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Acceleration, and Observers

  1. Dec 19, 2009 #1
    Alright, so I am going to try to structure this question as carefully as possible, since it has a lot of parts, but I may have to clarify more later.

    Anyway, first off, which of these scenarios (if any) is correct:

    The crew of a spaceship moving at near C to a planet 5 light years away observes the trip there and back as just slightly over 10 years long (excluding time it takes to accelerate, or turn around, etc.), and observers from Earth see this trip as taking longer.

    The crew of the spaceship sees the journey as nearly instantaneous, while observers from Earth see the trip as taking 10 years, though the people inside would appear to be moving VERY slowly.


    Now, if the first scenario is correct, then that means that the spaceship would appear to be moving at LESS than the speed of light to Earth, which would also mean that at some point, along the path from 0 to light speed, accelerating would, to an observer, make you move slower, or that there is a speed at which rather than accelerating forward, you appear to maintain the same speed, but slow down within the ship more and more. So, again, which would it be?

    If the SECOND scenario is correct, then moving at the speed of light, as observed by Earth, makes observers from inside the ship see themselves moving at much faster than light (hence traveling 10 light years in less than 10 years). If THIS is the case, then there has to be a point at which those on the spaceship would see themselves as moving at the speed of light. If so, what would this speed be to an Earth observer?


    Yes, a lot of... almost rambling there, but these things are really bothering me tonight. Please make me feel like I understand relativity again.
     
  2. jcsd
  3. Dec 19, 2009 #2

    Dale

    Staff: Mentor

    Hi Kaimana, welcome to PF!
    This is the correct scenario, the key to resolving your concern is understanding that lengths are relative. In other words, different frames will disagree on the distances involved. Let me work out a numerical example with specific numbers (I will change yours a little to make the numbers simpler) so that you can see how it works.

    Let's say that in the earth's reference frame the ship travels at 0.6c on the outbound leg and at -0.6c for the return and that the destination is 6 lightyears (ly) away. I will use the usual http://en.wikipedia.org/wiki/Four-vector" [Broken] (ct,x) where c=1 ly/years, t is measured in years, and x is measured in ly. So, in the earth's reference frame the takeoff event, E0, the arrive event E1, and the return event E2 have the spacetime coordinates:

    E0 = (0,0)
    E1 = (10,6)
    E2 = (20,0)

    On the outbound leg the ship travels 6-0 = 6 ly in 10-0 = 10 years for a velocity of 6/10 = .6 c and the ships clock records sqrt(10²-6²) = 8 years. On the inbound leg the ship travels 0-6 = -6 ly in 20-10 = 10 years for a velocity of -6/10 = -.6 c and the ship's clock records an additional sqrt(10²-(-6)²) = 8 years.

    Now, let's examine the frame where the earth is moving at -0.6 c (the first leg of the ship's journey is at rest). We will call this the primed frame and its coordinates are related to the unprimed frame via the http://en.wikipedia.org/wiki/Lorentz_transformation" [Broken].

    E0' = (0,0)
    E1' = (8,0)
    E2' = (25,-15)

    On the outbound leg the ship travels 0-0 = 0 ly in 8-0 = 8 years for a velocity of 0/8 = 0 c and the ships clock records sqrt(8²-0²) = 8 years. On the inbound leg the ship travels -15-0 = -15 ly in 25-8 = 17 years for a velocity of -15/17 = -.88 c and the ship's clock records an additional sqrt(17²-(-15)²) = 8 years.

    Finally, let's examine the frame where the earth is moving at 0.6 c (the last leg of the ship's journey is at rest). We will call this the double-primed frame.

    E0'' = (0,0)
    E1'' = (17,15)
    E2'' = (25,15)

    On the outbound leg the ship travels 15-0 = 15 ly in 17-0 = 17 years for a velocity of 15/17 = .88 c and the ships clock records sqrt(17²-15²) = 8 years. On the inbound leg the ship travels 15-15 = 0 ly in 25-17 = 8 years for a velocity of 0/8 = 0 c and the ship's clock records an additional sqrt(8²-0²) = 8 years.

    There is a lot in this example, so take your time looking over it. I am sure you will have follow-up questions which I will be glad to answer. But as you can see although the different frames disagree about the distance covered and the duration of each leg they all agree about the time the ships clock records. I particularly recommend following the two Wikipedia links I posted for background information on the Lorentz transform (the heart of relativity) and the spacetime coordinate system (also called four-vectors).
     
    Last edited by a moderator: May 4, 2017
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