DaleSpam said:
Yes, as I mentioned in post 504 I used v=0.6, c=1, and d=2 (or r=1) for this drawing. The exact times will be different for different v, c, or d, but it is always given by the Lorentz transform.
OK, thanks .6c is a good one to use.
Please evaluate the below and tell me what you think. Perhaps, I made a math error or
something.
Assume v = 3/5(c). Let r be one half to rest distance of the rod in O'
λ = 5/4, (c-v) = 2/5, (c + v) = 8/5
When t reaches r/(2c), the radius of the light sphere is ct = r/2 in O.
For the left light beam, its x location is -r/2 or x = -ct = -r/2.
t' = ( t - xv/c^2)λ
t' = ( t + tv/c )λ = t ( 1 + v/c )λ = t ( 1 + 3/5 )5/4 = 2t = r/c.
Thus, (ct')^2 = r^2 and therefore, the light sphere in O' strikes both points at +r and -r.
Then when t reaches 2r/c, x = 2r, so
t' = ( 2r/c - 2rv/c^2)λ = 2r/c(1 - v/c)λ = 2r/c(2/5)5/4 = r/c
Thus, (ct')^2 = r^2 and therefore, the light sphere in O' strikes both points at +r and -r.
Hence, there are two different times in O when O' sees the simultaneous strikes.
This is different from R of S.
Thus, time proceeds from 0 to t = r/(2c) then the light sphere in O' strikes the endpoints at the same time. Then time proceeds forward as the light sphere expands in O and then again, another simultaneous strikes occurs at t = 2r/c.