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Topic 3 of 10

Time Dilation

A clock moving relative to an observer runs slow by a factor of $\gamma$ — this is time dilation. It is not an illusion or measurement artifact; it is a real physical effect confirmed by muon lifetimes, atomic clocks on airplanes, and GPS satellites that must correct for it daily.

Key Concepts

Proper Time

\tau

is the time elapsed on a clock that travels with the object — measured in the object's rest frame. All observers agree on

\tau

; it is Lorentz invariant.

d\tau^2=dt^2-dx^2/c^2

Time Dilation Formula

If a clock is at rest in frame

S'

(which moves at

v

relative to

S

), then a time interval

\Delta\tau

(proper time) corresponds to

\Delta t=\gamma\Delta\tau

in frame

S

. Moving clocks run slow.

Twin Paradox

Twin A stays on Earth; twin B travels at

0.80c

for 5 yr and returns. A ages

5/0.80\times2=12.5

yr total; B ages

12.5/\gamma=12.5/1.667=7.5

yr. The asymmetry is resolved by B's acceleration (frame change) at turnaround.

Muon Lifetime

Muons produced by cosmic rays at altitude ~15 km reach Earth's surface despite proper lifetime

\tau_0=2.2

μs (

c\tau_0\approx660

m). In the lab frame their lifetime is

\gamma\tau_0\gg660

m — direct experimental confirmation of time dilation.

Key Equations

Time dilation

\Delta t = \gamma\,\Delta\tau

Lab frame time Δt > proper time Δτ; moving clocks run slow.

Proper time (infinitesimal)

d\tau = \frac{dt}{\gamma} = dt\sqrt{1-\beta^2}

Proper time always less than coordinate time; invariant interval.

Twin paradox result

\Delta\tau_{\rm traveler} = \frac{\Delta t_{\rm Earth}}{\gamma}

The traveling twin ages less. The effect is real and asymmetric.

Worked Example

Muon Survival to Earth Surface

Problem

A muon is produced at $h=15$ km altitude with $v=0.998c$ ( $\gamma\approx15.8$ ). Proper lifetime: $\tau_0=2.20$ μs. Does it reach Earth before decaying?

Solution

Without time dilation: maximum range $d_0=c\tau_0=3\times10^8\times2.2\times10^{-6}=660$ m $\ll 15$ km.

With time dilation: lab lifetime $\tau=\gamma\tau_0=15.8\times2.20=34.8$ μs.

d = v\tau = 0.998c\times34.8\,\mu\text{s} = 0.998\times3\times10^8\times3.48\times10^{-5} = 10{,}427\text{ m} \approx 10.4\text{ km}

More precisely, the mean range is ~10.4 km — a significant fraction reaches the surface. Observed flux confirms this.

Answer Yes — time dilation extends the lab lifetime to ~34.8 μs, giving mean range ~10.4 km.

Practice

Exercises

7 problems

1 of 7

A spaceship moves at $v=0.60c$ ( $\gamma=1.25$ ). The ship clock measures $\Delta\tau=10.0$ s. Find $\Delta t$ in the lab frame (in s).

Your answer

2 of 7

A muon has proper lifetime $\tau_0=2.20$ μs and moves at $\beta=0.90$ ( $\gamma=2.294$ ). Find its lab-frame lifetime (in μs).

Your answer

μs

3 of 7

Same muon ( $v=0.90c$ , $\tau_{\rm lab}=5.05$ μs). How far (in m) does it travel in the lab frame?

Unlock Exercise 3