Calculate time passed on a ship clock seen by Earth observers

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Homework Statement:
A spaceship is on its way to Alpha Centauri, 4.2 light years from Earth, at 0.6 c. The time observed by observers on Earth is 7 years. What time passes on the ship clock as observed by Earth observers during the trip?
Relevant Equations:
Tv = γTo
In this case, γ = 1/√(1-v^2/c^2) = √(1-0.6^2) = 0.8

However, I'm not sure if time observed by Earth is proper time or moving time. The definition of proper time in my textbook states that it is 'the time measured in a frame of reference where the events occur at the same points in space. I'm not sure which one applies here
 

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  • #2
anuttarasammyak
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Proper time belongs to a body. It is read of clock attached to the body. Earth proper time 7 yrs belongs to the Earth whose frame of reference is inertial frame of reference(IFR). The rocket proper time 0.8 yrs belongs to the rocket whose frame of reference is not ( at least a single ) IFR.

Special relativity is physics in IFR, so it tells the Earth, IFR, inhabitants have privilege to judge the "moving" rocket time less.
 
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PeterDonis
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What time passes on the ship clock as observed by Earth observers during the trip?

There is no need for the qualifier that I crossed out in the above quote. The time elapsed on the ship's clock between two given events--departure from Earth and arrival at Alpha Centauri--is an invariant, and will be the same for all observers.
 
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PeterDonis
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The rocket proper time 0.8 yrs belong to the rocket whose frame of reference is not ( at least a single ) IFR.

I don't see that in this scenario. This is not a twin paradox scenario, as far as I can tell; the ship only goes one way, Earth to Alpha Centauri, and it is always moving at the same speed relative to Earth. So it is always at rest in an inertial frame--that frame is just not the same as the frame in which Earth and Alpha Centauri are at rest.

Special relativity is physics in IFR

This is not correct. It is perfectly possible to do SR in non-inertial frames, as long as spacetime is flat (i.e., as long as there are no gravitating masses present, or we idealize them away--in this scenario, "the Earth" is idealized as just a point in space and its gravity is ignored). It so happens that no non-inertial frames are required in this scenario, but that is not a general requirement for SR.
 
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PeroK
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The definition of proper time in my textbook states that it is 'the time measured in a frame of reference where the events occur at the same points in space.
That's correct, but it's simpler to say that proper time is what a clock measures (a clock measures its own proper time).

If we go back to your problem:

What can you say about the ship clock with respect to the Earth's reference frame?
 

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