Can somebody walk me through this Special Relativity clock sync proble

[PsychonautQQ]

Homework Statement

A rocket that has a proper length of 700m is moving to the right at a speed of .9c. It has two clocks, one in the nose and one in the tail, that have been sync'd in the frame of the rocket. A clock on the ground and the clock in the nose of the rocket both read zero as they pass by each other.

a) At the instant the clock on the ground reads zero, what does the clock in the tail of the rocket read according to observers on the ground?

When the clock in the tail of the rocket passes the clock on the ground
b) what does the clock in the tail read according to observers on the ground
and
c) what does the clock in the nose read according to observers on the ground
and
d) what does the clock in the nose read according to observers in the rocket?

e) At the instant the clock in the nose of the rocket reads 1.h, a light signal is sent from the nose of the rocket to an observer standing by the clock on the ground. What does the clock on the ground read when the observer on the ground receives the signal?
f) when the observer on the ground receives the signal, he immediately sends a return signal to the nose of the rocket. What is the reading of the clock in the nose of the rocket when that signal is received at the nose of the rocket?

Homework Equations

ts = Lp v/c^2
Lp/γ=L
tpγ=t

The Attempt at a Solution

So far part A i ended up getting the correct solution by plugging the known variables into equation
ts = lp v/c^2.

Attempting part b of trying to figure out what the clock in the tail of the ship reads when it passes the clock on the ground according to the people on the ground, and I first found the length of the ship in their frame, then measured the time that it takes for the ship to travel it's contracted length. This gave me the wrong answer and I'm not sure why.

I feel like there's an error in my thinking about these time/event sync problems but I'm not sure where. Any feedback is appreciated :)

 

[Chestermiller]

This exercise seems to be designed to give you some experience in applying the Lorentz Transformation and the inverse Lorentz Transformation. Let x' = 0 represent the coordinate of the nose of the rocket within the rocket frame of reference, and let x' = -700m represent the coordinate of the tail of the rocket within the rocket frame of reference. Let x = 0 represent the coordinate of the ground clock that the nose of the rocket passes at ground time t = 0 (and rocket time t' = 0). What are the equations for the Lorentz Transformation in terms of x, x', t, and t' which satisfy the condition that "the ground clock and the clock in the nose of the rocket both read zero as they pass by each other?" What are the equations for the inverse Lorentz Transformation for this same system? Now show us how you apply these equations to obtain the solutions to each of the questions (a) - (f).

 

[PsychonautQQ]

So the in the rocket frame of reference, x'=0 for the nose and x'=-700 for the tail stays like that for any measurement of time?
also, does the x=0 become negative as time passes? else how will x ever equal x' for the tail of the rocket.

thanks by the way :D what you typed above really helped me out

edit: blah of course x=0 stays x=0 for all t.. x' for the tail of the rocket doesn't need to equal x ever it's all taken care of in the equations? i think I'm getting closer to a better conceptual understanding... is everyone on this website but me a genius?

edit2: I tried to get the answer in part B by taking the answer from part A and adding it to the time it would take a contracted rocket's tail to get to the ground clock.. This got me the wrong answer even though it makes great sense in my head.

I then got the CORRECT answer for part B by simply dividing the rockets length in it's own frame divided by the speed it was traveling... why would that possibly give the answer I was looking for in part b of 2.59 micro seconds?

 

[Chestermiller]

So the in the rocket frame of reference, x'=0 for the nose and x'=-700 for the tail stays like that for any measurement of time?
also, does the x=0 become negative as time passes? else how will x ever equal x' for the tail of the rocket.

thanks by the way :D what you typed above really helped me out

edit: blah of course x=0 stays x=0 for all t.. x' for the tail of the rocket doesn't need to equal x ever it's all taken care of in the equations? i think I'm getting closer to a better conceptual understanding... is everyone on this website but me a genius?

Don't be so hard on yourself. We all struggled with this stuff when we began studying SR. It definitely takes getting use to.

edit2: I tried to get the answer in part B by taking the answer from part A and adding it to the time it would take a contracted rocket's tail to get to the ground clock.. This got me the wrong answer even though it makes great sense in my head.

I then got the CORRECT answer for part B by simply dividing the rockets length in it's own frame divided by the speed it was traveling... why would that possibly give the answer I was looking for in part b of 2.59 micro seconds?

As reckoned from the rocket frame of reference, the clock on the ground is traveling backwards at 0.9c. The ground clock is at the nose of the rocket at t' = 0. So it must reach the tail of the rocket at t' = 700/0.9c.
What went wrong in your calculation? In A, you considered the event where the rocket clock at x' = 0 was passing the ground clock at x = 0, and where the time on the rocket clock at x' = 0 read t' = 0, and the time on the ground clock at x = 0 read t = 0. Suppose there were a second ground observer with a clock situated opposite the location of the tail of the rocket just as the nose of the rocket x' = 0 swept by location x = 0. His clock would also show t = 0. But what value of t' would he observe on the clock at the tail of the rocket x' = -700? Now add your answer to this value, and see what you get.

The easiest (and most bulletproof) way to do part b is to use the Lorentz Transformation equations directly. There are 4 parameters in the Lorentz Transformation equations x, x', t, and t', and 2 equations. But, for the event in part b, two of the parameters x and x' are specified: x=0 and x'= -700. This gives you enough information to determine both t and t'. Solve for them, and see what you get.

Chet

 

[PsychonautQQ]

The easiest (and most bulletproof) way to do part b is to use the Lorentz Transformation equations directly. There are 4 parameters in the Lorentz Transformation equations x, x', t, and t', and 2 equations. But, for the event in part b, two of the parameters x and x' are specified: x=0 and x'= -700. This gives you enough information to determine both t and t'. Solve for them, and see what you get.
Chet

okay so i set the equation up
x = γ(x'+vt')
x = 0
x' = -700
so i had to find where x' + vt' = 0 and solving for this i got 2.59 micro seconds which is the correct answer, but i solved for t' and i thought that was the time according to the spaceship when part B is asking for what people on the ground see? I think I got the correct answer through the wrong method? I was calculating for another value which was the same?

