Corrected exercises: special relativity (203-SN3-RE)

First postulate: the laws of physics are identical in all inertial reference frames. That one shocks nobody, since it is already Galileo's principle of relativity, three centuries old. Second postulate: light propagates at $c$ in every inertial frame, independently of the motion of the source AND of the observer. That one is intolerable, because it flatly contradicts the Galilean addition of velocities: if I chase a light ray at $0.9c$, I should see it recede at $0.1c$, and Einstein asserts that I see it recede at $c$. The rest of the chapter, dilation, contraction, the loss of simultaneity, is only the logical consequence of that refusal.

$\gamma=(1-\beta^{2})^{-1/2}$: $\gamma(0.100)=1.005$; $\gamma(0.600)=1.250$ exactly, since $\sqrt{1-0.36}=0.8$; $\gamma(0.800)=1.667$, since $\sqrt{1-0.64}=0.6$; $\gamma(0.950)=3.203$; $\gamma(0.990)=7.089$; $\gamma(0.999)=22.37$. The curve shows the essential point: $\gamma$ stays glued to 1 over the first two thirds of the axis, then takes off abruptly. Between $\beta=0$ and $\beta=0.5$ it gains only 15%; between $0.99$ and $0.999$ it triples. Relativity is not a gradual correction, it is a wall.

$\gamma=1.01$ gives $1-\beta^{2}=\dfrac{1}{1.01^{2}}=0.98030$, so $\beta^{2}=0.019704$ and $\beta=0.1404$. As a speed: $v=0.1404\times 3.00\times 10^{8}=4.21\times 10^{7}$ m/s, about 42,000 km/s. This is the real content of the phrase “classical mechanics is good enough at ordinary speeds”: it is good enough up to 14% of the speed of light, which covers absolutely everything an engineer will encounter, and nothing of what a particle physicist handles.

$\beta=\dfrac{250}{3.00\times 10^{8}}=8.33\times 10^{-7}$. At this scale a series expansion is mandatory, since a calculator will display $\gamma=1$ and nothing more: $\gamma-1\approx\dfrac{\beta^{2}}{2}=\dfrac{(8.33\times 10^{-7})^{2}}{2}=3.47\times 10^{-13}$. Over $12.0$ h $=43{,}200$ s, the lag is $43{,}200\times 3.47\times 10^{-13}=1.5\times 10^{-8}$ s, that is 15 nanoseconds. Hafele and Keating could measure it because they carried cesium atomic clocks, stable to better than a nanosecond, and above all because they flew the airplane eastward AND THEN westward: the rotation of the Earth changes the sign of the speed relative to the inertial frame, and it is the DIFFERENCE between the two flights, far larger than the uncertainty of each measurement, that they compared with the prediction. The calculation trap here is to type $1/\sqrt{1-\beta^{2}}$ into the calculator and read exactly 1: at the $10^{-13}$ level, the double precision of a pocket machine sees nothing.

In the car, the clock is at rest: the flash rises $1.50$ m and falls $1.50$ m, that is $3.00$ m at speed $c$. $\Delta t_{0}=\dfrac{2L}{c}=\dfrac{3.00}{3.00\times 10^{8}}=1.00\times 10^{-8}$ s, that is $10.0$ ns. It is the proper time because the two events that bound the tick, the departure of the flash and its return, happen AT THE SAME PLACE in this frame, on the bottom mirror. That is the only criterion, and it mentions neither speed nor preference: a single clock is enough to measure $\Delta t_{0}$, whereas two synchronized clocks are needed to measure $\Delta t$ on the platform.

Seen from the platform, during half a tick the flash travels the hypotenuse $\dfrac{c\,\Delta t}{2}$ while the car advances by $\dfrac{v\,\Delta t}{2}$; the vertical side is still $L$, because contraction acts only along the direction of motion. Pythagoras: $\left(\dfrac{c\Delta t}{2}\right)^{2}=L^{2}+\left(\dfrac{v\Delta t}{2}\right)^{2}$. Solving: $\dfrac{\Delta t^{2}}{4}(c^{2}-v^{2})=L^{2}$, so $\Delta t=\dfrac{2L}{\sqrt{c^{2}-v^{2}}}=\dfrac{2L}{c}\cdot\dfrac{1}{\sqrt{1-\beta^{2}}}$. The first factor is exactly $\Delta t_{0}$: we obtain $\Delta t=\gamma\,\Delta t_{0}$, and the Lorentz factor was not postulated, it FELL OUT of a right triangle. The only relativistic ingredient is having written $c$, and not $c$ plus something, for the speed of the flash seen from the platform.

