Special relativity transformations

AI Thread Summary
The discussion centers on a homework problem involving special relativity and Lorentz transformations, where two light flashes occur at different times and positions. The goal is to determine the speed v of an inertial frame in which the events are simultaneous. The initial calculations using the Lorentz transformations lead to a non-physical result, prompting the consideration of a possible misprint in the problem statement regarding the time units. Confirmation from the lecturer indicates that the original question indeed contained an error. The resolution highlights the importance of verifying problem details in physics homework.
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Homework Statement



Two light flashes occur on the laboratory x axis, the first at time t=0 and
position x=450 m, the second at time t=+1 ms, at the origin. In an inertial
frame moving along the x-axis with speed v, the events are simultaneous.
What is the speed v?

Homework Equations



From the question I have decided that the relevant equations are the Lorentz transformations. We know the position's in the rest frame and we know the time of the events. In the moving frame we know the the events are simultaneous. So we can't use the transforms for x as there would be two unknowns \Delta x'and v

The Attempt at a Solution



t_{1}' = \gamma (t_{1} - \dfrac{vx_{1}}{c^2})
t_{2}' = \gamma (t_{2} - \dfrac{vx_{2}}{c^2})

t_{2-1}' = \gamma (t_{2} -t_{1} - \dfrac{v}{c^2}(x_{2} - x_{1}))

0 = \gamma((1*10^-3) - 450\dfrac{v}{c^2})

0 =(1*10^-3) \gamma - \gamma 450\dfrac{v}{c^2}

\gamma 450\dfrac{v}{c^2} =(1*10^-3) \gamma

450\dfrac{v}{c^2} = (1*10^-3)

v = \dfrac{(1*10^-3)}{450} c^2 \mbox{WRONG!}

Not sure what I am doing wrong. Any help on how to tackle these questions would be great.
 
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Unless question sheet is a misprint and it should be micro rather than milli(seconds)?
 
Your reasoning seems to be correct, and using the numbers given does give a non-physical result, so you are perhaps correct in thinking that there is a misprint in the question.
 
Thanks, my lecturer got back to me and confirmed this :). I appreciate you taking the time.
 
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