Questions on time dilation and muon decay

Question 1

Define the proper time between two events.

Accepted Answer

The time between the events measured in the inertial frame in which both events occur at the same place, the time on a clock that rides along with them (1). Every other inertial observer measures a longer time (1).

Question 2

Define the proper length of an object.

Accepted Answer

The length measured in the object's own rest frame (1). An observer the object passes measures it shorter along the direction of motion, and unchanged across it (1).

Question 3

Using the idea of relativistic mass, explain why no object can be accelerated to the speed of light.

Accepted Answer

As v approaches c the mass m₀/√(1 − v²/c²) grows without limit (1), so each further gain in speed costs ever more energy: reaching c would require infinite energy (1).

Question 4

State what is meant by the rest energy of a particle.

Accepted Answer

The energy equivalent of the particle's rest mass, E₀ = m₀c², measured in the frame in which it is at rest (1).

Question 5

Calculate the rest energy of a proton. Mass of a proton = 1.67 × 10⁻²⁷ kg; c = 3.00 × 10⁸ m s⁻¹.

Accepted Answer

E = m₀c² (1) = 1.67 × 10⁻²⁷ × (3.00 × 10⁸)² = 1.50 × 10⁻¹⁰ J (1).

Question 6

A muon's lifetime in its rest frame is 2.2 μs. Calculate its lifetime as measured in a laboratory it passes at 0.95c.

Accepted Answer

t = t₀/√(1 − 0.95²) (1) = 2.2/0.3122 (1) t = 7.0 μs (1)

Question 7

A spacecraft of proper length 120 m passes Earth at 0.60c. Calculate its length as measured from Earth.

Accepted Answer

l = l₀√(1 − 0.60²) = 120 × 0.80 (1) l = 96 m (1) Contracted along the direction of motion only (1)

Question 8

Explain how the arrival of cosmic-ray muons at the Earth's surface provides evidence for time dilation.

Accepted Answer

Muons are created about 10 km up, moving near c, with a rest lifetime of 2.2 μs: one undilated lifetime covers only about 650 m, and the descent lasts around fifteen lifetimes, so almost none should survive (1). In fact a large fraction reach the ground (1). Their lifetime as measured from Earth is dilated roughly elevenfold (1), making the descent last barely one and a half lifetimes, and predicting the measured survival exactly (1).

Question 9

An electron (rest energy 8.2 × 10⁻¹⁴ J) travels at 0.98c. Calculate its total energy.

Accepted Answer

E = m₀c²/√(1 − 0.98²) (1) E = 5.0 × 8.2 × 10⁻¹⁴ (1) E = 4.1 × 10⁻¹³ J (1)

Question 10

A beam of pions has a rest-frame lifetime of 26 ns. In the laboratory their lifetime is measured as 52 ns. Determine the speed of the pions. c = 3.00 × 10⁸ m s⁻¹.

Accepted Answer

t = t₀/√(1 − v²/c²), so √(1 − v²/c²) = t₀/t = 26/52 = 0.50 (1). v²/c² = 1 − 0.25 = 0.75, so v = 0.87c (1) = 2.6 × 10⁸ m s⁻¹ (1).

Question 11

A muon created 10 km above the ground travels straight down at 0.996c. Calculate the thickness of the atmosphere below it as measured in the muon's rest frame, and state which frame measures the proper 10 km.

Accepted Answer

l = l₀√(1 − 0.996²) (1) = 10000 × 0.0894 = 894 m, about 890 m (1). The 10 km is the proper length, measured in the frame in which the atmosphere is at rest, the Earth frame; the muon measures the moving atmosphere contracted (1).

Question 12

An accelerator gives an electron a total energy of 4.1 × 10⁻¹³ J. The electron's rest energy is 8.2 × 10⁻¹⁴ J. Determine the electron's speed. c = 3.00 × 10⁸ m s⁻¹.

Accepted Answer

E/E₀ = 1/√(1 − v²/c²) = 4.1 × 10⁻¹³/(8.2 × 10⁻¹⁴) = 5.0 (1). So 1 − v²/c² = 1/25 (1), v = 0.98c (1) = 2.9 × 10⁸ m s⁻¹ (1).

Question 13

A spacecraft passes Earth at 0.80c. Its on-board clock records 60 s between two signals, and its proper length is 90 m. Calculate the time between the signals and the craft's length as measured from Earth.

Accepted Answer

The root is √(1 − 0.80²) = 0.60 (1) Time: 60 s is proper time, so Earth measures 60/0.60 (1) t = 100 s (1) Length: Earth measures 90 × 0.60 = 54 m (1)

Question 14

Describe Bertozzi's experiment and explain what it demonstrated directly.

Accepted Answer

Electrons were accelerated through known pds, fixing their kinetic energy from eV, checked by the heat delivered to the target (1), and their speed was measured directly by timing their flight over a measured distance (1). As the energy rose, the measured speed flattened out just below c instead of climbing without limit (1): direct evidence that kinetic energy varies with speed relativistically and that c is a limiting speed (1).

Question 15

Describe the shape of a graph of an object's mass against its speed, from rest towards the speed of light.

Accepted Answer

The curve starts at the rest mass and stays almost flat to a few tenths of c (1), then rises ever more steeply, approaching a vertical asymptote at v = c that it never reaches (1).

Question 16

The crew of a spacecraft travelling at 0.70c set off for a star 4.2 light years away, as measured from Earth. The mission plan requires the crew to age less than 4.5 years during the journey. Deduce whether the plan is met. (A light year is the distance light travels in one year; its value in metres is not needed.)

Accepted Answer

In the Earth frame the journey takes 4.2/0.70 = 6.0 years (1). The crew's ageing is the proper time: departure and arrival both happen at the spacecraft (1). √(1 − 0.70²) = 0.71 (1), so the crew age 6.0 × 0.71 = 4.3 years (1). 4.3 < 4.5, so the plan is met (1).

Question 17

Show that classical physics predicts an impossible speed for electrons accelerated from rest through 4.5 MV, and state what Bertozzi's measurements showed such electrons actually do. e/m = 1.76 × 10¹¹ C kg⁻¹; c = 3.00 × 10⁸ m s⁻¹.

Accepted Answer

Classically v = √(2(e/m)V) = √(2 × 1.76 × 10¹¹ × 4.5 × 10⁶) = 1.3 × 10⁹ m s⁻¹ (1). This is 4.2 times the speed of light, which nothing can reach (1). Bertozzi's time-of-flight measurements showed the speed instead levelling off just below c while the kinetic energy kept rising as supplied (1).

Question 18

A student sketches the kinetic energy of an electron against its speed as the classical curve E_k = ½m₀v². Describe how the correct relativistic curve differs.

Accepted Answer

At low speeds the two curves coincide: the classical formula is a good approximation (1). From a few tenths of c the relativistic curve rises above the classical one, the kinetic energy growing without limit (1). It has a vertical asymptote at v = c that it approaches but never reaches, whereas the classical curve passes through c with only finite energy (1).