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Proper time and proper length, $\Delta t = \gamma\Delta\tau$ and $L = L_0/\gamma$, the muon experiment, the twin paradox, and satellite clocks.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to identify proper times and lengths, apply time dilation and length contraction, and check a result from both frames.
From the last lesson you have the Lorentz transformation and the factor $\gamma$. From the first lesson you know how to time events carefully. This lesson draws the two most famous consequences out of the transformation: moving clocks run slow and moving rods are short. Both are small at everyday speeds and dramatic near $c$.
| Term | What it means |
|---|---|
| Proper time | $\Delta\tau$, the time between two events read by a single clock present at both. |
| Time dilation | $\Delta t = \gamma\Delta\tau$: a moving clock's interval, measured in another frame, is longer. |
| Proper length | $L_0$, an object's length measured in its own rest frame. |
| Length contraction | $L = L_0/\gamma$: a moving object is shorter along its motion. |
| Light clock | A clock ticking as light bounces between two mirrors, used to derive dilation. |
| Twin paradox | A traveling twin ages less than a stay-at-home twin; the asymmetry is the turnaround. |
| Mean lifetime | $\tau$, the average time an unstable particle survives at rest. |
The proper time $\Delta\tau$ between two events is what a clock present at both of them reads. In any other frame, where that clock moves at speed $v$, the time between the events is longer:
$$\Delta t = \gamma\,\Delta\tau.$$
An object's proper length $L_0$ is measured in its rest frame. In a frame where it moves along its length, it is shorter:
$$L = \frac{L_0}{\gamma}.$$
Nothing is wrong with the clock or the rod. Time and space themselves are measured differently by observers in relative motion, and each observer sees the other's clocks slow and rods short. The two effects fit together: a muon's slow clock in the ground frame and the short atmosphere in the muon's frame predict the same survival.
Another way: picture
Picture a light clock: a pulse bouncing between two mirrors a fixed distance apart. At rest, the pulse goes straight up and down. Seen from a frame where the clock moves sideways, the pulse must travel a longer, slanted zigzag at the same speed $c$, so each tick takes longer. The ratio of the paths is exactly $\gamma$.
Another way: steps
A light clock has mirrors a distance $d$ apart; at rest it ticks every $\Delta\tau = 2d/c$. Seen moving at $v$, during one tick the clock moves $v\Delta t$ sideways, so the light travels two slanted legs, each $\sqrt{d^2 + (v\Delta t/2)^2}$. Setting $c\Delta t$ equal to that total and solving gives $\Delta t = \gamma\Delta\tau$.
The same result follows from the Lorentz transformation: two ticks of a clock at rest at $x' = 0$ in S' are separated by $\Delta t'$ there, and $t = \gamma(t' + vx'/c^2)$ gives $\Delta t = \gamma\Delta t'$ in S. Every kind of clock, atomic, biological, radioactive, must run slow by the same factor, or the first postulate would fail.
Measuring a moving rod's length means marking both ends at the same time in your frame. A rod at rest in S' from $x' = 0$ to $x' = L_0$ has its ends at $x' = \gamma(x - vt)$; marking both at the same $t$ gives $L_0 = \gamma L$, so $L = L_0/\gamma$.
Only lengths along the motion contract; widths across it do not. Contraction is reciprocal: a spaceship's crew sees the Earth's meter sticks short along their direction of motion. There is no contradiction, because the two frames disagree about which marks are simultaneous, and a length measurement depends on simultaneity.
Muons made $15$ km up by cosmic rays live $2.2$ μs on average at rest. Even at nearly $c$, that is only about $660$ m of travel, and a descent of $15$ km takes about $50$ μs, more than twenty lifetimes. Without relativity, only one muon in ten billion would reach the ground.
In 1940 Bruno Rossi and David Hall measured the muon flux on Mount Evans in Colorado and at lower altitude near Denver, and later experiments on Mount Washington in New Hampshire repeated it with better timing. The survival matched dilated lifetimes. In the muon's frame, its clock is normal, but the atmosphere rushing past is contracted by $\gamma$ to a kilometer or two. Both frames predict the same count.
One twin stays on Earth; the other flies to a star at $0.8c$ and returns. Earth's clocks record $10$ years for a trip to a star $4$ light-years away and back; the traveler's clock records $6$. It seems each should see the other's clock slow, so who is younger?
The situations are not symmetric. The traveler changes frames at the turnaround, accelerating, while the Earth twin stays in one inertial frame. The traveler is younger. Experiments with atomic clocks flown around the world, by Joseph Hafele and Richard Keating in 1971 on commercial airliners, confirmed the asymmetry at the level of nanoseconds.
A GPS satellite moves at about $3.9$ km/s, so its clock loses about $7$ μs per day by special relativity. It also sits higher in the Earth's gravity, where general relativity makes clocks run faster, gaining about $45$ μs per day. The net gain of about $38$ μs per day would put positions off by about $11$ km per day if ignored.