$\gamma=\dfrac{1}{\sqrt{1-0.64}}=\dfrac{1}{0.6}=1.667$, so $\Delta t=1.667\times 10.0=16.7$ ns. Light path: $c\,\Delta t=3.00\times 10^{8}\times 1.667\times 10^{-8}=5.00$ m. Displacement of the car: $v\,\Delta t=0.800\times 5.00=4.00$ m. The full triangle therefore has sides $4.00$ m, $3.00$ m (the two times $L$) and $5.00$ m: it is the 3-4-5 triangle, which is no accident, it is what $\beta=0.8$ gives in this chapter. Remember it, since it instantly checks any calculation at $0.8c$.

If a mechanical watch and a light clock placed side by side in the car disagreed for an observer on the platform, then the traveller in the car could, by comparing them, detect their own uniform motion: that would violate the first postulate. Two clocks at rest relative to each other show the same time for everyone, so the slowing applies to TIME itself and not to a mechanism. This is why the light-clock argument, which looks like a contrived special case, actually proves the general case.

$\gamma=\dfrac{1}{\sqrt{1-0.360}}=\dfrac{1}{0.800}=1.250$. Length seen from Earth: $L=\dfrac{L_{0}}{\gamma}=\dfrac{120}{1.250}=96.0$ m. Width seen from Earth: $18.0$ m, UNCHANGED, because contraction acts only along the direction of motion. On board, the crew still measures $120$ m by $18.0$ m: in their own frame the ship is at rest, so they read the proper length, and nothing has changed for them. Neither party is wrong.

From Earth, the ship covers $6.00$ light-years at $0.600c$: $t=\dfrac{6.00}{0.600}=10.0$ years. Note that dividing a distance in light-years by a speed as a fraction of $c$ gives years directly: this unit system is what makes the chapter computable in your head.

For the crew, it is not the ship that moves, it is the star that approaches. The Earth-star distance is a proper length FOR EARTH, since both ends are at rest there; the crew sees it streaming past and measures it contracted: $L'=\dfrac{6.00}{1.250}=4.80$ light-years. Since the star approaches at $0.600c$, the trip lasts $t'=\dfrac{4.80}{0.600}=8.00$ years on board. Check by the other route: the onboard time is a proper time, the two events (departure, arrival) occurring at the same place in the ship, so $t=\gamma t'$, that is $t'=\dfrac{10.0}{1.250}=8.00$ years. The two methods agree, and this is the only check that counts: each frame uses ONLY ONE of the two effects. Counting contraction AND dilation in the same story divides the time by $\gamma$ twice, and it is the most frequent mistake of the chapter.

There is no force and no stress: the hull of the ship undergoes strictly nothing, and an onboard accelerometer measures no deformation. What produces the contraction is the measurement operation itself: measuring the length of a moving object means recording the position of its two ends AT THE SAME INSTANT. But “at the same instant” does not mean the same thing in the two frames, precisely because simultaneity is relative. The Earth observer marks the two ends simultaneously according to THEIR clocks; for the crew, these two readings were not made at the same time, the rear having been marked after the front, when the ship had already moved on. The whole contraction lies there, in a disagreement about simultaneity, never in a compression.

$\gamma=\dfrac{1}{\sqrt{1-0.9950^{2}}}=\dfrac{1}{\sqrt{0.009975}}=10.01$. In classical physics, the muon would travel in one mean lifetime $d_{0}=\beta c\,\tau_{0}=0.9950\times 3.00\times 10^{8}\times 2.20\times 10^{-6}=657$ m. The $15.0$ km then represent $\dfrac{15{,}000}{657}=22.8$ mean lifetimes, and the surviving fraction is $e^{-22.8}=1.2\times 10^{-10}$: out of ten billion muons created, a single one would arrive. The detector at sea level should be empty, and it is not: this is one of the oldest direct experimental proofs of relativity.

In the ground frame, the muon's clock is moving, so it runs slowly: its apparent mean lifetime is $\gamma\tau_{0}=10.01\times 2.20=22.0$ μs. The mean flight distance becomes $\beta c\gamma\tau_{0}=0.9950\times 3.00\times 10^{8}\times 2.20\times 10^{-5}=6.58\times 10^{3}$ m, that is $6.58$ km. The $15.0$ km are now only $\dfrac{15{,}000}{6{,}575}=2.28$ mean lifetimes, and the surviving fraction is $e^{-2.28}=0.102$, about 10%. One muon in ten reaches the ground, which is perfectly measurable.

