1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Time Dilation/Twin Paradox

  1. Sep 2, 2013 #1
    1. The problem statement, all variables and given/known data
    "A star is 5ly away. If an astronaut wants to travel there and back and only age half a year, at what speed β = v/c do they need to travel?"

    2. Relevant equations
    [tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]
    [tex]\Delta T = \gamma \Delta T_0[/tex]
    so [tex]\Delta T = \frac{\Delta T_0}{\sqrt{1 - v^2/c^2}}[/tex]

    3. The attempt at a solution
    I'm having problems finding the time elapsed on Earth (T0) without knowing the speed, v/c. I've tried to use [tex]\Delta T_0 = \frac{2l}{c}[/tex] and then put my answer into the equation for delta T, but my answer ended up being 20c which is definitely not right... I've tried it in both ly and in meters. I also tried equating 2d/c and 1/gamma * T but had too many unknowns to solve. Any help with this would be appreciated; this is the first question in a set of problems and whilst I've done the other 'harder' ones, I can't do this one. I feel like I'm missing something really obvious? I've tried to search for other equations for time elapsed on Earth but to no avail. Thank you.
  2. jcsd
  3. Sep 2, 2013 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Welcome to PF!

    Are you sure that ##T_0## is earth time? What is your justification for that?
  4. Sep 2, 2013 #3
    My lecturer gave me the equation:

    [tex]\Delta T' = \gamma\Delta T[/tex]

    Wouldn't T' be for the astronaut and normal T is elapsed on Earth?

    Upon reflection I tried it the other way around and did this:

    [tex]\Delta T = \frac{2D}{c} = \frac{10c-yr}{c} = 10[/tex]
    [tex]\Delta T = \frac{\Delta T'}{\sqrt{1 - β^2}}[/tex]
    [tex]10=\frac{0.5}{\sqrt{1 - β^2}}[/tex]
    [tex]0.5 = 10\sqrt{1 - β^2}[/tex]
    [tex]β^2 = 0.9975[/tex]
    [tex]∴ β = 0.99875 (i.e. v = 0.99875c)[/tex]

    I think this seems to be a reasonable answer? Since you can never actually travel at the speed of light. I'm not convinced I've done it correctly though.
  5. Sep 2, 2013 #4


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    It's important to understand the meaning of "proper time interval" and be able to identify which observer measures the proper time interval. The round trip "twin-paradox"" is a little tricky because the astronaut switches inertial reference frames when turning around. But think about just the trip out to the star where the astronaut stays in one inertial frame. Does the astronaut or the earth observer measure the proper time interval for the outgoing trip?

    Looks like you assumed the astronaut is traveling at the speed of light here. How would you correct that?
  6. Sep 2, 2013 #5
    The astronaut? Because they're on the same plane moving with the rocket for the duration as the trip so the proper time is the time measured by them. I think so, anyway.

    I don't know. This is the reason I've been stuck on this problem for the past 5 hours; if I put β into this equation then I can't find Delta T so I have two unknowns and can't complete the problem. I'm considering using simultaneous equations at the minute but so far I haven't been successful.
  7. Sep 2, 2013 #6


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Yes, that's right. The observer who measures the proper time between two events is the observer for whom the two events occur at the same place in their frame of reference. For the astronaut, the event of leaving the earth and the event of arriving at the star occur at the same place.

    You have two equations: ##\Delta T = \gamma \Delta T_0## and ## \Delta T = 2D/v##.

    Try combining these two equations to get an equation that involves just ##v, D## and ##\Delta T_0##. Then you'll have some algebra to do to solve for ##v##.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted