How to Calculate the Distance a Muon Travels Before Decaying

A muon is created 2.00 km above Earth’s surface and travels straight down at \(0.960c\). Its mean lifetime at rest is just \(2.20\ \mu\mathrm{s}\). Multiply those numbers together and you get a travel distance of only 634 m. So how could the muon reach a detector 2.00 km below?

The missing step is time dilation. From Earth’s reference frame, the muon’s mean lifetime is longer because it is moving at a speed close to the speed of light.

To calculate the mean distance a muon travels before decaying, use:

\[ \boxed{\bar d=\gamma v\tau_0} \]

where:

• \(\bar d\) is the mean distance travelled in Earth’s frame, in metres (m).
• \(\gamma\) is the Lorentz factor, which accounts for time dilation.
• \(v\) is the muon’s speed relative to Earth, in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(\tau_0\) is the muon’s mean proper lifetime, in seconds (s).

For our muon travelling at \(0.960c\), the calculation gives approximately 2.26 km, not 634 m.

That doesn’t mean every muon travels exactly 2.26 km. Radioactive decay is random, so some decay earlier and others much later. The calculation predicts a mean distance for many muons travelling at the same constant speed, assuming they aren’t stopped or absorbed by matter.

01Why the muon’s own lifetime is not the time measured on Earth

Imagine a muon moving through the atmosphere with an impossibly small clock attached to it.

The clock starts when the muon is created and stops when the muon decays.

In the muon’s reference frame, the muon stays at the same position throughout its lifetime. The surrounding atmosphere moves past it.

An observer on Earth describes the situation differently. They see the muon begin at one position and decay somewhere else.

That difference tells us which time is the proper time.

Identify proper time using two events

Proper time is the time interval between two events measured by a clock that is present at both events. For an object travelling at constant velocity, it is measured in the object’s rest frame, where the two events occur at the same position.

For a muon travelling at constant velocity:

The muon’s mean proper lifetime is approximately:

\[ \tau_0=2.20\times10^{-6}\ \mathrm{s} \]

That’s 2.20 microseconds.

The Particle Data Group’s more precise value is approximately \(2.19698\ \mu\mathrm{s}\), but 2.20 microseconds is suitable for most HSC calculations.

Here’s the distinction worth remembering: proper time is determined by the events you’re measuring, not by whichever clock happens to display a number.

For the muon’s creation and decay, a clock travelling with the muon measures its own elapsed lifetime. Synchronized clocks stationary on Earth measure a different, longer interval.

Calculate the Lorentz factor

The amount of time dilation depends on the relative speed.

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

Here:

• \(v\) is the muon’s speed relative to Earth.
• \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\) is the speed of light in a vacuum.
• \(\gamma\) is dimensionless, meaning it has no units.

For the muon travelling at \(0.960c\):

\[ \begin{aligned} \gamma &=\frac{1}{\sqrt{1-(0.960)^2}}\\ &=\frac{1}{\sqrt{0.0784}}\\ &=3.57 \end{aligned} \]

Earth therefore measures a mean lifetime of:

\[ \begin{aligned} \tau_{\mathrm{E}}&=\gamma\tau_0\\ &=(3.571)(2.20\ \mu\mathrm{s})\\ &=7.86\ \mu\mathrm{s} \end{aligned} \]

The symbol \(\tau_{\mathrm{E}}\) represents the mean lifetime measured in Earth’s frame.

The mean distance is then:

\[ \begin{aligned} \bar d&=v\tau_{\mathrm{E}}\\ &=(0.960)(3.00\times10^8)\\ &\quad\times(7.86\times10^{-6})\\ &=2.26\times10^3\ \mathrm{m} \end{aligned} \]

So the muon’s mean travel distance is 2.26 km.

Notice that the muon hasn’t gained any extra lifetime in its own frame. Its mean proper lifetime is unchanged. The longer time is measured by Earth’s clocks.

02The calculation method: distance, speed, and the correct clock

For constant-speed motion, distance is speed multiplied by elapsed time.

The difficulty is choosing the correct elapsed time.

