Why a Light Clock Uses a Diagonal Path | HSC Physics

Put a light clock on a train moving to the right. In the train’s frame, a pulse of light travels straight upwards to a mirror, reflects, and travels straight back down. But what does someone standing on the platform see?

The train moves sideways while the light is travelling. By the time the pulse reaches the upper mirror, that mirror has moved to the right. The pulse must therefore travel diagonally to reach it.

That is why a light-clock derivation uses a diagonal light path: the diagonal represents the actual path of the light pulse in the platform’s reference frame. Because light travels at the same speed, \(c\), in both frames, the longer path requires a longer time.

Before looking at the mathematics, predict what happens if the train moves faster. The diagonal becomes more pronounced, the light travels a greater distance during each half-trip, and the platform measures a longer clock tick. Light itself does not slow down.

01Start with the light clock

A light clock is an idealised clock made from two parallel mirrors facing each other.

The mirrors are separated by a distance \(L\). A short pulse of light travels between them, reflecting off each mirror.

One complete tick occurs when the pulse:

1. Leaves the lower mirror.
2. Reaches the upper mirror.
3. Reflects and returns to the lower mirror.

The clock measures time by counting these complete journeys.

This is a thought experiment rather than a report of a particular laboratory experiment. Its purpose is to isolate one important feature of special relativity: every inertial observer measures the same speed of light in a vacuum.

First, consider the clock’s own reference frame

Imagine sitting beside the light clock, moving with it.

The mirrors are stationary relative to you. The pulse travels straight upwards and then straight downwards.

Call the three events:

• A: the pulse leaves the lower mirror.
• B: the pulse reaches the upper mirror.
• C: the pulse returns to the lower mirror.

Notice something important about A and C.

Both events happen at the same position: the lower mirror.

A clock attached to that mirror can record both events. The time between them is therefore the proper time for this complete tick.

Proper time is the elapsed time between two events measured by one clock present at both events.

We write it as \(\Delta\tau\), measured in seconds.

The light travels a distance \(L\) upwards and \(L\) downwards. Therefore, its total distance is \(2L\).

Since its speed is \(c\):

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

This is the proper time for one complete light-clock tick.

An important detail: Why not use just the upward journey?

The upward journey from A to B takes:

\[ t_{AB}=\frac{L}{c} \]

However, A and B occur at different positions.

A clock attached to the lower mirror cannot record both events. Neither can a clock attached to the upper mirror.

Therefore, \(L/c\) is a one-way light-travel time, not the proper time between A and B.

This distinction matters when using the light-clock derivation. The proper time belongs to the complete A-to-C journey.

You can use the geometry of one half-trip to derive the result, but you must double the half-trip time to obtain the complete proper-time interval.

02Why does the platform observer see a diagonal path?

Now imagine watching the same clock from the platform.

The clock moves horizontally at a constant speed \(v\), which we take as positive in the rightward direction.

The mirrors are moving too.

When the light leaves the lower mirror at event A, the upper mirror is directly above it. But the upper mirror does not stay there.

During the light’s journey, the upper mirror moves horizontally.

The pulse must reach the mirror’s new position, not its original position.

Think about throwing a ball towards a teammate running sideways. You need to aim where the teammate will be, rather than where they were when you released the ball.

The light-clock situation has the same moving-target geometry. There is one crucial difference: the speed of light remains \(c\) in the platform frame. It does not increase by adding the train’s speed.

Picture the two directions of motion

During the upward half-trip:

• The light moves upwards and to the right.
• The upper mirror moves to the right.
• The light must cover the mirror’s horizontal displacement to reach it.

During the downward half-trip:

• The light moves downwards and to the right.
• The lower mirror also moves to the right.
• The light must again meet the mirror at its new position.

Both sections of the light path are diagonal in the platform frame.

