# How to incorporate non-synchronization into the Lorentz transformation?

Suppose a 2d space-time and two frames of reference moving with velocity $v$ relative to each other. The Lorentz transformation is:

$$t' = \gamma\left(t - \frac{v}{c^2}x\right)$$ $$x' = \gamma\left(x -vt\right)$$

However this seems to apply only if the origins "coincide" or "are synchronized". Citing Wikipedia for example:

At $t = t′ = 0$, the origins of both coordinate systems are the same, $(x, y, z) = (x′, y′, z′) = (0, 0, 0)$. In other words, the times and positions are coincident at this event.

So what if they are not? Suppose Alice $(t', x')$ is moving relative to Bob $(t, x)$ and Bob measures at $t = 0$ that Alice is at $x = x_0$. How do I incorporate this into the above Lorentz transformation?
Is it correct to assume that the origins will "coincide" at $t = -x_0 / v$ (as Bob will then measure Alice to be at $x = 0$) and to adjust all times by this value in order to compute $(t', x')$? That is using

$$t \rightarrow t + \frac{x_0}{v}$$

for the Lorentz transformation?

In the example you give, it's probably easier to think of replacing $x \rightarrow x - x_0$, but it should work out the same way- you're just trying to shift the time coordinate so that Alice and Bob share the same origin after all ;)
• Hi! Thanks for your explanation! However there seems to be a difference between the two translations. While for $x_0'$ indeed it doesn't matter, the result for $ct'$ is quite different. Applying the time translation I obtain $ct' = \gamma(ct + x_0/\beta - \beta x)$ and if I apply the spatial translation I obtain $ct' = \gamma(ct + \beta x_0 - \beta x)$. How do you explain this difference? Or do I have to apply an additional translation after the Lorentz transformation? – a_guest Apr 10 '17 at 11:49
• You can always choose your origin for both observers to be the same. Doing so basically means performing BOTH a time translation and a space translation to fix both $x_0$ and $t_0$. If either the time or spatial coordinates don't have the same origin, then this will show up in subsequent Lorentz boosts. – rwold Apr 12 '17 at 23:48