If you’re calculating the distance travelled in Earth’s frame, you must use the time measured in Earth’s frame.

\[ \bar d=v\tau_{\mathrm{E}} \]

Because \(\tau_{\mathrm{E}}=\gamma\tau_0\), this becomes:

\[ \bar d=\gamma v\tau_0 \]

You can use the following decision table.

The symbol \(\Delta\) means a change or interval. Here, \(\Delta t\) is an actual elapsed time in Earth’s frame, and \(\Delta\tau\) is the elapsed time experienced by the muon.

The distinction between a mean lifetime and an actual measured interval is important. If a problem tells you exactly when one muon decays, use that muon’s elapsed time to calculate its distance. Don’t replace it with the population’s mean lifetime.

03A hidden timing problem: observing a flash is not the same as timing an event

Suppose you place one light receiver at ground level.

A muon is created high in the atmosphere, producing a flash of light. Later, the muon decays closer to the ground, producing another flash.

The receiver detects both flashes.

Would the time between receiving them equal the muon’s actual flight time?

No. The two flashes travel different distances before reaching the receiver.

The second flash begins closer to the receiver, so it has less distance to travel. The recorded separation can be much shorter than the actual time between the events.

This is a light-travel delay, not relativistic time dilation.

How to correct a light-travel delay

The general correction for light signals in a stationary laboratory frame is:

\[ t_{\mathrm{arrival}} =t_{\mathrm{event}}+\frac{\ell}{c} \]

Here:

• \(t_{\mathrm{arrival}}\) is the time the receiver detects the flash.
• \(t_{\mathrm{event}}\) is when the flash was emitted.
• \(\ell\) is the distance travelled by the flash.
• \(c\) is the speed of light.

For two flashes arriving at the same receiver, subtract the two equations:

\[ \Delta t_{\mathrm{arrival}} =\Delta t+\frac{\ell_2-\ell_1}{c} \]

The distances \(\ell_1\) and \(\ell_2\) refer to the first and second light paths. All times and distances here are evaluated in Earth’s frame.

In our example, \(\ell_2-\ell_1=-800\ \mathrm{m}\), since the second flash starts closer to the receiver.

The correction explains the short observed delay without changing the muon’s actual flight time.

There is a particularly subtle point here. The two signal-arrival events happen at the same receiver, so the receiver measures a proper time interval between those arrival events. But those aren’t the muon’s creation and decay events.

Co-location only helps once you’ve identified the correct pair of events.

04How muon experiments measured the effect

The distance calculation isn’t just a mathematical trick. It helps explain why cosmic-ray muons can survive journeys through the atmosphere that would otherwise be very unlikely.

But there’s a measurement problem.

If you put a detector on a mountain and another near sea level, the number of detected muons will usually be different. How can you establish that the difference is caused by decay rather than by changes in particle energy, detector response, or interactions with matter?

The first idea: compare two altitudes

The simplest proposed experiment would be to count muons passing a high detector and compare them with the number passing a lower detector.

Without time dilation, the muons should decay according to their short mean rest lifetime even while moving.

With time dilation, the mean lifetime measured on Earth increases with speed, so a much larger fraction can survive the descent.

The problem is that cosmic-ray muons have a range of energies and speeds. Faster muons experience greater time dilation, and muons lose energy as they pass through the atmosphere.

A simple difference in total counts doesn’t isolate the relativistic effect.

The historical development

Bruno Rossi and David Hall reported experimental evidence in 1941 that the decay behaviour of fast cosmic-ray particles depended on their momentum, consistent with relativistic time dilation.

In 1963, David Frisch and James Smith carried out a particularly clear mountain-to-sea-level experiment using a more selective arrangement.

Their apparatus used a thick iron absorber above a plastic scintillator, with a photomultiplier connected to electronic recording equipment.

A muon travelled down through the iron. Muons with too little energy stopped in the iron and never reached the scintillator. Muons with too much energy passed straight through the scintillator.

A muon in the selected energy range stopped inside the plastic. It produced a flash of light, which the photomultiplier converted into an electrical pulse.

When that stopped muon decayed, a charged decay particle could produce a second flash, giving a second pulse on the oscilloscope.

The interval between those pulses measured the time from the muon stopping to its decay.