In the muon's frame, its mean lifetime is $2.20$ μs and nothing else: it is ITS clock, and it ticks normally. What changes is the atmosphere, which rushes toward it at $0.9950c$ and whose thickness is contracted: $L'=\dfrac{15{,}000}{10.01}=1.50\times 10^{3}$ m. The proper time of the trip is $\dfrac{1{,}498}{0.9950\times 3.00\times 10^{8}}=5.02$ μs, that is $\dfrac{5.02}{2.20}=2.28$ mean lifetimes: the SAME number as in b), hence the same fraction of 10%. This is the high point of the exercise. The two observers tell incompatible stories, one speaks of a slowed clock, the other of a flattened atmosphere, and yet they agree on the only verifiable thing: the number of muons that trigger the detector.

Because the number of mean lifetimes elapsed is a PHYSICAL quantity, countable, not a convention: it is the number of times the muon's internal clock has gone around, and everyone must agree on it. Applying both effects at once divides it by $\gamma$ twice and gives $0.228$ mean lifetimes, that is 80% survivors, which would contradict the detector's count. The rule that prevents this fits on one line and holds for the whole chapter: choose ONE frame, stay in it, and use only one effect there. Time dilation is what the ground says about the muon's clock; length contraction is what the muon says about the ground's ruler. These are two descriptions of the same disagreement, not two disagreements that add up.

$\gamma=1.250$ at $0.600c$, so the platform measures $L=\dfrac{300}{1.250}=240$ m: this is indeed the dimension on the figure. Be careful not to swap them: $300$ m is the PROPER length, measured in the train where the train is at rest, and it is always the largest.

Seen from the platform, the two flashes leave the two ends at the same instant and both travel at $c$. While they travel, the passenger, seated in the middle, MOVES toward the front of the train: they go to meet the front flash and run away from the rear flash. They receive the front flash first, a fact on which everyone agrees, since it concerns a single event, the reception. Now in THEIR frame, they are at rest and at equal distance from both ends, so light takes the same time to reach them from both sides. If they receive the front flash first even though the trips lasted equally long, then the FRONT flash was emitted first. In the train, the two strikes are not simultaneous, and the front one comes first.

Let $S$ be the platform and $S'$ the train, which moves at $+v$. Both strikes occur at $t=0$, at positions $x_{\text{rear}}$ and $x_{\text{front}}=x_{\text{rear}}+240$. The transformation $t'=\gamma\left(t-\dfrac{vx}{c^{2}}\right)$ gives $\Delta t'=t'_{\text{front}}-t'_{\text{rear}}=-\gamma\dfrac{v\,\Delta x}{c^{2}}$, negative: the front is indeed struck first, which confirms b). Numerically: $|\Delta t'|=1.250\times\dfrac{0.600\times 3.00\times 10^{8}\times 240}{9.00\times 10^{16}}=6.00\times 10^{-7}$ s, that is $600$ ns. Remarkable form: since $\Delta x=\dfrac{L_{0}}{\gamma}$, the $\gamma$ cancels and $|\Delta t'|=\dfrac{v L_{0}}{c^{2}}$, where $L_{0}$ is the proper length of the train. The $\gamma$ disappears, which is the best check of the whole exercise: $\dfrac{0.600\times 3.00\times 10^{8}\times 300}{9.00\times 10^{16}}=6.00\times 10^{-7}$ s.

On the platform: $\Delta t=0$ and $\Delta x=240$ m, so $s^{2}=(c\Delta t)^{2}-(\Delta x)^{2}=0-240^{2}=-5.76\times 10^{4}$ m$^{2}$. In the train: $\Delta t'=6.00\times 10^{-7}$ s and $\Delta x'=300$ m, so $c\Delta t'=180$ m and $s^{2}=180^{2}-300^{2}=32{,}400-90{,}000=-5.76\times 10^{4}$ m$^{2}$. Identical: the interval is an INVARIANT, and it is the only reliable numerical check available once Lorentz transformations have been used. Its value is negative, the interval is spacelike: the two flashes are too far apart for light to have time to go from one to the other, so neither can be the cause of the other. Their order has no absolute meaning, and it can indeed reverse depending on the frame, without any effect ever preceding its cause. Causality is protected only for pairs of timelike events, those that light can connect, and for those the order is the same in all frames.

