Time Dilation/Length Contraction

[Ionian32492]

Homework Statement

A Rocket Moves away from the Earth at a speed of (3/5)c. When a clock on the rocket says that one hour has elapsed, the rocket sends a signal back to the earth.

(A). According to clocks on the earth, when was the signal sent?
(B). According tot eh Clocks on the earth, how long after the rocket left did the signal arrive at the earth?
(C). According to an observer on the rocket, how long after the rocket left did the signal arrive at the earth?

Homework Equations

[itex] \Delta t = 3600s [/itex]

[itex] \beta = \frac{v}{c} = 0.6[/itex]

[itex] \gamma = \frac{1}{(1-\beta^2)^.5} [/itex]

[itex] \Delta t' = \Delta t \gamma [/itex]

The Attempt at a Solution

For part A:
[itex] \gamma = \frac{1}{(1-\beta^2)^.5} = 1.25 [/itex]

[itex] \Delta t' = \Delta t \gamma = 4500s [/itex]

This much I think is correct.

For part B:
[itex] \Delta t_t = \Delta t_1 + \Delta t_2 [/itex]

[itex] \Delta t_1 = \gamma \Delta t [/itex]

[itex] \Delta t_2 = \frac{d_{rocket}}{c} [/itex]

[itex]d_{rocket}=\gamma \beta c \Delta t = 2700c seconds [/itex]

[itex] \Delta t_t = \gamma \Delta t + \frac{d_{rocket}}{c} [/itex]

[itex] \Delta t_t = 4500s + 2700s = 7200s = 2hr [/itex]

I don't feel comfortable with this answer. I found an analogous question in my textbook regarding time dilation and length contraction, and it followed the same procedure I did, but I feel my answer should just be double the time the rocket observed.

I've yet to do part C, to be frank I'm so burnt out on the last two parts that I've yet to get to it. When I get to it, I will edit this.

EDIT: Part C

[itex] \Delta t_t = \Delta t_1 + \Delta t_2 [/itex]

[itex] \Delta t_1 = \Delta t = 3600s [/itex]

[itex] \Delta t_2 = \frac{\Delta x}{c} [/itex]

[itex]\Delta x=\beta c \Delta t = 2160c seconds [/itex]

[itex] \Delta t_t = \Delta t + \frac{\Delta x}{c} [/itex]

[itex] \Delta t_t = 3600s + 2160s = 5760s = 1hr 36 min [/itex]

Followed the same procedure, but from the reference point of the rocket. Comparing to part b, it seems to coincide with the results of the Twin Paradox, so I feel good about this one.

 

[Doc Al]

Your answer looks fine to me.

 

[Ionian32492]

Really? Thanks so much! I've been feeling really unsure about the class, mainly because we've yet to do any sort of problems until now, so I was getting worried. If you don't mind, can you look at my part C?

 

[Doc Al]

EDIT: Part C

[itex] \Delta t_t = \Delta t_1 + \Delta t_2 [/itex]

[itex] \Delta t_1 = \Delta t = 3600s [/itex]

[itex] \Delta t_2 = \frac{\Delta x}{c} [/itex]

[itex]\Delta x=\beta c \Delta t = 2160c seconds [/itex]

[itex] \Delta t_t = \Delta t + \frac{\Delta x}{c} [/itex]

[itex] \Delta t_t = 3600s + 2160s = 5760s = 1hr 36 min [/itex]

Followed the same procedure, but from the reference point of the rocket. Comparing to part b, it seems to coincide with the results of the Twin Paradox, so I feel good about this one.

The problem here is that you fail to take account of the fact that the Earth is moving away from the rocket (from the rocket's frame). What you call Δx is the distance to the Earth at the moment that the signal is sent--not the total distance the light travels.

 

[Ionian32492]

The problem here is that you fail to take account of the fact that the Earth is moving away from the rocket (from the rocket's frame). What you call Δx is the distance to the Earth at the moment that the signal is sent--not the total distance the light travels.

So would it be 2Δx then, to account for the distance from the Earth to the rocket and back?

 

[Doc Al]

So would it be 2Δx then, to account for the distance from the Earth to the rocket and back?

No. And the signal doesn't come back: It goes from rocket to Earth and that's it.

To find the time that the light takes to go from rocket to earth, you'll need to set up a kinematic equation that includes the fact that the Earth is moving. Give it a shot.

(There's an even easier way to solve this problem using time dilation.)

 

[Ionian32492]

[itex] x_0 = d_{rocket} = \gamma \beta c \Delta t = 2700c s [/itex]

[itex] \Delta x = x_0 + 0.6 c \Delta t_2 [/itex]

[itex] \Delta t_2 = \frac{\Delta x + x_0}{0.6c} [/itex]

This is my best idea, but I don't know what [itex]\Delta x[/itex] is

EDIT: I know [itex]\Delta x[/itex], its [itex] \beta c t[/itex]

 

[Doc Al]

[itex] x_0 = d_{rocket} = \gamma \beta c \Delta t = 2700c s [/itex]

You want to view things from the rocket's viewpoint, so there's no factor of gamma needed.

[itex] \Delta x = x_0 + 0.6 c \Delta t_2 [/itex]

Good. Note that Δx here is the distance the light travels in time Δt₂. Use that fact to solve for Δt₂.

 

[Ionian32492]

Thanks so much! I got [itex]\Delta t_t = 9000s = 2.5hr[/itex], which seems correct.

 

[Doc Al]

Thanks so much! I got [itex]\Delta t_t = 9000s = 2.5hr[/itex], which seems correct.

Good!

Now use the answer from part B to solve this in a different way (using time dilation).

 

[Ionian32492]

[itex] \Delta t_{t2} = 2.5hr = \gamma \Delta t_t [/itex]

Well that's painfully obvious now hahahaha. Thanks!