(Come) back to the future. Suppose that a father is 20 y older than his daughter. He wants to travel outward from Earth for 2 y and then back to Earth for another 2 y (both intervals as he measures them) such that he is then 20 y younger than his daughter. What constant speed is required for the trip?
Lol, I kinda learnt relativity at the start of the first term so I'm quite shaky :P I seem to remember that there are three frames you need to take into consideration... but first of all, do you know how to start at all?
using time dilation im getting v = 2.96 * 10^8 http://www.wolframalpha.com/input/?i=solve+24+%3D+4%2Fsqrt%281-%28v%2F%283*10^8%29%29^2%29+over+reals but im confused on this as i have been playing with relativity only for two days and i don't think i have understood this properly
Why numbers 24 and 4 sorry?
my question is something like this : In ship's frame, the earth would be moving in the other direction with the same speed. So, wouldn't father also ages more compared to the Earth ?
What is the grade level cause I don't understand so it could bee higher than my grade.
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Haha, the problem you have is called the Twin Paradox. Have you looked it up yet?
\[\Delta t' = \frac{ \Delta t }{ \sqrt{1-\frac{ v^{2} }{ c^{2} }} }\] So, \[40 = \frac{ 4 }{ \sqrt{1-\frac{ v^{2} }{ c^{2} }} }\] solve for v: \[v=c \sqrt{1-(4/40)^{2}}\]
father's travel time = 2+2 = 4 years father will be 4 years older the daughter needs to age 24 more years so that the difference is 20
v = ±2.95804×10^8
i saw one two videos on twin paradox not feeling satisfied yet... they talk about general relativity which i don't know
why 40 @aliqanber
actually: \[v=c \sqrt{1-(4/44)^{2}}=0.996c\]
Hmmm, let me think outloud here.
So it is important to get the frames right. So from the father's point of view, he travels for two years away and then two years back. Then by the time he has gotten back, in....his frame? he is is 20 years younger than his daugher.
http://www.physicspages.com/2015/02/28/the-twin-paradox-analyzed-using-lorentz-transformations/
I agree with your orginal answer @ganeshie8
The link above should help with your question.
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