealised special-relativity problems. Unless otherwise stated, ignore gravity, assume light travels through vacuum at \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\), and use the stated laboratory frame for positions and speeds.

Question 1

A spacecraft moves at \(0.10c\) relative to a laboratory.

Its onboard clock measures \(10.00\ \mu\mathrm{s}\) between two pulses produced at the same location inside the spacecraft.

Calculate:

a) The Lorentz factor.

b) The laboratory time interval between the pulses.

c) Which interval is the proper time.

Solution 1

The Lorentz factor is \(1.00504\), the laboratory interval is approximately \(10.05\ \mu\mathrm{s}\), and the onboard interval is the proper time.

a) Calculate the Lorentz factor.

\[ \begin{aligned} \gamma &=\frac{1}{\sqrt{1-(0.10)^2}}\\ &=1.00504 \end{aligned} \]

b) Calculate the laboratory interval.

Using \(\Delta t=\gamma\Delta\tau\),

\[ \begin{aligned} \Delta t &=1.00504(10.00)\\ &=10.0504\ \mu\mathrm{s}\\ &\approx10.05\ \mu\mathrm{s} \end{aligned} \]

c) Identify proper time.

The onboard clock is present at both emissions. In its rest frame, the pulses originate from the same position.

Its \(10.00\ \mu\mathrm{s}\) reading is therefore the proper time.

The tempting mistake is to treat \(\gamma\approx1\) as evidence that both intervals must be exactly equal. They are close, but not identical.

Question 2

A spacecraft travels along the laboratory \(x\)-axis at a constant \(0.10c\).

It emits three light pulses at these laboratory positions:

• Pulse 1: \(x=-150\ \mathrm{m}\)
• Pulse 2: \(x=0\)
• Pulse 3: \(x=+150\ \mathrm{m}\)

A stationary receiver is positioned \(200\ \mathrm{m}\) perpendicular to the spacecraft’s path, at \(x=0\).

Calculate:

a) The laboratory time between successive emissions.

b) The interval between the arrivals of pulses 1 and 2.

c) The interval between the arrivals of pulses 2 and 3.

d) The total proper time recorded by the spacecraft between pulses 1 and 3.

Explain why the receiver’s unequal pulse intervals do not mean that the spacecraft changed speed or its clock changed rate.

Solution 2

The receiver measures approximately \(4.83\ \mu\mathrm{s}\) between the first two pulses and \(5.17\ \mu\mathrm{s}\) between the next two. The spacecraft clock remains uniform and accumulates approximately \(9.95\ \mu\mathrm{s}\) between the first and third emissions.

a) Find the laboratory emission intervals.

Each pair of successive emission positions is separated by \(150\ \mathrm{m}\).

\[ \begin{aligned} \Delta t &=\frac{\Delta x}{v}\\ &=\frac{150}{0.10(3.00\times10^8)}\\ &=5.00\times10^{-6}\ \mathrm{s}\\ &=5.00\ \mu\mathrm{s} \end{aligned} \]

Therefore, successive emissions are \(5.00\ \mu\mathrm{s}\) apart in the laboratory frame.

b) Calculate the first reception interval.

The first pulse travels along the hypotenuse of a right triangle.

\[ \begin{aligned} d_1 &=\sqrt{150^2+200^2}\\ &=250\ \mathrm{m} \end{aligned} \]

The second pulse is emitted directly below the receiver, so

\[ d_2=200\ \mathrm{m} \]

The difference in arrival times is

\[ \begin{aligned} \Delta t_{\mathrm{rec},1} &=5.00+\frac{200-250}{300}\\ &=4.8333\ \mu\mathrm{s} \end{aligned} \]

Here, \(300\ \mathrm{m}\) is the distance light travels in one microsecond, so the distance ratio gives microseconds.

c) Calculate the second reception interval.

The third pulse also travels \(250\ \mathrm{m}\) because its emission position is symmetrical with the first.

\[ \begin{aligned} \Delta t_{\mathrm{rec},2} &=5.00+\frac{250-200}{300}\\ &=5.1667\ \mu\mathrm{s} \end{aligned} \]

d) Find the spacecraft’s proper time.

The total laboratory interval between pulses 1 and 3 is

\[ \Delta t=10.00\ \mu\mathrm{s} \]

The spacecraft is present at all three emission events, so

\[ \begin{aligned} \Delta\tau &=\frac{\Delta t}{\gamma}\\ &=10.00\sqrt{1-0.10^2}\\ &=9.9499\ \mu\mathrm{s} \end{aligned} \]

The spacecraft accumulates approximately \(9.95\ \mu\mathrm{s}\).