Since the muon was approximately at rest in the detector during that interval, this provided a measurement of its proper decay time. Repeating the measurement for many stopped muons allowed the researchers to determine a mean lifetime.

They then used the apparatus at Mount Washington and near sea level. They adjusted the amount of iron to compensate for energy lost while travelling through the intervening atmosphere, allowing a comparable population of muons to be selected at both sites.

This addressed an important weakness of a simple high-altitude versus low-altitude count: the two detectors needed to be comparing muons with equivalent initial speeds.

What was actually observed?

The selected muons travelled through approximately 1907 m of altitude difference at speeds near \(0.9952c\).

Their Earth-frame travel time was approximately:

\[ \begin{aligned} \Delta t&=\frac{1907}{0.9952c}\\ &=6.39\ \mu\mathrm{s} \end{aligned} \]

That’s almost three times the muon’s mean lifetime at rest.

The mean recorded counting rates were:

These are historical experimental results, not illustrative data. They are the reported raw counting rates, before the full background and experimental corrections.

For a one-hour comparison, the researchers estimated that a non-relativistic decay model would have left only about 27 counted muons at sea level, compared with 412 actually recorded.

The difference was far too large to explain using the short rest-frame lifetime without accounting for time dilation.

Their detailed analysis gave an observed time-dilation factor of approximately \(8.8\pm0.8\), consistent with the prediction for their selected muons and apparatus geometry.

Frisch and Smith didn’t watch one muon travel 1907 m and time its entire journey. They measured the distribution of decay times of stopped muons and compared counts of selected muon populations at two altitudes.

That distinction matters when interpreting the evidence.

The historical measurements and apparatus are described in Frisch and Smith, Measurement of the Relativistic Time Dilation Using μ-Mesons, American Journal of Physics, 31 (1963), 342-355.

For a broader explanation of why these experiments support relativity, see Muon Evidence for Time Dilation: HSC Physics Explained.

05Three mistakes that change the answer

Mistake 1: Multiplying speed by the proper lifetime in Earth’s frame

A student might write:

\[ d=v\tau_0 \]

That gives the distance travelled during one mean proper lifetime if that duration were also the Earth-frame duration. It ignores time dilation.

For a moving muon, the Earth-frame mean lifetime is \(\gamma\tau_0\).

Use:

\[ \bar d=\gamma v\tau_0 \]

Mistake 2: Treating a mean lifetime as an expiry time

A mean proper lifetime of \(2.20\ \mu\mathrm{s}\) does not mean every muon disappears exactly 2.20 microseconds after creation.

Some muons survive much longer.

A single muon travelling farther than the calculated mean distance doesn’t invalidate the model. The meaningful prediction concerns the distribution of decay distances across many muons.

Mistake 3: Applying one Lorentz factor to an interval when the motion changes

The simple expression \(\Delta\tau=\Delta t/\gamma\) assumes the muon’s speed stays constant over that interval.

If it slows down, stops, or changes speed, the value of \(\gamma\) changes.

You then need to treat the different stages separately.

This becomes especially important when studying a muon that enters a scintillator, stops, and only then decays.

06Questions and solutions

Use \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\) and \(\tau_0=2.20\ \mu\mathrm{s}\) throughout unless otherwise stated. Assume ideal timing equipment and negligible particle interactions except where specified.

Question 1

A muon is produced inside an evacuated straight beamline and travels at a constant speed of \(0.600c\) relative to the laboratory.

Calculate:

(a) its Lorentz factor,

(b) its mean lifetime in the laboratory frame, and

(c) its mean distance travelled before decaying.

Solution 1

The Lorentz factor is \(1.25\), the laboratory-frame mean lifetime is \(2.75\ \mu\mathrm{s}\), and the mean travel distance is 495 m.

