Relative Time Dilation and Aging in Space Travel

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Homework Help Overview

The discussion revolves around a problem in the context of special relativity, specifically focusing on time dilation experienced during space travel. The original poster presents a scenario involving twin brothers, one traveling at a significant fraction of the speed of light while the other remains on Earth, and seeks to determine the age difference upon the traveler's return.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the time dilation formula and question the correctness of the original poster's setup, particularly regarding the use of variables in the equation. There is an exploration of the implications of using the speed of light versus the traveler's speed in the calculations.

Discussion Status

The discussion is ongoing, with participants providing guidance on the correct application of the time dilation formula. There is an acknowledgment of a potential misunderstanding in the original poster's calculations, and some participants suggest clarifying the variables used in the equation.

Contextual Notes

Participants note the importance of distinguishing between proper time and dilated time in the context of the problem. There is also a mention of the speed of light as a critical constant in the calculations.

disneygirl828
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Homework Statement


I have a twin brother, and we're 30 years old. If i travel for three years at 2.55 X 10^8m/s (0.85c) while my brother is a rest on Earth, how much older than me will my brother be upon my return?

Homework Equations



t=To/√1-v^2/c^2

3. Try to work out
To=3
T=?
V= 2.55 X 10^8 m/s
C= (.85c) 2.55 X 10^8 m/s
T= 3/ (all of this under a square root sign -->)1-(2.55 x 10^8)^2/(2.55 x 10^8)^2
t=3/0
*i get stuck here because you can't have a zero in the denominator*
 
Last edited:
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disneygirl828 said:
T= 3/ (all of this under a square root sign -->)1-(2.55 x 10^8)^2/(2.55 x 10^8)^2
t=3/0
*i get stuck here because you can't have a zero in the denominator*

Hmm, you've done:
[tex]T= \frac{3}{\sqrt{1- \frac{v^2}{v^2}}}[/tex]
I'm guessing you've accidentally done this?
 
BruceW said:
Hmm, you've done:
[tex]T= \frac{3}{\sqrt{1- \frac{v^2}{v^2}}}[/tex]
I'm guessing you've accidentally done this?
Yes i used that equation, should I have used something different? If so what?
 
disneygirl828 said:
t=To/√1-v^2/c^2

This was the right one, i.e. :
[tex]\frac{3}{\sqrt{1- \frac{v^2}{c^2}}}[/tex]
c is the speed of light, but you used v instead, which is why the denominator came out as zero.
 
Last edited:
C = 3 x 10^8 m/s.
Remember this: the difference in ages will be the difference in the proper time and dilated time.
 

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