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An event occurs at x'=60m, t'= 8*10^-8s in frame S' (with y'=0 and z'=0) The frame S' has a velocity of 3c/5 in the x direction with respect to the frame S. The origins of S and S' coincide at t=0,t'=0. What are the space time coordinates of the event in S?

Just want to make sure I'm using the Lorentz formulas properly..

ɣ=(1 - v

^{2}/c

^{2})

^{-1/2}

x = ɣ(x'+vt) = (1-(9/25))

^{-1/2}*(60+(3c/5)*(8*10^-8))=93m

t = ɣ(t' + (vx'/c^2)) = (1-(9/25))

^{-1/2}*((8*10^-8)+(3c/5c^2)*(60))=45s (For some reason I think this is very wrong)

Question 2:

The space time coordinates of two events measured in the frame S are as follows:

Event 1: x

_{1}=x

_{o}, t

_{1}=x

_{o}/c

Event 2: x

_{2}=2x

_{o}, t

_{1}=x

_{o}/2c

(a) There exists a frame in which these events occur at the same time. Find the velocity of this frame with respect to S.

(b) What is the value of t at which both events occur in the new frame.

(a) Would doing the following work?

ɣ(t

_{1}+ (vx

_{1}/c

^{2}))=ɣ(t

_{2}+ (vx

_{2}/c

^{2}))

then solve for v which gives v=-c/2 (again I don't think this is right)

(b)

ɣ=(1 - v

^{2}/c

^{2})

^{-1/2}

Then using the value of v from a and the equation t'=ɣ(t - vx/c

^{2}) solve for t'

Any help on this would be greatly appreciated.... This is a lot more confusing than I first anticipated.