Kyouran said:
Hmm...why is ##t" = 0## not ##t = -\frac{vx}{c^2}##? Srry for being so confused guys :D
Kyouran said:
It's not that I have a physical reason for it, but it is an assumption and assumptions must be justified. In particular, I am thinking about the following step:
## x' = (x-vt) \gamma(v) ##
## t' = (t-\frac{vx}{c^2}) \gamma(v) ##
It might be clearer if we reserve ##v## for the generic velocity there and use ##v = u > 0## and ##v = -u < 0## for our two frames. Then we have:
## x' = (x-ut) \gamma(u) ##
## t' = (t-\frac{ux}{c^2}) \gamma(u) ##
## x'' = (x+ut) \gamma(-u) ##
## t'' = (t+\frac{ux}{c^2}) \gamma(-u) ##
Now, we consider points ##x = x_1 > 0## and ##x = x_2 = -x_1 < 0##:
## x'_1 = (x_1-ut_1) \gamma(u) ##
## t'_1 = (t_1-\frac{ux_1}{c^2}) \gamma(u) ##
## x''_2 = (x_2+ut_2) \gamma(-u) ##
## t''_2 = (t_2+\frac{ux_2}{c^2}) \gamma(-u) ##
At ##t' =0##, we have a relationship between the ##x_1## and ##t_1## coordinates:
##t_1 = \frac{ux_1}{c^2}##
And, at ##t'' = 0##, we have a relationship between the ##x_2## and ##t_2## coordinates.
##t_2 = -\frac{ux_2}{c^2} = +\frac{ux_1}{c^2}##
This gives us:
##x'_1 = (x_1-ut_1) \gamma(u) = x_1(1 - \frac{u^2}{c^2})\gamma(u)##
##x''_2 = (x_2+ut_2) \gamma(-u) = x_2(1 - \frac{u^2}{c^2})\gamma(-u) = -x_1(1 - \frac{u^2}{c^2})\gamma(-u)##
Now, we use the physical equivalence of ##(t_1, x_1)## in the S' frame and ##(t_2, x_2)## in the S'' frame and the chosen orientation of all axes to demand that:
##x''_2 = - x'_1##
And this requires ##\gamma(u) = \gamma(-u)##