Special Relativity and Time Dilation: A Physics Problem in Frames of Reference

[Rapier]

Homework Statement

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

Homework Equations

$\gamma$ = sqrt (1 - v^2/c^2)
d' = d$\gamma$

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.

 

[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

 

[Rapier]

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.

 

[Doc Al]

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.

 

[Rapier]

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.

 

[Doc Al]

Let's start over.

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?

 

[Liquidxlax]

(gamma)T_o= T

Sorry can't do anything fancy on my ipod.

T_o is the actual time and T is the teachers time

 

[Rapier]

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

In case you were about to ask, part b:
d = vt
d = (.9997417c)(1yr)
d = .9997414 ly

 

[Doc Al]

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'

 

[Rapier]

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.

 

[Doc Al]

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.

 

[Rapier]

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 realize 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!