The conceptual trap is to interpret each received interval as a clock-tick interval. Between the first two pulses, the signal path shortens; between the second and third, it lengthens.

Across all three pulses, the first and third signals travel equal distances, so the total reception interval is exactly the \(10.00\ \mu\mathrm{s}\) laboratory emission interval.

The changing reception rate comes entirely from geometry. The spacecraft’s constant speed gives a constant Lorentz factor.

Question 3

Two laboratory stations, A and B, are \(300\ \mathrm{m}\) apart along a straight line.

At laboratory time \(t=0\), both stations emit a flash simultaneously.

A stationary receiver at A detects its local flash immediately and subsequently detects the light from B.

Separately, a spacecraft travelling from A towards B at \(0.60c\) passes A at \(t=0\).

Determine:

a) Whether the two simultaneous flashes have a proper-time interval.

b) The time between the two flashes as measured by the receiver at A.

c) The time recorded by the spacecraft between crossing A and crossing B.

d) Why the three intervals in this problem must not be used interchangeably.

Solution 3

The simultaneous flashes do not define a proper-time interval. The receiver detects them \(1.00\ \mu\mathrm{s}\) apart, while the spacecraft records \(1.33\ \mu\mathrm{s}\) between crossing A and crossing B.

These measurements involve different pairs of events.

a) Examine the simultaneous flashes.

Both flashes occur at \(t=0\) in the laboratory frame, but at different positions.

Their laboratory time separation is zero and their spatial separation is \(300\ \mathrm{m}\).

No clock travelling at less than the speed of light can be present at both simultaneous, spatially separated events.

They therefore do not define a proper-time interval.

b) Calculate the receiver’s detection interval.

The flash from A reaches the receiver immediately. Light from B travels \(300\ \mathrm{m}\).

\[ \begin{aligned} \Delta t_{\mathrm{rec}} &=\frac{300}{3.00\times10^8}\\ &=1.00\times10^{-6}\ \mathrm{s}\\ &=1.00\ \mu\mathrm{s} \end{aligned} \]

This difference is caused by propagation delay, not time dilation.

c) Calculate the spacecraft’s proper time.

The spacecraft crosses the two stations \(300\ \mathrm{m}\) apart.

Its laboratory travel time is

\[ \begin{aligned} \Delta t &=\frac{300}{0.60(3.00\times10^8)}\\ &=1.667\ \mu\mathrm{s} \end{aligned} \]

For \(v=0.60c\),

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

Therefore,

\[ \begin{aligned} \Delta\tau &=\frac{1.667}{1.25}\\ &=1.333\ \mu\mathrm{s} \end{aligned} \]

The spacecraft records approximately \(1.33\ \mu\mathrm{s}\).

d) Explain the distinction.

The two simultaneous flashes are one event pair. The spacecraft crossing A and B is another. The detection of the flashes by the receiver is a third pair.

A tempting approach is to label any measured interval as \(\Delta t\), calculate a Lorentz factor, and immediately apply time dilation.

But time dilation only compares the appropriate time intervals between the same two events.

Here, switching event pairs changes the physical question. That’s why the numbers \(0\), \(1.00\), \(1.33\), and \(1.667\ \mu\mathrm{s}\) can all be meaningful without contradiction.

Question 4

A spacecraft is moving directly away from a stationary receiver.

It emits three light pulses, P, Q, and R. The receiver records both P-to-Q and Q-to-R arrival intervals as exactly \(11.0\ \mu\mathrm{s}\).

Between emissions P and Q, the spacecraft travels at \(0.10c\).

Immediately after Q, its speed increases to \(0.20c\), remaining constant until R. Assume the time taken to change speed is negligible.

Calculate:

a) The laboratory emission intervals P-to-Q and Q-to-R.

b) The proper-time intervals recorded by the spacecraft between each pair of emissions.

c) Whether the pulses could be successive ticks of a perfectly regular onboard clock with a constant tick period.

Explain why equal reception intervals do not imply equal elapsed times on the spacecraft.

Solution 4

The spacecraft accumulates approximately \(9.95\ \mu\mathrm{s}\) between P and Q and \(8.98\ \mu\mathrm{s}\) between Q and R. These are unequal proper-time intervals, even though the receiver records equal arrival intervals.

a) Find the emission intervals.