With $S'$ moving at $+v$ along $x$: $x'=\gamma(x-vt)$, $t'=\gamma\left(t-\dfrac{vx}{c^{2}}\right)$, $y'=y$, $z'=z$. The inverse transformations are obtained by swapping the roles of the two frames, which amounts to changing $v$ to $-v$ and exchanging primed and unprimed: $x=\gamma(x'+vt')$, $t=\gamma\left(t'+\dfrac{vx'}{c^{2}}\right)$. The principle of relativity demands it: since neither frame is privileged, the form of the laws must be the same in both directions, up to the sign of the velocity.

$u=\dfrac{u'+v}{1+\dfrac{u'v}{c^{2}}}=\dfrac{0.700+0.800}{1+0.560}=\dfrac{1.500}{1.560}=0.9615c$. The Galilean prediction would give $1.500c$, a faster-than-light speed, physically impossible. The denominator $1+\dfrac{u'v}{c^{2}}$ is the entire content of relativity here: it is practically 1 as long as one of the two motions is slow, which explains why Galilean law held for three centuries, and it brakes sharply as soon as both speeds are comparable to $c$. On the figure, the curve and the line visibly separate from $u'=0.2c$ on, and the curve approaches the asymptote $u=c$ without ever crossing it.

With $u'=c$: $u=\dfrac{c+0.800c}{1+\dfrac{0.800c\cdot c}{c^{2}}}=\dfrac{1.800c}{1.800}=c$. Light emitted by a source receding at $0.800c$ still arrives at $c$: this is exactly the second postulate, and the fact that the addition law respects it automatically is no coincidence, since it was built for that. Note that $u'=c$ gives $u=c$ whatever $v$ is: $c$ is a fixed point of the addition law.

Let us place ourselves in the first rocket. Earth recedes from it at $0.900c$, and the second rocket moves away from Earth at $0.900c$ in the same direction (seen from the first). $u=\dfrac{0.900+0.900}{1+0.810}=\dfrac{1.800}{1.810}=0.9945c$. The Galilean answer, $1.800c$, is wrong as a proper relative speed. Be careful, however, with the nuance that costs points: in Earth's frame, the gap between the two rockets does grow at the rate of $1.800c$ per second, and nothing forbids that, because this rate of separation is not the speed of any object. What is forbidden is for an object to exceed $c$ in a given frame, and neither rocket does.

Let $\beta_{1}=\dfrac{u'}{c}$ and $\beta_{2}=\dfrac{v}{c}$, both in $(-1,1)$, and assume them positive to fix ideas. Let us compute the gap to the barrier: $c-u=c-\dfrac{c(\beta_{1}+\beta_{2})}{1+\beta_{1}\beta_{2}}=c\cdot\dfrac{1+\beta_{1}\beta_{2}-\beta_{1}-\beta_{2}}{1+\beta_{1}\beta_{2}}=c\cdot\dfrac{(1-\beta_{1})(1-\beta_{2})}{1+\beta_{1}\beta_{2}}$. Factoring the numerator is the whole trick. Each factor $(1-\beta_{i})$ is strictly positive since $\beta_{i}<1$, and so is the denominator, hence $c-u>0$ strictly: the law can neither reach nor exceed $c$. The computation also shows how it approaches it: if both speeds are $0.99c$, the numerator contains $(0.01)^{2}$, and the gap to $c$ falls to $5\times 10^{-5}c$. The harder you push, the more the barrier resists.

An elementary charge accelerated through $2.00$ MV gains by definition $K=2.00$ MeV: that is the whole point of the electron-volt as a unit. From $K=(\gamma-1)mc^{2}$ we get $\gamma=1+\dfrac{K}{mc^{2}}=1+\dfrac{2.00}{0.511}=1+3.914=4.914$. Then $\beta=\sqrt{1-\dfrac{1}{\gamma^{2}}}=\sqrt{1-\dfrac{1}{24.15}}=\sqrt{0.9586}=0.9791$, that is $v=2.937\times 10^{8}$ m/s. Note the route followed: energy, then $\gamma$, then $\beta$. Going through the speed first is the classical reflex, and it is the surest way to get bogged down.

$\tfrac{1}{2}mv^{2}=K$ gives $\dfrac{v}{c}=\sqrt{\dfrac{2K}{mc^{2}}}=\sqrt{2\times 3.914}=\sqrt{7.828}=2.798$, that is $2.80c$. A speed almost three times that of light: the classical formula is not merely imprecise, it produces an absurdity. This is the physical reading of the dashed curve, which climbs straight up and crosses every value unimpeded, whereas the relativistic curve diverges at $\beta=1$: an INFINITE energy is needed to reach $c$, and that is the real reason nothing massive gets there.

