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 ;)

Generally speaking, translations extend the Lorentz group to the Poincaré group. You have to be clear whether you're applying the translation before or after your Lorentz transformation, as you get different results depending on the order.

  • $\begingroup$ 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? $\endgroup$ – a_guest Apr 10 '17 at 11:49
  • $\begingroup$ 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. $\endgroup$ – rwold Apr 12 '17 at 23:48

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