General Relativity Paradox?

Question

If I have a 1g rocketship (constant acceleration of 1 g all the time) and go tooling around exploring the solar system and then I come back to earth after 1 year, why, under gr, does my clock read slower than my twin brother back home. Haven't we both been experience the acceleration?

[Question inspired by seeing answers to a recent relativity question]

Alexander understands the question, which is both the rocket ship passenger and the man on earth experience the same acceleration (1g), which, according to General Relativity would cause comparable time dilation. Why is it that you can jetison this priniciple and simply apply special relativity to solve this problem?

Instead of trotting around the Solar System, our intrepid starship captain points his 1g rocketship away from the sun and spends 3 months jetting away (reaching a speed of ~ .25c -- and yes, this number is correct), then gets homesick and points the ship back towards earth accelerates back home for 6 months then applies then begins a 3 month period of braking. His flight was a year.

To earth bound observers, his velocity out was on average .125c and back it was also on average .125c which should give you a 1%+ time dialation. But under General Relativity's principle of equivalence, both the ground observer and our intrepid captain experienced a force of 1 g the whole time.

Wow, Alexander nailed it.

For some reason I'm not able to choose Alexander's answer as the Best Answer. I'll try back later.

Answer 1

People,
from the point of view of GR both astronaut in 1g ship and his brother in 1g earth gravity are in exacly the same conditions (locally).

Locally both of them can be either said to accelerate at 1g, or remain at rest in equivalent gravitation 1g.

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Let first consider simplified version:

Brother A sits in the chair on the surface of motionless planet.
Brother B is inside a hollow cannon ball fired vertically at speed 1km/s upward.
Brother C is inside a nose-down rocket fired vertically at speed 2km/s and rocket motor supplies extra 1g downwards.

After time 2 x 1km/s / 10 m/s² = 200 second all there brothers meet again at one 4-point.

Brother B was in free fall, he moved along geodesic line, and his clock is ahead.
Brothers A and C moved in symmetric trajectories, experienced equal accelerations 1g and their clocks are behind by equal amounts (they are now younger than C).

Brother B played the role of 'test particle' needed to establish geodesic. Here is what brother B saw:
http://i7.tinypic.com/6xg7twx.gif
Things played out nicely because experiment lasted for only 200 second, and at heights less than 100 km, where gravitation field is uniform

**********
Now lets set for a longer journey.

Brother A sits in the chair on the surface of motionless planet.
Brother B is inside a hollow cannon ball fired vertically at speed 11.2 km/s upward.
Brother C is inside a nose-down rocket fired vertically at speed c/2 and rocket motor supplies extra 1g downwards.

Brother B again serves as 'test particle' in free fall.
http://i1.tinypic.com/6y3u993.gif

After very short time brother B will found himself near apogee of his near-escape orbit, and brother A will stop accelerating. One year after the brothers will rejoin.

Once again brother B (geodesic) clock is ahead. Borother A is however not much farther behind, because his blue trajectory rather closely followed the red geodesic. The trajectory of brother C, however was very far from the geodesic, almost followed a photon and his clock is very far behind.

The apparent gravitation field of brother A was wrapped around earth, and his potential remained not very low.
Brother C, on the countrary, observed apperantly uniform gravitation field and there was no mass nearby to wrap this field around him.

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Local gravity of observer on earth in irrelevant over times longer than 84 minutes. Because there is a very close true geodesic just few radii of earth away.
Despite both brothers feel the same local gravty, the metrics of space-time abound them is different, because one has
nearby mass, and the the other is in open spece. RHS of equations of space-time cuvature are differnt. The two brothers are therefore in non-equivalent conditions, which becomes noticeble if the journey lasted longer than 84 minutes.

The brother who was on the rocket returns younger.

Answer 2

From a relative point of view, relative to the earth, only the rocket has accelerated away- the earth is the frame of reference and the rocket has been accelerated. so the answer is no you have not experienced the same acceleration.

Also on the return trip the rocket would be under different rates of accelerating as the planet moves about the sun (sun now being the frame of reference).

complex question. Very complex.

simple answer. Both ship and planet are fitted with atomic clocks- removing all relativistic effects.... hehe

Answer 3

If you are in a rocket ship with a 1g acceleration and the pull of the Earth is 1g aren't you just going to float a bit and not actually go anywhere.

Answer 4

It's not the acceleration per se that causes time dilation (this is a very common misconception), but the change in internal reference frame while you are in transit between destinations. Traveling near c (all over the place in the solar system, I presume), you travel a shorter distance than as perceived by people on earth monitoring your ship due to Lorentz contraction of the solar system itself in your reference frame. You, therefore, travel for a shorter time in your frame. Incidentally, this effect is described by Special Relativity; General is not required.