Total energy: $E=\gamma mc^{2}=4.914\times 0.511=2.511$ MeV, which is indeed $K+mc^{2}=2.00+0.511$, an immediate check. By the energy-momentum relation: $(pc)^{2}=E^{2}-(mc^{2})^{2}=2.511^{2}-0.511^{2}=6.305-0.261=6.044$, so $pc=2.459$ MeV and $p=2.459$ MeV/c. Check by the definition: $pc=\gamma\beta mc^{2}=4.914\times 0.9791\times 0.511=2.459$ MeV. Both paths coincide. The unit MeV/c is puzzling at first, but it avoids dragging $10^{-22}$ kg·m/s around through a whole problem.

Proton: $\gamma=1+\dfrac{2.00}{938.3}=1.002132$, so $\beta=\sqrt{1-\dfrac{1}{1.002132^{2}}}=0.06519$, that is $1.956\times 10^{7}$ m/s. Classical formula: $\beta=\sqrt{\dfrac{2\times 2.00}{938.3}}=0.06529$. The discrepancy is $0.16$%, negligible in almost any assessment. The same energy of $2.00$ MeV therefore makes an electron ultra-relativistic and leaves a proton in the classical regime, because what matters is never the absolute energy but the RATIO $\dfrac{K}{mc^{2}}$: $3.91$ for the electron, $0.0021$ for the proton. Before any calculation, form this ratio: if it is very small compared with 1, classical mechanics is enough; if it is 1 or more, it is disqualified.

$E$ and $p$ both change when we change frames, but the combination $E^{2}-(pc)^{2}$ keeps the same value for everyone, namely $(mc^{2})^{2}$, which depends only on the particle. It is the exact counterpart of the invariance of the interval $(c\Delta t)^{2}-(\Delta x)^{2}$ for spacetime. In practice, in a collision problem, we write the conservation of $E$ and $p$ in the laboratory frame, where the data are given, then compute the invariant of the entire system; since it has the same value in the center-of-momentum frame, where the total momentum is zero, we obtain at once the energy available to create new particles, without ever transforming a single speed. This is the standard method for finding a reaction threshold, and it saves pages of calculation.

Wrong in its vocabulary and in its conclusion. The mass $m$ is a relativistic INVARIANT: it has the same value in all frames, it is a property of the particle just like its charge. The notion of “relativistic mass” $\gamma m$ has been abandoned by modern physics precisely because it produces this confusion. What increases without limit are the energy $E=\gamma mc^{2}$ and the momentum $p=\gamma mv$. Correct statement: we cannot reach $c$ because an infinite energy would be needed, $\gamma$ diverging as $\beta\to 1$, and not because an object would become heavy.

Wrong, and the mistake is a classic because such a delay does exist. But it has another name, the Doppler effect, and it is ALREADY CORRECTED in any serious measurement: clocks are compared by accounting for the travel time of the signals, or by using two clocks synchronized on the spot, as in the muon experiment where the detector is on the ground, with no signal to propagate. The dilation remains after this correction. Correct statement: time dilation is a property of the time intervals themselves, independent of how the information reaches us, and it is symmetric whereas the Doppler effect is not, since it changes sign depending on whether we approach or recede.

The first half is true, the second is false, and the second is what matters. As long as both twins remain inertial, each does see the other's clock ticking slowly, with no contradiction, because they do not use the same definition of simultaneity. The symmetry is broken at the TURNAROUND: the traveller changes inertial frame, undergoes an acceleration that their twin does not, and experiment proves it, they are the only one of the two pressed against their couch. Correct statement: the traveller does come back younger, by a factor $\gamma$ on the duration of the trip, and there is no paradox because the two situations are not interchangeable.

The statement confuses two different things. In Earth's frame, the distance between the two rockets does grow at the rate of $1.800c$ per unit of time, and that violates nothing: this rate of separation is not the speed of any object, no information travels along it, and nothing can be carried at that rate. What relativity forbids is for an OBJECT to exceed $c$ in a given frame, and each of the two rockets moves at $0.900c$ there. Correct statement: the speed of one rocket measured in the frame of the other, which is the only relative speed in the proper sense, is $\dfrac{1.800}{1.810}c=0.9945c$, well below $c$.