The geometry of one half-trip

Let:

• \(L\) = vertical separation between the mirrors, in metres (m).
• \(v\) = horizontal speed of the clock relative to the platform, in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(t_h\) = time taken for one half-trip in the platform frame, in seconds (s).
• \(\Delta x\) = horizontal distance travelled by the clock during that half-trip, in metres (m).
• \(\ell\) = diagonal distance travelled by the light, in metres (m).
• \(c\) = speed of light in a vacuum, approximately \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).

During the half-trip, the clock moves horizontally:

\[ \Delta x=vt_h \]

The light travels a diagonal distance \(\ell\).

The vertical separation is \(L\), so Pythagoras gives:

\[ \ell^2=L^2+(vt_h)^2 \]

But the light travels at speed \(c\), meaning:

\[ \ell=ct_h \]

Substituting this into the geometry gives:

\[ (ct_h)^2=L^2+(vt_h)^2 \]

This is the key relationship in the derivation.

The light’s path is longer because the mirrors move sideways while the pulse is travelling.

03Deriving time dilation from the diagonal path

We can now combine the geometry with the constant speed of light.

Starting with:

\[ c^2t_h^2=L^2+v^2t_h^2 \]

Rearrange:

\[ c^2t_h^2-v^2t_h^2=L^2 \]

Factorise:

\[ t_h^2(c^2-v^2)=L^2 \]

Taking the positive square root:

\[ t_h=\frac{L}{\sqrt{c^2-v^2}} \]

Divide the numerator and denominator by \(c\):

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

The expression

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

is called the Lorentz factor.

It describes how strongly relativistic effects appear at a particular relative speed.

Therefore:

\[ t_h=\gamma\frac{L}{c} \]

This is the time for one half-trip measured on the platform.

However, remember that a complete clock tick requires two half-trips.

The platform measures:

\[ \Delta t=2t_h \]

Substituting:

\[ \Delta t=2\gamma\frac{L}{c} \]

From the clock’s own reference frame, we established that:

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

Therefore:

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

This is the time-dilation equation.

Here:

• \(\Delta t\) is the time between the same two events measured in the platform frame, in seconds.
• \(\Delta\tau\) is the proper time measured by the moving clock, in seconds.
• \(\gamma\) is the dimensionless Lorentz factor.
• \(v\) is the relative speed between the clock and the platform, in \(\mathrm{m\,s^{-1}}\).
• \(c\) is the speed of light in a vacuum, in \(\mathrm{m\,s^{-1}}\).

The platform measures a longer interval because the moving clock takes longer to complete its tick.

The important point is that light has not slowed down. Its speed remains \(c\). The difference comes from the geometry of its path and the different time intervals assigned to the same events.

04Time dilation is not the same as light-travel delay

Here is a distinction that causes confusion in relativity questions.

Suppose you watch a moving clock through a camera. You don’t receive light from the clock at the instant an event happens.

The light needs time to travel from the clock to your camera.

This creates a difference between:

1. The time between events occurring on the moving clock.
2. The time between the camera receiving light from those events.

These are not necessarily the same.

The diagonal light-clock derivation concerns the first situation. It compares the same two physical events in different reference frames.

A camera’s recorded arrival times introduce an additional effect: light-travel delay.

A contrasting case: A stationary source and receiver

Suppose a stationary laser is \(900\ \mathrm{m}\) from a stationary detector.

The laser emits two pulses \(2.00\ \mu\mathrm{s}\) apart.

Each pulse takes:

\[ t=\frac{900}{3.00\times10^8} \]

\[ t=3.00\ \mu\mathrm{s} \]

to reach the detector.

The first pulse arrives \(3.00\ \mu\mathrm{s}\) after emission. The second arrives \(3.00\ \mu\mathrm{s}\) after its emission.

The detector still receives the pulses \(2.00\ \mu\mathrm{s}\) apart.

Why?

Both pulses travel the same distance. The propagation delay shifts both arrival times by the same amount, so their separation remains unchanged.

