Age comparison to observer on Earth (time dilation)

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Homework Help Overview

The problem involves calculating the aging difference between a traveler moving at a significant fraction of the speed of light (0.780c) and an observer on Earth during a round trip to Alpha Centauri, which is approximately 4 light years away. The discussion centers around the concepts of time dilation and the associated equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of time dilation and Lorentz contraction equations to determine the aging of the traveler compared to the Earth observer. There are attempts to clarify the relationship between the time experienced by the traveler and the time measured by the observer on Earth.

Discussion Status

The discussion is ongoing, with participants exploring the calculations related to time experienced by both the traveler and the Earth observer. Some participants have provided insights regarding the expected outcomes based on the time dilation effect, but there is no explicit consensus on the final interpretation of the results.

Contextual Notes

Participants are navigating the complexities of relativistic effects while avoiding the use of Lorentz contraction in their reasoning. There is an acknowledgment of potential simplifications, such as ignoring general relativistic effects during acceleration phases of the journey.

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Homework Statement



Alpha Centauri is about 4 light years from the Earth. If you were to travel to Alpha Centauri and back at the speed of the electron that you calculated in question A (0.780c), how much would you age compared to an observer on Earth?

Homework Equations



There equations were not provided in the actual problem, but so far we've used these 2 (time dilation and Lorentz contraction) equations:

T' = (T) / sqrt[1 - (v^2/c^2) ]

L' = L * sqrt[1 - (v^2/c^2) ]

The Attempt at a Solution



v = 0.780c

Both ways:
L = 8cY

L = v*T

T = L / v

T = (8cY) / (0.780c)

T = 10.3Y

T' = ?

T' = (T) / sqrt[1 - (v^2/c^2) ]

T' = (T) / sqrt[1 - (v/c)^2 ]

T' = (T) / sqrt[1 - (0.780c/c)^2 ]

T' = (T) / sqrt[1 - (0.780)^2 ]

T' = (T) / sqrt[1 - (0.608) ]

T' = (T) / sqrt(0.392)

T' = (T) / sqrt(0.392)

T' = (T) / (0.626)

T' = (10.3Y) / (0.626)

T' = 16.5Y

From what I recall, T' is the change caused by motion of the moving object, relative to the observer.
In this case, relative to the person on Earth, the travel time to Alhpa Centauri and back would be 16.5 years.

There only other time variable left is T, so I assume this applies to the person/partice traveling to Alpha Centauri.

Am I correct? (I want to avoid using Lorentz contraction to ensure I understand the time dilation equation and use it properly.)

Thanks.
 
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Less time passes for the guy in the spaceship. Remember the guy in Heinlein's story who comes home and marries his great-great-grand niece? (Time For The Stars)
The Earth observer says 10.3 years.
The traveler says 10.3 * .626
(ignoring general relativistic effects during the accelerations)
 
OK. Doesn't that mean I did it right so far? Don't I have half of the answer (the time for the observer on Earth)?
 
You said 16.5 years for the Earth observer; I said 10.3.
You said 10.3 for the traveler; I said 10.3*.626.
 

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