The satellite clocks are built to tick slightly slow on the ground, at $10.22999999543$ MHz instead of $10.23$ MHz, so that in orbit they keep time with ground clocks. Relativity is not an exotic correction here but a daily necessity.
Checking an answer. The proper time must be the shortest time between the events, and the proper length the longest length of the object. Distance divided by time in any frame must give a speed below $c$. And both frames must agree about anything that can be counted, such as how many muons reach the ground.
Time dilation uses a single moving clock and two clocks at rest in the other frame, synchronized by that frame's own procedure. The asymmetry, one clock versus two, is what makes the formula apply one way round. Swapping which frame has the single clock swaps which time is proper.
Length contraction depends on marking both ends at once in the measuring frame. Mark them at different times and the "length" includes the object's motion between the marks, which is not a length at all. Every apparent paradox in relativity, including the pole in the barn, dissolves once simultaneity is tracked carefully in each frame.
Time dilation is not an effect of signal delays. The dilated time is what remains after every light-travel correction has been made, using clocks at rest in the observer's frame at the places where the events happen. The muons really do survive longer; the flown atomic clocks really do read less.
What a camera would photograph is different again: light from different parts of a fast object left at different times, and a sphere moving at nearly $c$ would look rotated rather than squashed. That effect, worked out by James Terrell and Roger Penrose in 1959, is about images, not measurements, and does not undo contraction. Keeping the two apart, what is seen and what is measured with synchronized clocks, is the first step in every relativity problem, and the step most often skipped by beginners.
At $30$ m/s, $\gamma - 1$ is about $5 \times 10^{-15}$: a lifetime of highway driving adds up to a few nanoseconds. Even astronauts on the International Space Station, at $7.7$ km/s, age only about $0.01$ s less than people on the ground over a six-month stay.
The effects become obvious only where $\beta$ is large: in accelerators, in cosmic rays, and in the electrons of heavy atoms, whose inner shells move at a large fraction of $c$, and in the particle beams of hospital radiation machines. Relativistic effects on gold's electrons shift its color from silver toward yellow; without them, gold would look like silver. Mercury, next to gold in the periodic table, owes its low melting point partly to the same relativistic shrinking of its electron shells, which is why it is liquid at room temperature.
The GPS satellites, operated from Schriever Space Force Base near Colorado Springs, orbit at about $20{,}200$ km and $3.87$ km/s. Special relativity makes their clocks lose about $7.2$ μs per day; general relativity, because they sit higher in Earth's gravity, makes them gain about $45.7$ μs. The net is a gain of $38.5$ μs a day.
At $300$ m per microsecond, that would be an error growing by about $11$ km every day. The satellite clocks are therefore set to run slow before launch, and receivers apply a further small correction for each satellite's slightly elliptical orbit. Every phone that finds its location relies on both theories of relativity.
Muons are the most common cosmic-ray particles at sea level: about one passes through each square centimeter every minute. Physics teachers across the country measure them with scintillator paddles and desktop detectors, some built in programs run by Fermilab near Chicago, and compare counts at different altitudes.
Students who take a detector up a mountain, or on an airplane, find the count falls far more slowly with depth of atmosphere than the muon's $2.2$ μs lifetime would allow without time dilation. The measurement, first made in Colorado in the 1940s, remains one of the clearest demonstrations that moving clocks run slow.
It is tempting to explain a moving clock's slowness as light from it taking longer and longer to arrive. But time dilation is what remains after all such delays are corrected, using synchronized clocks placed where the events happen. Muons really live longer in flight, and clocks flown around the world really read less on landing.
A related error is to apply the formulas backward, dividing the proper time by $\gamma$. The proper time is always the shortest interval between two events, and the proper length the longest length of an object.
Pions live $26$ ns at rest. A beam moves at $0.95c$. Find $\gamma$.
$\gamma = \dfrac{1}{\sqrt{1 - 0.9025}} = 3.20$
Definition.
Find the lifetime in the lab.
$\Delta t = 3.20 \times 26 = 83\ \text{ns}$
Dilated.
Find how far they go in that time.
$d = 0.95 \times 0.300 \times 83 = 23.7\ \text{m}$
$c = 0.300$ m/ns.
Compare with no dilation.
$0.95 \times 0.300 \times 26 = 7.4\ \text{m}$
Three times shorter.
Find the $23.7$ m length in the pion's frame.
$L = \dfrac{23.7}{3.20} = 7.4\ \text{m}$
Contraction gives the same story.
A ship $100$ m long at rest passes Earth at $0.8c$. Find $\gamma$.
$\gamma = \dfrac{1}{0.6} = 1.667$
At $0.8c$.
