Lorentz transformation help

[Nuclear#1]

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.

 

[George Jones]

Can you first find $\Delta T'$ by using invariance of the spacetime interval?

 

[Nuclear#1]

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.

 

[George Jones]

Great!

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

 

[rdl]

A: The Lorentz transformation is a key concept in modern physics, specifically in the theory of special relativity. It allows us to transform coordinates and time measurements between two different frames of reference that are moving relative to each other at a constant velocity. In your problem, you are given the spatial and temporal separations in one frame (S) and asked to find the velocity and temporal separation in another frame (S').

To solve this problem using the Lorentz transformation, we can start by writing out the equations for the transformation of spatial coordinates and time:

x' = γ(x-vt)
t' = γ(t-vx/c^2)

where γ is the Lorentz factor, given by γ = 1/√(1-v^2/c^2), and c is the speed of light.

Substituting in the values given in the problem, we get:

600m = γ(600m - vt)
1μs = γ(1μs - v(600m)/c^2)

We can then simplify these equations and solve for v in terms of γ:

600m = γ(600m - vt)
1μs = γ(1μs - 600mv/c^2)

Solving for γ in the first equation and substituting it into the second equation, we get:

1μs = (1/√(1-v^2/c^2))(1μs - 600mv/c^2)

Simplifying and rearranging, we get a quadratic equation in terms of v:

(600v/c^2)^2 - (1-1/γ^2)v^2 - 2v + 1 = 0

This is the proper quadratic formula to solve for v. Plugging in the values given in the problem (γ = 1.25, c = 3x10^8 m/s), we can solve for v and get the correct answer of v = 0.8c.

I hope this explanation helps you understand the Lorentz transformation better and how to apply it to solve problems like this one. Good luck on your exam!