Newbie question: Long trips and observed time

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Hello All,
I hope my question isn't too simplistic for this forum. I've been giving my kids this illustration for a couple years, and have begun to doubt my story a bit. Hoping someone here might be able to clarify for me.

It's simple really, here's the setup:
*You're going to a star 10 light years away, 20 light-years round trip.
*You're speed is 90% the speed of light "C"; we'll estimate 30yr round trip time due to only going 90% of C. (I don't think an exact measure matters for this illustration)
*We all know that as you approach C, 'time runs more slowly'. (whatever that really means)

What I've been telling the kids:
You'd return from this trip 30 years older and due to the final statement above, the Earth might be 1000 years older.

But now I'm not so sure.

Perhaps the opposite happens? Perhaps 'time slowing down' means you'd think you're getting there in only a week, and you'd return in exactly the pre-planned 30 Earth-years? Put a little differently, as the astronaut approaches C, can his perception of elapsed time drop BELOW the established 20 light-years, or can that perception only drop toward a fixed lower limit of the original 20-year estimate?

I'm not seeing the 1000 year return time for such a trip as plausible anymore. 'Relativity' means relative to the observers, and your 90% of C velocity results from fuel/acceleration calculations that were done on Earth. .9*C IS your velocity as Earth sees it, and the star's distance is unchanging.


Perhaps a distilled question would be: for a fixed distance, can the traveler experience far *less* time than the observer, and can velocities just below C remove the need for supplies/hibernation etc?

Many Thanks,
Tom
 
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Well, the real answers are:

When you returned, Earth would have passed 22.22 years since you left. You would have aged 9.687 years.
 
Look up time dilation http://en.wikipedia.org/wiki/Time_dilation. Let's work through your example:

- Alice is piloting the spaceship whilst Bob stays on Earth.
- Bob waves goodbye and Alice rockets off at .9c (let's ignore the acceleration for now)
- Alice pilots her ship towards a target star 10ly away
- At .9c the time dilation effect is 43.58%, this means that from Alice's perspective she will reach that star in 4.358 years.
- From Bob's perspective Alice takes 10 years to get there but everything on the ship is moving slowly (interestingly if Alice looks back at Bob she will see him acting slowly).

Does that answer your question?
 
PAllen said:
Well, the real answers are:

When you returned, Earth would have passed 22.22 years since you left. You would have aged 9.687 years.

Thanks PAllen!
So a device able to reach 'very nearly' C could traverse any distance with only perhaps a day's passing to the traveler? (so long as they realize millennia would pass on Earth)

Regardless of the technical hurdles of achieving that speed, this puts to bed conversations I've had with the kids regarding hibernation and supplies for extended travel.
 
tom_mi said:
Thanks PAllen!
So a device able to reach 'very nearly' C could traverse any distance with only perhaps a day's passing to the traveler? (so long as they realize millennia would pass on Earth)

Regardless of the technical hurdles of achieving that speed, this puts to bed conversations I've had with the kids regarding hibernation and supplies for extended travel.

Er you might want to continue those conversations. The "technical hurdles" are colossal, absolutely http://en.wikipedia.org/wiki/Nontrivial" . Also it will give the kids a better appreciation of the issues outside of speed i.e. how to build a stable ecology, how to pack an industry, how to build a society stable enough to span the journey, how to set up a colony on the other side, the ethics/politics/practicalities etc.
 
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Agreed Ryan & thanks for responding. Acceleration & social issues do matter; I just wanted to be sure I was drawing an accurate picture of what Relativity provides.
 
tom_mi said:
Perhaps a distilled question would be: for a fixed distance, can the traveler experience far *less* time than the observer...
Here's a formula to determine the speed (as a fraction of c) that you would have to travel at to go any distance, d (in light-years), in any time, t (in years):

v = 1 / √[1 + (t/d)²]

For more details, see post #7 on this thread:

https://www.physicsforums.com/showthread.php?p=3337763#post3337763
 
ghwellsjr said:
Here's a formula to determine the speed (as a fraction of c) that you would have to travel at to go any distance, d (in light-years), in any time, t (in years):
v = 1 / √[1 + (t/d)²]

Excellent! So to cross the galaxy (est 25,000 light-years) in 1 month and join up with the rebellion my son will need to go 0.99999999999444488888004628888933% of C.

He's going to need a faster bike. :)
 
tom_mi said:
Excellent! So to cross the galaxy (est 25,000 light-years) in 1 month and join up with the rebellion my son will need to go 0.99999999999444488888004628888933% of C.

He's going to need a faster bike. :)

FYI the galaxy is 100,000+ lightyears in diameter and Earth is 30,000 lightyears from the centre.
 
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ryan_m_b said:
FYI the galaxy is 100,000+ lightyears in diameter and Earth is 30,000 lightyears from the centre.

Oops, forgot I was thinking approx distance to center. :)
 
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