How to incorporate non-synchronization into the Lorentz transformation?
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Suppose a 2d space-time and two frames of reference moving with velocity vv relative to each other. The Lorentz transformation is:
t′=γ(t−vc2x)t′=γ(t−vc2x)
x′=γ(x−vt)x′=γ(x−vt)
However this seems to apply only if the origins "coincide" or "are synchronized". Citing Wikipediafor example:
At t=t'=0t=t′=0, the origins of both coordinate systems are the same, (x,y,z)=(x',y',z')=(0,0,0)(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′)(t′,x′) is moving relative to Bob (t,x)(t,x) and Bob measures at t=0t=0 that Alice is at x=x0x=x0. How do I incorporate this into the above Lorentz transformation?
Is it correct to assume that the origins will "coincide" at t=−x0/vt=−x0/v (as Bob will then measure Alice to be at x=0x=0) and to adjust all times by this value in order to compute (t′,x′)(t′,x′)? That is using
t→t+x0v
t′=γ(t−vc2x)t′=γ(t−vc2x)
x′=γ(x−vt)x′=γ(x−vt)
However this seems to apply only if the origins "coincide" or "are synchronized". Citing Wikipediafor example:
At t=t'=0t=t′=0, the origins of both coordinate systems are the same, (x,y,z)=(x',y',z')=(0,0,0)(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′)(t′,x′) is moving relative to Bob (t,x)(t,x) and Bob measures at t=0t=0 that Alice is at x=x0x=x0. How do I incorporate this into the above Lorentz transformation?
Is it correct to assume that the origins will "coincide" at t=−x0/vt=−x0/v (as Bob will then measure Alice to be at x=0x=0) and to adjust all times by this value in order to compute (t′,x′)(t′,x′)? That is using
t→t+x0v
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