For a source moving directly away from the receiver at constant speed,

\[ \Delta t_{\mathrm{rec}} =\Delta t_{\mathrm{emit}}(1+v/c) \]

For the P-to-Q interval,

\[ \begin{aligned} \Delta t_{PQ} &=\frac{11.0}{1+0.10}\\ &=10.0\ \mu\mathrm{s} \end{aligned} \]

For Q-to-R,

\[ \begin{aligned} \Delta t_{QR} &=\frac{11.0}{1+0.20}\\ &=9.1667\ \mu\mathrm{s} \end{aligned} \]

So the emissions are not equally spaced in laboratory time.

b) Calculate the proper-time intervals.

For P-to-Q,

\[ \begin{aligned} \Delta\tau_{PQ} &=\Delta t_{PQ}\sqrt{1-0.10^2}\\ &=10.0\sqrt{0.99}\\ &=9.9499\ \mu\mathrm{s} \end{aligned} \]

For Q-to-R,

\[ \begin{aligned} \Delta\tau_{QR} &=\Delta t_{QR}\sqrt{1-0.20^2}\\ &=9.1667\sqrt{0.96}\\ &=8.9815\ \mu\mathrm{s} \end{aligned} \]

The onboard intervals are therefore approximately \(9.95\ \mu\mathrm{s}\) and \(8.98\ \mu\mathrm{s}\).

c) Test the regular-clock explanation.

No. If these pulses were successive ticks of one perfectly regular onboard clock, equal numbers of onboard seconds would separate consecutive emissions.

Instead, the calculated proper-time intervals are different.

The spacecraft may still have a perfectly regular clock; its transmitter simply cannot have been programmed to emit these three pulses at consecutive equal proper-time intervals.

The equal arrival spacing hides two effects:

• The second stage has a shorter laboratory emission interval.
• The spacecraft also moves away faster, making the light-travel delay increase more rapidly.

These changes compensate.

The trap is assuming that equal received intervals imply equal emission intervals. They do not, particularly when the source changes speed.

Question 5

Two spacecraft, S and P, carry identical ideal clocks along the same straight route from station A to station B.

The stations are \(3000\ \mathrm{m}\) apart in an inertial laboratory frame.

Both spacecraft depart A at laboratory time \(t=0\) and arrive at B at \(t=100\ \mu\mathrm{s}\).

Their journeys differ:

• Spacecraft S travels at a constant \(0.10c\) for the entire journey.
• Spacecraft P remains stationary at A for the first \(50\ \mu\mathrm{s}\), then travels at \(0.20c\) for the remaining \(50\ \mu\mathrm{s}\).

Assume changes of velocity take negligible time and that the clocks record all elapsed proper time.

Both spacecraft transmit a light pulse at departure and another at arrival. A stationary receiver at A records these pulses.

Each spacecraft is also programmed to send one additional pulse when its onboard clock reaches exactly half of the total proper time it will have accumulated by arrival at B.

Determine:

a) Which spacecraft clock records more proper time between departure and arrival, and by how much?

b) The interval between the departure and arrival pulses received at A for each spacecraft.

c) When the receiver at A detects each spacecraft’s halfway-proper-time pulse.

d) Whether a receiver that only records the departure and arrival signals can determine which spacecraft accumulated more proper time.

A student argues that the spacecraft must accumulate identical proper times because they have the same average speed and share the same departure and arrival events. Evaluate the argument.

Solution 5

Spacecraft S accumulates approximately \(0.509\ \mu\mathrm{s}\) more proper time than P. Both produce the same \(110\ \mu\mathrm{s}\) departure-to-arrival reception interval at A, but their halfway-proper-time pulses arrive at different times.

The central issue is that elapsed proper time depends on the complete motion of the clock, not just its average speed or the positions of its endpoints.

a) Calculate the two proper times.

For S, the entire \(100\ \mu\mathrm{s}\) journey occurs at \(0.10c\).

\[ \begin{aligned} \Delta\tau_S &=100\sqrt{1-0.10^2}\\ &=99.4987\ \mu\mathrm{s} \end{aligned} \]

For P, the first \(50\ \mu\mathrm{s}\) is spent stationary in the laboratory frame.

Its proper time for that section is therefore \(50\ \mu\mathrm{s}\).

For the second section,

\[ \begin{aligned} \Delta\tau_{P,2} &=50\sqrt{1-0.20^2}\\ &=48.9898\ \mu\mathrm{s} \end{aligned} \]

Total proper time:

\[ \begin{aligned} \Delta\tau_P &=50+48.9898\\ &=98.9898\ \mu\mathrm{s} \end{aligned} \]

Therefore,

\[ \begin{aligned} \Delta\tau_S-\Delta\tau_P &=99.4987-98.9898\\ &=0.5089\ \mu\mathrm{s} \end{aligned} \]

S records about \(0.509\ \mu\mathrm{s}\) more time.

