Calculating Time Dilation for Space Travelers

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Homework Help Overview

The discussion revolves around calculating time dilation in the context of special relativity, specifically for a space traveler moving at relativistic speeds. The original poster presents a scenario involving time elapsed for both the traveler and an observer, raising questions about the interpretation of given distances and times.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of time dilation and length contraction, questioning the interpretation of the distance given in light-years versus time. There is discussion about whether the calculations presented are affected by rounding or if there are underlying assumptions that need clarification.

Discussion Status

Participants are actively engaging with the problem, with some clarifying the distinction between distance and time in the context of the problem. There is acknowledgment of the need to interpret the units correctly, and guidance has been offered regarding the use of light-years as a unit of distance.

Contextual Notes

There is a noted confusion regarding the initial conditions of the problem, specifically the interpretation of 6.5 as a distance in light-years rather than a time measurement. This has led to further exploration of relativistic effects on measurements from different frames of reference.

stunner5000pt
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Homework Statement
A space traveler moving at a speed of 0.70c with respect to Earth makes a trip to a distant star that is stationary relative to Earth. He measures the length of this trip to be 6.5ly. What would be the length of this same trip (in ly) as measured by a traveler moving at a speed of 0.90c with respect to Earth?
Relevant Equations
[tex] \Delta t = \gamma t_{0} [/tex]
If we took the perspective of the space traveller themselves, they are stationary and the whole universe goes past them at 0.7c. THen th elapsed time of 6.5 yr looking outside is

\Delta t =6.5 \frac{1}{\sqrt{1-0.7^2}} = 9.11 yrs

THen, when the traveller looks at the person travelling at 0.9c

the time taken should be

9.11 \times \frac{1}{\sqrt{1-0.9^2}} = 3.97 yrs

but the answer is 4.01 yr. Is this just down to rounding or is there something missing in the solution above?

thanks in advance for your help
 
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stunner5000pt said:
... elapsed time of 6.5 yr ...
The problem does not say anything about such time, 6.5 yr. It describes a distance, 6.5 ly.
 
Hill said:
The problem does not say anything about such time, 6.5 yr. It describes a distance, 6.5 ly.
Aha! missed that, thank you

ok so then that means the distance travelled according to the traveller is:

6.5c \times 24 \times 365 \times 60^2 = 6.14 \times 10^{16} m

when the person is in motion, how is the 6.5 ly being interpreted ? Is the noninertial observer seeing a shorter'length that the stationary length i.e. is the above number the length observed from rest?
 
No need to convert or to interpret ly. It is a unit of distance as good as any.
I don't see a "noninertial observer" in the problem. However, there is length contraction rather than time dilation here.
 
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stunner5000pt said:
ok so then that means the distance travelled
The distance travelled by the Earth and the destination planet.

And as has already been said, ly is a perfectly fine length unit. Preferable to meters here really as by definition c = 1 ly/y.
 

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