How Does Relativity Explain Shorter Travel Times at High Speeds?

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Relativity explains the shorter travel times at high speeds through the concept of length contraction. In the scenario of an astronaut traveling to Sirius, which is 8 light years away, the astronaut experiences a trip duration of only 6 years due to traveling at 0.8c. The distance measured in the astronaut's frame of reference is contracted, making it less than 8 light years. This contraction allows the time experienced by the astronaut to be shorter than the time it would take light to travel the full distance. Thus, the principles of special relativity reconcile the apparent discrepancy between distance and time experienced during high-speed travel.
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Relativity(Light years)? PLEASE HELP!

Homework Statement


Imagine an astronaut on a trip to Siris, which is 8 light years from Earth. On arrival at Siris,
the astronaut finds that the trip lasted 6 years. If the trip was made at a constant speed of .8c, how can the 8 light year distance be reconciled with the 6 year duration?


Homework Equations


E=mc^2


The Attempt at a Solution


I don't know how to solve this question...8 light years-> 6 years. 3/.8c??
Please help!
 
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Hi SAT2400! :smile:

(try using the X2 tag just above the Reply box :wink:)
SAT2400 said:
Imagine an astronaut on a trip to Siris, which is 8 light years from Earth. On arrival at Siris,
the astronaut finds that the trip lasted 6 years. If the trip was made at a constant speed of .8c, how can the 8 light year distance be reconciled with the 6 year duration?

(e = mc2 has nothing to do with it … and i think it's Sirius :wink:)

the astronaut's measurement of the time (on his own clock) is 6 years

what is his measurement of the distance?
 


The relevant equations are:

γ=1/root(1-v2/c2)
t'=γt
x=γx'

The 6 years is measured from his frame of reference, you need to measure the distance in the same frame as well.
 


hmm, can you explain more easily/??:(

Thanks !
 
Hi SAT2400! :wink:

Do you know the Lorentz transformation equations?

(you didn't mention them in your "Relevant equations")

If not, look them up in your book. :smile:
 


root of( 1-v^2/c^2) ...is this right??

hmm...could you please explain more in detail??

Sorry,,but I still don't know how to solve this question...:(
 
SAT2400 said:
root of( 1-v^2/c^2) ...is this right??

Sort-of …

but how are you going to use it? :smile:
 


I still have no idea...T_T

If I knew how to do this,,I would have not come to this website...

SO...

Please help!

THank you very much!
 
Use x and t for the coordinates in the Earth's frame (so Sirius is at x = 8, for all t).

Use x' and t' for the coordinates in the astronaut's frame.

Start the trip at (0,0) in both frames.

What do you get? :smile:
 
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For one such as yourself, who seems to have a very rudimentary understanding of SR, I wouldn't bother with Lorentz transforms. For the question all you need to utilize is Length-contraction.

To the people traveling on the rocket, the length between Earth and Sirius is contracted to: L=L_0\sqrt{1-\frac{v^2}{c^2}} The time that they measure is then simply: t=L/v

Now you will notice that since L is contracted, it is no longer 8 light years but something shorter, and hence t can be shorter than 8 years without the rocket traveling faster than c.
 
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