It may initially feel as though P should accumulate more time because it spends half the journey stationary.

But P must then move twice as fast as S to reach B at the same time.

The relativistic clock-rate reduction increases nonlinearly with speed. The extra reduction during P’s faster section outweighs the advantage gained while it is stationary.

b) Calculate the departure-to-arrival reception interval.

Both spacecraft emit their departure pulses at the same position and time: station A at \(t=0\).

Both also emit their arrival pulses at the same position and time: station B at \(t=100\ \mu\mathrm{s}\).

The final pulses each travel \(3000\ \mathrm{m}\) back to A.

\[ \begin{aligned} t_{\mathrm{light}} &=\frac{3000}{3.00\times10^8}\\ &=10.0\ \mu\mathrm{s} \end{aligned} \]

Both final signals reach A at

\[ 100+10.0=110\ \mu\mathrm{s} \]

The initial signals are received at \(t=0\).

Therefore, both spacecraft produce an identical departure-to-arrival reception interval of \(110\ \mu\mathrm{s}\).

This equality is exact under the stated assumptions, despite their different accumulated proper times.

c) Find the halfway-proper-time pulse arrivals.

For S, half the total proper time is accumulated halfway through its constant-speed journey.

It emits its additional pulse at laboratory time

\[ t_S=50.0\ \mu\mathrm{s} \]

Its position is

\[ \begin{aligned} x_S &=(0.10)(3.00\times10^8)\\ &\quad\times(50.0\times10^{-6})\\ &=1500\ \mathrm{m} \end{aligned} \]

Light takes another \(5.00\ \mu\mathrm{s}\) to return from that position to A.

So the receiver detects S’s halfway pulse at

\[ t_{S,\mathrm{rec}}=55.0\ \mu\mathrm{s} \]

For P, half its total proper time is

\[ \begin{aligned} \frac{\Delta\tau_P}{2} &=\frac{98.9898}{2}\\ &=49.4949\ \mu\mathrm{s} \end{aligned} \]

Here is the important hinge.

P remains stationary beside A for the first \(50\ \mu\mathrm{s}\) of laboratory time. Its proper time therefore advances at the same rate as laboratory time throughout that initial wait.

It reaches half of its eventual proper time before it starts moving.

The halfway pulse is emitted beside A at laboratory time \(49.4949\ \mu\mathrm{s}\). With negligible local propagation delay, the receiver detects it immediately.

Thus,

\[ t_{P,\mathrm{rec}}=49.4949\ \mu\mathrm{s} \]

d) Can endpoint signal arrivals reveal the clock difference?

No.

Both spacecraft transmit their first and last pulses from exactly the same spacetime events. The receiver therefore obtains identical endpoint arrival times.

The endpoint signals contain no information about how each spacecraft travelled between those events.

Their additional halfway-proper-time pulses distinguish the journeys because the pulses are emitted at different positions and laboratory times.

Even then, interpreting reception times requires knowledge of the signal paths and emission conditions.

Evaluate the student’s argument.

Both spacecraft have the same average laboratory speed:

\[ v_{\mathrm{avg}} =\frac{3000}{100\times10^{-6}} =0.10c \]

But average speed is not sufficient to determine accumulated proper time.

For motion consisting of constant-speed sections,

\[ \Delta\tau_{\mathrm{total}} =\sum_i\frac{\Delta t_i}{\gamma_i} \]

Here, \(\Delta t_i\) is the laboratory duration of section \(i\), and \(\gamma_i\) is its Lorentz factor.

The student incorrectly substitutes the common average speed into one Lorentz factor for both journeys.

That ignores how each clock’s speed changes during the trip.

For these two routes, S travels at constant speed between the shared events and accumulates more proper time. P’s change of speed reduces its total proper time.

Sharing the same start and finish events does not force two clocks following different paths through spacetime to record identical elapsed times.

06What this changes about comparing clocks

At low speeds, the Lorentz factor is extremely close to 1, so the physical time-dilation correction is small. That makes experimental interpretation more demanding, not less.

Before applying any time-dilation equation, identify the two events, find which clock is present at both, and distinguish emission times from signal arrival times. Then decide whether one constant speed describes the entire interval.

The next useful problem is comparing clocks with different motion histories. How to Compare Two Moving Atomic Clocks Using Time Dilation develops that comparison, where using the wrong relative speed or the wrong pair of events can change the conclusion even when the arithmetic is correct.