This is different from the receding-clock example, where the second flash travels a greater distance.

The useful distinction is:

• Time dilation changes the time interval assigned to the same two events in different reference frames.
• Light-travel delay changes when a signal reaches an observer.
• Both effects can occur in the same situation, but they must be calculated separately.

05Why use a light clock perpendicular to its motion?

You might wonder why the standard derivation uses mirrors stacked vertically rather than mirrors arranged horizontally.

The reason is that the transverse arrangement makes the geometry much simpler.

When the mirror separation is perpendicular to the clock’s motion, there is no length contraction along that separation.

The distance \(L\) remains the same in both reference frames.

This lets us use a straightforward right-angled triangle to derive time dilation.

If the mirrors were arranged parallel to the motion, the situation would be different.

The mirror separation would be affected by length contraction, and the two light journeys would take different amounts of time:

• On the outward journey, the light must catch up with the front mirror.
• On the return journey, the light meets the rear mirror as it moves towards it.

The paths are horizontal rather than diagonal, but time dilation still occurs.

This is an important point: the diagonal path is a convenient way to derive time dilation. It is not a requirement for time dilation to exist.

06A reliable way to approach light-clock questions

When a question involves a moving light clock, don’t immediately substitute numbers into the time-dilation equation.

First identify the physical events.

Use this decision process.

Step 1: Identify the two events.

Are you considering emission and reflection, or emission and return?

These are different event pairs.

Step 2: Check whether the events are co-located.

If one clock is present at both events, its elapsed time is the proper time.

For a complete light-clock tick, emission and return occur at the same lower mirror in the clock’s rest frame.

Step 3: Identify the reference frame.

Is the question asking for the time measured by the moving clock or the time assigned by a stationary platform?

Step 4: Choose the appropriate equation.

For a transverse light clock:

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

For the corresponding platform-frame interval:

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

where:

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

Step 5: Separate signal delays if necessary.

If the question involves a camera, detector, or observer receiving flashes, calculate the light-travel time separately.

Do not automatically treat the interval between received signals as the interval between the original events.

For a broader explanation of identifying the correct event pair, see How to Identify the Two Events Before Choosing Proper Time.

07Questions and solutions

Question 1

A transverse light clock has two mirrors separated by \(0.60\ \mathrm{m}\). The clock moves horizontally at \(0.80c\) relative to a stationary laboratory.

Calculate:

(a) The proper time for one complete tick.

(b) The time for the same tick measured in the laboratory.

Take \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\).

Solution 1

The proper time is \(4.00\ \mathrm{ns}\), and the laboratory measures \(6.67\ \mathrm{ns}\).

Step 1: Calculate the proper time.

The complete tick is a round trip between the lower mirror and the upper mirror.

Therefore:

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

Substitute:

\[ \begin{aligned} \Delta\tau &=\frac{2(0.60)}{3.00\times10^8}\\ &=4.00\times10^{-9}\ \mathrm{s} \end{aligned} \]

So:

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

Step 2: Calculate the Lorentz factor.

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

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

\[ \gamma=1.667 \]

Step 3: Calculate the laboratory time.

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

\[ \Delta t=1.667(4.00) \]

\[ \Delta t=6.67\ \mathrm{ns} \]

The laboratory measures a longer interval because the clock is moving relative to it.

The light still travels at \(c\). The additional time comes from the longer diagonal path in the laboratory frame.

Question 2

A transverse light clock has mirrors separated by \(1.20\ \mathrm{m}\). It moves at \(0.60c\) relative to a laboratory.

A pulse leaves the lower mirror at event A, reflects off the upper mirror at event B, and returns to the lower mirror at event C.

Calculate:

(a) The time between A and B in the clock’s rest frame.

(b) The time between A and C in the clock’s rest frame.

(c) The time between A and C in the laboratory frame.

