# Relativity and Time Dilation

by Rapier
Tags: dilation, relativity, time
 P: 68 1. The problem statement, all variables and given/known data A physics instructor is 22 years older than her student. She decides that instead she would like to be 22 years younger than her student and decides to use special relativity to accomplish this. She jumps in her super rocket travels away from earth for a period of 1 years (as measured by her) and then back to earth for the same period. a) Ignoring the brief period of acceleration, what is the speed at which the physics teacher is traveling during this time? vteacher = fraction of c * .99974 OK b) In the teacher's reference frame, how far does she travel away from the earth during the first portion of this period? Δxteacher = light-years * .99974 OK c) In the student's reference frame, how far does the teacher travel away from the earth during the first portion of this period? Δxstudent = light-years 43.8444 NO 2. Relevant equations $\gamma$ = sqrt (1 - v^2/c^2) d' = d$\gamma$ 3. The attempt at a solution I found that the teacher travels at .99974c, pretty easily. Since my v = .99974c, $\gamma$ = .022802. d' = d$\gamma$ .99974 ly = d (.022802) d = 43.844 ly But it is not fond of that answer and I don't see any other way to go about this.
 P: 318 you could use the lorrentz transformation matrix which is pretty simple x=x' y=y' z= (gamma)(z'-vt) t=(gamma)(t'-vz/c2) since you know t and t' and z it is easy to figure out
P: 68
 Quote by Liquidxlax you could use the lorrentz transformation matrix which is pretty simple x=x' y=y' z= (gamma)(z'-vt) t=(gamma)(t'-vz/c2) since you know t and t' and z it is easy to figure out
I must still be missing something, because I'm just not getting it. I think I might be confusing my reference frames. The prime reference frame is the 'proper', yes? In this case, the prime frame is the one with the teacher in it.

Mentor
P: 40,262

## Relativity and Time Dilation

 Quote by Rapier I must still be missing something, because I'm just not getting it. I think I might be confusing my reference frames. The prime reference frame is the 'proper', yes? In this case, the prime frame is the one with the teacher in it.
Yes, you're mixing up your reference frames. Think of the teacher as traveling from earth to planet X and back. The distance between earth and planet X, as seen by the teacher in her rocket, is contracted.
P: 68
 Quote by Doc Al Yes, you're mixing up your reference frames. Think of the teacher as traveling from earth to planet X and back. The distance between earth and planet X, as seen by the teacher in her rocket, is contracted.
z = .99974 ly
v = .99974c
t' = 1 yr
t = 22 yr
z= (gamma)(z'-vt)

(gamma) = sqrt(1 - v^2/c^2)
(gamma) = .022802

.99974 ly = (.022802)(z' - .99974 * 22 yrs)
z = 65.8387 ly

I'm sorry. I'm just not getting this. I am still not getting the correct answer.
Mentor
P: 40,262
Let's start over.

 Quote by Rapier a) Ignoring the brief period of acceleration, what is the speed at which the physics teacher is traveling during this time? vteacher = fraction of c * .99974 OK
How did you figure this? Note that she travels away from earth for 1 year, then back to earth for 1 year. How much time must pass on earth so that she comes back younger than her student? What must the gamma factor be?
 P: 318 (gamma)T_o= T Sorry cant do anything fancy on my ipod. T_o is the actual time and T is the teachers time
P: 68
 Quote by Doc Al Let's start over. How did you figure this? Note that she travels away from earth for 1 year, then back to earth for 1 year. How much time must pass on earth so that she comes back younger than her student? What must the gamma factor be?
Since she needs to go from +22 to -22, the Δt = 44 yrs.
The problem states that she travels away for 1 year, so Δt' = 1

Δt' = Δt$\gamma$
1 yr = 44 yrs * $\gamma$
$\gamma$ = .022727 <-- instead of recalcing $\gamma$ maybe I should just use this value?

sqrt(1 - v^2/c^2) = .022727
1 - v^2/c^2 = 5.165289e-4
v^2/c^2 = .99948347
v/c = .9997417
v = .9997417c

d = vt
d = (.9997417c)(1yr)
d = .9997414 ly
Mentor
P: 40,262
 Quote by Rapier Since she needs to go from +22 to -22, the Δt = 44 yrs. The problem states that she travels away for 1 year, so Δt' = 1
I read the problem as stating that she travels away for 1 year, then back for another year. Thus Δt' = 2 and Δt = 46 years.

 Δt' = Δt$\gamma$ 1 yr = 44 yrs * $\gamma$ $\gamma$ = .022727 <-- instead of recalcing $\gamma$ maybe I should just use this value?
Realize that gamma is always greater than 1: Δt = γΔt'
P: 68
 Quote by Doc Al I read the problem as stating that she travels away for 1 year, then back for another year. Thus Δt' = 2 and Δt = 46 years. Realize that gamma is always greater than 1: Δt = γΔt'
This is part of why I am so very frustrated with this unit. Using your values (Δt' = 2 and Δt = 46) I get v = .99905c. Which is also a valid answer (since it is within 1% of the correct answer and we are out far enough digits that I could just make them up and be correct).

Let me try again with the new v.

Part (C) asks how far the teacher travels during the first leg (so half of the 46 years), Δt = 23 yrs (because she travels out and instantly turns around and comes right back at the same velocity).

z = γ(z'-vt)
.99905437 ly = (1/23)(z' - .99905437c * 23 years)
z=45.9565 ly

and it still tells me no. Sigh.
Mentor
P: 40,262
 Quote by Rapier This is part of why I am so very frustrated with this unit. Using your values (Δt' = 2 and Δt = 46) I get v = .99905c. Which is also a valid answer (since it is within 1% of the correct answer and we are out far enough digits that I could just make them up and be correct).
OK.

 Let me try again with the new v. Part (C) asks how far the teacher travels during the first leg (so half of the 46 years), Δt = 23 yrs (because she travels out and instantly turns around and comes right back at the same velocity). z = γ(z'-vt) .99905437 ly = (1/23)(z' - .99905437c * 23 years) z=45.9565 ly
I don't understand what you're doing here. You have the speed and time as measured by earth observers. Just use distance = speed * time; no need for relativity.
P: 68
 Quote by Doc Al OK. I don't understand what you're doing here. You have the speed and time as measured by earth observers. Just use distance = speed * time; no need for relativity.
Oh geez. That did it. I guess I'm still having a problem with my frames of reference. I didn't realise that I had v and t for the same frame and could use my classical equations. I guess it's back to the book...actually I'll try some YouTubes.

Thanks SO MUCH for the help and patience!

 Related Discussions Introductory Physics Homework 2 Introductory Physics Homework 3 Special & General Relativity 3 Advanced Physics Homework 7 Introductory Physics Homework 1