A spaceship passes over an observation station on Earth....

A spaceship passes over an observation station on Earth. Just as the nose of the ship passes the station, a light in the nose of the ship flashes. As the tail of the ship passes the station, a light flashes in the ship's tail. According to an Earth observer, $50.0 \mathrm{~ns}$ elapses between the two events. In the astronaut's reference frame, the length of the ship is $12.0 \mathrm{~m}$. (a) How fast is the ship traveling according to an Earth observer? (b) What is the elapsed time between light flashes in the astronaut's frame of reference?

Key Concepts

Length Contraction

Length contraction is the phenomenon where the length of an object moving relative to an observer is measured to be shorter than its proper length, which is measured in the object's rest frame. This key concept is essential when comparing spatial measurements in different inertial frames.

Time Dilation

Time dilation is the effect in special relativity where a moving clock is observed to tick slower compared to a clock at rest in the observer's frame. This phenomenon is quantified by the Lorentz factor and is fundamental in determining how time intervals differ between frames.

Relativity of Simultaneity

Relativity of simultaneity describes the idea that events which appear simultaneous in one inertial frame may not be simultaneous in another frame moving relative to the first. This concept is crucial for understanding discrepancies between observers when events occur at different locations.

Inertial Reference Frames

In special relativity, inertial reference frames are perspectives in which observers move at constant velocity. Measurements of time intervals and lengths can differ between these frames, making it important to specify which frame is being considered when analyzing events.

Lorentz Transformation

The Lorentz transformation provides the mathematical framework used to relate space and time coordinates between two inertial frames in relative motion. It is critical for translating measurements like time intervals and distances, incorporating the effects of relative speed on these quantities.

Transcript

00:01 Okay, so for question a, when to determine the speed of the spaceship.

00:07 We know the speed can be equal to the length of a spaceship that was measured by the earth observer, divided by the time that was recorded by the earth observer.

00:19 And we know that l here, which is the lens that was measured by the people around the earth, is equal to l0 divided by gamma.

00:29 L0 is the length of the spaceship that was measured by an astronaut.

00:33 In a spaceship and gamma is the low -range factor, which is equal to 1 over square root 1 minus v square over c squared.

00:43 Okay, and then we can plug in this equation here, fact in here, okay? therefore, we have v is equal to l0, divo by gamma, divided by delta t, which we will have l0 over gamma times delta t, which is 1 over gamma times l0 over della t.

01:17 Therefore, we have v is equal to square root, 1 minus v square over c square, and then times l0 over della t.

01:32 Remember l0 was given from the question, which is 12 .0 meter.

01:40 Delta t was given as well, which is 15 .0 .t was given as well, which is, 50 .0 nanoseconds.

01:46 If we convert it to seconds, it's 5 times 10 to the power of negative 8 second, okay? because 1 nanoseconds is equal to 1 times 10 to the power of 99 seconds.

01:59 So therefore, if we plugging back to the equation here, we have v is equal to square rule 1 minus v square over c square.

02:10 And then times 12 .0 meter, over 5 times 10 to the power of negative 8 meter, i'm sorry, negative 8 seconds, okay? and this will give us fee is equal to square root 1 minus v square over c square and then times 2 .4 times 10 to the power of 8 meter per second.

02:52 And if we convert this part to c, we'll have f, is equal to square root 1 minus b square over c squared and then times 2 .4 times 10 to the power of 8 meter per second over 3 times 10 to power of 8 meter per second and then times c.

03:22 Okay? therefore we have b is equal to square root 1 minus v square square over c squared and then times 0 .8c.

03:38 And if we plug in back, let's turn back to the square root, we'll have v is equal to square root 1 minus v squared over c squared times 0 .8 c and then square.

04:00 And this will give us v is equal to square root 0 .64 c squared minus 0 .64 c square and n times b squared over c squared.

04:21 As you can tell, c squared and c squared can be canceled out...

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