Velocity out of the Lorentz factor

[Aureum]

System K:

x=a t=2a/c
x'=2a t'=3a/c

I need to figure out in what frame K' will these events appear to occur at the same time?

I know the answer is c/2 in the -x direction but finding I'm having a hard time finding the velocity.

I really confused on how to get the velocity out from Lorentz factor. That way I can solve for V. It's not just with this problem but often I find myself with an expression equaling velocity times the Lorentz factor.

When I try derive relativistic velocity from the relativistic energy equation

E=(ymvc)+(mc^2)
E=mc(yv+mc)
E/(mc)=yv+mc
(E/(mc))-mc=yv

How do I get the velocity out of the Lorentz factor so from the energy expression I can derive velocity?

E^2=(pc)^2+(mc^2)^2
to
v=pc^2/(√((pc)^2+(mc^2)^2))

 

[Goongyae]

So in system K you have two events x1=(a,2a/c) and x2=(2a,3a/c)

The difference is (a,a/c) with magnitude sqrt(a^2-c^2*a^2/c^2) = 0, meaning the interval is lightlike.

In a new coordinate system K' the events shall occur at x1'=(0,0) and x2'=(a',0). (I picked the zeroes arbitrarily). The events obviously occur at the same time in that system since both time coordinates are the same. Since the interval must be unchanged regardless of the coordinate change, a'^2 = 0, i.e. a' must be zero.

Thus in the coordinate system K' the events occur at the same place and time. From Ks point of view, K' travels a distance a in a time a/c and is moving at speed c.

Intuitively, this makes sense. Suppose event 1 is a someone turning on a flashlight and event 2 is the light hitting a screen. For someone moving along with the beam of light the events are basically coincident in space and time.

Does that help?

>often I find myself with an expression equaling velocity times the Lorentz factor

If you have yv = v/sqrt(1-v^2/c^2) = 1/sqrt(1/v^2 - 1/c^2) then simply compute v = 1/sqrt(1/(yv)^2 + 1/c^2) = cyv/sqrt(c^2 + (yv)^2) to extract the velocity

 

[Aureum]

Ya, that helps a lot thanks. I don't know what I was thinking earlier.