# Lorentz transformations

1. Sep 4, 2014

### Ascendant78

1. The problem statement, all variables and given/known data

Two events occur in an inertial system K as follows:
Event 1: x1 = a, t1 = 2a/c, y1 = 0, z1 = 0
Event 2: x2 = 2a, t2 = 3a/(2c), y2 = 0, z2 = 0

Is there a frame K' in which the two events described
occur at the same place? Explain.

2. Relevant equations

Lorentz transformation:
x' = (x-vt)/(sqrt(1-(v/c)^2)

3. The attempt at a solution

I set both equations up as x' (equal to each other), plugged in the values on each side for x1 and x2, and it resulted in me losing 'a' on both sides (cancels out). I'm assuming that this means there is no frame K' where they would occur in the same place, but I am a bit surprised at that answer and am not sure how to explain it. Can someone explain why I am losing 'a' on both sides of the equation and what that means when it occurs?

2. Sep 4, 2014

### Orodruin

Staff Emeritus
You are losing a on both sides because both t and both x are proportional to it. This is not the issue, the question is if you can find a v < c such that the equation you got is fulfilled. Can you?

3. Sep 4, 2014

### Ascendant78

Thanks for the information. As far as a v < c, the final answer I get is that v = -2c. Here is what I work with:

a - v(2a/c) = 2a - v(3a/(2c))

a - v(4a/(2c) = 2a - v(3a/(2c))

-a = va/(2c)

-2c = v

So, from what you're telling me, this basically means that it is impossible because the velocity would need to be 2x the speed of light in the -x direction?

4. Sep 4, 2014

### Orodruin

Staff Emeritus
If |v| > c, then the gamma factor of the transformation is no longer real. There simply are no Lorentz transformations that have |v| ≥ c. But you are correct in that there is no such transformation and therefore no such frame.

A different way of looking at it would be to simply check whether or not you can travel from one event to the other with a speed smaller than c. The statements are equivalent. This is easiest done by computing if |Δx| < c |Δt|.

5. Sep 4, 2014

### Ascendant78

Well thanks a lot for the information. I really appreciate it. Makes perfect sense.