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Relativity and Time Dilation 
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#1
Mar612, 12:34 AM

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 = lightyears * .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 = lightyears 43.8444 NO 2. Relevant equations [itex]\gamma[/itex] = sqrt (1  v^2/c^2) d' = d[itex]\gamma[/itex] 3. The attempt at a solution I found that the teacher travels at .99974c, pretty easily. Since my v = .99974c, [itex]\gamma[/itex] = .022802. d' = d[itex]\gamma[/itex] .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. 


#2
Mar612, 12:02 PM

P: 321

you could use the lorrentz transformation matrix which is pretty simple
x=x' y=y' z= (gamma)(z'vt) t=(gamma)(t'vz/c^{2}) since you know t and t' and z it is easy to figure out 


#3
Mar712, 04:40 PM

P: 68




#4
Mar712, 05:17 PM

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P: 41,304

Relativity and Time Dilation



#5
Mar712, 11:21 PM

P: 68

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. 


#6
Mar812, 06:32 AM

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P: 41,304

Let's start over.



#7
Mar812, 08:29 AM

P: 321

(gamma)T_o= T
Sorry cant do anything fancy on my ipod. T_o is the actual time and T is the teachers time 


#8
Mar812, 09:46 AM

P: 68

The problem states that she travels away for 1 year, so Δt' = 1 Δt' = Δt[itex]\gamma[/itex] 1 yr = 44 yrs * [itex]\gamma[/itex] [itex]\gamma[/itex] = .022727 < instead of recalcing [itex]\gamma[/itex] maybe I should just use this value? sqrt(1  v^2/c^2) = .022727 1  v^2/c^2 = 5.165289e4 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 


#9
Mar812, 10:05 AM

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#10
Mar812, 10:19 AM

P: 68

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. 


#11
Mar812, 10:24 AM

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#12
Mar812, 10:50 AM

P: 68

Thanks SO MUCH for the help and patience! 


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