Calculating Time Dilation for Space Travelers

AI Thread Summary
The discussion focuses on calculating time dilation for a space traveler moving at relativistic speeds. Initially, the traveler perceives the universe moving past them at 0.7c, leading to an elapsed time of approximately 9.11 years for a 6.5-year observation. A subsequent calculation for a traveler moving at 0.9c yields a time of 3.97 years, but the expected answer is 4.01 years, raising questions about rounding or missing elements in the calculations. Participants clarify that the problem describes a distance of 6.5 light-years, not time, and emphasize that light-years are a valid unit of distance. The conversation concludes with a consensus on the relevance of length contraction versus time dilation in the context of relativistic travel.
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.
 
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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