Wrong. No force acts, no stress appears in the ruler, and an observer travelling with it measures strictly no change, neither in length nor in interatomic distance. The contraction comes from the measurement operation: measuring the length of a moving object means recording its two ends at the same instant, and “at the same instant” does not mean the same thing in the two frames. Correct statement: length contraction is a consequence of the relativity of simultaneity, it is reciprocal, each observer seeing the other's ruler as shorter, which would be impossible for a material deformation.

$\beta=\dfrac{3.87\times 10^{3}}{3.00\times 10^{8}}=1.29\times 10^{-5}$. Typing $\gamma=(1-\beta^{2})^{-1/2}$ into a calculator, one reads exactly 1: the quantity sought is $10^{-10}$, far below the ten significant digits displayed. The series expansion is therefore not a convenience, it is the only way to get the result: $\gamma-1\approx\dfrac{\beta^{2}}{2}=\dfrac{(1.29\times 10^{-5})^{2}}{2}=8.32\times 10^{-11}$.

The satellite's clock is moving relative to the ground, so it runs slowly: it falls behind, at the relative rate $\dfrac{\beta^{2}}{2}$. Over $86{,}400$ s: $\Delta t=86{,}400\times 8.32\times 10^{-11}=7.19\times 10^{-6}$ s, that is a lag of $7.19$ μs per day. The sign is half the answer: special relativity always makes the moving clock LAG.

The two effects have opposite signs: gravitation makes the clock gain $45.9$ μs, speed makes it lag $7.19$ μs. Net offset: $45.9-7.19=+38.7$ μs per day, ahead. General relativity wins, by a factor of about six, because the satellite is very high, more than three Earth radii above the ground, while its orbital speed remains modest. A low-orbit satellite, faster and less high, would see the ratio reversed.

A timing error $\Delta t$ translates into a distance error $c\,\Delta t=3.00\times 10^{8}\times 38.7\times 10^{-6}=1.16\times 10^{4}$ m, that is about $11.6$ km after a single day. This is the figure that makes the argument unanswerable: without the two relativistic corrections, satellite navigation would drift by more than ten kilometres per day and would be unusable within a few hours. GPS is not a laboratory confirmation of relativity, it is a civilian infrastructure that would not work without it.

Since the satellite's clock will gain time once in orbit, it is made to tick a little too SLOWLY at launch, so that it ends up right once up there; correcting on the ground afterward would force every receiver to redo this calculation continuously, whereas a manufacturing offset is free and permanent. Check: the relative offset is $\dfrac{10.23-10.229\;999\;995\;4}{10.23}=\dfrac{4.6\times 10^{-9}}{10.23}=4.5\times 10^{-10}$. And the net offset per day is $\dfrac{38.7\times 10^{-6}}{86{,}400}=4.5\times 10^{-10}$: the two numbers coincide, the clock was detuned by exactly what relativity will make it gain.

$\gamma=\dfrac{E}{mc^{2}}=\dfrac{1{,}500}{139.6}=10.74$. Then $\beta=\sqrt{1-\dfrac{1}{\gamma^{2}}}=\sqrt{1-\dfrac{1}{115.5}}=\sqrt{0.99134}=0.99566$. The shortcut $\gamma=\dfrac{E}{mc^{2}}$ deserves to be remembered: whenever a TOTAL energy is given, $\gamma$ comes from a single division, without going through the kinetic energy.

Mean flight distance in the laboratory: $\ell=\beta c\,\gamma\tau_{0}=0.99566\times 3.00\times 10^{8}\times 10.74\times 26.0\times 10^{-9}=83.4$ m. Without relativity, one would write $\beta c\,\tau_{0}=7.77$ m. The ratio is exactly $\gamma=10.74$, which is the expected check: time dilation multiplies the range of the beam by the Lorentz factor, no more, no less. This is why a beam of unstable particles becomes usable once it is accelerated enough, and it is an accelerator design argument, not a theoretical curiosity.

$N/N_{0}=e^{-50.0/83.4}=e^{-0.599}=0.549$: about $54.9$% of the beam arrives at the target. Without dilation: $e^{-50.0/7.77}=e^{-6.44}=1.60\times 10^{-3}$, that is $0.16$%. The gain factor is $\dfrac{0.549}{0.00160}=343$: relativity gains a factor of three hundred in the available intensity. An experiment sized on the classical calculation would conclude that it needs three hundred times more beam and would be abandoned.

We look for $L$ such that $e^{-L/\ell}=0.0100$, that is $L=\ell\ln 100=83.4\times 4.605=384$ m. A line of nearly four hundred metres therefore remains usable with one percent of the beam, which matches the real orders of magnitude of secondary beam lines in the large laboratories.