(d) Identify which of these intervals is proper time. Explain why the A-to-B interval is not proper time, even though it is measured in the clock’s rest frame.

Solution 2

The A-to-B interval is \(4.00\ \mathrm{ns}\), the A-to-C proper time is \(8.00\ \mathrm{ns}\), and the laboratory measures \(10.0\ \mathrm{ns}\) for A to C.

Step 1: Calculate the A-to-B interval.

The light travels one mirror separation:

\[ t_{AB}=\frac{L}{c} \]

\[ t_{AB}=\frac{1.20}{3.00\times10^8} \]

\[ t_{AB}=4.00\ \mathrm{ns} \]

Step 2: Calculate the A-to-C interval.

The pulse travels upwards and then downwards.

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

\[ \Delta\tau=\frac{2(1.20)}{3.00\times10^8} \]

\[ \Delta\tau=8.00\ \mathrm{ns} \]

This is proper time because A and C occur at the same position in the clock’s rest frame.

Step 3: Calculate the Lorentz factor.

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

\[ \gamma=1.25 \]

Step 4: Calculate the laboratory interval.

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

\[ \Delta t=1.25(8.00) \]

\[ \Delta t=10.0\ \mathrm{ns} \]

Step 5: Explain the difference between A-to-B and A-to-C.

The A-to-B events occur at different positions, separated by the mirror distance.

There is no single clock present at both events. Therefore, \(4.00\ \mathrm{ns}\) is a one-way light-travel interval, not a proper-time measurement.

By contrast, A and C occur at the same lower mirror. One clock can record both events, making \(8.00\ \mathrm{ns}\) the proper time.

The key lesson is that measuring an interval in an object’s rest frame does not automatically make it proper time. The two events must also be co-located in that frame.

Question 3

An observer in a laboratory measures the time for one upward half-trip of a moving transverse light clock as \(4.00\ \mathrm{ns}\).

The vertical separation between the mirrors is \(0.90\ \mathrm{m}\).

Assuming light travels at \(3.00\times10^8\ \mathrm{m\,s^{-1}}\), calculate:

(a) The diagonal distance travelled by the light.

(b) The horizontal displacement of the clock during the half-trip.

(c) The clock’s speed as a fraction of \(c\).

(d) The proper time for one complete clock tick.

Solution 3

The light travels \(1.20\ \mathrm{m}\) diagonally, the clock moves horizontally by \(0.794\ \mathrm{m}\), and its speed is approximately \(0.661c\). The proper time for a complete tick is \(6.00\ \mathrm{ns}\).

Step 1: Calculate the diagonal light-travel distance.

The light travels at \(c\) for \(4.00\ \mathrm{ns}\).

\[ \ell=ct_h \]

\[ \ell=(3.00\times10^8)(4.00\times10^{-9}) \]

\[ \ell=1.20\ \mathrm{m} \]

Step 2: Calculate the horizontal displacement.

The vertical separation is \(0.90\ \mathrm{m}\).

Using Pythagoras:

\[ \ell^2=L^2+\Delta x^2 \]

Rearrange:

\[ \Delta x=\sqrt{\ell^2-L^2} \]

Substitute:

\[ \begin{aligned} \Delta x &=\sqrt{1.20^2-0.90^2}\\ &=\sqrt{1.44-0.81}\\ &=\sqrt{0.63}\\ &=0.794\ \mathrm{m} \end{aligned} \]

Step 3: Calculate the clock’s speed.

The clock moves horizontally by \(0.794\ \mathrm{m}\) during \(4.00\ \mathrm{ns}\).

\[ v=\frac{\Delta x}{t_h} \]

\[ v=\frac{0.794}{4.00\times10^{-9}} \]

\[ v=1.98\times10^8\ \mathrm{m\,s^{-1}} \]

Express this as a fraction of \(c\):

\[ \frac{v}{c} = \frac{1.98\times10^8}{3.00\times10^8} \]

\[ v\approx0.661c \]

Step 4: Calculate the proper time for one complete tick.

