Time Dilation at 0.97c: Calculating Travel Time to a Star

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Homework Help Overview

The discussion revolves around calculating the travel time to a star located 45.0 light-years away for a spacecraft traveling at 0.97c, focusing on the perspectives of both an observer on Earth and one on the spacecraft. Participants explore concepts related to time dilation and reference frames in the context of special relativity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the distinction between the time measured by an Earth observer and that measured by a spacecraft observer, questioning the application of time dilation and the correct use of reference frames. Some express confusion about which observer's time should be considered dilated.

Discussion Status

There is an ongoing exploration of the concepts of time dilation and reference frames. Some participants have offered insights into the calculations involved, while others are clarifying their understanding of the problem and correcting their initial misconceptions.

Contextual Notes

Participants are grappling with the definitions of time in different frames of reference and the implications of traveling at relativistic speeds. There is a noted confusion regarding the application of time dilation in this context.

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A certain star is 45.0 light-years away. How long would it take a spacecraft traveling at 0.97c to reach that star from Earth, as measured by an observer:
(a) on Earth?
(b) on the spacecraft ?

i'm not sure how to solve this question... simply i don't know which is tv and which is t0.
i was thinking that the person on Earth would be in rest frame, however relative to the spacecraft the person on Earth would be in tv and the person on the spacecraft would be in the rest frame (t0)... very confused :S
 
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The Earth bound observer will just measure t=d/v=45ly/(.97c). You can treat him as the stationary observer since the star isn't moving relative to the Earth (as far as this problem is concerned).

The spacecraft bound observer will measure t'=d'/v where d' is the Lorentz contracted distance.

This should be conceptually much simpler than doing the time dilation between t and t'.
 
Matterwave said:
The Earth bound observer will just measure t=d/v=45ly/(.97c). You can treat him as the stationary observer since the star isn't moving relative to the Earth (as far as this problem is concerned).

umm... can u simply put it into the equation t =d/v ? don't you need to use the time dilation equation? since the observer is not in the same frame of reference to the spacecraft (he is stationary therefore rest frame t0)?

[tex] <br /> Tv= \time \equiv \frac{t0}{\sqrt{{45 - v^2/c^2}}<br /> [/tex]
 
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Time dilation is only ever used to find the time that passes in ANOTHER frame from your frame.

If I'm on Earth, why would I dilate my own time? I would only dilate the spacecraft 's time.

(If I were on the spacecraft , I would only ever dilate the Earth's time. This leads to the Twin paradox)
 
OHHHHHHHH,, heheheh... i must have had a mental blank :), i get you now,, why would time dilate when your on Earth, when your not traveling at 0.97c the dilation that would occur would be the spacecraft , in which it is in the same frame of reference to 0.97c.

It's actually common sense, why would time dilate for someone that is NOT traveling 0.97c. (different frame of reference)

sorry i completely misunderstood question.

thank you, this was very helpful :D
 

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