How Do You Calculate Velocity in Lorentz Transformations?

AI Thread Summary
The discussion revolves around calculating velocity in Lorentz transformations for a physics problem involving two events with specific spatial and temporal separations. Initially, the user struggled with using the Lorentz transformation equation and sought help to derive the velocity and temporal separation in a moving frame. After some algebraic errors, they discovered that the quadratic equation simplifies to yield the correct velocity of 0.8c. It was noted that finding the temporal separation delta T' first is a more straightforward approach, eliminating the need for complex quadratic terms. The conversation highlights the importance of understanding the invariance of the spacetime interval in solving such problems.
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I'm studying for my modern physics final and this problem is giving me trouble;

Q: In a frame S, two events have spatial separation deltaX= 600m, delta y and delta z = 0, and a temporal separation deltaT= 1micro second. A second frame S' is moving along the same axis with nonzero speed v (0'x' is parallel to 0x). In S' it is found that the spatial separation is deltaX' is also 600m. What are v and deltaT'?

My attempts have been using x'=gamma(x-vt), now I plug in the known data and try to manipulate the equation into a quadratic I can solve, but every time I do it I wind up not getting anywhere close to being correct. What does the proper quadratic formula look like? The correct answers are v = 0.8C and T = -1 micro second. I just need help finding how to find the velocity. Please help, any thing would be helpful.
 
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Can you first find \Delta T' by using invariance of the spacetime interval?
 
Sorry but I figured it out, I was making algebraic errors. The quadratic turns into; 1.25v^2 - cv = 0. with v = 0 and v = 2.4E8, which is 0.8c. I didn't think you could find delta T' first, so I went about it doing it this way because it was the recommended method. Sorry for the inconvenience.
 
Great!

Actually, it is trivial to find delta T' first, which then eliminates the need to use terms quadratic in v.
 
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