Time Dilation and Length Contraction
For constant relative velocity, a moving clock records less elapsed time than the measuring frame's coordinate clocks. Moving lengths measured simultaneously are shorter along the motion. Atmospheric muon survival supports these predictions.
What a learner can do afterwards
- Derives the time dilation factor from a light clock
- Applies dilation and contraction consistently to the same situation from both frames
- Explains the survival of atmospheric muons from either frame and gets the same answer
1 · Read
A ship moves at constant speed v past the ground. Its light-clock mirrors are separated perpendicular to v. One tick is emission, reflection and return to the starting mirror. The ship records proper time Δτ between emission and return at one place. Ground time Δt refers to the same events after correcting signal delays. Light in a vacuum always travels at the same speed c in every inertial frame, whatever the motion of its source or the observer. At that speed c, light takes longer to travel the longer diagonal ground path.
Let D be the mirror gap, which is transverse and unchanged between these frames. In a ground half-tick, light travels cΔt/2 while the ship moves vΔt/2. These form a right triangle with legs D and vΔt/2. Pythagoras gives (cΔt/2)^2=D^2+(vΔt/2)^2. The ship has 2D=cΔτ, so D=cΔτ/2. Substitute and collect terms: (1−v²/c²)Δt²=Δτ². Take the positive square root to get Δt=Δτ/√(1−v²/c²).
The Lorentz factor is γ=1/√(1−v²/c²), for 0≤v<c. At v=0, γ=1; it grows with speed. For one clock's events, Δt=γΔτ. With γ=3, a 4 s clock interval gives 12 s in the other frame. Each frame measures the other's clocks as slow after correcting signal delays. The reciprocal comparisons use different simultaneous-event choices.
Proper length L0 is measured with the object at rest. For a moving rod, locate both endpoints at the same time in the measuring frame using its synchronized clocks. Along relative motion, L=L0/γ. With γ=3, a 9 m rest length is measured as 3 m. Perpendicular lengths are unchanged. This is a comparison between frames, not physical crushing.
Along motion, the same rod measures 9 m at rest and 3 m in a frame where γ=3.
Atmospheric muons form in cosmic-ray interactions and decay unpredictably. Survival describes a fraction of a group. The ground measures their moving decay clock's half-life as γ times its rest half-life. Fewer half-lives pass during the same ground travel time, so more survive than using the rest half-life as ground time would predict. No muon needs to exceed c; not every muon survives.
Let H be the vertical gap between two levels fixed in the ground frame, so H is their proper separation. Muons travel straight down along H at constant speed v. Ground flight time is H/v; the muon frame has gap H/γ and flight time H/(γv). For rest half-life h, ground half-life is γh; muon half-life is h. Divide each frame's flight time by its half-life. Equal counts give equal survival fractions: one half-life leaves 1/2 on average, two leave 1/4.
The ground approaches the stationary muon at speed v. Contract the gap first, then divide by v to find muon-frame flight time.
A half-life is the time for half a large group to decay on average. Two half-lives leave one quarter. Use a rounded muon rest half-life of 1.5 μs, γ=3 and ground flight time 4.5 μs. The ground half-life is 3×1.5=4.5 μs. The muon-frame flight time is 4.5÷3=1.5 μs. Each frame counts one half-life and predicts one half survives. This model ignores energy loss and speed changes.
For one clock's events, Δt=γΔτ. A moving length measured simultaneously is L0/γ along motion. Both frames predict the same muon survival fraction.
Lesson, questions and study cards written for this learning dataset. OpenStax relativity chapters were consulted for facts.
Readings and study cards written for this learning dataset. OpenStax chapters were consulted for facts. The linked video is a Khan Academy lesson.
More to read
Keeping the two tick times separate
Define one light-clock tick by emission from a mirror and return to that mirror after reflection. The mirror gap D is perpendicular to the ship's constant velocity v. The ship records proper time Δτ, so 2D=cΔτ.
Let Δt be ground coordinate time between those same events. During a ground half-tick, the ship moves vΔt/2 while light travels cΔt/2. The transverse gap D is unchanged. These distances form a right triangle with D and vΔt/2 as legs.
Pythagoras gives (cΔt/2)^2=D^2+(vΔt/2)^2. Substituting D=cΔτ/2 and collecting terms gives (1−v²/c²)Δt²=Δτ². Positive elapsed times give Δt=γΔτ, where γ=1/√(1−v²/c²).
This comparison uses corrected coordinate times. It is not the uncorrected rate at which images of the clock arrive at an observer.
A moving length needs one measuring-frame time
Proper length L0 is measured in the object's rest frame. To measure a moving rod in ground coordinates, record both endpoints at one ground time and subtract their positions.
If one endpoint is recorded later, the rod has moved during the delay. That procedure mixes length with motion. Endpoints recorded simultaneously in the rod frame are also a different event pair from endpoints recorded simultaneously in the ground frame.
Along relative motion, the measured length is L=L0/γ. Perpendicular components remain unchanged. This is a relation between coordinate measurements in inertial frames. It does not describe compression by a force, and a photograph does not directly supply simultaneous endpoint coordinates.
Muon survival is a group prediction
Cosmic-ray interactions in the atmosphere produce muons. Individual decay times vary. A half-life describes group behavior: after one half-life, half remain on average; after two, one quarter remain.
For constant-speed muons, the ground-frame half-life is γ times the rest half-life. During a fixed ground flight time, fewer half-lives pass than a calculation using the rest half-life as ground time would suggest.
Consider muons travelling straight down between a production level and the ground. Their motion is parallel to the vertical separation. In a muon's rest frame, the muon stays at one position while the ground approaches. The simultaneous vertical gap is contracted. The resulting flight time is shorter; the rest half-life stays unchanged.
Both frames predict the same surviving fraction for the same setup. Neither predicts that every muon survives. Detailed atmospheric models also need the distribution of speeds and energy losses.
Checking the two frame calculations
Use a model in which muons travel straight down at constant speed v from a production level to the ground. Both levels are fixed in ground coordinates, with vertical separation H parallel to the motion. H is their proper separation. If the muon rest half-life is h, its ground-frame half-life is γh. The ground flight lasts H/v, so the number of half-lives is H/(vγh).
In the muon frame, the production level and ground have simultaneous vertical separation H/γ. The ground approaches the stationary muon at speed v, reaching it after H/(γv). Dividing by the unchanged rest half-life h again gives H/(vγh).
For example, use rounded rest half-life 1.5 μs, γ=3 and ground flight time 4.5 μs. The ground half-life is 4.5 μs. The muon-frame flight lasts 1.5 μs. Both count one half-life and predict that half the group survives on average.
These predictions assume straight-down motion at constant speed, negligible energy loss and a large group. Agreement concerns the surviving fraction, not identical distances or times.
Explore more
2 · Watch
En anglaisTake it off screen
Where it sits
This opens up
Nothing builds on it yet.
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.