Lorentz transformation, special relativity problem

AI Thread Summary
The discussion revolves around solving a special relativity problem involving Lorentz transformations between two frames of reference, S and S'. The problem presents two events occurring in frame S' and asks for the relative velocity of the frames and the time in S when the events are observed. The participant expresses initial confusion about how to start the problem and notes that the events are simultaneous in frame S. After applying the Lorentz transformation equations and subtracting them, the participant confirms they arrived at the correct answer and acknowledges the need for more practice to gain confidence in solving such problems.
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Homework Statement


Frame S and S' are moving with respect to each other in the x-axis with some velocity. An event happens in S' at x'_1 = 1.0 c*year at t'_1 = 1.0 year. Another event happens at t'_2 = 2.0 c*year at t'_2 = 0.5 year. The two events are simultaneous at some point in S. The origins of S and S' are coincident at time t' = t = 0. Find the relative velocity of the frames, and the time in S when the events are noticed.


Homework Equations



t ' = \gamma \left(t - \frac{vx}{c^2}\right)
x ' = \gamma \left(x - vt \right)

The Attempt at a Solution



I really didn't even know where to start. I know that t_1 and t_2 as seen from S are equal, because the events were simultaneous. I tried listing out the equations, but I don't have enough equations for the amount of variables that I need to solve for. The assignment has already been turned in, and I know the answer of V (it was given in the back of the book), but I'd really like to know how to do the problem.
 
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You have

\begin{align*}<br /> t&#039;_1 &amp;= \gamma\left(t_1 - \frac{vx_1}{c^2}\right) \\<br /> t&#039;_2 &amp;= \gamma\left(t_2 - \frac{vx_2}{c^2}\right)<br /> \end{align*}

Subtract the first equation from the second. What do you get?
 
I get the right answer, thanks a ton!

I just need to do a lot more problems until I get comfortable with it, I think.

Thanks again!
 
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