Travel time to Alpha Centauri at 0.9C - need some help.

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AcesHigh
Posts: 655
Joined: Wed Mar 25, 2009 3:59 am

Travel time to Alpha Centauri at 0.9C - need some help.

Post by AcesHigh »

from another forum:
Given a spaceship with a power source capable of a constant 1 g acceleration, you can. Say you get to .9c. V_final/Acceleration gives 2.7*10^8 m/s / 9.81 m/s^2 = 2.75 * 10^7 seconds = 318 days at 1 g. At .9c, you get to alpha Centauri in 4.84 years (actually, more like 5.3, as the last 318 days you'll be decelerating.). Repeat the trip backwards, and your round trip takes 10.5 years or so of your life. The problem? .9c is a Lorentz factor of 2.29, so time on Earth is 10.5 years*2.29 = 24 years back on Earth (slightly less, as for nearly two of those years you are traveling slower than .9c). If you left today, you'll come back to Earth in the early 2030s, but your biological clock would read 2023. Time travel in a way is possible; it's just one way.

it didnt make sense to me. 24 years back on Earth???

I replied
you are calculating it wrong pal. Dont know how you judge a warp drive to be impossible but at the same time you dont know how to calculate lorentz factors.


time on Earth on that trip is 10.5 years. Its the time on the ship that is multiplied by the lorentz factor (after all, its the ship which is travelling near light speed, not Earth!)

ship time is 10.5 years multiplied by the lorentz factor of 0.453 = 4,7 years.



he replied
Actually, you are missing a step. Yes, Earth sees you gone for 10.5 years. But what you would see on Earth is a combination of two transformations: a Lorentz transformation into your new plane, and then a Lorentz transformation back from the new plane to Earth -- your time times gamma, 1/(1-beta^2)^(1/2), squared. It would be 10.5 years for you that Earth elapsed if you just stopped -- but you have to reverse and accelerate out of that frame back to Earth's frame to get back. That reversal is what kicks Earth so far ahead. Two frame shifts. Two different Lorentz factors from your point of view on the ship.

Another way to think about it: the Earth is (basically) in proper time (tau). As you speed up, you'll see gamma*tau go by -- 4.2 years. But from your perspective on the rocket, turning around to return to Earth's frame, Earth's time is sped up by tau/gamma -- 24 years.

You spend 4.7 years on the ship.
Earth sees you gone for 10.5 years.
You come back to a Earth 24 years in the future.

Relativity won't let you go between three different frames (Earth leaving, alpha Centauri, Earth returning, and the two Earth frames will be distinct from your point of view) with just one transformation.

is he right? I never heard of that before. Everytime I read about travel at relativistic speeds, the time outside the ship is calculated normally. So lets say, 5 ly at 0.9999C takes 5 years. For the ship´s passengers only it does takes less time.

jnaujok
Posts: 76
Joined: Wed Sep 16, 2009 6:19 pm
Location: Colorado Springs, CO
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He's wrong...

Post by jnaujok »

He's completely forgotten doppler shifting of the signal.

You proceed to Alpha Centauri at .9C (4.7 years)

The light from Earth (looking through a telescope at a laser clock) tells you that Earth has only seen about .5 years go by. This is because as you look at the clock, you are red shifting it (spreading the frequency of the clock ticks). By the time you get to Alpha Centauri, the light leaving earth has no time to catch up to you, so you see the Earth as it was 4.2 years ago, leaving only the 0.5 years (six months) since you left.

When you turn around and go back, you blue-shift the time, galloping through the 4.2 years of time still waiting to reach Alpha Centauri, and the the 4.7 years of time it takes to get back, meaning you arrive at 10.5 years (rounding error) that is exactly what we calculated the Earth would see go by.

You, however, experience only the 4.7 years or whatever the final Lorentz contraction factor is with turn-around and what have you.

He's also confusing reference frames. You never leave the Earth's reference frame for Earth time, you never leave the Ship's reference frame for Ship time. He's trying to apply the ships reference frame to the Earth's reference frame and ending up with a nonsense answer. It's a common mistake, and it's one of the hardest things to learn to correctly reconcile.

Basically he's missing the fact that, if he leaves today September 19th 2012, when he gets to Alpha Centauri, the apparent date on Earth from his ship (light speed lag) will be only March 19th, 2013 (more or less). In reality, it'd really be sometime in 2017. When he turns around, the pages rip off the calendar, and he arrives back at earth in 2023. All the math adds up, and he wonders why he's 6 years younger than all of his friends.

To get his answer, his travel would have to somehow artificially age the entire rest of the universe by an extra 14 years. That alone should demonstrate why he's wrong. Otherwise every time we fired up the LHC, the rest of the universe would age to 100,000,000,000 years before the particles made it around the loop once.

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