Confusion in special relativity - total newbie

AI Thread Summary
The discussion revolves around a beginner's confusion regarding special relativity calculations involving two frames of reference, S and S', with S' moving at 0.7c relative to S. The user attempts to calculate the position in S' using the Lorentz transformation but questions the validity of their result, which yields a negative position. Other participants clarify that the user should consider the relative positions of the frames at the time of the event and remind them that there is a direct formula for calculating the time in S'. The conversation emphasizes understanding the relationship between the two frames and correctly applying the Lorentz transformation equations. Overall, the focus is on resolving the user's misunderstanding of the transformation process in special relativity.
iamaelephant
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I'm sure it's a stupid question but this is my first crack at SR so help would be great.

Homework Statement


Frame S' has a speed of 0.7c along the x axis relative to frame S. Clocks are adjusted so t=t'=0 at x=x'=0.

An even occurs in S at t1 = 3 x 10^7s at x1 = 40m. At what time does the event occur in S'?


Homework Equations


\beta = 0.7
\gamma = 1.4003
x' = \gamma (x - t\beta c)


The Attempt at a Solution



x' = \gamma (x - t\beta c) = (1.4003)(40 - (3 * 10^7)0.7 c) = -32.21m

Is this answer correct? It seems to me it can't possibly be right! What am I doing wrong?
 
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iamaelephant said:

The Attempt at a Solution



x' = \gamma (x - t\beta c) = (1.4003)(40 - (3 * 10^7)0.7 c) = -32.21m

Is this answer correct? It seems to me it can't possibly be right! What am I doing wrong?

Why do you think this cannot be right? Think of where the origin of S' is wrt S when the event happens.

BTW, you have a direct formula for finding the time, which is the actual question.
 
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