The proper time is measured in the clock’s rest frame.

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

\[ \Delta\tau=\frac{2(0.90)}{3.00\times10^8} \]

\[ \Delta\tau=6.00\times10^{-9}\ \mathrm{s} \]

Therefore:

\[ \Delta\tau=6.00\ \mathrm{ns} \]

As a consistency check, the laboratory measures \(4.00\ \mathrm{ns}\) for each half-trip, or \(8.00\ \mathrm{ns}\) for the complete tick.

The ratio of the laboratory interval to the proper time is:

\[ \gamma=\frac{8.00}{6.00}=1.333 \]

This agrees with a speed of approximately \(0.661c\).

Question 4

A transverse light clock moves horizontally at \(0.60c\) relative to a laboratory.

The pulse leaves the lower mirror at event A, when the lower mirror is at \(x=4.00\ \mathrm{m}\). A stationary laboratory clock at that position reads \(2.00\ \mathrm{ns}\).

The pulse later returns to the lower mirror at event C, when it is at \(x=6.16\ \mathrm{m}\). A second stationary laboratory clock, synchronised with the first, reads \(14.0\ \mathrm{ns}\).

Calculate:

(a) The interval between A and C in the laboratory frame.

(b) The proper time between A and C.

(c) Explain why the laboratory can measure the interval using two different clocks, even though neither stationary clock is present at both events.

Solution 4

The laboratory interval is \(12.0\ \mathrm{ns}\), and the proper time is \(9.60\ \mathrm{ns}\). The laboratory can determine its coordinate time using synchronised clocks at different positions.

Step 1: Calculate the laboratory time interval.

Subtract the time at A from the time at C:

\[ \Delta t=14.0-2.00 \]

\[ \Delta t=12.0\ \mathrm{ns} \]

Step 2: Calculate the Lorentz factor.

The clock moves at \(0.60c\).

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

\[ \gamma=1.25 \]

Step 3: Calculate the proper time.

Rearrange the time-dilation equation:

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

Substitute:

\[ \Delta\tau=\frac{12.0}{1.25} \]

\[ \Delta\tau=9.60\ \mathrm{ns} \]

Step 4: Explain how the laboratory measures the interval.

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

The first stationary clock records event A. The second stationary clock records event C.

Because the two clocks are synchronised, their readings can be compared to establish the laboratory-frame time interval.

The laboratory does not need one stationary clock to be physically present at both events.

However, neither stationary laboratory clock measures the proper time between A and C. That proper time is measured by the moving clock itself, which is present at both events.

This is why it is important to distinguish a reference frame’s coordinate time from the reading of a single clock.

Question 5

Two identical ideal light clocks have the same proper period of \(8.00\ \mathrm{ns}\). Both move at \(0.80c\) relative to a laboratory.

However, their mirror arrangements are different.

• Clock T has its mirrors perpendicular to the direction of motion.
• Clock P has its mirrors parallel to the direction of motion.

A student makes the following argument:

“Clock T experiences time dilation because its light follows a diagonal path. Clock P has a straight horizontal light path, so it should not experience time dilation.”

Assess this argument.

Calculate the complete tick period measured in the laboratory for both clocks.

For Clock P, use the length-contraction relationship:

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

where \(L_0\) is the mirror separation in the clock’s rest frame and \(L\) is the separation measured in the laboratory.

Assume both clocks move at constant speed, have instantaneous reflections, and operate in a vacuum.

Solution 5

Both clocks have a laboratory-frame tick period of \(13.3\ \mathrm{ns}\). The student’s argument is incorrect because a diagonal light path is not required for time dilation.

The transverse clock uses diagonal geometry. The longitudinal clock instead has unequal outward and return travel times, together with length contraction.

Step 1: Calculate the Lorentz factor.

Both clocks move at \(0.80c\).