Find its length in Earth's frame.
$L = \dfrac{100}{1.667} = 60\ \text{m}$
Contracted.
Find how long it takes to pass a point on Earth, in Earth's frame.
$t = \dfrac{60}{0.8 \times 3.00 \times 10^8} = 0.25\ \mu\text{s}$
Short ship, speed $0.8c$.
Find how long the point takes to pass the ship, in the ship's frame.
$\tau = \dfrac{100}{2.4 \times 10^8} = 0.417\ \mu\text{s}$
Full proper length.
Identify the proper time.
$\text{Earth's } 0.25\ \mu\text{s: one Earth clock sees both ends pass}$
The shortest time between the two events.
Check the ratio.
$\dfrac{0.417}{0.25} = 1.667 = \gamma$
Consistent.
A twin travels at $0.8c$ to a star $4.0$ ly away and straight back. Find Earth's elapsed time.
$t = \dfrac{8.0\ \text{ly}}{0.8c} = 10\ \text{years}$
Earth frame.
Find the traveler's elapsed time.
$\tau = \dfrac{10}{1.667} = 6.0\ \text{years}$
Proper time of the traveler.
Find the distance in the traveler's frame.
$L = 4.0 \times 0.6 = 2.4\ \text{ly each way}$
Contracted.
Check the traveler's time from that distance.
$\dfrac{4.8\ \text{ly}}{0.8c} = 6.0\ \text{years}$
Agrees.
Find the age difference on return.
$10 - 6 = 4\ \text{years}$
The traveler is younger.
Explain the asymmetry.
$\text{only the traveler changes frames}$
The turnaround breaks the symmetry.
Find the difference at $0.99c$.
$\gamma = 7.09: \ \tau = \dfrac{8.08}{7.09} = 1.14\ \text{years vs } 8.08$
Nearly seven years apart.
Find the contraction factor.
$\sqrt{1 - 0.36} = 0.8$
$1/\gamma$.
Multiply the proper length.
$L = 2.0 \times 0.8$
Contracted along the motion.
Evaluate the length.
A starship travels at $0.6c$ from Earth to a distant station. The ship's clock records $16$ years for the trip. How long does the trip take by clocks on Earth and the station?
Complete the worked solution: a spaceship of proper length $720$ m passes a marker buoy at $0.8c$, with $c = 300$ m/μs. Find its length measured from the buoy's frame in m, the time it takes to pass the buoy in that frame in μs, and the time the buoy takes to pass along the ship in the ship's frame in μs.
Contract the ship's length.
$L = L_0\sqrt{1 - \beta^2} = 720 \times 0.6 =$ l
It moves past the buoy.
Time the pass in the buoy's frame.
$t = \dfrac{L}{v} =$ t
Its short length at speed $v$.
Time the pass in the ship's frame.
$\tau = \dfrac{L_0}{v} =$ p
The buoy moves along the full proper length.
Check the ratio of the times.
$\dfrac{\tau}{t} = \gamma$
Each frame sees the other's clock run slow.
Match each term to its meaning.
| time read by one clock present at both events | Δt = γΔτ | length in the object's rest frame | L = L₀/γ | |
|---|---|---|---|---|
| proper time | ||||
| time dilation | ||||
| proper length | ||||
| length contraction |
A muon, whose lifetime at rest is $2.2$ μs, moves through a lab at $0.6c$, where $\gamma = 1.25$. With $c = 300$ m/μs, fill in its lifetime in the lab in μs, the distance it travels in that time in m, and the length of a $100$ m lab corridor measured in the muon's frame in m.
| value | |
|---|---|
| lifetime in the lab (μs) | |
| distance traveled (m) | |
| corridor length in muon frame (m) |
A starship travels at $0.8c$ to a planet $D$ light-years away (in Earth's frame), stays $5$ years, and does not return. Write the total time recorded by the ship's clock, in years, as a formula in $D$.
Answer:
Muons made $15$ km up descend straight down at $0.995c$, where $\gamma = 10.01$. Their lifetime at rest is $2.2$ μs. What fraction of them survive to reach the ground?
Answer: fraction surviving
An engineer at the GPS master control station in Colorado Springs checks a satellite clock moving at $7.66$ km/s. Counting only special relativity, how many microseconds per day does it fall behind a clock at rest on the ground?
Answer: μs lost per day
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A starship travels at $0.8c$ to a planet $D$ light-years away (in Earth's frame), stays $7$ years, and does not return. Write the total time recorded by the ship's clock, in years, as a formula in $D$.
Answer:
You can apply time dilation and length contraction. Explain to someone why cosmic-ray muons reach the ground.
20. Your turn: a rod $2.0$ m long at rest moves along its length at $0.6c$. How long is it in the lab?, step 3
$L = 1.6\ \text{m}$
Its width is unchanged.