Lorentz Transformation, Time Dilation, Length Contraction

[Kunhee]

Homework Statement

[/B]
A spaceship is approaching Earth from the far side of the sun. The Earth and sun are 8 light minutes apart and the ship is traveling at .8c. Two events are indisputable. 1) the ship is at the sun 2) the ship is at the earth. Assume that the Earth and sun are at rest according to each other. The ship is detectable only after it passes the sun.

a) In the Earth's frame, how much time separates events 1 and 2?
b) In the Earth's frame, how much distance separates events 1 and 2?
c) How much time after first detecting the ship do humans on Earth have to prepare for its arrival?
d) In the ship's frame, how much time separates events 1 and 2?
e) In the ship's frame, how much distance separates events 1 and 2?
f) In the ship's frame, how far apart are Earth and sun?

The Attempt at a Solution

The Lorentz factor is 5/3 when we plug in .8c. I am having trouble setting up the problem.
What is the difference between a and c? Or e and f?

a) the time that separates the two events is simply t = d/v = 8c/.8c = 10 minutes
b) the distance that separates the two events is simply 8c = 8 light minutes.
c) Earth has 10 minutes to prepare for the ship's arrival.

d) t' = (5/3)(10) = 16.67 minutes
e) x' = (3/5)(8c) = 4.8 light minutes
f) Earth and Sun are 4.8 light minutes apart

 

[Chestermiller]

I have trouble understanding your answers. Can you give the numerical answers in minutes and in light-minutes for parts a and b?

 

[Kunhee]

For a, I did time dilation T' = (5/3)(To) where To is 8c/.8c just from t = d/v
T' = (5/3)(To)
T' = (5/3)(8c/.8c)
T' = (5/3)(10) = 16.67 minutes

For b, I did length contraction L' = (3/5)(Lo) where Lo is 8c
L' = (3/5)(8c)
L' = 4.8 light minutes

 

[Kunhee]

The Earth is at rest and the two events are Earth x1, = 0 and Sun, x2 = 8c. So when the question asks for time that separates these events in Earth's frame, it must mean how much dilated time it takes for the ship to get from x2 to x1. And when the question asks for the distance, it is asking for the contracted length?

Or because the Earth is at rest, none of this applies...?
I am a bit confused.

 

[Chestermiller]

The Earth is at rest and the two events are Earth x1, = 0 and Sun, x2 = 8c. So when the question asks for time that separates these events in Earth's frame, it must mean how much dilated time it takes for the ship to get from x2 to x1. And when the question asks for the distance, it is asking for the contracted length?

Or because the Earth is at rest, none of this applies...?
I am a bit confused.

Since the Earth and sun are at rest, and since the distance 8 light-minutes and the velocity 0.8c applies to this frame, you are correct in concluding that none of this Lorentz stuff applies.

 

[Kunhee]

I see. Thank you for the clarification.

a) the time that separates the two events is simply t = d/v = 8c/.8c = 10 minutes
b) the distance that separates the two events is simply 8c = 8 light minutes.
c) Earth has 10 minutes to prepare for the ship's arrival.

d) t' = (5/3)(10) = 16.67 minutes
e) x' = (3/5)(8c) = 4.8 light minutes
f) Earth and Sun are 4.8 light minutes apart

So what is the difference between questions a and c? Or e and f?

 

[Chestermiller]

I see. Thank you for the clarification.

a) the time that separates the two events is simply t = d/v = 8c/.8c = 10 minutes
b) the distance that separates the two events is simply 8c = 8 light minutes.
c) Earth has 10 minutes to prepare for the ship's arrival.

d) t' = (5/3)(10) = 16.67 minutes
e) x' = (3/5)(8c) = 4.8 light minutes
f) Earth and Sun are 4.8 light minutes apart

So what is the difference between questions a and c? Or e and f?

The answer to c should be 2 minutes.
The answer to e is not correct. Event 1 occurs at the same coordinate in the rocket frame (the nose of the ship) as Event 2 (the nose of the ship). So, in the rocket frame, what is the distance between the two events? Answer f is correct. Now, if, in the ship frame, the distance between the Earth and the sun is 4.8 light-minutes, and the Earth is traveling toward the ship at 0.8c, what should the answer to d be?

 

[Kunhee]

c) is just 10 minutes - 8 light minutes = 2 minutes?
This part is a bit confusing to me how it translates to how much time people have
after seeing event 2 until event 1 occurs.

e) the distance between the two events in the rocket frame is just 8 light minutes.
f) is therefore using 8c for length contraction so x' = (3/5)(8c) = 4.8 light minutes
d) is then t = d/v = 4.8c / .8c = 6 minutes

 

[Chestermiller]

c) is just 10 minutes - 8 light minutes = 2 minutes?
This part is a bit confusing to me how it translates to how much time people have
after seeing event 2 until event 1 occurs.

e) the distance between the two events in the rocket frame is just 8 light minutes.
f) is therefore using 8c for length contraction so x' = (3/5)(8c) = 4.8 light minutes
d) is then t = d/v = 4.8c / .8c = 6 minutes

Events 1 and 2 occur at the same location in the ship frame of reference. So the answer to.e is zero distance.

 

[Kunhee]

Could you explain why c is 2 minutes?

 

[Chestermiller]

Could you explain why c is 2 minutes?

Once the ship emerges from behind the sun, it takes 8 minutes for the light to reach the earth. It takes 10 minutes for the ship to reach the earth. So the people first see the ship two minutes before it arrives.

 

[Kunhee]

Ah I see. Thank you for the help!