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

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

\[ \gamma=1.667 \]

Step 2: Calculate the laboratory period of Clock T.

The proper period is:

\[ \Delta\tau=8.00\ \mathrm{ns} \]

Using time dilation:

\[ \Delta t_T=\gamma\Delta\tau \]

\[ \Delta t_T=1.667(8.00) \]

\[ \Delta t_T=13.3\ \mathrm{ns} \]

Clock T therefore takes \(13.3\ \mathrm{ns}\) per complete tick in the laboratory frame.

Step 3: Calculate the proper mirror separation of Clock P.

The proper period is a complete round trip, so:

\[ \Delta\tau=\frac{2L_0}{c} \]

Rearrange:

\[ L_0=\frac{c\Delta\tau}{2} \]

Substitute:

\[ L_0= \frac{(3.00\times10^8)(8.00\times10^{-9})}{2} \]

\[ L_0=1.20\ \mathrm{m} \]

Step 4: Apply length contraction.

The mirror separation is parallel to the direction of motion.

Therefore, its laboratory-frame length is:

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

\[ L=\frac{1.20}{1.667} \]

\[ L=0.720\ \mathrm{m} \]

This is the distance between the moving mirrors as measured in the laboratory.

Step 5: Calculate the outward light-travel time.

The light travels towards the front mirror, which is moving away from it.

The light’s speed is \(c\), while the front mirror moves at \(0.80c\).

The closing speed is therefore:

\[ c-v=c-0.80c=0.20c \]

The outward journey takes:

\[ t_{\mathrm{out}}=\frac{L}{c-v} \]

Substitute:

\[ t_{\mathrm{out}} = \frac{0.720}{0.20(3.00\times10^8)} \]

\[ t_{\mathrm{out}}=12.0\ \mathrm{ns} \]

Step 6: Calculate the return light-travel time.

After reflection, the light travels towards the rear mirror.

The rear mirror moves towards the returning light, so the closing speed is:

\[ c+v=c+0.80c=1.80c \]

The return journey takes:

\[ t_{\mathrm{return}}=\frac{L}{c+v} \]

Substitute:

\[ t_{\mathrm{return}} = \frac{0.720}{1.80(3.00\times10^8)} \]

\[ t_{\mathrm{return}}=1.33\ \mathrm{ns} \]

Step 7: Calculate the complete laboratory tick.

Add the outward and return times:

\[ \begin{aligned} \Delta t_P &=t_{\mathrm{out}}+t_{\mathrm{return}}\\ &=12.0+1.33\\ &=13.33\ \mathrm{ns} \end{aligned} \]

Therefore:

\[ \Delta t_P=13.3\ \mathrm{ns} \]

Both clocks have the same laboratory-frame period.

Why was the student’s argument tempting?

The student assumed that without a diagonal path, there would be no extra time.

But the horizontal clock has two competing effects:

• The outgoing pulse takes longer because it must catch a mirror moving away.
• The returning pulse takes less time because it meets a mirror moving towards it.

Length contraction also changes the mirror separation measured in the laboratory.

Ignoring length contraction would produce an incorrect result. For example, using \(1.20\ \mathrm{m}\) as the laboratory separation would give a total period of \(22.2\ \mathrm{ns}\), rather than the correct \(13.3\ \mathrm{ns}\).

The two arrangements are consistent. The transverse clock gives a simple diagonal-path derivation, while the longitudinal clock requires length contraction and different travel times in each direction.

The deeper conclusion is that time dilation is a relationship between reference frames, not a special property of diagonal light paths.

The next useful step is learning to identify the correct event pair before comparing clocks. The same reasoning applies to real moving atomic clocks and to particle lifetimes, including muons travelling through Earth’s atmosphere.

You can extend the topic with How to Compare Two Moving Atomic Clocks Using Time Dilation and How to Calculate the Distance a Muon Travels Before Decaying.