For part (a):

\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-(0.600)^2}}\\ &=\frac{1}{0.800}\\ &=1.25 \end{aligned} \]

For part (b), time dilation gives:

\[ \begin{aligned} \tau_{\mathrm{lab}}&=\gamma\tau_0\\ &=(1.25)(2.20\ \mu\mathrm{s})\\ &=2.75\ \mu\mathrm{s} \end{aligned} \]

For part (c):

\[ \begin{aligned} \bar d&=v\tau_{\mathrm{lab}}\\ &=(0.600)(3.00\times10^8)\\ &\quad\times(2.75\times10^{-6})\\ &=495\ \mathrm{m} \end{aligned} \]

The muon travels an average of 495 m before decaying, assuming constant speed and no other interactions.

The tempting shortcut \(v\tau_0\) would give only 396 m. That is incorrect for the laboratory-frame mean distance because it uses the muon’s proper lifetime without accounting for time dilation.

Question 2

An experiment uses two detectors, A and B, separated by 900 m along a straight evacuated beamline.

A muon is created at A, moves directly towards B, and decays at B. Each event immediately produces an electrical timing pulse, which travels along a cable to the same recorder.

The cable from A is 400 m long. The cable from B is 100 m long. Electrical signals travel through both cables at \(2.00\times10^8\ \mathrm{m\,s^{-1}}\).

The recorder receives B’s pulse 2.50 microseconds after A’s pulse.

Calculate:

(a) the actual Earth-frame time between creation and decay,

(b) the muon’s speed and Lorentz factor,

(c) the proper time experienced by the muon, and

(d) explain why the recorded pulse separation cannot be used directly as the muon’s lifetime.

Solution 2

The actual flight time is \(4.00\ \mu\mathrm{s}\), the speed is \(0.750c\), and the muon’s proper lifetime for this particular decay is \(2.65\ \mu\mathrm{s}\). The recorder’s 2.50-microsecond reading is shortened by unequal cable delays.

First find the cable delays.

For detector A:

\[ \begin{aligned} t_A&=\frac{400}{2.00\times10^8}\\ &=2.00\ \mu\mathrm{s} \end{aligned} \]

For detector B:

\[ \begin{aligned} t_B&=\frac{100}{2.00\times10^8}\\ &=0.500\ \mu\mathrm{s} \end{aligned} \]

A’s pulse spends 1.50 microseconds longer travelling through its cable.

Therefore, the actual time between the source events is:

\[ \begin{aligned} \Delta t&=2.50+2.00-0.500\\ &=4.00\ \mu\mathrm{s} \end{aligned} \]

The muon travels 900 m in that time:

\[ \begin{aligned} v&=\frac{900}{4.00\times10^{-6}}\\ &=2.25\times10^8\ \mathrm{m\,s^{-1}}\\ &=0.750c \end{aligned} \]

The Lorentz factor is:

\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-(0.750)^2}}\\ &=1.512 \end{aligned} \]

The muon experiences:

\[ \begin{aligned} \Delta\tau&=\frac{4.00}{1.512}\ \mu\mathrm{s}\\ &=2.65\ \mu\mathrm{s} \end{aligned} \]

Both detector signals reach the same recorder, but their arrival times are not the creation and decay times.

This is why measuring a time interval at one location does not automatically make it the muon’s proper lifetime. The event locations, signal paths, and relative motion must all be considered.

Finally, a 2.65-microsecond lifetime is entirely consistent with a mean lifetime of 2.20 microseconds. One muon can live longer than the mean.

Question 3

A muon is produced at an upper detector and travels directly towards a lower scintillator 720 m away at \(0.800c\).

It stops inside the lower scintillator. Its stopping time is negligible compared with the other times in the question.

The muon remains at rest in the scintillator for another \(1.50\ \mu\mathrm{s}\), measured by the scintillator’s clock, before decaying.

Calculate:

(a) the total time from creation to decay measured in Earth’s frame,

(b) the total elapsed time experienced by the muon,

(c) which parts of the journey have proper time measured in the muon’s moving frame and in Earth’s frame, and

(d) explain why dividing the total Earth-frame time by the original Lorentz factor gives an incorrect result.

Solution 3

Earth measures \(4.50\ \mu\mathrm{s}\) from creation to decay, but the muon experiences a total of \(3.30\ \mu\mathrm{s}\). The motion has two different stages, so one Lorentz factor cannot be used for the entire interval.

During the first stage, the muon travels 720 m at \(0.800c\).

