http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/veltran.html

Where does t' = ɣ (t- xv/c^2) come from?
Especially the (t- xv/c^2) part.

Question

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/veltran.html

Where does t' = ɣ (t- xv/c^2) come from?
Especially the (t- xv/c^2) part.

irishboy123 · Answer

you need to look at the derivation of the lorentz transformation. Galilean relativity is inconsistent with special reativity because the speed of light is a constant for observors on both of the typical Galilean ref frames. so the Galilean transform $x = x' + vt'$, and reverse transform $x' = x - vt $ are modified to $x = \gamma(x' + ct')$, and $x' = \gamma (x -ct) $, with x=ct and x' = ct'. this vid from 10:51 goes through one way of getting there, in quite some detail.... https://www.youtube.com/watch?v=6Tts3gxs_cM

masumanwar · Answer

the theory of special relativity and the time dilation due to relativistic motion between two reference frame where one clock is in motion with respect to other and measured time using lorentz transformation coming as t=to/{1-(v/c)^2}^-1/2 where t is the time in the clock in motion  with velocity v and