Twin paradox

Classical Mechanics Level 2

Dick is $20$ years old when he takes off on a space voyage at a speed of $0.80$ $c$ ( $c$ is the speed of light) to a star $20$ light-years away. His wife Jane, also $20$ years old, decides to wait for him to come back after successfully completing his mission. Finally, after $30$ years of voyage in his reference frame, Dick comes back. How old is Jane upon his return?

50 years old 70 years old 100 years old 30 years old

3 solutions

Chew-Seong Cheong
Jul 31, 2014

Time dilation is given by: $\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$ where, $\Delta t'$ and $\Delta t$ are the time periods of Jane and Dick in their reference frames respectively, $v$ is the speed Dick is travelling and $c$ the speed of light. Hence, $\Delta t' = \frac{30}{\sqrt{1-0.8^2}}=\frac{30}{0.6}=50 \space years$ Therefore, the age of Jane is $20+50=\boxed {70}$ years old.

Vsquare /C square is not clear,where we get this formulaa? K.K.GARG,India

Krishna Garg - 6 years, 10 months ago

$\frac {v}{c}$ is the ratio of the velocity of the ship to the speed of light, sometimes denoted as $\beta$ . It comes as part of the Lorentz factor $\gamma$ which is derived from the Lorentzian transformations. This factor $\gamma$ is equal to $\frac {1}{\sqrt {1 - \beta^2}}$ and is used in almost all equations regarding special relativistic effects (Time dilation, length contraction, relativistic mass).

Gavin Stolk - 6 years, 10 months ago

Alex Li
Feb 23, 2015

Note that Dick travelled $40$ lightyears (to the star and back) at a speed of $0.8c$ , which takes $50$ years. Therefore, Jane is $50+20=\boxed{70}$ years old when Dick returns.

Abdulmuttalib Lokhandwala
Jun 5, 2014

Refer the note by me here twin paradox

Twin paradox

3 solutions

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