Its Earth-frame travel time is:

\[ \begin{aligned} t_1&=\frac{720}{0.800(3.00\times10^8)}\\ &=3.00\ \mu\mathrm{s} \end{aligned} \]

The Lorentz factor is:

\[ \gamma=\frac{1}{\sqrt{1-0.800^2}} =1.667 \]

The muon’s proper elapsed time during this moving stage is:

\[ \begin{aligned} \Delta\tau_1&=\frac{3.00}{1.667}\\ &=1.80\ \mu\mathrm{s} \end{aligned} \]

For the second stage, the muon is stationary inside the scintillator.

The stopping and decay events occur at the same position in Earth’s frame. Earth’s measured interval is therefore also the muon’s proper elapsed time for this stage:

\[ \Delta\tau_2=1.50\ \mu\mathrm{s} \]

Add the Earth-frame times:

\[ \begin{aligned} t_{\mathrm{total}}&=3.00+1.50\\ &=4.50\ \mu\mathrm{s} \end{aligned} \]

Add the proper times experienced by the muon:

\[ \begin{aligned} \tau_{\mathrm{total}}&=1.80+1.50\\ &=3.30\ \mu\mathrm{s} \end{aligned} \]

The mistake would be to calculate:

\[ \frac{4.50}{1.667} =2.70\ \mu\mathrm{s} \]

That treats the muon as though it moved at \(0.800c\) for the entire 4.50 microseconds.

It didn’t. It spent the final 1.50 microseconds at rest relative to Earth.

For the moving stage, the muon is present at both endpoints in its initial inertial rest frame. For the stopped stage, the relevant events are co-located in Earth’s frame. There is no single inertial rest frame of the muon that applies unchanged to both stages.

This also illustrates why a stopped-muon decay measurement is different from measuring its journey through the atmosphere.

Question 4

A muon is produced at position \(x=0\) and travels horizontally to the right at \(0.600c\). It decays at \(x=600\ \mathrm{m}\).

An ideal flash of light is emitted at the creation event and another at the decay event.

Two stationary receivers record the flashes:

• Receiver L is at \(x=-600\ \mathrm{m}\).
• Receiver R is at \(x=1200\ \mathrm{m}\).

Take rightwards as positive. Assume all light travels directly through a vacuum at speed \(c\), and every flash is emitted without delay.

(a) Calculate the actual Earth-frame time between creation and decay, and the muon’s proper elapsed time.

(b) Find the interval between the two flashes arriving at each receiver.

(c) Two students disagree. One says the different receiver intervals prove the muon’s lifetime depends on which receiver watches it. The other says that, if the muon moved sufficiently quickly towards receiver R, the decay flash could arrive before the creation flash. Evaluate both claims.

Solution 4

The Earth-frame interval is \(3.33\ \mu\mathrm{s}\), and the muon’s proper elapsed time is \(2.67\ \mu\mathrm{s}\). Receiver L records a \(5.33\)-microsecond separation, while receiver R records only \(1.33\ \mu\mathrm{s}\). Neither interval changes the actual lifetime, and the decay flash cannot overtake the earlier creation flash in this arrangement.

First calculate the event interval.

\[ \begin{aligned} \Delta t&=\frac{600}{0.600c}\\ &=3.33\ \mu\mathrm{s} \end{aligned} \]

Since \(\gamma=1.25\):

\[ \begin{aligned} \Delta\tau&=\frac{3.33}{1.25}\\ &=2.67\ \mu\mathrm{s} \end{aligned} \]

Now consider receiver L.

The creation flash travels 600 m to L, taking:

\[ t_{L,1}=\frac{600}{c} =2.00\ \mu\mathrm{s} \]

The decay flash travels 1200 m to L. It is emitted at \(t=3.33\ \mu\mathrm{s}\), so it arrives at:

\[ \begin{aligned} t_{L,2}&=3.33+\frac{1200}{c}\\ &=3.33+4.00\\ &=7.33\ \mu\mathrm{s} \end{aligned} \]

Receiver L records:

\[ \Delta t_L=7.33-2.00 =5.33\ \mu\mathrm{s} \]

Now consider receiver R.

