Question bank: Theory of Relativity: Time Dilation

Explain what time dilation is and how it is related to Einstein's Special Theory of Relativity. Then, demonstrate how to calculate time dilation using the appropriate formula. Finally, present a numerical example where an object moves in relation to an observer at a speed close to the speed of light, showing how time dilation affects the time measured by the observer.

A scientist is conducting an experiment with extremely fast particles in his laboratory to study the effects of time dilation predicted by Einstein's special theory of relativity. He manages to accelerate a particle to a speed of 0.8c, where c is the speed of light. Considering that the time measured in the laboratory frame is 2 hours, answer: (a) What is the gamma factor (γ) of the moving particle relative to the laboratory frame? (b) What is the proper time, that is, the time measured in the frame of the moving particle? (c) What is the difference between the time measured in the laboratory frame and the proper time?

In a physics experiment, a group of students set out to empirically verify the time dilation predicted by Einstein's Special Theory of Relativity. They use an extremely precise clock capable of measuring time with nanosecond precision. They set up an experiment to measure the time it takes for a beam of light to travel a distance of 300 million meters at rest and aboard a hypothetical spaceship traveling at a constant speed of 0.8c, where c is the speed of light in a vacuum (c = 3 x 10^8 m/s). Considering the Lorentz factor (γ) as γ = 1/sqrt(1 - (v^2/c^2)), where v is the spaceship's velocity: (a) Calculate the time measured for the beam of light to travel the distance at rest and compare it with the time measured aboard the spaceship. (b) Explain how time dilation manifests in this experiment and the importance of this phenomenon for understanding the universe.

An astronaut travels in a spaceship at a constant speed of 0.8c, where c is the speed of light in a vacuum (c = 3.00 x 10^8 m/s), towards a space station. According to an observer at the station, the astronaut takes 60 Earth minutes to complete a specific task. Using the special theory of relativity and the Lorentz factor (γ = 1 / (1 - (v^2 / c^2))^0.5), calculate the dilated time that the astronaut observes to complete the task, considering that the space station is at rest relative to the Earth observer.

An astronaut travels in a spaceship at a constant speed of 0.8c, where c is the speed of light in a vacuum. He intends to conduct an experiment involving the emission of a light signal to a satellite in orbit. According to the special theory of relativity, the time for an observer at rest on Earth between the emission and reception of the light signal would be different from the time recorded by the astronaut in the spaceship. Considering the time dilation given by the Lorentz factor, where γ = 1 / sqrt(1 - (v^2 / c^2)), and that c = 3 x 10^8 m/s, calculate: 1) The γ factor for the astronaut's speed. 2) The time dilation, that is, the relationship between the time measured by the astronaut and the time that an observer on Earth would measure for the light signal to travel the round trip distance of 4.8 x 10^8 meters. Consider the speed of light as c = 3 x 10^8 m/s and respond in terms of c and the exact value of the γ factor, without rounding.