 

[PsychonautQQ]

As reckoned from the rocket frame of reference, the clock on the ground is traveling backwards at 0.9c. The ground clock is at the nose of the rocket at t' = 0. So it must reach the tail of the rocket at t' = 700/0.9c.
Chet

t' = 700/.9c is what the tail rocket clock will look like according to people in the rocket right? yet it still gives the answer for part B which is asking from the ground perspective

 

[TSny]

Something to keep in mind:

Suppose there is a clock C in some reference frame that happens to be located at some event. Then all observers in all reference frames will agree on the reading of clock C at the event. The reading of clock C at the event is a physical fact that all observers must agree with.

When the clock at the rear of the ship passes the clock on the ground at x = 0, both clocks are located at that event. So, all observers will agree on what the clock in the rear of the ship reads at that event and all observers will agree on what the clock on the ground reads at that event.

 

[PsychonautQQ]

Something to keep in mind:

Suppose there is a clock C in some reference frame that happens to be located at some event. Then all observers in all reference frames will agree on the reading of clock C at the event. The reading of clock C at the event is a physical fact that all observers must agree with.

When the clock at the rear of the ship passes the clock on the ground at x = 0, both clocks are located at that event. So, all observers will agree on what the clock in the rear of the ship reads at that event and all observers will agree on what the clock on the ground reads at that event.

Okay cool this helped,

So this thinking of mine is wrong:
When the tail of the ship passes the ground clock, the people on the ground will see a different time in the tail clock of the rocket than the people in the rocket because the rocket is traveling at relativistic speeds.

Correct way of thinking:
The clock in the rocket and the clock on the ground will be at the same location at the time of the event, so observers on the ground and observers in the rocket will agree with each other when comparing the times on each clock.

 

[TSny]

Okay cool this helped,

Correct way of thinking:
The clock in the rocket and the clock on the ground will be at the same location at the time of the event, so observers on the ground and observers in the rocket will agree with each other when comparing the times on each clock.

You got it.

 

[Chestermiller]

Okay cool this helped,

So this thinking of mine is wrong:
When the tail of the ship passes the ground clock, the people on the ground will see a different time in the tail clock of the rocket than the people in the rocket because the rocket is traveling at relativistic speeds.

Correct way of thinking:
The clock in the rocket and the clock on the ground will be at the same location at the time of the event, so observers on the ground and observers in the rocket will agree with each other when comparing the times on each clock.

Yes. TSny explained it very well. One more thing. Even though observers in both frames of reference who are physically present at a specific event will agree on what they see displayed on each others clocks, they will also agree (for the most part) that their individual clocks don't display the same value.

 

[PsychonautQQ]

Awesome! I've figured out every question except f is throwing me a bit.
e) At the instant the clock in the nose of the rocket reads 1.h, a light signal is sent from the nose of the rocket to an observer standing by the clock on the ground. What does the clock on the ground read when the observer on the ground receives the signal?
e = 4.36 hours

f) when the observer on the ground receives the signal, he immediately sends a return signal to the nose of the rocket. What is the reading of the clock in the nose of the rocket when that signal is received at the nose of the rocket?

So by problem e you figure out that by the time the signal gets back to the clock on the ground, 4.36 hours have passed according to the ground clock.

I believe that means that if the ground frame looks at the rocket clock it will see the amount of time 4.36*γ, and if the rocket frame views the rocket clock it will see an amount of time (4.36 / γ).
The rocket frame looking at the rocket clock at this point shows that 1.9 hours have passed.

So then I solve for x in the equation x = (x'+vt')γ with x' = 0 and t' = 1.9 hours. That x is how far away the ground has moved according to the rocket frame?

so I bet this next thought is incorrect non relativistic thinking, but I tried ct' = x + vt' and that was really just a desperation shot and I don't even know why i typed it lol..


Do I need calculus for this problem?

 

[Chestermiller]

My advice to you again is the same as before: Let the Lorentz Transformation be your friend.

You obtained the results for part e, and found that the signal from the nose of the rocket arrived at x = 0 at time t = 4.36 hours. According to the Lorentz Transformation, what were the coordinates of this same event x'_e,t'_e as reckoned from the frame of reference of the rocket? In part F, a signal is sent out from these coordinates, and travels at the speed of light in both the x,t frame of reference and the x',t' frame of reference. How far does the signal have to travel in the x' frame of reference to get from x'_e to the nose of the rocket (x' = 0)? How long (Δt') does it take to get there? At what time t' does it arrive there?

 

[PsychonautQQ]

"In part F, a signal is sent out from these coordinates, and travels at the speed of light in both the x,t frame of reference and the x',t' frame of reference."

Does it matter that the light travels through the S reference frame to get the answer? Should this be in my calculations somewhere? Or can I look at just the S' reference to figure out how long it takes to get from x'e to x'?

There are multiple questions in my textbook I'm struggling with the deal with a signal being sent and having to catch up to the rocket moving in the same direction as the signal.

 

[Chestermiller]

Does it matter that the light travels through the S reference frame to get the answer? Should this be in my calculations somewhere? Or can I look at just the S' reference to figure out how long it takes to get from x'e to x'?

The light travels through both frames of reference at the same speed c. In this case, since x' doesn't change, it is easier to use to use the S' frame of reference to get your answer. Once you get the arrival time t' of the signal at x' = 0, you can use the Lorentz Transformation to determine x and t.