The creation flash travels 1200 m, taking 4.00 microseconds.

The decay flash travels only 600 m, taking 2.00 microseconds after it is emitted.

Its arrival time is:

\[ t_{R,2}=3.33+2.00 =5.33\ \mu\mathrm{s} \]

Receiver R records:

\[ \Delta t_R=5.33-4.00 =1.33\ \mu\mathrm{s} \]

The receivers measure different arrival intervals because the second flash travels farther to L but less far to R.

Neither receiver directly measures the time between the source events without correcting the light-travel delays.

Now consider whether R could receive the decay flash first.

For a muon moving towards R at constant speed \(v\), the separation between arriving flashes is:

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

Because a massive muon must travel slower than light, \(v<c\). Therefore:

\[ \Delta t_R>0 \]

The separation can become extremely small as \(v\) approaches \(c\), but it cannot become negative.

The second student’s prediction is impossible under the stated assumptions. A muon cannot outrun light emitted earlier in the same direction.

The first student is also incorrect. The two receivers are detecting different photon arrival intervals, not different physical lifetimes for the same muon.

This is a useful experimental distinction: the ordering and timing of received signals can depend strongly on where the receiver is positioned, even when the underlying events are unchanged.

Question 5

A laboratory has an evacuated circular particle guide with circumference 1.20 km and a detector at point A.

Two muons are created simultaneously at A.

• Muon M travels around the circular guide at a constant speed of \(0.800c\), maintained by a magnetic field that changes its direction but not its speed. It completes exactly one full lap and decays as it passes A again.
• Muon S remains stationary at A and decays at exactly the same laboratory time as M.

Assume the creation and decay events occur at the same laboratory position A for both muons, the magnetic field does not directly alter decay behaviour, and relativistic effects depend on each muon’s accumulated proper time.

Answer the following:

(a) Calculate the laboratory time from creation to decay.

(b) Find the distance travelled, displacement, and total proper time experienced by each muon.

(c) A student argues: “Creation and decay occur at the same position in Earth’s frame, so Earth’s measured interval must be the proper lifetime of both muons.” Another argues: “Each muon is present at its own creation and decay, so each must measure the same proper interval.” Explain precisely why both arguments cannot be applied in that way.

(d) Imagine repeating the experiment with two large, equally sized populations of muons: one population kept stationary at A and the other circulating at \(0.800c\). Without calculating exponential decay probabilities, predict which population has the greater fraction surviving after \(5.00\ \mu\mathrm{s}\) on Earth’s clock. Explain.

(e) Is M’s actual travel distance consistent with the usual mean-distance calculation, even though it exceeds that predicted mean?

Solution 5

Both muons decay after \(5.00\ \mu\mathrm{s}\) according to Earth’s clock, but M experiences only \(3.00\ \mu\mathrm{s}\), while S experiences the full \(5.00\ \mu\mathrm{s}\). The difference comes from their paths through spacetime, not from their shared start and end positions.

This is an extension of the ordinary HSC constant-velocity calculation. The essential conceptual trap is that the travelling muon does not remain in one inertial reference frame throughout the circular journey.

Part (a): Laboratory time

Muon M travels one full circumference:

\[ d_M=1.20\times10^3\ \mathrm{m} \]

At \(0.800c\):

\[ \begin{aligned} \Delta t&=\frac{1200}{0.800c}\\ &=5.00\times10^{-6}\ \mathrm{s}\\ &=5.00\ \mu\mathrm{s} \end{aligned} \]

Muon S decays at the same laboratory time, so Earth’s clock measures 5.00 microseconds for both muons.

Part (b): Distance, displacement, and proper time

Muon M travels around the ring exactly once.

Its total distance is:

\[ d_M=1.20\ \mathrm{km} \]

But it returns to its starting position, so its displacement is:

\[ \Delta x_M=0 \]

More precisely, its displacement vector is zero.

Muon S never moves relative to the laboratory, so both its distance travelled and displacement are zero.

Now calculate their proper times.

