How Does Relativity Affect Time Perception for Astronauts and Earth Observers?

[Kaimana]

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.

 

[Dale]

Hi Kaimana, welcome to PF!

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.
...
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?

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" (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" .

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).

 

[Entropia]

Both scenarios are incorrect. According to the theory of relativity, time dilation and length contraction occur for observers in different frames of reference, but the laws of physics remain the same for all observers. This means that the crew of the spaceship and observers on Earth would both measure the same amount of time for the trip (10 years in this case), but the perceived time would be different due to their relative speeds.

In the first scenario, the crew of the spaceship would observe time passing slower on Earth, while observers on Earth would see the spaceship moving slower than the speed of light. This is because as an object approaches the speed of light, time slows down for that object. So while the crew may perceive the trip as taking slightly over 10 years, observers on Earth would see it taking longer.

In the second scenario, the crew of the spaceship would not see the journey as instantaneous, as nothing can travel faster than the speed of light. They would still experience time dilation, but to them, the trip would still take 10 years. Observers on Earth would also see the trip taking 10 years, but they would measure the spaceship as moving at a significant fraction of the speed of light.

In both scenarios, there is no point at which the crew of the spaceship would see themselves as moving at the speed of light. This is because it is impossible for anything with mass to reach the speed of light. As an object approaches the speed of light, its mass increases and it requires an infinite amount of energy to accelerate it further.

It's important to remember that these effects of time dilation and length contraction are only noticeable at speeds approaching the speed of light. In everyday life, these effects are negligible and do not significantly impact our daily experiences. But in extreme scenarios, such as space travel at near light speeds, these effects become more apparent.

I hope this helps to clarify the concept of relativity and how it applies to acceleration and observers. Remember, the laws of physics remain the same for all observers, but the perception of time and space can differ depending on the relative speeds of the observers.