An Einstein relativity question--assume astronauts leave the solar
system at a constant 1G subjective acceleration for the Alpha Centauri
system which we will assume is exactly 4 lightyears away. Halfway
there, the spaceship turns around and decelerates at a subjective 1G,
until it comes to a halt at 4 lightyears. Having forgotten their
toothbrushes, the astronauts immediately turn around and do the same
thing all over again until they are back in the Solar System. How long
will they have been gone by their own clock, and how long by the clock
of Earth observers?

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Request for Question Clarification by livioflores-ga on 18 Nov 2002 06:28 PST

Do you need the mathematical resolution of this question or you accept
as an answer only the result of this problem.
I canīt find how to solve this problem yet, but I found the result.
Thank you.

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Clarification of Question by davidlsmith-ga on 18 Nov 2002 10:45 PST

I'd like the mathematical formulas necessary to calculate the problem
if possible, but just the answer would go a long way towards
satisfying my curiosity. I'm 40 years past my last math course, so it
will have to be pretty simple math. See the comment by flajason-ga. 
David Smith

There is a site called The Relativistic Rocket that might get you the
answers you need. It is available here:
http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
And also appears to be mirrored in a couple of locations.
It has the formulas needed to calculate the times and distances along
with the time dilation.
Unfortunately my attempt to plug your numbers in came up with some
wierd results, so I don't trust my math skills to answer your
question.

Search used:
relativity constant 1g accelerate decelerate

An interesting question, but these calculations rapidly  get *very*
complicated and counterintuitive.

Duh, the formula for determining the time for the astronaut was right
there in front of me, and is actually fairly easy to figure.
The formula is also on that page.

Time that the astronauts age = 1.94 * (arccosh (n/1.94 +1)
where n = the distance in light years.

Here's how to work it on the Windows calculator.
(first make sure it's on Scientific under the View menu)
For n=4 light years to Alpha Centauri

Divide 4 by 1.94, then add 1
Check the boxes for Inv and Hyp, then click the COS button
Multiply that by 1.94

4/1.94 = 2.0618... + 1 = 3.0618...
arccosh(3.0618) = 1.78436...
1.78436 * 1.94 = 3.4616 years for the astronauts.

And of course for the return trip, it would just be doubled to 6.92
years for the time passed on the astronaut's clocks.

I'm still having difficulty getting the time from the Earth
perspective. If I have a breakthrough on that front, I'll let you
know.

Ok, pretty sure I've got it now...

Using the formula on that page and solving for t

t = c/a * (sinh (a/c)*T)

c = 1 ly
a = 1.03 ly/ly^2 (1G acceleration)
T = time as experienced by the astronaut's clock

Since they are accelerating halfway there and decelerating the rest of
the way at the same rate, we can divide their time in half as to how
long it takes to get there.
3.4616 years to Alpha Centauri / 2 = 1.7308 years to the halfway point
(T).
t = 1/1.03 * (sinh (1.03/1 * (1.7308))
t = 2.8 years Earth time for the astronauts to reach the halfway point
Double that for their deceleration = 5.6 years Earth time for the
astronauts to reach Alpha Centauri
Double again for their return trip to earth: 11.2 Earth years

So for our intrepid explorers, it took 6.92 years to go there and back
according to their clocks, while 11.2 years passed on Earth.

flajason-ga seems to have solved the whole thing, as you can see down
below. Can I call his comment my answer and give him the money? I
award him 5 stars, this is exactly what I wanted.  David Smith

David,

Thanks for the kind words. Unfortunately, I am not a Google Researcher
so I can't accept the money. I really enjoy this service becuase of
people like you who post interesting questions. It gives me a chance
to research new topics and ideas that I've never heard of before, or
ones that I used to know but have since forgotten. And if I can't find
the answer myself, I can usually count on the Google Researchers to
get the answer or at least get started in the right direction.
I like to learn at least one new thing a day, and this service is a
great tool to that end.
So this post is what I guess you would call a freebie (aside from the
$0.50 listing fee)

Regards,

Jason