Muon M has a constant speed of \(0.800c\), so its Lorentz factor has a constant magnitude:

\[ \gamma=\frac{1}{\sqrt{1-0.800^2}} =1.667 \]

Although its direction continuously changes, its instantaneous time-dilation factor remains the same. Its accumulated proper time is:

\[ \begin{aligned} \tau_M&=\frac{5.00}{1.667}\\ &=3.00\ \mu\mathrm{s} \end{aligned} \]

Muon S stays at rest relative to Earth’s clock throughout:

\[ \tau_S=5.00\ \mu\mathrm{s} \]

So both particles begin and end at the same laboratory location and time coordinates, but accumulate different amounts of proper time along their different paths.

Part (c): The apparent contradiction

The first student has correctly noticed that creation and decay are co-located in Earth’s inertial reference frame.

That means a stationary clock at A measures the proper time along the stationary worldline joining those events: 5.00 microseconds.

But that is not necessarily the proper time accumulated by a particle that travels away and returns.

Muon M doesn’t remain at rest in any single inertial reference frame during the circular journey. Its direction changes continuously.

We must distinguish:

• The proper time measured by the stationary clock at A between the shared endpoint events.
• The accumulated proper time along M’s moving trajectory.

Those are different physical clock paths.

The second student’s argument contains a related error. Being present at two events does not mean that all objects travelling between those events accumulate the same time. Their trajectories matter.

For straight, constant-velocity motion, the muon’s creation and decay are co-located in its inertial rest frame. That familiar HSC rule works because the muon has one unchanging rest frame throughout the journey.

For circular motion, there is no single inertial frame in which M remains stationary for the whole lap.

The correct comparison is:

If you’re unsure why M’s distance and displacement are different despite its return to A, revisit Distance and Displacement in HSC Physics Explained Clearly.

Part (d): Which population survives longer?

The circulating population has the greater fraction surviving after 5.00 microseconds of laboratory time.

By that laboratory time, the stationary muons have experienced 5.00 microseconds of proper time, while the circulating muons have experienced only 3.00 microseconds.

Muon decay probabilities are governed by the time experienced along the particle’s own trajectory.

The moving population has therefore undergone less accumulated proper time over which decay could occur. A greater fraction is expected to remain undecayed, assuming otherwise identical populations and no additional losses.

It would be incorrect to argue that the populations must have equal survival fractions merely because their creation and observation occur at the same laboratory location.

Part (e): Mean distance versus actual distance

The observed 1.20 km distance is consistent with the muon’s mean lifetime.

At \(0.800c\), the mean path length before decay under the stated constant-speed assumptions is:

\[ \begin{aligned} \bar d&=\gamma v\tau_0\\ &=(1.667)(0.800c)\\ &\quad\times(2.20\times10^{-6}\ \mathrm{s})\\ &=880\ \mathrm{m} \end{aligned} \]

That’s the mean distance travelled along the path, not a maximum distance or a prediction that every muon must decay after exactly 880 m.

Muon M happens to survive 1.20 km before decaying. Its accumulated proper lifetime is 3.00 microseconds, longer than the mean of 2.20 microseconds but perfectly possible.

The final lesson is more subtle than simply remembering to multiply by \(\gamma\): a decay-distance calculation must distinguish the path travelled, the frame measuring the motion, and the time accumulated by the particle.

06What to do when the HSC question changes the conditions

For the usual straight-line, constant-speed muon question, the method is short:

1. Identify the proper lifetime by checking that creation and decay occur at the same position in the muon’s inertial rest frame.
2. Calculate \(\gamma\) from the given speed.
3. Convert the mean proper lifetime into the Earth-frame mean lifetime using \(\tau_{\mathrm{E}}=\gamma\tau_0\).
4. Multiply by the speed to find the mean Earth-frame travel distance.
5. State that the result is a mean, not a fixed decay distance.

But when a question supplies detector readings, ask one extra question before calculating: are these the times of the physical events, or the times when signals from those events were received?

When the muon’s motion changes, separate the journey into stages rather than forcing one Lorentz factor onto the whole interval.

These two checks connect the standard muon-distance calculation to the deeper physics of measuring time: proper time belongs to a specified physical history, and an instrument’s reading only has meaning once you know exactly which events it records.