Special Relativity: Time Dilation Flashcards
(9 cards)
What is time dilation in the context of special relativity?
Time dilation is a consequence of special relativity in inertial frames. It means that time passes at different rates for observers depending on their relative motion—moving observers measure time as passing more slowly.
What is the difference between proper time (t0) and time (t) in time dilation?
Proper time t0 is the time measured by an observer stationary relative to the event (e.g. someone next to the clock). t is the time measured by an external observer who sees the clock in motion.
What is the equation used to calculate time dilation?
t = t0 / √(1 - v²/c²)
Where: t0 = proper time, v = relative velocity of the moving frame, c = speed of light.
In the time dilation equation, which time is always shorter: proper time or external time?
Proper time t0 is always shorter than t, since the denominator √(1 - v²/c²) is always less than 1 for v > 0.
Lucy is in a spaceship travelling at 0.9c and measures 1 hour on her clock. How much time passes on Earth?
Lucy measures the proper time t0 = 1 hr. t = 1 / √(1 - (0.9)²) ≈ 2.3 hours.
So, 2.3 hours pass on Earth.
How does muon decay provide evidence for time dilation?
Muons travel at very high speeds in the atmosphere. If special relativity weren’t true, they would decay too quickly to reach Earth’s surface. However, because of time dilation, they decay more slowly in the lab frame and can be detected at lower altitudes.
How is the muon decay experiment set up to test time dilation?
Two muon detectors are placed at different altitudes. The count rate is recorded at both detectors, and the muons’ velocity and the distance between detectors are measured to calculate expected decay times.
What is the expected muon count at the lower detector without accounting for special relativity?
Initial count rate: 100 s⁻¹. Time to travel 2 km at 0.996c: t = 2000 / (0.996 × 3 × 10⁸) ≈ 6.69 × 10⁻⁶ s. Number of half-lives: 6.69 × 10⁻⁶ / 1.5 × 10⁻⁶ ≈ 4.46. Final count rate: 100 × (1/2)⁴.46 ≈ 4.5 s⁻¹.
This is far lower than the actual observed count of 80 s⁻¹, suggesting the calculation is incorrect.
What is the expected muon count when special relativity is taken into account?
Same t = 6.69 × 10⁻⁶ s. Proper time: t0 = t / √(1 - (0.996c/c)²) = 6.69 × 10⁻⁶ × √(1 - 0.992) ≈ 6.0 × 10⁻⁷ s. Number of half-lives: 6.0 × 10⁻⁷ / 1.5 × 10⁻⁶ = 0.4. Final count rate: 100 × (1/2)⁰.4 ≈ 76 s⁻¹.
This value is much closer to the observed 80 s⁻¹, supporting the theory of time dilation.
